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http://www.m0ose.com/javascripts/particles/particles1.html | [
"number of particles\n\n## Web gl particle box\n\n### by Cody Smtih\n\nThis is a non-interacting particle simulator inspired by Ed Angels OpenGL example. Basically all that a particle can do is move in a straight line. but some trickery makes it look like it bounces around. Since its just a shader trick it can handle 20,000, or more, particles very easily.\n\n### how it works\n\nevery particle starts with an initial position and velocity. Modulo arithmatic allows the particle to bounce off of walls.",
null,
"here is a graph of the bounce function that makes the particles appear to bounce off walls. It is applied to the x,y, and z positions\n\n``` attribute vec3 aVertexPosition;\nattribute vec4 aVertexColor;\nattribute vec3 velocity;\n\nuniform mat4 uPMatrix;\nuniform mat4 viewMatrix;\nuniform float timeStep;\n\nvarying vec4 vColor;\n\nfloat bounce( float x)\n{\nfloat xa = abs(x);\nfloat x2 = mod(xa,2.0);\nfloat x3 = mod(xa,1.0);\nif( x2 > 1.0){\nx3 = 1.0 - x3;\n}\nreturn x3;\n}\n\nvoid main(void) {\n\nvec3 b = vec3( aVertexPosition - ( timeStep * velocity));\ngl_Position = vec4(bounce(b.x), bounce(b.y), bounce(b.z), 1.0);\n\ngl_Position = viewMatrix * gl_Position;\ngl_Position = uPMatrix * gl_Position;\nvColor = aVertexColor;\n//vColor = vec4( velocity, 1.0);\n}\n```\n\n### what i learned:\n\nThe trackball taught me a lot about quaternions. Some of which I understand. for example i,j,k are all different imaginary numbers. There is a multiplaication table that was invented for them that allows them to actually work. it also is very cool technique.\n\n### improvements:\n\nAdd a bounding box to make it visually appealing. Add slider to change number of particles.\n\n### limitations:\n\nI don't see an easy way to allow inter-particle interactions. seems to be limited to a cube environment. probably difficult to implement gravity or any other forces."
] | [
null,
"http://www.m0ose.com/javascripts/particles/bounce_graph.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93027157,"math_prob":0.9667596,"size":1101,"snap":"2020-45-2020-50","text_gpt3_token_len":235,"char_repetition_ratio":0.110300824,"word_repetition_ratio":0.0,"special_character_ratio":0.19891009,"punctuation_ratio":0.119266056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97989094,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T01:34:15Z\",\"WARC-Record-ID\":\"<urn:uuid:9c4ff091-3da9-4f0c-aa45-098ac7f704e6>\",\"Content-Length\":\"4211\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e39650f-8e7e-4f4e-8e9c-065884df5fbb>\",\"WARC-Concurrent-To\":\"<urn:uuid:48085560-7d14-4727-9e41-c5c3b22e9c53>\",\"WARC-IP-Address\":\"216.37.42.30\",\"WARC-Target-URI\":\"http://www.m0ose.com/javascripts/particles/particles1.html\",\"WARC-Payload-Digest\":\"sha1:BIDH3W4UD4MPODLU7KQ62QLHOQWUICLE\",\"WARC-Block-Digest\":\"sha1:K54MW2HJ6NRMIHVHUG5SHBXALPPZMCI6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141169606.2_warc_CC-MAIN-20201124000351-20201124030351-00612.warc.gz\"}"} |
https://www.ncatlab.org/nlab/show/Banach+manifold | [
"# nLab Banach manifold\n\nContents\n\n### Context\n\n#### Differential geometry\n\nsynthetic differential geometry\n\nIntroductions\n\nfrom point-set topology to differentiable manifolds\n\nDifferentials\n\nV-manifolds\n\nsmooth space\n\nTangency\n\nThe magic algebraic facts\n\nTheorems\n\nAxiomatics\n\ncohesion\n\ntangent cohesion\n\ndifferential cohesion\n\n$\\array{ && id &\\dashv& id \\\\ && \\vee && \\vee \\\\ &\\stackrel{fermionic}{}& \\rightrightarrows &\\dashv& \\rightsquigarrow & \\stackrel{bosonic}{} \\\\ && \\bot && \\bot \\\\ &\\stackrel{bosonic}{} & \\rightsquigarrow &\\dashv& \\mathrm{R}\\!\\!\\mathrm{h} & \\stackrel{rheonomic}{} \\\\ && \\vee && \\vee \\\\ &\\stackrel{reduced}{} & \\Re &\\dashv& \\Im & \\stackrel{infinitesimal}{} \\\\ && \\bot && \\bot \\\\ &\\stackrel{infinitesimal}{}& \\Im &\\dashv& \\& & \\stackrel{\\text{étale}}{} \\\\ && \\vee && \\vee \\\\ &\\stackrel{cohesive}{}& ʃ &\\dashv& \\flat & \\stackrel{discrete}{} \\\\ && \\bot && \\bot \\\\ &\\stackrel{discrete}{}& \\flat &\\dashv& \\sharp & \\stackrel{continuous}{} \\\\ && \\vee && \\vee \\\\ && \\emptyset &\\dashv& \\ast }$\n\nModels\n\nLie theory, ∞-Lie theory\n\ndifferential equations, variational calculus\n\nChern-Weil theory, ∞-Chern-Weil theory\n\nCartan geometry (super, higher)\n\n# Contents\n\n## Idea\n\nA notion of infinite-dimensional manifold. A Banach manifold is a manifold modelled on Banach spaces. By default, transition maps are taken to be smooth.\n\n## Properties\n\n### Embedding into the category of diffeological spaces\n\nThe category of smooth Banach manifolds has a full and faithful functor into the category of diffeological spaces. In terms of Chen smooth spaces this was observed in (Hain). For more see at Fréchet manifold – Relation to diffeological spaces.\n\nFor general references see at infinite-dimensional manifold.\n\nThe embedding into the category of diffeological spaces is discussed in\n\n• Richard Hain, A characterization of smooth functions defined on a Banach space, Proc. Amer. Math. Soc. 77 (1979), 63-67 (web, pdf)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8134204,"math_prob":0.97980034,"size":1515,"snap":"2019-35-2019-39","text_gpt3_token_len":351,"char_repetition_ratio":0.13765718,"word_repetition_ratio":0.010152284,"special_character_ratio":0.17161717,"punctuation_ratio":0.12133891,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9538379,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T23:11:34Z\",\"WARC-Record-ID\":\"<urn:uuid:41b48625-db39-4769-8ce1-b759aa28c1ee>\",\"Content-Length\":\"46639\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:386c6160-cb7b-4b1c-9a5d-678cdbae61e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:881b3080-5832-49eb-ae05-d656900261ca>\",\"WARC-IP-Address\":\"104.27.171.19\",\"WARC-Target-URI\":\"https://www.ncatlab.org/nlab/show/Banach+manifold\",\"WARC-Payload-Digest\":\"sha1:GWVAI2ON53DI3NEACFXCPJXCROTGHZYB\",\"WARC-Block-Digest\":\"sha1:7AOXZIGUWXW3UQ6SBVE5RJBRYYEIRQTP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573124.40_warc_CC-MAIN-20190917223332-20190918005332-00375.warc.gz\"}"} |
https://tex.stackexchange.com/questions/367135/tikz-takes-forever-to-compile-when-imported-into-main-file/367136 | [
"# Tikz takes forever to compile when imported into main file\n\nI used TikzEdt in order to generate some figures to use in my thesis. However, when I included the source code by copy-pasting/\\input in the main document under a figure environment my document took forever to compile without any errors and yielded a corrupted PDFoutput file after I cancel the compilation.\n\nAs far as I am concerned, I have all the necessary packages in the preamble and I have generated a minimal working example after inspecting the exported code from TikzEdt.\n\nThe non-compiling file is huge so the driving file is here, the class file is a modified book class and the included macros are basic.\n\nFurthermore TikzEdt complains that the following line with the message: \"Couldn't parse code. NoViableAltException in line 16 at position 44. Is some \\end{} command missing?\". A compiling MWE is here.\n\n\\draw[<->, thick] (-1,0)++({\\ang+90}:1) --++ (\\ang:{2*cos(\\ang)});\n\nEDIT The issue is resolved and the problem was my usage of \\ang which is an internal macro defined in the package siunitx. TikzEdt parser still complains, yet I do not know if it is a real problem.\n\nA screenshot from the issue in TikzEdt is below. The error message is displayed in red lines.",
null,
"• You want your example to produce the error but, if I've understood you correctly, the code you've posted will compile fine. Is that correct? Also, I'm confused about what you are doing with TkzEdt when it gives you the parsing error. I thought you exported the code from there? Are you saying it can't compile it either? If we need the other example, post it here instead of or as well as the existing one. – cfr Apr 28 '17 at 0:18\n• @cfr Both MWEs and TikzEdt compile and output a valid PDF. But TikzEdt complains, and I exported the code from there. I have a huge main thesis file with a big preamble and this is the file which does not compile. – Vesnog Apr 28 '17 at 0:21\n• @cfr After going through the preamble commenting out each comment to see if they are the culprit, I managed to make a minimal example with the same preamble as in the large document. This file compiled after commenting out \\siunitx package which is pretty surprising and it also switched to compatibility mode as evident from: Package caption Error: The subcaption' package does not work correctly(caption) in compatibility mode. \\begin{document}. – Vesnog Apr 28 '17 at 0:27\n• But you say in the question that the MWE does NOT compile. Now you say it does, which means that the problem is in code we don't have. We need an example which produces the error. All that is wrong with this one is a missing \\begin{document}. It compiles fine in either preview or non-preview mode. – cfr Apr 28 '17 at 0:29\n• Which error do you get from TeX when you compile? I don't have TkzEdt to compare. I can produce AN error by adding xparse but I don't know whether it is the right error. – cfr Apr 28 '17 at 0:33\n\nWell, you need a \\begin{document}, obviously, but it otherwise works fine converted to a regular document with the picture in a figure.\n\n\\documentclass{article}\n\\usepackage{tikz,amsmath, amssymb,bm,color}\n\\usetikzlibrary{shapes,arrows}\n\\usetikzlibrary{decorations.markings,math}\n\\usetikzlibrary{calc}\n\n\\begin{document}\n\\begin{figure}\n\\center\n\\begin{tikzpicture}\n\\tikzmath{\\ang = 45;};\n\\begin{scope}[thick,decoration={\nmarkings,\nmark=at position 0.5 with {\\arrow{latex}}}\n]\n\\filldraw[red] (-1,0) circle (2pt)\nnode[anchor=east, font = \\footnotesize] at (-1,-0.2) {$\\mathrm{E_0}$};\n\\draw[dashed, red] (-1,0) --++({\\ang+90}:1);\n\\filldraw[green] (1,0) circle (2pt)\nnode[anchor=west, font = \\footnotesize] at (1,-0.2) {$\\mathrm{E_0}$};\n\\draw[postaction={decorate}, red] (-1,0) --++ (\\ang:2);\n\\draw[postaction={decorate}, green] (1,0) --++ (\\ang:2);\n\\draw[<->, thick] (-1, -0.2) -- (1, -0.2);\n\\draw[dashed, green] (1,0) --++ ({\\ang+90}:2.5);\n\\draw[<->, thick] (-1,0)++({\\ang+90}:1) --++ (\\ang:{2*cos(\\ang)});\n% \\draw[<->, thick] (-1,0)++({\\ang+90}:1) --++ (\\ang:{2*cos(\\ang) });\n\\node[anchor=north, font = \\footnotesize] at (0, -0.2) {d=$\\lambda/2$};\n\\end{scope}\n\\draw[blue, thick, dash pattern= on 25 off 7 on 50 off 7 on 10] (-2,0) -- (2, 0);\n\\node[anchor=west, font = \\footnotesize, blue] at (1.5,-0.05) {z}; node[near start, auto] {true}\n\\draw[thick,blue,->] ([shift=(0:1)]1,0) arc (0:\\ang:1);\n% \\draw[thick,blue,->] (2,0) arc (0:\\ang:1);\n\\draw (1,0)++({\\ang/2}:1.2) node[rotate=\\ang, anchor=base, blue, font=\\normalsize]{$\\theta$};\n\\draw ({\\ang+90}:1.3) node[rotate=\\ang, anchor=base, black, font=\\normalsize]{$d\\cos\\theta$};\n\\end{tikzpicture}\n\\caption{Picture}\n\\end{figure}\n\\end{document}",
null,
"Possibly I don't understand the question.\n\n# EDIT\n\nAdding siunitx produces AN error, though whether it is YOUR error, I don't know. Specifically xparse complains that \\ang is not expandable. This is because the current code invokes \\ang, which is no longer an unassigned macro name. So it is trying to expand \\ang to its definition, but can't. (And, if it could, the results would not be good either.)\n\nThe solution is to use a new name such as \\myangle.\n\n\\documentclass{article}\n\\usepackage{tikz,amsmath,amssymb,bm}\n\\usetikzlibrary{shapes,arrows}\n\\usetikzlibrary{decorations.markings,math}\n\\usetikzlibrary{calc}\n\\usepackage{siunitx}\n\n\\begin{document}\n\\begin{figure}\n\\center\n\\begin{tikzpicture}\n\\tikzmath{\\myangle = 45;};\n\\begin{scope}[thick,decoration={\nmarkings,\nmark=at position 0.5 with {\\arrow{latex}}}\n]\n\\filldraw[red] (-1,0) circle (2pt)\nnode[anchor=east, font = \\footnotesize] at (-1,-0.2) {$\\mathrm{E_0}$};\n\\draw[dashed, red] (-1,0) --++({\\myangle+90}:1);\n\\filldraw[green] (1,0) circle (2pt)\nnode[anchor=west, font = \\footnotesize] at (1,-0.2) {$\\mathrm{E_0}$};\n\\draw[postaction={decorate}, red] (-1,0) --++ (\\myangle:2);\n\\draw[postaction={decorate}, green] (1,0) --++ (\\myangle:2);\n\\draw[<->, thick] (-1, -0.2) -- (1, -0.2);\n\\draw[dashed, green] (1,0) --++ ({\\myangle+90}:2.5);\n\\draw[<->, thick] (-1,0)++({\\myangle+90}:1) --++ (\\myangle:{2*cos(\\myangle)});\n% \\draw[<->, thick] (-1,0)++({\\myangle+90}:1) --++ (\\myangle:{2*cos(\\myangle) });\n\\node[anchor=north, font = \\footnotesize] at (0, -0.2) {d=$\\lambda/2$};\n\\end{scope}\n\\draw[blue, thick, dash pattern= on 25 off 7 on 50 off 7 on 10] (-2,0) -- (2, 0);\n\\node[anchor=west, font = \\footnotesize, blue] at (1.5,-0.05) {z}; node[near start, auto] {true}\n\\draw[thick,blue,->] ([shift=(0:1)]1,0) arc (0:\\myangle:1);\n% \\draw[thick,blue,->] (2,0) arc (0:\\myangle:1);\n\\draw (1,0)++({\\myangle/2}:1.2) node[rotate=\\myangle, anchor=base, blue, font=\\normalsize]{$\\theta$};\n\\draw ({\\myangle+90}:1.3) node[rotate=\\myangle, anchor=base, black, font=\\normalsize]{$d\\cos\\theta$};\n\\end{tikzpicture}\n\\caption{Picture}\n\\end{figure}\n\\end{document}\n\n\ncompiles to the output shown above.\n\n• Thanks for the answer, I think this is some sort of package clash issue as I explained in the comment under my answer. It seems to conflict with siunitx package. – Vesnog Apr 28 '17 at 0:28\n• Then you need to edit your question to provide code which really does not compile, else we can't possibly know what the problem might be. – cfr Apr 28 '17 at 0:31\n• You are right, it did not occur to me it was a package clash issue in the first place. I think I need to upload my class file and the main file together with the defined macros. – Vesnog Apr 28 '17 at 0:32\n• @Vesnog Please see my edit above first. You haven't said what error you get, so I don't know whether the error I got is your error, but presumably you will know whether it is or not. At any rate, this is A problem with the code, even if not the only one. – cfr Apr 28 '17 at 0:38\n• I am now trying to change that \\ang` variable and see if it resolves the problem. – Vesnog Apr 28 '17 at 0:42"
] | [
null,
"https://i.stack.imgur.com/XRsMG.png",
null,
"https://i.stack.imgur.com/yTTFP.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5678722,"math_prob":0.8650185,"size":4239,"snap":"2019-35-2019-39","text_gpt3_token_len":1620,"char_repetition_ratio":0.12491145,"word_repetition_ratio":0.35042736,"special_character_ratio":0.38806322,"punctuation_ratio":0.24528302,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99276406,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T02:11:35Z\",\"WARC-Record-ID\":\"<urn:uuid:7a2fc4e6-bec9-4539-bbd1-f7b0ca888393>\",\"Content-Length\":\"150413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5042945b-d958-433a-bfcd-ebf7e2050ef4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8994255-ccfe-4e3b-9a05-d20b280b6296>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/367135/tikz-takes-forever-to-compile-when-imported-into-main-file/367136\",\"WARC-Payload-Digest\":\"sha1:JFX7DOBDQQIMZN5DQAGAVNH2WD5WQIAG\",\"WARC-Block-Digest\":\"sha1:DG6D454YK773NOCYKFSBQFBQHAOLZG6T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573011.59_warc_CC-MAIN-20190917020816-20190917042816-00292.warc.gz\"}"} |
https://stats.stackexchange.com/users/24669/juho-kokkala?tab=topactivity | [
"68 Is there any *mathematical* basis for the Bayesian vs frequentist debate? 21 Probability that the minimum of a normal random sample will exceed the maximum of another? 20 Gelman and Rubin convergence diagnostic, how to generalise to work with vectors? 19 Example for a prior, that unlike Jeffreys, leads to a posterior that is not invariant 19 Variance of a bounded random variable\n\n### Reputation (7,373)\n\n +10 Why is the Kalman Filter a specific case of a (dynamic) Bayesian network? +10 Gelman and Rubin convergence diagnostic, how to generalise to work with vectors? +10 Variance of a bounded random variable +10 Generating random numbers from a multiplication of CDFs\n\n### Questions (0)\n\nThis user has not asked any questions\n\n### Tags (147)\n\n 172 probability × 28 62 normal-distribution × 10 155 bayesian × 21 50 distributions × 11 68 frequentist 44 markov-chain-montecarlo × 4 68 kolmogorov-axioms 41 mathematical-statistics × 9 68 philosophical 34 gaussian-process × 6\n\n### Bookmarks (0)\n\nThis user has no bookmarked questions."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71068233,"math_prob":0.8551243,"size":575,"snap":"2021-31-2021-39","text_gpt3_token_len":173,"char_repetition_ratio":0.16987741,"word_repetition_ratio":0.0,"special_character_ratio":0.34086958,"punctuation_ratio":0.041237112,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96353805,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T04:58:17Z\",\"WARC-Record-ID\":\"<urn:uuid:f62b908e-e89c-46b1-85c2-a9e3fff353a6>\",\"Content-Length\":\"151649\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10ff2d67-8b96-49a2-99cc-3b57683b3344>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ebc6149-f06a-4266-8745-a486a21f1408>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/users/24669/juho-kokkala?tab=topactivity\",\"WARC-Payload-Digest\":\"sha1:W4YXQO3I4LJPIDTHHPGMZCKNVUAUKOTD\",\"WARC-Block-Digest\":\"sha1:YON3JJY4UM22I3APYB3ZRW2ASUGCQEZT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780054023.35_warc_CC-MAIN-20210917024943-20210917054943-00579.warc.gz\"}"} |
https://wanweibaike.com/wiki-%E6%BF%95%E5%BA%A6 | [
"",
null,
"# 湿度本文重定向自 濕度\n\n## 测量\n\n### 绝对湿度\n\n“绝对湿度”指一定体积的空气中含有的水蒸气的质量，一般其单位是公克立方米。绝对湿度的最大限度是饱和状态下的最高湿度。\n\n$\\rho _{w}:={\\frac {e}{R_{w}\\cdot T}}={\\frac {m}{V}}$",
null,
"e蒸汽压，单位是帕斯卡（Pa）\n\n$R_{w}$",
null,
"–水的气体常数=461.52J/（kg K）\n\nT温度，单位是开尔文（K）\n\nm –在空气中溶解的水的质量，单位是千克（kg）\n\nV –空气的体积，单位是立方米（m3）。\n\n### 相对湿度（RH）\n\n“相对湿度”（RH）是绝对湿度与最高湿度之间的比，它的值显示水蒸气的饱和度有多高。相对湿度为100%的空气是饱和的空气。相对湿度是50%的空气含有达到同温度的空气的饱和点的一半的水蒸气。相对湿度超过100%的空气中的水蒸气一般凝结出来。随着温度的增高，空气中可以含的水就增多。也就是说，在同样多的水蒸气的情况下，温度降低，相对湿度就会升高；温度升高，相对湿度就会降低。因此在提供相对湿度的同时也必须提供温度的数据。透过最高湿度和温度也可以计算出露点。\n\n$\\varphi :={\\frac {\\rho _{w}}{\\rho _{w,max}}}\\cdot 100\\ \\%={\\frac {e}{E}}\\cdot 100\\ \\%={\\frac {s}{S}}\\cdot 100\\ \\%$",
null,
"$\\rho _{w}$",
null,
"–绝对湿度，单位是公克／立方米\n\n$\\rho _{w,max}$",
null,
"–最高湿度，单位是公克／立方米\n\ne–蒸汽压，单位是帕斯卡\n\nE饱和蒸汽压，单位是帕斯卡\n\ns–比湿，单位是公克／公斤\n\nS–最高比湿，单位是公克／公斤\n\n### 比湿\n\n$s={\\frac {m_{\\mathrm {water} }}{m_{\\mathrm {air\\ total} }}}={\\frac {m_{\\mathrm {water} }}{m_{\\mathrm {air\\ dry} }+m_{\\mathrm {water} }}}={\\frac {\\frac {m_{\\mathrm {water} }}{V_{\\mathrm {total} }}}{{\\frac {m_{\\mathrm {air\\ total} }}{V_{\\mathrm {total} }}}+{\\frac {m_{\\mathrm {Water} }}{V_{\\mathrm {total} }}}}}={\\frac {\\rho _{\\mathrm {Water} }}{\\rho _{\\mathrm {air\\ dry} }+\\rho _{\\mathrm {water} }}}={\\frac {\\rho _{\\mathrm {water} }}{\\rho _{\\mathrm {air\\ total} }}}$",
null,
"$s={\\frac {\\rho _{\\mathrm {Water} }}{\\rho _{\\mathrm {air\\ dry} }+\\rho _{\\mathrm {Water} }}}={\\frac {\\frac {e}{R_{w}\\cdot T}}{{\\frac {p-e}{R_{L}\\cdot T}}+{\\frac {e}{R_{w}\\cdot T}}}}={\\frac {e\\cdot M_{\\mathrm {Water} }}{{(p-e)}\\cdot {M_{\\mathrm {air\\ dry} }}+{e}\\cdot {M_{\\mathrm {Water} }}}}={\\frac {{\\frac {M_{\\mathrm {Water} }}{M_{\\mathrm {air\\ dry} }}}\\cdot e}{p-\\left(1-{\\frac {M_{\\mathrm {Water} }}{M_{\\mathrm {air\\ dry} }}}\\right)\\cdot e}}\\approx {\\frac {0{,}622\\cdot e}{p-0{,}378\\cdot e}}\\approx 0{,}622\\cdot {\\frac {e}{p}}$",
null,
"$\\rho _{\\mathrm {Water} }={\\frac {e}{R_{w}\\cdot T}}$",
null,
"$R_{w}={\\frac {R}{M_{\\mathrm {Water} }}}$",
null,
"$\\rho _{\\mathrm {air\\ dry} }={\\frac {p-e}{R_{L}\\cdot T}}$",
null,
"$R_{L}={\\frac {R}{M_{\\mathrm {air\\ dry} }}}$",
null,
"$S:={\\frac {m_{\\mathrm {water\\ saturate} }}{m_{\\mathrm {air\\ total} }}}={\\frac {\\rho _{\\mathrm {Water\\ saturate} }}{\\rho _{\\mathrm {air\\ total} }}}\\approx {\\frac {0{,}622\\cdot E}{p-0{,}378\\cdot E}}$",
null,
"$\\rho _{x}$",
null,
"–密度，单位为公克／立方米\n\n$V_{\\mathrm {total} }$",
null,
"–湿空气的总体积，单位为立方米\n\n$R_{w}$",
null,
"–水的气体常数，单位为焦耳／（公斤·开尔文）\n\n$R_{L}$",
null,
"–干空气的气体常数，单位为焦耳／（公斤·开尔文）\n\nT–温度，开尔文\n\n$M_{\\mathrm {Water} }$",
null,
"–水的摩尔质量=18.01528公克／摩尔\n\n$M_{\\mathrm {air\\ dry} }$",
null,
"–干空气的摩尔质量=28.9634公克／摩尔\n\ne–蒸汽压，单位是帕斯卡\n\np气压，单位为帕斯卡\n\nE饱和蒸汽压，单位为帕斯卡\n\n## 参考文献\n\n• Häckel H.（1999）: Meteorologie. 4. Aufl. Ulmer Verlag, Stuttgart; UTB 1338; 448 S. ISBN 3-8252-1338-2\n• Zmarsly E., Kuttler W., Pethe H.（2002）: Meteorologisch-klimatologisches Grundwissen. Eine Einführung mit Übungen, Aufgaben und Lösungen. Ulmer Verlag, Stuttgart. S ISBN 3-8252-2281-0\n• Hupfer P., Kuttler W.（1998）: Witterung und Klima. Teubner Verlag, Stuttgart/Leipzig. ISBN 3-443-07123-6\n• Weischet W.（2002）: Einführung in die Allgemeine Klimatologie. Borntraeger . ISBN 3-443-07123-6"
] | [
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https://www.cnblogs.com/laojie4321/p/8552213.html | [
"## python数据分析及展示（二）\n\n1. Matplotlib库的介绍\n\nhttp://matplotlib.org/gallery.html可查看Matplotlib库的效果\n\nMatplotlib库由各种可视化类构成，内部结构复杂，受Matlab启发\nmatplotlib.pyplot是绘制各类可视化图形的命令子库，相当于快捷方式\n\nimport matplotlib.pyplot as plt\n\nplt.plot([3,1,4,5,2])\n\nplt.plot()只有一个输入列表或数组时，参数被当作Y轴，X轴以索引自动生成\n\nplt.savefig('test',dpi=600)\n\nplt.savefig()将输出图形存储为文件，默认PNG格式，可以通过dpi修改输出质量\n\nplt.plot([0,2,4,6,8],[3,1,4,5,2])\n\nplt.plot(x,y)当有两个以上参数时，按照X轴和Y轴顺序绘制数据点\n\nplt.subplot(nrows, ncols, plot_number)\n\nplt.subplot(3,2,4)或plt.subplot(324)",
null,
"2. pyplot的plot()函数\n\nplt.plot(x, y, format_string, **kwargs)\n∙ x : X轴数据，列表或数组，可选\n∙ y : Y轴数据，列表或数组\n∙ format_string: 控制曲线的格式字符串，可选，由颜色字符、风格字符和标记字符组成\n∙ **kwargs : 第二组或更多(x,y,format_string)\n\n'b' 蓝色 'm' 洋红色magenta\n'g' 绿色 'y' 黄色\n'r' 红色 'k' 黑色\n'c' 青绿色cyan 'w' 白色\n'#008000' RGB某颜色 '0.8' 灰度值字符串\n\n'‐' 实线\n'‐‐' 破折线\n'‐.' 点划线\n':' 虚线\n'' ' ' 无线条\n\n'.' 点标记\n',' 像素标记(极小点)\n'o' 实心圈标记\n'v' 倒三角标记\n'^' 上三角标记\n'>' 右三角标记\n'<' 左三角标记\n\n'1' 下花三角标记\n'2' 上花三角标记\n'3' 左花三角标记\n'4' 右花三角标记\n's' 实心方形标记\n'p' 实心五角标记\n'*' 星形标记\n\n'h' 竖六边形标记\n'H' 横六边形标记\n'+' 十字标记\n'x' x标记\n'D' 菱形标记\n'd' 瘦菱形标记\n'|' 垂直线标记\n\ncolor : 控制颜色, color='green'\nlinestyle : 线条风格, linestyle='dashed'\nmarker : 标记风格, marker='o'\nmarkerfacecolor: 标记颜色, markerfacecolor='blue'\nmarkersize : 标记尺寸, markersize=20\n……\n\n3. pyplot的中文显示\n\n（1）pyplot并不默认支持中文显示，需要rcParams修改字体实现\n\nimport matplotlib.pyplot as plt\n\nimport matplotlib\n\nmatplotlib.rcParams['font.family']='SimHei'\n\n'font.family' 用于显示字体的名字\n'font.style' 字体风格，正常'normal'或斜体'italic'\n'font.size' 字体大小，整数字号或者'large'、'x‐small'\n\nplt.xlabel('横向：时间',fontproperties='SimHei', fontsize=20)\n\n4. pyplot的文本显示\n\nplt.xlabel() 对X轴增加文本标签\nplt.ylabel() 对Y轴增加文本标签\nplt.title() 对图形整体增加文本标签\nplt.text() 在任意位置增加文本\nplt.annotate() 在图形中增加带箭头的注解",
null,
"",
null,
"5. pyplot的子绘图区域\n\nplt.subplot2grid(GridSpec, CurSpec, colspan=1, rowspan=1)",
null,
"",
null,
"",
null,
"",
null,
"1. pyplot基础图表函数概述\n\nplt.plot(x,y,fmt,…) 绘制一个坐标图\nplt.boxplot(data,notch,position) 绘制一个箱形图\nplt.bar(left,height,width,bottom) 绘制一个条形图\nplt.barh(width,bottom,left,height) 绘制一个横向条形图\nplt.polar(theta, r) 绘制极坐标图\nplt.pie(data, explode) 绘制饼图\n\nplt.cohere(x,y,NFFT=256,Fs) 绘制X‐Y的相关性函数\nplt.scatter(x,y) 绘制散点图，其中，x和y长度相同\nplt.step(x,y,where) 绘制步阶图\nplt.hist(x,bins,normed) 绘制直方图\n\nplt.contour(X,Y,Z,N) 绘制等值图\nplt.vlines() 绘制垂直图\nplt.stem(x,y,linefmt,markerfmt) 绘制柴火图\nplt.plot_date() 绘制数据日期\n\n2. pyplot饼图的绘制",
null,
"",
null,
"3. pyplot直方图的绘制",
null,
"hist 中20表示直方的个数\n\n4. pyplot极坐标的绘制",
null,
"5. pyplot散点图的绘制",
null,
"posted on 2018-03-12 21:52 n哖苡逅 阅读(579) 评论(0编辑 收藏 举报"
] | [
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"https://images2018.cnblogs.com/blog/392443/201803/392443-20180312213724324-319022141.png",
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"https://images2018.cnblogs.com/blog/392443/201803/392443-20180312213813245-978989939.png",
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"https://images2018.cnblogs.com/blog/392443/201803/392443-20180312214104774-1784500235.png",
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"https://images2018.cnblogs.com/blog/392443/201803/392443-20180312214230185-373213359.png",
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https://www.mathvids.com/browse/college/discrete-math/logic/logic/610-lecture-2-boolean-algebra-and-formal | [
"# Lecture 2: Boolean Algebra and Formal Logic\n\nTaught by ArsDigita\n• Currently 4.0/5 Stars.\n12720 views | 2 ratings\nPart of video series\nLesson Summary:\n\nIn this lesson, the concept of proofs in mathematics is explored, and the process of refining a mathematical idea and turning it into a proof is discussed. The importance of having a solid, rigorous proof is emphasized, and the idea of a proof being sociological is introduced. The role of intuition in creating a proof is also discussed, as well as the importance of finding a balance between rigor and intuition. Overall, the lesson provides a valuable insight into the process of creating and refining mathematical proofs.\n\nLesson Description:\n\nLearn about boolean Algebra, what it is, and how it relates to logic. Also, learn about formal logic rules.\n\n• How do you prove that you can get a total of 61 or higher with some combination of 7s and 11s?\n• What is an example of an incorrect proof by induction where the base case is not covered?\n• What is Boolean Algebra?\n• What are some rules, identities, equivalences, and proofs using truth tables in logic?\n• What is Disjunctive Normal Form?\n• What is Conjunctive Normal Form?\n• How do you change between Disjunctive and Conjunctive Normal Form?\n• What is the NOR or NAND function in logic?\n• How can you rewrite a logical statement by replacing all not, and, and or operators with nor or nand?\n• What are binary numbers and how do you convert decimal numbers into binary?\n• How do you add numbers in binary?\n• What is a half adder circuit and how does it work?\n• How do computer circuits work on the inside?\n• What is a reduction in logic and problem solving?\n• What is an NP-complete problem and what does it mean for a problem to be NP-complete?\n• What is a non-deterministic polynomial algorithm?\n• What is an example of a P or polynomial-time algorithm?\n• What is the traveling salesman problem and why is it an NP problem?\n• Why is it important to be able to show that a problem is NP-complete?\n• Why is satisfiability hard and how can you show that it is hard using reduction?\n• How do you show that a problem is easy using reduction?\n• How does 2SAT reduce to finding strongly connected components in a graph?\n•",
null,
"#### Staff Review\n\n• Currently 4.0/5 Stars.\nThis video starts with an incorrect proof to a problem by induction. The idea and philosophy behind proofs is dealt with. This leads into a very rigorous discussion on Boolean Algebra carrying through what was started in the last lecture. Circuits are explained using their binary components. Finally, hard problems, P, NP, and NP-complete problems are analyzed along with reductions of these problems.\n•",
null,
"#### JMP92\n\n• Currently 4.0/5 Stars."
] | [
null,
"https://www.mathvids.com/assets/testimonial/artist-testimonial-avatar-01-c96ee2a66f1e941fb1528708979edbcecf5383e31bcadfe00817b03e5645bd16.jpg",
null,
"https://www.mathvids.com/assets/testimonial/artist-testimonial-avatar-02-89a3754134f148832d676aad61f1a05cad29a66df58339b5c09a21ffe1bd42da.jpg",
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https://metanumbers.com/558056 | [
"## 558056\n\n558,056 (five hundred fifty-eight thousand fifty-six) is an even six-digits composite number following 558055 and preceding 558057. In scientific notation, it is written as 5.58056 × 105. The sum of its digits is 29. It has a total of 5 prime factors and 16 positive divisors. There are 275,184 positive integers (up to 558056) that are relatively prime to 558056.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 29\n• Digital Root 2\n\n## Name\n\nShort name 558 thousand 56 five hundred fifty-eight thousand fifty-six\n\n## Notation\n\nScientific notation 5.58056 × 105 558.056 × 103\n\n## Prime Factorization of 558056\n\nPrime Factorization 23 × 79 × 883\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 139514 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 558,056 is 23 × 79 × 883. Since it has a total of 5 prime factors, 558,056 is a composite number.\n\n## Divisors of 558056\n\n16 divisors\n\n Even divisors 12 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 1.0608e+06 Sum of all the positive divisors of n s(n) 502744 Sum of the proper positive divisors of n A(n) 66300 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 747.031 Returns the nth root of the product of n divisors H(n) 8.41713 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 558,056 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 558,056) is 1,060,800, the average is 66,300.\n\n## Other Arithmetic Functions (n = 558056)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 275184 Total number of positive integers not greater than n that are coprime to n λ(n) 22932 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45820 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 275,184 positive integers (less than 558,056) that are coprime with 558,056. And there are approximately 45,820 prime numbers less than or equal to 558,056.\n\n## Divisibility of 558056\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 0 1 2 2 0 2\n\nThe number 558,056 is divisible by 2, 4 and 8.\n\n## Classification of 558056\n\n• Arithmetic\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n## Base conversion (558056)\n\nBase System Value\n2 Binary 10001000001111101000\n3 Ternary 1001100111202\n4 Quaternary 2020033220\n5 Quinary 120324211\n6 Senary 15543332\n8 Octal 2101750\n10 Decimal 558056\n12 Duodecimal 22ab48\n16 Hexadecimal 883e8\n20 Vigesimal 39f2g\n36 Base36 bylk\n\n## Basic calculations (n = 558056)\n\n### Multiplication\n\nn×i\n n×2 1116112 1674168 2232224 2790280\n\n### Division\n\nni\n n⁄2 279028 186019 139514 111611\n\n### Exponentiation\n\nni\n n2 311426499136 173793426401839616 96986464364105008746496 54123878357174984761034571776\n\n### Nth Root\n\ni√n\n 2√n 747.031 82.3302 27.3319 14.1038\n\n## 558056 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 1.11611e+06 3.50637e+06 9.78375e+11\n\n### Sphere\n\nRadius = n\n Volume 7.27984e+17 3.9135e+12 3.50637e+06\n\n### Square\n\nLength = n\n Perimeter 2.23222e+06 3.11426e+11 789210\n\n### Cube\n\nLength = n\n Surface area 1.86856e+12 1.73793e+17 966581\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 1.67417e+06 1.34852e+11 483291\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.39407e+11 2.04818e+16 455651\n\n## Cryptographic Hash Functions\n\nmd5 cc48ab8487879c96795cceae57cc12d8 bcf7567ce8b0042e549d90b5ee613b68ca6fbd01 ac1db65757c1989340c1f94a3dd334c78a810855834e2c4267f795eeb2500c11 bd8ec247e17090d071357797b27964b2a74fe8f8f07cf9608d2139ee18bd328c9bbf397e41131700745a30617fde888a9df9a5d26b17ff048e12a29ff72d701a fe22c21d09d6509a4c09396d33b71b336cd9aca4"
] | [
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http://www.sta.cuhk.edu.hk/events/a-unified-matrix-model-including-both-cca-and-f-matrices-in-multivariate-analysis-the-largest-eigenvalue-and-its-applications/ | [
"• Upcoming Events\n• Awards\n• Distinguished Lecture\n• Latest Seminars and Events\n• Others\n• Seminars\n• Workshop and Conference\n• Past Events\n• Student Issue\nUpcoming Events\n Topic: A Unified Matrix Model Including Both CCA and F Matrices in Multivariate Analysis: The Largest Eigenvalue and Its Applications Date: 27/02/2018 Time: 2:30 pm - 3:30 pm Venue: William M W Mong Engineering Building (ERB) - Room 404 Category: Seminars Speaker: Dr. HAN Xiao PDF: 20180227_HAN.pdf Details: Abstract We consider a general matrix model \\$\\bold{\\Omega}=(\\bbZ\\bbU_2\\bbU_2^T\\bbZ^T)^{-1}\\bbZ\\bbU_1\\bbU_1^T\\bbZ^T\\$, where \\$\\bbU_1\\$ and \\$\\bbU_2\\$ are two orthogonal isometries and \\$\\bbZ\\$ is the matrix of observed data. We establish the asymptotic Tracy-Widom distribution for the largest eigenvalue of \\$\\Omega\\$ under moment assumptions on the data \\$\\bbZ\\$. This result has wide applications in practice. For example, by appropriately choosing \\$\\bbU_1\\$ and \\$\\bbU_2\\$, our results can be used in deriving the asymptotic distribution of the maximum eigenvalues of the matrices used in canonical correlation analysis (CCA) and of F matrices (including centered and non-centered versions). Moreover, via appropriate matrices \\$\\bbU_1\\$ and \\$\\bbU_2\\$, our result on \\$\\bold{\\bbom}\\$ can be applied to some multivariate testing problems that cannot be done by both types of matrices. To see this, we consider two specific examples. One is in the multivariate analysis of variance (MANOVA) approach for testing the equivalence of several high-dimensional mean vectors, where \\$\\bbU_1\\$ and \\$\\bbU_2\\$ are chosen to be two nonrandom matrices. The other one is in the multivariate linear model for testing the unknown parameter matrix, where \\$\\bbU_1\\$ and \\$\\bbU_2\\$ are random. Extensive simulation studies strongly support the theoretical results.",
null,
"(+852) 3943-7931",
null,
"(+852) 2603-5188",
null,
"statdept@cuhk.edu.hk",
null,
"https://www.sta.cuhk.edu.hk/"
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https://dyuthi.cusat.ac.in/xmlui/handle/purl/3142 | [
"# On infinite graphs and related matrices\n\n Title: On infinite graphs and related matrices Author: Ancykutty, Joseph; Dr.Krishnamoorthy,A; Dr.Thrivikraman, T Abstract: This thesis Entitled On Infinite graphs and related matrices.ln the last two decades (iraph theory has captured wide attraction as a Mathematical model for any system involving a binary relation. The theory is intimately related to many other branches of Mathematics including Matrix Theory Group theory. Probability. Topology and Combinatorics . and has applications in many other disciplines..Any sort of study on infinite graphs naturally involves an attempt to extend the well known results on the much familiar finite graphs. A graph is completely determined by either its adjacencies or its incidences. A matrix can convey this information completely. This makes a proper labelling of the vertices. edges and any other elements considered, an inevitable process. Many types of labelling of finite graphs as Cordial labelling, Egyptian labelling, Arithmetic labeling and Magical labelling are available in the literature. The number of matrices associated with a finite graph are too many For a study ofthis type to be exhaustive. A large number of theorems have been established by various authors for finite matrices. The extension of these results to infinite matrices associated with infinite graphs is neither obvious nor always possible due to convergence problems. In this thesis our attempt is to obtain theorems of a similar nature on infinite graphs and infinite matrices. We consider the three most commonly used matrices or operators, namely, the adjacency matrix Description: Department of mathematics, Cochin University of Science and Technology URI: http://dyuthi.cusat.ac.in/purl/3142 Date: 2000\n\n\n## Files in this item\n\nFiles Size Format View Description\nDyuthi-T1116.pdf 1.906Mb PDF View/Open Pdf"
] | [
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https://www.souravsirclasses.com/xi-xii | [
"top of page",
null,
"",
null,
"This is a great place to achieve your dream\n\nHere, at Sourav Sir's Classes, we strive, so that you succeed. Our team of experienced faculty is dedicated to ensuring that you stay light years ahead of the competition. Our classes employ a tried and tested procedure, and, we have a legacy of producing students who not only crack the exam but excel at all of them. Our goal is to nurture the brightest minds of the present and future generations, and we are absolutely delighted to have you on board.\n\n##### WHERE DO WE COME IN\n\nSimply put, XI-XII Mathematics exam is a tough exam to crack and with our rigorous training and grooming, it becomes less so. Here at Sourav Sir's Classes, we fuel the passion of those who dare dream. With our well-documented lecture notes as well as our study material, XI-XII Mathematics, become less formidable, and we have an entire nation of students who can vouch for that!\n\nOur tuitions for the XI-XII Mathematics ensures, under the guidance of our prestigious and able faculty, ensures that the aspirant develops the necessary practices and habits for cracking the entrance exams. We believe that studying here, under the guidance of our faculty, is not just an educational journey that we undertake every year, but we strive to make it an experience for the student as well.\n\nCome taste success with the best XI-XII Mathematics tuitions of the nation!\n\n## Regular Online Classes\n\n• Regular live classes are conducted by our prestigious faculty over Skype as they go over numerous concepts and problem-solving tricks\n\n• Special doubt-clearing classes are conducted regularly as the student progresses through the course\n\n• Mock tests are conducted on a weekly basis to track the progress of the student\n\n• A thorough analysis of the tests are conducted so that the student gets to work on weak points",
null,
"## Recorded Lectures\n\n• Pre recorded video lectures by our esteemed faculty are available for those who cannot attend regular classes\n\n• Lectures are custom tailored according to the course requirement, adhering strictly to the latest syllabus\n\n• Special doubt clearing classes are conducted regularly as the student progresses through the course\n\n• Mock tests are conducted on a weekly basis to track progress of the student\n\n• A thorough analysis of the tests are conducted so that the student gets to work on weak points",
null,
"## Offline Classes\n\n• Offline live classes are held every day at our prestigious study centers, conducted by our esteemed faculty.\n\n• Regular doubt clearing sessions are conducted\n\n• Video backups are provided for students who have missed the classes\n\n• Weekly mock tests are conducted to track the progress of the students\n\n• A complete analysis of the tests are conducted so that the student gets to work on the weak points",
null,
"##### MOCK TESTS SERIES STRUCTURE\n• Complete Mock Tests ..\n\n• You can give the exams from your home using a phone or laptop or pc or a tablet.\n\n• After every exam complete solutions with marks will be provided to you.\n\n• Any doubts with any part you are free to ask us via what's app/email / call.\n\n• All the Mock Tests are Exam oriented .\n\n• You can also ask for any special topic or subject or section based specialised tests also --we will provide you that.\n\nSTUDY MATERIAL STRUCTURE\n\n• Materials will cover all the topics\n\n+\n\n• Special short cut tricks\n\n+\n\n• Complete model papers with solutions\n\n+\n\n• Special selective questions with complete solutions .\n\nThe Central Board of Secondary Education (CBSE) has released the 2019 board exam date sheet on its official website - cbse.nic.in - on Sunday, December 23. The CBSE Class 12 board exam 2019 will begin from February 15 and will conclude on April 3. The CBSE Class 10 board exam 2019 will be held from February 21 to March 29.\nThe CBSE Class 10 and Class 12 Board Exam date sheet was released on Sunday, December 23, on the official website of the Board. The students can visit the official website of the Board to check and download their board exam date sheet.\n\nThe CBSE Board has released the 2019 board exam date sheet seven weeks before the exam date so that students can get enough time for planning and preparation of the papers.\n\nWhile preparing the date sheet, the Board has also made sure that Board exam dates doesn't coincide with the dates of competitive exams. Last year, the exam date of Class 12 Physics paper had to be reschedule as it was clashing with JEE Main entrance exam date.\n\nThe CBSE Board will conduct 2019 board exams in morning session i.e. from 10.30 AM to 1.30 PM. The students will receive answer books at 10.00 AM and question paper will be distributed at 10.15 AM\n\nThe Board offers 40 vocational subjects in Class XII and 15 in Class X. The Board offers 240 subjects in total, out of which, this year Class 10 and 12 students have opted for 30,000 combinations of subjects.\n\nThe CBSE board exam results 2019 is likely to be released by the first week of June.\n\n##### CBSE Class 12 Exam\n\nThe Class 12 examination of CBSE board is very competitive. The 12th standard examination will begin in March 2019 and will end in April 2019. The board will release the admit card the exam and students can download it by visiting the official website. The CBSE 2019 Admit Card will contain important details like roll number, exam center, etc.\n\nThe class 12 results will be declared in the month of May – June (tentative). In 2018 the 12th standard results were declared on 26th May 2018. Candidates who will be appearing in the 12th standard exam must work hard with dedication to succeed in the exam. They must be committed to 2 years of hard work to master the basics for the Class 12 board exam.",
null,
"##### EXAM PATTERN\n\nStudents studying in CBSE board are assessed in two areas: Co-scholastic and Scholastic. The academic year of the Scholastic areas is divided into two terms which are Term 1 and Term 2 and two types of tests which are Formative Assessment and Summative Assessment are conducted to evaluate the academic subjects.\n\nFormative Assessment: In the primary classes, the formative assessment tests are in the form of oral tests, dictation, homework, class test, projects & assignments, storytelling, elocution, memory test, quiz, etc.\n\nSummative Assessment: Here students are tested internally. The Summative Assessment (SA) tests are in the form of pen and paper. The tests are conducted by the school. The Summative Assessment is conducted at the end of each term two times each year.\n\n##### MAKE THE JOURNEY EASY",
null,
"",
null,
"",
null,
"",
null,
"##### SUBJECTS\n\nUnits I. Relations and Functions\n\nII. Algebra\n\nIII. Calculus\n\nIV. Vectors and Three-Dimensional Geometry\n\nV. Linear Programming\n\nVI. Probability\n\nAppendix: 1. Proofs in Mathematics\n\n2. Mathematical Modelling\n\n• Chapters with Time Allocation\n\n• Relations and Functions Periods\n\n• Inverse Trigonometric Functions Periods\n\n• Matrices Periods\n\n• Determinants Periods\n\n• Continuity and Differentiability Periods\n\n• Applications of Derivatives Periods\n\n• Integrals Periods\n\n• Applications of the Integrals Periods\n\n• Differential Equations Periods\n\n• Vectors Periods\n\n• Three-dimensional Geometry Periods\n\n• Linear Programming Periods\n\n• Probability Periods\n\nUnit I: Relations and Functions 1. Relations and Functions Types of relations: Reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. Binary operations. 2. Inverse Trigonometric Functions Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions.\n\nUnit II: Albegra 1. Matrices Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric and skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple properties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants Determinant of a square matrix (up to 3 × 3 matrices), properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.\n\nUnit III: Calculus\n\n1. Continuity and Differentiability Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse trigonometric functions, derivative of implicit function. Concept of exponential and logarithmic functions and their derivatives. Logarithmic differentiation. Derivative of functions expressed in parametric forms. Second order derivatives. Rolle’s and Lagrange’s Mean Value Theorems (without proof) and their geometric interpretations.\n\n2. Applications of Derivatives Applications of derivatives: Rate of change, increasing/decreasing functions, tangents and normals, approximation, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).\n\n3. Integrals Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, only simple integrals of the type ± + + ± – + + ? ? ? ? ? 2 2 2 2 2 2 2 2 , , , , , dx dx dx dx dx x a ax bx c x a a x ax bx c ++ ±- ++ ++ ? ? ? ? 2 2 2 2 2 2 ( ) ( ) ,, px q px q dx dx a x dx and x a dx ax bx c ax bx c to be evaluated. Definite integrals as a limit of a sum. Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.\n\n4. Applications of the Integrals Applications in finding the area under simple curves, especially lines, arcs of circles/parabolas/ ellipses (in standard form only), area between the two above said curves (the region should be clearly identifiable).\n\n5. Differential Equations Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solution of differential equations by method of separation of variables, homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type: P Q,dy y dx += where P and Q are functions of x.\n\nUnit IV: Vectors and Three-Dimensional Geometry\n\n1. Vectors Vectors and scalars, magnitude and direction of a vector. Direction cosines/ratios of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors.\n\n2. Three-dimensional Geometry Direction cosines/ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.\n\nUnit V: Linear Programming Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constrains).\n\nUnit VI: Probability Multiplication theorem on probability. Conditional probability, independent events, total probability, Baye’s theorem. Random variable and its probability distribution, mean and variance of haphazard variable. Repeated independent (Bernoulli) trials and Binomial distribution.\n\nAppendix 1. Proofs in Mathematics Through a variety of examples related to mathematics and already familiar to the learner, bring out different kinds of proofs: direct, contrapositive, by contradiction, by counter-example. 2. Mathematical Modelling Modelling real-life problems where many constraints may really need to be ignored (continuing from Class XI). However, now the models concerned would use techniques/results of matrices, calculus and linear programming.",
null,
"##### NOTES AND STUDY MATERIALS\n\nSTUDY MATERIALS ARE PROVIDED free of cost to students who have enrolled in our programs\n\nOur faculty of highly qualified teachers and researchers have combed through several books and journals, having collected and processed all the information so that it becomes easily accessible to the student. The study material is prepared in a time tested fashion with the student kept in mind in such a fashion that the student can catch up on the various topics in the event of missed lectures. Besides the theory being extensively discussed, an enormous collection of problems has been added so that the student can learn to apply the concepts over a diverse spectrum of difficulty. All the problems have been worked out thoroughly so that in case the student gets stuck, we've got you covered !",
null,
"## Email: souravsirclasses@gmail.com",
null,
"## US\n\n8/2C, Kashi Ghosh Lane,\n\nKolkata 700006",
null,
""
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null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9208088,"math_prob":0.86810064,"size":14285,"snap":"2023-40-2023-50","text_gpt3_token_len":3045,"char_repetition_ratio":0.12380085,"word_repetition_ratio":0.0540305,"special_character_ratio":0.20630032,"punctuation_ratio":0.11601563,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.963893,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T04:20:21Z\",\"WARC-Record-ID\":\"<urn:uuid:c5ea3f3e-83c8-43c8-9896-19941f9962bf>\",\"Content-Length\":\"1050453\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97c75991-d5f1-4741-b368-d6322acbadca>\",\"WARC-Concurrent-To\":\"<urn:uuid:689971f6-4e6f-476b-9a26-800ec25adb60>\",\"WARC-IP-Address\":\"146.75.33.84\",\"WARC-Target-URI\":\"https://www.souravsirclasses.com/xi-xii\",\"WARC-Payload-Digest\":\"sha1:TVISU7PGQIBKFEHD6GWNUP3IQIARYFFE\",\"WARC-Block-Digest\":\"sha1:COL2J2KECK65ODKQLJ5EAKDO3VQ44P7W\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100484.76_warc_CC-MAIN-20231203030948-20231203060948-00357.warc.gz\"}"} |
http://blog.cambridgecoaching.com/an-insiders-tip-to-prepping-for-the-sat-math-section-plug-in-numbers | [
"",
null,
"Preparing for both mathematics sections on the SAT can be a bit intimidating. You can’t expect yourself to know every topic that might come up, and the time limit adds to the stress. Much more efficient than trying to learn everything you might come across is to start with what you have already learned in high school and use examples to apply it more widely in the SAT. The most classic sort of example is what teachers have been drilling for ages: plug in numbers. However, mathematics is all about generalizing rules and strategies, so I want to talk about how to expand plugging in numbers to the art of creating your own examples.\n\nFirst, we are going to look at a normal example of plugging in numbers. An example SAT question might be:",
null,
"Now, if you are all brushed-up on your exponent rules, this problem is nice and algebraic and maybe even pretty. But if you are a bit rusty, we can get around the hard part of the problem. We just plug some numbers in for x or y to eliminate answers.\n\nA few tips on plugging in:\n\n1. Pick easy numbers. 0 and 1 are much easier to work with than 7 or 12.\n2. Don’t over-plug--we want to remove variables, but leave the problem possible. For instance, in this question, we plug in a number for either or and find the other value algebraically. If we plug in for both, the expression breaks down.\n3. If unsure, plug in a second number.\n\nUsing these rules, we can plug in y = 0, so 3x = 12 or = 4. Then the expression becomes",
null,
"We can multiply this out by hand to find its actual value, 4096. This rules out B) and C), which multiply out to 256 and 64 respectively. A) also multiplies out to 4096, but we need to feel confident that D) isn’t right too. Since we aren’t sure, we follow rule 3 and plug in again. This time, we will do y = 3 (we pick a value of y so that x is a whole number, following rule 1). Then we get:",
null,
"This is the same thing we got last time, which makes us much more confident on A) as the right answer.\n\nIt’s worth noting that we are not entirely certain through this method; these values might have been coincidences, but it seems unlikely at this point and a confident guess is much better than leaving the question blank. We don’t want to spend too much time on an individual question, though, so we are going to leave it at that.\n\n# Using Examples\n\nNow, we are going to expand on that method. No longer do we just plug in numbers, we plug in whole examples instead! The sample problem here is question 29 from the calculator section of the same practice test.",
null,
"Almost every student to whom I have shown this problem has been a bit confused, and for good reason! This question examines topics that are not frequently covered in a math curriculum—synthetic division and the properties of polynomials. However, we can make a few example polynomials that p(x) might clarify the problem. For example, p(x) = x- 5 satisfies the condition (i.e. p(3) = 3 - 5 = -2) and rules out B) and C). But we still don’t know if A) or D) is valid, so by rule 3, we try another example. We might pick p(x) = 2x - 8 = 2(x - 4). This example shows that none of the listed linear terms are factors of p(x) so D) is the right answer. Even though we might not even know what the answer is saying, we were able to rule out all of the other possible answers that we did understand.\n\n# Going Beyond the Exam\n\nAnything I teach in preparation for a standardized exam I try to make applicable outside of the exam. Using examples like these will serve you for generally learning mathematics. To flesh out a concept that you have recently learned, it can be helpful to pick out and test unusual examples. For instance, a topic that frequently comes up with students soon after the SAT is “limits”. These tools consider the continuous properties of functions. Instead of looking at the value of a function at a point (i.e. f(2) = 5), we look at the value the function tends to at a point, which we write:",
null,
"Frequently these two italicized phrases line up. An example might be f(x) = 2x . When we plug in 2, we get f(2) = 2 * 2 = 4. The function is a line, so as we plug in points closer and closer to 2, we get values closer and closer to 4:",
null,
"Now we can use another example to understand when the two italicized quantities are different to get a full sense of the properties of limits. We will create an excessively didactic example: a function, g, that is equal to 1 everywhere except for when x = 0, where g(0) = 0. If we plug points into that are closer and closer to 0 (but not 0), we keep getting 1. So the value the function tends to at 0 is 1:",
null,
"which is different from the value we get when we plug in zero itself! We illustrated the properties of limits simply by taking examples.\n\nAnd you can do this at home. Whenever you are struggling to understand a concept, try to make some examples to examine it. This is an art, and it isn’t always obvious what examples to pick. But skill and practice that you build up will have applications across all of the math you study (and the SAT!).\n\nAre you interested in getting set up with an SAT mathematics tutor?",
null,
"(this sample problem is question 14 on the no-calculator section of the sample test available here\n\nTags: math"
] | [
null,
"https://blog.cambridgecoaching.com/hs-fs/hubfs/how%20to%20prepare%20for%20the%20SAT%20math%20section.jpg",
null,
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null,
"https://no-cache.hubspot.com/cta/default/174241/80cb5e58-6cbe-4a0b-bec4-a3cdd6f96b76.png",
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http://www.dailyfreecode.com/Chat/Programmer1516.aspx | [
"",
null,
"Search:\n\n### GuestBook of Hildegarde Miller\n\nEnter your message (HTML allowed after you earn 150 points)",
null,
"Victor Slade from United States Mar 24 Hi,How're you?My name's Victor. I studying Computer Engineering.I need your help with something.My teacher gave a homework but I can't.The subject of homework is parallel matrix multiplication using multi-threading.You are given two NxN matrices and you will multiply these matrices in parallel.In this project you are required to solve this problem with p threads. Each thread would be responsible for finding multiplication result of N/p rows.For example, if matrices are 8x8 and number of threads is 4, then first thread computes the result of 1st and 2nd rows, second thread computes the result of 3rd and4th rows and so on"
] | [
null,
"http://www.dailyfreecode.com/Images/Logo/logo.gif",
null,
"http://www.gravatar.com/avatar/0da4d7d4057e39a64bf3001d549a9d50",
null
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https://www.bioregionalassessments.gov.au/assessments/261-surface-water-numerical-modelling-galilee-subregion/26162-results-analysis | [
"# 2.6.1.6.2 Results analysis\n\nPage 29 of 37\n\n## Annual flow\n\nFigure 13 shows the hydrological changes to the annual flow (AF) at 61 model nodes. The biggest impact occurs at node 3 (the uppermost model node in Sandy Creek), where the median pmax is −21%. That is, of the reductions in streamflow between the baseline and CRDP from the 347 replicates, the median of the predicted maximum changes is 21%. There is a tightly constrained distribution of pmax values around this median value. At all other model nodes in Sandy Creek (tributary 10) the median reduction in pmax exceeds 9%.",
null,
"Figure 13 Predictive distribution of (a) maximum raw change (amax), (b) maximum percentage change (pmax) and (c) year of maximum change (tmax) for annual flow (AF) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\nElsewhere, there are median reductions of between 12% and 18% in North Creek (tributary 8), up to 12% in Bully Creek (tributary 5) and up to 9% at node 1 which is on a tributary of Native Companion Creek (tributary 11). In the Belyando River (tributary 2) the median impact on pmax is less than 2%, while further downstream in the Suttor River (tributary 1) the median impact is less than 1% and corresponds to a decrease in flow of about 75 ML/day (or 28 GL/year). The additional coal resource development has no impact at all on AF in tributaries 3, 4, 6 and 7.\n\nThe maximum changes in AF occur in 2040 at most of the model nodes in tributaries 11, 10, 2 and 1 and there is little uncertainty in these years of maximum change. There is greater uncertainty in tmax at the model nodes in tributaries 9, 8 and 5, where the median tmax values occur between 2040 and 2051.\n\nNodes 3 and 1 have relatively small catchment areas which both contain parts of the South Galilee Coal Project. The projected changes in pmax at these locations are similar in magnitude to the proportion of their catchment areas that are included in the mine footprint (24% and 10%, respectively, Table 4 in Section 2.6.1.3). The model nodes in tributary 8 are downstream of the proposed Carmichael Coal Mine Project and the China Stone Coal Project, while most of the model nodes in tributary 5 are downstream of the Hyde Park Coal Project.\n\nAmong the most heavily impacted model nodes, the largest uncertainties in pmax occur in tributaries 5 and 8 (Carmichael River and North Creek, respectively). The impact on pmax at node 35 (Native Companion Creek) is very small for most replicates, but the 5th percentile change is 4% and the year of maximum change (the median tmax is in 2093) is much later than at other model nodes. This model node is upstream of all coal mines, but does have interaction with the groundwater model, so the only source of flow perturbation is through reductions in baseflow associated with groundwater drawdown.\n\n## Interquartile range\n\nFigure 14 shows the changes to the interquartile range (IQR) in AF. As for AF, the largest median pmax values occur in tributaries 10 and 8. The median reductions in IQR vary between 10% and 21% in tributary 10 (Sandy Creek) and between 17% and 19% in tributary 8 (North Creek). Although the median changes in tributaries 2 and 11 are smaller, there is greater uncertainty at some model nodes with some extreme replicates having predicted IQR reductions in excess of 60%.\n\nThe median tmax values for IQR are almost all in 2040, though some of the model nodes in tributary 11 (Native Companion Creek) have median tmax values of 2043 and are associated with considerable uncertainty.",
null,
"Figure 14 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for interquartile range (IQR) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## 99th percentile\n\nFigure 15 shows the decrease in daily streamflow rate at the 99th percentile (P99) in the 61 model nodes. Again, the biggest median reductions are in tributaries 10 (21% at node 3) and 8 (19% at node 15). However, there are also substantial impacts predicted for nodes 19 and 1 with reductions of 12% and 9%, respectively. The median year of maximum change in P99 remains 2040 in tributaries 10 and 11, but is slightly earlier (2038) in some model nodes in tributaries 1, 2 and 5, and slightly later in tributaries 8 (2042) and 9 (2051).",
null,
"Figure 15 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the daily streamflow rate at the 99th percentile (P99) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## Flood (high-flow) days\n\nFigure 16 shows the changes to the number of flood (high-flow) days (FD) at the 61 model nodes. There are reductions in median amax of up to 31 days in tributary 10, up to 24 days in tributary 8 and up to 15 days in tributary 11. However, there is much greater uncertainty around changes in the number of high-flow days (and in the timing of the maximum impacts) than there is for changes in annual flow.",
null,
"Figure 16 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the number of flood (high-flow) days (FD) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## Zero-flow days\n\nThe remaining results figures characterise the changes for the low-streamflow hydrological response variables. Figure 17 shows the changes to the number of zero-flow days (ZFD) at the 61 model nodes. The biggest changes are in the lower parts of the Belyando River (tributaries 1 and 2) where the median number of ZFD increase by more than 75 days per year in the reaches between nodes 34 and 32. The increases exceed 65 days further downstream in the Suttor River (tributary 1). Elsewhere, there are predicted median increases in ZFD of up to 16 days in Native Companion Creek (tributary 11) with the increases getting larger with distance downstream. In all model nodes with substantial increases in ZFD, there is considerable uncertainty in the amax values.\n\nThe increases in ZFD in tributaries 1, 2 and 11 tend to occur relatively late in the simulation period, with median predicted tmax values occurring later than 2080. In other model nodes with less substantial increases in ZFD, the median year of maximum change is earlier and occurs between 2030 and 2050.",
null,
"Figure 17 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the number of zero-flow days (ZFD) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## Low-flow days\n\nFigure 18 shows the predicted changes to low-flow days (LFD), the number of days per year the flow is less than the long-term 10th percentile, at 61 model nodes. As for ZFD, the largest changes in the frequency of LFD occur in tributaries 1 and 2, where median increases in the predicted maximum annual difference between the two development pathways are between 100 and 200 days. Increases of LFD of more than 40 days per year are predicted for some model nodes in Sandy Creek and Native Companion Creek (tributaries 10 and 11, respectively).\n\nYears of maximum change in LFD reflect those for ZFD, with later median tmax values being predicted for tributaries 1, 2 and 11 contrasting with earlier tmax values for model nodes in tributaries 8 and 10.",
null,
"Figure 18 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the number of low-flow days (LFD) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## Low-flow spells\n\nFigure 19 shows the changes to the number of low-flow spells (LFS) at the 61 model nodes. The maximum annual number of LFS across the modelling domain is simulated to increase by a median of up to 10 events. Median maximum increases of more than five events per year are predicted for some model nodes in tributaries 1, 2, 5 and 10, with smaller median increases in tributaries 8, 9 and 11. Maximum changes in LFS are likely to occur before 2050 in all parts of the subregion, except tributary 11, where median tmax values of between 2061 and 2075 are predicted.",
null,
"Figure 19 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the number of low-flow spells (LFS) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\n## Longest low-flow spell\n\nFigure 20 shows the maximum changes to the length of the longest low-flow spell (LLFS) at the 61 model nodes. The longest low-flow spell is projected to increase in length by about 70 days between nodes 34 and 33 in the Belyando River (tributary 2). Other substantial increases in LLFS (by more than 20 days) are predicted for tributaries 1, 8, 10 and 11. The large changes in LLFS are likely to occur between 2065 and 2081 in tributaries 1, 2 and 11, but between 2032 and 2043 in tributaries 8 and 10.",
null,
"Figure 20 Predictive distribution of (a) maximum raw change (amax), (b) maximum percent change (pmax) and (c) year of maximum change (tmax) for the length of the longest low-flow spell (LLFS) at the 61 model nodes in the Galilee subregion\n\nThe alternating grey and white shaded zones indicate which surface water model nodes belong to each of the 11 tributaries that comprise the surface water modelling domain in the Galilee subregion. These are numbered at the top of the uppermost figure and are described in more detail in Table 4 and shown in the map on Figure 9. The circle indicates the median prediction, the thick orange vertical line spans the 25th and 75th percentiles and the thin orange vertical line spans the 5th and 95th percentiles.\n\nData: Bioregional Assessment Programme (Dataset 1)\n\nLast updated:\n3 January 2019",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8952067,"math_prob":0.8902797,"size":13797,"snap":"2020-45-2020-50","text_gpt3_token_len":3384,"char_repetition_ratio":0.17610382,"word_repetition_ratio":0.45031983,"special_character_ratio":0.23534101,"punctuation_ratio":0.06666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96865076,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T21:12:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5d1c2318-a0a6-4ac2-9d1e-f4cd4e862b15>\",\"Content-Length\":\"63126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b2a6f6f-b0b9-452d-965e-db85d4226a45>\",\"WARC-Concurrent-To\":\"<urn:uuid:53e0d715-b031-4e32-9fff-c94c153ed78f>\",\"WARC-IP-Address\":\"54.66.240.82\",\"WARC-Target-URI\":\"https://www.bioregionalassessments.gov.au/assessments/261-surface-water-numerical-modelling-galilee-subregion/26162-results-analysis\",\"WARC-Payload-Digest\":\"sha1:M3MUC4XAW7S5UPDO6FCTBPL4HDV2FNG2\",\"WARC-Block-Digest\":\"sha1:5MP2LHTDOXIEU6W7KSCU7MKSCGMWNEV6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141681524.75_warc_CC-MAIN-20201201200611-20201201230611-00671.warc.gz\"}"} |
https://www.varsitytutors.com/sat_math-help/how-to-find-an-angle-in-a-hexagon | [
"## Example Questions\n\n200\n\n210\n\n170\n\n190\n\n180\n\n### Example Question #2 : How To Find An Angle In A Hexagon\n\nIf a triangle has 180 degrees, what is the sum of the interior angles of a regular octagon?",
null,
"",
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"",
null,
"",
null,
"",
null,
"",
null,
"Explanation:\n\nThe sum of the interior angles of a polygon is given by",
null,
"where",
null,
"= number of sides of the polygon. An octagon has 8 sides, so the formula becomes",
null,
"### Example Question #1 : How To Find An Angle In A Hexagon\n\nFind the sum of all the inner angles in a hexagon.",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Explanation:\n\nTo solve, simply use the formula to find the total degrees inside a polygon, where n is the number of vertices.\n\nIn this particular case, a hexagon means a shape with six sides and thus six vertices.\n\nThus,",
null,
"### All SAT Math Resources",
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] | [
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"https://vt-vtwa-app-assets.varsitytutors.com/assets/problems/og_image_practice_problems-9cd7cd1b01009043c4576617bc620d0d5f9d58294f59b6d6556fd8365f7440cf.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6632162,"math_prob":0.9480566,"size":2770,"snap":"2020-45-2020-50","text_gpt3_token_len":643,"char_repetition_ratio":0.27982646,"word_repetition_ratio":0.12061404,"special_character_ratio":0.21010831,"punctuation_ratio":0.15471698,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99976844,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,10,null,10,null,null,null,10,null,10,null,null,null,10,null,10,null,10,null,4,null,4,null,8,null,4,null,8,null,4,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T20:56:59Z\",\"WARC-Record-ID\":\"<urn:uuid:90edaea9-8b38-4011-8b7c-0919b9f5ea6d>\",\"Content-Length\":\"199295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:553875c1-99ff-4f40-8902-674f04d70e77>\",\"WARC-Concurrent-To\":\"<urn:uuid:8aafce19-4e2c-42e3-8549-8ae5bdb52b36>\",\"WARC-IP-Address\":\"13.32.182.30\",\"WARC-Target-URI\":\"https://www.varsitytutors.com/sat_math-help/how-to-find-an-angle-in-a-hexagon\",\"WARC-Payload-Digest\":\"sha1:GTJEXAQJTXJVXHP77LAUGTSR3ZHFHJFI\",\"WARC-Block-Digest\":\"sha1:WT45WGLEDRYOQQG25L5UMTJPBX5C4P3F\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141732696.67_warc_CC-MAIN-20201203190021-20201203220021-00033.warc.gz\"}"} |
http://slideplayer.com/slide/763417/ | [
"",
null,
"# Motion can be plotted on a graph by labeling the X- axis with TIME and the Y-axis DISTANCE.\n\n## Presentation on theme: \"Motion can be plotted on a graph by labeling the X- axis with TIME and the Y-axis DISTANCE.\"— Presentation transcript:\n\nMotion can be plotted on a graph by labeling the X- axis with TIME and the Y-axis DISTANCE.\n\nA point on a graph can show the location and speed of an object at a particular time. Those points are: x (time) y (distance) DONT WRITE THIS\n\nConsider a car moving with a constant, rightward (+) velocity - say of +10 m/s. If the position-time data for such a car were graphed, the resulting graph would look like the graph at the right. Note that a motion described as a constant, positive velocity results in a line of constant and positive slope when plotted as a position-time graph. www.glenbrook.k12 DONT WRITE THIS\n\nConsider a car moving with a rightward (+), changing velocity - that is, a car that is moving rightward but speeding up or accelerating. If the position-time data for such a car were graphed, then the resulting graph would look like the graph at the right. Note that a motion described as a changing, positive velocity results in a line of changing and positive slope when plotted as a position-time graph. www.glenbrook.k12 DONT WRITE THIS\n\nactiveart/motion_graphs/motion_graphs.html Another example…\n\nSlope tells you how one variable changes in relation to the other variable or the rate of change. DONT WRITE THIS\n\nSpeed and slope: Speed is the rate of change in distance in relation to time. The faster the object the steeper the slope.\n\nThis object is increasing speed. As time goes by, distance increases.\n\nA straight line on a graph is linear or has constant speed. A non-linear line illustrates increasing or decreasing speed\n\nRise Divided by Run Rise Run Slope = x (time) y (distance) OR…\n\ny 1 y 2 - 2 1 x x - SLOPE =\n\nDONT WRITE THIS\n\nTry to calculate the slope of the linear line... Rise is 20-50 meters Run is 4- 10 seconds 410 20 50\n\n1. 15m/s 2. 10m/s 3. 3m/s\n\n1. 15m/s 2. 10m/s 3. 18m/s\n\n1. 15m/s 2. 10m/s 3. 18m/s Time (seconds) Distance (meters)\n\n1. 15m/s 2. 10m/s 3. 18m/s Time (seconds) Distance (meters)\n\nThe shapes of the distance vs. time graphs for constant velocity and accelerated motion (i.e., changing velocity) - reveal an important principle. The principle is that the slope of the line shows useful information about the velocity of an object. Whatever characteristics the velocity has, the slope will exhibit the same. If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). This can be applied to any motion. Importance of Slope and Motion www.glenbrook.k12\n\nDownload ppt \"Motion can be plotted on a graph by labeling the X- axis with TIME and the Y-axis DISTANCE.\"\n\nSimilar presentations"
] | [
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"http://slideplayer.com/static/blue_design/img/slide-loader4.gif",
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https://www.nag.com/numeric/nl/nagdoc_24/nagdoc_fl24/html/f07/f07fuf.html | [
"F07 Chapter Contents\nF07 Chapter Introduction\nNAG Library Manual\n\n# NAG Library Routine DocumentF07FUF (ZPOCON)\n\nNote: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.\n\n## 1 Purpose\n\nF07FUF (ZPOCON) estimates the condition number of a complex Hermitian positive definite matrix $A$, where $A$ has been factorized by F07FRF (ZPOTRF).\n\n## 2 Specification\n\n SUBROUTINE F07FUF ( UPLO, N, A, LDA, ANORM, RCOND, WORK, RWORK, INFO)\n INTEGER N, LDA, INFO REAL (KIND=nag_wp) ANORM, RCOND, RWORK(N) COMPLEX (KIND=nag_wp) A(LDA,*), WORK(2*N) CHARACTER(1) UPLO\nThe routine may be called by its LAPACK name zpocon.\n\n## 3 Description\n\nF07FUF (ZPOCON) estimates the condition number (in the $1$-norm) of a complex Hermitian positive definite matrix $A$:\n $κ1A=A1A-11 .$\nSince $A$ is Hermitian, ${\\kappa }_{1}\\left(A\\right)={\\kappa }_{\\infty }\\left(A\\right)={‖A‖}_{\\infty }{‖{A}^{-1}‖}_{\\infty }$.\nBecause ${\\kappa }_{1}\\left(A\\right)$ is infinite if $A$ is singular, the routine actually returns an estimate of the reciprocal of ${\\kappa }_{1}\\left(A\\right)$.\nThe routine should be preceded by a call to F06UCF to compute ${‖A‖}_{1}$ and a call to F07FRF (ZPOTRF) to compute the Cholesky factorization of $A$. The routine then uses Higham's implementation of Hager's method (see Higham (1988)) to estimate ${‖{A}^{-1}‖}_{1}$.\nHigham N J (1988) FORTRAN codes for estimating the one-norm of a real or complex matrix, with applications to condition estimation ACM Trans. Math. Software 14 381–396\n\n## 5 Parameters\n\n1: UPLO – CHARACTER(1)Input\nOn entry: specifies how $A$ has been factorized.\n${\\mathbf{UPLO}}=\\text{'U'}$\n$A={U}^{\\mathrm{H}}U$, where $U$ is upper triangular.\n${\\mathbf{UPLO}}=\\text{'L'}$\n$A=L{L}^{\\mathrm{H}}$, where $L$ is lower triangular.\nConstraint: ${\\mathbf{UPLO}}=\\text{'U'}$ or $\\text{'L'}$.\n2: N – INTEGERInput\nOn entry: $n$, the order of the matrix $A$.\nConstraint: ${\\mathbf{N}}\\ge 0$.\n3: A(LDA,$*$) – COMPLEX (KIND=nag_wp) arrayInput\nNote: the second dimension of the array A must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}\\right)$.\nOn entry: the Cholesky factor of $A$, as returned by F07FRF (ZPOTRF).\n4: LDA – INTEGERInput\nOn entry: the first dimension of the array A as declared in the (sub)program from which F07FUF (ZPOCON) is called.\nConstraint: ${\\mathbf{LDA}}\\ge \\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}\\right)$.\n5: ANORM – REAL (KIND=nag_wp)Input\nOn entry: the $1$-norm of the original matrix $A$, which may be computed by calling F06UCF with its parameter ${\\mathbf{NORM}}=\\text{'1'}$. ANORM must be computed either before calling F07FRF (ZPOTRF) or else from a copy of the original matrix $A$.\nConstraint: ${\\mathbf{ANORM}}\\ge 0.0$.\n6: RCOND – REAL (KIND=nag_wp)Output\nOn exit: an estimate of the reciprocal of the condition number of $A$. RCOND is set to zero if exact singularity is detected or the estimate underflows. If RCOND is less than machine precision, $A$ is singular to working precision.\n7: WORK($2×{\\mathbf{N}}$) – COMPLEX (KIND=nag_wp) arrayWorkspace\n8: RWORK(N) – REAL (KIND=nag_wp) arrayWorkspace\n9: INFO – INTEGEROutput\nOn exit: ${\\mathbf{INFO}}=0$ unless the routine detects an error (see Section 6).\n\n## 6 Error Indicators and Warnings\n\nErrors or warnings detected by the routine:\n${\\mathbf{INFO}}<0$\nIf ${\\mathbf{INFO}}=-i$, the $i$th parameter had an illegal value. An explanatory message is output, and execution of the program is terminated.\n\n## 7 Accuracy\n\nThe computed estimate RCOND is never less than the true value $\\rho$, and in practice is nearly always less than $10\\rho$, although examples can be constructed where RCOND is much larger.\n\nA call to F07FUF (ZPOCON) involves solving a number of systems of linear equations of the form $Ax=b$; the number is usually $5$ and never more than $11$. Each solution involves approximately $8{n}^{2}$ real floating point operations but takes considerably longer than a call to F07FSF (ZPOTRS) with one right-hand side, because extra care is taken to avoid overflow when $A$ is approximately singular.\nThe real analogue of this routine is F07FGF (DPOCON).\n\n## 9 Example\n\nThis example estimates the condition number in the $1$-norm (or $\\infty$-norm) of the matrix $A$, where\n $A= 3.23+0.00i 1.51-1.92i 1.90+0.84i 0.42+2.50i 1.51+1.92i 3.58+0.00i -0.23+1.11i -1.18+1.37i 1.90-0.84i -0.23-1.11i 4.09+0.00i 2.33-0.14i 0.42-2.50i -1.18-1.37i 2.33+0.14i 4.29+0.00i .$\nHere $A$ is Hermitian positive definite and must first be factorized by F07FRF (ZPOTRF). The true condition number in the $1$-norm is $201.92$.\n\n### 9.1 Program Text\n\nProgram Text (f07fufe.f90)\n\n### 9.2 Program Data\n\nProgram Data (f07fufe.d)\n\n### 9.3 Program Results\n\nProgram Results (f07fufe.r)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77427137,"math_prob":0.998748,"size":3528,"snap":"2021-43-2021-49","text_gpt3_token_len":990,"char_repetition_ratio":0.111521,"word_repetition_ratio":0.034013607,"special_character_ratio":0.25141722,"punctuation_ratio":0.13190185,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99953616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T09:07:28Z\",\"WARC-Record-ID\":\"<urn:uuid:f1aa9484-87ef-4274-beda-bbe6f3745495>\",\"Content-Length\":\"18601\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b522faf2-9c1d-42a7-b9eb-8df5b2e6c648>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ba4f3a0-a7f4-400e-8187-6e0075c4dba0>\",\"WARC-IP-Address\":\"78.129.168.4\",\"WARC-Target-URI\":\"https://www.nag.com/numeric/nl/nagdoc_24/nagdoc_fl24/html/f07/f07fuf.html\",\"WARC-Payload-Digest\":\"sha1:BVLSNQA4QJVXD5WSP3M6VG4P2AWXD2CF\",\"WARC-Block-Digest\":\"sha1:LOO43AKF7EAF27WHFNG5VJWFFO7UFLSW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585504.90_warc_CC-MAIN-20211022084005-20211022114005-00423.warc.gz\"}"} |
https://niniloos.com/volume-of-prisms-and-cylinders-worksheets/ | [
"# Volume of prisms and cylinders worksheets information\n\n» » Volume of prisms and cylinders worksheets information\n\nYour Volume of prisms and cylinders worksheets images are available. Volume of prisms and cylinders worksheets are a topic that is being searched for and liked by netizens now. You can Find and Download the Volume of prisms and cylinders worksheets files here. Download all royalty-free vectors.\n\nIf you’re looking for volume of prisms and cylinders worksheets pictures information connected with to the volume of prisms and cylinders worksheets interest, you have visit the right site. Our website always gives you suggestions for seeking the maximum quality video and picture content, please kindly hunt and locate more enlightening video articles and images that fit your interests.\n\nVolume Of Prisms And Cylinders Worksheets. 2 4 ft 3 ft 35 ft 126 ft. 09092019 The Corbettmaths Practice Questions on the Volume of a Cylinder. Volume Of Prisms And Cylinders Practice A. This is superb thank you.",
null,
"Pin On Geometry Worksheets From pinterest.com\n\nMore Volumes of prisms and cylinders interactive worksheets. This is superb thank you. Prisms including Cavalieris Principle Notes and Practice3 pages total. Take a break from typical worksheets with this cut and paste activity. Find the volume of rectangular prisms and cylinders. Some of the worksheets displayed are Volume Volume 10 volume of prisms and cylinders Volume of prisms and cylinders work answers Volume and surface area of rectangular prisms and cylinders Volume cones spheres and cylinders Practice volumes of prisms and cylinders answers Practice volumes of prisms and cylinders.\n\n### Find the volume and surface area of rectangular prisms and cylinders.\n\n1 7 km 8 km 14074 km. Some of the worksheets for this concept are Volumes of prisms Volumes of solids Volume Answer key volume of rectangular prisms Volume and surface area work 10 volume of prisms and cylinders Volume and surface area of rectangular prisms and cylinders Volume of triangular prism es1. Volume Of rectangular prism. Volume of a Prism Worksheets. Some of the worksheets displayed are Volume Volume 10 volume of prisms and cylinders Volume of prisms and cylinders work answers Volume and surface area of rectangular prisms and cylinders Practice volumes of prisms and cylinders answers Practice volumes of prisms and cylinders. Take a break from typical worksheets with this cut and paste activity.",
null,
"Source: pinterest.com\n\nJust what I. This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres and hemispheres L-blocks and mixed. Therefore we get textvolume of cone dfrac13pitimes42times55dfrac883pi. Cavalieris principle is also introduced. Some of the worksheets displayed are Volume Volume 10 volume of prisms and cylinders Volume of prisms and cylinders work answers Volume and surface area of rectangular prisms and cylinders Practice volumes of prisms and cylinders answers Practice volumes of prisms and cylinders.",
null,
"Source: pinterest.com\n\nVolume of a Cube by simon13101980. Videos worksheets 5-a-day and much more. Cavalieris principle is also introduced. This is superb thank you. Volume Of Cylinders And Prisms Word Problems.",
null,
"Source: pinterest.com\n\nOct 16 2020 - By Mary Higgins Clark Best Book Volume Of Prisms And Cylinders Worksheet Answers PDF ePub eBook volume of prisms and cylinders worksheet answers contains important information and a detailed explanation about PDF ePub eBook volume of prisms and cylinders worksheet answers its contents of the package names of things and. Math Quiz by jegb07. Round your answers to the nearest tenth if necessary. The vertical height 55 mm and the radius 4 mm. 14112014 Finding the volume of prisms and cylinders.",
null,
"Source: pinterest.com\n\nTake a break from typical worksheets with this cut and paste activity. I think the PPT is outstanding and the worksheets progressive and flexible for differentiation. Volume prism worksheets pgapickemclub 100767 prisms worksheets wanderersyouthclub 100768 Volume And Surface Area Of Triangular Prisms C Grade Worksheets. Volume of a Cube by simon13101980. Just what I.",
null,
"Source: gr.pinterest.com\n\nFind the volume and surface area of rectangular prisms and cylinders. Volume of a Cube by simon13101980. Find the volume and surface area of rectangular prisms and cylinders. Some of the worksheets for this concept are Volumes of prisms Volumes of solids Volume Answer key volume of rectangular prisms Volume and surface area work 10 volume of prisms and cylinders Volume and surface area of rectangular prisms and cylinders Volume of triangular prism es1. Volume Prisms And Cylinders.",
null,
"Source: pinterest.com\n\nRound your answers to the nearest tenth if necessary. Some of the worksheets for this concept are Volume word problems cylinders cones spheres Math word problems volume of cubework Volume of cones word problems Volume of a triangular pyramid word problems Volume or rectangular prisms customary units Volume of prisms and cylinders. Take a break from typical worksheets with this cut and paste activity. Great resource pack for basic volume. This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres and hemispheres L-blocks and mixed.",
null,
"Source: pinterest.com\n\nTo downloadprint click on pop-out icon or print icon. Featured here are innumerable exercises to practice finding the volume of prisms using dimensions of varying base faces expressed as. Volume Of Cylinders And Prisms Word Problems. Prisms including Cavalieris Principle Notes and Practice3 pages total. I think the PPT is outstanding and the worksheets progressive and flexible for differentiation.",
null,
"Source: pinterest.com\n\nTwo pages of notes and one page of practiceOn the 2 pages of notes students will practice finding the volume of right and oblique prisms and cylinders. Some of the worksheets for this concept are 10 volume of prisms and cylinders Practice volumes of prisms and cylinders answers Volumes of prisms and cylinders Volume of prisms and cylinders work answers Volume of prisms and cylinders work answers Practice volumes of prisms and cylinders answers Volume Volume word problems cylinders. This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres and hemispheres L-blocks and mixed. Volume Of rectangular prism. Find the surface area of rectangular prisms and cylinders.",
null,
"Source: pinterest.com\n\n4 8 in 8 in 55 in 880 in. Volume of a Prism Worksheets. Lower ability full lesson 521f on volume of prisms to be taught over one or two lessons depending on previous knowledgeEx4 and Ex5 are a suggested starting. 15052019 TWO FULL LESSONS on finding the volume of prisms. To downloadprint click on pop-out icon or print icon.",
null,
"Source: pinterest.com\n\nVolume of Cylinders. Found worksheet you are looking for. Math Quiz by jegb07. Volume Of Cylinders And Prisms Word Problems. 09092019 The Corbettmaths Practice Questions on the Volume of a Cylinder.",
null,
"Source: pinterest.com\n\nFound worksheet you are looking for. Math Quiz by jegb07. Volume Of Cylinders And Prisms Word Problems. Cavalieris principle is also introduced. Found worksheet you are looking for.",
null,
"Source: pinterest.com\n\nVolume of a Prism Worksheets. Next we have to work out the volume of the cone. Some of the worksheets displayed are Volume Volume 10 volume of prisms and cylinders Volume of prisms and cylinders work answers Volume and surface area of rectangular prisms and cylinders Practice volumes of prisms and cylinders answers Practice volumes of prisms and cylinders. Volume Of rectangular prism. Volume Of Prisms And Cylinders Practice A.",
null,
"Source: pinterest.com\n\nGreat resource pack for basic volume. Showing top 8 worksheets in the category - Volume Of Prisms And Cylinders. This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres and hemispheres L-blocks and mixed. Higher ability full lesson 521h on volume of prisms could be taught over one or two lessons depending on previous knowledge. Prisms including Cavalieris Principle Notes and Practice3 pages total.",
null,
"Source: pinterest.com\n\nHigher ability full lesson 521h on volume of prisms could be taught over one or two lessons depending on previous knowledge. 09092019 The Corbettmaths Practice Questions on the Volume of a Cylinder. This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres and hemispheres L-blocks and mixed. Some of the worksheets displayed are Volume Volume 10 volume of prisms and cylinders Volume of prisms and cylinders work answers Volume and surface area of rectangular prisms and cylinders Practice volumes of prisms and cylinders answers Practice volumes of prisms and cylinders. Find the volume and surface area of rectangular prisms and cylinders.",
null,
"Source: pinterest.com\n\nTherefore we get textvolume of cone dfrac13pitimes42times55dfrac883pi. Volume of a Cube by simon13101980. I think the PPT is outstanding and the worksheets progressive and flexible for differentiation. Some of the worksheets for this concept are 10 volume of prisms and cylinders Practice volumes of prisms and cylinders answers Volumes of prisms and cylinders Volume of prisms and cylinders work answers Volume of prisms and cylinders work answers Practice volumes of prisms and cylinders answers Volume Volume word problems cylinders. Higher ability full lesson 521h on volume of prisms could be taught over one or two lessons depending on previous knowledge.\n\nThis site is an open community for users to submit their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.\n\nIf you find this site value, please support us by sharing this posts to your preference social media accounts like Facebook, Instagram and so on or you can also save this blog page with the title volume of prisms and cylinders worksheets by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website."
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http://radio.radiotrician.org/2019/04/now-some-z-matching-questions.html | [
"## Sunday, April 28, 2019\n\n### Now some Z matching questions\n\nThe rule of thumb for small signal amps states that Rl should equal Rc. As you can see The transistor Z which is Zgen in my circuit is equal Rc WHEN Vc is set to one half Vcc. Rc and Z transistor form a voltage divider and therefore are equal when Vc is equal 1/2 Vcc. The shot ahead of this one shows the output is higher when Rl is larger than Rc.\n\nSo the question to be asked before designing the circuit is do we want voltage or power gain?\n\nHow much collector current shall we have?\n\nWhat will the load Z be?"
] | [
null
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https://en.m.wikibooks.org/wiki/More_C%2B%2B_Idioms/Covariant_Return_Types | [
"# Covariant Return Types\n\n### Intent\n\nAvoid unnecessary casting of the derived value returned from an overridden method in a derived class.\n\n### Motivation\n\nIn C++ a base class determines the method signatures of virtual functions that derived classes might override. The type of the return value in the overridden function is generally same as that of the base class's function. However, this can be limiting if the type returned by the overridden function is substitutable (sub-class) for the type of the base function. Consider the following example.\n\n```class Base {\npublic:\nvirtual Base * clone() const {\nreturn new Base(*this);\n}\n};\nclass Derived : public Base {\npublic:\nvirtual Base * clone() const override {\nreturn new Derived(*this);\n}\n};\nDerived *d1 = new Derived();\nBase * b = d1->clone();\nDerived *d2 = dynamic_cast<Derived *>(b);\nif(d2) {\n// Use d2 here.\n}\n```\n\nThe caller of clone knows that the run-time type of the object returned by the function is Derived. However, a dynamic_cast is necessary to downcast the Base pointer to Derived if functions specific to Derived are to be used.\n\n### Solution and Sample Code\n\nC++ allows the derived type to implement an overridden function with return type that is a (pointer to a) sub-type of the return type of the base function. Here's an updated example.\n\n```class Base {\npublic:\nvirtual Base * clone() const {\nreturn new Base(*this);\n}\n};\nclass Derived : public Base {\npublic:\nvirtual Derived * clone() const override {\nreturn new Derived(*this);\n}\n};\nDerived *d1 = new Derived();\nDerived *d2 = d1->clone();\nif(d2) {\n// Use d2 here.\n}\n```\n\nThis alternative is direct and saves unnecessary casts."
] | [
null
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https://pro.arcgis.com/en/pro-app/latest/tool-reference/image-analyst/summarize-categorical-raster.htm | [
"# Summarize Categorical Raster (Image Analyst)\n\nAvailable with Image Analyst license.\n\n## Summary\n\nGenerates a table containing the pixel count for each class, in each slice of an input categorical raster.\n\n## Usage\n\n• Use this tool to calculate the number of pixels in each category for every slice in a multidimensional, categorical raster dataset. For example, calculate the number of pixels in each land cover class for a multidimensional raster containing 30 years of land cover data.\n\n• The input raster dataset must have a raster attribute table. To generate a raster attribute table, use the Build Raster Attribute Table tool.\n\n• If the input raster has a Class_Name or ClassName field, the output table will use the names listed in that field. Otherwise, the output table will use class values from the Class_Value or ClassValue field. The field names are not case sensitive.\n\n• Supported multidimensional raster datasets include Cloud Raster Format (CRF), multidimensional mosaic datasets, or multidimensional raster layers generated by netCDF, GRIB, or HDF files.\n\n## Parameters\n\n Label Explanation Data Type Input Categorical Raster The input multidimensional, categorical raster. Raster Dataset; Raster Layer; Mosaic Dataset; Mosaic Layer; Image Service; String Output Summary Table The output summary table. Geodatabase, database, text, Microsoft Excel, and comma-separated value (CSV) tables are supported. Table Dimension (Optional) The input dimension to use for the summary. If there is more than one dimension and no value is specified, all slices will be summarized using all combinations of dimension values. String Area Of Interest (Optional) The polygon feature layer containing the area or areas of interest to use when calculating the pixel count per category. If no area of interest is specified, the entire raster dataset will be included in the analysis. Feature Layer Area Of Interest ID Field (Optional) The field in the polygon feature layer that defines each area of interest. Text and integer fields are supported. Field\n\n`SummarizeCategoricalRaster(in_raster, out_table, {dimension}, {aoi}, {aoi_id_field})`\n Name Explanation Data Type in_raster The input multidimensional, categorical raster. Raster Dataset; Raster Layer; Mosaic Dataset; Mosaic Layer; Image Service; String out_table The output summary table. Geodatabase, database, text, Microsoft Excel, and comma-separated value (CSV) tables are supported. Table dimension(Optional) The input dimension to use for the summary. If there is more than one dimension and no value is specified, all slices will be summarized using all combinations of dimension values. String aoi(Optional) The polygon feature layer containing the area or areas of interest to use when calculating the pixel count per category. If no area of interest is specified, the entire raster dataset will be included in the analysis. Feature Layer aoi_id_field(Optional) The field in the polygon feature layer that defines each area of interest. Text and integer fields are supported. Field\n\n### Code sample\n\nSummarizeCategoricalRaster example 1 (Python window)\n\nThis example generates a table containing the pixel count for each land cover category in 20 years of land cover data in the Boston area, within an area of interest.\n\n``````# Import system modules\nimport arcpy\nfrom arcpy.ia import *\n\n# Check out the ArcGIS Image Analyst extension license\narcpy.CheckOutExtension(\"ImageAnalyst\")\n\narcpy.ia.SummarizeCategoricalRaster(\"BostonLandCover2000_2020.crf\",\n\"C:\\Data\\MyData.gdb\\BostonLandCoverSummary\", \"StdTime\", \"C:\\Data\\MyData\\AOI\",\n\"Districts\")``````\nSummarizeCategoricalRaster example 2 (stand-alone script)\n\nThis example generates a table containing the pixel count for each fire risk class in yearly data, within an area of interest.\n\n``````# Import system modules\nimport arcpy\nfrom arcpy.ia import *\n\n# Check out the ArcGIS Image Analyst extension license\narcpy.CheckOutExtension(\"ImageAnalyst\")\n\n# Define input parameters\ninputRaster = \"C:/Data/YearlyFireRisk.crf\"\noutputTable = \"C:/Data/FireRiskSummary.csv\"\ndimension = \"StdTime\"\naoi = \"C:/Data/MyData.gdb/SanBernardinoMountainRange\"\naoi_id_field = \"WATERSHEDS\"\n\n# Execute\n\narcpy.ia.SummarizeCategoricalRaster(inputRaster, outputTable, dimension, aoi, aoi_id_field)``````\n\n## Licensing information\n\n• Basic: Requires Image Analyst\n• Standard: Requires Image Analyst\n• Advanced: Requires Image Analyst"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5604185,"math_prob":0.84060514,"size":4218,"snap":"2022-40-2023-06","text_gpt3_token_len":935,"char_repetition_ratio":0.1238728,"word_repetition_ratio":0.48287672,"special_character_ratio":0.19393077,"punctuation_ratio":0.140625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97989315,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T22:38:50Z\",\"WARC-Record-ID\":\"<urn:uuid:f4a0ea41-ece3-4ea1-93ca-fb0e27bc5819>\",\"Content-Length\":\"22791\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d97293c3-4a7e-40e5-8f83-8202194ce70d>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9b92fdc-bfd6-4b5d-8571-3197bf666067>\",\"WARC-IP-Address\":\"198.102.60.60\",\"WARC-Target-URI\":\"https://pro.arcgis.com/en/pro-app/latest/tool-reference/image-analyst/summarize-categorical-raster.htm\",\"WARC-Payload-Digest\":\"sha1:PFY32UWIZ6JMZ7KBMV7SSYOLWGXAVWAA\",\"WARC-Block-Digest\":\"sha1:KXFGY3MJDQHGDLDO2JZO23SCXECOQZCL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337529.69_warc_CC-MAIN-20221004215917-20221005005917-00545.warc.gz\"}"} |
https://36vultures.net/6th-grade-math-worksheets-common-core.html | [
"# 6th Grade Math Worksheets Common Core",
null,
"### 6th Grade Common Core Math Assessment Short Form A 10 Questions",
null,
"### Geometry Sixth Grade Common Core Math Worksheets All Standards",
null,
"### Comparing And Ordering Integers Worksheet Free Ccss 6 Ns C 7",
null,
"### 6th Grade Math Common Core Expressions And Equations Worksheets",
null,
"### Common core grade 6 math 6 g 1.\n\n6th grade math worksheets common core. Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade and more. 6th grade common core science. 6 ee 1 expressions and equations. Math 6th grade common core printables printable worksheets at internet 4 classrooms fun activities learning games and educational resources for prek 12th grade.\n\nEasier to grade more in depth and 100 free. Displaying all worksheets related to 6th grade common core science. Apply and extend previous understandings of arithmetic to algebraic expressions. 3 write interpret and use.\n\nClick on the common core topic title to view all available worksheets. Solve real world and mathematical problems involving area surface area and volume. Common core math video lessons math worksheets and games. Browse through the list of common core standards for grade 6 math.\n\n2 completing understanding of division of fractions and extending the notion of number to the system of rational numbers which includes negative numbers. The best source for free math worksheets. In sixth grade students will focus on four areas. Common core grade 6 math.\n\nGrade 6 introduction print this page. Common core math worksheets by grade level. In grade 6 instructional time should focus on four critical areas. Set the tone for effective learning with our printable common core worksheets with adequate exercises in both math and english catering to the needs of students in kindergarten through grade 8.\n\nEasier to grade more in depth and 100 free. The worksheets are aligned to the ccss but blend well into any curriculum without a hitch. 1 connecting ratio and rate to whole number multiplication and division and using concepts of ratio and rate to solve problems. Common core and math in sixth grade.\n\nWorksheets are graduated cylinders name answers science content standards a look at intermediate level science core curriculum grades 5 8 fourth grade common core state standards for english language arts michigan k 12 standards science fifth grade. 1 connect ratio and rate to whole number multiplication and division and use concepts of ratio and rate to solve problems.",
null,
"### Common Core Math Worksheets Sixth Grade Math Integer Preview",
null,
"### Common Core Math Worksheets Sixth Grade Math Integer Preview",
null,
"### Fractions 3nf 4nf 5nf 6rpa All Fraction Standards Common Core",
null,
"### 6 Ee Expressions And Equations All Standards Sixth Grade Common",
null,
"### 6 Ns C 5 6 Ns C 6 6 Ns C 7 6 Ns C 8 Sixth Grade Common Core",
null,
"### 6th Grade Math Differentiated Worksheet Bundle For Math Centers",
null,
"### 6 G A 2 Geometry Word Problems 6th Grade Common Core Math",
null,
"### 6 Ee B 7 Word Problems Common Core Math Worksheets Common Core",
null,
"### K 6 Common Core Math Assessment Bundle 7 Tests Each With 10",
null,
"### 3nf 4nf 5nf 6nf Fraction And Ratio Word Problems All Standards",
null,
"### Ratio Worksheets With Images Middle School Math Worksheets",
null,
"### 7th Grade Math Common Core Worksheet Bundle 5 Worksheets And",
null,
"### 5 Nf Fractions All Standards Fifth Grade Common Core Math\n\nSource : pinterest.com"
] | [
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"https://i.pinimg.com/originals/cd/87/3f/cd873f6666c4c997bceb19f3e5eed5bb.jpg",
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"https://i.pinimg.com/originals/ea/09/cd/ea09cd0d1c56eb88fa8c00c03cd493b8.jpg",
null,
"https://i.pinimg.com/474x/d9/99/75/d9997550d2475c5cb9f2cbb254389eef.jpg",
null,
"https://i.pinimg.com/originals/2d/ff/d6/2dffd62e4068a793b7a4467cf6dc9deb.jpg",
null,
"https://i.pinimg.com/originals/2c/a9/37/2ca937d053daccc8f6f90f176b06ae62.jpg",
null,
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null,
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null,
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"https://i.pinimg.com/originals/07/d4/15/07d4151e69bc10a4d35066bc611ebe8e.jpg",
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null,
"https://i.pinimg.com/originals/72/fa/6f/72fa6f4fa5913ac7fd2be7b7c1a03faa.jpg",
null,
"https://i.pinimg.com/originals/b8/2c/44/b82c44bee77b1e7092d3e49014155adb.jpg",
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"https://i.pinimg.com/originals/0e/6c/76/0e6c76d601803afc5aaf7eb714c0f727.jpg",
null,
"https://i.pinimg.com/originals/56/e9/72/56e97284c4cc8a515757c1c52a9f8983.jpg",
null
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https://mathoverflow.net/questions/202332/diameter-of-sum-graph-over-a-meager-set | [
"# Diameter of sum-graph over a meager set\n\nWe say that $S\\subseteq \\mathbb{N}$ is meager if $$\\text{lim sup}\\frac{S\\cap\\{1,\\ldots, n\\}}{n} = 0.$$\n\nGiven $S\\subseteq \\mathbb{N}$, we associate to $S$ the sum-graph $G_S = (\\mathbb{N}, E)$ where $$E = \\big\\{\\{m,n\\}: m,n \\in \\mathbb{N} \\text{ and } m+n\\in S\\big\\}.$$\n\nIs there a set $S\\subseteq \\mathbb{N}$ such that\n\n• $S$ is meager;\n• $G_S$ is connected;\n• $\\text{diam}(G_S)$ is infinite?\n\nLet $S=\\{2^k-1: k \\in \\mathbf{N}\\}=\\{1_2,(11)_2,(111)_2,...\\}$.\nClearly $S$ is sparse.\nWriting $m,n \\in \\mathbf{N}$ in their binary expansion, $m$ is connected to $n$ in $G_S$ if $m>n$ and $n$ is obtained by bitwise inverting $m$. For example, $(11001)_2$ is connected to $(00110)_2=(110)_2$. This shows that $G_S$ is connected by repeatedly applying this procedure to an arbitrary element to find a path to 1.\nTo show that there is no bound on the diameter of $G_S$, consider the sequence $(a_k)$ given by $1,(101)_2, (10101)_2, (1010101)_2,...=\\{(2^{2k}-1)/3: k \\in \\mathbf{N}\\}.$ Consider a path from $a_n$ to 1. Whenever a digit of a number on this path is flipped, all digits to the right are flipped too. This means that if the left-most digit of $a_n$, which is initially 1, is flipped $r$ times, then the second digit, which is initially 0, must be flipped at least $r+1$ times. By induction, $a_n$ has distance precisely $2n-2$ from 1."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8591826,"math_prob":1.0000045,"size":944,"snap":"2020-10-2020-16","text_gpt3_token_len":312,"char_repetition_ratio":0.1,"word_repetition_ratio":0.0,"special_character_ratio":0.3845339,"punctuation_ratio":0.16589862,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000072,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-22T04:55:52Z\",\"WARC-Record-ID\":\"<urn:uuid:0a0dde04-454b-421b-a2c7-f12f91ed183b>\",\"Content-Length\":\"118787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c778fb21-c92b-44fa-9663-57d1e69f75dd>\",\"WARC-Concurrent-To\":\"<urn:uuid:90fd8106-202a-4a1e-b7a3-ce0e87957d0c>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/202332/diameter-of-sum-graph-over-a-meager-set\",\"WARC-Payload-Digest\":\"sha1:MHFVOFE4DLGVS2KA6QKWVJ5SI75YOTHI\",\"WARC-Block-Digest\":\"sha1:F6MHOY5RBZWUDHC2IJDGKNY75JTEMERT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145648.56_warc_CC-MAIN-20200222023815-20200222053815-00101.warc.gz\"}"} |
http://dict.cnki.net/h_1813090022.html | [
"",
null,
"全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多",
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"多元非线性 在 金属学及金属工艺 分类中 的翻译结果: 查询用时：0.159秒",
null,
"在分类学科中查询 所有学科 金属学及金属工艺 计算机软件及计算机应用 数学 自动化技术 医药卫生方针政策与法律法规研究 更多类别查询",
null,
"历史查询",
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"",
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"多元非线性",
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"multivariant nonlinear\n The optimal arithmetic is selected. By the dummy experiment design, the multivariant nonlinear regression model between the blade final forging shape optimal objective function and the coordinate of the control points of the B-spline to describe the preform shape is built. 借助于虚拟实验设计,建立了叶片终锻件形状优化目标函数与表示预成形毛坯形状的三次B样条曲线的控制点坐标之间的相关关系——多元非线性回归预测模型。 短句来源 According to the content of this experiment, regression model with 4 factorial polynomial is applied to make a multivariant nonlinear regression analysis of data, multivariant nonlinear regression equation of electrical parameters to roughness and machining time is established, and practical technological parameters are acquired. This study provides proof of theoretical analysis for advanced master of technological rule of MWEDM. 根据实验内容采用 4阶多项式回归模型对实验数据进行了多元非线性回归分析 ,建立了电参数对表面粗糙度和加工时间的多元非线性回归方程 ,获得了实用的工艺参数 ,为进一步掌握微细电火花线切割加工工艺规律提供了理论分析依据。 短句来源 A mathematic model,which deals with the relationship between maximum temperature of thick plate butt multipass welding and weld heat input,interpass temperature,ambient temperature and the distance of measuring points to heat source center,was founded using multivariant nonlinear regression analysis. 采用多元非线性回归的方法,建立了厚板对接多道焊的最高温度与焊接线能量、层间温度、环境温度以及测点到热源中心的距离之间的数学模型. 短句来源",
null,
"“多元非线性”译为未确定词的双语例句\n ZWZY5 system for prescription-design is of manifeld functions,such as drawing up a prescription plan for orthogonal design test,doing orthogonal regression and optimization operations, deriving regression equations of property targets,obtaining the optimization prescription,predicting properties of welding materials,working out the nonlinear regressions containing lots of unknown numbers,and gaining dozens of welding material prescriptions of a certain known property target. ZWZY5配方设计系统具有拟正交试验配方方案、进行正交回归设计、最优化计算、建立焊接材料性能指标回归方程，求得最优化配方，预报焊接材料性能，求解多元非线性回归方程，从而获得某一已知性能指标的多个焊接材料配方等功能。 短句来源 In this paper, the mathematic model of copper yield stress in hot rolling was established by means of multiple non-linear curve fitting method. 应用多元非线性曲线拟合的方法，建立了铜热轧变形抗力数学模型； 短句来源 Linear, nonlinear regression analysis and RBF neural network were used to predict nugget strength based on the input vector which was constructed by factors inspected of electrode displacement signal. 基于电极位移信号特征向量集合建立多元线性回归、多元非线性回归、RBF神经网络焊点接头强度预测模型,采用交叉有效性检验方法,检验预测模型的有效性。 短句来源 By means of multiple non-linear curve fitting method, was established the mathematic model of rib reduction, which can be applied to production of rib reduction and size selection of cold-drawing materials. 采用多元非线性曲线拟合的方法,建立了凸筋拉缩的数学模型,可用于凸筋拉缩量的预测及管坯规格的选择。 短句来源 On the basis of the data obtained on Gleeble1500 Thermal Simulator,the predicting models for the relation between flow stress and deformation strain,strain rate and temperature for 50CrV4 have been developed with Artificial Neural Network method. Comparison with nonlinear regression method,the neural network gives better results. 以Gleeble＿1500热模拟机得到的实验数据为基础，采用人工神经网络方法建立了50CrV4钢变形抗力与应变、应变速率和温度对应关系的预测模型，并与多元非线性回归模型比较，具有较高的精度。 短句来源 更多",
null,
"查询“多元非线性”译词为用户自定义的双语例句\n\n我想查看译文中含有：的双语例句",
null,
"没有找到相关例句",
null,
"The optimistically calculating pouring parameters of castings are carried out using the multielement non-linear regression analyse. The regressive equation for ingate design are also established. Considering optimum pouring time to obtain perfect castings, a design program to determine the gating system of casting was developed. 采用多元非线性回归分析方法对铸件浇注工艺参数进行了优化设计,提出了铸件内浇道优化设计公式.在考虑能获得最佳铸件质量的浇注时间基础上,编制了铸件浇注系统辅助设计程序. In this paper, stress conditions of rib reduction are presented and the influence of the process conditions on rib reduction is investigated. By means of multiple non-linear curve fitting method, was established the mathematic model of rib reduction, which can be applied to production of rib reduction and size selection of cold-drawing materials. 根据凸筋拉缩变形的力学条件,研究了工艺因素对凸筋拉缩的影响规律。采用多元非线性曲线拟合的方法,建立了凸筋拉缩的数学模型,可用于凸筋拉缩量的预测及管坯规格的选择。 An experiment was conducted for the hot metal deformation resistance of 09cuPRE,16Mn,Q235 grades using the thermal simulation testing machine.According to the effect of deformation temperature,velocity and degree on the deformation resistance,the computation model of deformation resistance was selected.The model parameters were determined through multi-element non-linear regression data processing,resulting in a model with higher fitting accuracy.Charts and tables worked out in accordance with the model can... An experiment was conducted for the hot metal deformation resistance of 09cuPRE,16Mn,Q235 grades using the thermal simulation testing machine.According to the effect of deformation temperature,velocity and degree on the deformation resistance,the computation model of deformation resistance was selected.The model parameters were determined through multi-element non-linear regression data processing,resulting in a model with higher fitting accuracy.Charts and tables worked out in accordance with the model can be directly referred to. 介绍了０９ＣｕＰＲＥ，１６Ｍｎ，Ｑ２３５钢在热模拟试验机上完成热金属变形抗力试验的情况。就试验中变形温度、变形速度、变形程度对变形抗力的影响规律，选定了变形抗力计算模型，并通过多元非线性回归数据处理确定模型参数，该模型具有较高的拟合精度。根据模型整理出的图表可供直接查找使用。 << 更多相关文摘",
null,
"相关查询\n\n CNKI小工具 在英文学术搜索中查有关多元非线性的内容 在知识搜索中查有关多元非线性的内容 在数字搜索中查有关多元非线性的内容 在概念知识元中查有关多元非线性的内容 在学术趋势中查有关多元非线性的内容\n\n CNKI主页 | 设CNKI翻译助手为主页 | 收藏CNKI翻译助手 | 广告服务 | 英文学术搜索",
null,
"2008 CNKI－中国知网",
null,
"2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社"
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https://www.saving.org/height-calculator/ft/964 | [
"### Convert 80 feet and 4 inches to centimeters and meters\n\nFeet\nInches\nCM\nMeters\n\n#### Calculate Height\n\nWhat's the conversion? Use the above calculator to calculate height. Convert feet and inches to meters and centimeters.\n\nHow much is 80'4 in cm and meters?\n\nHow tall is 80'4? How far? How long? How wide? How big? How short? How narrow?\n\nConvert 80 feet 4 inches to centimeters\n80'4 = 2,448.56 cm\n\nConvert 80 feet 4 inches to meters\n80'4 = 24.49 meters\n\nConvert 80 feet 4 inches to inches\n80'4 = 964 inches\n\nConvert 80 feet 4 inches to feet\n80'4 = 80.33 ft"
] | [
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https://www.softmath.com/math-com-calculator/adding-matrices/how-to-solve-large.html | [
"",
null,
"Search Engine visitors came to this page yesterday by entering these keyword phrases :\n\nTurning expressions into polynomials, impact mathematics course 2 answer key, algebra time and distance word problem solver, Maths Formula look up book, printable ks2 maths work sheets.\n\nThird standard maths online books, study guide for mcdougal algebra 1 chapter 9.1, factoring a binomial cubed.\n\nWhat's the least common multiple of 36 42 and 98, interval notation calculator, free lesson plans for 9th grade stuents, Simplifying Radical Expressions Worksheet, 6th grade multiply, divide, add, subtract worksheet, algebra problems, recursive equations, free math worksheets integers.\n\nSolving linear equations that contain decimals, how to use the TI-83 slope, technique in getting the least common denominator.\n\nHow to find suare root of a number, simplifying expressions by combiniing simalar terms, problems in rational algebraic expressions, table of simplifying square roots, pre algebra calculator online, how to use casio calculator.\n\nHow do you cube on a TI-83 plus calculator, algebra calculator that shows the steps, free ged work, expression factor calculator, integer+expanded form, solve the radical expression, trigonometry poems.\n\nYear 8 maths online test, java program that would square an integer, matlab, solve differential equation, hungerford solution algebra, t-83 calculator online.\n\nMcGraw hill activity worksheet answers, 9th taks worksheets, free help solving linear equations, ks2 mathematics free book, sums on combinations.\n\nEquations square, indiana prentice hall Chemistry chapter 8 study guide, prentice hall science worksheets grade 5.\n\nFormula math chart grade 7, mcdougal littell algebra 1 texas edition, how to do cube root on ti 83, free teacher's manual geometry, glencoe algebra one solving systems of linear equations, square roots converted to fractions, elementary math trivia.\n\nFree primary revision paper, algebra II practice for factoring binomials, Free Optional Tests +Past Papers.\n\nMultiply and divide polynomials software, McDougel Littel algebra II answer key, sample algabra problems, scale factor + seventh grade math, functions percentage charts formula sheet, qudratic function graph.\n\nWebsites for free books of cost accounting, solving system of equations with 3 variables, quadratic equation and percents.\n\nHomogeneous linear system equations solver, how do you add fractions in C++ programs, solvng linear programming equations, define multiply a rational expression, systems of linear equations ti89, PDF Application and Concepts Course 2 Fractions Student Edition Answer Key, Mathematics Exercises for 1st graders.\n\nAlgebra i work sheet intercept slope, quadratic functions with ti-89, free online sats practice year 6, fluid mechanics mcq, evaluate algebraic expressions calculator, real life linear algebra equations, how to find the LCM in +algabraic expressions.\n\nEquations that can not be factored, 3rd grade fraction word problems, permutation and combination aptitude questions, why we factor algebraic expressions to solve for an equation.\n\nCode to plot linear curves in java, answer key world pass t184, Example of Polar equation using matlab, math trivia geometry, polynomials and synthetic division free problem solver, learn algebra software, matlab solve second order trig.\n\nSquare root of a variable in an equation, finding the LCM with fractions, aptitude question PERMUTATION & COMBINATION, middle school math with pizzazzi book E answers, algebra aptitude, Algebra Worksheet solve y, solving equations with variables in excel easy.\n\n5th grade math proportion worksheets, elimination with multiplication poems, easy alegbra, maths year 8 online problem solver, simplifying mixed number radical, Inequality Graphing Calculator calculator online, math algebra solver software.\n\nTi-89 decimal to fraction, hyperbola solver, Exclusive Algebra Software Solving System, solving algebra simultaneous equations calculator, factoring exercises, multi-step inequalities calculator.\n\nFinding domain range equation, boolen expresions computer, rules for solving non linear single variable equations.\n\nSlope formula t1-83, simplifying quotients with radicals, examples of radical and power math problems, combinations for pre algebraic help free online, free online class 8 maths questions papers.\n\nFirst ordedr partial differential equation, convert lineal metre to square metre calculate, free computer programs for graphing linear equations, rudin chapter 7 solutions, fractions with root for denominator, solve my vertex form quadratic equation.\n\nPizzazz probability, equation of parabola, PRE- ALGEBRA 3. 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] | [
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https://mmcm.bmstu.ru/authors/83/ | [
"#### 62-752 Modeling loads on compound elastic shells by means of the initial approximation method\n\n##### Dubrovin V. M. (Bauman Moscow State Technical University), Butina T. A. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2017-2-2838\n\nThe article presents a method for computing loads (such as strain or torque) on a compound shell consisting of elastically linked external and internal shells for the case when the external shell is subjected to transverse loading (bending moment, shear forces and distributed inertial loads). To demonstrate the application of our method, we investigated the effect the rigidity properties of the external shell have on the internal shell loading.\n\nDubrovin V.M., Butina T.A. Modeling loads on compound elastic shells by means of the initial approximation method. Маthematical Modeling and Coтputational Methods, 2017, №2 (14), pp. 28-38\n\n#### 629.7.015.4 Modeling of stressed-deformed state of rotation shells under conditions of material creep\n\n##### Butina T. A. (Bauman Moscow State Technical University), Dubrovin V. M. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2019-2-314\n\nOne of the main properties of structural materials is creep. The prob-lem of determining the stress-strain state of axisymmetrically loaded shells of rotation at creep is considered\n\nБутина Т.А., Дубровин В.М. Моделирование напряженно-деформированного состояния оболочек вращения в условиях ползучести материала. Математическое моделирование и численные методы, 2019, № 2, с. 3–14.\n\n#### 539.3 Modeling of the dynamic stability of a cylindrical shell under the axial compressive load\n\n##### Dubrovin V. M. (Bauman Moscow State Technical University), Butina T. A. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2015-2-4657\n\nThe article describes a method for calculating the dynamic stability of cylindrical shell under axial compressive time-varying load. The case of linearly varying load was con-sidered as an example.\n\nDubrovin V., Butina T. Modeling of the dynamic stability of a cylindrical shell under the axial compressive load. Маthematical Modeling and Coтputational Methods, 2015, №2 (6), pp. 46-57\n\n#### 22.251 Modeling of the process of interaction of the shock wave with cylindrical shell considering wave radiation effect\n\n##### Dubrovin V. M. (Bauman Moscow State Technical University), Butina T. A. (Bauman Moscow State Technical University), Polyakova N. S. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2015-4-3852\n\nThe objective of this research is to examine the shock wave with cylindrical shell and to propose a method for calculating its dynamic stability under axial compressive timevarying load. For weak shock waves we conducted comparative analysis of the exact solution and the existing approximate solutions. We evaluated the wave radiation effect after the shell deformation. The case of linearly varying load was considered as an example.\n\nDubrovin V., Butina T., Polyakova N. Modeling of the process of interaction of the shock wave with cylindrical shell considering wave radiation effect. Маthematical Modeling and Coтputational Methods, 2015, №4 (8), pp. 38-52\n\n#### 539.384 Modeling the stability of compressed and twisted rods in precise problem statement\n\n##### Dubrovin V. M. (Bauman Moscow State Technical University), Butina T. A. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2015-3-316\n\nThe article describes the method for calculating the stability of a rod under simultaneous action of axial force and torque, considering changing the torsion of the rod when it’s bent. The method is based on the use of the complete system of equations. The following cases are considered: end clamped rod, rod with a hinged support, the rod in the form of compressed and twisted console. Diagrams of dependence of the critical axial force versus the critical torque are obtained, i.e., the range of rod stability for the case of loading is determined.\n\nDubrovin V., Butina T. Modeling the stability of compressed and twisted rods in precise problem statement. Маthematical Modeling and Coтputational Methods, 2015, №3 (7), pp. 3-16\n\n#### 539.3 Simulation of dynamic stability of a cylindrical shell under cyclic axial impact\n\n##### Dubrovin V. M. (Bauman Moscow State Technical University), Butina T. A. (Bauman Moscow State Technical University)\n\ndoi: 10.18698/2309-3684-2016-3-2432\n\nIn this article we suggest a method for calculating the dynamic stability of a cylindrical shell with its axial compressive time-varying load, and cyclic axial load, which varies according to a certain law. As an example, we consider the axial load, changing linearly and the cyclic load, which varies according to the harmonic law. To show the cyclic load, we use Ince — Strutt diagram, defining the stable and unstable regions of the shell fluctuations.\n\nDubrovin V., Butina T. Simulation of dynamic stability of a cylindrical shell under cyclic axial impact. Маthematical Modeling and Coтputational Methods, 2016, №3 (11), pp. 24-32"
] | [
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https://v1.archmathsci.org/conferences-and-workshops/askloster/2009-2/abstracts-of-2009-talks/ | [
"# Abstracts of 2009 Talks\n\nNovel mathematical structures in physics – Non- commutative geometry.\n\nThe Discussion of possible relationships between the foundations of mathematics and the foundations of physics was also encouraged.\n\n### 1. Higher dimensional algebra: origins and prospects\n\nABSTRACT: In the second half of the 1960s I convinced myself that the use of groupoids rather than groups in 1-dimensional homotopy theory led to more powerful theorems with simpler proofs. The objects of a groupoid add a spatial element to group theory which enables wider applications.\nThis led to the question of the putative uses of groupoids in higher homotopy theory, and the existence of a higher dimensional group(oid) theory, for example higher order symmetry. Now we can see higher dimensional algebra’ (HDA) as being about partially defined algebraic operations whose domains of definitions are defined by geometric conditions. For this reason, HDA gives a range of noncommutative, higher dimensional methods for local-to-global problems, and can give precise results seemingly unobtainable by other means.\nThe talk will give some of the intuitions, background and ideas essential for the theory. The main gap in the methods seems to be the link to analogues of representation theory.\n\n2. Category theory, higher dimensional algebra, groupoid atlases: prospective descriptive tools in theoretical neuroscience\nIn neuroscience, and systems theory, one deals with hierarchical systems, with problems of describing their structure and the interactions and communications at local, global and hierarchical levels. This speculative presentation will discuss what seems available, and what might be needed in the future. There will be a mention of the Ehresmann-Vanbremeersch theory of Memory Evolutive Systems.\n\nReferences:\n\n136. (with T. Porter), Category theory and higher dimensional algebra: potential descriptive tools in neuroscience’, Proceedings of the International Conference on Theoretical Neurobiology, Delhi, February 2003, edited by Nandini Singh, National Brain Research Centre, Conference Proceedings 1 (2003) 80-92. arXiv:math/0306223\n\n149. (with Bak, A., Minian, G., and Porter, T.), Global actions, groupoid atlases and applications’, J. Homotopy and Related Structures, 1 (2006) 101-167. Memory Evolutive Systems; Hierarchy, Emergence, Cognition’, by A Ehresmann and J.-P. Vanbremeersch; Elsevier, 2009.\n\n### What can we learn from relations?\n\nAmong other things I will talk about hidden variable theories and properties of quantum correlations under time reversal.\n\n### States and measures\n\nIn classical physics, the most general kind of state of a physical system is given by a probability measure on the state space of the system. This corresponds to an integral on the commutative algebra of physical quantities. Quantum systems, on the other hand, do not have a state space description. Quantum states are positive linear functionals (i.e., integrals) on the non-commutative algebra of physical quantities. In this talk, it will be shown that quantum states can also be understood as probability measures which live not on a (non-existent) state space, but\non a presheaf associated to the quantum system. This presheaf is a generalised space. Its non-commutative character is expressed by the fact that it has no points.\n\n### 1. The Geometric Structure of Quantum Mechanics.\n\nThe traditional approach to quantum mechanics using Hilbert space obscures the geometric structure of the theory, yet the appearance of a Clifford algebraic structure in the Dirac theory. David Bohm proposed an approach based on a notion of ‘structure process’. I will present some of the background ideas which motivates an approach to quantum theory based entirely within the Clifford structure thus exposing the geometric background.\n\n### 2. The Bohm Model of the Dirac Particle.\n\nUsing the ideas discussed in the first talk I will present the complete Bohm model of the Dirac particle, thus showing for the first time that this model can be made fully Lorentz invariant. Bob Callaghan and I have obtained an expression for the quantum potential which reduces to the corresponding expression for the Pauli particle in the non-relativistic limit, and to that for the Schroedinger particle when the spin is zero. I am informed by Olival Freire, a science historian form Sao Paulo, that Bohm once wrote that “a good treatment of Dirac equation in his own approach would deserve a champagne case”! Why didn’t we complete this before he passed on!\n\n### 1. Infinity and renormalization in quantum physics and computer science.\n\nThe main observable quantities in Quantum Field Theory, correlation functions, are expressed by the celebrated Feynman path integrals which are not well defined mathematical objects. Perturbation formalism interprets such an integral as a formal series of finite–dimensional but generally divergent integrals, indexed by Feynman graphs, the list of which is determined by the Lagrangian of the theory. Renormalization is a prescription that allows one to systematically “subtract infinities’’ from these divergent terms producing an asymptotic series for quantum correlation functions. On the other hand, graphs treated as “flowcharts’’, also form a\ncombinatorial skeleton of the abstract computation theory and various operadic formalisms in abstract algebra. In this role of descriptions of various (classes of) computable functions, such as recursive functions, functions computable by a Turing machine with oracles etc., graphs can be used to replace standard formalisms having linguistic flavor, such as Church’s $\\lambda$–calculus and various programming languages.\nThe functions in question are generally not everywhere defined due to potentially infinite loops and/or necessity to search in an infinite haystack for a needle which is not there. In this paper I argue that such infinities in classical computation theory can be addressed in the same way as Feynman divergences, and that meaningful versions of renormalization in this context can be devised. Connections with quantum computation are also touched upon.\n\n### 2. Foundations of Mathematics: from discrete to continuous and backwards\n\nAbstract: In this talk, the scope of which will be more of a round table contribution, I will discuss some historical and philosophical issues related to the continuous/ discrete dichotomy. I will also argue that an emerging shift in foundations is related to homotopy theory and proceeds from the continuous to the discrete rather than in the reverse direction that has dominated the consciousness of working mathematicians for the last century or so.\n\n### 1. Topological Quantum Information Theory\n\nAbstract: This talk is about using topological structures such as unitary representations of the braid group and unitary solutions to the Yang-Baxter Equation in studying entanglement, quantum computing and quantum algorithms. A motivation for this project is the question: Is there a relationship betweem topological entanglement (linking and knotting) and quantum entanglement (indecomposable states, non-locality)? We will give numerous examples and speculate about the eventual role of topological structures in this aspect of physics and communication.\n\n### 2. Reflexivity, Eigenforms, Iterants and Discrete Physics\n\nAbstract: A reflexive domain is a domain D such that there is a 1-1 correspondence between D and the mappings [D,D] from D to D. Letting I:D —–> [D,D] be this 1-1 correspondence. We have the Fixed Point Theorem: For any F:D —-> D, there is an element E of D such that F(E) =E.\nProof. Define Gx = F(I(x)x). Then G = I(g) for some g in D and so I(g)x = F(((x)x), whence I(g)g =F(I(g)g) and I(g)g is the desired fixed point. Q.E.D. (The reader may note that this arguement is a generalization of Cantor’s Diagonal argument, so that if we were in a classical logical algebra and F(x) = ~ x represented negation, then we would conclude the existence of fixed points for negation; non-standard or imaginary logical values must appear in a reflexive logical space). This talk will explore the analogies of fixed points, observations and observables, eigenvectors and recursive processes in relation to foundations of physics. In particular we shall re-open the books on the complex numbers and view them in terms of recursion and reflexivity, finding new and natural ways to think about their roles in physical theory. To give a hint, think of the oscillatory process generatoed by R(x) = -1/x. The fixed point is i with i^2 = -1, but the processes generated over the real numbers must be directly related to the idealized i. We shall let {+1,-1} stand for an undisclosed alternation or ambiguity between +1 and -1 and call {+1,-1} an <iterant>. There are two <iterant views>: [+1,-1] and [-1,+1]. These, seen as points of view of alternating process will become the square roots of negative unity. We introduce a temporal shift operator # such that [a,b]# = #[b,a] and ## = 1 so that contcatenated observations can include a time step of one-half period of the process …abababab… . We combine iterant views term-by-term as in [a,b][c,d] = [ac,bd]. Then we have, with i = [1,-1]# (i is view/operator), ii = [1,-1]#[1,-1]# = [1,-1] [-1,1]## = [-1,-1] = -1. This gives rise to a new process-oriented construction of the complex numbers, quaternions, and in fact of all of matrix algebra. We relate this point of view to thinking about the role of complex numbers in quantum mecahnics, the role of temporal shift operators in discrete physics and our previous work on Non- Commutative Worlds that begins with the understanding that temporal shift operators in discrete physics allow the corresponding calculus to be represented in a non- commutative (Lie algebraic) context where all derivatives are represented by commutators. We also relate these ideas of reflexivity and fixed points to left or right distributive non-associative algebras and their relationships with knot theory and with approoaches to wholeness in physics such as the work of Barbara Piechosinska. A <magma> is a non-associative algebra with a single binary operation that is left- associative: a*(b*c) = (a*b)*(a*c). Note that this axiom says that every element A of the magma is a structure preserving mapping of the magma to itself: A(x*y) = (A*x)* (A*y). So the notion of a magma is another view of what should be a self-reflexive domain. We will raise questions about the relationship of magmas and reflexive domains, illustrate the remarkable and deep relationships between magmas and knots and braids and raise the question of the relationships of left distributivity and the Jacobi identity where a(bc) = (ab)c + b(ac) so that left distributivity is replaced by derivation satisfying the Leibniz rule. It is, at the very least, mysterious, that this level\nof reflexivity (after all a Lie algebra is an algebra satisfying the Jacobi identity and anticommuativity) is so fundamental to mathematics and physics.\n\n### The Schroedinger equation, classical or quantum?\n\nABSTRACT: For the vast majority of quantum physicists Schroedinger’s equation cannot be derived mathematically from first principles; it should be seen as a postulate, in fact a physical law governing the evolution of de Broglie’s matter waves. We will show that Schroedinger’s equation can be rigorously derived using the notion of density of trajectories introduced in 1928 by Van Vleck. The physical content of the solutions to this equation is then made clear applying the principle of the symplectic camel, which is a topological reformulation of the uncertainty principle of quantum mechanics. The relation with the so-called Bohmian trajectories is briefly discussed.\n\n### Sheaves as modules\n\nabstract: A quantale is a monoid in the monoidal category Sup of complete lattices and supremum-preserving morphisms, so by general categorical principles we can study modules on a quantale. But each hom-set of Sup is itself a complete lattice, making Sup a so-called 2- category. Using this 2-dimensional structure I shall define “principally generated modules” on Q, to then show that these provide a non- commutative generalisation of sheaf theory.Georg"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89809227,"math_prob":0.92601633,"size":12357,"snap":"2021-31-2021-39","text_gpt3_token_len":2696,"char_repetition_ratio":0.12086133,"word_repetition_ratio":0.0021008404,"special_character_ratio":0.20166707,"punctuation_ratio":0.10559567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9788537,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T01:02:09Z\",\"WARC-Record-ID\":\"<urn:uuid:b241d2a3-1d25-4cac-885a-e09e72ae62d1>\",\"Content-Length\":\"26529\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38d4ea1d-b1b5-4ff9-b4b4-6c34f3f11127>\",\"WARC-Concurrent-To\":\"<urn:uuid:5ecf804a-7c58-4f01-b05d-1b953cd9d5f6>\",\"WARC-IP-Address\":\"46.43.0.247\",\"WARC-Target-URI\":\"https://v1.archmathsci.org/conferences-and-workshops/askloster/2009-2/abstracts-of-2009-talks/\",\"WARC-Payload-Digest\":\"sha1:KSW2R2IZBQI7Z4JKGQSUN4FXCU532EHP\",\"WARC-Block-Digest\":\"sha1:EXUAZD73FQCZWAA43RNSYPVAPRUVCUXN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057479.26_warc_CC-MAIN-20210923225758-20210924015758-00322.warc.gz\"}"} |
https://free.netcurso.net/algebraforalevelmaths | [
"Introduction to 'Algebra for A-Level Maths'\n• Welcome\nSolving Equations\n• Solving equations with the unknown on both sides\n• Time to Practise - Solving an equation with the unknown on both sides\n• Check your answers - Solving an equation with the unknown on both sides\n• Time to Practise - Solving equations with fractions\n• Solving Equations Test Yourself\n• Time to Practise - Solving a quadratic equation\n• Time to Practise - Sketching a quadratic\n• The Formula\n• Time to Practise - The Formula\nCompleting the Square\n• Completing the square\n• Time to Practise - Completing the square"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8292776,"math_prob":0.9964634,"size":1332,"snap":"2020-24-2020-29","text_gpt3_token_len":308,"char_repetition_ratio":0.16114458,"word_repetition_ratio":0.42009133,"special_character_ratio":0.2117117,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99966884,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-02T15:46:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a078fb7e-f583-469f-aa34-3e3576368f93>\",\"Content-Length\":\"32236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e1fc898-e72b-4fe8-bf3e-b80576166dc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:4583d859-f52e-4dd6-835b-1ccce0ebe2e1>\",\"WARC-IP-Address\":\"104.31.92.83\",\"WARC-Target-URI\":\"https://free.netcurso.net/algebraforalevelmaths\",\"WARC-Payload-Digest\":\"sha1:BAHZFKE44MA2VFVCUSEFOMRWS2JR3QR2\",\"WARC-Block-Digest\":\"sha1:4P5QJDKXZ4ERAVQEHIY4LMOXYRAT5F3P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655879532.0_warc_CC-MAIN-20200702142549-20200702172549-00040.warc.gz\"}"} |
https://electronics-fun.com/soft-starter/ | [
"# single phase soft starter circuit diagram and working\n\nSome electrical component comes with inrush current problems (like motor, capacitor, buck/boost converter etc.). It basically means that these devices draw very large current during starting. Let’s say you have connected these devices with some safety circuit and power for power it up. But because of its inrush current as you power it, power supply cut the power due to over current. In that case we need to limit its inrush current. For that we need a circuit called soft starter. In this project I have designed a single phase soft starter circuit diagram. This means this circuit will work only for single phase appliances.\n\n## Limit the inrush current\n\nFor limit the current we can use resistor in series. It will limit the inrush current but it will create power loss. We need such resistor who decreases its resistance over time after powering up. Such resistor does exist and called thermistor.\n\nIt is a type of resistor whose value decrease or increase depending on temperature. If we connect it in series with device which cause the inrush current, it will limit the inrush current. But now when current is started flowing, it will heat up and decrease its resistance. So very less power will be lost.\n\nBut there is still one problem. Let’s say we have powered the device with thermistor previously and turned it off. Since thermistor heated up during previous power up, it will take some time to cool down. If we want to turn the device on again immediately, there will not any inrush current limiting because thermistor is still hot. If you are working with such a device which doesn’t need to power off and on frequently, you can go with the thermistor.\n\n## Single phase soft starter functional block diagram\n\nSo, we have to use the resistor with some additional logic. We need to design such circuit which let firstly current pass through resistor but after some time it should bypass the current flow.\n\nFigure shown above is functional diagram of a single phase soft starter circuit. For switch we will obviously choose the relay by which we will short the resistor after 1 second. For turning relay ON we will use the MOSFET as switch. For 1 second delay we need a RC (resistor and capacitor) circuit with 1 second time constant.\n\nIn this figure a RC circuit is shown.\n\nRC time constant (T) = RxC\n\nAs you can see from graph one RC time constant is the time at which voltage across capacitor is 63% of its input voltage. In above case voltage across capacitor after one time constant will be 5×0.63 = 3.15 volt. This time constant depends upon value of capacitor and resistor.\n\n## Single phase soft starter circuit\n\nSo, let’s build previously described circuit.\n\nWe have created the circuit we discussed earlier. Here, to bypass current from resistor we have used the relay and to turn ON relay we have used the MOSFET as switch. To turn switch ON after 1 second we have used the RC circuit with 1 second time constant. Diode D1 is connected in parallel with relay’s coil to prevent the high voltage spikes at source side of MOSFET.\n\nBut there is still have a problem in this circuit which related discharging of capacitor. We need to discharge the capacitor every time circuit is turned off. Otherwise next time when we power the device relay will activate initially because capacitor holds it charge for long period of time. So, we need to add another resistor which will pull the capacitor to ground.\n\nFor build complete soft starter circuit we need following components –\n\n1. A relay\n2. Power resistor (10Ω)\n3. Diode 1N4007 or 1N4002\n4. MOSFET (IRFZ44N)\n5. 10 mF capacitor\n6. 2 Resistors (100kΩ)\n\nIn this circuit as we have connected the resistor R3, it will create voltage divider with resistor R2. So, voltage across capacitor will reach up to 6 volt. Which is enough to activate the MOSFET."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9418075,"math_prob":0.9451557,"size":3781,"snap":"2022-40-2023-06","text_gpt3_token_len":814,"char_repetition_ratio":0.15170771,"word_repetition_ratio":0.0091047045,"special_character_ratio":0.20867495,"punctuation_ratio":0.082987554,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97168434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T18:47:10Z\",\"WARC-Record-ID\":\"<urn:uuid:bff4ba63-7807-4503-861f-ef4c45325a0f>\",\"Content-Length\":\"92484\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b248def-0d59-47e2-819d-a9dfb6ff286a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9d33893-b78b-47c9-bf54-5b16358dab39>\",\"WARC-IP-Address\":\"104.21.31.83\",\"WARC-Target-URI\":\"https://electronics-fun.com/soft-starter/\",\"WARC-Payload-Digest\":\"sha1:E7CYG2QD4I55ATYKQ7RUZNXRK3X6IZGV\",\"WARC-Block-Digest\":\"sha1:5AFVBJRF34FM2GHWYCZOT2G6AYBAZMJT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499826.71_warc_CC-MAIN-20230130165437-20230130195437-00384.warc.gz\"}"} |
https://math.stackexchange.com/questions/2414023/sequence-ffnfn1-2n/2414075 | [
"# Sequence : $f(f(n))+f(n+1)=2n$\n\nDoes there exist a function $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that $f(f(n))+f(n+1)=2n$, $\\forall n \\in \\mathbb{Z}^+$ ?\n\nMy attempt :\n\nSubstitute $n=1$, $f(f(1))+f(2)=2$ so $f(f(1))=f(2)=1$\n\nSubstitute $n=2$, $f(f(2))+f(3)=4$ so $f(1)+f(3)=4$\n\nSubstitute $n=3$, $f(f(3))+f(4)=6$\n\nAssume that there exists $k >1$ such that $f(k) \\geq k$,\n\nthen $f(f(k-1))+f(k)=2k$, we have $f(f(k-1))<k$\n\nFor a positive integer $n$, let $P(n)$ denote the condition that $f\\big(f(n)\\big)+f(n+1)=2n$. The condition $P(1)$ implies that $f(2)=1$. Therefore, $P(2)$ leads to $f(1)\\in\\{1,2,3\\}$. Hence, there are only three functions satisfying the condition, and they are listed below.\n\nIf $f(1)=1$, then $P(2)$ gives $f(3)=3$. Then, $P(3)$ gives $f(4)=3$. By induction on the positive integer $k$, we can easily see that $$f(2k-1)=2k-1\\text{ and }f(2k)=2k-1$$ for every $k=1,2,3,\\ldots$. In other words, $$f(n)=2\\,\\left\\lfloor\\frac{n}{2}\\right\\rfloor+1$$ for all $n\\in\\mathbb{Z}_{>0}$.\n\nIf $f(1)=2$, then $P(2)$ implies $f(3)=2$. Then, $P(3)$, $P(4)$, and $P(5)$ lead to $f(4)=5$, $f(5)=4$, and $f(6)=5$. It can be proven again by induction that $$f(3k-2)=3k-1\\,,\\,\\,f(3k-1)=3k-2\\,,\\text{ and }f(3k)=3k-1$$ for all $k=1,2,3,\\ldots$. In other words, $$f(n)=3\\,\\left\\lceil\\frac{n}{3}\\right\\rceil-1-\\left\\lfloor\\frac{n\\mod 3}{2}\\right\\rfloor$$ for $n=1,2,3,\\ldots$.\n\nIf $f(1)=3$, then $P(1)$ gives $f(3)=1$. Then, $P(3)$, $P(4)$, and $P(5)$ yield $f(4)=5$, $f(5)=4$, and $f(6)=5$, respectively. We claim that, for all positive integers $k$, $$f(3k+1)=3k+2\\,,\\,\\,f(3k+2)=3k+1\\,,\\text{ and }f(3k+3)=3k+2\\,.$$ However, this claim can be easily verified by induction on $k$. Thus, $$f(n)=3\\,\\left\\lceil\\frac{n}{3}\\right\\rceil-1-\\left\\lfloor\\frac{n\\mod 3}{2}\\right\\rfloor$$ for $n=4,5,6,\\ldots$, with $f(1)=3$, $f(2)=1$, and $f(3)=1$.\n\n• I got it. Thank you for your thorough solution. – carat Sep 2 '17 at 11:57\n• The $f(1)=3$ case does not look right: looks like a copy of the solution for $f(1)=2$. Indeed, I think one solution is $f(1)=3$, $f(2)=1$, $f(3)=1$, $f(4)=3$, $f(4m+k)=4m+f(k)$. – Einar Rødland Sep 2 '17 at 14:54\n\nYes there exists such a function.\nLet's define $g(x)$ as follows:\n\n$$g(n) = \\left\\{ \\begin{array}{lcc} \\ \\ \\ n \\ \\ \\ \\ \\ \\ \\ , & \\mbox{ if } n \\mbox{ is odd } , \\\\ \\\\ n-1 \\ \\ \\ , & \\mbox{ if } n \\mbox{ is even } . \\end{array} \\right.$$\n\nOne can check easilly that $g(n)$ satisfies the relation $g(g(n))+g(n+1)=2n$, $\\forall n \\in \\mathbb{N}$ ."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5255782,"math_prob":1.00001,"size":1414,"snap":"2020-10-2020-16","text_gpt3_token_len":650,"char_repetition_ratio":0.10496454,"word_repetition_ratio":0.012903226,"special_character_ratio":0.48019803,"punctuation_ratio":0.19143577,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T13:55:19Z\",\"WARC-Record-ID\":\"<urn:uuid:2993569d-91bb-4133-8754-899383dfacf3>\",\"Content-Length\":\"149467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd242e87-6b1d-4623-8cd9-97ea3cb2f9c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:77f40e0a-2b03-407b-a7af-f299154cd1fc>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2414023/sequence-ffnfn1-2n/2414075\",\"WARC-Payload-Digest\":\"sha1:VQKPHMV5BZZPJKDFK72UZRTQZ56C4Y24\",\"WARC-Block-Digest\":\"sha1:QUYED5V3OOY5VWVIQ6ZRAYFIKGBKLW34\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875149238.97_warc_CC-MAIN-20200229114448-20200229144448-00504.warc.gz\"}"} |
http://codeforces.com/blog/entry/94009 | [
"### MrPaul_TUser's blog\n\nBy MrPaul_TUser, history, 2 months ago,",
null,
"Ideas: MikeMirzayanov\n\n1560A - Dislike of Threes\n\nTutorial\nSolution\n\n1560B - Who's Opposite?\n\nTutorial\nSolution\n\n1560C - Infinity Table\n\nTutorial\nSolution\n\n1560D - Make a Power of Two\n\nTutorial\nSolution\n\n1560E - Polycarp and String Transformation\n\nTutorial\nSolution\n\n1560F1 - Nearest Beautiful Number (easy version)\n\nTutorial\nSolution\n\n1560F2 - Nearest Beautiful Number (hard version)\n\nTutorial\nShort solution\nLong solution",
null,
"Tutorial of Codeforces Round #739 (Div. 3)",
null,
"Comments (131)\n » 2 months ago, # | ← Rev. 5 → Very Good Round with Balanced Problemsets! Problem C was very interesting to me. Thanks!\n• » » How dare you say something wholesome here? No wonder you got downvoted. /s\n• » » » It seems that you get downvotes too.):\n• » » » » They are just spoilt children don't mind them.\n• » » C was pretty similar to a CSES Problem-https://cses.fi/problemset/task/1071\n• » » » Thanks Man!\n » Great problemset and quick editorial.\n » Thank you for the round! Problems were interesting and of the right difficulty for a Div. 3.For problem F1, there's a simpler solution: Simply generate all beautiful numbers for $k = 2$, my program counts about $80,000$. Link\n• » » Can you please explain what i,j,k,l,m denote in your code?\n• » » » i and j are the two digits that are in the number. k is the number of digits in the number. I use l as a bitmask to determine which digits are i and which are j. Then I loop over the digits of the bitmask with m and construct the new number, which I insert into the set. I don't let i == j by setting j = i + 1 since that is already covered when the bitmask is all 0 or 1.\n• » » » » Got it. Thanks!\n » Prob C was very interesting ngl.\n• » » 2 months ago, # ^ | ← Rev. 2 → Completely agree ! Pre-computing all powers of 2 upto 10^18\n• » » » That was D i guess\n• » » 2 months ago, # ^ | ← Rev. 2 → One great observation was that the leading row in the table was forming an arithmetic progression with common difference of 2.Great problem\n• » » » 2 months ago, # ^ | ← Rev. 3 → Well, not the row elements, but the differences between consecutive elements form an arithmetic progression with a common difference of 2. This is because $\\sum\\limits_{i=0}^{k}2k+1 = k^2$ and every element in first row will obviously be $k^2 + 1$ and numbers $2k+1$ form an arithmetic progression with a common difference of 2.Using this, we can also solve it in less than $\\mathcal O(\\sqrt k)$. The time complexity will depend on how you calculate the square root.\n• » » » » Great approach can you please give me this particular solution....if you don't mind pls\n• » » » » » Pardon the late response. Here's the pseudocode. n = floor(sqrt(k)) d = k - n * n if d = 0: return (n, 1) else if (d <= n + 1): return (d, n + 1) else return (n + 1, (n + 1 - (d - (n + 1))) This works in $\\mathcal O(1)$ once you calculate square root.\n• » » gives me old spoj problems vibe.\n » 2 months ago, # | ← Rev. 2 → I liked all problems. For F I made a dp with bitmasks (normal dp on digits), you could've set the constraints to disallow this solution, because it was pretty brainless (no cases to think about, just \"brute-force\")O(T * 2^LEN * LEN), where LEN is the number of digits of N.\n• » » But don't you think this thing is above Div3 level for like 98% official particpants of Div3?\n• » » » idk I never went to Div3 before, thinking carefully about many cases can be harder and more tricky.\n• » » » » yeah, that's your tunnel vision right there.\n• » » I implemented a similar solution (126326225) with complexity O(LEN * 1024) per test case. However, I feel considering a Div. 3 round, it wouldn't be brainless for official participants to think of a Digit DP there. However, what could have been changed was that instead of bombarding with 10^4 test cases, the length of the number could have been increased to let's say N <= 10^1000.\n• » » Could you elaborate on how the recurrence of the dp worked?\n• » » 2 months ago, # ^ | ← Rev. 2 → Isn't it O(T*2^LEN)?EDIT: No, there is popcount in every recursion\n• » » » You can avoid popcount by having a 4th argument (which you don't need to use for memoization). So, ll dp[1<<10] but the function ll solve(int mask, int pos, bool flag, int popcount) { ... } . But this does not change the complexity\n• » » I did the same solution and it got hacked. Did yours pass?\n• » » 2 months ago, # ^ | ← Rev. 3 → linkret I see no do major difference in our approach ( sorry , if there is) Can you tell why mine is TLE ? Here is my submission 126427896\n• » » » I think it's better to always count the bits, and immediately return INF if there are too many, instead of just doing it in the end.\n• » » » » Hey [user:linkert] even after that , for clearing the array the whole complexity is (10^4 * 2^10 * 10 * 2) . Shouldn't it be TLE ?\n• » » » » » 2 months ago, # ^ | ← Rev. 2 → 2*10^8 is fast enough for Codeforces, even 10^9 is, too. My solution is about 100ms.\n• » » » » » » Counting bits worked and AC\n• » » Hey , can you explain the approach a bit too?\n• » » » 2 months ago, # ^ | ← Rev. 2 → DP (memoized recursion). Use a bitmask to remember which digits from 0 to 9 have already been used. The number of 1s in this bitmask will tell you the number of different digits you had used: it must not exceed K. In the DP, try out every possible digit for the current position, and move to the next position, adding it to the bitmask. Also you will need to keep track of a 3rd state, weather the number you are currently building (from left to right) is already larger than the target number N, or not. Whenever you choose a digit larger than the corresponding digit of N, from then on you will always be larger and those dp states are allowed to choose any digits. But if you are not already larger than N, then it is not allowed to choose the current digit that is LESS than the corresponding digit of N. And if you choose the same digit, then the next recursion call still has to be careful (since it is not yet larger). You can google \"digit dp\" to get a better explanation about this part.\n• » » » » Thanks alot , great explanation\n » 2 months ago, # | ← Rev. 2 → Just curious, how did you come up with problem E? What motivated this particular string construction?It was a nice problem :)\n » 2 months ago, # | ← Rev. 3 → A clean solution for C#include using namespace std; #define int long long #define forn(i, a, n) for (int i = a; i < n; i++) #define MAXN 200100 #define MOD 1000000007 void solve() { int k; cin >> k; int n = ceil(sqrt(k)); int diff = n*n - k; if(diff >= n) { cout << 2*(n-1) - diff + 1 << \" \" << n << endl; } else { cout << n << \" \" << diff + 1 << endl; } } int32_t main() { int t; cin >> t; while(t--) {solve();} return 0; } \n• » » $sqrt()$ runs in $O(logK)$\n » 2 months ago, # | ← Rev. 2 → I'm not pretty good at math, so I solved C using binary search.I noticed that the last number of the $i_{th}$ row is $i^2$, so I do a binary search to find the row which $n$ belongs to. Then the starting number of the new column is $i^2 + 1$ (let's call this $start$), thus the number $corner$ at the position $(i + 1, i + 1)$ is $start + i$. I finally check whether $n$ exceeds $corner$ or not, if it does then the answer is $(i + 1, i + corner + 1 - n)$, else the answer is $(n - start + 1, i + 1)$Here's my submission\n• » » pretty smart vision :)\n• » » I solved using the same approach!Also, binary search was not necessary for finding the nearest square since the constraints allowed a linear search algorithm. :)\n » 2 months ago, # | ← Rev. 2 → Why does the short solution in F2 work in m^2 ? Couldn't figure it out:(Btw problems are pretty good!!\n• » » Suppose n = 789152352 and k = 3. Then the algorithm takes you to:789200000789300000789400000789500000789600000789700000Now this is a special point in the algorithm. The fourth character of the number is now among the 3 previously used digits, and now you just have to fill the 0's.This act of \"filling in the 0's\" is O(M^2). (For each of the 0's, you have to do O(M * B) work where B the base, in this case the number is base 10 so B = 10. And there are O(M) 0's. So in total it is O(M^2 * B). Or just O(M^2) since B = 10, is a constant.)There is another case, where you have to \"loop around\" because the character that was being incremented (in this case the 1 got incremented to 2, 3, .., then 7) hit 9 and then looped back around to 0. For example, if:n = 456713538, k = 3. Then you have to do:456800000456900000457000000But you still end up with a state where now you just have to \"fill in the 0's\". Again O(M^2).\n » Example test cases were really handy for giving AC in one go.. DigitForces\n » My O(1) solution to the problem C. CLICK\n• » » it is O(40000 * 100)\n• » » » Oh my bad, I thought anything that goes on before any input is taken comes into compile time. So I mistook my O(40000*100) as O(1) as it didn't depend on n. Thanks for helping, a newbie here :)\n• » » As a newbie I just wanted to show my solution and hopefully get some feedback ( which apparently I got ). Maybe my rating is what made you all to downvote. I was working hard, but now I'll double my hard work. Ciao :)\n » 2 months ago, # | ← Rev. 7 → Problem A could be solved in $\\mathcal{O}(1)$ time complexity: The unit digit is cyclic, and the cycle's length is 10. The sequence of remainders when dividing by 3 is cyclic, and the cycle's length is 3. Let's define $f(x)$ to be $1$ if Polycarp likes $x$, and $0$ otherwise. Then, $f(x)$ is cyclic, and its cycle's length is $lcm(3, 10) = 30$. We can compute the number of liked numbers among the first $30$ positive integers, let's call it $M$. Let's also say that there are $T$ contiguous ranges of size 30 between 1 and the number we're looking for (excluding that number).Then $T \\cdot M < K$, which means that $T = \\left\\lceil\\frac{K}{M}\\right\\rceil - 1$. Now, we just need to find the $(K-T\\cdot M)$-th liked number strictly greater than $30\\cdot T$, which is less than 30 steps away from $30\\cdot T$.\n• » » great!!!\n » Can anyone tell me that in the problem B can there be any testcase like:- 1 2 2as 2 people are in the circle 1 is next to 2 and we have to output the person next to 2.While I am hacking my solution by myself it's showing invalid input\n• » » No a,b,c should be distinct\n• » » » Dhanyavad Mitr.\n• » » 2 months ago, # ^ | ← Rev. 5 → input 1 1 2 2 output 1 but in the problem it is mention that : Each test case consists of one line containing three distinct integers a,b,c.\n » I think it was a good round, at least for new users like me.\n » MrPaul_TUser the solutions to problems A, B and C are possible in O(1) too. Will it be fine if you can please publish my O(1) solutions?\n » Problemset was great. First 3 questions were easy to solve. I got stuck at D and wasn't able to solve.\n » I solved D 30s after contest got over. Worst feeling ever :/\n• » » I submitted E after a min, and man I felt sed cause my rank was pretty low and I hadn't solved D\n• » » » E was easier to observe, but I wasted all my time on F1 and couldn't even solve it. Bad luck for me.\n• » » » » atleast you solved D orz ;)\n » \" Let's iterate a prefix of n starting from the empty one so that the prefix will be the prefix of x. This prefix must contain only the digits a and b\" can someone explain what this means. Problem F1\n » Look at my submission for problem D. Link: https://codeforces.com/contest/1560/submission/126381278Why this is getting TLE\n• » » I submitted your code in pypy3 it passedhttps://codeforces.com/contest/1560/submission/126414861\n• » » » Oh! Thanks alot from now on I'll be submitting in pypy3\n » MathForces\n » For problem С is possible to generate arithmetic progression, and in solution use lower_bound\n• » » Problem C can be easily done, first get the square root of k (bcz every row starts with square number), then start counting from top most box of next column.\n• » » » What difficulty of your solution?\n• » » » » You mean complexity, it will be whatever time you need to calculate the sqrt.\n » Guys i had submitted solution for E problem with the same logic. It ran fine in my local IDE but codeforces gave wrong answer on test 1. Is this some bug in the codeforces or is there some part of the code which is wrong. Please do let me know.Thank you.\n• » » it gave WA on my machine too, it is very messy so I couldn't find the mistake, maybe you could add some comments.\n » Speed of editorial was quick in the round! Thank!\n » My soln got stuck for f1 as I was unable to place the right characters to go for can someone give explanation of f1\n » I can't see anyone mentioning it here, but many people have solved A using this method: Precompute an array with liked integers ranging from 1 to 1666, and then just output the k-1th element of the array. Are there any drawbacks to this method? It does take more space compared to the other solution, but the time complexity is much lower, O(1). (Correct me if I'm wrong).\n• » » No, this method works fine but we can calculate till 1000th number instead of storing from 1 to 1666 Hope thats clears it\n » I solved F1 and F2! At 2 minutes after contests finished. :(\n• » » ehh, that sucks, but you know that you are good enough to even Fs in Div3, now you can try increasing speed:)\n• » » » Thanks for nice comment :)\n• » » » You did it very fast from A to D. Any tips on how to increase speed?\n• » » » » I try giving virtual contests, would recommend that, or you can do mashup of problems within time limit. Basically, solving problems in a timed environment helps ig.\n » For problem E, this submission gives different output on CF judge and local machine for some inputs, for e.g. for the i/p string vOutput of the above submission on OJ is -1, but when I run locally, the o/p is v v.Can anyone please help in figuring out why this is happening.Thank you.\n• » » I submitted your solution with GNU G++17 compiler and it gave Runtime error on test 2. I have faced problem like yours on codeforces too and submitting on GNU G++17 compiler always helped me.\n » 2 months ago, # | ← Rev. 2 → This is my code for F2.\n• » » » It's just to adjust each digit from high to low.\n• » » Please explain why you did this way\n• » » Very clean code, thanks!\n » It's a pity that I didn't attend.\n » This was a great round!\n » 2 months ago, # | ← Rev. 3 → I have another solution for Problem F2.First, take the longest prefix of $n$, such that there are at most $k$ distinct digits. Let's call that prefix $p$.If $p = n$, the answer is $n$.Otherwise, let's call $i$ as the first index after $p$ (for example: if $n = 199754$, $k = 3$, then $p$ will be $1997$ and $i$ will be $4$ (indexes start at $0$)).Now find the minimum $x$ such that $p$ contains $x$ and $x$ is greater than $i-th$ digit of $n$ ($x$ can't be equal to $i-th$ digit of $n$, because in that case you can choose longer prefix than $p$).Now there are $2$ cases: If $x$ exists, change the $i-th$ digit of $n$ into $x$. Then change the all remaining digits of $n$ (from $i+1$ to the last) into the smallest number in $p$. Otherwise, you can see that all digits of $p$ are less than $9$ (because in that case the $i-th$ digit of $n$ can't be bigger than all digits of $p$). Now you have to change one of the digits of $p$ into a bigger one. It's easy to see that you have to change the digit which have the biggest index. Lets call that index $j$. At fist $j$ will be $i-1$. Now if you want to change $j-th$ digit (let's call it $a$), insert all digits before $j$ (from $0$ to $j-1$) into a $multiset$. Let's call that $multiset$ $s$. Now there are $3$ cases: If $s$ doesn't contain $a$, change $a$ into $a+1$ and insert $a+1$ into $s$. Now if $s$ contain less than $k$ distinct digits, change all remaining digits into $0$, otherwise, change them into the smallest digit in $s$. If $s$ contains $a$, check if there is a digit in $s$ greater than $a$, find that minimum digit (let's call it $b$), change $a$ into $b$, then insert $b$ into $s$. Now if $s$ contain less than $k$ distinct digits, change all remaining digits into $0$, otherwise, change them into the smallest digit in $s$. Otherwise $j$ will become $j-1$ and the digit before $a$ will be deleted from $s$ (after this $s$ may contain the digit before $a$). Time complexity: $O(len$ $\\mathrm{log}$ $len)$, where $len$ is the number of digits of $n$. This means that this solution will also work for $n \\leq 10^{10^5}$ or even for bigger $n$.My code\n » May I ask you 1 thing ? Does this contest rates for people who are having rating under 1600? - If yes, why I have been rated yet ? - I need an answer.\n• » » If there's no blog updates or notifications saying it's unrated,it will be rated for you. Just wait :)\n• » » » Thank you very much. I received the scores. :)\n » Great Contest, I became pupil for the first time.\n• » » Congratulations!\n » 2 months ago, # | ← Rev. 2 → Why my $O(t2^{10}mk)$ solution got $Accepted$ for problem F2...upd : got hacked lol\n• » » Ah you get hacked,orz!\n• » » I think we have the same approach but I break when the value will only rise, that will only take $O(2^k \\times m \\times 2)$ memory but $O(mk)$ query\n » For problem C, I got my answer wrong on just one number: 999950883 . Can someone please tell why did this happen. Is this number special in some terms? My submission link: https://codeforces.com/contest/1560/submission/126314408\n » 2 months ago, # | ← Rev. 2 → Disliked this contest, 7 constructive/greedy approaches, really? Nothing educational/interesting. Your div.3 #734 was far more better: it had cool dp and graph problems.\n• » » I have the same opinion.\n » Very good Editorial,and very good problem!especially for problem F2,a short solution and a long solution are very good for readers.And it is pretty suitable for beginner coders!:)\n » is there any other way to guess that 2^18 thing(size of set of powers of 2) for problem D? Thanks\n » how does problem rating system works I dont think C would be 800 atleast 900, I guess since a lot of people have solved it , it is given 800.\n » Problem C is similar to the CSES Number Spiral, here is the O(1) solution comparatively easy from the editorial. int n; cin >> n; int y = sqrt(n); if(y*y == n){ cout << y << ' ' << 1 << '\\n'; return; } int z = y+1; if(n > z*z - z){ cout << z << ' ' << z*z - n + 1 << '\\n'; return; } cout << n - y*y << ' ' << y+1 << '\\n'; \n• » » 2 months ago, # ^ | ← Rev. 3 → sqrt() doesn't work in $O(1)$. It works in $O(\\mathrm{log}$ $n)$.\n• » » » 2 months ago, # ^ | ← Rev. 2 → why down vote isn't asymptotic complexity of this library function $sqrt()$ is $\\mathcal O(logn)$\n » Problem E has the sorting and binary search tags, can someone explain where it can be used in the solution?\n » Can anyone please explain to me...why in problem D..we are only considering up to power of 2 equal to 18?\n• » » yes please anyone?\n• » » it is $2\\cdot10^{18}$\n• » » » Yes...sorry for the mistake..do you know why this upper limit is taken!\n• » » » » 2 months ago, # ^ | ← Rev. 2 → In worst we may have to remove either all digit of n which can be atmost 10 operation as $limit$ of n is $10^9$ or we may have to append 9 digit in right of n which increases value of n equal $10^{18}$ is equivalent to $2^{60}$ at most. Checking for all 60 power of 2 and and finding min operation to convert it to perfect power of $2$.\n » In Editorial for D problem I am unable to understand why upper limit is 10^18\n• » » 2 months ago, # ^ | ← Rev. 2 → Observe that n's upper bound means there's a limit to your answer which is the number you'd get from deleting all digits and adding a single-digit power of two. The question then is how large of a power-of-2 could you construct with any max-digit-length n + that many add-a-digit-to-the-right operations?The last case in the sample was helpful in that regard: if you bounded your powers of two according to n itself, you'd get an incorrect result for it.Extra: consider why 10^9 and 10^18 are commonly seen limits...Disclosure: I didn't think of this during the contest either, womp womp.edit: uhhhh, this was in the editorial... in which case, sorry for being repetitive...\n » I really liked this contest, genuinely enjoyed doing problems C,D,E,F.\n » what kind of error is this?\"wrong answer order too long\"...everything seems alright...getting all the outputs that are visible, matching...can someone help? string trans problem https://codeforces.com/contest/1560/submission/126609563\n » 2 months ago, # | ← Rev. 2 → Which is 189 number in 5 test case in 1560F1 - Nearest Beautiful Number (easy version) problem?MrPaul_TUser please give that case because there are a huge number of submissions failed on that case and it is interesting to me to know why\n » in problem c i didn't understand the logic/intution behind statement \"the number belongs to the i-th row and the (i−(m−i))-th column. if(m>i)\" can someone explain it please\n » Hey guys, this was my 2nd competition abd I could solve only A, B and C, one more than my first competition. But I notice that for most of the question I form the logic almost equal to the editorial but screw up in small details. Any tips and suggestions for me??\n » This is my approach for problem C. Although this showed WA during the contest but it got accepted now. 126677915\n » Can anyone please help me with understanding why this code gives TLE though it's mostly similar to the solution given in editorial. https://codeforces.com/contest/1560/submission/126656613\n » Please someone help me.... why my answer is the wrong test3 for F1. 126757323 above is my submission. also tell me, how to improve my code.\n » 5 weeks ago, # | ← Rev. 2 → For problem F,May I get the test case for 204th number? I could not know where my submission got the answer wrong. Thanks in advancehttps://codeforces.com/contest/1560/submission/129336848\n » How I solved C problem..... (1,2,5,10,........The first thing as mentioned in the editorial the numbers are increasing in Arithmetic Progression .) see the increment follows(+1,+3,+5,......)... If you notice the max number for a particular layer is minimum in that layer + 2*col -2 (col=column number 1-based indexing..) For eg consider the layer that contains 2 .....(2,3,4) here max is 4 which is equal to 2+2*2-2 .... Now we will run a while loop till we find mx,mn such that mx>=k and mn<=k ...... when will hit this condition we can simply increment mn upto(column num -1) and check if we get mn==k ........if we still dont get mn==k it means we have to find k towards left sol we will decrement col num ........Finally just print row num and col num Here my submission (https://codeforces.com/contest/1560/submission/129896567)\n » Can anyone please provide more beginer friendly tutorial for problem D. I understand the whole, but didn't understand how we come to consider power of two that are less than 10^18. I understand why it is needed but don't know how to approach so that figure out power should be less than 10^18(20digits) long.\n » my cpp code... #include using namespace std; typedef long long ll; int main() { int t; cin >> t; while(t--) { int n; cin>>n; int a=1, d=1, cnt=0; while (a<=n) { a+=d; cnt+=1; d+=2; } int p = a-d+2; int mc = p+cnt-1; if ( n <= mc) { cout<\n » How to solve E by binary search? (It is one of the problem tags)."
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https://www.mathworks.com/matlabcentral/answers/1874637-get-back-variables-out-of-function-inside-a-for-loop | [
"# Get back variables out of function inside a for loop\n\n1 view (last 30 days)\nEnzo on 9 Dec 2022\nAnswered: Dyuman Joshi on 9 Dec 2022\nHello everyone,\nI have extracted the values stored in 2 different matrices and used them in order to create a line (plot). One point is always the same (I1,M1) where I1 is the x coordinate and M1 is the Y.\ninside J and WZ matrices, I have, let's say, 3 values each (but they could be more or less) and I have used them in order to create 3 different lines with one point constant (I1, M1) while the others change. PLease note that, I have added 2500 to all the values stored inside WZ.\nRunning the for loop, ends up in having created a slope matrix, with 3 values inside, which reprenset 3 different slopes. I have found the minimun value among them (-0.0168), and now, I would like to getting back from that value (the minimum), and to call back the corresponding WZ and J values associated\nJ = [2.5 27 56];\nWZ = [12.2 23.2 33];\nM1 = 22\nM1 = 22\nI1 = 2600\nI1 = 2600\nfor ii = 1:numel(WZ)\nslope(ii) = (J(ii)-M1)/(WZ(ii)+3750-I1);\nslope_1 = min(slope);\nend\ndisp(slope_1)\n-0.0168\n-0.0168 0.0043 0.0287\n##### 2 CommentsShowHide 1 older comment\nEnzo on 9 Dec 2022\n@Stephen23 yes, it works even better. Thanks a lot\n\nAlan Stevens on 9 Dec 2022\nHere's one way\nJ = [2.5 27 56];\nWZ = [12.2 23.2 33];\nM1 = 22;\nI1 = 2600;\nfor ii = 1:numel(WZ)\nslope(ii) = (J(ii)-M1)/(WZ(ii)+3750-I1);\nslope_1 = min(slope);\nend\ndisp(slope_1)\n-0.0168\nindx = find(slope==slope_1)\nindx = 1\nWZ_1 = WZ(indx)\nWZ_1 = 12.2000\nJ_1 = J(indx)\nJ_1 = 2.5000\n##### 2 CommentsShowHide 1 older comment\nStephen23 on 9 Dec 2022\nUsing the second output from MIN() is simpler than calling superfluous FIND().\n\nDyuman Joshi on 9 Dec 2022\nJ = [2.5 27 56];\nWZ = [12.2 23.2 33];\nM1 = 22;\nI1 = 2600;\nslope=(J-M1)./(WZ+3750-I1)\nslope = 1×3\n-0.0168 0.0043 0.0287\n[minval,minid]=min(slope)\nminval = -0.0168\nminid = 1\nJval=J(minid)\nJval = 2.5000\nWZval=WZ(minid)\nWZval = 12.2000\n\n### Categories\n\nFind more on Logical in Help Center and File Exchange\n\nR2022b\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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https://docs.panda3d.org/1.10/cpp/reference/panda3d.core.LMatrix3d | [
"# LMatrix3d\n\nclass LMatrix3d\n\nThis is a 3-by-3 transform matrix. It typically will represent either a rotation-and-scale (no translation) matrix in 3-d, or a full affine matrix (rotation, scale, translation) in 2-d, e.g. for a texture matrix.\n\nInheritance diagram\n\nclass CRow\nCRow(LMatrix3d::CRow const&) = default\n\nDefines a row-level constant accessor to the matrix.\n\nint size(void)\n\nReturns 3: the number of columns of a `LMatrix3`.\n\nclass Row\n\nThese helper classes are used to support two-level operator [].\n\nRow(LMatrix3d::Row const&) = default\n\nDefines a row-level index accessor to the matrix.\n\nint size(void)\n\nReturns 3: the number of columns of a `LMatrix3`.\n\nLMatrix3d(void)\nLMatrix3d(LMatrix3d const &other)\nLMatrix3d(double, double, double, double, double, double, double, double, double)\nLMatrix3d(LVecBase3d const&, LVecBase3d const&, LVecBase3d const&)\n\nConstructs the matrix from three individual rows.\n\nstd::size_t add_hash(std::size_t hash, double threshold) const\n\nAdds the vector into the running hash.\n\nbool almost_equal(LMatrix3d const &other, double threshold) const\nbool almost_equal(LMatrix3d const &other) const\n\nReturns true if two matrices are memberwise equal within a default tolerance based on the numeric type.\n\nReturns true if two matrices are memberwise equal within a specified tolerance.\n\nint compare_to(LMatrix3d const &other) const\nint compare_to(LMatrix3d const &other, double threshold) const\n\nThis flavor of compare_to uses a default threshold value based on the numeric type.\n\nSorts matrices lexicographically, componentwise. Returns a number less than 0 if this matrix sorts before the other one, greater than zero if it sorts after, 0 if they are equivalent (within the indicated tolerance).\n\nvoid componentwise_mult(LMatrix3d const &other)\nstatic LMatrix3d const &convert_mat(CoordinateSystem from, CoordinateSystem to)\n\nReturns a matrix that transforms from the indicated coordinate system to the indicated coordinate system.\n\ndouble determinant(void) const\n\nReturns the determinant of the matrix.\n\nvoid fill(double fill_value)\n\nSets each element of the matrix to the indicated fill_value. This is of questionable value, but is sometimes useful when initializing to zero.\n\nvoid generate_hash(ChecksumHashGenerator &hashgen) const\nvoid generate_hash(ChecksumHashGenerator &hashgen, double threshold) const\n\nAdds the vector to the indicated hash generator.\n\ndouble get_cell(int row, int col) const\n\nReturns a particular element of the matrix.\n\nstatic TypeHandle get_class_type(void)\nLVecBase3d get_col(int col) const\n\nReturns the indicated column of the matrix as a three-component vector.\n\nLVecBase2d get_col2(int col) const\n\nReturns the indicated column of the matrix as a two-component vector, ignoring the last row.\n\ndouble const *get_data(void) const\n\nReturns the address of the first of the nine data elements in the matrix. The remaining elements occupy the next eight positions in row-major order.\n\nstd::size_t get_hash(void) const\nstd::size_t get_hash(double threshold) const\n\nReturns a suitable hash for phash_map.\n\nint get_num_components(void) const\n\nReturns the number of elements in the matrix, nine.\n\nLVecBase3d get_row(int row) const\nvoid get_row(LVecBase3d &result_vec, int row) const\n\nthese versions inline better\n\nReturns the indicated row of the matrix as a three-component vector.\n\nStores the indicated row of the matrix as a three-component vector.\n\nLVecBase2d get_row2(int row) const\n\nReturns the indicated row of the matrix as a two-component vector, ignoring the last column.\n\nLMatrix3d const &ident_mat(void)\n\nReturns an identity matrix.\n\nThis function definition must appear first, since some inline functions below take advantage of it.\n\nbool invert_from(LMatrix3d const &other)\n\nComputes the inverse of the other matrix, and stores the result in this matrix. This is a fully general operation and makes no assumptions about the type of transform represented by the matrix.\n\nThe other matrix must be a different object than this matrix. However, if you need to invert a matrix in place, see `invert_in_place`.\n\nThe return value is true if the matrix was successfully inverted, false if there was a singularity.\n\nbool invert_in_place(void)\n\nInverts the current matrix. Returns true if the inverse is successful, false if the matrix was singular.\n\nbool invert_transpose_from(LMatrix3d const &other)\nbool invert_transpose_from(LMatrix4d const &other)\n\nSimultaneously computes the inverse of the indicated matrix, and then the transpose of that inverse.\n\nbool is_identity(void) const\n\nReturns true if this is (close enough to) the identity matrix, false otherwise.\n\nbool is_nan(void) const\n\nReturns true if any component of the matrix is not-a-number, false otherwise.\n\nvoid multiply(LMatrix3d const &other1, LMatrix3d const &other2)\n\nthis = other1 * other2\n\nvoid output(std::ostream &out) const\n\nReads the matrix from the Datagram using get_stdfloat().\n\nReads the matrix from the Datagram using get_float32() or get_float64(). See `write_datagram_fixed()`.\n\nLMatrix3d rotate_mat(double angle)\nLMatrix3d rotate_mat(double angle, LVecBase3d const &axis, CoordinateSystem cs = ::CS_default)\n\nReturns a matrix that rotates by the given angle in degrees counterclockwise.\n\nReturns a matrix that rotates by the given angle in degrees counterclockwise about the indicated vector.\n\nLMatrix3d rotate_mat_normaxis(double angle, LVecBase3d const &axis, CoordinateSystem cs = ::CS_default)\n\nReturns a matrix that rotates by the given angle in degrees counterclockwise about the indicated vector. Assumes axis has been normalized.\n\nLMatrix3d scale_mat(LVecBase2d const &scale)\nLMatrix3d scale_mat(double sx, double sy)\nLMatrix3d scale_mat(LVecBase3d const &scale)\nLMatrix3d scale_mat(double sx, double sy, double sz)\n\nReturns a matrix that applies the indicated scale in each of the two axes.\n\nReturns a matrix that applies the indicated scale in each of the two axes.\n\nReturns a matrix that applies the indicated scale in each of the three axes.\n\nReturns a matrix that applies the indicated scale in each of the three axes.\n\nLMatrix3d scale_shear_mat(LVecBase3d const &scale, LVecBase3d const &shear, CoordinateSystem cs = ::CS_default)\nLMatrix3d scale_shear_mat(double sx, double sy, double sz, double shxy, double shxz, double shyz, CoordinateSystem cs = ::CS_default)\n\nReturns a matrix that applies the indicated scale and shear.\n\nvoid set(double e00, double e01, double e02, double e10, double e11, double e12, double e20, double e21, double e22)\nvoid set_cell(int row, int col, double value)\n\nChanges a particular element of the matrix.\n\nvoid set_col(int col, LVecBase3d const &v)\nvoid set_col(int col, LVecBase2d const &v)\n\nReplaces the indicated column of the matrix from a three-component vector.\n\nReplaces the indicated column of the matrix from a two-component vector, ignoring the last row.\n\nvoid set_rotate_mat(double angle)\nvoid set_rotate_mat(double angle, LVecBase3d const &axis, CoordinateSystem cs = ::CS_default)\n\nFills mat with a matrix that rotates by the given angle in degrees counterclockwise.\n\nFills mat with a matrix that rotates by the given angle in degrees counterclockwise about the indicated vector.\n\nvoid set_rotate_mat_normaxis(double angle, LVecBase3d const &axis, CoordinateSystem cs = ::CS_default)\n\nFills mat with a matrix that rotates by the given angle in degrees counterclockwise about the indicated vector. Assumes axis has been normalized.\n\nvoid set_row(int row, LVecBase3d const &v)\nvoid set_row(int row, LVecBase2d const &v)\n\nReplaces the indicated row of the matrix from a three-component vector.\n\nReplaces the indicated row of the matrix from a two-component vector, ignoring the last column.\n\nvoid set_scale_mat(LVecBase2d const &scale)\nvoid set_scale_mat(LVecBase3d const &scale)\n\nFills mat with a matrix that applies the indicated scale in each of the two axes.\n\nFills mat with a matrix that applies the indicated scale in each of the three axes.\n\nvoid set_scale_shear_mat(LVecBase3d const &scale, LVecBase3d const &shear, CoordinateSystem cs = ::CS_default)\n\nFills mat with a matrix that applies the indicated scale and shear.\n\nvoid set_shear_mat(LVecBase3d const &shear, CoordinateSystem cs = ::CS_default)\n\nFills mat with a matrix that applies the indicated shear in each of the three planes.\n\nvoid set_translate_mat(LVecBase2d const &trans)\n\nFills mat with a matrix that applies the indicated translation.\n\nLMatrix3d shear_mat(LVecBase3d const &shear, CoordinateSystem cs = ::CS_default)\nLMatrix3d shear_mat(double shxy, double shxz, double shyz, CoordinateSystem cs = ::CS_default)\n\nReturns a matrix that applies the indicated shear in each of the three planes.\n\nint size(void)\n\nReturns 3: the number of rows of a `LMatrix3`.\n\nLMatrix3d translate_mat(LVecBase2d const &trans)\nLMatrix3d translate_mat(double tx, double ty)\n\nReturns a matrix that applies the indicated translation.\n\nvoid transpose_from(LMatrix3d const &other)\nvoid transpose_in_place(void)\nbool validate_ptr(void const *ptr)\nvoid write(std::ostream &out, int indent_level = 0) const\nvoid write_datagram(Datagram &destination) const\n\nWrites the matrix to the Datagram using add_stdfloat(). This is appropriate when you want to write the matrix using the standard width setting, especially when you are writing a bam file.\n\nvoid write_datagram_fixed(Datagram &destination) const\n\nWrites the matrix to the Datagram using add_float32() or add_float64(), depending on the type of floats in the matrix, regardless of the setting of `Datagram::set_stdfloat_double()`. This is appropriate when you want to write a fixed-width value to the datagram, especially when you are not writing a bam file.\n\nLVecBase3d xform(LVecBase3d const &v) const\n\n3-component vector or point times matrix.\n\nvoid xform_in_place(LVecBase3d &v) const\n\n3-component vector or point times matrix.\n\nLVecBase2d xform_point(LVecBase2d const &v) const\n\nThe matrix transforms a 2-component point (including translation component) and returns the result. This assumes the matrix is an affine transform.\n\nvoid xform_point_in_place(LVecBase2d &v) const\n\nThe matrix transforms a 2-component point (including translation component). This assumes the matrix is an affine transform.\n\nLVecBase2d xform_vec(LVecBase2d const &v) const\nLVecBase3d xform_vec(LVecBase3d const &v) const\n\nThe matrix transforms a 2-component vector (without translation component) and returns the result. This assumes the matrix is an affine transform.\n\nThe matrix transforms a 3-component vector and returns the result. This assumes the matrix is an orthonormal transform.\n\nIn practice, this is the same computation as `xform()`.\n\nLVecBase3d xform_vec_general(LVecBase3d const &v) const\n\nThe matrix transforms a 3-component vector (without translation component) and returns the result, as a fully general operation.\n\nvoid xform_vec_general_in_place(LVecBase3d &v) const\n\nThe matrix transforms a 3-component vector (without translation component), as a fully general operation.\n\nvoid xform_vec_in_place(LVecBase2d &v) const\nvoid xform_vec_in_place(LVecBase3d &v) const\n\nThe matrix transforms a 2-component vector (without translation component). This assumes the matrix is an affine transform.\n\nThe matrix transforms a 3-component vector. This assumes the matrix is an orthonormal transform.\n\nIn practice, this is the same computation as `xform()`."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6644016,"math_prob":0.9791233,"size":10685,"snap":"2022-05-2022-21","text_gpt3_token_len":2535,"char_repetition_ratio":0.2058796,"word_repetition_ratio":0.38380513,"special_character_ratio":0.21778193,"punctuation_ratio":0.12152186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99589866,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T07:45:38Z\",\"WARC-Record-ID\":\"<urn:uuid:3076d705-7866-4664-a1d7-2af123382300>\",\"Content-Length\":\"398969\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ba838d1-b017-41df-b0dc-b0327e3b2bfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1da782c-4a51-46bc-9101-01d350affac3>\",\"WARC-IP-Address\":\"128.2.236.121\",\"WARC-Target-URI\":\"https://docs.panda3d.org/1.10/cpp/reference/panda3d.core.LMatrix3d\",\"WARC-Payload-Digest\":\"sha1:DYXCSJPX3IPUU3GNBAFN4CX3NXJ5NP65\",\"WARC-Block-Digest\":\"sha1:GBPNXI3H324ZOTMVMGNBOPFFEU2SBTN2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531762.30_warc_CC-MAIN-20220520061824-20220520091824-00388.warc.gz\"}"} |
https://eccc.weizmann.ac.il/eccc-reports/2000/TR00-056/index.html | [
"",
null,
"",
null,
"Under the auspices of the Computational Complexity Foundation (CCF)",
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"",
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"",
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"",
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"",
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"REPORTS > DETAIL:\n\n### Paper:\n\nTR00-056 | 20th July 2000 00:00\n\n#### On Pseudorandomness with respect to Deterministic Observers.",
null,
"TR00-056\nAuthors: Oded Goldreich, Avi Wigderson\nPublication: 24th July 2000 11:12\nKeywords:\n\nAbstract:\n\nIn the theory of pseudorandomness, potential (uniform) observers\nare modeled as probabilistic polynomial-time machines.\nIn fact many of the central results in\nthat theory are proven via probabilistic polynomial-time reductions.\nIn this paper we show that analogous deterministic reductions\nare unlikely to hold. We conclude that randomness of the observer\nis essential to the theory of pseudorandomness.\n\nWhat we actually prove is that the hypotheses\nof two central theorems (in the theory of pseudorandomness)\nhold unconditionally when stated with\nrespect to deterministic polynomial-time algorithms.\nThus, if these theorems were true for deterministic\nobservers, then their conclusions would hold\nunconditionally, which we consider unlikely.\nFor example, it would imply (unconditionally)\nthat any unary language in BPP is in P.\n\nThe results are proven using diagonalization and\npairwise independent sample spaces.\n\nISSN 1433-8092 | Imprint"
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"https://eccc.weizmann.ac.il/resources/gf/logoNew.png",
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"https://eccc.weizmann.ac.il/resources/gf/subtitle.png",
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"https://eccc.weizmann.ac.il/resources/txt2img/6477d27f5652481c8709ce20804beef47000ddfacb628eb8f3e1424aa319da92d9706a1b9969c3beea6d0d0579f8c3574dfe71145b9a63e0f4cc7e59723f9d59-000000-13.png",
null,
"https://eccc.weizmann.ac.il/resources/txt2img/734cc234b69ec76be631e268baeba4246056bc255901fd92951a0836428e49f37084bbbfa3c4e253e31cc4d576b67f6cd530e1bb77f0ecc98955de6ba9eb86c4-000000-13.png",
null,
"https://eccc.weizmann.ac.il/resources/txt2img/112d9535943722a86180ae44a5a638b4f2c8b88f2b35a2161475927f703e4959e03e1c231f19ff9bb2aff902b0183e2db60085b49f5c3b501624b17f86a1b036-000000-13.png",
null,
"https://eccc.weizmann.ac.il/resources/txt2img/734cc234b69ec76be631e268baeba4246056bc255901fd92951a0836428e49f37084bbbfa3c4e253e31cc4d576b67f6cd530e1bb77f0ecc98955de6ba9eb86c4-000000-13.png",
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"https://eccc.weizmann.ac.il/resources/txt2img/94b19a5a52ca45c018ed1cb67a8f8a31a33b54a97551ad8a99f802714d157423de95da10ff4cd42551e3995e26f1c3c4c437b5c95fd23fd10cb8195fe86f48f1-000000-13.png",
null,
"https://eccc.weizmann.ac.il/resources/gf/rss.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91131514,"math_prob":0.7902911,"size":981,"snap":"2021-43-2021-49","text_gpt3_token_len":195,"char_repetition_ratio":0.10951894,"word_repetition_ratio":0.0,"special_character_ratio":0.16411825,"punctuation_ratio":0.084415585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9572238,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T06:33:08Z\",\"WARC-Record-ID\":\"<urn:uuid:d35dd5b6-e8e6-4025-95ea-e2a5e98c8f32>\",\"Content-Length\":\"20698\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f662fc3e-acb0-4b90-81c5-23121020167b>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e3f8fc1-f736-4b1c-9675-fed58b99e5bc>\",\"WARC-IP-Address\":\"132.77.150.87\",\"WARC-Target-URI\":\"https://eccc.weizmann.ac.il/eccc-reports/2000/TR00-056/index.html\",\"WARC-Payload-Digest\":\"sha1:IGKP73UW6TZZOI77H2563IBGI6H2LYIX\",\"WARC-Block-Digest\":\"sha1:GKS3KER5FSDQ7DQHGTZUPSEOCWOGKTZR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363290.39_warc_CC-MAIN-20211206042636-20211206072636-00455.warc.gz\"}"} |
https://www.studymite.com/cpp/examples/program-to-find-k-largest-elements-in--array-cpp | [
"# Program to find k largest elements in a given array of integers in C++\n\nGiven – array of integers and we have to print k number of largest elements from the array.\n\nExample:\n\nGiven array is [12, 20, 14, 26, 30, 1, 70, 56]\n\nWe have to find largest 4 elements i.e., k = 4\n\nTherefore our program should print 70, 56, 30 and 26.\n\nAlgorithm:\n\n1. Sort the given array in ascending order using the `sort` function.\n2. Print the first `k` elements of the sorted array using a loop.\n\nCode:\n\n``````#include <bits/stdc++.h>\nusing namespace std;\n\nvoid kLarge(int array[], int size, int k)\n{\n// Sorting given array in reverse order\nsort(array, array + size, greater<int> ());\n\n// Printing first kth largest elements\nfor (int i = 0; i < k; i++)\ncout << array[i] << \" \";\n}\n\n// driver program\nint main()\n{\nint array[] = { 12, 20, 14, 26, 30, 1, 70, 56 };\nint size = sizeof(array) / sizeof(array);\nint k = 4;\nkLarge(array, size, k);\nreturn 0;\n}\n\n``````\n\nOutput\n\n``````70 56 30 26\n``````"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.53871846,"math_prob":0.99360096,"size":934,"snap":"2023-40-2023-50","text_gpt3_token_len":283,"char_repetition_ratio":0.13870968,"word_repetition_ratio":0.022727273,"special_character_ratio":0.36509636,"punctuation_ratio":0.20574163,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99659234,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T00:25:58Z\",\"WARC-Record-ID\":\"<urn:uuid:689ed712-d08a-4e9f-865c-34a28f5c8d93>\",\"Content-Length\":\"58095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a26cea6a-e908-4bb7-ae65-866808cb19a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:02a68c23-efc3-4db8-acff-b5477dd149a7>\",\"WARC-IP-Address\":\"3.210.81.252\",\"WARC-Target-URI\":\"https://www.studymite.com/cpp/examples/program-to-find-k-largest-elements-in--array-cpp\",\"WARC-Payload-Digest\":\"sha1:WDPK52A5RVOUKLFNRVHWXHJAA7MXHKYH\",\"WARC-Block-Digest\":\"sha1:AZTTEATABQJIFFBQTMCXVO4AOS7YVRLT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510529.8_warc_CC-MAIN-20230929222230-20230930012230-00283.warc.gz\"}"} |
https://math.stackexchange.com/questions/23724/set-theory-problems-well-ordered-sets-countability-zorns-lemma | [
"# set theory problems (well-ordered sets, countability, Zorn's Lemma, …)\n\n1. Let $A \\subseteq P(\\omega)$, where $\\omega$ is the set of all natural numbers and $P(\\omega)$ is the power set of $\\omega$. If $\\langle A,\\subseteq\\rangle$ is a well ordered set how can you prove that $A$ is a countable set.\n\n2. Let $A$ be a set which elements are closed sets of real numbers. If $\\langle A,\\subseteq\\rangle$ is a well ordered set how can you prove that $A$ is a countable set.\n\n3. How can you prove that Zorn's lemma (so and the axiom of choice) is equivalent to that for every partially ordered set $\\langle A,\\le\\rangle$ that satisfies Zorn's condition, for every $b\\in A$ there exists a maximum element $a$ for which $b\\le a$.\n\n• I'll wager 3 has to be \"maximal element $a$\"; otherwise, the statement would assert that Zorn's condition implies the existence of a maximum. – Arturo Magidin Feb 25 '11 at 16:37\n• And please say what you've tried or where you are stuck. – Arturo Magidin Feb 25 '11 at 16:38\n\n1. Since $A$ is well-ordered, we can order-biject it with an ordinal $\\beta$, $f\\colon \\beta\\to A$. Now, you can use $f$ to define an injection from $\\beta$ to $\\omega$ by letting $g(\\alpha)$ be the least element of $f(\\alpha+1)-f(\\alpha)$. Show that $g$ is an injection, and therefore $\\beta$ is countable.\n2. Consider the set of complements, which are open, and well ordered by $\\supseteq$. Show that there is always a rational that is in one set but not in its $\\supseteq$-successor, and that you can pick distinct rationals at each stage (that is, define an injection from $A$ to $\\mathbb{Q}$).\n3. Replacing \"maximum element $a$\" with \"maximal element $a$\", assuming Zorn's Lemma and given $\\langle A,\\leq\\rangle$ and $b$, look at $\\mathcal{C}=\\{x\\in A\\mid x\\geq b\\}$ and apply Zorn's Lemma to it. For the converse, pick an arbitrary $b\\in A$ and conclude the original set has maximal elements."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8927847,"math_prob":0.9996815,"size":1860,"snap":"2019-43-2019-47","text_gpt3_token_len":532,"char_repetition_ratio":0.11637931,"word_repetition_ratio":0.11764706,"special_character_ratio":0.28387097,"punctuation_ratio":0.09358289,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99996626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T03:22:48Z\",\"WARC-Record-ID\":\"<urn:uuid:7b4406ca-f856-42a3-a9fa-5b764f40c789>\",\"Content-Length\":\"140300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f8fa39c6-118c-4586-8933-91f99616db0e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c7ac355-8355-4e8e-9340-94d2db423134>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/23724/set-theory-problems-well-ordered-sets-countability-zorns-lemma\",\"WARC-Payload-Digest\":\"sha1:HCTITMRRE4BBTMR6RI5PG7ZI7FSLKZIX\",\"WARC-Block-Digest\":\"sha1:CURK4KPXGSTDLNG7ZJSHLRF7IV6QRGGL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668561.61_warc_CC-MAIN-20191115015509-20191115043509-00516.warc.gz\"}"} |
http://export.arxiv.org/abs/math/0209183 | [
"math\n\n# Title: Ramification of surfaces: Artin-Schreier extensions\n\nAuthors: Igor Zhukov\nAbstract: Let $A$ be a regular 2-dimensional local ring of characteristic $p>0$, and let $L/K$ be a cyclic extension of degree $p$ of its field of fractions such that the corresponding branch divisor is normal crossing. For each $\\gp\\in\\Spec A$ of height 1 such that $A/\\gp$ is regular, consider the ramification jump $h_\\gp$ of the extension of the residue field at $\\gp$. In this paper the semi-continuity of $h_\\gp$ with respect to Zariski topology of suitable jet spaces is proved. The asymptotic of $h_\\gp$ with respect to intersection multiplicity of the prime divisor defined by $\\gp$ and the branch divisor is also addressed.\n Comments: LaTeX, 9 pages Subjects: Algebraic Geometry (math.AG) MSC classes: Primary 14E22; Secondary 11S15 Journal reference: Contemporary mathematics 300 (2002), 211--220 Cite as: arXiv:math/0209183 [math.AG] (or arXiv:math/0209183v1 [math.AG] for this version)\n\n## Submission history\n\nFrom: Igor B. Zhukov [view email]\n[v1] Sun, 15 Sep 2002 18:40:47 GMT (10kb)\n\nLink back to: arXiv, form interface, contact."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86392313,"math_prob":0.9086362,"size":976,"snap":"2020-34-2020-40","text_gpt3_token_len":259,"char_repetition_ratio":0.09465021,"word_repetition_ratio":0.01369863,"special_character_ratio":0.25512296,"punctuation_ratio":0.113513514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9625947,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T21:39:30Z\",\"WARC-Record-ID\":\"<urn:uuid:5d95011a-b548-4042-9549-4b1083f2a979>\",\"Content-Length\":\"15510\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdf74fa9-a87f-441a-80d9-2a1cd0198819>\",\"WARC-Concurrent-To\":\"<urn:uuid:e214ff29-ab47-400c-a2f3-57f2d6265962>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"http://export.arxiv.org/abs/math/0209183\",\"WARC-Payload-Digest\":\"sha1:2MLXIAYKIAI5QD4TTJZF7FUAOXRM7VUL\",\"WARC-Block-Digest\":\"sha1:CBWWE736J5X6QUPKQQT6PHIV555CAOFE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738699.68_warc_CC-MAIN-20200810205824-20200810235824-00021.warc.gz\"}"} |
https://quantummoxie.wordpress.com/tag/data-pipelining-inequality/ | [
"Well, I’ve been lax in posting as I’ve been busy getting ready for the March Meeting of the APS. But I have an idea that’s been percolating in my head in relation to the March Meeting and it forms the basis of an idea that I have that is the at the core of a hypothesis I’m presenting there.\n\nThe short end of it is that I have a theory based around the idea that Bell’s inequalities are merely another statement of the Second Law of Thermodynamics, the latter actually not a fundamental law, but rather merely a strong argument about probabilities (see, for example, Dan Schroeder’s Thermal Physics). In short (and there will be a paper up on the arXiv about this soon), here’s argument:\n\n1. The entropy of mixing represents the entropy created when mixing two systems. It is always zero or positive since it is essentially another way of counting the configurations of the system. It is zero when the two systems are the same. As such it is a measure of ‘separation’ of the probability distributions of the two systems.\n\n2. The relative (Shannon) entropy and its other forms the conditional and mutual entopies (see Nielsen and Chuang’s Quantum Computation and Quantum Information are also measures of the separation of probability distributions of two systems.\n\n3. The data pipelining inequality implies that, if X -> Y -> Z is a Markov chain, then based on fundamental properties of Shannon entropies (again see Nielsen and Chuang) we can write H(X:Y) ≥ H(X:Z) as well as H(Y:Z) ≥ H(X:Z)\n\n4. It follows trivially from 3. that\n\nH(X:Y) + H(Y:Z) ≥ H(X:Z)\n\nIt is possible to form inequalities of the type derived by Cerf and Adami from this inequality. In addition, if we speak entirely in terms of relative information, by dividing through by the total relative information available, represented by a sum of each of these, we can write an inequality similar to Wigner’s form of Bell’s inequalities (see Wigner’s paper or Sakurai’s Modern Quantum Mechanics):\n\nPr(X:Y) + Pr(Y:Z) ≥ Pr(X:Z)\n\nIs it possible Bell’s inequalities, which are entirely classical obviously (which is why they are violated), are just another way of writing the second law? The Shannon entropies are purely classical and the inequality above represents, essentially, the positivity condition for entropy inherent in the second law.\n\nThe thing is that, for quantum systems, we would want to speak in terms of the von Neumann entropy which is not wholly classical. In that sense it is not clear that we could even write such an inequality in these terms. But this begs the question, then: shouldn’t the definition of entropy be consistent across regimes?\n\nThat’s the question I want feedback on."
] | [
null
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https://meta.m.wikimedia.org/wiki/Wikix | [
"# Wikix\n\nWikix is a 'C' based program written by Jeffrey Vernon Merkey that will read any XML dump provided by the foundation, extract all image names from the XML dump which it may reference, then generate a series of BASH or Bourne Unix style scripts which can be invoked to download all images from Wikimedia Commons and Wikipedia.\n\nThe program relies on cURL, an automated web spider, to download referenced images. The program will also convert text based utf8 characters into actual utf8 strings for those dumps which may contain improperly formatted names for specific images. The program can be configured to generate 16 parallel scripts which will download all images from Wikipedia. The program includes Jeff Bezanson's utf8 libraries.\n\nAs of March 24, 2008, using a cable modem, the entire set of Wikipedia images can be downloaded in about 96 hours using this program (420 GB as of 3/24/08).\n\n## Compiling and Installation\n\n### On Ubuntu\n\n1. Extract the contents of wikix.tar.gz. Suppose the source code is extracted into /home/you/wikix.\n2. Start your terminal program. e.g. Konsole (in KDE)\n3. You need to install some packages before you compile Wikix. Type in your terminal:-\n`sudo aptitude install libssl-dev build-essential curl`\n4. Now goto the directory that contains the extracted source code, e.g. /home/you/wikix, by typing\n`cd /home/you/wikix`\n5. Now type in your terminal.\n`sudo make`\nNow if the compilation and linking of Wikix completes without errors then you will have a brand new executable - wikix, in your /home/you/wikix (in this example) directory.\n\nIn case of any problems, please report it in the discussion page.\n\n## Options\n\n```# ./wikix -h\nUSAGE: wikix -htrciop < file.xml [ > script.out ]\n-h this help screen\n-t use xml dump to strip from tree\n-r wikipedia path\n-c commons path\n-i image path\n-o output path\n-p parallel (16 process) mode\n```\n\n## Example\n\nThe program would typically be invoked in a directory that you wish to use to host the images. Wikix will construct a MediaWiki style directory structure which can be quickly imported into a MediaWiki Wikipedia installation (e.g., via `php rebuildImages.php --missing`):\n\n```wikix -p < name_of_xml_file.xml > script.out &\n```\n\nThat is, the xml file is fed via stdin.\n\nThe -p option tells wikix to create parallel scripts. If you omit the -p option, it will create one very large file. By default, the program is set up to mirror the English Wikipedia. You can override the default settings by substituting path information for commons and the target Wikipedia site through the command line options.\n\nThe program will create a series of scripts as:\n\n```image_sh\nimage00\nimage01\nimage02\nimage03\nimage04\nimage05\nimage06\nimage07\nimage08\nimage09\nimage10\nimage11\nimage12\nimage13\nimage14\nimage15\n```\n\n```\\$./image_sh\n```\n\nIn case you get the following error.\n\n```./image_sh: <line no.>: Syntax error: Bad fd number\n```\n\nThen please open the `image_sh` all the `imagexx` files and change the topmost line from\n\n```#!/bin/sh\n```\n\nto\n\n```#!/bin/bash\n```\n\nthen re-type\n\n```\\$./image_sh\n```\n\n## Source Code\n\nThe full source code with build scripts can be download from those web servers:\n\n### Source Code Files\n\nMakefile\n\n```#CFLAGS = -g\n#CFLAGS_LIB = -g -c\nCFLAGS = -Wno-pointer-sign -g\nCFLAGS_LIB = -Wno-pointer-sign -g -c\nCC = gcc\nLD = ld\nAR = ar\n\nall: wikix\n\nlibcutf8.so: utf8.o\n\\$(LD) -shared -lc -o libcutf8.so utf8.o /usr/lib/libc.a\n\nlibcutf8.a: utf8.o\n\\$(AR) r libcutf8.a utf8.o\n\nwikix: wikix.c libcutf8.a\n\\$(CC) \\$(CFLAGS) wikix.c -o wikix libcutf8.a -lssl\n\nclean:\nrm -f *.o wikix\n\ninstall: all\ninstall -m 755 wikix /usr/bin\ninstall -m 644 libcutf8.so /usr/lib\ninstall -m 644 libcutf8.a /usr/lib\n```\n\nplatform.h\n\n```#define LINUX\n```\n\nutf8.h\n\n```#include <stdarg.h>\n\n/* is c the start of a utf8 sequence? */\n#define isutf(c) (((c)&0xC0)!=0x80)\n\n/* convert UTF-8 data to wide character */\nint u8_toucs(u_int32_t *dest, int sz, char *src, int srcsz);\n\n/* the opposite conversion */\nint u8_toutf8(char *dest, int sz, u_int32_t *src, int srcsz);\n\n/* single character to UTF-8 */\nint u8_wc_toutf8(char *dest, u_int32_t ch);\n\n/* character number to byte offset */\nint u8_offset(char *str, int charnum);\n\n/* byte offset to character number */\nint u8_charnum(char *s, int offset);\n\n/* return next character, updating an index variable */\nu_int32_t u8_nextchar(char *s, int *i);\n\n/* move to next character */\nvoid u8_inc(char *s, int *i);\n\n/* move to previous character */\nvoid u8_dec(char *s, int *i);\n\n/* returns length of next utf-8 sequence */\nint u8_seqlen(char *s);\n\n/* assuming src points to the character after a backslash, read an\nescape sequence, storing the result in dest and returning the number of\ninput characters processed */\n\n/* given a wide character, convert it to an ASCII escape sequence stored in\nbuf, where buf is \"sz\" bytes. returns the number of characters output. */\nint u8_escape_wchar(char *buf, int sz, u_int32_t ch);\n\n/* convert a string \"src\" containing escape sequences to UTF-8 */\nint u8_unescape(char *buf, int sz, char *src);\n\n/* convert UTF-8 \"src\" to ASCII with escape sequences.\nif escape_quotes is nonzero, quote characters will be preceded by\nbackslashes as well. */\nint u8_escape(char *buf, int sz, char *src, int escape_quotes);\n\n/* utility predicates used by the above */\nint octal_digit(char c);\nint hex_digit(char c);\n\n/* return a pointer to the first occurrence of ch in s, or NULL if not\nfound. character index of found character returned in *charn. */\nchar *u8_strchr(char *s, u_int32_t ch, int *charn);\n\n/* same as the above, but searches a buffer of a given size instead of\na NUL-terminated string. */\nchar *u8_memchr(char *s, u_int32_t ch, size_t sz, int *charn);\n\n/* count the number of characters in a UTF-8 string */\nint u8_strlen(char *s);\n\nint u8_is_locale_utf8(char *locale);\n\n/* printf where the format string and arguments may be in UTF-8.\nyou can avoid this function and just use ordinary printf() if the current\nlocale is UTF-8. */\nint u8_vprintf(char *fmt, va_list ap);\nint u8_printf(char *fmt, ...);\n```\n\nutf8.c public domain\n\n```/*\nBasic UTF-8 manipulation routines\nby Jeff Bezanson\nplaced in the public domain Fall 2005\n\nThis code is designed to provide the utilities you need to manipulate\nUTF-8 as an internal string encoding. These functions do not perform the\nerror checking normally needed when handling UTF-8 data, so if you happen\nto be from the Unicode Consortium you will want to flay me alive.\nI do this because error checking can be performed at the boundaries (I/O),\nwith these routines reserved for higher performance on data known to be\nvalid.\n*/\n#include <stdlib.h>\n#include <stdio.h>\n#include <string.h>\n#include <stdarg.h>\n#ifdef WIN32\n#include <malloc.h>\n#else\n#include <alloca.h>\n#endif\n\n#include \"utf8.h\"\n\nstatic const u_int32_t offsetsFromUTF8 = {\n0x00000000UL, 0x00003080UL, 0x000E2080UL,\n0x03C82080UL, 0xFA082080UL, 0x82082080UL\n};\n\nstatic const char trailingBytesForUTF8 = {\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\n1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,\n2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2, 3,3,3,3,3,3,3,3,4,4,4,4,5,5,5,5\n};\n\n/* returns length of next utf-8 sequence */\nint u8_seqlen(char *s)\n{\nreturn trailingBytesForUTF8[(unsigned int)(unsigned char)s] + 1;\n}\n\n/* conversions without error checking\nonly works for valid UTF-8, i.e. no 5- or 6-byte sequences\nsrcsz = source size in bytes, or -1 if 0-terminated\nsz = dest size in # of wide characters\n\nreturns # characters converted\ndest will always be L'\\0'-terminated, even if there isn't enough room\nfor all the characters.\nif sz = srcsz+1 (i.e. 4*srcsz+4 bytes), there will always be enough space.\n*/\nint u8_toucs(u_int32_t *dest, int sz, char *src, int srcsz)\n{\nu_int32_t ch;\nchar *src_end = src + srcsz;\nint nb;\nint i=0;\n\nwhile (i < sz-1) {\nnb = trailingBytesForUTF8[(unsigned char)*src];\nif (srcsz == -1) {\nif (*src == 0)\ngoto done_toucs;\n}\nelse {\nif (src + nb >= src_end)\ngoto done_toucs;\n}\nch = 0;\nswitch (nb) {\n/* these fall through deliberately */\ncase 3: ch += (unsigned char)*src++; ch <<= 6;\ncase 2: ch += (unsigned char)*src++; ch <<= 6;\ncase 1: ch += (unsigned char)*src++; ch <<= 6;\ncase 0: ch += (unsigned char)*src++;\n}\nch -= offsetsFromUTF8[nb];\ndest[i++] = ch;\n}\ndone_toucs:\ndest[i] = 0;\nreturn i;\n}\n\n/* srcsz = number of source characters, or -1 if 0-terminated\nsz = size of dest buffer in bytes\n\nreturns # characters converted\ndest will only be '\\0'-terminated if there is enough space. this is\nfor consistency; imagine there are 2 bytes of space left, but the next\ncharacter requires 3 bytes. in this case we could NUL-terminate, but in\ngeneral we can't when there's insufficient space. therefore this function\nonly NUL-terminates if all the characters fit, and there's space for\nthe NUL as well.\nthe destination string will never be bigger than the source string.\n*/\nint u8_toutf8(char *dest, int sz, u_int32_t *src, int srcsz)\n{\nu_int32_t ch;\nint i = 0;\nchar *dest_end = dest + sz;\n\nwhile (srcsz<0 ? src[i]!=0 : i < srcsz) {\nch = src[i];\nif (ch < 0x80) {\nif (dest >= dest_end)\nreturn i;\n*dest++ = (char)ch;\n}\nelse if (ch < 0x800) {\nif (dest >= dest_end-1)\nreturn i;\n*dest++ = (ch>>6) | 0xC0;\n*dest++ = (ch & 0x3F) | 0x80;\n}\nelse if (ch < 0x10000) {\nif (dest >= dest_end-2)\nreturn i;\n*dest++ = (ch>>12) | 0xE0;\n*dest++ = ((ch>>6) & 0x3F) | 0x80;\n*dest++ = (ch & 0x3F) | 0x80;\n}\nelse if (ch < 0x110000) {\nif (dest >= dest_end-3)\nreturn i;\n*dest++ = (ch>>18) | 0xF0;\n*dest++ = ((ch>>12) & 0x3F) | 0x80;\n*dest++ = ((ch>>6) & 0x3F) | 0x80;\n*dest++ = (ch & 0x3F) | 0x80;\n}\ni++;\n}\nif (dest < dest_end)\n*dest = '\\0';\nreturn i;\n}\n\nint u8_wc_toutf8(char *dest, u_int32_t ch)\n{\nif (ch < 0x80) {\ndest = (char)ch;\nreturn 1;\n}\nif (ch < 0x800) {\ndest = (ch>>6) | 0xC0;\ndest = (ch & 0x3F) | 0x80;\nreturn 2;\n}\nif (ch < 0x10000) {\ndest = (ch>>12) | 0xE0;\ndest = ((ch>>6) & 0x3F) | 0x80;\ndest = (ch & 0x3F) | 0x80;\nreturn 3;\n}\nif (ch < 0x110000) {\ndest = (ch>>18) | 0xF0;\ndest = ((ch>>12) & 0x3F) | 0x80;\ndest = ((ch>>6) & 0x3F) | 0x80;\ndest = (ch & 0x3F) | 0x80;\nreturn 4;\n}\nreturn 0;\n}\n\n/* charnum => byte offset */\nint u8_offset(char *str, int charnum)\n{\nint offs=0;\n\nwhile (charnum > 0 && str[offs]) {\n(void)(isutf(str[++offs]) || isutf(str[++offs]) ||\nisutf(str[++offs]) || ++offs);\ncharnum--;\n}\nreturn offs;\n}\n\n/* byte offset => charnum */\nint u8_charnum(char *s, int offset)\n{\nint charnum = 0, offs=0;\n\nwhile (offs < offset && s[offs]) {\n(void)(isutf(s[++offs]) || isutf(s[++offs]) ||\nisutf(s[++offs]) || ++offs);\ncharnum++;\n}\nreturn charnum;\n}\n\n/* number of characters */\nint u8_strlen(char *s)\n{\nint count = 0;\nint i = 0;\n\nwhile (u8_nextchar(s, &i) != 0)\ncount++;\n\nreturn count;\n}\n\n/* reads the next utf-8 sequence out of a string, updating an index */\nu_int32_t u8_nextchar(char *s, int *i)\n{\nu_int32_t ch = 0;\nint sz = 0;\n\ndo {\nch <<= 6;\nch += (unsigned char)s[(*i)++];\nsz++;\n} while (s[*i] && !isutf(s[*i]));\nch -= offsetsFromUTF8[sz-1];\n\nreturn ch;\n}\n\nvoid u8_inc(char *s, int *i)\n{\n(void)(isutf(s[++(*i)]) || isutf(s[++(*i)]) ||\nisutf(s[++(*i)]) || ++(*i));\n}\n\nvoid u8_dec(char *s, int *i)\n{\n(void)(isutf(s[--(*i)]) || isutf(s[--(*i)]) ||\nisutf(s[--(*i)]) || --(*i));\n}\n\nint octal_digit(char c)\n{\nreturn (c >= '0' && c <= '7');\n}\n\nint hex_digit(char c)\n{\nreturn ((c >= '0' && c <= '9') ||\n(c >= 'A' && c <= 'F') ||\n(c >= 'a' && c <= 'f'));\n}\n\n/* assumes that src points to the character after a backslash\nreturns number of input characters processed */\n{\nu_int32_t ch;\nchar digs=\"\\0\\0\\0\\0\\0\\0\\0\\0\\0\";\nint dno=0, i=1;\n\nch = (u_int32_t)str; /* take literal character */\nif (str == 'n')\nch = L'\\n';\nelse if (str == 't')\nch = L'\\t';\nelse if (str == 'r')\nch = L'\\r';\nelse if (str == 'b')\nch = L'\\b';\nelse if (str == 'f')\nch = L'\\f';\nelse if (str == 'v')\nch = L'\\v';\nelse if (str == 'a')\nch = L'\\a';\nelse if (octal_digit(str)) {\ni = 0;\ndo {\ndigs[dno++] = str[i++];\n} while (octal_digit(str[i]) && dno < 3);\nch = strtol(digs, NULL, 8);\n}\nelse if (str == 'x') {\nwhile (hex_digit(str[i]) && dno < 2) {\ndigs[dno++] = str[i++];\n}\nif (dno > 0)\nch = strtol(digs, NULL, 16);\n}\nelse if (str == 'u') {\nwhile (hex_digit(str[i]) && dno < 4) {\ndigs[dno++] = str[i++];\n}\nif (dno > 0)\nch = strtol(digs, NULL, 16);\n}\nelse if (str == 'U') {\nwhile (hex_digit(str[i]) && dno < 8) {\ndigs[dno++] = str[i++];\n}\nif (dno > 0)\nch = strtol(digs, NULL, 16);\n}\n*dest = ch;\n\nreturn i;\n}\n\n/* convert a string with literal \\uxxxx or \\Uxxxxxxxx characters to UTF-8\nexample: u8_unescape(mybuf, 256, \"hello\\\\u220e\")\nnote the double backslash is needed if called on a C string literal */\nint u8_unescape(char *buf, int sz, char *src)\n{\nint c=0, amt;\nu_int32_t ch;\nchar temp;\n\nwhile (*src && c < sz) {\nif (*src == '\\\\') {\nsrc++;\n}\nelse {\nch = (u_int32_t)*src;\namt = 1;\n}\nsrc += amt;\namt = u8_wc_toutf8(temp, ch);\nif (amt > sz-c)\nbreak;\nmemcpy(&buf[c], temp, amt);\nc += amt;\n}\nif (c < sz)\nbuf[c] = '\\0';\nreturn c;\n}\n\nint u8_escape_wchar(char *buf, int sz, u_int32_t ch)\n{\nif (ch == L'\\n')\nreturn snprintf(buf, sz, \"\\\\n\");\nelse if (ch == L'\\t')\nreturn snprintf(buf, sz, \"\\\\t\");\nelse if (ch == L'\\r')\nreturn snprintf(buf, sz, \"\\\\r\");\nelse if (ch == L'\\b')\nreturn snprintf(buf, sz, \"\\\\b\");\nelse if (ch == L'\\f')\nreturn snprintf(buf, sz, \"\\\\f\");\nelse if (ch == L'\\v')\nreturn snprintf(buf, sz, \"\\\\v\");\nelse if (ch == L'\\a')\nreturn snprintf(buf, sz, \"\\\\a\");\nelse if (ch == L'\\\\')\nreturn snprintf(buf, sz, \"\\\\\\\\\");\nelse if (ch < 32 || ch == 0x7f)\nreturn snprintf(buf, sz, \"\\\\x%hhX\", (unsigned char)ch);\nelse if (ch > 0xFFFF)\nreturn snprintf(buf, sz, \"\\\\U%.8X\", (u_int32_t)ch);\nelse if (ch >= 0x80 && ch <= 0xFFFF)\nreturn snprintf(buf, sz, \"\\\\u%.4hX\", (unsigned short)ch);\n\nreturn snprintf(buf, sz, \"%c\", (char)ch);\n}\n\nint u8_escape(char *buf, int sz, char *src, int escape_quotes)\n{\nint c=0, i=0, amt;\n\nwhile (src[i] && c < sz) {\nif (escape_quotes && src[i] == '\"') {\namt = snprintf(buf, sz - c, \"\\\\\\\"\");\ni++;\n}\nelse {\namt = u8_escape_wchar(buf, sz - c, u8_nextchar(src, &i));\n}\nc += amt;\nbuf += amt;\n}\nif (c < sz)\n*buf = '\\0';\nreturn c;\n}\n\nchar *u8_strchr(char *s, u_int32_t ch, int *charn)\n{\nint i = 0, lasti=0;\nu_int32_t c;\n\n*charn = 0;\nwhile (s[i]) {\nc = u8_nextchar(s, &i);\nif (c == ch) {\nreturn &s[lasti];\n}\nlasti = i;\n(*charn)++;\n}\nreturn NULL;\n}\n\nchar *u8_memchr(char *s, u_int32_t ch, size_t sz, int *charn)\n{\nint i = 0, lasti=0;\nu_int32_t c;\nint csz;\n\n*charn = 0;\nwhile (i < sz) {\nc = csz = 0;\ndo {\nc <<= 6;\nc += (unsigned char)s[i++];\ncsz++;\n} while (i < sz && !isutf(s[i]));\nc -= offsetsFromUTF8[csz-1];\n\nif (c == ch) {\nreturn &s[lasti];\n}\nlasti = i;\n(*charn)++;\n}\nreturn NULL;\n}\n\nint u8_is_locale_utf8(char *locale)\n{\n/* this code based on libutf8 */\nconst char* cp = locale;\n\nfor (; *cp != '\\0' && *cp != '@' && *cp != '+' && *cp != ','; cp++) {\nif (*cp == '.') {\nconst char* encoding = ++cp;\nfor (; *cp != '\\0' && *cp != '@' && *cp != '+' && *cp != ','; cp++)\n;\nif ((cp-encoding == 5 && !strncmp(encoding, \"UTF-8\", 5))\n|| (cp-encoding == 4 && !strncmp(encoding, \"utf8\", 4)))\nreturn 1; /* it's UTF-8 */\nbreak;\n}\n}\nreturn 0;\n}\n\nint u8_vprintf(char *fmt, va_list ap)\n{\nint cnt, sz=0;\nchar *buf;\nu_int32_t *wcs;\n\nsz = 512;\nbuf = (char*)alloca(sz);\ntry_print:\ncnt = vsnprintf(buf, sz, fmt, ap);\nif (cnt >= sz) {\nbuf = (char*)alloca(cnt - sz + 1);\nsz = cnt + 1;\ngoto try_print;\n}\nwcs = (u_int32_t*)alloca((cnt+1) * sizeof(u_int32_t));\ncnt = u8_toucs(wcs, cnt+1, buf, cnt);\nprintf(\"%ls\", (wchar_t*)wcs);\nreturn cnt;\n}\n\nint u8_printf(char *fmt, ...)\n{\nint cnt;\nva_list args;\n\nva_start(args, fmt);\n\ncnt = u8_vprintf(fmt, args);\n\nva_end(args);\nreturn cnt;\n}\n```\n\nwikix.c\n\n```#include \"platform.h\"\n\n#ifdef WINDOWS\n\n#define strncasecmp strnicmp\n\n#include \"windows.h\"\n#include \"winioctl.h\"\n#include \"winuser.h\"\n#include \"stdarg.h\"\ntypedef UCHAR BYTE;\ntypedef USHORT WORD;\n#include \"stdio.h\"\n#include \"stdlib.h\"\n#include \"ctype.h\"\n#include \"conio.h\"\n\n#endif\n\n#ifdef LINUX\n\n#include <unistd.h>\n#include <stdio.h>\n#include <stdlib.h>\n#include <fcntl.h>\n#include <ctype.h>\n#include <string.h>\n//#include <ncurses.h>\n#include <termios.h>\n#include <sys/ioctl.h>\n#include <sys/stat.h>\n#include <sys/types.h>\n#include <sys/socket.h>\n#include <netinet/in.h>\n#include <net/if.h>\n#include <stdio.h>\n#include <errno.h>\n#include <stdlib.h>\n#include <string.h>\n#include <unistd.h>\n#include <sched.h>\n#include <ctype.h>\n#include <openssl/md5.h>\n\n#endif\n\n#define NAME_HASH_SIZE 8192\n\ntypedef struct _hash\n{\nstruct _hash *next;\nstruct _hash *prior;\nunsigned long len;\nchar *text;\n} hash;\n\ntypedef struct _hash_list {\nhash *tail;\n} hash_list;\n\nunsigned char buffer[8192 * 4];\nunsigned char ImagePath;\nunsigned char OutputPath;\nunsigned char iPath;\nunsigned char cPath;\nunsigned char md5_out;\nunsigned char md5_ulout;\nunsigned char wk;\nunsigned char final1;\nunsigned char final2;\nunsigned char ulwk;\nunsigned char fwk;\nunsigned char expand;\nunsigned char html;\nFILE *fpl;\nint pmode = 0, tree = 0;\nint lobj = 0;\nFILE *imagelog = NULL, *imagereject = NULL, *fragmentlog = NULL;\n\nunsigned long shash(char *v, unsigned long len, unsigned long M)\n{\nregister unsigned long h = 0, a = 127, i;\n\nfor (i = 0; i < len && *v; v++, i++)\nh = ((a * h) + tolower(*v)) % M;\n\nreturn h;\n}\n\nunsigned long add_to_hash(hash_list *top, hash *name)\n{\nregister unsigned long Value;\nregister hash_list *HashTable;\n\nValue = shash(name->text, name->len, NAME_HASH_SIZE);\nif (Value == (unsigned long) -1)\nreturn -1;\n\nHashTable = (hash_list *) top;\nif (HashTable)\n{\n{\nHashTable[Value].tail = name;\nname->next = name->prior = 0;\n}\nelse\n{\nHashTable[Value].tail->next = name;\nname->next = 0;\nname->prior = HashTable[Value].tail;\nHashTable[Value].tail = name;\n}\nreturn 0;\n}\nreturn -1;\n}\n\nunsigned long remove_from_hash(hash_list *top, hash *name)\n{\nregister unsigned long Value;\nregister hash_list *HashTable;\n\nValue = shash(name->text, name->len, NAME_HASH_SIZE);\nif (Value == (unsigned long) -1)\nreturn -1;\n\nHashTable = (hash_list *) top;\nif (HashTable)\n{\n{\nelse\nHashTable[Value].tail = NULL;\n}\nelse\n{\nname->prior->next = name->next;\nif (name != HashTable[Value].tail)\nname->next->prior = name->prior;\nelse\nHashTable[Value].tail = name->prior;\n}\nif (lobj)\nlobj--;\nreturn 0;\n}\nreturn -1;\n}\n\nvoid free_hash(void)\n{\nregister int i;\nregister hash_list *HashTable;\nregister hash *tmp, *name;\n\n{\nfor (i=0; i < NAME_HASH_SIZE; i++)\n{\nwhile (name)\n{\ntmp = name;\nname = name->next;\nfree((void *)tmp);\n}\n}\n}\n\n{\nfor (i=0; i < NAME_HASH_SIZE; i++)\n{\nwhile (name)\n{\ntmp = name;\nname = name->next;\nfree((void *)tmp);\n}\n}\n}\n\n}\n\nhash_list *init_hash_list(void)\n{\nlearn_list_head = (hash_list *) malloc(sizeof(hash_list) * NAME_HASH_SIZE);\nreturn NULL;\n\nname_list_head = (hash_list *) malloc(sizeof(hash_list) * NAME_HASH_SIZE);\nreturn NULL;\n\n}\n\nhash *search_name_hash(hash_list *top, char *text, unsigned long len)\n{\nregister unsigned long Value;\nregister hash *name;\nregister hash_list *HashTable;\n\nValue = shash(text, len, NAME_HASH_SIZE);\n\nHashTable = (hash_list *) top;\nwhile (name)\n{\nif (len == name->len)\n{\nif (!strncasecmp(name->text, text, len))\nreturn (hash *) name;\n}\nname = name->next;\n}\nreturn NULL;\n\n}\n\nint learn(char *s, int len)\n{\nregister hash *name;\n\nif (name)\nreturn 1;\n\nname = malloc(sizeof(hash) + len + 2);\nif (!name)\nreturn 1;\n\nmemset(name, 0, sizeof(hash) + len);\nname->text = (char *)((unsigned long)name + sizeof(hash));\nname->len = len;\nstrncpy(name->text, s, len);\n\n{\nfree(name);\nreturn 1;\n}\nlobj++;\nreturn 0;\n\n}\n\nint imagename(char *s, int len)\n{\nregister hash *name;\n\nif (name)\nreturn 1;\n\nname = malloc(sizeof(hash) + len + 2);\nif (!name)\nreturn 1;\n\nmemset(name, 0, sizeof(hash) + len);\nname->text = (char *)((unsigned long)name + sizeof(hash));\nname->len = len;\nstrncpy(name->text, s, len);\n\n{\nfree(name);\nreturn 1;\n}\nlobj++;\nreturn 0;\n\n}\n\nunsigned char *nprintf(char *s, int len, FILE *fp)\n{\nregister int i;\n\nif (!s || !*s)\nreturn s;\n\nfor (i=0; *s && (i < len); i++)\nputc(*s++, fp);\n\nreturn s;\n}\n\nunsigned char *str8rchr(const char * s, int c1, int c2, int c3, int c4,\nint c5, int c6, int c7, int c8)\n{\nconst char *p = s + strlen(s);\ndo {\nif ((*p == (char)c1) || (*p == (char)c2) || (*p == (char)c3) ||\n(*p == (char)c4) || (*p == (char)c5) || (*p == (char)c6) ||\n(*p == (char)c7) || (*p == (char)c8))\nreturn (char *)p;\n} while (--p >= s);\nreturn NULL;\n}\n\nunsigned char *str5rchr(const char * s, int c1, int c2, int c3, int c4,\nint c5)\n{\nconst char *p = s + strlen(s);\ndo {\nif ((*p == (char)c1) || (*p == (char)c2) || (*p == (char)c3) ||\n(*p == (char)c4) || (*p == (char)c5))\nreturn (char *)p;\n} while (--p >= s);\nreturn NULL;\n}\n\nchar *strnstr(const char * s1,const char * s2)\n{\nint l1, l2;\n\nl2 = strlen(s2);\nif (!l2)\nreturn (char *) s1;\nl1 = strlen(s1);\nwhile (l1 >= l2) {\nl1--;\nif (!strncasecmp(s1,s2,l2))\nreturn (char *) s1;\ns1++;\n}\nreturn NULL;\n}\n\nunsigned char *imagetypes[]=\n{\n// 7\n\".svg+xml\",\n\".xcf.bz2\",\n\n// 6\n\".bitmap\",\n\".xcfbz2\",\n\n// 5\n\".xcfgz\",\n\".alpha\",\n\".dicom\",\n\".matte\",\n\".xjtgz\",\n\n// 4\n\".aifc\",\n\".aiff\",\n\".fits\",\n\".icon\",\n\".im24\",\n\".im32\",\n\".jpeg\",\n\".midi\",\n\".mpeg\",\n\".xwav\",\n\".mpga\",\n\".tiff\",\n\".djvu\",\n\n// 3\n\".aif\",\n\".als\",\n\".apm\",\n\".bmp\",\n\".bz2\",\n\".cel\",\n\".dcm\",\n\".eps\",\n\".fit\",\n\".flc\",\n\".fli\",\n\".gbr\",\n\".gif\",\n\".gih\",\n\".gpb\",\n\".ico\",\n\".im1\",\n\".im8\",\n\".jpe\",\n\".jpg\",\n\".kar\",\n\".mid\",\n\".mov\",\n\".mp2\",\n\".mp3\",\n\".mp4\",\n\".mpa\",\n\".mpg\",\n\".ogg\",\n\".ogm\",\n\".pcc\",\n\".pcx\",\n\".pdf\",\n\".pdm\",\n\".pgm\",\n\".pix\",\n\".png\",\n\".pnm\",\n\".ppm\",\n\".psd\",\n\".psp\",\n\".ras\",\n\".rgb\",\n\".sgi\",\n\".svg\",\n\".swf\",\n\".tga\",\n\".tif\",\n\".tub\",\n\".wav\",\n\".wmf\",\n\".xbm\",\n\".xcf\",\n\".xjt\",\n\".xpm\",\n\".xwd\",\n\".pov\",\n\".wma\",\n\".dia\",\n\".fig\",\n\".jif\",\n\".pgn\",\n\".art\",\n\".djv\",\n\n// 2\n\".bw\",\n\".ps\",\n\".g3\",\n\".js\",\n\".rs\",\n};\n\nunsigned char *strip_image_info(unsigned char *s, char *title)\n{\nregister int i;\nunsigned char *p, *j;\nFILE *fp = stdout;\nunsigned char ch = '\\0';\n\nwhile (*s && (isspace(*s))) s++;\n\nif (!strncasecmp(s, \"no image\", 8))\nreturn s;\n\np = s;\nwhile (*s)\n{\nif ((!strncasecmp(s, \"image\", 5) ||\n!strncasecmp(s, \"map\", 3)) && !isalnum(ch))\n{\nunsigned char *fragment, *end;\n\nfragment = s;\nif (!strncasecmp(s, \"image\", 5))\ns += 5;\nelse\nif (!strncasecmp(s, \"map\", 3))\ns += 3;\n\nif (*s)\n{\nwhile (*s && isalnum(*s)) s++;\n\nend = s;\nwhile (*s && isspace(*s)) s++;\n\nif (*s && *s == '=' || *s == ':')\n{\nmemset(&fwk, 0, 256);\nmemmove(&fwk, fragment, (end - fragment));\nif (!learn(&fwk, end - fragment))\n{\nif (*title)\nfprintf(fragmentlog, \"[%s] %s\\n\", title, &fwk);\nelse\nfprintf(fragmentlog, \"%s\\n\", &fwk);\nfflush(fragmentlog);\n}\n\ns++;\ns = strip_image_info(s, title);\nch = '\\0';\n}\n}\ncontinue;\n}\n\nif ((*s == '|') || (*s == ']') || (*s == '\\n'))\n{\nregister int y;\nunsigned char ch = '\\x22', *l;\nunsigned char dir1, dir2;\nunsigned char *lp, *lw, *lo, *delim, *blp;\nunsigned char *ulp, *fname, *bulp;\nregister int cnvt = 0, bcnvt = 0, unicnvt = 0, invl = 0;\n\nlp = &wk;\nj = lp;\nwhile (*p && (p < s))\n{\n// skip self referencing images\nif (!strncasecmp(p, \"{{\", 2))\nreturn s;\n\nif (!memcmp(p, \""\", 6))\n{\np += 6;\n*j++ = '\\x22';\n}\nif (!memcmp(p, \"&\", 5))\n{\np += 5;\n*j++ = '&';\n}\nif (!memcmp(p, \"<\", 4))\n{\np += 4;\nwhile (*p)\n{\nif (!memcmp(p, \">\", 4))\n{\np += 4;\nbreak;\n}\np++;\n}\n}\n\nif (*p == '\\n')\np++;\n\nif (!memcmp(p, \"[[\", 2))\nbreak;\n\n*j++ = *p++;\n}\n*j = '\\0';\ns++;\n\nfor (j=NULL, y=0; y < (sizeof(imagetypes) / sizeof (char *)); y++)\n{\nj = strnstr(lp, imagetypes[y]);\nif (j)\n{\nregister int ilen = strlen(imagetypes[y]);\n\nj += ilen;\n*j = '\\0';\nbreak;\n}\n}\n\nif (!j)\n{\nif (*lp && isalpha(*lp))\n{\nunsigned char *sp = strchr(lp, '.');\nunsigned char *sj, *slp = lp;\n\nif (sp)\n{\nunsigned char *sllp = sp, *meter;\n\nsllp++;\nif ((*sllp != ' ') && (isalpha(*sllp)))\n{\nsj = str8rchr(slp, ':', '/', '\\\\', '{', '\\n', '&',\n'=', '>');\nif (sj)\nslp = ++sj;\n\nmeter = sllp;\nwhile (*sllp)\n{\nif (!isalpha(*sllp))\n{\n*sllp = '\\0';\nbreak;\n}\nsllp++;\n}\n\nif (*slp &&\n(((sllp - meter) >= 3) && ((sllp - meter) <= 5)))\n{\nif (*title)\nfprintf(imagereject, \"[%s] %s\\n\", title, slp);\nelse\nfprintf(imagereject, \"%s\\n\", slp);\nfflush(imagereject);\n}\n}\n}\n}\nreturn s;\n}\n\nj = str5rchr(lp, ':', '/', '\\\\', '{', '\\n');\nif (j)\nlp = ++j;\n\nif (!*lp)\nreturn s;\n\n#ifdef UNICODE_EXPANSION\n// filename string extracted. convert xml control character tags\nl = &expand;\nulp = lp;\nwhile (*ulp)\n{\nif (!strncasecmp(ulp, \"&\", 5))\n{\nulp += 5;\n*l++ = '&';\ncontinue;\n}\nif (!strncasecmp(ulp, \"<\", 4))\n{\nulp += 4;\n*l++ = '<';\ncontinue;\n}\nif (!strncasecmp(ulp, \">\", 4))\n{\nulp += 4;\n*l++ = '>';\ncontinue;\n}\nif (!strncasecmp(ulp, \""\", 6))\n{\nulp += 6;\n*l++ = '\\\"';\ncontinue;\n}\nif (!strncasecmp(ulp, \"'\", 6))\n{\nulp += 6;\n*l++ = '\\'';\ncontinue;\n}\nif (!strncasecmp(ulp, \" \", 6))\n{\nulp += 6;\n*l++ = ' ';\ncontinue;\n}\nif (!strncasecmp(ulp, \"–\", 6))\n{\nulp += 6;\n*l++ = '-';\ncontinue;\n}\nif ((ulp == '&') && (ulp != '&'))\n{\nunsigned char *sc = strchr(ulp, ';'), *slp;\nunsigned char unicode;\nunsigned char unidest;\nunsigned short uni;\n\nif (sc)\n{\nslp = ulp;\nslp++;\nwhile (*slp != ';')\n{\nif ((*slp == '#') || (*slp == '-') ||\n(*slp == 'x') || (*slp == 'X') ||\nisxdigit(*slp))\nslp++;\nelse\n{\ninvl = 1;\nbreak;\n}\n}\n\nif (!invl)\n{\nint unilen = sc - ulp;\nint slen = sc - ulp;\n\nslp = ulp;\nslp++;\nunilen--;\nif (*slp == '#')\n{\nunilen--;\nslp++;\n}\n\nif (unilen < 31)\n{\nmemset(unicode, 0, 32);\nstrncpy(unicode, slp, unilen);\nuni = atoi(unicode);\n\nfprintf(imagelog, \"UNI1: %s (#%d) %s \\n\",\nunicode, (int)uni,\nlp);\n\nunicode = '\\0';\nsprintf(unicode, \"\\\\u%04X\", uni);\n\nunilen = u8_unescape(l, 32, unicode);\n\nfprintf(imagelog, \"UNI2: %s unilen %d slen %d\\n\",\nunicode, (int)unilen, (int)slen);\n\nulp += slen;\nl += unilen;\nulp++;\n\nunicnvt = 1;\ncontinue;\n}\n}\n}\n}\n*l++ = *ulp++;\n}\n*l = '\\0';\nlp = &expand;\n#endif\n\n// convert spaces to underline characters in image names\nulp = &ulwk;\nmemmove(ulp, lp, strlen(lp) + 1);\nulp = toupper(ulp);\n{\nl = ulp;\nwhile (*l)\n{\nif (*l == ' ')\n{\n*l = '_';\ncnvt = 1;\n}\nl++;\n}\n}\n\nif (learn(lp, strlen(lp)))\nreturn s;\n\nif (cnvt && learn(ulp, strlen(ulp)))\nreturn s;\n\nmemset(md5_out, 0, 16);\nlp = toupper(lp);\n\n#ifdef UNICODE_EXPANSION\nif (unicnvt || invl)\n{\nif (invl)\nfprintf(imagelog, \"INVL: %s -> %s\\n\", wk, lp);\nelse\nfprintf(imagelog, \"%s -> %s\\n\", wk, lp);\nfflush(imagelog);\nif (invl)\nreturn s;\n}\nelse\nreturn s;\n#else\nfprintf(imagelog, \"%s\\n\", lp);\nfflush(imagelog);\n#endif\n\nMD5(lp, strlen(lp), md5_out);\ndir1 = '\\0';\nsprintf(dir1, \"%x/%02x/\", (md5_out >> 4), md5_out);\n\nif (cnvt)\n{\nmemset(md5_ulout, 0, 16);\nulp = toupper(ulp);\nMD5(ulp, strlen(ulp), md5_ulout);\ndir2 = '\\0';\nsprintf(dir2, \"%x/%02x/\", (md5_ulout >> 4), md5_ulout);\n}\n\n// add trailing \\\\ characters to bash control chars\nfname = &final1;\nblp = lp;\nwhile (*blp)\n{\nif ((*blp == '\\\"') || (*blp == '\\'') || (*blp == '`'))\n{\nbcnvt = 1;\n*fname++ = '\\\\';\n}\nelse\nif ((*blp == ' ') || (*blp == '(') || (*blp == ')') ||\n(*blp == '{') || (*blp == '}') || (*blp == '[') ||\n(*blp == ']') || (*blp == '&') || (*blp == '-') ||\n(*blp == ';'))\n*fname++ = '\\\\';\n\n*fname++ = *blp++;\n}\n*fname = '\\0';\nblp = &final1;\n\n// add trailing \\\\ characters to bash control chars\nfname = &final2;\nbulp = ulp;\nwhile (*bulp)\n{\nif ((*bulp == '\\\"') || (*bulp == '\\'') || (*bulp == '`'))\n{\nbcnvt = 1;\n*fname++ = '\\\\';\n}\nelse\nif ((*bulp == ' ') || (*bulp == '(') || (*bulp == ')') ||\n(*bulp == '{') || (*bulp == '}') || (*bulp == '[') ||\n(*bulp == ']') || (*bulp == '&') || (*bulp == '-') ||\n(*bulp == ';'))\n*fname++ = '\\\\';\n\n*fname++ = *bulp++;\n}\n*fname = '\\0';\nbulp = &final2;\n\n// debug of control characters\n// if (!bcnvt)\n// return s;\n\nif (tree)\n{\nif (pmode)\nfp = fpl[(md5_out >> 4) % 16];\n\nfprintf(fp, \"if [ -a \\$IMAGE./%s%s ]; then\\n\",\ndir1, blp);\n\nfprintf(fp, \"\\t/bin/mkdir -p \\$OUTPUT./%s\\n\", dir1);\nfprintf(fp, \"\\tcp -f \\$IMAGE./%s%s \\$OUTPUT./%s%s\\n\",\ndir1, blp, dir1, blp);\nfprintf(fp, \"\\techo ./%s%s copied to \\$OUTPUT./%s%s >> \"\n\"copied.log\\n\", dir1, blp, dir1, blp);\n\nif (cnvt)\n{\nfprintf(fp, \"elif [ -a \\$IMAGE./%s%s ]; then\\n\",\ndir2, bulp);\nfprintf(fp, \"\\t/bin/mkdir -p \\$OUTPUT./%s\\n\", dir2);\nfprintf(fp, \"\\tcp -f \\$IMAGE./%s%s \\$OUTPUT./%s%s\\n\",\ndir2, bulp, dir2, bulp);\nfprintf(fp, \"\\techo ./%s%s copied to \\$OUTPUT./%s%s >> \"\n\"copied.log\\n\", dir2, bulp, dir2, bulp);\n}\nfprintf(fp, \"else\\n\");\nfprintf(fp,\nblp);\nfprintf(fp, \"fi\\n\\n\");\n}\nelse\n{\nif (pmode)\nfp = fpl[(md5_out >> 4) % 16];\n\nfprintf(fp, \"if [ -a \\$IMAGE./%s%s ]; then\\n\",\ndir1, blp);\nfprintf(fp, \"\\techo %s%s already exists >> exists.log\\n\",\ndir1, blp);\n\nif (cnvt)\n{\nfprintf(fp, \"elif [ -a \\$IMAGE./%s%s ]; then\\n\",\ndir2, bulp);\nfprintf(fp, \"\\techo %s%s already exists >> exists.log\\n\",\ndir2, bulp);\n}\nfprintf(fp, \"else\\n\");\n\nfprintf(fp, \"\\tcurl --retry 7 -f -O \\$IMAGEPATH./%s%s\\n\",\ndir1, blp);\nfprintf(fp, \"\\tif [ -a \\$IMAGE./%s ]; then\\n\", blp);\nfprintf(fp, \"\\t\\t/bin/mkdir -p \\$OUTPUT./%s\\n\", dir1);\nfprintf(fp, \"\\t\\t/bin/mv ./%s \\$OUTPUT./%s\\n\",\nblp, dir1);\ndir1, blp);\n\nfprintf(fp, \"\\telse\\n\");\nfprintf(fp, \"\\t\\tcurl --retry 7 -f -O \\$COMMONSPATH./%s%s\\n\",\ndir1, blp);\n\nfprintf(fp, \"\\t\\tif [ -a \\$IMAGE./%s ]; then\\n\",\nblp);\nfprintf(fp, \"\\t\\t\\t/bin/mkdir -p \\$OUTPUT./%s\\n\", dir1);\nfprintf(fp, \"\\t\\t\\t/bin/mv ./%s \\$OUTPUT./%s\\n\",\nblp, dir1);\ndir1, blp);\nfprintf(fp, \"\\t\\telse\\n\");\n\nif (cnvt)\n{\n\nfprintf(fp, \"\\t\\t\\tcurl --retry 7 -f -O \\$IMAGEPATH./%s%s\\n\",\ndir2, bulp);\nfprintf(fp, \"\\t\\t\\tif [ -a \\$IMAGE./%s ]; then\\n\",\nbulp);\nfprintf(fp, \"\\t\\t\\t\\t/bin/mkdir -p \\$OUTPUT./%s\\n\",\ndir2);\nfprintf(fp, \"\\t\\t\\t\\t/bin/mv ./%s \\$OUTPUT./%s\\n\",\nbulp, dir2);\n\nfprintf(fp, \"\\t\\t\\telse\\n\");\nfprintf(fp, \"\\t\\t\\t\\tcurl --retry 7 -f -O \\$COMMONSPATH./%s%s\\n\",\ndir2, bulp);\n\nfprintf(fp, \"\\t\\t\\t\\tif [ -a \\$IMAGE./%s ]; then\\n\",\nbulp);\nfprintf(fp, \"\\t\\t\\t\\t\\t/bin/mkdir -p \\$OUTPUT./%s\\n\",\ndir2);\nfprintf(fp, \"\\t\\t\\t\\t\\t/bin/mv ./%s \\$OUTPUT./%s\\n\",\nbulp, dir2);\nfprintf(fp, \"\\t\\t\\t\\telse\\n\");\nfprintf(fp, \"\\t\\t\\t\\t\\techo ./%s%s failed >> failed.log\\n\",\ndir1, blp);\nfprintf(fp, \"\\t\\t\\t\\t\\techo ./%s%s failed >> failed.log\\n\",\ndir2, bulp);\nfprintf(fp, \"\\t\\t\\t\\tfi\\n\");\nfprintf(fp, \"\\t\\t\\tfi\\n\");\n}\nelse\n{\nfprintf(fp,\n\"\\t\\t\\techo ./%s%s failed >> failed.log\\n\", dir1,\nblp);\n}\nfprintf(fp, \"\\t\\tfi\\n\");\nfprintf(fp, \"\\tfi\\n\");\nfprintf(fp, \"fi\\n\\n\");\n}\n\nreturn s;\n}\nch = *s;\ns++;\n}\nreturn s;\n}\n\nint main(int argc, char *argv[])\n{\nregister int i, r, inpage = 0;\nunsigned char *s, *j, fname, *buffer, *title, *title_p;\nFILE *fl;\n\nImagePath = '\\0';\nOutputPath = '\\0';\n\niPath = '\\0';\ncPath = '\\0';\n\nfor (i=0; i < argc; i++)\n{\n// remote path\nif (!memcmp(argv[i], \"-h\", 2))\n{\nprintf(\"USAGE: wikix -htrciop < file.xml [ > script.out ]\\n\");\nprintf(\" -h this help screen\\n\");\nprintf(\" -t use xml dump to strip from tree\\n\");\nprintf(\" -r wikipedia path\\n\");\nprintf(\" -c commons path\\n\");\nprintf(\" -i image path\\n\");\nprintf(\" -o output path\\n\");\nprintf(\" -p parallel (16 process) mode\\n\");\nexit(1);\n}\n\n// remote path\nif (!memcmp(argv[i], \"-t\", 2))\n{\ntree = 1;\n}\n\n// remote path\nif (!memcmp(argv[i], \"-r\", 2))\n{\ni++;\nif (argv[i])\nstrncpy(iPath, argv[i], 256);\n}\n\n// commons\nif (!memcmp(argv[i], \"-c\", 2))\n{\ni++;\nif (argv[i])\nstrncpy(cPath, argv[i], 256);\n}\n\n// image tree\nif (!memcmp(argv[i], \"-i\", 2))\n{\ni++;\nif (argv[i])\nstrncpy(ImagePath, argv[i], 256);\n}\n\n// output image tree\nif (!memcmp(argv[i], \"-o\", 2))\n{\ni++;\nif (argv[i])\nstrncpy(OutputPath, argv[i], 256);\n}\n\nif (!memcmp(argv[i], \"-p\", 2))\n{\npmode = 1;\n}\n}\n\nmemset(&fwk, 0xFF, 256);\n\nif (!init_hash_list())\n{\nprintf(\"wikix: could not allocate workspace\\n\");\nexit(1);\n}\n\nbuffer = malloc(0x10000);\nif (!buffer)\n{\nprintf(\"gfdl-wikititle: could not allocate buffer workspace\\n\");\nexit(1);\n}\nbuffer = '\\0';\n\ntitle = malloc(0x10000);\nif (!title)\n{\nprintf(\"gfdl-wikititle: could not allocate namespace\\n\");\nexit(1);\n}\ntitle = '\\0';\n\nif (!pmode)\n{\nprintf(\"#!/bin/sh\\n\\n\");\n\nprintf(\"IMAGE=%s\\n\", ImagePath);\nprintf(\"OUTPUT=%s\\n\", OutputPath);\nprintf(\"IMAGEPATH=%s\\n\", iPath);\nprintf(\"COMMONSPATH=%s\\n\\n\", cPath);\n\nprintf(\"/bin/mkdir -p \\$OUTPUT./thumb\\n\");\nprintf(\"/bin/chmod 777 \\$OUTPUT./thumb\\n\");\nprintf(\"/bin/mkdir -p \\$OUTPUT./temp\\n\");\nprintf(\"/bin/chmod 777 \\$OUTPUT./temp\\n\");\nprintf(\"/bin/mkdir -p \\$OUTPUT./tmp\\n\");\nprintf(\"/bin/chmod 777 \\$OUTPUT./tmp\\n\\n\");\n\n}\nelse\n{\nfl = fopen(\"image_sh\", \"w\");\nif (!fl)\n{\nprintf(\"FILE error could not create image_sh\\n\");\nexit(1);\n}\n\nchmod(\"image_sh\", 0755);\n\nfprintf(fl, \"#!/bin/sh\\n\\n\");\n\nfprintf(fl, \"IMAGE=%s\\n\", ImagePath);\nfprintf(fl, \"OUTPUT=%s\\n\", OutputPath);\nfprintf(fl, \"IMAGEPATH=%s\\n\", iPath);\nfprintf(fl, \"COMMONSPATH=%s\\n\\n\", cPath);\n\nfprintf(fl, \"/bin/mkdir -p \\$OUTPUT./thumb\\n\");\nfprintf(fl, \"/bin/chmod 777 \\$OUTPUT./thumb\\n\");\nfprintf(fl, \"/bin/mkdir -p \\$OUTPUT./temp\\n\");\nfprintf(fl, \"/bin/chmod 777 \\$OUTPUT./temp\\n\");\nfprintf(fl, \"/bin/mkdir -p \\$OUTPUT./tmp\\n\");\nfprintf(fl, \"/bin/chmod 777 \\$OUTPUT./tmp\\n\\n\");\n\nfor (r=0; r < 16; r++)\n{\nfname = '\\0';\nsprintf(fname, \"image%02d\", r);\nfpl[r] = fopen(fname, \"w\");\nif (!fpl[r])\n{\nprintf(\"FILE error could not create [%s]\\n\", fname);\nexit(1);\n}\n\nchmod(fname, 0755);\n\nfprintf(fpl[r], \"#!/bin/sh\\n\\n\");\nfprintf(fpl[r], \"\\nIMAGE=%s\\n\", ImagePath);\nfprintf(fpl[r], \"OUTPUT=%s\\n\", OutputPath);\nfprintf(fpl[r], \"IMAGEPATH=%s\\n\", iPath);\nfprintf(fpl[r], \"COMMONSPATH=%s\\n\\n\", cPath);\n\nfprintf(fl, \"./%s >& imagelog.%02d &\\n\", fname, r);\n}\nfclose(fl);\n}\n\nimagelog = fopen(\"image.log\", \"wb\");\nif (!imagelog)\n{\nprintf(\"FILE error could not create image log\\n\");\n}\n\nimagereject = fopen(\"reject.log\", \"wb\");\nif (!imagereject)\n{\nprintf(\"FILE error could not create reject log\\n\");\n}\n\nfragmentlog = fopen(\"fragment.log\", \"wb\");\nif (!fragmentlog)\n{\nprintf(\"FILE error could not create image name fragment log\\n\");\n}\n\nwhile (s = fgets(buffer, 8192 * 4, stdin))\n{\nunsigned char ch = '\\0';\n\nif (strstr(s, \"<page>\"))\n{\ninpage++;\nif (*title)\n*title = '\\0';\ncontinue;\n}\n\nif (strstr(s, \"</page>\"))\n{\nif (inpage)\ninpage--;\nif (*title)\n*title = '\\0';\ncontinue;\n}\n\ntitle_p = strstr(s, \"<title>\");\nif (inpage && title_p)\n{\nregister char *ts, *tp;\n\nts = title_p;\nts += 7;\ntp = strstr(ts, \"</title>\");\nif (tp)\n{\nif (tp - ts)\n{\nstrncpy(title, ts, tp - ts);\ntitle[tp - ts] = '\\0';\n}\n}\n}\n\nwhile (*s)\n{\nif (inpage && !strncasecmp(s, \"<title>\", 7))\n{\nregister char *ts, *tp;\n\ns += 7;\nts = s;\ntp = strstr(ts, \"</title>\");\nif (tp)\n{\nif (tp - ts)\n{\nstrncpy(title, ts, tp - ts);\ntitle[tp - ts] = '\\0';\n}\n}\n}\n\nif ((!strncasecmp(s, \"image\", 5) ||\n!strncasecmp(s, \"map\", 3)) &&\n!isalnum(ch))\n{\nunsigned char *fragment, *end;\n\nfragment = s;\nif (!strncasecmp(s, \"image\", 5))\ns += 5;\nelse\nif (!strncasecmp(s, \"map\", 3))\ns += 3;\n\nif (*s)\n{\nwhile (*s && isalnum(*s)) s++;\n\nend = s;\nwhile (*s && isspace(*s)) s++;\n\nif (*s && (*s == '=' || *s == ':'))\n{\nmemset(&fwk, 0, 256);\nmemmove(&fwk, fragment, (end - fragment));\nif (!imagename(&fwk, end - fragment))\n{\nif (*title)\nfprintf(fragmentlog, \"[%s] %s\\n\", title, &fwk);\nelse\nfprintf(fragmentlog, \"%s\\n\", &fwk);\nfflush(fragmentlog);\n}\n\ns++;\ns = strip_image_info(s, title);\nch = '\\0';\n}\n}\ncontinue;\n}\nch = *s;\ns++;\n}\n}\n\nif (pmode)\n{\nfor (r=0; r < 16; r++)\n{\nif (!fpl[r])\nfclose(fpl[r]);\nfpl[r] = NULL;\n}\n}\nfclose(fragmentlog);\nfclose(imagelog);\nfclose(imagereject);\nfree(title);\nfree(buffer);\nfree_hash();\nreturn 0;\n\n}\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.50602937,"math_prob":0.9917084,"size":36931,"snap":"2020-45-2020-50","text_gpt3_token_len":13011,"char_repetition_ratio":0.15249005,"word_repetition_ratio":0.18112554,"special_character_ratio":0.4248734,"punctuation_ratio":0.25730923,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96171266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T13:05:11Z\",\"WARC-Record-ID\":\"<urn:uuid:225b69aa-f311-4d5c-a41e-a1c20113be5a>\",\"Content-Length\":\"288975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4acba10-de19-4f2e-a25d-7c27eb76d794>\",\"WARC-Concurrent-To\":\"<urn:uuid:82b9f265-e1b9-4bb0-949d-3808bcb22cf3>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://meta.m.wikimedia.org/wiki/Wikix\",\"WARC-Payload-Digest\":\"sha1:S2JQW5EQF4GJCAFSYSII6GRL5OZSSDWQ\",\"WARC-Block-Digest\":\"sha1:ESO567TCTLH2QZGJT5KGXYI5UGRYVLBX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141747774.97_warc_CC-MAIN-20201205104937-20201205134937-00408.warc.gz\"}"} |
https://urtextacourseinmiracles.com/audio-daily-lessons-1-50/ | [
"",
null,
"# AUDIO Daily Lessons 1- 50\n\nIn the first paragraph of the Introduction of the workbook lessons it says:\n\n`A theoretical foundation such as the text is necessary as a background to make these exercises meaningful. Yet it is the exercises which will make the goal possible. An untrained mind can accomplish nothing. It is the purpose of these exercises to train the mind to think along the lines which the course sets forth.`\n\n## THE AUDIO’s OF THE FIRST 50 LESSONS\n\nIntroduction and Lesson 1\n\nLesson 2\n\nLesson 3\n\nLesson 4\n\nLesson 5\n\nLesson 6\n\nLesson 7\n\nLesson 8\n\nLesson 9\n\nLesson 10\n\nLesson 11\n\nLesson 12\n\nLesson 13\n\nLesson 14\n\nLesson 15\n\nLesson 16\n\nLesson 17\n\nLesson 18\n\nLesson 19\n\nLesson 20\n\nLesson 21\n\nLesson 22\n\nLesson 23\n\nLesson 24\n\nLesson 25\n\nLesson 26\n\nLesson 27\n\nLesson 28\n\nLesson 29\n\nLesson 30\n\nLesson 31\n\nLesson 32\n\nLesson 33\n\nLesson 34\n\nLesson 35\n\nLesson 36\n\nTo be continued."
] | [
null,
"https://urtextacourseinmiracles.files.wordpress.com/2022/03/workbook-.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92420393,"math_prob":0.56133914,"size":2297,"snap":"2022-40-2023-06","text_gpt3_token_len":573,"char_repetition_ratio":0.18578282,"word_repetition_ratio":0.11441647,"special_character_ratio":0.22594689,"punctuation_ratio":0.022075055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9926876,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T18:47:47Z\",\"WARC-Record-ID\":\"<urn:uuid:410da191-4c0c-42c5-bfda-bad07c6919b8>\",\"Content-Length\":\"182061\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7b28ef9-5048-485a-a13f-d9ed24da5aec>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0cabc5d-ac33-455d-8d7d-6100390f6ea3>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://urtextacourseinmiracles.com/audio-daily-lessons-1-50/\",\"WARC-Payload-Digest\":\"sha1:CV42QSG3KOADONUW4VMMJYDDMJZWMRML\",\"WARC-Block-Digest\":\"sha1:OFSX4UGAC6ETMEH3I45H4MQXQU5I2NPJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337663.75_warc_CC-MAIN-20221005172112-20221005202112-00456.warc.gz\"}"} |
http://manuals.opensound.com/developer/morse2.c.html | [
"",
null,
"",
null,
"Web www.opensound.com manuals.opensound.com developer.opensound.com\n\n# Open Sound System OSS 4.x Programmer's Guide\n\n Do you have problems with sound/audio application development? Don't panic! Click here for help!\n\n# morse2.c\n\nAnother morse code program that uses select\n\n## Description\n\nCopyright (C) 4Front Technologies, 2002-2004. Released under GPLv2/CDDL.\n\nThis program is a significantly more complicated version of morse.c. It uses the select system call to be able to handle keyboard input and audio output at the same time. It shows how to prevent output underruns by writing silent samples to the audio device when there is no \"payload\" signal to play.\n\nWhat the program does is playing randomly selected morse symbols (based on the command line options). It then waits until the users hits a key. If the input character matches the morse output then the program continues by playing another morse character. If there was an error then the next character is repeated. The Esc key is used to stop the program.\n\nParts on this program are common with morse.c and commented only there.\n\nThe morse3.c program is a slightly different version of this one.\n\n```#include <stdio.h>\n#include <unistd.h>\n#include <stdlib.h>\n#include <string.h>\n#include <fcntl.h>\n#include <sys/soundcard.h>\n#include <math.h>\n#include <termios.h>\n#include <signal.h>\n#include <sys/time.h>\n#include <sys/types.h>\n\n#define BUFSZ (16*1024)\n#define SRATE 48000\n#define ATTACK 100\n#define CHARDELAY 3\n\nchar randomlist;\nint nrandom = 0;\nint done = 0;\nint chars_to_play = 100;\n\nint dotsize;\nint audiofd = -1;\nstatic int ncodes;\nstatic int playc;\nstatic int terminal_fd = 0;\nstatic struct termios ti, saved_ti;\nstatic int totalchars = 0, totalerrors = 0, errors = 0;\nstatic time_t t0;\n\ndouble a, step;\n#include \"charlist.h\"\n\nstatic void\nterminate (int sig)\n{\ntime_t t;\n\nt = time (0) - t0;\n\nif (terminal_fd == -1)\nreturn;\n\nif (tcsetattr (terminal_fd, TCSAFLUSH, &saved_ti) == -1)\n{\nperror (\"tcgetattr\");\nexit (-1);\n}\n\nprintf (\"\\n\\nTotal characters: %d\\n\", totalchars);\nprintf (\"Errors: %d\\n\", totalerrors);\nprintf (\"Elapsed time: %d seconds\\n\", t);\nif (errors == 0) totalchars--;\nif (totalchars <= totalerrors) totalchars = 0;\nelse\n{\ntotalchars -= totalerrors;\ntotalchars += totalchars / 5;\n}\n\nprintf (\"Characters per minute: %g\\n\",\nt > 0?(float) totalchars / (float) t * 60.0:0);\nexit (sig);\n}\n\nstatic int\ngenpulse (short *buf, int w, int state)\n{\nint i, l;\na = 0.0;\n\nl = w * dotsize;\n\nfor (i = 0; i < ATTACK; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\ntmp *= (double) (i) / (double) ATTACK;\n\n*buf++ = (int) tmp *state;\n\na += step;\nif (a > 360.0)\na -= 360.0;\n}\n\nfor (i = ATTACK; i < l - ATTACK; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\n\n*buf++ = (int) tmp *state;\n\na += step;\nif (a > 360.0)\na -= 360.0;\n}\n\nfor (i = l - ATTACK; i < l; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\n\ntmp *= (double) (l - i) / (double) ATTACK;\n\n*buf++ = (int) tmp *state;\n\na += step;\nif (a > 360.0)\na -= 360.0;\n}\n\nreturn l;\n}\n\nstatic void\nplayerror (void)\n{\nshort buffer, *buf = buffer;\nint i, l;\na = 0.0;\n\nl = dotsize;\n\nfor (i = 0; i < ATTACK; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\ntmp *= (double) (i) / (double) ATTACK;\n\n*buf++ = (int) tmp;\n\na += step * 2.0;\nif (a > 360.0)\na -= 360.0;\n}\n\nfor (i = ATTACK; i < l - ATTACK; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\n\n*buf++ = (int) tmp;\n\na += step * 2.0;\nif (a > 360.0)\na -= 360.0;\n}\n\nfor (i = l - ATTACK; i < l; i++)\n{\ndouble tmp = 0x7fff * cos (a * M_PI / 180.0);\n\ntmp *= (double) (l - i) / (double) ATTACK;\n\n*buf++ = (int) tmp;\n\na += step * 2.0;\nif (a > 360.0)\na -= 360.0;\n}\n\nwrite (audiofd, buffer, 2 * l);\n}\n\nstatic int\ngenmorse (short *buf, char c)\n{\nint l = 0, i;\nconst char *s;\n\nif (c == ' ')\nreturn genpulse (buf, 4, 0);\n\nfor (i = 0; i < ncodes; i++)\nif (Chars[i] == c)\n{\ns = Codes[i];\n\nwhile (*s)\n{\nif (*s++ == '.')\nl += genpulse (&buf[l], 1, 1);\nelse\nl += genpulse (&buf[l], 3, 1);\n\nl += genpulse (&buf[l], 1, 0);\n}\n\nl += genpulse (&buf[l], CHARDELAY, 0);\nreturn l;\n}\n\nprintf (\"<?>\");\nfflush (stdout);\nreturn 0;\n}\n\nstatic void\nplaychar (char c)\n{\nshort buf[16 * BUFSZ];\nint l;\n\nl = 0;\n\nif (c <= 'Z' && c >= 'A')\nc += 32;\n\nswitch (c)\n{\ncase ' ':\ncase '\\n':\ncase '\\t':\nl = genmorse (buf, ' ');\nbreak;\n\ncase '\\r':\nbreak;\n\ncase 'Å':\nl = genmorse (buf, 'å');\nbreak;\ncase 'Ä':\nl = genmorse (buf, 'ä');\nbreak;\ncase 'Ö':\nl = genmorse (buf, 'ö');\nbreak;\ncase 'Ü':\nl = genmorse (buf, 'ü');\nbreak;\n\ndefault:\nl = genmorse (buf, c);\n}\n\n```\n\nNext write whatever we have in the buffer. Note that this write will block but that time is at most the time required to play one morse code character. There is no need to avoid blocking because it doesn't cause any annoying delays.\n\nThis is often the case with audio application. While many programmers think blocking is evil it's actually programmer's best friend.\n\n``` if (write (audiofd, buf, 2 * l) != 2 * l)\n{\nperror (\"write audio\");\nterminate (15);\n}\n}\n\n```\n\nThe randomplay() routine picks a character from the list of characters selected for practice.\n\n```static void\nrandomplay (void)\n{\nint old = playc;\nif (totalchars == chars_to_play)\n{\ndone = 1;\nreturn;\n}\n\n// while (playc==old)\n{\nint i, x, tmp;\n\nx = random () % nrandom;\nplayc = randomlist[x];\n#if 1\nif (x != nrandom - 1)\n{\ntmp = randomlist[x];\nfor (i = x; i < nrandom - 1; i++)\nrandomlist[i] = randomlist[i + 1];\nrandomlist[nrandom - 1] = tmp;\n}\n#endif\n}\n\nplaychar (playc);\ntotalchars++;\n}\n\nstatic int\nfindcode (char c)\n{\nint i;\n\nfor (i = 0; i < ncodes; i++)\nif (Chars[i] == c)\nreturn i;\n\nreturn 0;\n}\n\nint\nmain (int argc, char *argv[])\n{\nchar *devname = \"/dev/dsp\";\nshort buf[32 * BUFSZ];\nchar line;\nint i, parm;\nint l, speed, wpm = 12;\n\nif (argc > 1)\ndevname = argv;\n\nif (argc > 2)\n{\nwpm = atoi (argv);\nif (wpm == 0)\nwpm = 12;\n}\n\nncodes = strlen (Chars);\n\nsrandom (time (0));\n\nif (argc > 3)\n{\nfor (i = 3; i < argc; i++)\nparse_charlist (argv[i]);\n}\nelse\n{\nstrcpy (randomlist, Chars);\nnrandom = strlen (randomlist);\n}\n\nif (nrandom < 2)\n{\nexit (-1);\n}\n\nrandomlist[nrandom] = 0;\n\nprintf (\"Practicing codes: %s\\n\", randomlist);\nfor (i = 0; i <= nrandom; i += 4)\n{\nint j, k;\nchar line, tmp;\nmemset (line, ' ', 80), line = 0;\n\nfor (j = 0; j < 4; j++)\nif (i + j <= nrandom)\n{\nint ix;\n\nix = findcode (randomlist[i + j]);\n\nsprintf (tmp, \"%c %s\", randomlist[i + j], Codes[ix]);\nfor (k = 0; k < strlen (tmp); k++)\nline[j * 20 + k] = tmp[k];\n}\n\nprintf (\"%s\\n\", line);\n}\n\nspeed = wpm;\n\nprintf (\"Words per minute %d. Characters per minute %d\\n\", wpm, wpm * 5);\n\ndotsize = SRATE / speed;\n\nif ((audiofd = open (devname, O_WRONLY, 0)) == -1)\n{\nperror (devname);\nexit (-1);\n}\n\nparm = 0x0003000a;\nioctl (audiofd, SNDCTL_DSP_SETFRAGMENT, &parm);\n\nparm = AFMT_S16_LE;\nif (ioctl (audiofd, SNDCTL_DSP_SETFMT, &parm) == -1)\n{\nperror (\"SETFMT\");\nclose (audiofd);\nexit (-1);\n}\n\nif (parm != AFMT_S16_LE)\n{\nprintf\n(\"Error: 32/24 bit sample format is not supported by the device\\n\");\nprintf (\"%08x/%08x\\n\", parm, AFMT_S16_LE);\nclose (audiofd);\nexit (-1);\n}\n\nparm = SRATE;\nif (ioctl (audiofd, SNDCTL_DSP_SPEED, &parm) == -1)\n{\nperror (\"SPEED\");\nclose (audiofd);\nexit (-1);\n}\n\nif (parm != SRATE)\n{\nprintf\n(\"Error: %d Hz sampling rate is not supported by the device (%d)\\n\",\nSRATE, parm);\nclose (audiofd);\nexit (-1);\n}\n\nif (tcgetattr (terminal_fd, &saved_ti) == -1)\n{\nperror (\"tcgetattr\");\nexit (-1);\n}\n\nsignal (SIGINT, terminate);\n\n```\n\nSet up the terminal (stdin) for single character input.\n\n```\nif (tcgetattr (terminal_fd, &ti) == -1)\n{\nperror (\"tcgetattr\");\nexit (-1);\n}\n\nti.c_lflag &= ~(ECHO | ICANON | IEXTEN | ISIG);\nti.c_iflag &= ~(BRKINT | ICRNL | INPCK | ISTRIP | IXON);\nti.c_cflag &= ~(CSIZE | PARENB);\nti.c_cflag |= CS8;\nti.c_oflag &= ~(OPOST);\n\nti.c_cc[VMIN] = 1;\nti.c_cc[VTIME] = 1;\n\nif (tcsetattr (terminal_fd, TCSAFLUSH, &ti) == -1)\n{\nperror (\"tcgetattr\");\nexit (-1);\n}\n\na = 0.0;\n\n```\n\nPlay some \"extra\" audio data to delay the startup. This just gives some extra time to the user to prepare before we start.\n\n```\nstep = 360.0 * 600.0 / parm;\n\nl = 0;\nl += genpulse (&buf[l], 1, 0);\nwrite (audiofd, buf, l * 2);\nmemset (buf, 0, 4096);\nfor (i = 0; i < 2; i++)\nwrite (audiofd, buf, 4096);\n\n```\n\nThe actual playback starts here\n\n```\nrandomplay ();\n\nt0 = time (0);\nwhile (!done)\n{\nint n;\n\n```\n\nSet up select for output events on the audio device and input events on the keyboard.\n\n``` FD_ZERO (&readfds);\nFD_ZERO (&writefds);\n\nFD_SET (audiofd, &writefds);\n\n```\n\nCall select with no timeouts\n\n``` if ((n = select (audiofd + 1, &readfds, &writefds, NULL, NULL)) == -1)\n{\nperror (\"select\");\nexit (-1);\n}\n\nif (n == 0)\ncontinue;\n\nif (FD_ISSET (0, &readfds))\t/* 0 means stdin */\n{\n\n```\n\nHandling of keyboard input. Check if the answer was right.\n\n```\t if (read (0, line, 1) == 1)\n{\nif (*line == 27)\t/* ESC */\nterminate (SIGINT);\nif ((unsigned char) *line != (unsigned char) playc)\n{\nint x;\n\ntotalerrors++;\nchars_to_play += 4;\nrandomlist[nrandom++] = playc;\nrandomlist[nrandom++] = playc;\nfor (x = 0; x < nrandom; x++)\nif (randomlist[x] == *line)\n{\nrandomlist[nrandom++] = *line;\nbreak;\n}\nplayerror ();\nif (++errors > 3)\n{\nprintf (\"It is '%c' not '%c'\\r\\n\", playc, *line);\nfflush (stdout);\n}\nplaychar (playc);\n}\nelse\n{\nerrors = 0;\nrandomplay ();\n}\n}\n}\n\nif (FD_ISSET (audiofd, &writefds))\n{\n\n```\n\nThe audio device is ready to accept more data. Keep the device happy by writing some silent samples to it.\n\nNote that the real \"playload\" signal will be played by the playchar() routine.\n\n```\t memset (buf, 0, 1024);\nwrite (audiofd, buf, 1024);\n}\n}\n\n```\n\nEverything done. Restore the teminal settings and exit.\n\n``` terminate (15);\n\nclose (audiofd);\n\nexit (0);\n}\n```"
] | [
null,
"http://manuals.opensound.com/gifs/logo.jpg",
null,
"http://www.google.com/logos/Logo_25wht.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5046846,"math_prob":0.9837513,"size":9353,"snap":"2019-51-2020-05","text_gpt3_token_len":3163,"char_repetition_ratio":0.11819446,"word_repetition_ratio":0.1859384,"special_character_ratio":0.3998717,"punctuation_ratio":0.22788066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9911755,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T22:01:04Z\",\"WARC-Record-ID\":\"<urn:uuid:e4a05295-c02e-4a1d-9acb-dcdc7d12e3ca>\",\"Content-Length\":\"32166\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a840b670-86b2-4efd-9c82-47811a2eeeaf>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b612478-8978-4cd0-98f0-bbdbeb6071c3>\",\"WARC-IP-Address\":\"5.9.116.58\",\"WARC-Target-URI\":\"http://manuals.opensound.com/developer/morse2.c.html\",\"WARC-Payload-Digest\":\"sha1:J5JULRDRBD3R3B5IOPJ5ATKDXUWGAFK6\",\"WARC-Block-Digest\":\"sha1:RAOYH67AELDZ6LSBIHJ4ACAA7ZD66LLX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251728207.68_warc_CC-MAIN-20200127205148-20200127235148-00262.warc.gz\"}"} |
http://xahlee.info/MathGraphicsGallery_dir/dense/dense1.html | [
"# Math: Density Plots of Trig Expressions\n\nBy Xah Lee. Date: . Last updated: .\n\nThe following are density plots of arbitrary expressions of one term involving trig functions. The expression is printed on the lower right corner of the image.\n\n## Mathematica Code\n\nThe following is Mathematica code that generates all possible equations of one term. (tweak the funList and nesting level to define what “all possible” means. if nesting level is 2, it takes about 20 minutes and returns a list of some 2876 terms on a 2002 personal computer.\n\n```<< DiscreteMath`Combinatorica`\nfunList = {Sin, Tan, Power[#, -1] &};\nNest[Module[{li = #},\n(Union[#, SameTest -> (Module[{arg1 = #1, arg2 = #2},\nIf[(*both expression contains both x and y*)\nAnd @@ (((((StringMatchQ[#, \"*x*\"] &&\nStringMatchQ[#, \"*y*\"]) &)@\nToString@#) &) /@ {arg1, arg2})\n, SameQ[arg1, arg2 /. {x -> y, y -> x}],\nSameQ[arg1, arg2]]\n] &)] &)@\nUnion@Flatten@(Table[(Times @@ # &) /@ KSubsets[#, i], {i, 1,\n2}] &)@Flatten@{li, Outer[#1@#2 &, funList, li]}\n] &, {x, y}, 1];\nSelect[%, (StringMatchQ[ToString@#, \"*x*\"] &&\nStringMatchQ[ToString@#, \"*y*\"]) &]\n```\n\nHere are the first few items of the result, with nesting level at 2:\n\n```{1/(x^2*y^2), 1/(x*y^2), x/y^2, 1/(x*y), x/y, x^2/y, x*y, x^2*y,\nx^2*y^2, Cos[x]/y^2, Cos[x]/y, Cos[x]/(x*y), (x*Cos[x])/y, y*Cos[x],\n(y*Cos[x])/x, x*y*Cos[x], y^2*Cos[x], Cos[x]*Cos[y], Cot[x]/y^2,\nCot[x]/(x*y^2)}\n```\n\nThe problem is to enumerate all possible single-term expressions of 2 variables x and y, using all the trig functions, multiplication and its inverse, integer power and its inverse (roots), and nesting."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6378654,"math_prob":0.9882212,"size":1547,"snap":"2021-31-2021-39","text_gpt3_token_len":517,"char_repetition_ratio":0.10758263,"word_repetition_ratio":0.0,"special_character_ratio":0.37168714,"punctuation_ratio":0.19161677,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99856585,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T15:17:54Z\",\"WARC-Record-ID\":\"<urn:uuid:550d8bfa-1d4f-45b9-9ca1-12125d3b14b7>\",\"Content-Length\":\"4868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8915ef4-7755-4422-bd40-d14dadc666bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:17587877-20a1-46d0-a316-032a676668f5>\",\"WARC-IP-Address\":\"74.208.236.186\",\"WARC-Target-URI\":\"http://xahlee.info/MathGraphicsGallery_dir/dense/dense1.html\",\"WARC-Payload-Digest\":\"sha1:OXYJLM6NTHTX5KFLRUAOR7VCTWSYLU4G\",\"WARC-Block-Digest\":\"sha1:4RTNYXFKE6WWFKREOCHEZQ5KFNIWFCVY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056890.28_warc_CC-MAIN-20210919125659-20210919155659-00484.warc.gz\"}"} |
https://jumbotutors.com/reviewing-the-case-study-of-starbucks-corporation-and-completing/ | [
"Reviewing the case study of starbucks corporation and completing\n\nAssignment 1: Cash Flows and Dividend Valuation In this Assignment you will be reviewing the case study of Starbucks Corporation and completing questions and exercises relating to financial reports, financial statement analysis and financial valuation. This will show you how investors and analysts understand the present profitability and risk of firms and forecast future investment decisions. Questions and Exercises: Income and Cash Flows Answer the following questions from your textbook: (view grading rubric):\n\n1.Chapter 3: Questions and Exercises: 3.2 (p.196): Articulation of the Statement of Cash Flows with Other Financial Statements. Describe how the statement of cash flows is linked to each of the other financial statements (income statement and balance sheet). Also review how the other financial statements are linked with each other.\n\n2.Chapter 3: Integrative Case 3.1 Starbucks, Cash Flows: f (p. 219): Refer to the income statement for Starbucks in Exhibit 1.27). Compute the amount of EBITDA for 2010, 2011, and 2012.\n\n3.Chapter 11: Integrative Case 11.1 Starbucks, Dividend Valuation; a, b (p. 903): a. Use the CAPM to compute the required rate of return on equity capital for Starbucks. b. Compute the weighted-average cost of capital for Starbucks as of the start of Year +1. At the start of Year +1, Starbucks had \\$550 million in outstanding interest-bearing debt on the balance sheet and no preferred stock. Assume that the balance sheet value of Starbucks’ debt is approximately equal to the market value of the debt. Assume that at the start of Year +1, Starbucks will incur interest expense of 6.25% on debt capital and that Starbucks’ average tax rate is 33.0%.\n\nsee attached Rructions"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8583402,"math_prob":0.8086238,"size":1776,"snap":"2022-05-2022-21","text_gpt3_token_len":389,"char_repetition_ratio":0.14221218,"word_repetition_ratio":0.01459854,"special_character_ratio":0.22297297,"punctuation_ratio":0.13946587,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.968529,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T22:31:42Z\",\"WARC-Record-ID\":\"<urn:uuid:e08d8f3b-c40f-4573-819f-c272d0b03597>\",\"Content-Length\":\"43432\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3194f573-fbd8-4dc5-8509-2112f2053354>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a5dc6fb-2ded-400c-aa56-375714199812>\",\"WARC-IP-Address\":\"68.65.121.156\",\"WARC-Target-URI\":\"https://jumbotutors.com/reviewing-the-case-study-of-starbucks-corporation-and-completing/\",\"WARC-Payload-Digest\":\"sha1:LNWVARZMFXRC7OSODGY7NK5LI2O4CSTP\",\"WARC-Block-Digest\":\"sha1:HO4EAUG5KUW7GEXJCU7JUVQQL5OYX5SW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301063.81_warc_CC-MAIN-20220118213028-20220119003028-00067.warc.gz\"}"} |
https://cosmocoffee.info/viewtopic.php?f=2&t=3294&p=9013&sid=9b8bcb65e1ba8f7854181d4b2e126025 | [
"## [1911.09417] First simulations of axion minicluster halos\n\n Authors: Benedikt Eggemeier, Javier Redondo, Klaus Dolag, Jens C. Niemeyer, Alejandro Vaquero Abstract: We study the gravitational collapse of axion dark matter fluctuations in the post-inflationary scenario, so-called axion miniclusters, with N-body simulations. Largely confirming theoretical expectations, overdensities begin to collapse in the radiation-dominated epoch and form an early distribution of miniclusters with masses up to $10^{-12}\\,M_\\odot$. After matter-radiation equality, ongoing mergers give rise to a steep power-law distribution of minicluster halo masses. The density profiles of well-resolved halos are NFW-like to good approximation. The fraction of axion DM in these bound structures is $\\sim 0.75$ at redshift $z=100$. [PDF] [PS] [BibTex] [Bookmark]\n\nDiscussion related to specific recent arXiv papers\nPosts: 22\nJoined: September 13 2019\nAffiliation: N/A\nContact:\n\n### [1911.09417] First simulations of axion minicluster halos\n\nThis paper was commented on through Cosmo Comments. The following comments can also be viewed as annotations on the paper via Hypothesis.\n\nThis is a very interesting paper that simulates an era of QCD axion dark matter that hasn’t previously been simulated (at least at this resolution and depth). The era is the first gravitational growth of structures (“mini clusters” and their haloes) during radiation domination through until after matter-radiation equality (down to $z=99$).\n\nDuring this time, very small distance scales become non-linear and have some potentially interesting effects. Firstly, the initial conditions aren’t the usual adiabatic ones, nor are they Gaussian, due to their relevance during the earlier evolution through the Peccei-Quinn symmetry breaking phase and the QCD phase transition. Secondly, the structures that form here form the background for potential growth of “axion stars”, which are already well studied.\n\nThe results and machinery developed are relevant for a range of reasons. e.g., determining what dark matter is, potential observational evidence of the QCD axion, studies of the evolution of the non-linear Schrödinger equation (i.e. the evolution of axion stars) and it is relevant to people who simulate cold dark matter because their expertise and tools would be immediately translatable for use here.\n\nI have some questions:\n1. The y-axis label of figure 4 is confusing to me. An uppercase \"N\" is used, indicating it is the actual number of halos of a given mass, rather than the number density per $\\log(M)$, as in figure 2. But the description indicates it isn't the total number greater or smaller than a given mass (as in figure 3) – and the plot wouldn't make sense in this context anyway. Is it the total number of halos within a logarithmic mass bin that is plotted (normalised to total number of subhalos of any mass)? Or is it something else?\n2. My personal, a priori, belief for the profiles in figure 5 is that the lines should be closer to NFW than power-law, which is what is claimed in the text. However, if I try not to give into confirmation bias, the evidence isn't overwhelming in the paper. The main issues are that no comparison fit was given to a power-law as an alternative and that the relative error for the fit to NFW gets as large as ~0.5 in the bottom panel. I would be interested in seeing an additional set of dashed lines in the lower panel showing the relative error to the best fit power-law profile. Of course, for the smallest mass bin this would actually give a better fit because the NFW scale radius is smaller than the smallest radius being fitted to and therefore the \"NFW\" fit is actually to an $r^{-3}$ power-law, but this is explained in the text already. It is the largest mass bin that would yield the interesting comparison between NFW and power-law.\n3. Do the authors have any idea or speculation as to why the low- and medium-mass profiles are under-dense compared to NFW at larger r? Are these haloes simply not fully virialised yet for some reason? Or is there a lack of additional mass to accrete to these haloes in their outer regions?\n4. Is there any expectation for what the value of $\\alpha$ (the fit to the slope of the mini cluster halo mass function) should be. Or, is $\\alpha=-0.7$ a pure empirical fit at this stage (to be explained later)?\n5. It isn't ever mentioned (that I can see) how many haloes are in each mass bin for Table 1 and Figure 5. Also, in Figure 5 why only stack 20 profiles in each bin and not all of them?\n6. In the appendices, I didn't fully understand why the velocities of the N-body particles needed to be set to zero and why this is OK physically. Surely, if the authors evolve the output of the early universe simulation forward with linear theory, they know the rate of change of the relevant quantities, which can then be used to define a velocity field alongside the density field? As they state in the text, setting the velocities to zero removes the initial logarithmic growth, but wouldn't including this growth be important? Or is it actually the correct physics to set these to zero because these modes aren't coming from adiabatic initial conditions? But if this is the correct physics, what is the use of the linear \"evolution\" step? Clarification would be really helpful (for me)!\nThe last sentence of S1 is: \"The velocities of the particles are set to zero, which is compatible with the last smoothing procedure.\" I don't understand what this sentence means. It might be a clue as to what I'm missing for answering the last question.\n\n[These comments were shared with us by a member of the cosmology community. They do not necessarily reflect the opinion of the Cosmo Comments team.]"
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https://www.jpost.com/israel/study-arabic-education-reduces-stereotyping-by-jewish-kids | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
] | [
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https://skhane.org/giza-blueprint/ | [
"# Giza complex Blueprint {3}\n\nPetrie with reference to the 1st Pyramid\n\nNorth: 9069.4East: 9067.7South: 9069.5West: 9068.6Mean: 9068.8\n\nRe the height Petrie states: “The mean base being 9068.8 ± .5 inches, this yields a height of 5776.0 ± 7.0 inches”.\n\nHowever, I favour a 1st Pyramid base side length of 9072, the Petrie: 9068.8 ± .5 inches will obtain: 9068.8 +.05 = 9069.3 hence 9072 minus 9069.3 = 2.7, a slender difference, equally Gematria value Only-begotten son: 1176 x 7.714285 = 9072\n\nThe 7.714285 x 746.666 = 5760 my preferred 1st Pyramid height, the 746.666 x 765.45 = 571536, the 1st Pyramid base area {756 x 756} furthermore the 765.45 being 486 x 1.575 which is 9072 ÷ 5760 the 3rd PWS member is 486\n\nThen again actual Pi minus 0.1176 x 3000 = 9071.97796…tolerable as 9072, in addition to 9071.97796… ÷ 1.575 = 5759.986006… tolerable as 5760\n\nOr else Phi plus 19 = 20.618033988749…this multiplied by 440 = 9071.93495….otherwise 20.618033988749… x 279.3636 = 5759.9289…The 279.3636 is 162.592592 Egyptian cubits of 1.71818 due to the cyclic 592, the Platonic Precession: 25920 x 162.592592 = 4214400, which is 279.3636 x 15085.714285 this is 2.857142 miles of 5280\n\nThe 2.857142 is one seventh of 20, subsequently the Schooldays Pi {22 ÷ 7} of 3.142857 is 1.1 x 2.857142 the 19 pertains to the Metonic cycle, otherwise the British Admiralty {or as I dub it} “Sea” mile of 6080 is 19 x 320 or Gem: Weapons of god: 665 is 19 x 35, etc\n\nPetrie with reference to the 2nd Pyramid\n\nNorth: 8471.9\n\nEast: 8475.2\n\nSouth: 8476.9\n\nWest: 8475.5\n\nMean: 8474.9\n\nHowever the mean ought to be: 8471.9 + 8475.2 + 8476.9 + 8475.5 = 33899.5 this divided by 4 = 8474.875 for this reason we employ 8474.625 in view of the fact that the Petrie: 8474.875 minus 8474.625 = 0.25, minute dissimilarity\n\nPetrie with reference to the 3rd Pyramid:\n\nNorth: …\n\nEast: 4149.2\n\nSouth: 4157.8\n\nWest: 4153.9\n\nMean: 4153.6\n\nPetrie mentions: “The N. end of the W. side could not be reached, after several attempts; and hence the lack of knowing the length of the N. or azimuth of the W. side”\n\nSubsequently with the data we possess: 4149.2 + 4157.8 + 4153.9 = 12460.9 moreover this divided by 3 = 4153.6333…whereas the Petrie mean indicates: 4153.6, hence I will employ 4154.625 In view of the fact that 4154.625 minus the 4153.6333 = 0.991666 which is one third of 2.975 that minus 1.4 = 1.575, this is the 1st Pyramid ratio: 9072 ÷ 5760 = 1.575.\n\nLikewise the 2nd Pyramid 8474.625 minus the 2.975 = 8471.65, comparable to Petrie with reference to the 2nd Pyramid: North: 8471.9, subsequently I consider the entire is reasonable\n\nOr else re the 3rd Pyramid: whilst we employ a North to match the South I.E. 4157.8\n\nSubsequently\n\nNorth: 4157.8\n\nEast: 4149.2\n\nSouth: 4157.8\n\nWest: 4153.9\n\nSum: 16618.7\n\nIn that case 16618.7÷ 4 = 4154.675 which minus a mere 0.05 = 4154.625 accordingly I believe this 4154.625 is satisfactory to employ\n\nThe Petrie data re the three Giza Pyramids being:\n\nN to S E to W\n\nCentre of 1st to centre of 2nd Pyramid: 13931.6 13165.8\n\nCentre of 2nd to centre of 3rd Pyramid 15170.4 9450.2\n\nCentre of 1st to centre of 3rd Pyramid: 29102 22616\n\nTherefore to obtain the total lengths we need to add half the base side length of the 1st Pyramid and half the base side length of the 3rd Pyramid\n\nAs a result 1st Pyramid as 9072 ÷ 2 = 4536\n\nAs a result 3rd Pyramid as 4154.625 ÷ 2 = 2077.3125\n\nHence N to S Centre of 1st to centre of 3rd Pyramid: 29102 + 4536 + 2077.3125 = 35715.3125 which is 12 x 2976.27604… feet, therefore I will employ exactly: 2976.278125 Since: 2976.278125 minus 2976.27604… = 0.00208333,\n\nFurthermore the 1st Pyramid height of 5760 ÷ 0.00208333 = 2764800 this is the seconds {86400 in one day} in 32 days, otherwise 1 ÷ 0.00208333 = 480 which is the 1st Pyramid height {5760 = 12 x 480}\n\nHence E to W Centre of 1st to centre of 3rd Pyramid 22616 + 4536 + 2077.3125 = 29229.3125 which is 12 x 2435.77604… feet, therefore I will employ exactly 2435.778125 since 2435.778125 minus 2435.77604… = 0.00208333, as prior\n\nRegarding North to South\n\nThe distance between the 1st and 2nd Pyramid is 5158.2875\n\nThe distance between 2nd and 3rd Pyramid is 8855.75\n\nThe 1st Pyramid base side length is 9072 ÷ 2 = 4536\n\nThe 3rd Pyramid as 4154.625 ÷ 2 = 2077.3125\n\nTherefore 4536 + 5158.2875 + 8474.625 + 8855.75 + 2077.3125 = 29101.975 which is the Petrie 29102 plus 0.025 which is 12 x 0.00208333 subsequently the Petrie 29102 + 4536 + 2077.3125 = 35715.3125 which is 12 x 2976.27604… that plus 0.00208333 = 2976.278125\n\nRegarding East to West\n\nThe distance between the 1st and 2nd Pyramid is 4392.4875\n\nThe distance between 2nd and 3rd Pyramid is 3135.55\n\nThe 1st Pyramid base side length is 9072 ÷ 2 = 4536\n\nThe 3rd Pyramid as 4154.625 ÷ 2 = 2077.3125\n\nTherefore 4536 + 4392.4875 + 8474.625 + 3135.55 + 2077.3125 = 22615.975 which is the Petrie 22616 plus 0.025 which is 12 x 0.00208333 Subsequently the Petrie 22616 + 4536 + 2077.3125 = 29229.3125 this is 12 x 2435.77604… that plus 0.00208333 = 2435.778125\n\nHence 2435.778125 plus 540.5 = 2976.278125, the 540.5, in any case the 540.5 minus a reduced mile of 528 = 12.5 this is one eighth of 100\n\nFor this reason we comprise a rectangle of: 2976.278125 x 2435.778125 = 7249553.150791015625 square feet that divided by the square feet in one acre: 43560 = 166.426840008… this divided by the 1st Pyramid {Great Pyramid} Tan: 480 ÷ 378 of 1.269841 = 131.06113650706..\n\nThis 131.06113650706…divided by the Greek foot {441:440} of 1.0022727 = 130.7639457213… this divided by the Rhind Papyrus fraction {256:81} of 3.160493827 = 41.37452970089…\n\nThis 41.37452970089… divided by the Egyptian unit of measure {12 ÷ 7} of 1.714285 = 24.135142325…The ancient Comma of Pythagoras is 531441 ÷ 524288, the musical scale: Pythagorean Limma is 256:243 that I denote as FP {Fraction of Plato}\n\nThe 24.135142325… x 524288 x 256 = 3239363967.888 exactly, which is the 7249553.150791015625 x 446.83636 moreover merely regarding brevity this 446.83636 divided by the Egyptian cubit of 1.71818 = 260.063492 that multiplied by 63 = 16384 which opens the door to many more units of measure, likewise the Musical Scales\n\nThe prior 166.426840008… multiplied by √2 x √3 x Pi ÷ 3.5065454 = 365.2324799… a tolerable Earth year that multiplied by 360000 ÷ 5280 = 24902.21454022… a tolerable Earth Equatorial circumference {miles} The 3.5065454 divided by the Egyptian foot of 1.14545 = 3.06126984 this divided by the 1st Pyramid {Great Pyramid} Tan of 480 ÷ 378 of 1.269841 = 2.41075\n\nThen 1881 x 2.41075 ÷ 6080 = 0.74582578125 then 3.5065454 ÷ 0.74582578125 = 4.70156106519…this multiplied by 9265.0078125 = 43560 the square feet in one acre: The 9265.0078125 multiplied by the musical scale: Pythagorean Limma {256:243} then by the 1st Pyramid height of 5760 = 56221440 this is 10648 modern-day miles of 5280, the 10648 divided by the square feet in one acre: 43560 = 0.2444 that multiplied by 45 = 11 And so forth\n\nThe data presented via Petrie being as follows\n\n1st “The mean base being 9068.8 ± .5 inches, this yields a height of 5776.0 ± 7.0 inches”\n\n2nd “Hence the height will be 5,664 ± 13 inches”\n\n3rd “Hence the height of the Pyramid would have been 2564 ± 15; or 2580.8 ± 2.0 by the granite courses”\n\n{End of Petrie data}\n\nVia my calculations the Pyramid measurements are as follows {modern day inches}\n\nThe 1st Pyramid base side length of 9072……height of 5760\n\nThe 2nd Pyramid base side length of 8474.625……height of 5670\n\nThe 3rd Pyramid base side length of 4154.625 ……height of 2578.8\n\nVia these measurements: The 1st Pyramid Tan will be\n\n5760 ÷ 4536 = 1.269841 Tan, which produces a Slant Angle of 51.779567945149… {4536 x 2 = 9072} this 51.779567945149… converts to: 51° 46′ 46.4442″\n\nOtherwise Base side: 9072 ÷ 24 = 378 Height: 5760 ÷ 12 = 480 subsequently: 480 ÷ 378 = 1.269841\n\nWhereas with reference to the Petrie data:\n\n“Casing stones, in situ, N. side, by theodolite 51º 46′ 45 ± 2’…”\n“On the whole, we probably cannot do better than take 51º 52′ ± 2′ as the nearest approximation to the mean angle of the Pyramid, allowing some weight to the South side”\n\n{End of Petrie data}\n\nHence my calculated 51° 46′ 46.4442″ is quite reasonable\n\nThe 2nd Pyramid Tan will be:\n\nHeight of 5670 ÷ 12 = 472.5 Base side length of 8474.625 ÷ 24 = 353.109375 In that case 472.5 ÷ 353.109375 = 1.33811230585… Tan, which produces a Slant Angle of: 53.22845004102..., that converts to: 53° 13′ 42.42″\n\nSeeing as there are 90 degrees in a right angle, in that case deduct the 53.22845004102…from 90 to obtain 36.77154995897…afterwards click the Tan button to obtain 0.74732142857 that multiplied by the prior Tan of 1.33811230585... will obviously obtain 1 {one}\n\nIn this case the 0.74732142857 x 7 = 5.23125 that multiplied by the FP {256:243} then by 202.5 = 1116 this is: Gem: Bereshit Bara, the opening two words of the Hebrew Torah that translate {roughly} as “In the beginning [god] created”\n\nThe 1st Pyramid Tan 1.269841 divided by the 2nd Pyramid Tan of: 1.33811230585… = 0.9489795918367…that multiplied by Gem: Founder of the City: 1225 = 1162.5 this multiplied by 0.96 = 1116 Then again the 5.23125 multiplied by the FP obtains 5.5111 that plus 144.6 = 150.111 one ninth of 1351 that is: 1116 + 235 possibly inferring the Earth tilt?\n\nThe Maya Tzolkins {5256} reduced to 525.6 minus the 144.6 = 381, which multiplied by 8 = 3048, the feet-metres ratio being 1 ÷ 0.3048 In that case 3048 + 1116 + 1116 = 5280 the modern-day mile, also as we observe the 1162.5 x 0.96 = 1116 Bereshit Bara\n\nThe 1162.5 ÷ 20.61818 = 56.38227513 this divided by 2.8282 then multiplied by 1016064 = 20255400 which is 5280 x 3836.25 which is 1116 x 3.4375 that multiplied by Gem: Mary: 192 = 660 the modern-day furlon, the 2.8282 x 2.475 = 7, the Greek furlong of 633.6 is 2.475 x 256. The 1016064 is 1013760 x 1.0022727 {441:440} we have observed these units prior\n\nThis 20255400 is the 2nd Pyramid base side length: 8474.625 x 2390.123456790 that multiplied by 81 = 193600 that divided by the square feet in one acre: 43560 = 4.444 one ninth of 40 {Please see 1162.5 below}\n\nThe 3rd Pyramid Tan will be:\n\nHeight of 2578.8 ÷ 12 = 214.9 Base side length of 4154.625 ÷ 24 = 173.109375 In that case 214.9 ÷ 173.1093751.241411679754Tan which produces a Slant Angle of: 51.147355523771..., that converts to: 51 8′ 50.47″\n\nHebrew Torah\n\nAs indicated prior, the opening two words of the Hebrew Torah obtained the Gematria value: 1116 the words are: Bereshit Bara, that roughly translates to “in the beginning {god} created} The 1st Pyramid base side length is 9072 inches, otherwise 756 feet accordingly the base perimeter is 4 x 756 = 3024 in addition to one half of 756 is 378\n\nThe norm regarding feet-metres is: 1 ÷ 0.3048 = 3.2808398950131 feet in one metre However 1116 + 1116 + 3048 = 5280, the mile\n\nSeeing as there is {approximately} an 11 day difference between 12 lunar months and one solar year, the Hebrew calendar varies via a repeating 19 year Metonic cycle of 235 lunar months Then 1116 + 3024 + 3780 = 7920 that multiplied by the Egyptian foot of 1.14545 = 9072 the 1st Pyramid\n\nThen 1116 divided by Gem: City of god: 882 = 1.26530612244… Currently we perceive Gem: Divine Circle: 1224 attached to 4489 which is 1116 + 3373 which is Gem: Founder of the City: 1225 + 2148 which is 1037 + 1111 etc Therefore: 1225 x 1.26530612244… = 1550\n\nSubsequently 1550 minus 1037 = 513 that minus 360 = 153 the “fishes” landed via Simon Peter {Gem: 1925} upon the Sea of Galilee however 153 x 12 = 1836, the sloping roofline length of the magnificent Grand Gallery within the 1st Pyramid\n\nOtherwise 1550 + 1116 minus 1111 minus 1111 = 444, then 1116 + 235 = 1351 Then 1351 ÷ 9 150.111 this divided by Pi then by Phi obtains a Synodic month of 29.5308077734… that multiplied by 235 = 6939.7398267… which is 19 x 365.24946456…a tolerable Earth year that, as per norm, multiplied by 360000 ÷ 5280 = 24903.372584… a tolerable Earth Equatorial circumference {miles}\n\nWhereas the year I employ is: 1881 x 6080 x 30 ÷ 299008 ÷ Pi = 365.2421989187… this multiplied by 19 = 6939.601779456…this multiplied by Phi x Pi then divided by the 150.111 = 234.99532531218…possibly referring to the Earth “tilt” of 23.5?\n\nThe 26th PWS member is 4608 The: 1360800 days: Mayan Long Count {9.9.0.0.0} divided by 46080 = 29.53125 that multiplied by 12.8 = 378 the 1360800 days indicating 400 Saturn and 378 Jupiter Synodic orbits Then 69.12675 x 923.520923 = 63840 that divided by the Sea mile of 6080 = 10.5\n\nThe 923.5209233.16374609375 = 2921.7857142 one seventh of 20452.5 this minus 17931.105790406… = 2521.39420959…The 3.16374609375 divided by exactly 9.999 x 256 = 81\n\nThen 2521.3942095936936 x Pi x Pi x Pi x 5280 ÷ 360000 x 1881 ÷ 50 afterwards divided by exactly 24904.6220231138783518548889657 = √3 The 24904.6220231138783518548889657 is a tolerable Earth Equatorial circumference {miles}\n\n……………………………….\n\nThe 3rd Pyramid base side length of 4154.675 ÷ 20.61818 = 201.5054012345679\n\nThe 3rd Pyramid height 2578.8 ÷ 20.61818 = 125.074074 this multiplied by 243 = 30393 that divided by the mile of 52805.75625 The Pythagorean Limma {256:243} multiplied by the 5.75625 x 7776 = 47155.2 which is the 3rd Pyramid height of 2578.8 x 18.285714 that multiplied by 21 = 384; this is the 1st PWS member and 7776 is the 32nd PWS member\n\nThe 3nd Pyramid Tan will be:\n\n4154.625 ÷ 24 = 173.109375…….2578.8 ÷ 12 = 214.9 Hence: 214.9 ÷ 173.1093751.2414116797544904… the Tan, that produces a Slant Angle of. 51.147355523771….that converts to: 51 8′ 50.47″\n\nSeeing as there are 90 degrees in a right angle, in that case deduct the 51.147355523771...from 90 to obtain. 38.8526444762. Afterwards click the Tan button to obtain. 0.8055345509539…..this multiplied by the 3rd Pyramid Tan of. 1.2414116797544904…will obviously obtain 1 {one}\n\nThe 1st Pyramid Tan of 1.269841 divided by the 3rd Pyramid Tan of 1.2414116797544904 = 1.02290101708435…{please observe below}\n\nAs indicated prior\n\n“The 2nd Pyramid Tan will be:\n\nHeight of 5670 ÷ 12 = 472.5 Base side length of 8474.625 ÷ 24 = 353.109375 In that case 472.5 ÷ 353.109375 = 1.338112305854241 Tan, which produces a Slant Angle of: 53.228450041025553..., that converts to: 53° 13′ 42.42″”\n\nThe 2nd Pyramid Tan 1.33811230585424 divided by the 3rd Pyramid Tan 1.2414116797544904… = 1.0778956954222…Subsequently the above 2nd Pyramid produced: 1162.5 x.1.077895695422 ÷ 1225 = 1.0229133274735...as we observe prior {Gem: Founder of the City is 1225}\n\nThe Three Pyramid complex area {William Flinders Petrie data}:\n\nN to S E to W\n\nCentre of 1st to centre of 2nd Pyramid: 13931.6 13165.8\n\nCentre of 2nd to centre of 3rd Pyramid 15170.4 9450.2\n\nCentre of 1st to centre of 3rd Pyramid: 29102 22616\n\n{End of Petrie data}\n\nThen exactly 22142.6 divided by the sloping floor line length of the Grand Gallery namely: 1881 = 11.77171 Then Phi x 100 x 3.20338332501… = 518.3183098861…\n\nThen 1 ÷ Pi + 518 = 518.3183098861… this multiplied by Pi afterwards by the 11.77171 = 19168.4167339…this is the square root of 367428200.0858346… that minus 432 = 367427768.08583… tolerable as 367427768.2\n\nThe 13931.6 x 13931.6 = 194089478.56\n\nThe 13165.8 x 13165.8 = 173338289.64\n\nThen 194089478.56 plus 173338289.64367427768.2\n\nThe 3.20338332501… x 86400 = 276772.319280917762022816 exactly this divided by the NC Earth mean circumference of 24883.211.12285876739… that divided by the 3.20338332501… = 3.47222 that multiplied by 0.288 = 1\n\nThe square root of the Rhind Papyrus {256:81} of 3.160493827 is 1.777 However 3.47222 multiplied by the 22142.6 = 76884.02777 which is 1.777 x 43247.265625 that multiplied by the FP {Pythagorean Limma {256:243} then by 7776 = 354281600 which is 5280 x 67098.7878 which divided by the 11.77171 = 5700 this multiplied by 0.33 = 1881\n\nOr the 11.12285876739… x 1881 ÷ 3.20338332501..= 6531.25 that divided by Gem: Simon Peter: 1925 = 3.39285714 that multiplied by 5.6 = 19 Gem: Eve furthermore 19 x 99 = 1881 The Sea mile of 6080 ÷ 3.39285714 = 1792 which is Gem: Mary: 192 x 9.333 that multiplied by 324 = 3024\n\nThe 1792 is also Gem: Weapons of god: 665 x 2.694736842105… that multiplied by 1881 = 5068.8 which is the Greek mile. Or 19 x 2.694736842105… = 51.2 that multiplied by 118.75 = 6080, the 4th PWS member is 512, as a result the end member: 10368 ÷ 2.694736842105… = 3847.5\n\n……………………………..\n\nAgain merely as an example since the options are abundant, the 3847.5 multiplied by the FP {256:243} obtains 4053.333 that divided by the aforementioned 3.47222 = 1167.36 this divided by the longer value Greek foot of 1.01376 = 1151.5151 this multiplied by the 1881 = 2166000 this divided by 6080 = 356.25 {the FP is the Pythagorean Limma: 256:243}\n\nThen 356.25 divided by Gem: 665 = 0.53571428 one seventh of 3.75 which is 1.125 x 3.333 one third of 10, the 1.125 x 8 = 9 {Gem: Weapons of god: 665} Or the Greek cubit of 1.52064 x 2.694736842105…x 1881 = 7707.820032, which, for example, divided by the Roman furlong of 608.256 = 12.672 that multiplied by 381.8181 = 4838.4 which is the Roman mile\n\nThe 381.8181 x 11 = 4200, hence ample scope since 381.8181 multiplied by the yards in one mile: 1760 = 672000 and this divided by the 1st Pyramid base side length of 9072 = 74.074074 Subsequently Gem: Lord Jesus Christ: 3168 x 74.074074 = 234666.666 that multiplied by 1368 = 321024000 which is 6080 x 52800 {and more besides}\n\nThe 1368 divided by Gem: Divine Circle: 1224 then multiplied by 1052.21052631578… = 1176, Gem: Only-begotten son This 1052.21052631578… divided by the 2.694736842105… = 390.46875 which multiplied by the prior 11.77171 = 4596.487689393, this will produce additional numbers of import, given that it is exactly 931.947072 x 4.932133838 as we will observe presently\n\n…………………………….\n\nNonetheless I imagine the options are before now, apparent, in view of the fact that all the foregoing numbers are perfectly interwoven within a marvellously rich tapestry of number, hence a final few examples afterwards we move progress\n\nThe prior 22142.6 divided by the Sea mile of 6080 = 3.641875 this added to 0.008125 = 3.65, Gem: Abraxas, Mithras is 365 likewise 3.65 obtains: 584…14602190 etc the 0.008125 x 1600 = 13 that obtains: 26, 117, 234, 312, 1625, 1366560 etc Then again the 3.641875 multiplied by the FP {256:243} x 7776 = 29834.24, this divided by the 22142.6 = 1.34736842105… that multiplied by Gem: Eve: 19 = 25.6\n\nOr the Greek cubit of 1.52064 x 19 x 1.34736842105.. = 38.928384 exactly that multiplied by the aforementioned: 4.932133838 = 192, Gem: Mary, the 4.932133838 is exactly 1.52587890625 x 3.2323 {6080 ÷ 1881} the 1.52587890625 x 25.6 x 256 = 10000\n\nOr the NC Earth radius of 3960 x 1881 x 1.34736842105… = 10036224 this is 1650.69473684… Sea miles of 6080, the 1650.69473684…x 2.9320987654 = 4840, the square yards in one acre The 2.9320987654 x 641.52 = 1881, the 641.52 divided by Gem: Mary: 192 = 3.34125, then Gem: Lord Jesus Christ: 3168 ÷ 3.34125 = 948.148148 this is 900 x FP {256:243} the highlighted 684 x 8.888 = 6080\n\n……………………………….\n\nVia my calculations the measurements are as follows {modern day inches}\n\nRegarding North to South\n\nI indicated prior: “Subsequently the Petrie 29102 + 4536 + 2077.3125 = 35715.3125 which is 12 x 2976.276041666 that plus 0.00208333 = 2976.278125 feet”\n\nRegarding East to West\n\nI indicated prior: Subsequently the Petrie 22616 + 4536 + 2077.3125 = 29229.3125 this is 12 x 2435.776041666 that plus 0.00208333 = 2435.778125\n\nThe Metres-Feet ratio: 1 ÷ 0.3048 = 3.280839895013123… feet in one metre, that multiplied by 12 = 39.3700787401 inches in one metre Then Pi x 39.3700787401… x 19.69090 = 2435.465170964…whereas above we observe 2435.778125 feet\n\nThe 19.69090 x 55 = 1083 that minus a Colel {1} obtains 1082, Gem: Prophecy, yet Gem: Lord Jesus Christ: 3168 minus 1083 minus 1083 minus 666 {Mr. Nasty} = 336 this is the 1st Pyramid Grand Gallery height, moreover 336 x 27 = 9072 the 1st Pyramid base side length\n\nThen Pi x 39.3700787401… x 24.0666 = 2976.679653… whereas above we observe 2976.278125 feet, the 24.0666 x 45 = 1083 Otherwise Pi x 24.0666 x 120 = 9072.919583…again tolerable as 9072 and 9072.919583… ÷ 1.575 = 5760.58386…tolerable as 5760\n\n………………………………….\n\nAgain, via my calculations the measurements are as follows {modern day inches}\n\nThe 1st Pyramid base side length of 9072……height of 5760\n\nThe 2nd Pyramid base side length of 8474.625……height of 5670\n\nThe 3rd Pyramid base side length of 4154.675……height of 2578.8\n\nThe data presented via Petrie being as follows\n\n1st The mean base being 9068.8 ± .5 inches, this yields a height of 5776.0 ± 7.0 inches.\n\n2nd Hence the height will be 5,664 ± 13 inches.\n\n3rd Hence the height of the Pyramid would have been 2564 ± 15; or 2580.8 ± 2.0 by the granite courses.\n\n{End of Petrie data}\n\nRe the 1st Pyramid\n\nI plumped for a 1st Pyramid base side length of 9072 since the Petrie 9068.8 ± .5 will obtain 9069.3 hence my 9072 is 2.7 inches over and a 1st Pyramid height of 5760, since the Petrie: 5776.0 ± 7.0 will obtain 5769 hence a difference of 9 inches Then again Gem {Gematria Value} Only-begotten Son is 1176 As prior: actual Pi minus 0.1176 x 3000 = 9071.97796076… that divided by 1.575 = 5759.98600683… tolerable as 9072 and 5760\n\nThen 0.1176 x 7.714285 = 0.9072 or 0.1176 x 4.89795918367… = 0.576 the 7.714285 x 7 = 54 accordingly there is ample scope regarding 7.714285 Please note the highlighted 1836 within 4.89795918367…; this is the sloping roof line length of the awe inspiring Grand Gallery, literally “entombed” within the 1st Pyramid {Great Pyramid} and Gem: Founder of the City: 1225 x 4.89795918367… = 6000 The lunar, Yin-Yang number: 1080 x 1.7 = 1836\n\nSubsequently 1.7 x 1106.4705882352 = 1881, which is the sloping floor line length, however within the underlined cyclic we espy: 882: Gem: City of god and 2352, this being 10 x 235.2 the Gallery 7 corbelled perpendicular height\n\nLikewise the Gallery leads to the supposed Kings Chamber, wall height: 235.2 and 235.2 is 0.8 x 294 We also observe 294: Gem: Church…..the aforementioned 1176 “affixed” to 1764 which is 1080 plus 684, or 1764 plus 72 = 1836, or 1764 minus 540 = 1224 Gem: Divine Circle which multiplied by 1.5 = 1836\n\nAnd since Church: 294 obtains: 882…1176…1764…2352 whichever of the “numbers” will suffice, we employ 1176, as a result 1176 divided by the {abridged}: 1106.4705882352… = 1.06283891547049… Gem: Divine Circle 1224 divided by the 1.06283891547049… then multiplied by Gem: Founder of the City: 1225 = 1410750 which, for example, divided by the Sea mile of 6080 = 232.03125 which, to be brief, multiplied by 8.10666 = 1881, furthermore 8.10666 x 750 = 6080\n\nOr else Gem: Mary 192 divided by 8.10666 then multiplied by Gem: Eve: 19 = 450, the 19 x 99 = 1881 Otherwise 192 divided by 8.10666 afterwards multiplied by the 6080 = 144000\n\n………………………………….\n\nRhind Papyrus “Pi”\n\nAn Egyptian Scribe named Ahmes wrote upon the Rhind Papyrus {Circa 1650 B.C.E} “Cut off 1/9th of a diameter and construct a square upon the remainder…”\n\nHence we may state that exactly 1774092.72 ÷ 0.7 ÷ 5760 = 440.003154761…Subsequently the 1st Pyramid 9072 ÷ 440.003154761 = 20.6180339886… whereas Phi plus 19 = 20.6180339887\n\nThe 440.003154761… divided by the Rhind Papyrus {256:81} of 3.160493827 = 139.219748186… that divided by the Schooldays Pi {22 ÷ 7} of 3.142857 = 44.29719260475…\n\nThis 44.29719260475…divided by the Greek foot {441:440} of 1.0022727 = 44.196745455… that multiplied by Gem: Founder of the City: 1225 = 54141.01318359375 this divided by the 1774092.72 = 0.030517578125 that multiplied by 256 = 7.8125 that, again, multiplied by 256 = 2000\n\nOtherwise 1774092.72 ÷ 54141.01318359375 = 32.768 which is 0.12 x 273.0666 that divided by a reduced Platonic Precession {25920} of 259.2 obtains: 1.05349794238683… this is the FP that as indicated prior is the musical scale:Pythagorean Limma {256:243} however the FP is, in addition, one third of 3.160493827 that, as indicated is the Rhind Papyrus fraction {256:81}\n\nThe 2nd Pyramid\n\nWith regard to the 2nd Pyramid I deduced a height of: 5670 in view of the fact that the 1st Pyramid height of 5760 ÷ 5670 = 1.015873 which, as indicated prior is the most apt Eye of Horus fraction {63:64} of 0.984375 reversed as 64 ÷ 631.015873\n\nLikewise the Petrie: 5,664 ± 13 inches obtains\n\n{1} 5664 + 13 = 5677 which is 7 inches difference regarding my preferred 5670\n\n{2} 5664 minus 13 = 5651 which is 19 inches difference with regard to 5670 furthermore 19 is Gem: Eve, likewise 19 x 320 = 6080 that as indicated prior is the Sea mile, otherwise Gem: Weapons of god: 665 is 19 x 35 that multiplied by 173.714285 = 6080\n\nThe 173.714285 is the Schooldays Pi of 3.142857 x 55.2727 this divided by the Egyptian cubit of 1.71818 = 32.169312 this multiplied by the Gem: 665 then by the Platonic Precession: 25920 = 554496000 which is the Sea mile of 6080 x 91200 this is Gem: Mary: 192 x 475 that divided by the 32.169312 afterwards multiplied by 256 = 3780, otherwise 475 ÷ 173.7142852.734375\n\nThe Sea mile of 6080 is 684 x 8.88, this for example, is 5 x 1.777 moreover this is the square root of the Rhind Papyrus fraction {256:81} of 3.160493827 Then 684 ÷ 192 = 3.5625 that divided by the 2.734375 = 1.30285714 which, merely to be brief, multiplied by 70000 = 91200\n\nThe 32.169312 x 19987.30953947..= 642978 this divided by the mile of 5280 = 121.77613636 which is 121.5 x 1.0022727 which is the Greek foot {441:440} the 121.5 x 2 = 243 which is a component of the Pythagorean Limma {256:243} which plays a lead role within the Giza complex likewise strategically sited wondrous monuments World-Wide\n\nThe 19987.30953947.. multiplied by Gem: Eve: 19379758.88125 which divided by the 642978 = 0.590625 that multiplied by the other Limma component, namely 256 = 151.2, the shorter value Greek cubit is 1.512, nonetheless this 151.2 x 60 = 9072 the 1st Pyramid base side length Otherwise the Greek cubit of 1.5206419987.30953947..x 19 = 577476.545184 exactly, which again to be brief given that the options are numerous, divided by the 379758.88125 = 1.52064\n\nAs we observe prior: “35 that multiplied by 173.714285 = 6080…”…….However: the 173.714285 is akin to the presently observed North to South distance\n\n“The North to South 2976.278125 which is one twelfth of 35715.3375 that divided by 20.61818 = 1732.22536375661…”\n\nTherefore 1732.22536375661 ÷ 173.714285 x 19 = 189.46214916… this divided by the above 32.169312 then multiplied by 1245184 = 7333549.3 this divided by the 35715.3375 = 205.333 that is Gem: Simon Peter: 1925 ÷ 9.375 this is the mile of 5280 divided by exactly 563.2\n\nThe 1245184 divided by the musical Scale: Pythagorean Minor Third {32:27} of 1.185185 = 1050624 which is exactly 172.8 Sea miles of 6080 Then 563.2 ÷ 173.714285 x 19 = 61.6 this divided by the Greek cubit of 1.52064 = 40.50925925 subsequently owing to the cyclic: 925 the 40.50925925 multiplied by the Platonic Precession: 25920 = 1050000 that divided by Gem: Simon Peter: 1925 = 545.4545 which is one eleventh of 6000\n\nEqually Simon Peter landed the 153fishes” upon the Sea of Galilee, the Grand Gallery sloping roofline length of 1836 is 12 x 153 and 153 x 8 = 1224 Gem: Divine Circle, also Gem: “The Net” Seeing as Simon Peter “gained” 153 it follows that: 1925 x 153 = 294525 that divided by Gem: Founder of the City: 1225 = 240.428571 that multiplied by 7.6363 = 1836\n\nThe 240.428571 x 7 = 1683 which divided by 1836 = 0.91666 one twelfth of 11 Then again 1836 ÷ 1683 = 1.0909 one eleventh of 12, a case of: As above-So below Accordingly 7.6363 obtains: 2523365041008…11761344…1512…1764…2016…2940 et cetera\n\nIn addition to the 1st Pyramid base side length of 9072 ÷ 7.6363 = 1188 which is the mile of 5280 ÷ 4.444 one ninth of 40 The 1st Pyramid height of 5760 ÷ 7.6363 = 754.285714 which is one seventh of 5280\n\nRe the 3rd Pyramid\n\nRe Petrie data: “Hence the height of the Pyramid would have been 2564 ± 15; or 2580.8 ± 2.0 by the granite courses”\n\n{End of Petrie data}\n\nAccordingly 2580.8 + 2 = 2582.8 or 2580.8 minus 2 = 2578.8 And 2578.8 multiplied by the reversed Eye of Horus fraction {64:63} of 1.015873 = 2619.7333 In that case the 1st Pyramid height of 5760 ÷ 2619.7333 = 2.1986970684The 1st Pyramid base side length of 9072 divided by this 2.1986970684… = 4126.08 that divided by the 2nd Pyramid height of 5670 = 0.727703703 that multiplied by 3600 = 2619.7333\n\nThere will be various times we acquire cyclic patterns within the digits for example, the Schooldays Pi {22 ÷ 7} of 3.142857 the 142857 is cyclic, these cyclic patterns will be underlined, but for the cyclic being too extended as to signify something, likewise within the cyclic, more often than not, will be found the appropriate “numbers” to employ. Such as within the 2.1986970684… I obviously highlighted the 684; the Sea mile of 6080 is 684 x 8.888 one ninth of 80. Likewise Gem: Mary is 192, the 684 is Gem: Mary: 192 x 3.5625\n\n………………………….\n\nThe “Cubit”\n\nWe will employ a “cubit” of 20.61818 which is 12 Egyptian cubits of 1.71818 However to indicate the numerous “cubits” or “inches” we may possibly employ, the following ought to shed some light on the matter\n\nAnother “Royal” cubit is 12 ÷ 7 = 1.714285\n\n{A} 1.714285 x 280 = 480 that multiplied by 12 = 5760\n\n{B} 1.714285 x 441 = 756 that multiplied by 12 = 9072\n\n{C} 1.714285 x 1.0022727 = 1.71818 the Egyptian cubit, the employed 1.0022727 being the Greek foot {441 ÷ 440}\n\n{D} 1.714285 x 1.008 = 1.728, another version pertaining to an Egyptian cubit, the 1.008 being one tenth of a Greek foot of 10.08, furthermore 10.08 x 75 = 756\n\n{E} the Geographical rendition of 1.714285 namely 1.73787428571 being 1.728 ÷ 0.99431818 this being an Ancient unit of measure ratio: 175 ÷ 176 and the 1.73787428571 ÷ 1.714285 = 1.01376 which is the longer value Greek foot Consequently, as would appear apparent, the entire being merely varying ratios of each other, so to speak, given that the foregoing are merely a few, furthermore I stress the word, few, of the numerous options available\n\n1st Pyramid base side length of 9072 ÷ 20.61818 = 440\n\n1st Pyramid height of 5760 ÷ 20.61818 = 279.365079 this multiplied by 1512 = 422400 which is 80 miles of 5280 feet, otherwise 422400 is 69.47368421… Sea miles of 6080, this 69.47368421… x 1881 = 130680, which is 24.75 miles of 5280 feet, the 130680, is 27 x 4840, the square yards in one acre\n\nThe shorter value Greek cubit is 1.512 that multiplied by 6000 = 9072 1st Pyramid base side length Otherwise 1.512 x 3809.523809 = 5760 the 1st Pyramid height The 3809.523809 divided by the Rhind Papyrus {256:81} of 3.160493827 = 1205.3571428 that divided by the above 1.714285 = 703.125 that multiplied by 12.9024 = 9072\n\nThe 12.9024 divided by the longer value Greek foot of 1.01376 = 12.7272 {14 ÷ 1.1} in addition to 12.7272 x 452.571428 = 5760 the 1st Pyramid height, the 452.571428 x 7 = 3168, Gem: Lord Jesus Christ, however 3168 x 1.666 = 5280 the 1.666 x 3 = 5\n\n…………………………………..\n\nAs indicated prior, the opening two words of the Hebrew Torah: Bereshit Bara, translate {roughly} as “In the beginning {god} created”: …Bereshit Bara obtained the Gematria Value: 1116\n\nThe 1st Pyramid height of 5760 ÷ 20.61818 = 279.365079 The 1st Pyramid Tan of: 1.269841 x 878.85 = 1116 the 878.85 ÷ 20.61818 = 42.625 The 878.85 ÷ 279.365079 = 3.145883522727 which is 3.13875 Greek feet of 1.0022727 {441:440} and the 1116 ÷ 3.13875 = 355.555 one ninth of 3200\n\nThe 2nd Pyramid\n\nThe 2nd Pyramid base side length: 8474.625 ÷ 20.61818 = 411.0267857142 which divided by the Schooldays Pi {22:7} of 3.142857 x 130.78125 that divided by the prior 3.13875 = 41.666 one third of 125\n\nThe 2nd Pyramid height 5670 ÷ 20.61818 = 275, the 2nd Pyramid Tan will be: Base side length of: 8474.625 ÷ 24 = 353.109375 Height of: 5670 ÷ 12 = 472.5 Subsequently: 472.5 ÷ 353.109375 = 1.338112305854241 the Tan, which produces a Slant Angle of: 53.228450041025..., that converts to: 53° 13′ 42.42″\n\n“The direct measures by goniometer and level on the granite in situ gave 53º 12′ ± 2′,\n\n“Both of granite and limestone, lying around the Pyramid, the mean is 53º 14′ ± 5′;”\n\n{End of Petrie data}\n\nYet the 1st Pyramid Tan: 1.269841 divided by the 2nd Pyramid Tan of 1.338112305854241 then multiplied by Gem: Founder of the City: 12251162.5 that divided by the 411.0267857142 = 2.8282 that multiplied by the Grand Gallery 1881 = 5320 that multiplied by 1.0657894736842…= 5670\n\nThe 1.0657894736842… multiplied by Gem: Eve: 19 = 20.25 possibly another version re the “cubit” given that 5670 ÷ 20.25 = 280 or the 2nd Pyramid base side length: 8474.625 ÷ 5320 = 1.5929746240601…\n\nThen 5670 ÷ 1.5929746240601… ÷ 1.338112305854241…= 2660, which divided by the Sea mile of 6080 = 0.4375 that multiplied by 20736 = 9072 the 1st Pyramid base side length The 20736 is 2 x 10368 the last PWS member, or 20736 ÷ 52803.92727 that divided by the 2.8282 = 1.38857142 one seventh of 9.72 the 8th PWS member is 972\n\nThen again the 2660 is 4 x 665, Gem Weapons of god Or 3.92727 x 11 = 43.2 Or 3.92727 x 440 = 1728 the 14th PWS member\n\nThe Canonical distance Earth to Sun is: 93312000 divided by the Canonical distance Earth to Moon of 237600 = 392.7272 the prior 1162.5 x 0.96 = 1116 Bereshit Bara The 1162.5 ÷ 20.61818 = 56.38227513 this divided by the 2.8282 then multiplied by 1016064 = 20255400 which is 5280 x 3836.25 which is the 1116 x 3.4375 that multiplied by Gem: Mary: 192 = 660 the modern-day furlong\n\nThis 20255400 is the 2nd Pyramid base side length: 8474.625 multiplied by 2390.123456790 that multiplied by 81 = 193600 that divided by the square feet in one acre: 43560 = 4.444 one ninth of 40\n\nThe 3rd Pyramid\n\nThe 3rd Pyramid base side length of 4154.625 ÷ 20.61818201.5029761904 this is 63.7568010602… x 3.160493827 the Rhind Papyrus fraction {256:81} the 63.7568010602… is the Schooldays Pi {22 ÷ 7} of 3.142857 x 20.2862548828125\n\nThis 20.2862548828125 multiplied by a Comma of Pythagoras {531441:524288} component, namely 524288, then by the 1st Pyramid height of 480 = 5105203200 which is the 3rd Pyramid base side length of 4154.625 x 1228800.\n\nAmple options here, seeing as 1228800 is, for example 1166400 times the FP {256:243} the 1166400 is the seconds in 24 hours: 86400 x 13.5. otherwise the 1166400 is the mile of 5280 x 220.9090 this multiplied by 3.3 = 729 which is the square root of the Comma component {531441:524288} namely 531441\n\nThe 3rd Pyramid height 2578.8 ÷ 20.61818 = 125.074074 this multiplied by 243 = 30393 that divided by the mile of 52805.75625 The Pythagorean Limma {256:243} multiplied by the 5.75625 x 7776 = 47155.2 which is the 2578.8 x 18.285714 that multiplied by 21 = 384, the 1st PWS member and 7776 is the 32nd PWS member\n\nThe 3rd Pyramid Tan will be:\n\n4154.625 ÷ 24 = 173.109375 then 2578.8 ÷ 12 = 214.9……..In that case 214.9 ÷ 173.1093751.2414116797544904 Tan which produces a Slant Angle of: 51.147355523771... that converts to: 51 8′ 50.47″\n\nWith regard to the Petrie data…By 9 pieces of limestone casing (brought to England} 51º 58′ ± 15″…”\n\n“But from a consideration of the granite courses (see below), the angle would be 51º 10′ 30″ ± 1′ 20″; and this might well be adopted, as being close to the very uncertain result from the measured angles”\n\n{End of Petrie data}\n\nThe 1st Pyramid Tan of: 1.269841 divided by the 3rd Pyramid Tan 1.2414116797544904 … then multiplied by 196.991666 = 201.5029761904 as we observe prior The 196.991666 x 13.0909 = 2578.8 the 3rd Pyramid height, the 13.0909 x 11 = 144\n\nThe Giza area etc\n\nEmploying the identical “cubit” as prior: 20.61818 which is 12 x 1.71818 the Egyptian cubit\n\nThe 1st Pyramid height of 5760 ÷ 20.61818 = 279.365079\n\nThe 2nd Pyramid height of 5670 ÷ 20.61818 = 275\n\nThe 3rd Pyramid height of 2578.8 ÷ 20.61818 = 125.074074\n\nThen 279.365079 + 275 + 125.074074 = 679.439153 {Please see below}\n\nThe 1st Pyramid base side length of 9072 ÷ 20.61818 = 440\n\nThe 2nd Pyramid base side length: 8474.625 ÷ 20.61818 = 411.0267857142\n\nThe 3rd Pyramid base side length of 4154.625 ÷ 20.61818201.5029761904\n\nSubsequently 440 + 411.0267857142 + 201.5029761904 = 1052.529761904 this minus the prior 679.439153 = 373.090608465 that multiplied by 273030912 = 101865269088 this is 16754156.1 Sea miles of 6080\n\nThe 16754156.1 is 720896 x 23.24073944091…this multiplied by a Comma of Pythagoras {531441:524288} component, namely: 524288 = 12184840.8 this divided by the 373.090608465 then multiplied by 2520 = 82301184 the square root of this is 9072\n\nThe employed 273030912 is, for example, 1881 x 145152, the shorter value Roman cubit is 1.45152 The 1052.529761904 is 333.02699497…x 3.160493827, which is the Rhind Papyrus fraction {256:81}\n\nThis 333.02699497… is, for example the Egyptian unit {12 ÷ 7} of 1.714285 x 194.2657470703125 this divided by 1.375 then multiplied by Pi = 443.857340974…this is 1.215154474…x …365.268243926 admittedly a rather shoddy Earth year, nevertheless passable given that we employ actual Pi\n\nThe 1.215154474…x 720896 = 876000 this is 5760 x 152.08333 that divided by the 1.375 = 110.6060, the 720896 is the seconds in 24 hours: 86400 x 8.34370370 that multiplied by 690.340909 = 5760, there are many options re this 690.340909 Gem: Abraxas, Mithras 365 ÷ 110.6060 = 3.3, the Maya Synodic period of Venus 584 ÷ 110.6060 = 5.28, the Sothic cycle: 1460 ÷ 110.6060 = 13.2\n\n……………………………………\n\nThe 1.375 divided by 567 then added to the 1052.529761904 = 1052.5321869488… this multiplied by Gem: Tree of Life: 1625 x 9072 = 15516429500 this multiplied by 2.153846 = 33420002000 Gem: Tree of Life: 1625 x 9072679.43915310016292000\n\nPrior we see:”Then 279.365079 + 275 + 125.074074 = 679.439153 {Please see below}…”\n\nGem: Founder of the City: 1225 multiplied by the Platonic: 25920 = 31752000 this multiplied by the 679.439153 = 21573552000 which is the 10016292000 x 2.153846 this multiplied by Gem: YHWH: 26 = 56, ample scope\n\nThe 31752000 x 1052.5321869488… = 33420002000 this divided by the square feet in one acre; 43560 = 767217.6767 this divided by 728.925619834… = 1052.5321869488…\n\nThe 767217.6767 ÷ 20459.138047 = 37.5 that divided by √3 = 21.6506350946… hence Gem: City of god: 882 divided by the 21.6506350946… then divided by 23.52 = √3 The 20459.138047 x 2079 = 42534548 this is the 1052.5321869488… x 40411.6363 this is 40320 Greek feet {441:440} of 1.0022727, the 2079 x 4.3636 = 9072, the 4.3636 x 1320 = 5760 1st Pyramid\n\nThe 23.52 relates to the Grand Gallery 7-corbelled perpendicular height of 235.2, which is apt, in view of the fact that the Gallery leads to the hypothetical “Kings Chamber” of wall height: 235.2\n\nOr 1st Pyramid base side length of 9072 ÷ 21.650635094… ÷ 241.92 = √3 Otherwise the 1st Pyramid height of 5760 ÷ 21.650635094… ÷ 153.6 = √3 the 13th PWS member is 1536\n\nThen Gem: YHWH: 261731.971340388…÷ 84.0021373204…= 536.07272 one eleventh of 5896.8 that divided by the Platonic Precession of 25920 = 0.2275 that multiplied by 1600 = 364, pertinent to the Maya temple steps: there being 91 each side in addition to 91 x 4 = 364 + 1 {the top platform-step} obviously obtains 365\n\nThe 679.439153 multiplied by Gem: YHWH: 26 x 377622 = 6670850472 this is the prior 10016292000 x 0.666 Mr. Nasty: 666 albeit reduced The 377622 is 666 x 567 one tenth of the 2nd Pyramid height of 5670 furthermore the 1st Pyramid height of 5760 is 10 x 576 Gem: Eagle, Pneuma {Breath}\n\nAccordingly most apt is that: 576 ÷ 567 = 1.015873 which is the Eye of Horus fraction reversed {63:64} to read as 64 ÷ 63 = 1.015873 The 1.015873 multiplied by 1.25 = 1.269841 the 1st Pyramid Tan {480 ÷ 378}\n\nThe 728.925619834… x 19 x 1881 x 660 = 17193708000 which is the seconds in 24 hours 86400 x 199001.25 which divided by Gem: Weapons of god: 665 = 299.25, this 299.25 x 120 = 35910 this is the Sea mile of 6080 x 5.90625\n\nThe 5.90625 x 1536 = 9072, above we observe: 153.6 furthermore the 13th PWS member is 1536, the 5.90625 x 975.238095 = 5760 The 975.238095 divided by the Rhind Papyrus {256:81} of 3.160493827 = 308.571428 one seventh of 2160\n\nNorth to South distance\n\nAs we observe prior:\n\n“As a result 679.439153 + 1052.53218694885361552028 = 1731.971340388007054…”\n\nHence North to South = 2976.278125 feet divided by the 1731.971340388007054… = 1.718433819079… this multiplied by the cubic inches in one cubic yard: 46656 then divided by 219.512512 = 365.2422695…a tolerable Earth year that multiplied by 360000 ÷ 5280 = 24902.8820142…a tolerable Earth Equatorial circumference {miles}\n\nThe 219.512512 multiplied by the 4th PWS member: 5121595602.8 = 179330447155.2 this divided by the square feet in one acre: 43560 = 4116860.58666 which is one third of 12350581.76 this divided by the Greek mile of 5068.8 = 2436.5888 which is 11.1 x 219.512512\n\nThen 75917.648648 x 1731.971340388007054… = 131487191.6891057724…this is 5280 x 24902.8772138…a tolerable Earth Equatorial circumference {miles}\n\nOr 131487191.6891057724… is 360000 x 365.2421991…a tolerable Earth year that multiplied by 5280 ÷ 360000 = 24902.8772138… another tolerable Earth Equatorial circumference The 75917.648648 x 14519.98548 = 1102323156.05412 this divided by the above 1595602.8 = 690.8506027027 that multiplied by 109.890109 = 75917.648648\n\nThe mile of 5280 ÷ 109.890109 = 48.048 exactly which is 48 x 1.001 The employed 14519.98548 is 10.01 x 666 x 2.178 that multiplied by 20000 = 43560, square feet in one acre\n\n………………………………\n\nAs we observe prior:\n\nI calculated the three Giza Pyramids enclosed within a “rectangle” of distance\n\nNorth to South: 2976.278125 feet\n\nEast to West: 2435.778125 feet\n\nHence area of: 2976.278125 x 2435.778125 = 7249553.150791015625 square feet this creates a diagonal of: 3845.938968780189997… ths multiplied by exactly 34188.58 = 131487192.1092590281…which divided by the mile of 5280 obtains another Earth Equatorial circumference of: 24902.8772934… that divided by 68.1818 {360000 ÷ 5280} obtains an Earth year of 365.2422003…\n\nThere are 4840 square yards in one acre\n\nThe 7249553.150791015625 ÷ 4840 = 1497.8415600807..that divided by the Egyptian cubit of 1.71818 = 871.75963814.. this divided by the Rhind Papyrus {256:81} of 3.160493827 afterwards divided by the Schooldays Pi {22 ÷ 7} of 3.142857 = 87.76415391098… this divided by the Egyptian foot of 1.14545 = 76.61949944609…\n\nIn that case the 7249553.150791015625 ÷ 76.61949944609… = 94617.6 this divided by the mile of 5280 = 17.92 that divided by Gem: Founder of the City: 1225 = 0.0146285714 which is one seventh of 0.1024, the 9th PWS member is 1024\n\nConversely the 1st Pyramid base side length of 9072 + 1 = 9073 ÷ 3870 = 2.3444 which is exactly 1.31875 x 1.777 which is the square root of the “Rhind” {256:81} and the “Rhind” divided by the Eye of Horus fraction reversed: {64:63} of 1.015873 = 3.111 that multiplied by 2.25 = 7\n\nThe 1.31875 is 1.125 x 1.17222 that multiplied by 7740 = 9073 the 1.125 x 8 = 9, otherwise the Comma of Pythagoras {531441:524288} of exactly 1.0136432647705078125 multiplied twice by the Pythagorean Limma {256:243} also obtains 1.125\n\nThe employed 3870 divided by half the 1st Pyramid base side length {feet} of 378 = 10.238095 that divided by the “Rhind” of 3.160493827 = 3.2393973314…which is 3.142857 x 1.030717329.. that divided by the Egyptian foot of 1.14545 = 0.89983258928… that divided by 1.714285 {12 ÷ 7} obtains: 0.52490234375\n\nThe Comma component, namely 524288 x 0.52490234375 = 275200 that divided by the employed 3870 = 71.111 which is one ninth of 640, the acres in one square mile The 71.111 2.1109375 = 150.111 this we observed prior, in addition to the 2.1109375 minus the: 0.52490234375 = 1.58603515625 that multiplied by 256 = 406.025 Please make a note of the 2490234375 within this 0.52490234375 given that we will employ this presently\n\nThe 406.025 + 5.88125 = 411.90625 as we will observe below The 5.88125 minus 5.405 = 0.47625 that multiplied by 6400 = 3048, which is 10000 x 0.3048, this is the feet-metres ratio number: 1 ÷ 0.3048 = 3.28083989501312…feet in one metre\n\nThe difference with reference to East to West = 2435.778125 feet in addition to North to South2976.278125 is 540.5 feet this we observe above albeit reduced as: 5.405 it would be trouble-free had we obtained exactly 540, nonetheless we cannot alter the measurements mid stream In any case the 540.5 minus a reduced mile of 528 = 12.5 this is one fourth of 50 Otherwise 540.5 ÷ 259.2 + 3 = 5.08526234567901 that multiplied by 81 = 411.90625 as we observe above\n\nI mentioned prior: “Please note the 2490234375 within this 0.52490234375…”\n\nSubsequently 540.5 minus the 0.52490234375 = 539.97509765625 Followed by 540 minus 539.97509765625 = 0.02490234375 that multiplied by 1000000 = 24902.34375 a tolerable Earth Equatorial circumference {miles} that multiplied by 5280 then divided by 360000 = 365.234375 a tolerable Earth year Then again the 365.234375 x 360000 = 131484375 that minus the NC {Numerical Canon} Earth mean of 131383296 = 101079 this plus 6921 = 108000 an increased lunar, Yin-Yang number, the employed 6921 is a reversed 1296 which is the 11th PWS member\n\nTherefore 6921 minus 1296 = 5625 which is 360 ÷ 0.064 otherwise 5625 divided by the Maya Tzolkins: 5256 = 1.07020547945 The prior Earth year of 365.234375 divided by this 1.07020547945 = 341.275 that multiplied by the FP {256:243} and Rhind {256:81} component 256 = 87366.4 The NC Earth mean circumference is 24883.2, this 87366.4 minus 24883.2 = 62483.2 that minus 24883.2 = 37600 which is 1600 x 23.5, the Earth tilt?\n\nThen again, to be brief {if feasible} the 31st PWS member is 6912, the 35th member is 9216 moreover as is evident the above 6921 is these members rearranged as it were Accordingly 9216 minus the 6921 = 2295 the Platonic Precession {25920} reduced to 2592 minus 2295 = 297 this multiplied by 800 = 237600, this is the NC Earth to Moon distance\n\nThe NC Earth to Sun distance: 93312000 ÷ 297 = 314181.8181 a “Pi” version of 3.141818 reduced in addition to 3.141818 x 27500 = 86400 seconds in 24 hours Or 479001600 {=12!} divided by 297 = 1612800 which is the 1st Pyramid base side length of 9072 x 177.777 which is 100 x 1.777 which is the square root of: 3.160493827, the Rhind Papyrus {256:81} A straightforward method is 6921 minus 6912 = 9\n\nLikewise the aforementioned: 71.111 + 79 = 150.111 then again 1.777 6921 = 12304 that plus 86.4 = 12390.4 This 12390.4 ÷ 12250 = 1.01146122448… an ancient ratio of measure in view of the fact that its square root is 1.00571428 {7.04 ÷ 7}\n\nPrior we observe “9072 + 1 = 9073 ÷ 38702.3444\n\nThe Sothic Cycle: 1460\n\nThe 2.3444 x 1460 x Pi x 21.702946785… = 233378.7878 this multiplied by 666 x 11 = 1709733000 this divided by the Maya Calendar: 1360800 = 1256.4175485008… that divided by the 71.111 = 17.668371775793650 this multiplied by the Maya Synodic period of Venus: 584 afterwards by the Gallery 1881 x 7 = 135861439.484375\n\nThis 135861439.484375 divided by the Maya “super number” 1366560 then by the 2.3444 = 42.406024639… that multiplied by Gem: YHWH: 26 = 1102.556640625 which multiplied by 256 = 282254.5 which is Gem: Abraxas, Mithras: 365 x 773.3 this minus the Maya Synodic period of Venus: 584 = 189.3 that multiplied by 48 = 9086.4, this minus the 1st Pyramid base side length of 9072 = 14.4\n\nThe 150.111 divided by Pi ÷ Phi29.530807773461… a tolerable Synodic Month, or 150.111 divided by 5.083203692307 = 29.5308077735…another Synodic Month The 5.083203692307 is Pi x 1.61803398874…this is Phi correct to 11 decimal places the 5.083203692307 multiplied by Gem: YHWH: 26 = 132.163296 this minus 0.78 = 131.383296 which is the Canonical mean Earth circumference: of 131383296 reduced\n\nAs a result 233378.7878 ÷ 1460 = 159.8484 that divided by the 2.3444 = 68.1818 {360000 ÷ 5280} Obviously 68.1818 ÷ Pi = 21.702946785… The 159.8484 divided by the 71.111 = 2.24786931818 one eleventh of exactly 24.7265625\n\nWe observe prior: “2.3444 x 1460 x Pi 21.702946785… = 233378.7878…”\n\nOtherwise Gem: Abraxas, Mithras: 365 x Pi 21.702946785… = 24886.3636 one eleventh of 273750 which is the mile of 5280 x 51.84659090 that divided by the Egyptian foot of 1.14545 = 45.2628968253 that multiplied by the 365 then by the 1st Pyramid base side length of 9072149878125\n\nOr the 24886.3636 minus the NC Earth of 24883.2 = 3.16363 that multiplied by 797.5 = 2523, this multiplied twice by Pi = 24901.011903… a tolerable Earth Equatorial circumference The 797.5 minus 365 twice = 67.5 that multiplied by 16 = 1080 the lunar, Yin-Yang number This 149878125 divided by Gem: Simon Peter: 1925 then multiplied by the Grand Gallery sloping floor line length of 1881 x 7 = 1025166375 which is the Sothic Cycle of 1460 x 702168.75 that divided by the 365 = 1923.75 and this plus 1.25 = 1925\n\n………………………………\n\nAs we observe prior: “17.668371775793650 this multiplied by the Maya Synodic period of Venus: 584 then by the Gallery 1881 x 7 = 135861439.484375”\n\nGem: Tree of Life: 1625 x 17.668371775793650 … ÷ 2.3444 = 12246.44252232… Gem: Founder of the City: 1225 x 12246.44252232…= 15001892.08984375 exactly, which is the Earth of 24886.3636 x 602.8157552…one third of 1808.447265625\n\nThen 1225 ÷ 602.8157552…= 2.032130 that multiplied by exactly 259.259526.848 the longer value Egyptian foot of 1.152 plus 526.848 = 528 one tenth of a modern day mile\n\nThe 1808.447265625 multiplied by the FP {256:243} then by the Tetractys number total: 8436 = 16072229.012345679 that multiplied by 81 = 1301850550 this divided by the 602.8157552… = 2159616 which divided by 5280 = 409.01818 that divided by 68.1818 = 5.9989333\n\nThe 68.1818 x 5280 = 360000, the 1808.447265625 divided by this 5.9989333 then multiplied by 19 = 5727.7679443359375\n\nWe observe prior that: 68.1818 ÷ Pi = 21.7029467852...this multiplied by 16.829152397…= 365.2421989187… this being the Earth year I normally employ obtained via: 343094400 ÷ 299008 ÷ Pi = 365.2421989187… this multiplied by the 21.7029467852…= 7926.83200686…\n\nThe 16.829152397… x 299008 = 5032051.2 which is the Sea mile of 6080 x 827.64 In that case 1881 ÷ 827.64 = 2.2727 that multiplied by 0.44 = 1 {one} This 7926.83200686… is the square root of 62834665.6650482… that divided by the Pythagorean Minor Third {32:27} of 1.185185 = 53016749.1548844… that divided by Pi afterwards by the Sea mile of 6080 = 2775.617662718…which is 7.6 x 365.21285035\n\nAdmittedly a somewhat shoddy year Nonetheless 365.21285035… x 68.1818 = 24900.8761607… a tolerable Earth Equatorial circumference {miles} the 7.6 x 800 = 6080 the Sea mile\n\nNC {Numerical Canonical} Earth\n\nThe NC Earth mean diameter is 24883.2 The 1st Pyramid base perimeter of 3024 x 3 = 9072 otherwise 3024 x 120 = 362880 Subsequently 1881 x Pi ÷ Phi x 36000 ÷ 5280 = 24901.161571654…a tolerable Earth Equatorial circumference {miles}\n\nThen 28 ÷ Pi x 2793.90379497… = 24901.161571654…The 2793.90379497… x Phi ÷ Pi ÷ Pi = 458.03571428 which is one seventh of 3206.25 Then 362880 ÷ 3206.25 x 1881 = 212889.6 which is the 24883.2 x 8.555 that multiplied by exactly 0.9 = 7.7\n\nThe 1st Pyramid base side length of 9072 divided by a “cubit” of 20.625 = 439.85454 which divided by the lead PWS member: 384 = 1.14545 the Egyptian foot Otherwise 439.85454 ÷ 256 = 1.71818 the Egyptian cubit\n\n………………………………..\n\nPi in the company of Phi\n\n5.083203692307 ÷ Pi = 1.61803398874Phi correct to 11 decimal places\n\n5.083203692307 ÷ Phi = 3.14159265358Pi correct to 11 decimal places\n\nThe 5.083203692307 multiplied by Gem: YHWH: 26 and then by 5256 Maya Tzolkins obtains 694650.283776 which is the NC Moon diameter of: 11404800 ÷ 16.4180455494892…\n\nThe 16.4180455494892… x 8049879.969063…= 132163296 exactly which is 5280 x 25030.92727 this divided by the 5.083203692307 = 4924.2424 that multiplied by Gem: Eve: 19 afterwards by the modern-day furlong of 660 = 61750000 which is Gem: Tree of Life: 1625 x 38000 which is the Sea mile of 6080 x 6.25 which is one fourth of 25\n\nIn closing, the Nineveh Constant reduced to 1959552\n\nSubsequently 1959552 x 1881 ÷ 6080 = 606236.4 which divided by a reduced Maya: 13068 = 46.39090 that multiplied by 5256 Maya Tzolkins obtains: 243830.61818 that divided by the NC Earth of 24883.2 = 9.79900568181 that divided by the Egyptian cubit of 1.71818 obtains exactly 5.703125 this multiplied by 256 = 1460, the Sothic cycle\n\nOtherwise the Maya Synodic period of Venus: 584 ÷ 5.703125 = 102.4, the 9th PWS member is 1024 Or else Gem: Abraxas, Mithras: 365 ÷ 5.703125 = 64 Otherwise Gem: Son of man: 2190 ÷ 5.703125 = 384 the lead PWS member\n\nUntil the next essay"
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http://www.dg7ybn.de/AG/KF2YN_AG_correction_online_calc.htm | [
"DG7YBN / Average Gain Correction per KF2YN Online Calculator Last Update Nov. 13th 2019\n\n.......\n\nEnter Numbers\n\nEnter Gain in dBi or dBD\n\nLossless Data\n T-pattern [K] ∗ AG Gain\n\nLossy Data\n T-pattern [K] ∗ AG Gain\n\nUse T_pattern, not T_total in here!\n\nCorrected Data\n\n Gain [dBi] Gain [dBD] T-loss [K] T-ant, total G/T [dB]\n\nSet of formulas author is KF2YN\nsee Dubus 4/2010\n\nFor using this calculator Javascript must be activated\n\n∗ ) pure gain correction is fine without antenna temperatures filled in\n\nUse numbers produced by TANT\n\nAG = Average Gain\n\nLossless = data from simulaton with no losses in wires\n\nLossy = data from simulaton with wire losses included\n\n ....... Fundamental knowledge about Average Gain • Average Gain (AG) is factorised Radiation Efficiency • If Radiation Efficiency is 98.7% then AG is 0.987 • Radiation Efficiency is related to sum of gain per radiaton patterns segments over Phi and Theta • Phi (= φ) and Theta (= θ) are the radation patterns azimuth and elevation referencing angles • Radiation Efficiency is percentage of power found in whole 3D radiation pattern against fed in power • Ideal simulation outcome produces 100 percent of Radiation Efficiency when using lossless wires • Thus aim of the \"AG-Test\" (AGT) is to find out if the lossless design busts 100% of Radiation Efficiency = AG > 1.000 • AG_lossless = AGT > 1.000 would mean that we radiate more than we feed; which can not be ... • Which shows that the model, as computed by the NEC-kernel has a convergence issue • Most simulations show slighty diverted outcome i.e. Power Gain > or < 100 percent • Using KF2YN's AG correction allows to standardises actual simulation outcome to 100.0 percent = AG = 1.000 • Thus key figures like Antenna Gain and Temperature, which result in G/T can be lifted on a standardised AG-level • This we appreciate when comparing antenna designs like in the VE7BQH G/T tables The set of formulas put up by KF2YN: The set of formulas shown by KF2YN in Dubus 4/2010 in the way I write them today:",
null,
"Note: • The last one for T_loss.corr was added by him later, was not part of the set shown in Dubus. • num = numerical value • lossless = means value from antenna model with wire losses put to zero • lossy = menas value for antenna model with wire losses put to specific used material • AVG = Average Gain as numerical value What does a typical AG overshoot look like? Neither NEC2 nor NEC4 will get the tapering that right that the AG fits to 1.000 absolutely. Not so for the untapered model. So this experiment offers a view on how much numbers go stray on a design with an 'imperfection' that can be switched on/off by just editing the Ø of the middle section of the DE. The image shows same Yagis pattern with tapered and not tapered DE",
null,
"Do not get confused reading the table below, AG(T) is specified for computed with lossless wires. The shown numbers are for comparism only. Typical AGs for Yagi designs without convergence issue when using lossy wires are around 0.97x ... 0.98x. The delta to an AG(T) = 1.000 relects wire losses, we find these expressed as T_loss in Antenna Temperature computations like when using TANT. T_loss is derived by the difference between Radiation Efficiency with and without wire losses applied. A negative T_loss means a gain. That can not be. This is yet another signal for the convergence overshooting the 100%. ``` tapered not taperd AG (lossless) 1.051 1.000 AG (Aluminum) 1.030 0.981 Gain (lossless) 15.89 dBi 15.68 dBi Gain (Aluminum) 15.81 dBi 15.60 dBi Single Yagis Antenna Temperature and G/T - ratio T_loss -8.8 K 5.6 K T_ant 230.3 K 233.2 K G/T -7.81 dB -8.08 dB and with KF2YN's correction applied on the tapered version T_loss = 5.6 K T_ant = 233.16 G/T = -8.09 ```",
null,
"What designs may hold a convergency issue? • any tapering is highly likely to cause an AGT > 1.03 • any sharp bends are likely to cause an AGT > 1.01 Unless manipulated segment densities are used; such as using much lower segmentation density on the bends, which (a) is not recommended acc. NEC manual and (b) just works out since the convergence of whatever model is less at lower overall segmentation density (refering to less than approx. 12 segments per half wave). How to check the Average Gain number • EZNEC by Roy Levallen, W7EL Edit 'Wire Loss' to 'zero', run 3 dimesional pattern. EZNEC shows the AG when doing a 3 D pattern in its main windows bottom line",
null,
"• 4nec2 by Arie Voors Edit NEC LD-card, set wire losses to zero and run 'Full Far Field Pattern'. Using older versions of 4nec2 we have to chose calculate in the main menu, then NEC output-data and chose Far Field pattern, opt for Full pattern, click expert settings and select option 1: average gain. From version 5.8.11 on 4nec2 shows the Radiation Efficiency with one digit behind the decimal separator (tnx Arie!). This Radiation Efficiency is equal to Average Gain when divided by 100.",
null,
"The example shows a design with 101.4% of Radiation Efficiency equalling an AG of 1.014 How to edit the LD card: Set 4nec2 to use the \"NEC editor (new)\" as Geometry Editor Menu -> Settings -> NEC editor (new) // Ctrl+F4;",
null,
"Set all wires to \"Perfect\" = lossless for AGT or ... use AGT function, which will set all wire losses to zero during the test run automatically",
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"",
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"",
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"73, Hartmut, DG7YBN",
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http://h14s.p5r.org/2012/10/difference.html | [
"# Difference\n\nThis is my second post in a series about Babbage’s mechanical calculating engines. The first post was about why it was so important in the early 1800s to be able to produce accurate arithmetic tables and ended with what I’ll be generous and call a cliffhanger: that Babbage’s difference engine was a groundbreaking solution to that problem. In this post I’ll explain how the difference engine works, how to “code” it and how to interpret what you get back.\n\nBefore we get to the engine itself there’s a bit of math to explain the basic principles it works by. To give you a taste of the engine itself before that here is a small JavaScript emulator1 that performs a simple calculation. It’s tiny compared to the real engine but works by the same principles just on a smaller scale.\n\nTry stepping it a few times. The model is actually quite straightforward and the computations you can use it for are based on some basic mathematical principles. You can treat it as an exercise before reading the rest if you like: try to work out what it’s calculating, how it does it, and how you can make it calculate other things. Of course you can also just read on an I’ll tell you.\n\n## Differences\n\nThe difference engine is a very simple device in principle. It’s an adding machine, that’s all it is2. Part of what makes it such an impressive accomplishment is that it took a difficult problem, calculating complex functions, and solved it using a simple enough approach that it could be implemented mechanically using contemporary technology3.\n\nSo, how do you calculate complex functions using just addition? Obviously you can’t in general so let’s start off easy by looking just at polynomials.\n\nConsider this simple polynomial, the square function:",
null,
"$f(x) = x^2$\n\nThe first few values are",
null,
"$\\begin{tabular}{lcl} f(0) & = & 0 \\\\ f(1) & = & 1 \\\\ f(2) & = & 4 \\\\ f(3) & = & 9 \\\\ & \\vdots & \\end{tabular}$\n\nThe difference engine is based on a technique called divided differences. Divided differences is similar to differentiation but simpler and it’s based on simple arithmetic. It works as follows. You take the values of your polynomial at a fixed interval, here we’ll use the first four values from before:",
null,
"$\\begin{tabular}{l} 0 \\\\ \\\\ 1 \\\\ \\\\ 4 \\\\ \\\\ 9 \\end{tabular}$\n\nThen you find the distance between each successive pair of them:",
null,
"$\\begin{tabular}{l|l} 0 \\\\ & 1 \\\\ 1 \\\\ & 3 \\\\ 4 \\\\ & 5 \\\\ 9 \\end{tabular}$\n\nThese are the first differences. Then you find the distance between the first differences the same way:",
null,
"$\\begin{tabular}{l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 \\\\ 4 && 2\\\\ & 5 \\\\ 9 \\end{tabular}$\n\nThese are the second differences. And one last time to get the third differences:",
null,
"$\\begin{tabular}{l|l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 && 0\\\\ 4 && 2\\\\ & 5 \\\\ 9 \\end{tabular}$\n\nYou can see the similarity to differentiation: it’s a polynomial of degree 2 so the first differences increase linearly just like the first derivative, the second differences are constant just like the second derivative, and so on. We don’t actually need the third differences, they’ll all be 0 anyway, so I’ll leave those out below.\n\nWhat’s neat is that once you have these values you can extend the table using nothing but addition. You know the difference between the first derivatives is fixed, 2, so you can get the next first derivative by adding 2 to the previous one. And you know the difference between the function values is the first differences so you can get the next value just by adding the next first difference to the previous function value. Okay maybe it’s easier to explain with a concrete example:",
null,
"$\\begin{tabular}{l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 \\\\ 4 && 2 \\\\ & 5 \\\\ 9 \\end{tabular} \\quad\\to\\quad \\begin{tabular}{l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 \\\\ 4 && 2 \\\\ & 5 & \\tiny{+0} \\\\ 9 && \\bf{2} \\end{tabular} \\quad\\to\\quad \\begin{tabular}{l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 \\\\ 4 && 2 \\\\ & 5 \\\\ 9 & \\tiny{+2} & \\bf{2} \\\\ & \\bf{7} \\\\ \\end{tabular} \\quad\\to\\quad \\begin{tabular}{l|l|l} 0 \\\\ & 1 \\\\ 1 && 2\\\\ & 3 \\\\ 4 && 2 \\\\ & 5 \\\\ 9 && \\bf{2} \\\\ \\tiny{+7} & \\bf{7} \\\\ \\bf{16} \\\\ \\end{tabular}$\n\nNotice that we don’t need the full table at this point, we only need that for calculating the initial values. All we need to generate more values is the last of each of the differences:",
null,
"$\\begin{tabular}{l|l|l} \\multicolumn{1}{r}{}&&2 \\\\ & 7 & \\tiny{+0}\\\\ 16 &\\tiny{+2}& 2 \\\\ \\tiny{+9} & 9 \\\\ \\bf{25} \\\\ \\end{tabular} \\quad\\to\\quad \\begin{tabular}{l|l|l} \\multicolumn{1}{r}{}&&2 \\\\ & 9 & \\tiny{+0}\\\\ 25 &\\tiny{+2}& 2 \\\\ \\tiny{+11} & 11 \\\\ \\bf{36} \\\\ \\end{tabular} \\quad\\to\\quad \\begin{tabular}{l|l|l} \\multicolumn{1}{r}{}&&2 \\\\ & 11 & \\tiny{+0}\\\\ 36 &\\tiny{+2}& 2 \\\\ \\tiny{+13} & 13 \\\\ \\bf{49} \\\\ \\end{tabular} \\quad\\to\\quad\\dots$\n\nThis provably works for any polynomial. To generate a sequence of values for a polynomial of degree n all you need is n+1 initial values; from the values you calculate the table of differences using subtraction, and from there on you can calculate as many successive values as you like using just addition. You don’t even need to know the closed form of the polynomial as long as you can evaluate the initial values at fixed intervals.\n\nThis is the basic principle of the difference engine, and what it’s named after. The engine has 8 integer registers called columns that can each hold a 31-digit integer value which represent the current value of the function and the first to the seventh difference. By cranking the handle those values are added together from right to left. Here is a another mini-emulator, this one calculating the square function using the differences we just calculated:\n\nYou can see the values are being added together from right to left and the current function value in the leftmost column is printed for each emulated crank on the handle. Printing was also a part of the original difference engine. A common source of errors in mathematical tables was typesetting so to avoid that step the engine would automatically stamp its output in a soft material that could be used directly for printing, as well as print a log on paper.\n\nBeing able to evaluate an integer polynomial is just the beginning though. First of all, integers aren’t enough, we need to be able to evaluate real-valued functions. Secondly, so far we’ve only seen positive values, we also need negatives. Finally, polynomials can be useful in themselves but we’re really more interested in more complex functions like logarithms and trigonometric or astronomical functions. But with a few tricks the difference engine can handle all those things.\n\n## Precision\n\nFirst off: how do we use this to evaluate real-valued functions? You use fixed-point decimal numbers. For instance, say we want to plot the square function from before but this time in steps of 0.25:",
null,
"$\\begin{tabular}{lcl|l|l} f(0) & = & 0 \\\\ &&& 0.0625 \\\\ f(0.25) & = & 0.0625 && 0.125 \\\\ &&& 0.1875 \\\\ f(0.5) & = & 0.25 \\end{tabular}$\n\nThese are fractional numbers but if you multiply them by 105 we’re back to integers",
null,
"$\\begin{tabular}{lcl|l|l} 10000 f(0) & = & 0 \\\\ &&& 625 \\\\ 10000 f(0.25) & = & 625 && 1250 \\\\ &&& 1875 \\\\ 10000 f(0.5) & = & 2500 \\\\ \\end{tabular}$\n\nNow we’re back to something the engine can calculate:\n\nI’ve added a line between the digits to mark where the decimal point goes. It also gets added to the output (I don’t believe the original engine did this). But the decimal point is purely a matter of interpretation by the operator, the adding mechanism is not aware of it, it’s operating purely on 6-digit integer values.\n\nIn this case we were lucky because there was a relatively small factor of 10 we could multiply onto the values to get integers without losing precision. That’s unlikely to be the case in general. If you can’t use this trick you multiply the values with as large a factor of 10 as you have precision for and just bite the bullet, round the scaled values to the nearest integers and lose some accuracy. That’s not necessarily as bad as it sounds. The original design had 31 digit precision in base 10 which corresponds to roughly 103 bits, well beyond the already quite large 64-bit integers on most modern machines. So you can afford to scale the values up quite a bit before rounding. We’ll see an example in a bit of how long it takes for errors to accumulate.\n\n## Negative values\n\nTo represent negative values we use the exact same trick as with binary numbers, just in base 10: tens complement. A negative number d is represented as 10n – d where n is the number of decimal digits, so -1 is represented by 999…999 and so forth. The adding mechanism itself has no concept of negative values, just like with twos complement the correct behavior just falls out of how overflow works. It’s up to the operator or printer to interpret the output correctly as signed or unsigned values.\n\nHere is an example of a function that starts off positive, peaks, and then descends into the negative numbers.\n\nYou’ll notice that the numbers in the columns are all positive but the ones that represent negative values are printed as negative. As with the decimal point that’s a hack I added in the printer which makes it print smaller integer values as positive and larger ones as negative. But it’s purely a matter of interpretation, the calculating part of the engine is oblivious.\n\n## Polynomial approximation",
null,
"Okay, now we’re getting close to the goal: being able to produce accurate arithmetic tables of general functions.\n\nIn my previous post the main example of how important mathematical tables were was how astronomical tables were used by mariners to navigate by lunar distance. I don’t actually know the math underlying those astronomical tables so here I’ll use an example I do understand: trigonometric functions. On the right here is a table of trigonometric functions from 1785. It gives 7 digits of the values of 8 different trignonmetric functions, sin, cos, tan, etc., for each arcminute between 0° and 45°. There’s 60 arcminutes to one degree so that’s 2700 values for each function, 21,600 values in total. The author of this table, Charles Hutton, said about this edition in a later one that it had so many errors that is was basically useless:\n\nFinding, as well from the report of others, as from my own experience, that those editions […] were so very incorrectly printed, the errors being multiplied beyond all tolerable bounds, and no dependence to be placed on them for any thing of real practice […]\n\nFor this last part about how to calculate complex functions I’ll show how to replicate one column, sin, of this table.\n\nSince we know how to evaluate polynomials the solution that first springs to mind is to approximate the function we want to calculate using a polynomial. Taylor polynomials were well known at this time so that’s an obvious approach. Taylor’s theorem says that for an infinitely differentiable function f (which sin is),",
null,
"$f(x) = \\sum_{k=0}^{\\infty}\\frac{f^{(k)}(a)}{k!}(x-a)^k$\n\nwhere f(k) means f differentiated n times and a is any point on the function. Since the engine only has 8 columns and we need n+1 columns for a polynomial of degree n we have to limit ourselves to at most the first 7 terms. And in fact, since I want to demonstrate this with the emulator in a little bit and 8 columns of 31 digits takes up an enormous amount of space we’ll limit ourselves even more in this example to 4 columns of 13 digits. This means that we’ll use only the first 3 terms of the Taylor polynomial. For sin those are:",
null,
"$\\sin(x) \\approx x - \\frac{x^3}{3!}$\n\n(Conveniently all the even degree terms are 0). This approximates sin quite well around 0 so we’ll use that as the basis for generating the table.\n\nTo calculate the differences we first need to produce n+1 values at fixed intervals. The generated table should have an entry per arcminute so we’ll start at 0′ and do steps of 1′:\n\nx sin(x)\n0′ 0\n1′ 2.909 10-4\n2′ 5.818 10-4\n3′ 8.727 10-4\n\nThen we need to find the nth differences:\n\nx sin(x) Δ1 Δ2 Δ3\n0′ 0 2.909 10-4 -2.461 10-11 -2.461 10-11\n1′ 2.909 10-4 2.909 10-4 -4.923 10-11\n2′ 5.818 10-4 2.909 10-4\n3′ 8.727 10-4\n\nAll of this is a matter of evaluating a polynomial so that’s not super hard to do by hand with as many decimals as you need, as long as you only need a few of them. From this table we take the last of each of the differences and that’ll be the starting point for the calculation:\n\n0.000872664515235\n0.000290888130722\n-0.000000000049228\n-0.000000000024614\n\nAt this point we need to decide how much accuracy we want, that is, where we want the fixed decimal point to be. We have 13 digits which gives us room enough to multiply by 1013 before rounding. That gives us these integer values:\n\n8726645152\n2908881307\n-492\n-246\n\nAnd now we’re ready to get tabulating:\n\nIf you follow along with the original table you can see that it generates exactly the same values. The values generated by the engine continue to match the table values until 1° 1′, 57 values in, where the table says 0.0177432 and the engine produces 0.0177433, and after that it continues to produce matching values up until 1° 53′, more than 100 values in.\n\nNot bad right? And remember, this is a simplified emulator that can only calculate the third degree approximation where the original could go up to the seventh, and only with 13 digits of precision where the original had 31.",
null,
"So what’s the source of the deviation? There’s two: the approximating polynomial and the accumulating error of the engine. Let’s first look at the polynomials.",
null,
"The first plot on the right is of how quickly the approximating polynomials of different degrees deviate from the true sine function. At 1° the approximations are still well within 0.01 billionth of the correct value. Like I said, near 0 these polynomials approximate sin really well.\n\nThis suggests that the main source of inaccuracy is the engine itself, the precision we had to discard when fixing the decimal point, and as you can see in the second graph, it is. The engine loses precision faster than the polynomial by a large factor. This makes sense because the inaccuracy accumulates for every step.\n\nLuckily in this case the polynomial deviates slowly enough we can use it to calculate new almost-accurate initial values at fixed intervals, for instance for each degree, and reset the machine to those. However, eventually the polynomial itself will deviate too much and at that point we can use the fact that the Taylor polynomial has an a parameter that specifies the point around which we’re approximating. So say the polynomial that approximates around 0° becomes too inaccurate at 6° we can derive a Taylor polynomial around 6° and use that to continue the calculation. Indeed, since the polynomial approximates equally well on both sides of the point we might as well approximate around 9° and use it for all values between 6° and 12°.\n\nSin is a relatively easy function to approximate in this way, a function such as log is harder but the same basic principles apply to harder functions. It’s a matter of how often you need to reset to get rid of the accumulating errors and how long the same approximating polynomial remains accurate.\n\nOne of the weak points of the engine is that even though it requires less manual work than producing a table completely manually, there’s still a fair amount of manual analysis and computation to be done. That’s not a killer in itself though. Even if it took just as much work to operate, which it surely wouldn’t have, just having a different way to create these tables would have been immensely valuable since two different approaches are likely to produce different errors and hence can be used to cross-check each other. But as this illustrates, even if the engine had been built it was definitely not a matter of just plugging in a few values and then going to lunch, using it involved a fair amount of work4.\n\n## The revolution that didn’t happen\n\nHere’s a video of the only difference engine ever built which was completed in 2002.\n\nBabbage tried to build it and ultimately gave up for a number of reasons including family problems, financial problems, problems working with his engineer, and his gradual change of focus to the analytical engine. Despite what you often hear the technology did exist to build it; the modern one was built from his designs and only with technology that would have been available back then.\n\nIt also appears that Babbage’s interests and those of the English government who paid for the whole thing were just too far apart. Babbage was interested in the machine itself whereas his sponsors just wanted accurate tables, whatever way Babbage could produce them. It’s a shame really. It seems from what I’ve read that the difference engine was a failure not of vision or technology but of product management. The technology was so promising that if a successful prototype had been built and he’d delivered the tables the English government wanted it’s not unlikely that they would have continued to fund research in more advanced engines. The ground would have been fertile for mechanical calculation on a larger scale by the mid 1800s. Obviously that wouldn’t have meant a mechanical iPad in every home by 1900 but it would certainly have been a better outcome than what happened, that the designs went in the drawer.\n\nUltimately Babbage moved on to the analytical engine, the first ever programmable computer. My next post will be about that and in particular the first ever software programs which were written for it.\n\n## Sources\n\nFor more information about the difference engine a quick introduction is given in the first part of Menabrea’s Sketch of the Analytical Engine from 1842. A more detailed description, including of the underlying mechanics of the engine, can be found in Lardner’s Babbage’s Calculating Engine from the July 1834 edition of the Edinburgh Review.\n\n## Notes\n\n1: While the emulator performs the same type of calculations as the difference engine it actually looks nothing like it. I made a point to give the emulator a mechanical feel but it’s inspired more by the Curta calculator, the mechanical calculator I know best, not the difference engine. Note also that I have only tested it on the browsers I have easy access to, Chrome, Firefox, Opera, and Safari on mac. If it doesn’t work on your platform the code lives on github and patches are welcome.\n\n2: Ada Lovelace in her famous notes about it is almost contemptuous of how simple it is compared to the analytical engine:\n\nThe Difference Engine can in reality […] do nothing but add; and any other process […] can be performed by it only just to the extent in which it is possibly, by judicious mathematical arrangement and artifices, to reduce them to a series of additions.\n\n3: Incidentally, much of the underlying math was developed before Babbage by J. H. Müller.\n\n4: I wonder if it’s possible to automate some of the manual work involved in operating the engine using the engine itself, like calculating initial values.\n\n### 7 Responses to Difference\n\n1.",
null,
"Randall Neff\n\nThere is a second Babbage Difference Engine (design #2) constructed by the London Science Museum, built 2002 to 2008, which belongs to Nathan Myhrvoid. It is on loan to the Computer History Museum in Mountain View, Ca (Silicon Valley), and is demonstrated daily. The Engine works with 31 decimal digits and can compute with up to seven differences. It is cranked at one complete cycle ( seven 31 digit adds) every eight seconds.\nSee:\n3.",
null,
"Nathan Lisgo\n•",
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"Christian Plesner Hansen"
] | [
null,
"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
null,
"http://books.google.com.au/books",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://h14s.p5r.org/wp-content/uploads/2012/10/chart_1-2.png",
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"http://h14s.p5r.org/wp-content/uploads/2012/10/chart_1-5.png",
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"http://0.gravatar.com/avatar/96a999e058ec24132a619c225a7f8fe0",
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"http://0.gravatar.com/avatar/966dc0c871a9bd9bafd063dfb40418d4",
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http://lagrange.math.siu.edu/Kocik/const.htm | [
"",
null,
"COSMIC CONSTANTS\nand units via logarithm\n\nThe fundamental units used in physics are\n\n• meter (m) for distance (space)\n• second (s) for duration (time)\n• kilogram (kg) for mass (inertia)\n• The choice of these is determined by the nature of the human perception of reality. On the other hand, there are three fundamental \"cosmic\" constants. Ignore their values and look at the units they are expressed in the table below left:\n\n [c] = m/s [h] = m2 kg/s [G] = m3 / s2kg [length] = (h G / c3)1/2 [time] = (h G / c5)1/2 [mass] = (h c / G)1/2\n\nThis is a well known idea that one can use these constants to construct the \"natural\" units of length, time and mass. They are called Planck's units (displayed in the table above right). The meaning of the Planck units are not fully understood yet.\n\nHow is their existence possible (arithmetically)? Here is my remark: express the cosmic constants in a \"logarithmic way,\" say:\n\n c = m - s h = 2m + kg - s G = 3m - 2s -kg\n\nThink of these linear equations as a transformation. The matrix of the transformation of units (see the left table below) is nondegenerate, hence its inverse exists and reveals the right powers of the cosmic constants (right table below) when defining the Planck units of length, time and mass.\n\nConstant's matrix\nm s kg\nc 1 -1 0\nh 2 -1 1\nG 3 -2 -1\n\nThe inverse matrix\nc h G\nm -3/2 1/2 1/2\ns -5/2 1/2 1/2\nkg 1/2 1/2 -1/2\n\nYou may think of two bases in the same three-dimensional space in which every physical magnitude is represented by a vector of its units (understood---as above---in a \"logarithmic way).\n\n UNIT SPACE to be added soon\n\nHomework: locate basic physical magnitudes as vectors in the Unit Space, such as\n\n speed = m - s acceleration = m - 2s force = m - 2s + kg pressure = -m - 2s + kg density = -3m + kg\n\nHave fun. Add electric field, magnetic field, dielectric constant, current, action, etc.",
null,
"GR space QM 3 time TR 2 -2 mass 1 -1 -1 -1 1 0",
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"",
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""
] | [
null,
"http://lagrange.math.siu.edu/Kocik/img/const.gif",
null,
"http://lagrange.math.siu.edu/Kocik/img/rainbows.gif",
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"http://lagrange.math.siu.edu/Kocik/img/rainbows.gif",
null,
"http://lagrange.math.siu.edu/Kocik/img/constr17.gif",
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https://www.physicsforums.com/threads/partial-fraction-of-integrand.54183/ | [
"# Partial fraction of integrand\n\nAs part of finding the integral of z/(z^2 -1), I'm stuck on getting the partial fraction for it. 1/2 [(1/(z-1) - 1/(z+1)] gives 1/(z^2-1). What should I do to get the z in the numerator. Any hints welcome.\nRegards\n\nshmoe\n$$\\frac{z}{z^2-1}=\\frac{a}{z-1}+\\frac{b}{z+1}$$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82745594,"math_prob":0.9984804,"size":221,"snap":"2021-43-2021-49","text_gpt3_token_len":78,"char_repetition_ratio":0.09677419,"word_repetition_ratio":0.0,"special_character_ratio":0.3800905,"punctuation_ratio":0.09259259,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99705154,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T21:37:02Z\",\"WARC-Record-ID\":\"<urn:uuid:f6be966e-577c-4794-a3e9-4fdde35d86e4>\",\"Content-Length\":\"62563\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e087b5b6-743f-442c-b41d-f654078c4509>\",\"WARC-Concurrent-To\":\"<urn:uuid:436da48a-f48b-4d62-9054-e297416a4c4a>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/partial-fraction-of-integrand.54183/\",\"WARC-Payload-Digest\":\"sha1:4XXYACD42EJYXFPKOAJFDQO2NRA7Q75X\",\"WARC-Block-Digest\":\"sha1:URCANUZIBXXVN4I5IIJTC2KR3U4J3OOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358842.4_warc_CC-MAIN-20211129194957-20211129224957-00481.warc.gz\"}"} |
https://www.transtutors.com/questions/consider-the-rotating-disk-with-an-offset-slot-as-shown-in-figure-9-28-inside-the-sl-1910129.htm | [
"# Consider the rotating disk with an offset slot, as shown in Figure 9.28. Inside the slot is a... 1 answer below »\n\nConsider the rotating disk with an offset slot, as shown in Figure 9.28. Inside the slot is a particle P of mass m connected to a spring with spring constant k. The disk has mass M and moment of inertia I about the center axis (not including the small mass m). The disk is free to rotate about its axis by an angle θ. You can assume that the unstretched length of the spring is zero and at t = 0 the spring is stretched by an amount x0",
null,
"a. Show that the equations of motion for the mass in the slot and the angle θ of the disk are",
null,
"b. Is energy conserved?\n\nc. What is the total angular momentum",
null,
"of the system? Is it conserved?\n\nd. Suppose we start the system by offsetting the mass an amount x0 = 60 cm. If the disk has a radius of 1 m, I = 2.5 kg-m2, l = 75 cm, m = 0.25 kg, and k = 1N/m, use matlab to simulate the motion of the mass and the disk over the time interval [0, 20] s. Plot x versus t, θ versus t, and",
null,
""
] | [
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"https://files.transtutors.com/book/qimg/63c53090-0023-46b2-9543-8583f5f5aa90.png",
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"https://files.transtutors.com/book/qimg/7f4b986a-0ba3-4a3a-9da5-812235e760c9.png",
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"https://files.transtutors.com/book/qimg/a2c126f8-382c-466b-aded-e3199656b9e5.png",
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"https://files.transtutors.com/book/qimg/188eeb57-6e16-487d-9a74-788a53df7329.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.898223,"math_prob":0.9957892,"size":2205,"snap":"2020-10-2020-16","text_gpt3_token_len":544,"char_repetition_ratio":0.12857792,"word_repetition_ratio":0.014117647,"special_character_ratio":0.2430839,"punctuation_ratio":0.11885246,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99858856,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-19T11:32:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a7f3bfdc-a161-415c-9cc3-d7eefa996e66>\",\"Content-Length\":\"81298\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6be19181-291a-48f1-b247-e17afc404490>\",\"WARC-Concurrent-To\":\"<urn:uuid:352987b9-1e87-469a-96e5-139d0441fcea>\",\"WARC-IP-Address\":\"35.199.55.187\",\"WARC-Target-URI\":\"https://www.transtutors.com/questions/consider-the-rotating-disk-with-an-offset-slot-as-shown-in-figure-9-28-inside-the-sl-1910129.htm\",\"WARC-Payload-Digest\":\"sha1:JHWCBOCE27GGPTZSXHIEI7K4K5U5YTAU\",\"WARC-Block-Digest\":\"sha1:VXFN7SB32IDRQ7PYBILJGSEFHLOTBN4E\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144111.17_warc_CC-MAIN-20200219092153-20200219122153-00409.warc.gz\"}"} |
https://motovique.com/motorbike/often-asked-what-is-the-average-velocity-of-the-motorbike-during-first-5-s.html | [
"# Often asked: What Is The Average Velocity Of The Motorbike During First 5 S?\n\n## How do you find average velocity?\n\nThe sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.\n\n## What does it mean when an object moves at a constant speed of 6m S?\n\nAn object moves at a constant speed of 6 metre per second this means that the object A. increases its speed by 6m / s every second. decreases its speed 6m / s every second. C.\n\n## How do you find average velocity from a graph?\n\nAverage velocity can be calculated from a position-time graph as the change in position divided by the corresponding change in time.\n\n## How do you find average velocity from acceleration and time?\n\nWhen a velocity is changing as a result of a constant acceleration, the average velocity can be found by adding the initial and final velocities, and dividing by 2. The unit for velocity is meters per second (m/s). Note that this formula applies for constant acceleration only.\n\nYou might be interested: Question: When You Just Want To Wash Your Motorbike In Peace?\n\n## What is difference between speed and velocity?\n\nSpeed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object’s movement.\n\n## What is the difference between velocity and average velocity?\n\nSpeed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.\n\n## What is constant speed?\n\nDefinition: When the speed of an object remains the same – it does not increase or decrease – we say it is moving at a constant speed.\n\n## What does 6m S mean?\n\nAn object moves at a constant speed of 6 m / s. This means that the object. (a) Decreases its speed by 6 m / s every second.\n\n## What is the acceleration of the object between 0 s and 2 s?\n\nBetween 0 and 2 seconds the acceleration is +2 m/s2. Between 2 and 3 seconds the acceleration is zero.\n\n## What’s the slope of a velocity time graph?\n\nThe slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.\n\n## What does a constant velocity graph look like?\n\nConstant velocity means the velocity graph is horizontal, equal to 11.11 m/s at all times. A constant velocity means the position graph has a constant slope (of 11.11 m/s). It’s a straight line sloping up, and starting below the origin. The displacement is the area under the curve of the velocity graph.\n\nYou might be interested: What Is The Most Reliable Motorbike In The World?\n\n## How do you interpret a velocity vs time graph?\n\nThe principle is that the slope of the line on a velocity – time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). If the acceleration is positive, then the slope is positive (i.e., an upward sloping line).\n\n## What is the formula for velocity and acceleration?\n\nKey Equations\n\nDisplacement Δx=xf−xi\nInstantaneous acceleration a(t)=dv(t)dt\nPosition from average velocity x=x0+–vt\nAverage velocity –v=v0+v2\nVelocity from acceleration v=v0+at(constanta)\n\n13\n\n## What is the formula for constant acceleration?\n\nThe first equation of motion Since the acceleration is constant, we have a=v−ut. This gives the first equation of motion, v=u+at.\n\n## What is the equation to find the final velocity?\n\nFinal velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth’s gravitational force as the acceleration rate of an object."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90574104,"math_prob":0.98821104,"size":4528,"snap":"2021-31-2021-39","text_gpt3_token_len":1020,"char_repetition_ratio":0.22259063,"word_repetition_ratio":0.19503106,"special_character_ratio":0.2329947,"punctuation_ratio":0.09801762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99982363,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T16:38:10Z\",\"WARC-Record-ID\":\"<urn:uuid:52419bd7-a95c-4e32-95fc-9607a9f575b6>\",\"Content-Length\":\"39613\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc0ce0a5-9cd1-4db1-93a8-7f6cb0d1541a>\",\"WARC-Concurrent-To\":\"<urn:uuid:16a1f925-be5c-423e-80b7-b539178f92c8>\",\"WARC-IP-Address\":\"104.21.31.40\",\"WARC-Target-URI\":\"https://motovique.com/motorbike/often-asked-what-is-the-average-velocity-of-the-motorbike-during-first-5-s.html\",\"WARC-Payload-Digest\":\"sha1:RGSWSNYU7IVSAABO2GWLW4PTIZLTATYV\",\"WARC-Block-Digest\":\"sha1:R4WDBIFXBTYEYMXUI6QB5GW4UW3VPOMK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056548.77_warc_CC-MAIN-20210918154248-20210918184248-00648.warc.gz\"}"} |
https://manpages.org/gpiv_rr | [
"",
null,
"gpiv_rr(1) interrogates an image (pair) in order to obtain displacements of particles\n\n## SYNOPSIS\n\ngpiv_rr [--cf int] [--cl int] [--cp int] [-g] [--gauss] [-h | --help] [--ifit 0/1/2/3] [--ischeme int] [--ia_size_i int] [--ia_size_f int] [--ia_shift int] [--linec int int int] [--liner int int int] [-o] [-p | --print] [--peak int] [--p_piv] [--point int int] [--rf int] [--rl int] [--rp int] [-s float] [--spof] [-v | --version] [--val_r int] [--val_s int] [--val_t float] [filename] < stdin > stdout\n\n## DESCRIPTION\n\ngpiv_rr interrogates an image or image pair that is obtained from a fluid flow by the so-called Digital Particle Image Velocimetry (DPIV) technique. Therefore, image(s) are sub-divided into interrogation areas on a rectangular grid. At each interrogation area the mean (or most probable) particle displacement is estimated. This is done by correlating the belonging interrogation areas of an image-pair by means of the Fast Fourier Transformation (FFT) technique, resulting into a two-dimensional correlation function. The location of the highest function peak, then, represents the mean or most probable displacement of the particle images that have been resident within the interrogation areas. Estimation of the correlation peak at sub-pixel level may be calculated by different interpolation schemes. The program allows cross-correlation of single-exposed images on different frames or auto-correlation of a multi-exposed single-frame image. Interrogation areas of arbitrary sizes may be used in order to obtain an optimum spatial resolution. Adaptive sizes of interrogation areas allow for large dynamic ranges of the particle displacements. Zero offsetting of the interrogation areas by an iterative interrogation process results into higher accuracy/lower biases of the particle image displacements. A central differential interrogation scheme than might be applied. This may result into improved estimators, compared with the 'classical' forward interrogation scheme, especially in case of strong shear strain and vorticity of the flow. Most accurate results, however, are obtained by deforming the images towards each other by using the PIV estimators. As a convergence criterium for these iterative procedures, the cumulative difference (defined by GPIV_CUM_RESIDU_MIN = 0.25) between the PIV estimators from the current and the previous iteration is used. After each interrogation the PIV estimators are validated. Before outliers are substituted as defined by the VALID parameters, it will be tried if the PIV estimator from the second or third highest correlation peak will be valid.\n\ngpiv_rr is fed with images of the following formats: Portable Network Graphics (filename.png), raw binary data (filename.r) that is accompanied by an ASCII header file (filename.h), HDF5 (filename.hdf), tif, gif, pgm, bmp and LaVision's (tm) uncompressed image format (filename.img). For cross-correlation the second image frame has to be concatenated after the first one into a single image file. This might be performed by gpiv_combing. Image parameters are read from the header or from other parameter resources (containing the key IMG) in case they are absent in the image header.\n\nThe configuration parameters (containing the EVAL or VALID) key may be overruled by the command line options, as explained below.\n\n## Options\n\n--cf N\nSpecify the first column N in the image to interrogate. If --ad_int has been used, the first column has to be equal or larger than (int_size_2 - int_size_1)/2.\n\n--cl N\nSpecify the last column N in the image to interrogate.\n\n--cp N\nPre-shift of N columns. This can be used if there is a common mean flow in x-direction in the area of observation. Relative small interrogation areas (allowing a high spatial resolution) may be used in that case with conservation of a high probability in finding the correct displacement peaks.\n\n-g\nGraphic visualisation of the output with gnuplot. Can only be used in combination with filename).\n\n--gauss\nGauss weighting of interrogation area to reduce high spatial frequency signal generated by the boundaries.\n\n-h | --help\nPrint a help message on standard output and exit successfully.\n\n--ifit 0/1/2/3\nThree-point interpolation model for peak maximum estimation at sub-pixel level. 0: none, 1: Gauss, 2: Parabolic or 3 Center of Gravity.\n\n--ischeme 0/1/2/3/4\nInterrogation scheme: no correction (0),\n\nlinear kernel weighting (1); This is applied to the calculation of the correlation function; the weighting of the image data decreases towards the edges of the interrogation region. Kernel weighting compensates this effect. Will be disabled if interrogation area size of image 2 differs from image 1.\n\nzero offset (2); Searches (iteratively) the correlation peak by zero offsetting the interrogation area's, until the peak maximum lies between (-1,-1) and (1,1). The images are interrogated by the 'classic' forward differential scheme. During the last iteration step, sub-pixel displacement will be calculated as defined with -ifit.\n\nZero offset with central differential (3); The images are interrogated by the central differential scheme. This is done by displacing the interrogation area of the first image with half the (integer) magnitude of the pre-shift value in negative direction and displacing the interrogation area of the second image in positive direction (of identic magnitude).\n\nImage deformation (4); The images of a pair are deformed following the particle displacements obtained from the initial PIV estimators or from the previous iteration step. The first image is deformed in positive direction with half the (float) magnitudes of the estimators and the second image in negative direction. In this way, both deformed images will show the particle positions at the moment in-between the recordings. After the iteration has been converged and -p option has been used, the deformed images are stored (defined by GPIV_DEFORMED_IMG_NAME = gpiv_defimg) in TMPDIR (/tmp for UNIX-like systems), which may be used as a check.\n\n--ia_size_i N\nInitial size of the interrogation area's N. N must be equal or larger than ia_size_f.\n\nThe sizes must be chosen in such a way that the particle displacements remain within 1/4th of the interrogation area's in order to keep the in-plane errors at minimum.\n\nChoosing larger sizes of the initial interrogation area's allows high dynamic ranges of the estimators. In that case, the largest particle displacements may contribute adequately to the calculation of the estimators, while the estimators of the smallest flow scales are not smoothed by the large, initial, dimensions of the interrogation area's. The dimensions of the interrogation area's of the first and second image start with ia_size_i. For each next image interrogation, the sizes will be halved until they will be equal to the final ia_size_f value. The estimator will be used as a local pre-shift (zero offsetting, as defined by --ischeme). In case ia_size_f and/or ia_size_i is not a power of two, the sizes of the interrogation area's will be reduced with the appropriate factor during the last (iterative) interrogation in order to set them equal to ia_size_f. During the last interrogation, the estimator will be between (-1,-1) and (1,1). Then, sub-pixel displacement will be calculated as defined by --ifit.\n\n--ia_size_f N\nFinal size of the interrogation area's N, expressed in pixels. May be chosen arbitrarily.\n\n--ia_shift N\nShift of adjacent interrogation areas N, expressed in pixels.\n\n--linec COL RF RL\nselects a vertical line at column COL to interrogate from the first row RF to the last row RL\n\n--liner ROW CF CL\nselects an horizontal line at row ROW to interrogate from the first column CF to the last column CL\n\n-p | --print\nPrint parameters, command line options, progress of calculation and eventually used input and output filenames to stdout. The output is identic of filename.par, in case -f is used.\n\n--peak N\nFind the N-th maximum of the correlation peak. In case of auto-correlation, the second peak is taken by default, as the first peak denotes the zero-shift of the particle displacements.\n\n--p_piv\nPrint the piv results to stdout, even if -f has been specified.\n\n--point COL ROW\nSelect a single area in the image to interrogate at location COL ROW. This option might be useful for substitution of erroneous displacement vectors. A new estimation of the particle displacement with --peak, then, may give a correct result. Mind to use --p_piv if -f is used; else the original data file will be overwritten with a single point analyses.\n\n--rf N\nSpecify the first row N in the image to interrogate. If -ad_int has been used, the first row has to be equal or larger than (int_size_2 - int_size_1)/2.\n\n--rl N\nSpecify the last row N in the image to interrogate.\n\n--rp N\nPre-shift of N rows. This can be used if there is a common mean flow in y-direction. Relative small interrogation areas (allowing a high spatial resolution) may be used in that case with conservation of a high probability in finding the correct displacement peaks.\n\n-s S\nScale factor for graphical output with gnuplot. This will only affect the length of the vectors that represent the particle image displacement magnitude, but not the PIV data itself. In order to adapt the scaling of the data, see gpiv_scale.\n\n--spof\nApplies symmetric phase only filtering. This option may drasticly improve the SNR with higher and thinner covariance peak. Especially useful when there is flare or high reflections (from boundaries, for example) in one of the two image frames from a PIV image pair. (Werner, Meas. Sci. Technol., 16, 601-618).\n\n-v | --version\nPrint version information on standard output then exit successfully.\n\n--val_r N\nValidation parameter to define residue type: Signal to Noise Ratio (N = 0), median value from surroundings (N = 1) or normalised median (N = 2).\n\n--val_s N\nValidation parameter to substitute an erroneous vector by: nothing (N = 0), local mean from the surroundings (N = 1), the median of the surroundings (N = 2) or the estimation from the next highest correlation peak (N = 3).\n\n--val_t F\nValidation parameter representing the threshold value (float number) of residues to be accepted.\n\nfilename\nUsing the full filename of the input image overrides stdin and stdout. Output will be written to filename.piv. Parameters are stored in filename.par and may be used for future use by including them in ./gpivrc. If stdin and stdout are used, the input is expected to be a PNG formatted image.\n\n## NOTES\n\ngpiv_rr has been tested with artificial images and with PIV images from gas and liquid flows. Comparison with PIV data, obtained from different algorithms, and with literature results that similar data reliability and accuracy may be obtained with this program.\n\n## AUTHOR\n\nGerber Van der Graaf\n\n## BUGS\n\nThe program seems to work fine. Though as the PIV technology itself is subject of research, this program is constantly under development."
] | [
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"https://manpages.org/assets/menu-533ee2b7609b0e242c51d7f42ca18b7f.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82521677,"math_prob":0.855379,"size":10924,"snap":"2022-40-2023-06","text_gpt3_token_len":2495,"char_repetition_ratio":0.16126373,"word_repetition_ratio":0.0710631,"special_character_ratio":0.22528377,"punctuation_ratio":0.09439834,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9501161,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T00:14:46Z\",\"WARC-Record-ID\":\"<urn:uuid:14d11ca2-f784-45e6-a7b1-7be2a50f4b2d>\",\"Content-Length\":\"21438\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e93299a3-32be-46c0-8491-37cb69bc2b19>\",\"WARC-Concurrent-To\":\"<urn:uuid:c12ebd3c-4724-4f09-8d9c-a0e18dc7741b>\",\"WARC-IP-Address\":\"104.21.82.31\",\"WARC-Target-URI\":\"https://manpages.org/gpiv_rr\",\"WARC-Payload-Digest\":\"sha1:LMZHXHTSRDKVASYZLTXGXTLCTPTMLXLB\",\"WARC-Block-Digest\":\"sha1:CLDOEMFJSR7MCYASGOSXVBMBUZ6KW5LI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335396.92_warc_CC-MAIN-20220929225326-20220930015326-00394.warc.gz\"}"} |
https://math.stackexchange.com/questions/934805/analysis-problem-showing-statements-are-true | [
"Analysis Problem - Showing statements are true.\n\nProve the following.\n\nLet $\\{A_n \\}_{n \\in \\mathbb{N}}$ and $\\{ B_n\\}_{n \\in \\mathbb{N}}$ be sequences of sets with $$A_1 \\subset A_2 \\subset A_3 \\dots \\subset A_n \\dots$$ $$B_1 \\subset B_2 \\subset B_3 \\dots \\subset B_n \\dots$$ then $\\left( \\bigcup_{n=1}^{\\infty} A_n\\right ) \\cap \\left ( \\bigcup_{n=1}^{\\infty} B_n\\right ) = \\bigcup_{n=1}^{\\infty}A_n \\cap B_n$\n\nSimilarly, prove that $$A_1 \\supset A_2 \\supset A_3 \\dots \\supset A_n \\dots$$ $$B_1 \\supset B_2 \\supset B_3 \\dots \\supset B_n \\dots$$ then $\\left( \\bigcap_{n=1}^{\\infty} A_n\\right ) \\cup \\left ( \\bigcap_{n=1}^{\\infty} B_n\\right ) = \\bigcap_{n=1}^{\\infty}A_n \\cup B_n$\n\nCan anyone help with these two exercises? I've been stuck on them for awhile but haven't been able to make any significant progress in my notebook.\n\n• Hint: replace every occurence of $\\infty$ by $k \\in \\mathbb{N}$ and work on that. What can you say about for example $\\bigcup_{n=1}^{k} A_n ?$ – Shakespeare Sep 17 '14 at 9:23\n• Or even just $(A_1\\cup A_2)\\cap(B_1\\cup B_2)$ – David Peterson Sep 17 '14 at 9:36\n\nIf $x\\in\\left(\\bigcup_{n=1}^{\\infty}A_{n}\\right)\\cap\\left(\\bigcup_{n=1}^{\\infty}B_{n}\\right)$ then $x\\in A_{m}$ and $x\\in B_{k}$ for positive integers $m,k$.\nNote that $x\\in A_{m}\\subset A_{n}$ and $x\\in B_{k}\\subset B_{n}$ if $n$ exceeds $k$ and $m$."
] | [
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http://unistudyguides.com/wiki/Gauss%27s_Law | [
"# Gauss's Law\n\nThis is a topic from Higher Physics 1B\n\n## Introduction\n\nGauss’s law is an expression of the general relationship between the net electric flux through a closed surface (Gaussian surface) and the charge enclosed by the surface. In order to understand this law one must first define and examine the term ‘flux’\n\n## Electric Flux\n\nIn general, ‘flux’ refers to the rate of flow per area, and so electric flux refers to the rate of flow of electric field lines through a given area. Thus electric flux is the product of the magnitude of the electric field and the surface area perpendicular to it, as given by the equation:\n\nΦE = EA\n\nWhere ΦE represents electric flux in Voltmeters (Vm) or Newton-metres-squared-per-coulomb (Nm2C-1)\n\nE represents the electric field (NC-1)\nA represents perpendicular area (m2)\n• The perpendicular area is given by the product of area and the cosine of the angle between the surface and the normal to the field lines. This means that flux is zero when the surface is parallel to the field lin (as Cos90° = 0)\n• For more complicated and non-uniform surfaces, the net flux is calculated as the sum of the fluxes through small areas of the surface ΔA. The method of summing these fluxes is called surface integration, and uses the slightly different integration symbol:\n• For a closed surface an electric field line leaving the surface causes positive flux, while field lines entering the surface cause negative flux\n• Thus the net flux is proportional to the total number of field lines leaving the object minus the total number of field lines entering the object\n\n## Statement and Equation\n\nGauss’s law states that the net electric flux through a closed surface surrounding an inclosed charge is proportional to the magnitude of the charge, and is given by\n\nΦE = q / εo and given that\nΦE = E.dA it shows that\nE.dA = q / εo\n\nWhere ΦE represents electric flux in Voltmeters (Vm) or Newton-metres-squared-per-coulomb (Nm2C-1)\n\nQ represents the inclosed charge (C)\nE represents the electric field (NC-1)\nA represents perpendicular area (m2)\nεo represents the permittivity of free space (8.8542 x 10-12 C2N-1m-2)\n\n## Selecting a Gaussian Surface\n\n• Gauss’s law states that regardless of which hypothetical Gaussian surface is chosen to calculate the flux the result is the same, and so the equation is independent of the shape of the surface provided it is closed\n• For the purpose of making calculations easier though one should pick surfaces where sides are either parallel to the field lines (so flux is zero) or perpendicular the field lines (flux is given by E dA).\n• For a point charge or spherical charged object the ideal surface for calculations is a larger sphere, as every point on its surface is perpendicular to field lines\n• For a line of charge or wire the ideal surface for calculations is a cylinder with the line at its axis such that the two circular faces are parallel to field lines and the curved face is perpendicular to field lines\n• For a charged plane the ideal surface for calculations is a cylinder with its circular faces parallel to the plane such that the two circular faces are perpendicular to field lines and the curved face is parallel to field lines\n\n## Derivation\n\nThis derivation is included for the purpose of bettering one’s understanding of Gauss’s Law and is not necessarily examinable. Choosing a sphere as the Gaussian surface for the reasons mentioned above,\n\nΦE = E.dA ....................... (By definition)\n= EA ..................................... (Due to the properties of the sphere)\n= E(4π r2)\n= (keq / r2) (4π r2)\n= ke4π q\n= q/εo .................................... (By the definition of εo)\n\n## Electrostatic Equilibrium\n\nElectrostatic equilibrium is when there is no net motion of charge within a conductor. A conductor in electrostatic equilibrium has the following properties:\n\n1. The electric field is zero everywhere within the conductor, regardless of whether is it solid or hollow\n2. If the conductor is isolated, any charge it carries resides on the surface\n3. If the conductor is charged, the electric field at a point just outside and perpendicular to its field lines has magnitude σ/εo\n4. If the conductor is irregularly shaped the surface charge density is greatest at locations where the radius of curvature is the smallest (to over-simplify, the ‘pointy bits’) as seen in the figure below\n5. The surface of the conductor has a constant Electric Potential and as such is an equipotent surface\n\nThe first two properties have simple justifications which may assist in learning but are not necessarily examinable:\n\nProperty 1 – If the internal electric field of a conductor were not zero then the free electrons inside would undergo a force by Coulomb’s law. They would then accelerate, undergoing a net motion, which would mean the conductor is not in electrostatic equilibrium\nProperty 2 – Imagine a Gaussian surface extremely close to the charged conductor’s surface. There is no net flux through the Gaussian surface, thus there is no enclosed charge. Therefore the charge must reside outside the Gaussian surface, on the surface of the conductor itself\n\nThe remaining properties do have justifications which can be found in the Textbook."
] | [
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http://scripts.downloadroute.com/f2matlab-c0527bc4.html | [
"f2matlab 1.0",
null,
"OS : Windows / Linux / Mac OS / BSD / Solaris",
null,
"Script Licensing : Freeware",
null,
"Created : Oct 3, 2007",
null,
"Downloads : 30",
null,
"Thank you for voting...\n\nf2matlab is aimed at converting numerical Fortran90 ...\n\nmatlab by Ben Barrowes is aimed at converting numerical Fortran90 code to Matlab m-files. Accordingly, only basic data types and constructions are recommended. f2matlab 1.0 can handle: all numeric types (handled by Matlab interpreter) most string functions comparisons, branches, loops, etc. basic read/write/print statements (if it's not too fancy. . . )\nf2matlab [f2matlab1.0.exe] LIMITATIONS: f2matlab [f2matlab1.0.exe] can not handle some features of fortran90 yet. These include: can't handle complex read and write statements, derived-typed variables etc.\nDemands:\n• MATLAB Release: R14\n\nf2matlab 1.0 scripting tags: matlab, write, matlab tools, types, statements, handle, fmatlab, f2matlab, development, development tools, basic. What is new in f2matlab 1.0 software script? - Unable to find f2matlab 1.0 news. What is improvements are expecting? Newly-made f2matlab 1.1 will be downloaded from here. You may download directly. Please write the reviews of the f2matlab. License limitations are unspecified."
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"http://scripts.downloadroute.com/themes/2010/ico_licensing.png",
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http://www.cfm.brown.edu/people/dobrush/am33/Mathematica/residue.html | [
"# Preface\n\nThis tutorial was made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel.\n\nFinally, the commands in this tutorial are all written in bold black font, while Mathematica output is in normal font. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.\n\n# Residue Method\n\nNow we define the inverse Laplace transform using residue method, assuming that singular points are at most of multiplicity 3. Although the residue method is widely used in complex analysis, it was adapted by Vladimir Dobrushkin to embrace previously discussed methods: partial fraction decomposition and the convolution formula. It is applied to the expression of the form: P(λ)/Q(λ). The output is a collection of residues for each singular point.\n\n$f(t) = {\\cal L}^{-1} \\left[ F(\\lambda ) \\right] = {\\cal L}^{-1} \\left[ \\frac{P(\\lambda )}{Q(\\lambda )} \\right] = \\sum_{i=1}^N \\,\\mbox{Res}_{\\lambda = \\lambda_i} \\, F(\\lambda ) \\, e^{\\lambda \\, t} ,$\nwhere sum ranges over all zeroes of the equation $$Q(\\lambda ) =0 ,$$ and the function f must be multiplied by the Heaviside function to ensure that f(t) is identically zero for negative argument. The residures at each null are evaluated according to the following rule.\n\nSuppose that a function $$F(\\lambda ) = \\frac{P(\\lambda )}{Q(\\lambda )}$$ is a ratio of two irreducible polynomials (or entire functions). Let $$\\lambda = \\lambda_i , \\ i=1,2,\\ldots , N$$ be nulls of the denominator ($$N = \\infty$$ if Q(λ) is an entire function). Recall that if $$Q(\\lambda_i ) =0, Q' (\\lambda_i ) =0, \\ldots , Q^{(m_i -1)} (\\lambda_i ) =0, Q^{(m_i )} (\\lambda_i \\ne 0 ,$$ then we say that the denominator has zero at $$\\lambda = \\lambda_i$$ of multiplicity mi. In this case, we say that the function F(λ) has poles at $$\\lambda = \\lambda_i , \\quad i=1,2,\\ldots , N .$$ Note that the numerator is not zero at these poles.\n\nIf $$\\lambda = \\lambda_i$$ is a simple root (meaning that its multiplicity is 1) of the equation $$Q(\\lambda ) =0 ,$$ then\n\n$\\mbox{Res}_{\\lambda = \\lambda_i} \\, F(\\lambda ) \\, e^{\\lambda \\, t} = \\frac{P(\\lambda_i )}{Q' (\\lambda_i )} \\, e^{\\lambda_i t} .$\n\nIf $$\\lambda = \\lambda_i$$ is a double root (meaning that $$Q(\\lambda_i ) = Q' (\\lambda_i ) =0, \\quad\\mbox{but} \\quad Q'' (\\lambda_i ) \\ne 0 ,$$ ), then\n\n$\\mbox{Res}_{\\lambda = \\lambda_i} \\, F(\\lambda ) \\, e^{\\lambda \\, t} = \\lim_{\\lambda \\mapsto \\lambda_i} \\, \\frac{\\text d}{{\\text d}\\lambda} \\left[ (\\lambda - \\lambda_i )^2 \\,\\frac{P(\\lambda_i )}{Q (\\lambda_i )} \\, e^{\\lambda_i t} \\right] .$\nIn general, when λi is an n-fold root of $$Q(\\lambda ) =0 ,$$ then\n$\\mbox{Res}_{\\lambda = \\lambda_i} \\, F(\\lambda ) \\, e^{\\lambda \\, t} = \\frac{1}{(n-1)!} \\,\\lim_{\\lambda \\mapsto \\lambda_i} \\, \\frac{{\\text d}^{n-1}}{{\\text d}\\lambda^{n-1}} \\left[ (\\lambda - \\lambda_i )^n \\,\\frac{P(\\lambda_i )}{Q (\\lambda_i )} \\, e^{\\lambda_i t} \\right] .$\n\nNote that if $$\\lambda_i = \\alpha + {\\bf j} \\beta$$ is a complex null of the denominator Q(λ) (the root of the polynomial equation with real coefficients $$Q(\\lambda ) =0$$ ), then $$\\overline{\\lambda_i} = \\alpha - {\\bf j} \\beta$$ is also a null of Q(λ). In this case, we don't need to calculate the residue at the complex conjugate because we know that the output will be a complex conjugate of the residue at $$\\alpha + {\\bf j} \\beta .$$ Therefore, if the denominator Q(λ) has two complex conjugate roots $$\\lambda = \\alpha \\pm {\\bf j} \\beta ,$$ then the sum of residues at these points is just double value of the real part of one of them:\n\n$\\mbox{Res}_{\\lambda = \\alpha + {\\bf j} \\beta} \\, F(\\lambda ) \\, e^{\\lambda \\, t} + \\mbox{Res}_{\\lambda = \\alpha - {\\bf j} \\beta} \\, F(\\lambda ) \\, e^{\\lambda \\, t} = 2\\,\\Re \\, \\mbox{Res}_{\\lambda = \\alpha + {\\bf j} \\beta} \\, F(\\lambda ) \\, e^{\\lambda \\, t} .$\nIn the case of simple complex roots, we get\n$\\mbox{Res}_{\\lambda = \\alpha + {\\bf j} \\beta} \\, F(\\lambda ) \\, e^{\\lambda \\, t} + \\mbox{Res}_{\\lambda = \\alpha - {\\bf j} \\beta} \\, F(\\lambda ) \\, e^{\\lambda \\, t} = 2\\,e^{\\alpha \\,t} \\left\\{ A\\,\\cos \\beta t - B\\,\\sin \\beta t \\right\\} ,$\nwhere\n$A = \\Re \\,\\mbox{Res}_{\\lambda = \\alpha + {\\bf j} \\beta} \\, F(\\lambda ) = \\mbox{Re} \\, \\frac{P(\\alpha + {\\bf j} \\beta )}{Q' (\\alpha + {\\bf j} \\beta )} , \\qquad B = \\Im \\,\\mbox{Res}_{\\lambda = \\alpha + {\\bf j} \\beta} \\, F(\\lambda ) = \\mbox{Im} \\, \\frac{P(\\alpha + {\\bf j} \\beta )}{Q' (\\alpha + {\\bf j} \\beta )} .$\n\nExample: Let us find the inverse Laplace transform of the function\n\n$F(\\lambda ) = \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda +1)(\\lambda -2)(\\lambda -3)}$\nThe denominator has three simple roots $$\\lambda = -1, \\ 2, \\ 3 .$$ Therefore, its inverse Laplace transform is the sum of residues:\n${\\cal L}^{-1} \\left[ \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda +1)(\\lambda -2)(\\lambda -3)} \\right] = \\left( \\mbox{Res}_{\\lambda =-1} + \\mbox{Res}_{\\lambda = 2} + \\mbox{Res}_{\\lambda =3} \\right) F(\\lambda )\\, e^{\\lambda \\,t} .$\nWe calculate each residue separately.\n\\begin{align*} \\mbox{Res}_{\\lambda =-1} \\, F(\\lambda )\\, e^{\\lambda \\,t} &= \\lim{\\lambda \\mapsto -1} \\,\\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda -2)(\\lambda -3)} \\, e^{\\lambda\\,t} = -\\frac{1}{6}\\, e^{-t} , \\\\ \\mbox{Res}_{\\lambda =2} \\, F(\\lambda )\\, e^{\\lambda \\,t} &= \\lim{\\lambda \\mapsto 2} \\,\\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda +1)(\\lambda -3)} = -\\frac{13}{3}\\, e^{2t} , \\\\ \\mbox{Res}_{\\lambda =3} \\, F(\\lambda )\\, e^{\\lambda \\,t} &= \\lim{\\lambda \\mapsto 3} \\,\\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda +1)(\\lambda -2)} = \\frac{11}{2}\\, e^{3t} . \\end{align*}\nAdding all residues, we find the inverse Laplace transform\n${\\cal L}^{-1} \\left[ \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda +1)(\\lambda -2)(\\lambda -3)} \\right] = \\left( \\frac{11}{2}\\, e^{3t} - \\frac{13}{3}\\, e^{2t} - \\frac{1}{6}\\, e^{-t} \\right) H(t) .$\n\nExample: Let us find the inverse Laplace transform of the function that has multiple poles\n\n$F(\\lambda ) = \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda -2)^3 (\\lambda -3)} ,$\nwhere singular point $$\\lambda =2$$ has multiplicity 3. Calculating residues, we get\n\\begin{align*} \\mbox{Res}_{\\lambda =2} \\,F(\\lambda ) \\, e^{\\lambda\\, t} & = \\frac{1}{2}\\,\\lim_{\\lambda \\mapsto 2} \\, \\frac{{\\text d}^2}{{\\text d} \\lambda^2} \\, \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda -3)} \\, e^{\\lambda \\,t} = - \\left( 22+ 21\\,t + \\frac{13}{2}\\,t^2 \\right) e^{2t} , \\\\ \\mbox{Res}_{\\lambda =3} \\,F(\\lambda ) \\, e^{\\lambda\\, t} & = \\lim_{\\lambda \\mapsto 3} \\, \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda -2)^3} \\, e^{\\lambda\\, t} = 22\\, e^{3t} . \\end{align*}\nAdding these residues, we find the inverse Laplace transform\n${\\cal L}^{-1} \\left[ \\frac{\\lambda^2 +4\\,\\lambda +1}{(\\lambda -2)^3 (\\lambda -3)} \\right] = 22\\, e^{3t} \\, H(t) - e^{2t} \\left( 22+ 21\\,t + \\frac{13}{2}\\,t^2 \\right) H(t) .$\n\nWe present Mathematica code to calculate the residue.\n\n(* use variable \"s\" to define p and q *)\n\nNatalia[p_, q_] :=\n(g = ToRules[Roots[q == 0, s]];\nu = s /. {g};\nh = Tally[u];\nj = Length[r];\nRes1 = {s - s0}*{p/{Factor[q, Extension -> I]}}*Exp[s*t];\nRes2[s_] = {s - s0}^2*{p/{Factor[q, Extension -> I]}}*Exp[s*t];\nRes3[s_] = {s - s0}^3*{p/{Factor[q, Extension -> I]}}*Exp[s*t];\nr = Table[h[[i, 2]], {i, 1, Length[h]}];\nDo[If[r[[i]] == 1, Res = Res1 /. s0 -> h[[i, 1]];\nRes = ComplexExpand[Res /. s -> h[[i, 1]]]; Print[Res],\nIf[r[[i]] == 2, ResII[s_] = Res2[s] /. s0 -> h[[i, 1]];\nResII[s_] = ComplexExpand[ResII'[s] /. s -> h[[i, 1]]];\nPrint[ResII[s]],\nIf[r[[i]] == 3, ResIII[s_] = Res3[s] /. s0 -> h[[i, 1]];\nResIII[s_] = {1/2}*{ComplexExpand[\nResIII''[s] /. s -> h[[i, 1]]]}; Print[ResIII[s]]]]], {i, 1,\nj}])\n\nExample: We demonstrate application of the our code to find the inverse Laplace transform of\n\n$F(\\lambda ) = \\frac{2\\lambda -1}{(\\lambda^2 +1)(\\lambda -2)^3} .$\nFirst, we use the standard Mathematica command\nInverseLaplaceTransform[2*s-1)/(s^2+1)/(s-2)^3 , s, t]\nto get\n${\\cal L}^{-1} \\left[ \\frac{2\\lambda -1}{(\\lambda^2 +1)(\\lambda -2)^3} \\right] = \\frac{1}{125} \\left[ 7\\,\\cos t + 24\\,\\sin t \\right] H(t) - \\frac{1}{250} \\, e^{2t} \\left[ 14 + 20t +75t^2 \\right] H(t) .$\nNatalia[2*s - 1, (s^2 + 1)*(s - 2)^3]\n{{{1/2 (-((14 E^(2 t))/125)-4/25 E^(2 t) t+3/5 E^(2 t) t^2)}}}\n{{(7 Cos[t])/250+I (-((12 Cos[t])/125)+(7 Sin[t])/250)+(12 Sin[t])/125}}\n{{(7 Cos[t])/250+I ((12 Cos[t])/125-(7 Sin[t])/250)+(12 Sin[t])/125}}\n\nThere is another option, which we demonstrate on a similar example for the function\n\n$F(\\lambda ) = \\frac{2\\lambda -1}{(\\lambda^2 +1)^2 (\\lambda -2)^3} .$\nNow we have two double complex conjugate roots $$\\lambda = \\pm {\\bf j} .$$\n\nP = 2 s - 1 (* Defines polynomial P *)\nOut= -1 + 2 s\nQ = (s^2 + 1)^2 (s - 2)^3 (* Defines polynomial Q *)\nOut= (-2 + s)^3 (1 + s^2)^2\ns /. Solve[Q == 0] (* Finds the roots of Q(s)=0 *)\nOut= {-I, -I, I, I, 2, 2, 2}\nTally[%] (* Finds the number of repeats of each root (multiplicity)*)\nOut= {{-I, 2}, {I, 2}, {2, 3}}\nMatrixForm[%] (* Creates solutions matrix with roots in one column and multiplicities in another *)\nOut=\n(-i 2 )\n( i 2 )\n( 2 3 )\nDimensions[%] (* Measures the dimensions of the matrix (2x2),(3x2),(6x2), etc*)\nOut= {3, 2}\nv = Part[%, 1] (* Finds the length of the matrix (v=number of rows) *)\nOut= 3\nHere we start the Do loop to find residue for each root, where each row of the solutions matrix contains a root and its multiplicity\nDo[T := Tally[s /. Solve[Q == 0]];\nM := MatrixForm[T]; (* Recreates the solutions matrix for use in Do loop *)\nr := Extract[Extract[Extract[M, 1], l], 1]; (* Extracts the root (r) from row of the matrix *)\nm := Extract[Extract[Extract[M, 1], l], 2]; (* Extracts multiplicity (m) of root from row of the matrix *)\nf = ((((s - r)^m) (E^(s*t)) (P))/Q);\nF = Simplify[f]; (* Simplifies f into nicely factored form *)\nd = (1/(m - 1)!) D[F, {s, If[m > 1, m - 1, 0]}]; (* it is the equation for residue *)\nRes = d /. s -> r; (* Calculates residue at a particular root (r=root) *)\nPrint[Y = ComplexExpand[Re[Res], {I}]], {l, v}]\nIt takes the Euler expansion of previous line of code and extracts all real parts of that expression. Final output will contain several terms, that are functions of t, that will be the total residue when added. Note that the Res function computes the residue for each root by taking roots and multiplicities from row, where l ranges from 1 to v, the length of the solutions matrix.\nOut= -((17 Cos[t])/625)-6/125 t Cos[t]+(31 Sin[t])/625+7/500 t Sin[t] -((17 Cos[t])/625)-6/125 t Cos[t]+(31 Sin[t])/625+7/500 t Sin[t] (34 E^(2 t))/625-14/125 E^(2 t) t+3/50 E^(2 t) t^2\n\nTo check the answer, we type:\n\nInverseLaplaceTransform[(2 s - 1)/((s^2 + 1)^2 (s - 2)^3), s, t]\nOut=\n(1/1250)(68 E^(2 t) - 140 E^(2 t) t + 75 E^(2 t) t^2 - 68 Cos[t] -\n120 t Cos[t] + 124 Sin[t] + 35 t Sin[t])\n\nTo check, we make manual calculations.\n\\begin{align*} \\mbox{Res}_{\\lambda = 2} \\, F(\\lambda )\\, e^{\\lambda\\,t} &= \\left. \\frac{1}{2}\\,\\frac{{\\text d}^2}{{\\text d}\\lambda^2} \\, \\frac{2\\lambda -1}{(\\lambda^2 +1)}\\, e^{\\lambda\\,t} \\right\\vert_{\\lambda =2} = \\lim_{\\lambda \\mapsto 2} \\frac{e^{\\lambda\\,t}}{2(1+\\lambda^2 )^3} \\left[ 2 + 4t-t^2 + 2\\lambda^5 t^2 -\\lambda^4 t(4+t) -2\\lambda^2 (3+t^2) +4\\lambda^3 (1+t+t^2 ) +2\\lambda (2t+t^2 -6) \\right] \\\\ &= \\frac{1}{250}\\, e^{2t} \\left[ 75t^2 -20t -14 \\right] , \\\\ \\mbox{Res}_{\\lambda = {\\bf j}} \\, F(\\lambda )\\, e^{\\lambda\\,t} &= \\left. \\frac{{\\text d}}{{\\text d}\\lambda} \\,\\frac{2\\lambda -1}{(\\lambda -2)^3}\\, e^{\\lambda\\,t} \\right\\vert_{\\lambda ={\\bf j}} = \\lim_{\\lambda \\mapsto {\\bf j}} \\frac{e^{\\lambda\\,t}}{(\\lambda -2)^4} \\left[ -1 + 2t + 2\\lambda^2 t - \\lambda (4+5t) \\right] \\\\ &= \\frac{24{\\bf j} -7}{625} \\, e^{{\\bf j}t} \\left[ -1 - {\\bf j} (4+5t) \\right] . \\end{align*}\nExtracting the real part and multiplying by 2, we obtain\n$2\\,\\mbox{Re}\\, \\mbox{Res}_{\\lambda = {\\bf j}} \\, F(\\lambda )\\, e^{\\lambda\\,t} = \\frac{1}{625} \\left[ 14\\,\\cos t -14 (4+5t) \\,\\sin t + 48\\,\\sin t + 48\\,(4+5t)\\,\\cos t \\right] .$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79975045,"math_prob":0.9999919,"size":5377,"snap":"2020-34-2020-40","text_gpt3_token_len":1565,"char_repetition_ratio":0.12004467,"word_repetition_ratio":0.051652893,"special_character_ratio":0.32899386,"punctuation_ratio":0.100652374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000042,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T18:49:33Z\",\"WARC-Record-ID\":\"<urn:uuid:672edd8d-1c5e-40e9-be97-ec440ad20d3b>\",\"Content-Length\":\"29771\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:edfca29d-ac58-401f-bb48-cf020b441a13>\",\"WARC-Concurrent-To\":\"<urn:uuid:549ec3b9-36de-4b19-adb5-466c00c271dc>\",\"WARC-IP-Address\":\"138.16.167.39\",\"WARC-Target-URI\":\"http://www.cfm.brown.edu/people/dobrush/am33/Mathematica/residue.html\",\"WARC-Payload-Digest\":\"sha1:DUC3CK6G4JOKGROMUQRS62YRCBSTZMHX\",\"WARC-Block-Digest\":\"sha1:V7CV3PF5XQMSOVO7VGWGAHXNFKVZN7PI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400202007.15_warc_CC-MAIN-20200921175057-20200921205057-00537.warc.gz\"}"} |
http://mathonline.wikidot.com/linear-spaces | [
"Linear Spaces\n\n# Linear Spaces\n\n Definition: A Linear Space (or Vector Space) over $\\mathbb{R}$ (or over $\\mathbb{C}$) is a set $X$ with the binary operation of addition on elements of $X$, and a well-defined operation of scalar multiplication $\\cdot$ on numbers in $\\mathbb{R}$ (or $\\mathbb{C}$) with elements in $X$, such that the following properties are satisfied: 1) If $x, y \\in X$ then $(x + y) \\in X$ (Closure under $+$). 2) $x + (y + z) = (x + y) + z$ for all $x, y, z \\in X$ (Associativity of $+$). 3) $x + y = y + x$ for all $x, y \\in X$ (Commutativity of $+$. 4) There exists an element $0 \\in X$ such that $x + 0 = 0 = x = x$ (Existence of an additive identity). 5) For all $x \\in X$ there exists an element $-x \\in X$ such that $x + (-x) = (-x) + x = 0$ (Existence of additive inverses). 6) For all $\\alpha \\in \\mathbb{R}$ (or $\\mathbb{C}$), if $x \\in X$ then $\\alpha x \\in X$ (Closure under $\\cdot$). 7) $1x = x1 = x$ for all $x \\in X$. 8) For all $\\alpha, \\beta \\in \\mathbb{R}$ (or $\\mathbb{C}$), if $x \\in X$ then $\\alpha (\\beta x) = (\\alpha \\beta)x$. 9) For all $\\alpha \\in \\mathbb{R}$ (or $\\mathbb{C}$), if $x, y \\in X$ then $\\alpha (x + y) = \\alpha x + \\alpha y$. 10) For all $\\alpha, \\beta \\in \\mathbb{R}$, if $x \\in X$ then $(\\alpha + \\beta) x = \\alpha x + \\beta x$.\n\nLet $A \\subseteq \\mathbb{R}$ and let $X$ be the set of functions $f : A \\to \\mathbb{R}$. With the binary operation $+$ of function addition defined for all $f + g \\in X$ by:\n\n(1)\n\\begin{align} \\quad (f + g)(x) = f(x) + g(x) \\end{align}\n\nAnd scalar multiplication $\\cdot$ defined for all $\\alpha \\in \\mathbb{R}$ and for all $f \\in X$ by:\n\n(2)\n\\begin{align} \\quad (\\alpha f)(x) = \\alpha f(x) \\end{align}\n\nThen the set $X$ with these operations forms a linear space.\n\n Definition: If $X$ is a linear space over $\\mathbb{R}$ (or $\\mathbb{C}$) with the operations of addition and scalar multiplication and if $Y \\subseteq X$ then $Y$ is a Linear Subspace (or Vector Subspace) of $X$ if $Y$ is a linear space with the operations of addition and scalar multiplication.\n\nWe now state a relatively simple criterion for determining whether a subset $Y$ of a linear space $X$ is a linear subspace or not.\n\n Theorem 1: Let $X$ be a linear space over $\\mathbb{R}$ (or $\\mathbb{C}$) and let $Y \\subseteq X$. Then $Y$ is a linear subspace of $X$ if: 1) For all $x, y \\in Y$ we have that $(x + y) \\in Y$. 2) For all $\\alpha \\in \\mathbb{R}$ (or $\\mathbb{C}$), if $x \\in Y$ then $\\alpha x \\in Y$. 3) $0 \\in Y$.\n• Proof: We are given that (1), (4), and (6) from the definition of a linear space are given from the hypotheses of the theorem.\n• Properties (2), (3), (7), (8), (9), and (10) from the definition of a linear space $X$ are all inherited properties to $Y$ from the operations of addition and scalar multiplication.\n• All that remains to show is that (5) holds. Let $x \\in Y$. Then since $-1 \\in \\mathbb{R}$ (or $\\mathbb{C}$) we have that $-1x = -x \\in Y$. So indeed (5) holds.\n• Hence $Y$ is a linear subspace of $X$. $\\blacksquare$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63068485,"math_prob":1.0000095,"size":3019,"snap":"2023-40-2023-50","text_gpt3_token_len":1036,"char_repetition_ratio":0.17313433,"word_repetition_ratio":0.108391605,"special_character_ratio":0.37330243,"punctuation_ratio":0.08974359,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T01:49:25Z\",\"WARC-Record-ID\":\"<urn:uuid:acfb11e3-334d-4b0a-bdbb-5f2b29970a7d>\",\"Content-Length\":\"19219\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3668f9a-2be2-4081-bf70-1bd795567839>\",\"WARC-Concurrent-To\":\"<urn:uuid:401c66ad-c61e-448a-a27c-01a3143c50dc>\",\"WARC-IP-Address\":\"107.20.139.176\",\"WARC-Target-URI\":\"http://mathonline.wikidot.com/linear-spaces\",\"WARC-Payload-Digest\":\"sha1:NPMBEUBKC6VXRXN7HUDHYUX4TTKTRBVS\",\"WARC-Block-Digest\":\"sha1:JYVW3Z7XBTHFBX7ZO2HY36U7JJYQW5NH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100164.15_warc_CC-MAIN-20231130000127-20231130030127-00456.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/84-5-plus-20-60 | [
"Solutions by everydaycalculation.com\n\n1st number: 16 4/5, 2nd number: 20/60\n\n84/5 + 20/60 is 257/15.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 60 is 60\n2. For the 1st fraction, since 5 × 12 = 60,\n84/5 = 84 × 12/5 × 12 = 1008/60\n3. Likewise, for the 2nd fraction, since 60 × 1 = 60,\n20/60 = 20 × 1/60 × 1 = 20/60\n1008/60 + 20/60 = 1008 + 20/60 = 1028/60\n5. After reducing the fraction, the answer is 257/15\n6. In mixed form: 172/15\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84971446,"math_prob":0.99736285,"size":716,"snap":"2019-51-2020-05","text_gpt3_token_len":288,"char_repetition_ratio":0.15589887,"word_repetition_ratio":0.0,"special_character_ratio":0.50558656,"punctuation_ratio":0.09756097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99746794,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T12:07:10Z\",\"WARC-Record-ID\":\"<urn:uuid:cfdd07bc-d580-4f69-9230-6b655714708a>\",\"Content-Length\":\"7608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c312145-f5c3-4c9e-b6e0-641b156a2e96>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbc392ab-1f0b-482a-a19d-6ce8835ca8ce>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/84-5-plus-20-60\",\"WARC-Payload-Digest\":\"sha1:NKNSI5AX3HJQXTGBFTDFZIFQ2K733ZXO\",\"WARC-Block-Digest\":\"sha1:SUJDPWZ5RNWUS2RJW5N6BNZBKK2RA6UX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250619323.41_warc_CC-MAIN-20200124100832-20200124125832-00312.warc.gz\"}"} |
https://physicscatalyst.com/heat/changes-of-phases.php | [
"",
null,
"# Change of Phases and Latent Heat\n\n## Change of phases:\n\n• There are three phases of matter i.e, Solid, Liquid and Gas.\n• Substance for example, H2O exists in solid phase as Ice,in liquid phase as Water and in gas state as Steam.\n• Transition from one phase to another phase are accompanied by absorption or liberation of heat and usually by change in volume even at constant T.\n• Change of phase from solid to liquid is called melting , from liquid to solid is called fusion and from liquid to gas is called vaporisation\n• Once the temperature for phase change is reached( e,g melting or boiling temperature) no further temperature change occurs until all the substance has undergone phase change.\n\nMelting Point:-Temperature at which solid and liquid phase are in thermal equilibrium with each other\nBoiling point:-Temperature at which liquid and vapour phase are in thermal equilibrium with each other\n\n• Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. . Draw a graph between temperature and time . You will observe no change in the temperature so long as there is ice in the beaker. In the above process, the temperature of the system does not change even though heat is being continuously supplied. The heat supplied is being utilised in changing the state from solid (ice) to liquid (water).",
null,
"• Change of phase from solid state to Vapour state without passing through liquid state is called sublimation for e.g Dry Ice and Iodine sublimes.\n\n## Latent Heat:\n\n• The amount of heat per unit mass that must be transferred as heat when a sample completely undergoes phase change is called latent Heat of the substance for the process.\n• Thus when a sample of mass m completely undergoes phase change,the total energy transferred is\nQ=Lm\nwhere,\nL- Latent heat and is characteristics of a substance\n• Its unit is JKg-1.\n• Latent Heat for a solid liquid change is called Latent Heat of Fusion Lf.\n• Latent Heat for a liquid gas change is called Latent Heat of Vaporisation Lv.\n• The plot of Temperature and heat energy is given below",
null,
"We can observe that water requires Latent Heat of Fusion for conversion from ice to liquid at constant temperature 0°C and Latent Heat of Vaporisation for conversion from liquid to gas at constant temperature 100° C\nAnother graph could be like below",
null,
"## Solved Example\n\nQuestion 1\nWhen 0.15 kg of ice of 0 °C mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Calculate the heat of fusion of ice. ($s_{water} = 4186 J kg^{-1} K^{-1}$)\nSolution\nHeat lost by water = $ms_w \\Delta T$\n$= (0.30 kg ) (4186 (50.0 - 6.7 )$\n$= 54376.14$ J\nHeat required to melt ice = $m_iL_f = (0.15 kg) L_f$\nHeat required to raise temperature of ice water to final temperature = $m_is_w \\Delta T$\n$= (0.15 kg) (4186 ) (6.7 - 0 )$\n$= 4206.93 J$\nHeat lost = heat gained\n$54376.14 = (0.15 kg) L_f + 4206.93$\n$L_f = 3.34 \\times 10^5 J /kg$\n\nQuestion 2\nHow much energy does the freezer has to remove from 1.5 kg of water at 20° C to make ice at -10°C? ($s_{water} = 4186 J kg^{-1} K^{-1}$ and $L_f = 3.34 \\times 10^5 J /kg$, $s_{ice} = 2100 J kg^{-1} K^{-1}$ )\nSolution\nHeat Removed = Heat removed from 20° C to 0° C + Latent heat of Fusion + Heat removed from 0° C to -10° C\n$= 1.5 \\times 4186 \\times (20 -0) + 1.5 \\times 3.34 \\times 10^5 + 1.5 \\times 2100 \\times (0 - (-10))$\n$=125480 + 501000+31500$\n$=658 kJ$"
] | [
null,
"https://physicscatalyst.com/image/logo200.png",
null,
"https://physicscatalyst.com/heat/change-of-phase-water-temperature-time-graph.png",
null,
"https://physicscatalyst.com/heat/temperature-vs-heat-water.png",
null,
"https://physicscatalyst.com/heat/change-of-phase-water-temperature-time-graph-1.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88219917,"math_prob":0.99303156,"size":3598,"snap":"2021-43-2021-49","text_gpt3_token_len":978,"char_repetition_ratio":0.14579855,"word_repetition_ratio":0.036585364,"special_character_ratio":0.29683158,"punctuation_ratio":0.082633056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99699706,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T03:19:51Z\",\"WARC-Record-ID\":\"<urn:uuid:b6da3359-c53f-465e-9cc2-7f82a6747abe>\",\"Content-Length\":\"28844\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2f23564-d52d-4648-ad68-0c5661142825>\",\"WARC-Concurrent-To\":\"<urn:uuid:652fd9be-373f-4487-a5f8-6a187017acfa>\",\"WARC-IP-Address\":\"172.64.142.37\",\"WARC-Target-URI\":\"https://physicscatalyst.com/heat/changes-of-phases.php\",\"WARC-Payload-Digest\":\"sha1:I4N4B5HSOZ5NE653TZHQFX3BBI6U4LLV\",\"WARC-Block-Digest\":\"sha1:GW3TZLX4RI55TKC6LQKIOZIXPB3SUVQI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363437.15_warc_CC-MAIN-20211208022710-20211208052710-00443.warc.gz\"}"} |
http://mizugadro.mydns.jp/t/index.php?title=Pu-241&diff=26057&oldid=26056 | [
"# Difference between revisions of \"Pu-241\"\n\nPu-241 is isotope of Plutonium.\n\nmass: 241.057 amu\n\n## Production\n\nU-238 + n $\\rightarrow$ Np-239 + $\\beta^-$\n\nNp-239 $\\xrightarrow[~]{\\beta^-}$ Pu-239\n\nPu-239 + n $\\rightarrow$ Pu-240\n\nPu-240 + n $\\rightarrow$ Pu-241\n\nMost or nuclear reactors have significant concentration of U-238 in the active rods; so, production of plutonium happens automatically, whether it is wanted of not.\n\nAt accidents with nuclear plants, isotopes of Plutonium appear as one of main contaminants. Then, it decays to Americium, that becomes the main contaminant at the scale larger than a century.\n\n## Using\n\n$\\rm ^{241}_{~94}Pu \\xrightarrow[15.4~y]{\\beta^-,~ 21\\,KeV} {} ^{241}_{~95}Am \\xrightarrow[432.7\\,y]{\\alpha,\\, 4.959\\,MeV} {}^{237}_{~96}Np \\xrightarrow[2.4 \\times 10^6\\,y]{\\alpha,~4.958 \\,MeV} \\mathrm{^{233}_{91}Pa} ~.\\,.\\,.$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6291422,"math_prob":0.9829027,"size":1690,"snap":"2020-45-2020-50","text_gpt3_token_len":565,"char_repetition_ratio":0.12930012,"word_repetition_ratio":0.35772356,"special_character_ratio":0.41183433,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98768985,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T10:31:38Z\",\"WARC-Record-ID\":\"<urn:uuid:5837670e-3ef1-4a96-8600-d0cbef3384de>\",\"Content-Length\":\"24908\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bef35c62-5fb6-47f8-9a38-d9f7d2b6ea68>\",\"WARC-Concurrent-To\":\"<urn:uuid:225e0d3d-b16d-4a4d-b2e5-dbe9cc54ebe8>\",\"WARC-IP-Address\":\"123.98.230.254\",\"WARC-Target-URI\":\"http://mizugadro.mydns.jp/t/index.php?title=Pu-241&diff=26057&oldid=26056\",\"WARC-Payload-Digest\":\"sha1:ERU7RXY62KYRRFQZLHC6A6LN5DMXHSE6\",\"WARC-Block-Digest\":\"sha1:DLBFEADGM36LOY5NQDVLWTHK3LY4TTZT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197593.33_warc_CC-MAIN-20201129093434-20201129123434-00229.warc.gz\"}"} |
https://www.imooc.com/wenda/detail/523884 | [
"/ 猿问\n\n# C语言中swap的作用和用法?\n\n2019-02-06 16:11:17\n\n#include<stdio.h>void main() { void swap(int * p1,int * p2; int a,b; int * pointer_1,* pointer_2; scanf(\"%d,%d\",&a,&b); pointer_1=&a;pointer_2=&b if(a<b) swap(pointer_1,pointer_2); printf(\"\\n%d,%d\\n\",a,b); }void swap(int *p1,int *p2) 输入 5，9 输出9，5 {int temp; temp=*p1; *p1=*p2 ; *p2=temp;\n\n## 3 回答",
null,
"swap函数一般是一个程序员自定义函数。通常是实现两个变量数值的交换。比如\nint a = 2;\nint b =3;\nswap(a,b); //一般用到变量数值交换，交换后a=3 b = 2;\n\n1、通过使用临时变量实现交换。\nvoid swap1（int x,int y）\n{\nint temp;\ntemp=x;\nx=y;\ny=temp;\n}\n2、通过临时指针变量实现交换。\n\nvoid swap2（int *x,int *y）\n{\nint *temp;\ntemp=x;\nx=y;\ny=temp;\n}\n3、借助指针加入临时变量来实现交换。\n\nvoid swap3（int *x,int *y）\n{\nint temp;\ntemp=*x;\n*x=*y;\n*y=temp;\n}",
null,
"swap函数一般是一个程序员自定义函数。通常是实现两个变量数值的交换。比如\n\n 123 `int` `a = 2;``int` `b =3;``swap(a,b); ``//一般用到变量数值交换，交换后a=3 b = 2;`\n\n实现的方法多种多样。比如下面几种写法：\n\n1、通过使用临时变量实现交换。\n\n 1234567 ` ``void` `swap1（``int` `x,``int` `y）`` ``{ `` ``int` `temp; `` ``temp=x;`` ``x=y;`` ``y=temp;`` ``}`\n\n2、通过临时指针变量实现交换。\n\n 1234567 ` ``void` `swap2（``int` `*x,``int` `*y）`` ``{`` ``int` `*temp;`` ``temp=x;`` ``x=y;`` ``y=temp;`` ``}`\n\n3、借助指针加入临时变量来实现交换。\n\n 1234567 ` ``void` `swap3（``int` `*x,``int` `*y）`` ``{`` ``int` `temp;`` ``temp=*x;`` ``*x=*y;`` ``*y=temp;`` ``}`",
null,
"swap函数一般是一个程序员自定义函数。\n\n1. 通常是实现两个变量数值的交换,用法比较广泛。\n\n2. 可使用临时变量实现交换；可通过临时指针变量实现交换；可借助指针加入临时变量来实现交换。\n\n3. return 0;\n\n4. }\n\n5. swap1: x:4,y:3\n\n6. swap2: x:4,y:3\n\n7. swap3: x:3,y:4\n\n8. swap4: x:4,y:3\n\n9. swap5: x:3,y:4\n\n10. swap6: x:4,y:3\n\n11. swap7:ppx:3,ppy:4",
null,
"• 3 回答\n• 0 关注\n• 446 浏览\n\n0/150"
] | [
null,
"https://img4.mukewang.com/545868cd00013bbb02200220-100-100.jpg",
null,
"https://img4.mukewang.com/545862aa0001f8da02200220-100-100.jpg",
null,
"https://img2.mukewang.com/545863aa00014aa802200220-100-100.jpg",
null,
"https://img.mukewang.com/5c9497ca0001481005110314.jpg",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.6553067,"math_prob":0.99498105,"size":1404,"snap":"2019-51-2020-05","text_gpt3_token_len":899,"char_repetition_ratio":0.14785714,"word_repetition_ratio":0.3151515,"special_character_ratio":0.37749287,"punctuation_ratio":0.26801154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99451727,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-05T14:50:51Z\",\"WARC-Record-ID\":\"<urn:uuid:e874fd42-2e72-4e22-8b41-26e591e7b6ae>\",\"Content-Length\":\"38896\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31647aa6-1522-4c8e-87e5-2e707ea31daa>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ccb7eea-4e56-4c8c-b088-d126371042c6>\",\"WARC-IP-Address\":\"117.121.101.40\",\"WARC-Target-URI\":\"https://www.imooc.com/wenda/detail/523884\",\"WARC-Payload-Digest\":\"sha1:ZTKPG3KNMNYJBSNJXXPM2C6W6KL4V45H\",\"WARC-Block-Digest\":\"sha1:YG7J6DTEV33X6C3ZC7PYC445H2YD3W35\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540481076.11_warc_CC-MAIN-20191205141605-20191205165605-00054.warc.gz\"}"} |
https://www.khanacademy.org/math/mappers/operations-and-algebraic-thinking-220-223/x261c2cc7:exponents/v/powers-of-fractions | [
"If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## MAP Recommended Practice\n\n### Course: MAP Recommended Practice>Unit 17\n\nLesson 10: Exponents\n\n# Powers of fractions\n\nRaising fractions to a power is a fundamental math skill. To do this, simply multiply the numerators and denominators separately. For example, when raising a fraction like 2/3 to the third power, multiply the numerators (2 × 2 × 2 = 8) and the denominators (3 × 3 × 3 = 27) to get the result 8/27. This method applies to all powers, making it easy to understand and apply. Created by Sal Khan.\n\n## Video transcript\n\nLet's go through more exponent examples. So to warm up, let's think about taking a fraction to some power. So let's say I have 2/3, and I want to raise it to the third power here. Now, we've already learned there are two ways of thinking about this. One way is to say let's take three 2/3's. So that's one 2/3, two 2/3's, and three 2/3's. So that's one, two, three, 2/3. And then we multiply them. And we will get-- let's see, the numerator will be 2 times 2 times 2, which is 8. And the denominator's going to be 3 times 3 times 3 times 3, which is equal to 27. Now, the other way of viewing this is you start with a 1, and you multiply it by 2/3 three times. So you multiply by 2/3 once, twice, three times. You will get the exact same result here. So let's do one more example like that. So lets say I had 4/9, and I want to square it. When I raise something to the second power, people often say, you're squaring it. Also, raising something to the third power, people sometimes say, you're cubing it. But let's raise 4/9 to the second power. Let's square it. And I encourage you to pause the video and work this out yourself. Well, once again, you could view this as taking two 4/9's and multiplying them. Or you could view this as starting with a 1, and multiplying it by 4/9 two times. Either way, your numerator is going to be 4 times 4, which is 16. And your denominator is going to be 9 times 9, which is equal to 81."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9422182,"math_prob":0.93884134,"size":4564,"snap":"2023-40-2023-50","text_gpt3_token_len":1332,"char_repetition_ratio":0.12434211,"word_repetition_ratio":0.022857143,"special_character_ratio":0.2971078,"punctuation_ratio":0.10308285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921814,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T09:15:24Z\",\"WARC-Record-ID\":\"<urn:uuid:b4d960e4-71bd-437a-a000-d8a2466251b9>\",\"Content-Length\":\"1049637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d08261f-e0fa-4517-89c1-e7595d533dda>\",\"WARC-Concurrent-To\":\"<urn:uuid:036f5608-8cf0-4426-bd7e-695235d5fe5d>\",\"WARC-IP-Address\":\"146.75.29.42\",\"WARC-Target-URI\":\"https://www.khanacademy.org/math/mappers/operations-and-algebraic-thinking-220-223/x261c2cc7:exponents/v/powers-of-fractions\",\"WARC-Payload-Digest\":\"sha1:MV6HVZRCOR3D74WPLYVJB3EGESQQCRHP\",\"WARC-Block-Digest\":\"sha1:OTWTTR2ZEL62MIEMP7ZBL2WLILDT2NVX\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103810.88_warc_CC-MAIN-20231211080606-20231211110606-00489.warc.gz\"}"} |
https://www.geeksforgeeks.org/half-adder-in-digital-logic/ | [
"# Half Adder in Digital Logic\n\nThe addition of 2bits is called Half adder the input variebles are augent and addent bits and output variebles are sum&carry bits. A and B are the two input bits",
null,
"Truth Table",
null,
"Logical Expression\n\nSum = A XOR B\nCarry = A AND B\n\nImplementation",
null,
""
] | [
null,
"https://media.geeksforgeeks.org/wp-content/cdn-uploads/gq/2015/07/halfadder11-300x128.png",
null,
"https://media.geeksforgeeks.org/wp-content/cdn-uploads/gq/2015/07/ha_truth-300x235.png",
null,
"https://media.geeksforgeeks.org/wp-content/cdn-uploads/Half_Adder.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71228355,"math_prob":0.9461806,"size":1336,"snap":"2020-34-2020-40","text_gpt3_token_len":305,"char_repetition_ratio":0.22372372,"word_repetition_ratio":0.018099548,"special_character_ratio":0.19685629,"punctuation_ratio":0.033333335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9890342,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,7,null,10,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T23:58:57Z\",\"WARC-Record-ID\":\"<urn:uuid:dd830c9a-a250-4f9f-b860-5526275585b5>\",\"Content-Length\":\"95613\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76bdb1a1-83e5-4cc4-b1ae-64072cdd4a26>\",\"WARC-Concurrent-To\":\"<urn:uuid:1778781d-f1bf-44da-bc13-cf8b354deee4>\",\"WARC-IP-Address\":\"23.199.71.8\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/half-adder-in-digital-logic/\",\"WARC-Payload-Digest\":\"sha1:7VYDC3JUJKD6PFDPAACGQLAJGEWJ6PT6\",\"WARC-Block-Digest\":\"sha1:M6D2KM4RE5MPFYXI45MQVX7YY5R2B2Q4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738366.27_warc_CC-MAIN-20200808224308-20200809014308-00512.warc.gz\"}"} |
https://vdocuments.mx/education/sdt-topic-08.html | [
"# sdt topic - 08\n\nPost on 18-Jan-2015\n\n477 views\n\nCategory:\n\n## Education\n\nTags:\n\n• #### single number\n\nEmbed Size (px)\n\nDESCRIPTION\n\nTRANSCRIPT\n\n• 1. Topic 8 : Array Software Development Techniques\n\n2. Finding the Highest Number of 3 inputs 3. Finding the Highest Number of 6 inputs 4. Introduction Until now we have been working with single units of data. A whole number contains a single number, no more. Today we are going to introduce a new kind of data structure. The Array This is a list of similarly types variables. A list of whole numbers A list of characters A list of string 5. The Problem So Far Scalability We can write a program to discover the largest number in a set of five numbers. But that becomes a much more time-consuming prospect if we want to find the largest in a set of thousand numbers. The array is used to manage this problem. 6. The Array We decide on the size of an array. We decide on the type of things it will store. In our pseudocode, we can create an array using the following syntax: data myArray as array of (10) whole numbers This creates an array of whole numbers, and provides ten compartments starting the compartment number from 0. So, compartment number of first compartment is 0. Compartment number five is the sixth compartment in an array. 7. The Array We can put something into a compartment like so: myArray = 100 This puts the value 100 into the compartment marked as 5 It is the sixth compartment. myArray = 10 myArray = 40 We can get information out too: output myArray 8. The Array The number that indicates a compartment is known technically as an index. The compartment is known as the element. 9. The Structure of an Array Index Value 0 10 1 50 2 203 3 56 4 79 This is an array of size five. The contents of myArray would be 203 The square brackets indicate the index 10. Question Draw a diagram of an array called myNums which has been initialized to hold 5 whole numbers. Your diagram should indicate the structure of the array, with the different parts clearly labelled. 11. Finding the Highest Number of 3 inputs 12. Finding the Highest Number of 3 inputs 13. Passing an Array to a Function When we pass an array to a function, we do not specify its size. There is a size of function that gives you the size of the array. But not how many elements have been filled. 14. Passing an Array to a Function Pseudocode to pass data as array of whole number in function and display it. 15. Main Program 16. Desk-Checking Main Function Line Num numbers Remarks 1 0, 0, 0 2 20, 0, 0 3 20, 30, 0 4 20, 30, 15 5 20, 30, 15 Call display(numbers) 17. Desk-Checking an Array Line Num nums i len Output Remarks 1 20, 30, 15 0 2 20, 30, 15 0 0 3 20, 30, 15 0 3 4 20, 30, 15 0 3 Loop while i is less than len(3) 5 20, 30, 15 0 3 20 6 20, 30, 15 1 3 7 20, 30, 15 1 3 Goto line 4 4 18. Returning an Array You can return arrays from your functions just like they were our usual variables. There are some complications though If you return an array, it will overwrite in every the array you are using to hold it. It will overwrite the size. It will overwrite existing contents. You can also declare an array without giving a size. But that sets it as null in your desk-check. 19. Returning an Array 20. Main Program 21. Desk-Checking Main Function Line Num b i sizeof(b) output Remarks 1 null 2 null 0 3 20, 40, 57 0 0 Returning from function 4 20, 40, 57 0 3 Loop while i is less than sizeof b (3). 5 20, 40, 57 0 3 20 6 20, 40, 57 1 3 7 20, 40, 57 1 3 Goto 4 4 22. Desk-Checking Function Line Num a Remarks 1 0, 0, 0 2 20, 0, 0 3 20, 40, 0 4 20, 40, 57 5 20, 40, 57 return a 23. Voting Scenario Write pseudo code and source code for following scenario. The number of different parties standing in the election. This must be a whole number between 2 and 10 inclusive. Take the name of parties. The number of votes cast for each party standing in the election. These must be whole numbers between 0 and 600,000 inclusive. 24. Two Dimensional Arrays The concepts are identical. You just need two indexes, and two sets of sizes. These are grids of data. Two dimensional arrays allow for several kinds of programs that would otherwise be very difficult. Especially games 25. Two Dimensional Arrays Creating data myGrid as array (10, 10) of whole numbers Or Data myGrid as array (10)(10) of whole numbers Setting myGrid[10, 5] = 100 Or myGrid = 100 26. Two Dimensional Arrays Getting Output myGrid [10, 5] Or Output myGrid 27. Two Dimensional Arrays Example 1. data i as whole number 2. data j as whole number 3. data str as array(2)(2) of String 4. str = \"Nepal\" 5. str = \"America\" 6. str = \"Australia\" 7. str = \"England\" 28. Two Dimensional Arrays Example 8. i = 0 9. loop while i is less than sizeof(str) 10. j = 0 11. loop while j is less than sizeof(str[i]) 12. Output str[i][j] 13. j = j + 1 14. end loop 15. i = i + 1 16.end loop 29. End of Topic 8 Software Development Techniques\n\nRecommended"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83165574,"math_prob":0.96749246,"size":4955,"snap":"2021-04-2021-17","text_gpt3_token_len":1443,"char_repetition_ratio":0.13694203,"word_repetition_ratio":0.052742615,"special_character_ratio":0.31422806,"punctuation_ratio":0.12969589,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.95174843,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-20T15:53:09Z\",\"WARC-Record-ID\":\"<urn:uuid:76dc9bed-05cf-4303-b0d5-c2075c6461ce>\",\"Content-Length\":\"84220\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9470027e-e000-4e3b-a04e-333679416eaf>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7182f21-5785-4248-9f45-45ad29d8f516>\",\"WARC-IP-Address\":\"5.196.140.100\",\"WARC-Target-URI\":\"https://vdocuments.mx/education/sdt-topic-08.html\",\"WARC-Payload-Digest\":\"sha1:ZPM4E6GQJFQ4XHKIRKSEPDMJ3Z7BCPKN\",\"WARC-Block-Digest\":\"sha1:O622WKDIFXY6FUE2NGOEVCVMVFK5LRA3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703521139.30_warc_CC-MAIN-20210120151257-20210120181257-00026.warc.gz\"}"} |
https://discuss.kotlinlang.org/t/issue-with-mapto-on-a-sequence/21832 | [
"",
null,
"# Issue with mapTo() on a sequence\n\nHi,\n\nwhen writing code like\n\n``````fun main() {\nval a = mutableListOf<Int>()\nval b = setOf(1, 2, 3).asSequence().mapTo(a) { it * 2 }.find { it == 4 }\nprintln(a)\nprintln(b)\n}\n``````\n\nI would have assumed that printing `a` would result in only `[2, 4]` whereas it actually results in `[2, 4, 6]`. Why is that? Isn’t the whole point a sequence to only evaluate those mappings that are needed to make `find` return true?\n\nWhatever the answer is, any hints on how to actually get the behavior that only those mappings land in `a` until `find` succeeds?\n\nThanks in advance for any insights!\n\nmapTo() is a terminal operation; it forces the complete sequence to be evaluated, and then returns the destination — the list `a`, which then gets scanned in the usual way.\n\nNormally in this sort of case I’d expect to see a simple `map()` (which is an intermediate operation, i.e. not terminal).⠀I’m guessing you’re using `mapTo()` in order to see the sequence behaviour — but unfortunately, it changes that behaviour!\n\nOne way to see how the sequence behaves would be to use a straight `map()` (ignoring `a` completely), and add side-effects in the lambdas, e.g.:\n\n``````val b = setOf(1, 2, 3).asSequence()\n.map{ println(\"Doubling \\$it\"); it * 2 }\n.find{ println(\"Checking \\$it\"); it == 4 }\nprintln(\"Result: \\$b\")\n``````\n\nThis prints:\n\n``````Doubling 1\nChecking 2\nDoubling 2\nChecking 4\nResult: 4\n``````\n\nThis shows that only the first two elements of the set get evaluated, doubled, and checked — so the sequence is behaving lazily, as expected.\n\n1 Like\n\nBut from an API design point of view, what’s the reason for `mapTo()` being terminal, but `map()` not? To me that design seems counter-intuitive, as the only difference between those functions is the way the target collection is specified.\n\nI’m not only using `mapTo()` here for demonstration purposes, but because I’m really interested in all mapped values until a specific value was found in some code that follows this block, and I really do not want to map unnecessary original values, as in my case the mapping is an expensive operation.\n\nBecause with mapTo you ask “transform this and put this in the collection and after filter” and with map only “transform this and filters”\n\nYou can emulate the behaviour that you want with this:\n\n``````\nfun main() {\nval a = mutableListOf<Int>()\nval b = setOf(1, 2, 3).asSequence().map { it * 2 }.onEach{ a.add(it) }. find { it == 4 }\nprintln(a)\nprintln(b)\n}\n\n``````\n2 Likes\n\nThanks, finally a good use for `onEach()`",
null,
"2 Likes"
] | [
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"https://aws1.discourse-cdn.com/business5/uploads/kotlinlang/original/2X/6/6ec18b0a413f764371c55b3fc4fe21387eaddc1d.svg",
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"https://emoji.discourse-cdn.com/twitter/slight_smile.png",
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https://samacheer-kalvi.com/samacheer-kalvi-11th-computer-science-guide-chapter-14/ | [
"Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 14 Classes and Objects Text Book Back Questions and Answers, Notes.\n\n## Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 14 Classes and Objects\n\n### 11th Computer Science Guide Classes and Objects Text Book Questions and Answers\n\nBook Evaluation\n\nPart I\n\nQuestion 1.\nThe variables declared inside the class are known as data members and the functions are known as\na) data functions\nb) inline functions\nc) member functions\nd) attributes\nc) member functions\n\nQuestion 2.\nWhich of the following statements about member functions are True or False?\ni) A member function can call another member function directly with using the dot operator.\nii) Member function can access the private data of the class.\na) i-True, ii-True\nb) i-False, ii-True\nc) i-True, ii-False\nd) i-False, ii-False\nb) i-False, ii-True",
null,
"Question 3.\nA member function can call another member function directly, without using the dot operator called as\na) sub function\nb) sub member\nc) nesting of member function\nd) sibling of member function\nc) nesting of member function\n\nQuestion 4.\nThe member function defined within the class behave like\na) inline functions\nb) Non inline function\nc) Outline function\nd) Data function\na) inline functions\n\nQuestion 5.\nWhich of the following access specifier protects data from inadvertent modifications?\na) Private\nb) Protected\nc) Public\nd) Global\na) Private",
null,
"Question 6.\nclass x\n{\ninty;\npublic:\nx(int z)\n{\ny=z;\n}\n} x1;\nintmain( )\n{\nx x2(10);\nreturn 0;\n}\nHow many objects are created for the above program?\na) 10\nb) 14\nc) 5\nd) 2\nc) 5\n\nQuestion 7.\nState whether the following statements about the constructor are True or False.\ni) constructors should be declared in the private section.\nii) constructors are invoked automatically when the objects are created.\na) True, True\nb) True, False\nc) False, True\nd) False, False\nc) False, True",
null,
"Question 8.\nWhich of the following constructor is executed for the following prototype ?\na) Default constructor\nb) Parameterized constructor\nc) Copy constructor\nd) Non Parameterized constructor\nc) Copy constructor\n\nQuestion 9.\nWhat happens when a class with parameterized constructors and having no default constructor is used in a program and we create an object that needs a zero- argument constructor?\na) Compile-time error\nb) Domain error\nc) Runtime error\nd) Runtime exception\na) Compile-time error\n\nQuestion 10.\nWhich of the following create a temporary instance?\na) Implicit call to the constructor\nb) Explicit call to the constructor\nc) Implicit call to the destructor\nd) Explicit call to the destructor\nb) Explicit call to the constructor",
null,
"Part – II\n\nQuestion 1.\nWhat are called members?\nThe class comprises members. Members are classified as Data Members and Member functions. Data members are the data variables that represent the features or properties of a class. Member functions are the functions that perform specific tasks in a class.\n\nQuestion 2.\nDifferentiate structure and class though both are user-defined data types.\nThe only difference between structure and class is the members of the structure are by default public whereas it is private in class.\n\nQuestion 3.\nWhat is the difference between the class and object in terms of oop?\nObject:\n\n• Object is an instance of a class.\n• Object is a real-world entity such as pen, laptop, mobile, chair, etc.\n• Object allocates memory when it is created.\n\nClass:\n\n• Class is a blueprint or template from which objects are created.\n• Class is a group of similar objects.\n• Class doesn’t allocate memory when it is created.\n\nQuestion 4.\nWhy it is considered a good practice to define a constructor though a compiler can automatically generate a constructor?\nA user-defined constructor is the best method of initialise array of objects and normal objects.",
null,
"Question 5.\nWrite down the importance of the destructor.\nThe purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object.\n\nPart – III\n\nQuestion 1.\nRewrite the following program after removing the syntax errors if any and underline the errors:\n#include<iostream>\n#include<stdio.h>\nclassmystud\n{ intstudid =1001;\nchar name;\npublic\nmystud( )\n{ }\nvoid register ( ) {cin>>stdid;gets(name);\n}\nvoid display ( )\n{ cout<<studid<<“: “<<name<<endl;}\n}\nint main( )\n{ mystud MS;\nregister.MS( );\nMS.display( );\n}\nMODIFIED PROGRAM:\n#include<iostream>\n#include<stdio.h>\nclass mystud\n{\nint studid;\nchar name;\npublic:\nmystud( )\n{\nstudid=1001;\n}\nvoid register ( )\n{\ncin>>stdid;\ngets(name);\n}\nvoid display ( )\n{\ncout<<studid<<“: “<<name<<endl;\n}\n};\nint main( )\n{\nmystud MS;\nMS.reqister( );\nMS.display( );\n}",
null,
"Question 2.\nWrite with an example of how will you dynamically initialize objects?\nDynamic initialization of Objects:\nWhen the initial values are provided during runtime then it is called dynamic initialization.\nProgram to illustrate dynamic initialization\n#include\nusing namespace std;\nclass X\n{\nint n;\nfloat avg;\npublic:\nX(int p,float q)\n{\nn=p;\navg=q;\n}\nvoid disp( )\n{\ncout<<“\\n Roll numbe:-” <<n;\ncout<<“\\nAverage :-“<<avg;\n}\n};\nint main( )\n{\nint a ; float b;\ncout<<“\\nEnter the Roll Number”;\ncin>>a; .\ncout<<“\\nEnter the Average”;\ncin>>b;\nX x(a,b); // dynamic initialization\nx.disp( );\nreturn 0;\n}\nOutput\nEnter the Roll Number 1201\nEnter the Average 98.6\nRoll number:- 1201\nAverage :- 98.6",
null,
"Question 3.\nWhat are advantages of declaring constructors and destructor under public access ability?\n\nWhen constructor and destructor are declared under public:\n\n1. we can initialize the object while declaring it.\n2. we can explicitly call the constructor.\n3. we can overload constructors and therefore use multiple constructors to initialize objects automatically.\n4. we can destroy the objects at the end of class scope automatically (free unused memory).\n\nHowever, some C++ compiler and Dev C++ do not allow to declare constructor and destructor under private section. So it is better to declare constructor and destructor under public section only.\n\nQuestion 4.\nGiven the following C++ code, answer the questions (i) & (ii).\nclass TestMeOut\n{\npublic:\n~TestMeOut( ) //Function 1\n{\ncout<<“Leaving the examination hall”<<endl;\n}\nTestMeOut( ) //Function 2\n{\ncout<<“Appearing for examination'<<endl;\n}\nvoid MyWork( ) //Function 3\n{\ncout<<“Attempting Questions//<<endl;\n}\n};\ni) In Object-Oriented Programming, what is Function 1 referred to as and when does it get invoked/called?\nFunction 1 is called a destructor. It will be automatically invoked when the object goes out of scope (ie. at the end of a program).\n\nii) In Object-Oriented Programming, what is Function 2 referred to as, and when does it get invoked/called?\nFunction2 is called a constructor. It will be automatically invoked when an object comes into scope.(ie. at the time of object creation).",
null,
"Question 5.\nWrite the output of the following C++ program code:\n#include<iostream>\nusing namespace std;\nclass Calci\n{\nint Bonus;\npublic:\nCalci( )\n{\nBonus=0;\n}//ascii value of A=65\nvoid Down(int G)\n{\n}\nvoid Up(int G)\n{\nBonus++;\n}\nvoid Show( )\n{\n}\n};\nint main( )\n{\nCalci c;\nc.Down(3);\nc.Show( );\nc.Up(7);\nc.Show( );\nc.Down(2);\nc.Show( );\nreturn 0;\n}\nOutput",
null,
"",
null,
"Part – IV\n\nExplain In Detail\n\nQuestion 1.\nExplain nested class with example.\nWhen one class becomes a member of another class then it is called Nested class and the relationship is called containership. When a class is declared within another class, the inner class is called a Nested class (i.e. the inner class) and the outer class is known as the Enclosing class. The nested class can be defined in private as well as in the public section of the Enclosing class.\n\nClasses can be nested in two ways:\n\n1. By defining a class within another class\n2. By declaring an object of a class as a member to another class\n3. By defining a class within another class\n\nC++ program to illustrate the nested class\n#include<iostream>\nusing namespace std;\nclass enclose\n{\nprivate:\nint x;\nclass nest\n{\nprivate :\nint y;\npublic:\nint z;\nvoid prn( )\n{\ny=3;z=2;\ncout<<“\\n The product of”\n< <y< <‘*'< <z<<“= “< <y*z< <“\\n”;\n}\n}; //inner class definition over\nnest m1;\npublic:\nnest n2;\nvoid square( )\n{\nn2.prn( ); //inner class member function is called by its object\nx=2;\nn2.z=4;\ncout<<“\\n The product of” <<n2.z<<‘*'<<n2.z<<“=”n2.z*n2.z<<“/n”;\ncout<<“\\n The product of” <<x<<‘*'<<x<<“= “<<x*x;\n}\n}; //outer class definition over\nint main( )\n{\nenclose e;\ne.square( ); //outer class member function is called\n}\nOutput\nThe product of 3*2=6\nThe product of 4*4=16\nThe product of 2*2=4\n\nIn the above program the inner class nest is defined inside the outer class enclose, nest is accessed by enclose by creating an object of nest",
null,
"Question 2.\nMention the differences between constructor and destructor.\n\n CONSTRUCTOR DESTRUCTOR The name of the constructor must be same as that of the class. The destructor has the same name as that of the class prefixed by the tilde character A constructor can have parameter list. The destructor cannot have arguments. The constructor function can be overloaded. Destructors cannot be overloaded i.e., there can be only one destructor in a class. Constructor cannot be inherited but a derived class can call the base class constructor. Destructor cannot be inherited. The constructor is executed automatically when the object is created. The destructor is executed automatically when the control reaches the end of class. Allocated memory space for the object. Destroy the object.\n\nQuestion 3.\nDefine a class RESORT with the following description in C++ :\nPrivate members:\nRno // Data member to storeroom number\nName //Data member to store user name\nCharges //Data member to store per day charge\nDays //Data member to store the number of days\nCompute ( ) // A function to calculate total amount as Days * Charges and if the\n//total amount exceeds 11000 then total amount is 1.02 * Days *Charges\n\nPublic member:\ngetinfo( ) // Function to Read the information like name , room no, charges and days\ndispinfo ( ) // Function to display all entered details and total amount calculated\n//using COMPUTE function\nPROGRAM\nusing namespace std;\n#include<iostream>\nclass RESORT\n{\nprivate:\nint Rno,Days,Charges;\nchar Rname;\nint compute( )\n{\nif (Days * Charges >11000)\nreturn (Days * Charges * 1.02);\nelse\nreturn(Days * Charges);\n}\npublic:\ngetinfo( )\n{\ncout<<“\\nEnter customer name : “;\ncin>>Rname;\ncout<<‘nEnter charges per day : “;\ncin>>Charges;\ncout< <‘nEnter Number of days : “;\ncin>>Days;\ncout<<‘n Enter Room Number : “;\ncin>>Rno;\n}\ndispinfo( )\n{\ncout<<‘nRoom Number :\n“<<Rno;\ncout<<‘nCustomer name :\n“<<Rname;\ncout<<‘nCharges per day :\n“<<Charges;\ncout<<‘nNumber of days :\n“<<Days;\ncout<<‘nTotal Amount :\n“<<compute( );\n}\nint main( )\n{\nRESORT Obj;\nObj,getinfo( );\nObj.dispinfo( );\nOutput",
null,
"",
null,
"Question 4.\nWrIte the output of the following:\n#include<iostream>\n#indude<stdio.h>\nusing namespace std;\nclass sub\n{\nint day, subno;\npublic :\nsub(int,int); // prototype\nvoid printsub( )\n{\ncout<<” subject number: “<<subno;\ncout<<” Days : ” <<day;\n}\n};\nsub::sub(int d=150,int sn=12)\n{\ncout<<endl<<“Constructing the object\n“<<endl;\nday=d;\nsub no=sn;\n}\nclass stud ‘\n{\nint rno;\nfloat marks; public:\nstud( )\n{\ncout<< “Constructing the object of\nstudents “<<endl;\nrno=0;\nmarks=0.0;\n}\nvoid getval( )\n{\ncout<<“Enter the roll number and the marks secured”; cin>>rno>>marks;\n}\nvoid printdet( )\n{\ncout<<“Roll no : “<<rno<<“Marks : “<<marks<<endl; .\n}\n};\n{\nsub obj;\nstud objone;\nfloat fees; ,\npublic :\n{\ncout<< “Constructing the object of admission “<<endl;\nfees=0.0;\n}\nvoid printdet( )\n{\nobjone.printdet( );\nobj.printsub();\ncout<<“fees : “<<fees<<endl;\n}\n};\nint main( )\n{\nsystem (“cls”);\ncout<<endl<< ?’Back in main ()”;\nreturn 0;\n}\nOutput",
null,
"",
null,
"Question 5.\nWrite the output of the following.\n#indude<iostream>\n#include<stdio,h>\nusing namespace std;\nclass P\n{\npublic:\nP( )\n{\ncout<< “\\nConstructor of class P }\n~P( )\n{\ncout< < “\\nDestructor of class P\n}\n};\nclass Q\n{\npublic:\nQ( )\n{\ncout< <“\\nConstructor of class Q “;}\n~ Q( )\n{\ncout<< “\\nDestructor of class Q\n}\n};\nclass R\n{\nP obj1, obj2;\nQ obj3;\npublic:\nR( )\n{\ncout<< “\\nConstructor of class R “;}\n~R( )\n{\ncout<< “\\nDestructor of class R\n}\n};\nint main ( )\n{\nR oR; :\nQ oq;\nPop;\nreturn 0;\nOutput",
null,
"",
null,
"### 11th Computer Science Guide Classes and Objects Additional Questions and Answers\n\nChoose The Correct Answer (1 Mark)\n\nQuestion 1.\nThe most important feature of C++ is ………………..\n(a) object\n(b) class\n(c) public\n(d) All the above\n(b) class\n\nQuestion 2.\nHow many features are commonly present in OOP languages?\na) 3\nb) 2\nc) 4\nd) 5\nc) 4\n\nQuestion 3.\nCalling a member function of an object is also known as ……………….. to object.\n(a) call function\n(b) call by value\n(c) call by reference\n(d) sending message\n(d) sending message",
null,
"Question 4.\n……………… is a way to bind the data and its associated functions together,\na) Class\nb) Array\nc) Structure\nd) All the above\na) Class\n\nQuestion 5.\nWhen one class become a member of another class, the relationship is called ………………..\n(a) containership\n(b) partnership\n(c) friendship\n(d) all the above\n(a) containership\n\nQuestion 6.\nThe body of the class is defined inside the ………………. brackets.\na) Angle < >\nb) Square [ ]\nc) Curly { }\nd) None of these\nc) Curly { }",
null,
"Question 7.\n……………….. can be defined either in private or in the public section of a class.\n(a) Object\n(b) Data type\n(c) Memory\n(d) constructor\n(d) constructor\n\nQuestion 8.\nThe members of the structure are by default ………………..\na) Private\nb) Public\nc) Protected\nd) None of these\nb) Public\n\nQuestion 9.\nThere are ……………….. ways to create an object using the parameterized constructor.\n(a) 3\n(b) 2\n(c) 1\n(d) 4\n(c) 1\n\nQuestion 10.\nThe class body contains ………………….\na) Data members\nb) Member functions\nc) Both A and B\nd) None of these\nc) Both A and B",
null,
"Question 11.\nThe class body has………………… access specifiers.\na) Three\nb) Four\nc) Two\nd) Five\na) Three\n\nQuestion 12.\nThe class body has…………….. access specifiers.\na) Private\nb) Public\nc) Protected\nd) All the above\nd) All the above\n\nQuestion 13.\n…………………. is a visibility label.\na) Private\nb) Public\nc) Protected\nd) All the above\nd) All the above\n\nQuestion 14.\n……………….. allows preventing the functions of a program to access directly the internal representation of a class type.\na) Data Hiding\nb) Data Capturing\nc) Data Processing\nd) None of these\na) Data Hiding",
null,
"Question 15.\nThe access restriction to the class members is specified by ……………. section within the class\nbody.\na) Private\nb) Public\nc) Protected\nd) All the above\nd) All the above\n\nQuestion 16.\nA …………………. member is accessible from where outside the class but within a program.\na) Private\nb) Public\nc) Protected\nd) All the above\nb) Public\n\nQuestion 17.\nWe can set and get the value of public data members using ………………… function.\na) Member\nb) Nonmember\nc) Either A or B\nd) None of these\nc) Either A or B",
null,
"Question 18.\nA …………….. member cannot be accessed from outside the class.\na) Private\nb) Public\nc) Protected\nd) All the above\na) Private\n\nQuestion 19.\nOnly the class member functions can access ……………… members.\na) Private\nb) Public\nc) Protected\nd) All the above\na) Private\n\nQuestion 20.\n……………….. members can be accessed in child classes.\na) Private\nb) Public\nc) Protected\nd) All the above\nc) Protected",
null,
"Question 21.\nIf all members of the class are defined as …………….. then the class become frozen.\na) Private\nb) Public\nc) Protected\nd) All the above\na) Private\n\nQuestion 22.\nIf all members of the class are defined as ………………….. then the object of the class can not access anything from the class,\na) Private\nb) Public\nc) Protected\nd) All the above\na) Private\n\nQuestion 23.\n………………. are the data variables that represent the features or properties of a class.\na) Data members\nb) Member functions\nc) Both A and B\nd) None of these\na) Data members\n\nQuestion 24.\n………………… are the functions that perform specific tasks in a class.\na) Data members\nb) Member functions\nc) Both A and B\nd) None of these\nb) Member functions",
null,
"Question 25.\nMember functions are called as …………………..\na) Methods\nb) Attributes\nc) Properties\nd) None of these\na) Methods\n\nQuestion 26.\nData members are also called as ……………….\na) Methods\nb) Attributes\nc) Properties\nd) None of these\nb) Attributes\n\nQuestion 27.\nClasses contain a special member function called as ……………………\na) Constructors\nb) Destructors\nc) Both A and B\nd) None of these\nc) Both A and B",
null,
"Question 28.\nThe member functions of a class can be defined in ……………….. ways.\na) Two\nb) Three\nc) Four\nd) Five\na) Two\n\nQuestion 29.\nThe member functions of a class can be defined in ………………. way.\na) Inside the class definition\nb) Outside the class definition\nc) Either A or B\nd) None of these\nc) Either A or B\n\nQuestion 30.\nWhen a member function is defined Inside a class, it behaves like ………………. functions.\na) Inline\nb) General\nc) Local\nd) None of these\na) Inline",
null,
"Question 31.\nIf a function is inline, the compiler places a copy of the code of that function at each point where the function is called at ………………….\na) Run Time\nb) Compile time\nc) Both A and B\nd) None of these\nb) Compile time\n\nQuestion 32.\nWhen Member function defined outside the class, and then it is be called as ………………….\nmember function.\na) Outline\nb) Non-inline\nc) Either A or B\nd) None of these\nc) Either A or B\n\nQuestion 33.\nWhen Member function defined outside the class using …………….. operator.\na) Scope resolution\nb) Membership\nc) Reference\nd) Conditional\na) Scope resolution",
null,
"Question 34.\nThe class variables are called ………………….\na) Object\nb) Attributes\nc) Procedures\nd) None of these\na) Object\n\nQuestion 35.\nObjects are also called as …………………. of class.\na) Instant\nb) Instance\nc) Attributes\nd) None of these\nb) Instance\n\nQuestion 36.\nObjects can be created in ………………… methods.\na) Three\nb) Four\nc) Two\nd) Five\nc) Two",
null,
"Question 37.\nObjects can be created as ……………….\na) Global object\nb) Local object\nc) Either A or B\nd) None of these\nc) Either A or B\n\nQuestion 38.\n…………….. objects can be used by any function in the program.\na) Global object\nb) Local object\nc) Either A or B\nd) None of these\na) Global object\n\nQuestion 39.\nIf an object is declared outside all the function bodies or by placing their names immediately after the closing brace of the class declaration then it is called as ……………….\na) Global object\nb) Local object\nc) Either A or B\nd) None of these\na) Global object",
null,
"Question 40.\nIf an object is declared within a function then it is called ……………….\na) Global object\nb) Local object\nc) Either A or B\nd) None of these\nb) Local object\n\nQuestion 41.\n……………….. object can not be accessed from outside the function.\na) Global\nb) Local\nc) Either A or B\nd) None of these\nb) Local\n\nQuestion 42.\nNo separate space is allocated for …………………. when the objects are created.\na) Member functions\nb) Data members\nc) Both A and B\nd) None of these\na) Member functions",
null,
"Question 43.\nMemory space required for the ……………….\na) Member functions\nb) Data members\nc) Both A and B\nd) None of these\nb) Data members\n\nQuestion 44.\nThe members of a class are referenced (accessed) by using the object of the class followed by the ………………. operator.\na) Scope resolution\nb) Conditional\nc) Dot (membership)\nd) None of these\nc) Dot (membership)\n\nQuestion 45.\nCalling a member function of an object is also known as ……………….\na) Sending a message to object\nb) Communication with the object\nc) Either A or B\nd) None of these\nc) Either A or B",
null,
"Question 46.\nAn array which contains the class type of element is called …………………\na) Array of objects\nb) Structure Objects\nc) Block of objects\nd) None of these\na) Array of objects\n\nQuestion 47.\nThe ………………… of the outline member function given in a class specification, instructs the compiler about its visibility mode.\na) Name\nb) Prototype\nC) Data type\nd) None of these\nb) Prototype\n\nQuestion 48.\nA member function can call another member function of the same class directly without using the dot operator is called ………………………\na) Nesting of the member function\nb) Invariant Members\nc) Variant Members\nd) None of these\na) Nesting of the member function",
null,
"Question 49.\nA member function can call another member function of the same class for that you do not\nneed a(n) …………………..\na) Member function\nb) Data Member\nc) Object\nd) None of these\nc) Object\n\nQuestion 50.\nA member function can access ………………. functions.\na) Public\nb) Private\nc) Both A and B\nd) None of these\nc) Both A and B\n\nQuestion 51.\n…………………. operator will reveal the hidden file scope(global) variable.\na) Membership\nb) Conditional\nc) Scope resolution\nd) All the above\nc) Scope resolution",
null,
"Question 52.\nWhen an object is passed by ……………… the function creates its own copy of the object and works on it.\na) Value\nb) Reference\nc) Either A or B\nd) None of these\na) Value\n\nQuestion 53.\nWhen an object is passed by …………….. changes made to the object inside the function do not affect the original object.\na) Value\nb) Reference\nc) Either A or B\nd) None of these\na) Value\n\nQuestion 54.\nWhen an object is passed by ……………….. its memory address is passed to the function so the called function works directly on the original object used in the function call.\na) Value\nb) Reference\nc) Either A or B\nd) None of these\nb) Reference",
null,
"Question 55.\nWhen an object is passed by ………………………. any changes made to the object inside the function definition are reflected in the original object.\na) Value\nb) Reference\nc) Either A or B\nd) None of these\nb) Reference\n\nQuestion 56.\nMember Functions can ………………….\na) Receive object as an argument\nb) Return an object\nc) Both A and B\nd) None of these\nc) Both A and B\n\nQuestion 57.\nWhen one class become a member of another class then it is called ……………. class.\na) Nested\nb) Inline\nc) External\nd) Global\na) Nested",
null,
"Question 58.\nWhen one class becomes a member of another class then the relationship is called ………………….\na) Containership\nb) Nesting\nc) Parent-Child\nd) None of these\na) Containership\n\nQuestion 59.\nClasses can be nested in ……………….. ways.\na) Three\nb) Two\nc) Four\nd) Five\nb) Two\nQuestion 60.\nClasses can be nested in ………………..way.\na) By defining a class within another class\nb) By declaring an object of a class as a member to another class\nc) Either A or B\nd) None of these\nc) Either A or B",
null,
"Question 61.\nWhen a class is declared within another class, the inner class is called a Nested class (ie the inner class) and the outer class is known as …………………… class.\na) Enclosing\nb) Abstract\nc) Transit\nd) None of these\na) Enclosing\n\nQuestion 62.\nThe nested class can be defined in ………………….. section of the Enclosing class.\na) Private\nb) Public\nc) Either Private or Public\nd) None of these\nc) Either Private or Public\n\nQuestion 63.\nWhenever an object of a class is declared as a member of another class it is known as a _____ class.\na) Abstract\nb) Container\nc) Literal\nd) None of these\nb) Container",
null,
"Question 64.\nInstantiating object is done using ……………….\na) Constructor\nb) Destructor\nc) Data abstraction\nd) Data hiding\na) Constructor\n\nQuestion 65.\nA(n) ………………..in C++ can be initialized during the time of their declaration.\na) Array\nb) Structure\nc) Array or Structure\nd) None of these\nc) Array or Structure\n\nQuestion 66.\nMember function of a class can access all the members irrespective of their associated …………………..\na) Access specifier\nb) Data type\nc) Return type\nd) Argument\na) Access specifier",
null,
"Question 67.\nWhen an instance of a class comes into scope, a special function called the ……………. gets executed.\na) Constructor\nb) Destructor\nc) Data abstraction\nd) Data hiding\na) Constructor\n\nQuestion 68.\nThe constructor function name has the same name as the …………….. name.\na) Object\nb) Class\nc) Data member\nd) None of these\nb) Class\n\nQuestion 69.\nThe constructors return ………………\na) int\nb) char\nc) float\nd) nothing\nd) nothing",
null,
"Question 70.\n…………………. are not associated with any data type\na) Constructor\nb) Data member\nc) Data abstraction\nd) Member functions\na) Constructor\n\nQuestion 71.\n……………… can be defined either inside class definition or outside the Class definition.\na) Constructor\nb) Destructor\nc) Data abstraction\nd) Member functions\na) Constructor\n\nQuestion 72.\nA constructor can be defined in ……………… section of a class.\na) Private\nb) Public\nc) Either Private or Public\nd) None of these\nc) Either Private or Public",
null,
"Question 73.\nIf a constructor is defined in ………………… section of a class, then only its object Can be created in any function.\na) Private\nb) Public\nc) Either Private or Public\nd) None of these\nb) Public\n\nQuestion 74.\nThe main function of the constructor is ……………………….\na) To allocate memory space to the object\nb) To initialize the data member of the class object\nc) Either A or B\nd) None of these\nc) Either A or B\n\nQuestion 75.\nA constructor that accepts no parameter is called …………………… constructor.\na) Null\nb) Default\nc) Empty\nd) None of these\nb) Default",
null,
"Question 76.\nIdentify the correct statement from following with respect to constructor.\na) If a class does not contain an explicit constructor (user defined constructor) the compiler automatically generate a default constructor implicitly as an inline public member.\nb) In the absence of user defined constructor the compiler automatically provides the default constructor. It simply allocates memory for the object.\nc) Parameterized constructor is achieved by passing parameters to the function.\nd) All the above\nd) All the above\n\nQuestion 77.\nA constructor which can take arguments is called ……………….. constructor.\na) Parmeterized\nb) Default\nc) Empty\nd) None of these\na) Parmeterized\n\nQuestion 78.\n……………….. type of constructor helps to create objects with different initial values.\na) Parmeterized\nb) Default\nc) Empty\nd) None of these\na) Parmeterized",
null,
"Question 79.\nDeclaring a constructor with arguments hides the ……………………\na) Data members\nb) Compiler generated constructor\nc) Member functions\nd) None of these\nb) Compiler generated constructor\n\nQuestion 80.\n………………… Constructor is used to creating an array of objects.\na) Default\nb) Parameterized\nd) None of these\na) Default\n\nQuestion 81.\nThere are ………………. ways to create an object using the parameterized constructor.\na) Three\nb) Two\nc) Four\nd) Five\nb) Two",
null,
"Question 82.\n……………… is a way to create an object using the parameterized constructor,\na) Implicit call\nb) Explicit call\nc) Either A or B\nd) None of these\nc) Either A or B\n\nQuestion 83.\nIn …………….. method, the parameterized constructor is invoked automatically\nwhenever an object is created.\na) Implicit call\nb) Explicit call\nc) Either A or B\nd) None of these\na) Implicit call\n\nQuestion 84.\nIn the………………….. method, the name of the constructor is explicitly given to invoking the parameterized constructor.\na) Implicit call\nb) Explicit call\nc) Either A or B\nd) None of these\nb) Explicit call",
null,
"Question 85.\n………………… method is the most suitable method as it creates a temporary object\na) Implicit call\nb) Explicit call\nc) Either A or B\nd) None of these\nb) Explicit call\n\nQuestion 86.\nThe chance of data loss will not arise in ………………….. method.\na) Implicit call\nb) Explicit call\nc) Either A or B\nd) None of these\nb) Explicit call\n\nQuestion 87.\nA ……………….. object lives in memory as long as it is being used in an expression.\na) Temporary\nb) Nested\nc) Inline\nd) None of these\na) Temporary",
null,
"Question 88.\nA constructor having a reference to an already existing object of its own class is called ………………….. constructor.\na) Reference\nb) Value\nc) Copy\nd) Move\nc) Copy\n\nQuestion 89.\nA copy constructor is called ……………..\na) When an object is passed as a parameter to any of the member functions\nb) When a member function returns an object\nc) When an object is passed by reference to an instance of its own class\nd) All the above\nd) All the above\n\nQuestion 90.\nThe constructors are executed in the …………………. of the object declared.\na) Order\nb) Reverse order\nc) Either A or B\nd) None of these\na) Order",
null,
"Question 91.\nWhen the initial values are provided during runtime then it is called ………………….. initialization.\na) Static\nb) Dynamic\nc) Run time\nd) None of these\nb) Dynamic\n\nQuestion 92.\n…………………. constructor can have parameter list.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\na) Constructor\n\nQuestion 93.\nNo return type can be specified for ………………….\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nc) Both A and B",
null,
"Question 94.\nThe ……………….. function can be overloaded.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\na) Constructor\n\nQuestion 95.\nThe compiler generates a ………………… in the absence of a user-defined.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\na) Constructor\n\nQuestion 96.\nCompiler generated constructor is ………………….. member function.\na) private\nb) protected\nc) public\nd) None of these\nc) public",
null,
"Question 97.\nThe ………………. is executed automatically,\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nc) Both A and B\n\nQuestion 98.\nThe ……………. is executed automatically when the object is created.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\na) Constructor\n\nQuestion 99.\nWhen a class object goes out of scope, a special function called the ………………. gets executed.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nb) Destructor",
null,
"Question 100.\nThe destructor has the same name as the class tag but prefixed with a …………………\na) ~ (tilde)\nb) #\nc) @\nd) None of these\na) ~ (tilde)\n\nQuestion 101.\nA …………………… is a special member function that is called when the lifetime of an object ends.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nb) Destructor\n\nQuestion 102.\nThe ………………… cannot have arguments,\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nb) Destructor",
null,
"Question 103.\nThere can be ……………… destructor in a class.\na) Two\nb) Three\nc) Only one\nd) Four\nc) Only one\n\nQuestion 104.\n_____ cannot be inherited.\na) Constructor\nb) Destructor\nc) Both A and B\nd) None of these\nc) Both A and B\n\nQuestion 1.\nDefine methods of a class and write its types.\nThe class comprises members. Member functions are called methods. The member functions of a class can be defined in two ways.\n\n1. Inside the class definition\n2. Outside the class definition\n\nQuestion 2.\nWhy classes are needed?\nClasses are needed to represent real-world entities that not only have data type properties but also have associated operations. It is used to create user-defined data types.",
null,
"Question 3.\nWhat is called as nesting of member functions?\nOnly the public members of a class can be accessed by the object of that class, using the dot operator. However, a member function can call another member function of the same class directly without using the dot operator. This is called as nesting of member functions.\n\nQuestion 4.\nWhat are the visibility labels of a class body?\nThe class body has three visibility labels viz., private, public, and protected. The Visibility labels are also called as access specifiers.\n\nQuestion 5.\nWhat is a parameterized constructor?\nA constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.\n\nQuestion 6.\nWhat happened if all the members of a class are defined as private?\nIf all members of the class are defined as private, then the class becomes frozen.\nThe object of the class can not access anything from the class.",
null,
"Question 7.\nA class specification just defines the properties of a class. To make use of a class specified, the variables of that class type have to be declared. The class variables are called objects. Objects are also called an instance of the class.\n\nFor example:\nstudent s;\nIn the above statement ‘s’ is an instance of the class student.\n\nQuestion 8.\nHow many ways objects can be created for a class? Give its types.\nObjects can be created in two methods:\n\n1. Global object\n2. Local object\n\nQuestion 9.\nWhat do you mean by an array of objects?\nAn array which contains the class type of element is called an array of objects. It is declared and defined in the same way as any other type of array.\n\nExample:\nclass stock\n{\nint itemno;\nfloat price; public:\n}s;\nHere s is an array of objects.",
null,
"Question 10.\nWhat is a nested member function?\nA member function can cal! another member function of the same class directly without using the dot operator. This is called as nesting of member functions.\n\nQuestion 11.\nHow many ways objects can be passed to function argument?\nObjects can also be passed in both ways\n\n• Pass By Value\n• Pass By Reference\n\nQuestion 12.\nWhat is a container class?\nWhenever an object of a class is declared as a member of another class it is known as a container class. In the container-ship, the object of one class is declared in another class.\n\nQuestion 13.\nWhat is the need for a constructor in a class?\nInstantiating object is done using constructor. An array or a structure in C++ can be initialized during the time of their declaration using constructor. The constructor function initializes the class object.",
null,
"Question 14.\nWhat are the functions of a constructor?\nThe main functions of the constructor are:\n\n• To allocate memory space to the object and\n• To initialize the data member of the class object.\n\nQuestion 15.\nWhat is a default constructor?\nDefault constructor:\nA constructor that accepts no parameter is called the default constructor.\nFor example in the class Data program Data::Data( ) is the default constructor.\nUsing this constructor objects are created similar to the way the variables of other data types are created.\n\nExample:\nint num; //ordinary variable declaration\nData d1; // object declaration\nIf a class does not contain an explicit constructor the compiler automatically generates a default constructor implicitly as an inline public member.\n\nQuestion 16.\nWhat is the significance of default constructor?\nDefault constructors are very useful to crate objects without having specific initial value. It is also used to create array of objects.\n\nQuestion 17.\nHow many ways a constructor can be invoked?\nThere are two ways to create an object using parameterized constructor:\n\n1. Implicit call\n2. Explicit call",
null,
"Question 18.\nWhat is copy constructor?\nA constructor having a reference to an already existing object of its own class is called copy constructor.\nIn other words Copy Constructor is a type of constructor which is used to create a copy of an already existing object of a class type.\n\nQuestion 19.\nWhat is the order of constructor invocation?\nThe constructors are executed in the order of the object declared. (If it is in same statement left to right)\n\nFor example:\nTest t1;\nTest t2; // the order of constructor execution is first for t1 and then for t2.\nConsider the following example\nSample s1,s2,s3 ; //The order of construction is s1 then s2 and finally s3\n\nQuestion 20.\nWhat do you mean by dynamic initialization of Object?\nWhen the initial values are provided during runtime then it is called dynamic initialization.\n\nQuestion 21.\nWrite a note on the destructor.\n\n• When a class object goes out of scope, a special function called the destructor gets executed.\n• The destructor has the same name as the class tag but prefixed with a ~(tilde).\n• The destructor function also returns nothing and it does not associate with any data type",
null,
"Question 22.\nWhat is the need for a destructor in a class?\nThe purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object\n\nQuestion 23.\nDefine destructor.\nA destructor is a special member function that is called when the lifetime of an object ends and destroys the object constructed by the constructor. Normally it is declared under the public visibility of a class.\n\nQuestion 1.\nExplain the local object with an example.\nIf an object is declared within a function then it is called a local object.\nIt cannot be accessed from outside the function.\n# include\n# include\nusing namespace std\n{\nint a,b; public:\nint sum; void\ngetdata()\n{\na = 5; b = 10; sum\n= a + b;\n}\n} a1;\nint main()\n{\na1.getdata(); //global object\na2.getdata(); //global object\na3.getdata();\ncout << a1 .sum; //Local object for a global class\ncout << a2.sum;\ncout << a3.sum;\nreturn 0; //public data member accessed from outside the class\n}\nOutput:\n151515",
null,
"Question 2.\nWrite about private, protected, and public members of a class.\nThe Public Members:\nA public member is accessible from anywhere outside the class but within a program.\n\nThe Private Members:\nA private member cannot be accessed from outside the class. Only the class member functions can access private members. By default, all the members of a class would be private.\n\nThe Protected Members:\nA protected member is very similar to a private member but they can be accessed in child classes which are called derived classes (inherited classes).\n\nQuestion 3.\nWhat is a constructor?\nThe definition of a class only creates a new user-defined data type. The instances of the class type should be instantiated (created and initialized). Instantiating objects is done using the constructor. An array or a structure in C++ can be initialized during the time of their declaration.\n\nThe initialization of a class type object at the time of declaration similar to a structure or an array is not possible because the class members have their associated access specifiers (private or protected or public). Therefore Classes include special member functions called constructors. The constructor function initializes the class object.",
null,
"Question 4.\nExplain memory allocation of objects.\nMemory allocation of objects:\nAll the objects belonging to that class use the same member function, no separate space is allocated for member functions when the objects are created.\nMemory space required for the member variables are only allocated separately for each object because the member variables will hold different data values for different objects.\nMemory for Objects for p1 and p2 is illustrated:",
null,
"Question 5.\nExplain the default constructor with an example.\nA constructor that accepts no parameter is called the default constructor. For example in the class data program Data::Data() is the default constructor. Using this constructor objects are created similar to the way the variables of other data types are created.\n\nExample:\nint num; //ordinary variable declaration\nData d1; // object declaration",
null,
"Question 6.\nHow will you refer members of the class? Give its syntax and an example.\nThe members of a class are referenced (accessed) by using the object of the class followed by the dot (membership) operator and the name of the member.\n\nThe general syntax for calling the member function is:\nObject_name.function_name (actual parameter); For example consider the following illustration:",
null,
"Question 7.\nExplain the different methods of passing an object to the function argument.\nPass By Value:\nWhen an object is passed by value the function creates its own copy of the object and works on it. Therefore any changes made to the object inside the function do not affect the original object.\n\nPass By Reference:\nWhen an object is passed by reference, its memory address is passed to the function so the called function works directly on the original object used in the function call. So any changes made to the object inside the function definition are reflected in the original object.\n\nQuestion 8.\nWhen an instance of a class comes into scope, a special function called the constructor gets executed. The constructor function name has the same name as the class name. The constructors return nothing. They are not associated with any data type. It can be defined either inside class definition or outside the class definition.",
null,
"Question 9.\nWhat is a parameterized constructor?\nParameterized Constructors:\nA constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.\n\nExample:\nclass simple\n{\nprivate:\nint a,b;\npublic:\nsimple(int m, int n)\n{\na= m ;\nb= n;\ncout< < “\\n Parameterized Constructor of class-simple\n}\n};\n\nQuestion 10.\nWhat do you mean by the implicit and explicit call of a constructor?\nImplicit call:\nIn this method, the parameterized constructor is invoked automatically whenever an object is created.\nFor example, simple s1(10,20); in this for creating the object si parameterized constructor is automatically invoked\n\nExplicit call:\nIn this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.\n\nFor example:\nsimple s1=simple(10,20); //explicit call",
null,
"Question 11.\nWhen copy constructor Is executed? Give examples.\nA copy constructor is called\n\n• When an object is passed as a parameter to any of the member functions\nExample: void simple: :putdata(simple x);\n• When a member function returns an object\nExample: simple get data( ) { }\n• When an object is passed by reference to an instance of its own class\nFor example: simple1, s2(s1); // s2(s1) calls copy constructor\n\nExplain in Detail (5 Marks)\n\nQuestion 1.\nExplain how to define class members?\nDefinition of class members:\nClass comprises of members. Members are classified as Data Members and Member functions.\n\n• Data members are the data variables that represent the features or properties of a class.\n• Member functions are the functions that perform specific tasks in a class.\n• Member functions are called methods, and data members are also called attributes.\n\nExample:",
null,
"Defining methods of a class:\nWithout defining the methods (functions), class definition will become incomplete. The member functions of a class can be defined in two ways.\n\n• Inside the class definition\n• Outside the class definition\n\nInside the class definition:\nWhen a member function is defined inside a class, it behaves like inline functions. These are called Inline member functions.\n\nOutside the class definition:\nWhen Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.\n\nThe syntax for defining the outline member function is:\nreturn_type class_name :: function name (parameter list)\n{\nfunction definition",
null,
"Class using Inline and Outline member function:\n# include<iostream>\nusing namespace std;\nclass Box\n{\n// no access specifier mentioned\ndouble width;\npublic:\ndouble length;\n//inline member function definition\nvoid printWidth( )\n{\ncout<<“\\n The width of the box is…”<<width;\n}\n//prototype of the function\nvoid setWidth(double w);\n};\n// outline member function definition\nvoid Box :: setWidth(double w)\n{\nwidth=w;\n}\nint main( )\n{\n// object for class Box\nBox b;\n// Use member function to set the width.\nb.setWidth(10.0);\n//Use member function to print the width.\nb.printWidth( );\nreturn 0;\nOutput\nThe width of the box is… 10",
null,
"Question 2.\nWhat are the ways to create an object using the parameterized constructor with an example?\nThere are two ways to create an object using the parameterized constructor:\n1. Implicit call: In this method, the parameterized constructor is invoked automatically whenever an object is created. For example, simple s1( 10,20); in this, for creating the object s1 parameterized constructor is automatically invoked.\n\n2. Explicit call: In this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.\n\n#include\nusing namespace std;\nclass simple\n{\nprivate:\nint a, b;\npublic:\nsimple(int m,int n)\n{\na = m;\nb = n;\ncout << “\\n Constructor of class – simple invoked for implicit and explicit call” << endl;\n}\nvoid putdata()\n{\ncout << “\\n The two integers are…” << a << ‘\\t’ << b << endl;\ncout << “\\n The sum of the variables” << a << “+” << b << “=” << a + b;\n}\n};\nint main()\n{\nsimple s1(10,20); //implicit call\nsimple s2 = simple(30,45); //explicit call\ncout << “\\n\\t\\tObject 1\\n”;\ns1.putdata();\ns2.putdata();\nreturn 0;\n}\nOutput:\nConstructor of class – simple invoked for the implicit and explicit call\nConstructor of class-simple invoked for the implicit and explicit call\n\nObject 1\nThe two integers are… 10 20\nThe sum of the variables 10 + 20 = 30\n\nObject 2\nThe two integers are… 30 45.\nThe sum of the variables 30 + 45 = 75\n\nQuestion 3.\nWhat are the characteristics of a destructor?\nCharacteristics of destructors:\n\n• The destructor has the same name as that of the class prefixed by the tilde character\n• The destructor cannot have arguments.\n• It has no return type.\n• Destructors cannot be overloaded i.e., there can be only one destructor in a class.\n• In the absence of a user-defined destructor, it is generated by the compiler.\n• The destructor is executed automatically when the control reaches the end of the class scope to destroy the object.\n• They cannot be inherited.",
null,
"Evaluate Yourself\n\nQuestion 1.\nDefine a class in general and in C++’s context.\nClasses represent real-world entities that not only have data type properties but also have associated operations.\nIn C++ class is a way to bind the data and its associated functions together. It is a user-defined data type.\n\nQuestion 2.\nWhat is the purpose of a class specifier?\nData hiding is one of the important features of Object Oriented Programming which allows preventing the functions of a program to access directly the internal representation of a class type.\nThe access restriction to the class members is specified by class specifies like public, private, and protected sections within the class body.\n\nQuestion 3.\nCompare a structure and a class in C++ context.\nThe only difference between structure and class is the members of structure are by default public where as it is private in class.\n\nQuestion 4.\nCompare private and public access specifier.\nPublic members:\nA public member is accessible from anywhere outside the class but within a program. We can set and get the value of public data members even without using any member function.\n\nPrivate members:\nA private member cannot be accessed from outside the class. Only the class member functions can access private members. By default all the members of a class would be private.",
null,
"Question 5.\nWhat is a non-inline member function? Write its syntax.\nWhen Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.\nThe syntax for defining the outline member function is:\nSyntax:\nreturn_type class_name :: function_name (parameter list)\n{\nfunction definition\n}",
null,
"Activity – 1\nState the reason for the invalidity of the following code fragment.\n\n (i) (ii) class count { int first int second; public: int first }; class item { int prd; }; int prdno;\n\n• Data member first is duplicated and it is defined with two scopes( both private and public). It is invalid.\n• Object name prefix with the only class name. No data type allowed in between class name and object name.",
null,
"Activity – 2\nclass area\n{\nint s;\npublic:\nvoid calc( );\n};\nWrite an outline function definition for calc( ); which finds the area of a square\nint area :: calc( ) .\n{\nreturn(s * s);\n}\n\nActivity – 3\nIdentify the error in the following code fragment\nclass A\n{\nfloat x;\nvoid init( )\n{\nAa1;\nX1.5=1; .\n}\n};\nvoid main( )\n{\nA1.init( );\n}\nError:\nLocal object can not be accessed from outside the function. Al is the local object, so it can not be accessed in main( );",
null,
"Activity – 4\nWhat is the size of the objects s1, s2?\nclass sum\n{\nint n1,n2;\npublic:\n{\nint n3=10;n1=n2=10;\n}\n} s1,s2;\nThe size of the object SI and S2 is 8 bytes each in Dev C++, In Turbo C++ 4 bytes each.\nProgram to test the memory requirement:\nclass sum\n{\nint n1,n2;\npublic:\n{\nint n3-10;\nn1=n2=10;\n}\n} s1,s2;\nusing namespace std;\n#include<iostream>\nint main( )\n{\ncout<<sizeof(s1)<< “”<<sizeof(s2);\n}\nOutput",
null,
"",
null,
"Activity – 5\ni) Write member function called display with no return.\nclass objects.\nii) Try the output of the above coding with the necessary modifications.\nPROGRAM\n#indude<iostream>\nusing namespace std;\nclass compute\n{\nint n1, n2;\npublic :\nvoid init (int a, int b)\n{\nn1 = a;\nn2 = b;\n}\nint n;\nreturn (n1+n2);;\n{\nint prd ( )\n{\nreturn (n1*n2);\n}\n};\ncompute c1, c2;\nvoid display(compute &objl,compute &obj2)\n{\nc1.init(12,15);\nc2.init(8,4);\ncout<<“\\n Sum of object-1 “<<obj1.n;\ncout<<“\\n Sum of object-2 “<<obj2.n;\ncout<<“\\n Sum of the two objects are”<<obj1.\nn+obj2.n;\nc1.init(5,4);\nc2.init(2,5);\nobj1.n = obj1.prd( );\nobj2.n = obj2.prd( );\ncout<<“\\n Product of object-1 “<<objl.n;\ncout<<“\\n Product of object-2 “<<obj2.n;\ncout<<“\\n Product of the two objects are “<<objl.n*obj2.n;\n}\nint main( )\n{\ndisplay(c1,c2);\nreturn 0;\n}\noutput",
null,
"",
null,
"Activity – 6\n#include<iostream>\nusing namespace std;\nclass Sample\n{\nint i,j;\npublic :\nint k;\nSample( )\n{\ni=j=k=0;//constructor defined inside the class\n}\n};\nint main( )\n{\nSample s1;\nreturn 0;\n}\nOutput\nIn the above program justify your reason for no output.\nConstructor alone is defined without output statement. When the above program is executed, the constructor executed. But no output on the screen because of missing cout",
null,
"Hands-On Practice\n\nQuestion 1.\nDefine a class employee with the following specification.\nprivate members of class Employee\nempno- integer\nename – 20 characters\nbasic-float\nnetpay, hra, da, – float\ncalculate ( ) – A function to find the basic+hra+da with float return type\n\npublic member functions of class employee\nhavedata( ) – A function to accept values for empno, ename, basic, hra,\nda and call calculate( ) to compute netpay\ndispdata( ) – A function to display all the data members on the screen\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<iomanip>\nclass Employee\n{\nprivate :\nint empno;\nchar ename;\nfloat basic,hra,da,netpay;\nfloat calculate( )\n{\nreturn (basic+hra+da);\n}\npublic:\nvoid have data( )\n{\ncout<<setw(35)<<“Enter Employee number :”;\ncin>>empno;\ncout<<setw(35)<<“Enter Employee name :”;\ncin>>ename;\ncout<<setw(35)<<“Enter Basic pay :”;\ncin>>basic; ,\ncout<<setw(35)<<“Enter House Rent Allowance (HRA):”;\ncin>>hra; .\ncout<<setw(35)<<“Enter Dearness Allowance (DA):”;\ncin>>da;\nnetpay = calculate( );\n}\nvoid dispdata( )\n{\ncout<<“\\nEMPLOYEE DETAILS\\n\\n”;\ncout<<setw(35)<<“Employee number :”<<empno<<endl;\ncout<<setw(35)<<“Employee name :”<<ename<<endl;\ncout<<setw(35)<<“Basic pay :”<<basic<<endl;\ncout<<setw(35)<<“House Rent Allowance (HRA) :”<<hra<<endl; cout<<setw(35)<<“Dearness Allowance (DA) :”<<da<<endl;\ncout<<setw(35)< <“Netpay :”<<netpay<<endl;\n}\n};\nint main( )\n{\nEmployee e;\ne.havedata( );\ne.dispdata( );\n}\nOutput",
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"",
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"Question 2.\nDefine a class MATH with the following specifications.\nprivate members:\nnum1, num2, result – float\ninit( ) function to initialize num1, num2 and result to zero .\n\nprotected members:\nadd( ) function to add num1 and num2 and store the sum in result\ndiff( ) function to subtract num1 from num2 and store the difference in the result\n\npublic members:\ngetdata( ) function to accept values for num1 and num2\n2. Subtract…\ninvoke add() when the choice is 1 and invoke prod when the choice is 2 and also display the result.\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<iomanip>\nclass MATH\n{\nprivate:\nfloat num1,num2,result;\ninit( )\n{\nnum1=0;\nnum2=0;\nresult=0;\n}\n{\nresult = num1+num2;\n}\nvoid diff( )\n{\nresult = num1 – num2;\n}\npublic:\ngetdata( )\n{\ncout<<“\\nEntertwo numbers “;\ncin>>num1>>num2;\n{\nint choice;\ncout<<“\\n2.Subtract …….”;\nswitch(choice)\n{\ncase 1: getdata( );\nbreak;\ncase 2: getdata( );\ndiff( );\ncout<<“\\nSubtracted value is “<<result;\nbreak;\ndefault: cout<<“\\End”;\n}\n}\n};\nint main( )\n{\nMATH m;\n}\nOutput",
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"",
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"Question 3.\nCreate a class called Item with the following specifications.\nprivate members:\ncode, quantity- Integer data type\nprice – Float data type\ngetdata( )-function to accept values for all , data members with no return\n\npublic members:\ntaxt – float\ndispdata( ) member function to display code,quantity,price and tax .The tax is calculated as if the quantity is more than 100 tax is 2500 otherwise 1000.\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<iomanip>\nclass Item\n{\nprivate:\nint code,quantity;\nfloat price;\nvoid getdata( )\n{\ncout<<“\\nEnter product code “; cin>>code;\ncout<<“\\nEnter quantity “; cin>>quantity;\ncout<<“\\nEnter price “; cin>> price;\n}\npublic:\nfloat tax;\nvoid display( )\n{\ngetdata( );\nif(quantity>100)\ntax = 2500;\nelse\ntax = 1000;\ncout<<endl<<setw(25)<< “Product code : “<<code<<endl<<endl;\ncout<<setw(25)<<“Quantity : ” <<quantity<<endl<<endl;\ncout<<setw(25)<<“Unit price :” <<price<<endl<<endl;\ncout<<setw(25)<<“Total Amount: ” < cout<<setw(25)<<“Net Bill amount : ”\n<<quantity* price+tax<<endl<<endl;\n}\n};\nint main( )\n{\nItem i; i.display( );\n}\nOutput",
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"",
null,
"Question 4.\nWrite the definition of a class FRAME in C++ with the following description.\nPrivate members:\nFramelD – Integer data type\nHeight, Width, Amount – Float data type\nSetAmount( ) -Member function to calculate and assign amount as 10*Height*Width\n\nPublic members:\nGetDetail( ) Afunction to allow user to entervalues of FramelD, Height, Width. This function should also call SetAmount() to calculate the amount.\nShowDetail( ) A function to display the values of all data members.\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<iomanip>\nclass FRAME\n{\nprivate:\nint FrameId;\nfloat Height, Width, Amount;\nvoid SetAmount()\n{\nAmount = 10 * Height * Width;\n}\npublic:\nvoid Getdetails( )\n{\ncout<<“\\nEnter Frame Id : “; cin>> FrameId;\ncout<<“\\nEnter Frame Height: cin>> Height;\ncout<<“\\nEnter Frame Width : “; cin>>Width;\nSetAmount( );\n}\nvoid ShowDetaiis( )\n{\ncout<<endl<setw(25)<<“Frame Id :” <<FrameId<<endl<<endl; cout<<setw(25)<<“Frame Height:” <<Height<<endl<endl;\ncout<<setw(25)<<“Frame Width :”<<Width<<endk<endl;\ncout<<setw(25)<<“Total Amount:”\n<<Amount< <endl< <endl;\n}\nint main( )\n{\nFRAME F;\nF.Getdetails( );\nF.ShowDetails( );\n}\nOutput",
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"",
null,
"Question 5.\nDefine a class RESORT in C++ with the following description:\nPrivate Members:\nRno //Data member to store Room No\nRName //Data member to store customer name\nCharges //Data member to store per day charges Days //Data member to store a number of days of stay\nCOMPUTE( ) //A function to calculate and return Amount as\n//Days*Chagres and if the value of Days*Charges is more than 5000 then as 1.02*Days*Charges\n\nPublic Members:\nGetinfo( ) //A function to enter the content Rno, Name, Charges //and Days Displayinfo( ) //A function to display Rno, RName, Charges, Days and\n// Amount (Amount to displayed by calling function COMPUTE())\nPROGRAM\nusing namespace std;\n#include<iostream>\nclass RESORT\n{\nprivate:\nint Rno,Days,Charges;\nchar Rname;\nint compute( )\n{\nif (Days * Charges >5000)\nreturn (Days * Charges * 1.02);\nelse\nreturn(Days * Charges);\n}\npublic:\ngetinfo( )\n{\ncout<<“\\nEnter customer name :”;\ncin>>Rname;\ncout<<“\\nEnter charges per day :”;\ncin>>Charges;\ncout<<“\\nEnter Number of days :”;\ncin>>Days;\ncout<<“\\nEnter Room Number :”;\ncin>>Rno;\n}\ndispinfo( )\n{\ncout<<“\\nRoom Number : “<<Rno;\ncout<<“\\nCustomer name :\n“<<Rname;\ncout«”\\nCharges per day :\n“<<Charges;\ncout<<“\\nNumber of days :\n“<<Days;\ncout<<“\\nTotal Amount :\n“<<compute( );\n}\n};\nint main( )\n{\nRESORT Obj;\nObj.getinfo( );\nObj.dispinfo( );\n}\nOutput",
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"",
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"Question 7.\nstruct pno\n{\nint pin;\nfloat balance;\n}\nCreate a BankAccount class with the following specifications\n\nprotected members\npno_obj //array of 10 elements\ninit(pin) // to accept the pin number and initialize it and initialize\n// the balance amount is 0\n\npublic members\ndeposit(pin, amount):\nIncrement the account balance by accepting the amount and pin. Check the pin number for matching. If it matches increment the balance and display the balance else display an appropriate message withdraw(self, pin, amount):\nDecrement the account balance by accepting the amount and pin. Check the pin number for matching and balance is greater than 1000 and amount is less than the balance. If it matches withdraw the amount and display the balance else display an appropriate message\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<iomanip>\nstruct pno\n{\nint pin;\nfloat balance;\n};\nclass BankAccount\n{\npublic:\npno pno_obj;\nvoid deposit(int pn,float amt)\n{\nfor(int i=0;i<10;i++)\nif(pno_obj[i].pin == pn)\n{\npno_obj[i].balance = pno_obj[i].balance + amt;\ncout<<“\\nTransaction successful!\ncout<<“\\nBalance amount in your account is”<< pno_obj[i].balance; break;\n}\nvoid withdraw(int self,int pn,float amt)\n{\nfor(int i=0;i<10;i++)\n{\nif(pno_obj[i].pin== pn)\n{\nif (pno_obj[i].balance>1000 && amt < pno_obj[i].balance)\n{\npno_obj[i].balance=pno_obj[i]. balance- amt;\ncout<<“\\nTransaction successful”;\ncout<<“\\nBalance amount in your account is “<< pno_obj[i].balance; break;\n}\n} }\n}\n};\nint main( )\n{\nint pin_no, tamt;\nBankAccount b;\n// initialization of objects with pin and balance amount as 0\nfor(int i=0;i<10;i++)\n{\nb.pno_obj[i].pin=i+l;\n}\nint choice;\nwhile(choice !=3)\n{\ncout <<“\\n1. Deposit”;\ncout <<“\\n2.Withdrawal”;\ncout<<“\\n3.Exit”;\ncout<<“\\nEnter your choice “; cin >>choice;\nswitch(choice)\n{\ncase 1:\ncout<<“\\nEnter PIN\ncin>>pin_no;\ncout<<“\\nEnter Deposit amount”;\ncin>>tamt;\nb.deposit(pin_no,tamt); break;\n\ncase 2:\ncout<<“\\nEnter PIN cin>>pin_no;\ncout<<“\\nEnter Withdrawal amount”;\ncin>>tamt;\ncout<<“\\nEnter 1 for Self 2 for Others :”;\nint type;\ncin>>type;\nb.withdraw(type,pin_no,tamt); break;\ndefault: cout<<“\\nTransaction completed”;\n}\n}\n}\nOutput",
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"",
null,
"Question 8.\nDefine a class Hotel in C++ with the following description:\nPrivate Members:\nRno //Data member to store Room No\nName //Data member t store customer name\nCharges //Data member to store per day charges\nDays //Data member to store number of days of stay\nCalculate() //A function to calculate and return Amount as\n//Days*Chagres and if the value of Days*Charges is more than 12000 then as 1.2*Days*Charges\n\nPublic Members:\nHotel( ) //to initialize the class members\nGetinfo( ) //A function to enter the content Rno, Name, Charges //and Days\nShowinfo( ) //A function to display Rno, RName, Charges, Days and\n//Amount (Amount to displayed by calling function CALCULATE( ))\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<string.h>\nclass Hotel\nprivate:\nint Rno,Days,Charges;\nchar Name;\n1 int Calculate( )\n{\nif (Days * Charges >12000)\nreturn (Days * Charges * 1.02);\nelse\nreturn(Days * Charges);\n}\npublic:\nHotel( )\n{\nRno=0;\nDays=0;\nCharges=0;\nstrcpy(Name,””);\n}\nvoid Getinfo( )\n{\ncout<<“nEnter customer name :”;\ncin>>Name;\ncout<<“nEnter charges per day : “;\ncin>>Charges;\ncout<<“\\nEnter Number of days :”;\ncin>>Days;\ncout<<“\\nEnter Room Number :”;\ncin>>Rno;\n}\nvoid Showinfo( )\n{\ncout<<“\\nRoom Number: “<<Rno;\ncout<<“\\nCustomer name : “<<Name;\ncout<<“\\nCharges per day : “<<Charges;\ncout<<“\\nNumber of days : “<<Days;\ncout<<“\\nTota! Amount: “<<Calculate( );\n}\n};\nint main( )\n{\nHotel obj;\nobj.Getinfo( );\nobj.Showinfo( )\n}\nOutput",
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"",
null,
"Question 9.\nDefine a class Exam in C++ with the following description:\nPrivate Members:\nRollno – Integer data type\nCname – 25 characters\nMark – Integer data type\n\npublic:\nExam(int,char[],int) //to initialize the object ~Exam() // display message “Result will be intimated shortly”\nvoid Display( ) // to display all the details if the mark is above 60 other wise display “Result Withheld”\nPROGRAM\nusing namespace std;\n#include<iostream>\n#include<string.h>\nclass Exam\n{\nprivate:\nint Rollno,Mark;\nchar Cname;\npublic:\nExam(int r,char n,int m)\n{\nRollno = r;\nMark = m;\nstrcpy(Cname,n);\n}\n~Exam( )\n{\ncout<<“\\n\\nResult will be intimated shortly”;\n}\nvoid Display( )\n{\nif (Mark>60)\n{\ncout<<“\\n\\nRoll Number : “<<Rollno;\ncout<<“\\nCandidate name : “<<Cname;\ncout<<“\\nMark :”<<Mark;\n}\nelse\n{\ncout<<“\\n\\nRoll Number : “<<Rollno;\ncout<<“\\nCandidate name : “<<Cname;\ncout<<“\\nResult Withheld”;\n}\n}\n};\nint main( )\n{\nExam obj 1(1011,”SURYA”,78),obj2( 1012,”JOHN”,44);\nobjl.Display( );\nobj2. Display( );\n}\nOutput",
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"",
null,
"Question 10.\nDefine a class Student in C++ with the following specification:\nPrivate Members:\nA data member Rno(Registration Number) type long\nA data member Cname of type string A data member Agg_marks (Aggregate Marks) of type float\nA data member Grade of type char\nA member function setGrade () to find the grade as per the aggregate marks obtained by the student. Equivalent aggregate marks range and the respective grade as shown below.\n>=90 – A\nLess than 90 and >=75 – B\nLess than 75 and >=50 – C\nLess than 50 – D\nPublic members:\nA constructor to assign default values to data members:\nA copy constructor to store the value in another object\nRno=0, Cname=”N.A” Agg_marks=0,0\nA function Getdata ( ) to allow users to enter values for Rno.Cname, Aggjnarks and call functionsetGrade ( ) to find the grade.\nA function dispResult( ) to allow user to view the content of all the data members.\nA destructor to display the message “END”\nPROGRAM .\nusing namespace std;\n#include<iostream>\n#include<string.h>\n#include<iomanip>\nclass Student\n{\nprivate:\nlong Rno;\nfloat Agg_marks;\n{\nif (Ag g_ma rks >=90)\nelse if(Agg_marks>=75)\nelse if(Agg_marks>=50)\nelse\n}\npublic:\nStudent( )\n{\nRno = 0;\nAgg_marks = 0;\nstrcpy(Cname,””);\n}\nStudent(Student &s) .\n{\nRno = s.Rno;\nAgg_marks = s.Agg_marks;\nstrcpy(Cname,s.Cname);\n\n~Student( )\n{\ncout<<“\\nEND”; >\nvoid Getdata( )\n{\ncout<<“\\nEnter Register Number”;\ncin>>Rno;\ncout<<“\\nEnter Candidate Name\ncin>>Cname;\ncout<<“\\nEnter Aggrigate Mark”;\ncin>>Agg_marks;\n}\nvoid dispResult( )\n{\ncout<<setw(30)<<“Candidate Register Number\n“<<Rno<<endl<<endl;\ncout< <setw(30)<< “Candidate Name : “<<Cname<<endk<endI;\ncout<<setw(30)<<“Aggrigate Mark : “<<Agg_marks«endk<endl;\n}\n};\nint main( )\n{\nStudent s1;\ns1.Getdata( );\nStudent s2(s1);\nCOut<<“\\nFIRST CANDIDATE DETAIL \\n\\n”; s1.dispResult( );\ncout<<“\\nSECOND CANDIDATE DETAIL \\n\\n”;\ns2.dispResult( );\n}\nOutput",
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https://file.scirp.org/Html/14-7502495_61511.htm | [
" Mathematical Physics in Diffusion Problems\n\nJournal of Modern Physics\nVol.06 No.14(2015), Article ID:61511,36 pages\n10.4236/jmp.2015.614217\n\nMathematical Physics in Diffusion Problems\n\nTakahisa Okino\n\nDepartment of Applied Mathematics, Faculty of Engineering, Oita University, Oita City, Japan",
null,
"",
null,
"",
null,
"Received 14 October 2015; accepted 24 November 2015; published 27 November 2015",
null,
"ABSTRACT\n\nUsing the divergence theorem and the coordinate transformation theory for the general Fickian second law, fundamental diffusion problems are investigated. As a result, the new findings are obtained as follows. The unified diffusion theory is reasonably established, including a self-diffusion theory and an N (N ³ 2) elements system interdiffusion one. The Fickian first law is incomplete without a constant diffusion flux corresponding to the Brown motion in the localized space. The cause of Kirkendall effect and the nonexistence of intrinsic diffusion concept are theoretically revealed. In the parabolic space, an elegant analytical method of the diffusion equation is mathematically established, including a nonlinear diffusion equation. From the Schrödinger equation and the diffusion equation, the universal expression of diffusivity proportional to the Planck constant is reasonably obtained. The material wave equation proposed by de Broglie is also derived in relation to the Brown motion. The fundamental diffusion theories discussed here will be highly useful as a standard theory for the basic study of actual interdiffusion problems such as an alloy, a compound semiconductor, a multilayer thin film, and a microstructure material.\n\nKeywords:\n\nBrown Particle, Boltzmann Factor, Markov Process, Parabolic Law, Error Function",
null,
"1. Introduction\n\nFirst of all, we state that the basic diffusion equation of the general nonlinear Fickian second law is discussed in accordance with the fundamental mathematical physics in the present work. The extended diffusion equations in detail are not thus discussed. Nevertheless, the new findings, which are extremely dominant in the diffusion study, are reasonably obtained. In the diffusion history, the problems relevant to the coordinate transformation of diffusion equation had not been discussed in accordance with the Gauss divergence theorem until recently. That is just a reason why the new diffusion theories are discussed in the present study. It will be gradually clarified in the text that the coordinate transformation theory is essentially indispensable for the diffusion study. It is obvious that analyzing the extended diffusion equation must be based on the fundamental diffusion theory. The new fundamental findings different from the existing diffusion theories obtained here will thus exert a great influence on the actual diffusion problems in detail, just because of fundamental ones.\n\nA great many phenomena in various science fields are expressed by using the well-known evolution equations. The diffusion equation is one of them and mathematically corresponds to the Markov process in relation to the normal distribution rule . In other words, the motion of diffusion particles corresponds to the well-known Brown movement satisfying the parabolic law . It is widely accepted that the Brown problem is a general term of investigating subjects in various science fields relevant to the Markov process, such as material science, information science, life science, and social science - .\n\nIn physics, we can also understand the diffusion equation in accordance with the Gauss divergence theorem . If we apply the divergence theorem to the diffusion problem for a material under the condition of no sink and source of the material, it is found that the material conservation law is valid for the diffusion particles, regardless of a thermodynamic state of material. The diffusion equation is also called “the continuous equation” and is extremely fundamental one in physics. In history, the heat conduction equation, which is mathematically equivalent to the diffusion equation, was proposed by Fourier, regardless of the Markov process and the divergence theorem .\n\nIn accordance with the industrial requirement, the solid materials, such as alloys, semiconductors, and multilayer materials, have been widely fabricated. The heat treatment is indispensable for their fabrication processes then. The migration of particles in a material is caused by the heat treatment. In relation to the migration of their particles, the diffusion problems of various solid materials have been thus widely investigated . Therefore, the diffusion problem is a fundamental study subject in the materials science including the cases of liquid and gas states.\n\nIn the present work, the fundamental problems of the general Fickian second law where a driving force affects the diffusion system are discussed in accordance with the mathematical theory. The present analytical method is applicable to interdiffusion problems of an N elements system of every material in an arbitrary thermodynamic state. Although the physical validity of the present method is investigated by using the diffusion data concerning the solid metals, the mathematical generality discussed here is still kept.\n\nThe heat conduction equation proposed by Fourier in 1822 has been applied to investigating the temperature distribution in materials . In 1827, the so-called Brown motion was found, where the self-diffusion of water was visualized by pollen micro particle motions . In 1855, Fick applied the heat conduction equation to diffusion phenomena as it had been . Nevertheless, the Brown motion had not been recognized as a diffusion problem until the Einstein theory of Brown motion in 1905, although it was a typical diffusion problem .\n\nAlthough the concentration of diffusion particles is a real quantity in physics, the temperature is a thermodynamic state quantity. As far as the shape of heat conduction material is unchangeable during a thermal treatment, the coordinate system of heat conduction equation set in a material is a fixed one, since the coordinate system is not influenced by variations of the material internal structure. On the other hand, strictly speaking, the coordinate system of diffusion equation set in the diffusion field (solvent) is a moving one, since it is generally influenced by such variations.\n\nWhen the Fickian first and second laws (F1 law and F2 law) were proposed, the Gauss divergence theorem had been already reported in 1840 . Nevertheless, the problem of coordinate system of diffusion equation was not mathematically investigated in accordance with the divergence theorem.\n\nIn general, it is indispensable for understanding the diffusion problems to discuss their coordinate systems, since it is, strictly speaking, considered that the diffusion particles, solvent particles and also the diffusion region space simultaneously move against the experimentation system in the diffusion region outside.\n\nAlthough the Fickian laws were still widely applied to various diffusion phenomena as essential equations in physics, the problem between coordinate systems was not discussed. Recently, the diffusion equation was thus mathematically investigated in accordance with the divergence theorem and the coordinate transformation theory - . It is revealed in the text that the diffusion flux should be determined by taking account of the concerned coordinate system of diffusion equation.\n\nUsing the corresponding diffusion flux to the coordinate system of diffusion equation for interdiffusion, one way diffusion, impurity diffusion and self-diffusion, they are uniformly discussed in the text. As a result, we found that the foundation of diffusion problems is included in interdiffusion problems. The interdiffusion theory of an",
null,
"elements system applicable to every material was thus reasonably established . In the analysis of interdiffusion problems, the only difference between a binary system and an N elements system is whether the solvent material is one element or",
null,
"elements.\n\nThe coordinate transformation theory reveals that the corresponding F1 law to the F2 law is incomplete without a constant diffusion flux relevant to the Brown motion in the localized space. Further, it was also found that the Kirkendall effect (K effect) is caused by a shift between the coordinate systems of the diffusion equation like the Doppler effect relevant to a wave equation is caused by a shift between the fixed coordinate system and the moving one for the wave equation - . At the same time, this means that the concept of intrinsic diffusion is unnecessary for understanding the K effect.\n\nAs far as the sink and source of a material is nonexistent in the diffusion system, the Gauss divergence theory shows that the general F2 law satisfies the material conservation law even if a driving force affects the diffusion system. The driving force affects entropy and jump frequencies of diffusion particles in the material. The influence of a driving force is thus all incorporated into the diffusivity of diffusion equation then. In the existing diffusion theory, the diffusion fluxes influenced by a driving force are discussed incorporating not influence of both entropy and jump frequencies but only that of jump frequencies on the diffusivity into them, for example such as a drift velocity. However, such concept as a drift velocity is not necessary for analyzing the diffusion equation, since we can solve a nonlinear diffusion equation in accordance with the fundamental theory in the mathematical physics, regardless of the discussion about diffusion fluxes.\n\nThe diffusivity is obtained from the Taylor expansion of a probability density function, satisfying the parabolic law . This indicates that the analysis of diffusion equation is essentially possible in the parabolic space. The diffusivity is defined by an interaction between a diffusion particle and the diffusion field near the diffusion particle itself. This indicates that the diffusivity should be essentially investigated in the quantum mechanics, since the behavior of a micro particle should be investigated by analyzing the Schrödinger Equation (S equation) .\n\nBased on the suggestions mentioned above, the elegant analytical method in the parabolic space is reasonably established in the text . As a result, the solutions of the Boltzmann transformation equation for nonlinear interdiffusion phenomena were, for the first time, obtained as analytical expressions in the parabolic space .\n\nFrom applying the diffusion equation to a problem of diffusion elementary process, we derived the S equation . It is revealed in the text that the diffusivity corresponds to the angular momentum operator in the quantum mechanics. As a result, the universal expression of diffusivity, which is applicable to every material in an arbitrary thermodynamic state, was obtained in proportional to the Planck constant",
null,
".\n\nIt was also found that the well-known material wave relation proposed by de Broglie in 1923, which is the most fundamental one in materials science, is obtained from a relation between diffusivity expressions . This gives evidence for the theory discussed in the present study.\n\nIn the present work, we review the fundamental diffusion theories relevant to the general F2 law, where they are systematically reframed in points of view different from the previous works, adding some new discussions to them. The new findings obtained here will be widely applicable to fundamental problems as a standard theory in various actual diffusion phenomena.\n\n2. Fundamental Theory of Diffusion Equation\n\nAs far as a material is conserved in the given region, the divergence theory shows that the diffusion equation is applicable to diffusion phenomena of every material in an arbitrary thermodynamic state. The diffusion information of a material, such as crystal material or amorphous material, and/or solid, liquid and gas states, is all incorporated into the diffusivity in the given diffusion equation. The diffusion equation with such an arbitrary diffusivity is thus investigated in the following.\n\nIn the present study, an abbreviate differential notation for an arbitrary independent variable",
null,
"and the well- known Dirac’s bracket notation for an arbitrary vector are used as follows ;",
null,
"If an operator Q is Hermite one,",
null,
"is valid in the Hermite conjugate",
null,
". Here, the notation",
null,
"is thus defined as",
null,
"because of",
null,
".\n\n2.1. Diffusion Equation\n\nThe function",
null,
"is defined as a normalized concentration where a diffusion particle in the initial state",
null,
"exists in the time and space state",
null,
"after j times jumps. A diffusion particle moves at random. It is, therefore, considered that the jump frequency",
null,
"and jump displacement",
null,
"are equivalent in probability to their mean values of all diffusion particles in the collective system.\n\nSince it is also considered that the probability of diffusion-jump from the state of to is equivalent to one from the same state to in the isotropic space, the relation of\n\n(2-1)\n\nis thus valid. The Taylor expansion of both sides of Equation (2-1) yields\n\n(2-2)\n\n(2-3)\n\nThe substitution of Equations (2-2) and (2-3) into Equation (2-1) gives\n\n(2-4)\n\nSince the averaged and are physically finite, it is considered that becomes a constant value given by\n\n(2-5)\n\nsatisfying the well-known parabolic law . Substituting Equations (2-5) into Equations (2-4), the diffusion equation is obtained as\n\n(2-6)\n\nwhere is rewritten as D when the diffusion-jump direction is not isotropic resulting from the existence of a driving force in the diffusion system.\n\nThe existence of a driving force affects diffusivity and the diffusivity D becomes a function of the independent variables t and. In that case, Equation (2-6) gives the diffusion flux yielding\n\n(2-7)\n\nwhere is a diffusivity caused by a driving force.\n\nOn the other hand, using a drift velocity caused by a driving force F, the approximate diffusion flux is expressed as\n\n(2-8)\n\nin the usual textbooks (see Appendix A) . However, the diffusion flux discussed here is different from one given by\n\n(2-9)\n\nIn the present study, the discussion about diffusion fluxes is not necessary for analyzing the diffusion equation, since we can solve the nonlinear diffusion equation in accordance with the fundamental theory in the mathematical physics, regardless of diffusion fluxes.\n\nIn the following, therefore, Equation (2-6) is directly solved in accordance with the mathematical physics. It is in the nature of things that its analytical method is applicable to the diffusion problems relevant to a driving force then.\n\nFor a collective system composed of N elements in the region within the closed surface, the divergence theorem yields\n\n(2-10)\n\nwhere and are a concentration and a diffusivity for an element j under the condition of no sink and source of the element. The integral calculation of in Equation (2-10) means an inflow flux from the surface outside through the surface element, and y and z are then accepted as constant values because of the multiple integral calculation. Therefore, Equation (2-10) yields the diffusion equation of\n\n(2-11)\n\nin the diffusion field. Defining the outflow flux as a plus value, the diffusion flux is\n\n(2-12)\n\nIn mathematics, Equation (2-12) yields\n\nfor (2-13)\n\nwhere is the F1 law and is an integral constant against x. In physics, means a movement of a diffusion region space caused by a movement of the diffusion field. In the coordinate system of diffusion region inside, it should be thus physically accepted as. However, must be taken into account in the coordinate system of diffusion region outside. The intrinsic diffusion flux independent of the time and space is relevant to the Brown motion in the localized space. It plays an important role for understanding a self-diffusion mechanism.\n\n2.2. Coordinate System of Diffusion Equation\n\nAs discussed later, the unified theory of diffusion problems shows that the foundation of diffusion is in the interdiffusion problems. In the interdiffusion problems, solvent particles as well as diffusion particles move in the diffusion region space. When a micro particle in a material jumps from a site M to an interstice space N, if we call such an interstice space “micro hole” including a vacancy in case of a crystal material, the micro hole of site N is annihilated and a new micro hole is generated at the site M after jumping.\n\nThe diffusion region space interacts with the space of diffusion system outside in accordance with the annihilation and/or generation of micro holes like the diffusion system becomes a thermal equilibrium state resulting from an antinomy between the principle of increase of entropy and that of minimum of free energy.\n\nThe coordinate system of the basic diffusion equation must be set in the diffusion field, since the diffusivity depends on an interaction between a diffusion particle and the diffusion field near the diffusion particle itself. In general, therefore, the diffusion problem should be investigated among three coordinate systems, where the origins of two coordinate systems are a point P of the coordinate system set in the diffusion field (solvent) and a point Q of the coordinate system set in the diffusion region space composed of micro holes and the third origin is a point R of the coordinate system set in the diffusion region outside.\n\nIn the present work, we investigate an interdiffusion problem applicable to every material in an arbitrary thermodynamic state. It is thus investigated using the original diffusion equation (2-11) of the coordinate system and the diffusion equations transformed into the coordinate systems of and under the condition of.\n\nThe origins P and Q are set at a point of solvent material and at a point of micro holes on the initial interface of a diffusion couple. The origin R is set at a point of the diffusion system outside. The x axis is perpendicular to the interface then. The axis and also axis are parallel to the x axis under the condition of at.\n\nIt is generally considered that P and Q move with a velocity and with a velocity against R, respectively and also that P moves with a velocity against Q, including a possibility of. The relation among these velocities of\n\n(2-14)\n\nmust be physically valid.\n\nHereafter, we use the suffix P, Q and R for a physical quantity relevant to, and. Equation (2-13) is thus rewritten as\n\nfor . (2-13)\n\nWhen the coordinate origin of moves from that of with a velocity, the relations of\n\n, for (2-15)\n\nare valid under the condition of at. The relations of\n\n, (2-16)\n\nbetween differential operators are obtained then.\n\nUsing Equation (2-16) for Equation (2-11), the diffusion equations of is obtained as\n\n(2-17)\n\nwhere and. Equation (2-17) leads to the diffusion flux of\n\n, (2-18)\n\nsince is in the diffusion region inside.\n\nIn the same manner of the above, the diffusion equation of is obtained as\n\n(2-19)\n\nSince is in the diffusion region outside, the diffusion flux becomes\n\n, , (2-20)\n\nwhere is the diffusion flux of micro holes caused by the movement of the diffusion field against.\n\n3. Interdiffusion Problems\n\nThe interdiffusion problems of an N elements system in a diffusion couple composed of arbitrary materials A and B are discussed as follows.\n\nIn an amorphous material, we then estimate a specimen cross section perpendicular to the x axis where each cross section interval l corresponds to the averaged jump distance in the solid state or to the averaged collision distance in the gas or liquid state. In a crystal material, we estimate the jump distance l between the nearest neighbor crystal cross sections perpendicular to the x axis.\n\nWhen materials A and B are in the solid state during a thermal diffusion, we conceive that the cross section S of the material A is uniform and equal to that of the material B. If materials A and B are in the fluid state, we conceive that they are in the receptacle corresponding to the case of solid materials. The interface of diffusion couple between materials A and B is smoothly joined. In that case, the origin of coordinate system is set at a point on the initial interface in the diffusion field and the orientation of x axis is defined as.\n\nThe material A is composed of N elements and we define the normalized concentration for in the initial state, also including a possibility of for. The normalized concentration in the material B is similarly defined as.\n\nThe concentration and diffusivity of an element j are and at a diffusion time t in the diffusion region. The boundary values of concentration are and.\n\nThe remained elements are considered to be a solvent, also including a possibility where becomes a solvent itself in accordance with a given diffusion system.\n\nThe cross section of diffusion system is uniform and the shape variation of diffusion couple is usually negligible during the thermal diffusion. In that case, the relation of normalized concentrations yielding\n\n(3-1)\n\nis usually accepted on each cross section between. Here, Equation (3-1) is physically valid regardless of a coordinate system.\n\n3.1. Correlation of Diffusion Fluxes with Coordinate Systems\n\nEquations (2-17) and (3-1) yield\n\n(3-2)\n\nbecause of. In mathematics, the partial differential equation (3-2) of and is rewritten as\n\nusing an arbitrary function of because of. Here, means a diffusion flux dependent on and. On the other hand, means a diffusion flux only dependent only on. Therefore, Equation (3-2) means for an arbitrary. The interdiffusion coefficient is thus defined as\n\n(3-3)\n\nin the diffusion region. We can easily confirm that Equation (3-3) does not depend on a coordinate system. Their boundary values are thus obtained as\n\nand (3-4)\n\nUsing Equation (3-3) for the F1 law, the following relation of\n\n(3-5)\n\nis valid regardless of a coordinate system. The relation of\n\n(3-6)\n\nis also valid as discussed later in the self-diffusion theory.\n\nUsing Equations (3-5) and (3-6) for Equation (2-20), the relation of\n\nfor (3-7)\n\nis obtained. Here, means the total diffusion flux of micro holes against the origin of. In that situation, is valid in the present coordinate system because of, using a velocity of the origin of against the origin of. Equation (3-7) is thus rewritten as\n\n(3-7)\n\nIn the same manner, the following relations\n\n(3-8)\n\n(3-9)\n\nare also obtained from Equations (2-13) and (2-18). Equations (3-7)-(3-9) show that the diffusion flux independent of the space coordinate corresponds to the velocity between the coordinate systems. The relation of\n\n(3-10)\n\nmust be physically valid because of. Equation (3-10) then yields the previous equation of\n\n(2-14)\n\nbecause of Equations (3-7)-(3-10).\n\nIn the following, is investigated by using the concentration difference and the diffusion junction depth between and. Taking account of their concentration gradients, the flux of diffusion field for an element k () is obtained as\n\n(3-11)\n\nwhere the suffix means if or if and is tentatively adopted. Here, is adopted instead of as an approximate equation because of.\n\nEquation (3-11) yields\n\n(3-12)\n\nEquation (3-12) shows the movement of against. Substituting Equation (3-12) into Equation (2-14) yields\n\n(3-13)\n\nbecause of.\n\n3.2. Kirkendall Effect in Interdiffusion Problems\n\nIn the metallurgy field, the interface of is the so-called Matano interface (M interface) and that of is the Kirkendall interface (K interface) . In 1947, Kirkendall found in the binary system interdiffusion experimentation that the inert marker set on the K interface in the initial state moves from the M interface during a thermal diffusion and also that the M interface does not move during thermal diffusion, i.e.,. Since then, the displacement between the M interface and the K interface has been called the Kirkendall effect (K effect). Using an experimental value m, the K effect is obtained as\n\n(3-14)\n\nregardless of the space coordinate and it satisfies the parabolic law.\n\nAs far as the material shape of a diffusion couple is unchangeable, the K effect also reveals that the total quantity of element I diffusing across the K interface is different from that of element II. In other words, for the micro hole concentration and its thermal equilibrium value between, the relation of\n\nis valid between if is valid between during thermal diffusion, and vice versa. The value of becomes gradually zero in accordance with the thermodynamic principle.\n\nIn the binary system interdiffusion problems, the F1 law shows that the relation of\n\nis valid because of, and is valid then for an interdiffusion coefficient. On the other hand, it was also considered that the K effect is caused by a difference between the diffusion flux of element I and that of element II across the K interface. In other words, the so-called intrinsic diffusion coefficient satisfying the relation of was newly conceived to understand the K effect then.\n\nHowever, the definition of diffusivity shows that the diffusivity is proportional to a statistical jump frequency of a diffusion particle at a point occupied by the diffusion particle itself. It is, therefore, physically difficult to accept what a diffusion particle of the element I or II has two diffusivities and or and at the same point in the concerned diffusion field under the condition of, since the statistical jump frequency value of a diffusion particle should be physically only one in the concerned diffusion field.\n\nIn the present diffusion theory, the behavior of micro holes is visualized by the movement of an inert marker in the diffusion region because of the inert characteristic. Therefore, the K effect means a shift between the coordinate systems and after the thermal diffusion. In other words, Equation (3-12) yields the displacement between the K interface and the M interface because of in accordance with the experimental results. The K effect is thus obtained as\n\n(3-15)\n\nsatisfying the parabolic law.\n\nThe concept of intrinsic diffusion is thus unnecessary for understanding the K effect. Further, the K effect occurs in the interdiffusion not only of metal crystal but also of every material, since it is caused by a shift between the coordinate systems.\n\nEquation (3-15) gives evidence that the K effect is caused by a shift between the coordinate systems, since it does not depend on the space coordinate. Substituting into Equation (3-15) and comparing it with Equation (3-14) give\n\n(3-16)\n\nHere, we must notice that is tentatively adopted in the diffusion junction depth in the present case. Therefore, the value in the relation of\n\nmust be concretely investigated so as to be suitable for the concerned interdiffusion problems, since the m value is experimentally obtained.\n\n3.3. Unified Theory of Diffusion Problems\n\nThe experimental results show that the relation of is valid in the metallurgy field. This means that the shape of diffusion system is unchangeable before and after diffusion treatment. In that case, Equation (2-19) is thus rewritten as\n\n(3-17)\n\nEquation (3-17) of is thus equivalent to Equation (2-11) of. However, the diffusion flux of Equation (2-20) is rewritten as\n\n,\n\nIn the following, various diffusion problems are systematically investigated by applying the diffusion flux of Equation (2-18) to them for, under the condition of\n\n.\n\nEquation (2-18) is then expressed as\n\n(3-18)\n\n(a) Interdiffusion:\n\nUnder the condition of, Equation (3-18) becomes\n\n(i)\n\n(ii).\n\n(b) One-way diffusion:\n\nUnder the condition of, Equation (3-18) becomes\n\n(i),\n\n(ii).\n\n(c) Impurity diffusion:\n\nUnder the condition of and, Equation (3-18) becomes\n\n(i), (ii).\n\n(d) Self-diffusion:\n\nUnder the condition of and, Equation (3-18) becomes\n\n(i), (ii).\n\nIf we rewrite and as\n\nand for\n\nthe above theory is still valid in the N elements system. In the diffusion problems, it was thus found that the behavior of diffusion region space plays an important role for understanding diffusion mechanisms.\n\n3.4. Self-Diffusion Theory\n\nThe diffusivity expression of Equation (2-5) is valid regardless of pure materials . The Einstein theory of Brown motion shows that the self-diffusion of water is visualized by diffusion phenomena of pollen micro particles. It is thus obvious that the self-diffusion occurs in a pure material. However, even if we apply the F2 law to the self-diffusion problems in a pure material, we cannot understand whether the self-diffusion occurs or not, since the concentration is unchangeable in the self-diffusion system.\n\nLangevin theoretically investigated not the diffusion equation of collective motion but the motion equation of a diffusion particle with a viscosity resistance in a pure liquid material . It was thus revealed that the diffusion particle moves in accordance with the normal distribution rule.\n\nThe theories of Langevin as well as Einstein are relevant to diffusion particles with a driving force in a pure material. In other words, the diffusion phenomena of a pure material itself are not directly investigated. However, Equation (2-5) shows that the self-diffusion occurs in a pure material even if there is no driving force in the diffusion system.\n\nThe self-diffusion problems have been also experimentally investigated as an impurity diffusion problem by using small quantities of the isotope material. Strictly speaking, however, the self-diffusion problems are obviously different from the impurity ones, as can be seen from the difference of (i) between (c) and (d) in Section 3.3.\n\nAs discussed in Section 2.1, the corresponding F 1 law to the F2 law is incomplete without the intrinsic diffusion flux. In the following, therefore, the self-diffusion problems of a pure material are directly investigated in accordance with the intrinsic diffusion flux, regardless of a driving force.\n\n(1) Application of intrinsic diffusion flux to self-diffusion problem\n\nWe cannot experimentally confirm in Equation (2-13). The theory of Brown motion reveals that the diffusion phenomena occur in a material even if the macro concentration gradient is zero. The concept of intrinsic diffusion flux should be thus valid in a localized region space. The coordinate notation is, for convenience, used instead of and/or in the following.\n\nUsing the coordinate notation for the equation of relevant to the self-diffusion, the relation of\n\n(3-19)\n\nis obtained. If the concentration difference between and is defined as, Equation (3-19) is rewritten as\n\n. (3-20)\n\nTaking account of in Equation (3-19) and using Equation (2-5) for Equation (3-20), we have the relation of\n\n(3-21)\n\nEquation (3-21) shows an instantaneous concentration gradient in the localized time and space in the cause of the thermal fluctuation, even if the macro concentration gradient is zero. In other words, although the concentration gradient of a pure material is zero in macro behavior, it is nonzero in the minute region.\n\nThe behavior of concentration distribution in the local space is thus obtained as\n\n(3-22)\n\nIn a pure material or a material in the thermal equilibrium state, Brownian particles move in accordance with Equation (3-22). If, the Brownian particles get a standstill state inconsistent with the Einstein theory of Brown motion. The intrinsic diffusion flux is thus an important concept in the fundamental diffusion theory. At the same time, the F1 law is incomplete without.\n\nIf we apply Equation (3-22) to the interdiffusion problems, the relation of\n\n(3-23)\n\nis valid in accordance with Equation (3-3). Equation (3-23) reveals that Equation (3-6) is valid because of\n\n.\n\nUsing the notation for Equation (3-22), the concentration behavior of an element j in the thermal equilibrium state is expressed as\n\n(3-24)\n\nunder the condition of in the localized space.\n\nThe diffusion region of self-diffusion problem of a pure material is, for convenience, divided into 2 regions as a binary system interdiffusion problem composed of a pure material I of and the pure material II of at. The diffusion region is between at a time then. The boundary values are for and for under the condition of.\n\nUsing Equation (3-24) for, the relations of\n\nand (3-25)\n\nare obtained, where. Equation (3-25) shows that and depend on and that is valid if. If we pay attention only to one side of or, it is obvious that the particle migration occurs via the random movement in the pure material. In other words, the diffusion occurs in a pure material, even if the macro concentration gradient is zero.\n\nOn the other hand, if we pay attention to both sides of and in Equation (3-25), seems to be valid, since we cannot actually distinguish their diffusion particles between whether they were ones of or in the initial state at.\n\n(2) Physical meaning of intrinsic diffusion flux\n\nHere, we estimate the value in the following. When the averaged jump distance or the averaged collision distance is written as l in a pure material, the substitution of and into Equation (3-25) yields\n\n(3-26)\n\nwhere is rewritten as and corresponds to the concentration gradient. It is considered that the order of l is several angstroms and the diffusion junction depth in a usual experiment is several micrometers. Therefore, the absolute value of is considerably small, compared with that of.\n\nWe cannot actually observe in the macro state because of Equation (3-6). However, we can approximately estimate it by using the relation of\n\nwhere is a self-diffusion coefficient obtained from the tracer diffusion experiment for the concerned material, and l corresponds to the averaged jump distance or the averaged collision distance of the diffusion particle.\n\nEquation (3-25) independent of means the time-averaged concentration profile caused by the Brown motion in the localized region space. The diffusion occurs even in a pure material, resulting from the chain reaction of instantaneous jumping of diffusion particle into micro holes caused by a thermal fluctuation in a material minute region. Equation (3-26) thus gives evidence that the Brown motion occurs in a pure material, even if an external force does not exist in the diffusion system.\n\nFrom the historical point of view, if the F1 law were investigated at an early stage in accordance with the Gauss divergence theory, we might understand the Brown motion behavior before the Einstein theory .\n\n3.5. Diffusion Equation of Micro Holes\n\nThe space coordinate dependence of micro holes has been neglected in the diffusion theory mentioned above. In the metallurgy field, however, the experimental results of multiple markers show that their movements depend on the space coordinate in accordance with a marker position set in the initial diffusion couple - . This indicates that we must investigate the space coordinate dependence of micro holes in the diffusion study in detail, since the behavior of micro holes is visualized by the characteristic of inert markers.\n\nIn that case, Equation (3-1) is rewritten as\n\n(3-27)\n\nwhere is the normalized concentration of micro holes between. In general, there is a difference between the thermal equilibrium concentration of micro holes in the\n\nmaterial A and that of in the material B. When the difference is not negligible, the diffusion of micro holes occurs between.\n\nFor the thermal equilibrium concentration of micro holes in the diffusion region, the generation/annihilation term of the diffusion equation is expressed as\n\n(3-28)\n\nwhere is a chemical reaction constant. Taking account of Equation (3-28), the diffusion equation of micro holes is expressed as\n\n(3-29)\n\nwhere is a diffusivity of micro holes between.\n\nThe K-effect shows that the concentration of micro holes yields a supersaturated section and an unsaturated one in the diffusion region. Therefore, the averaged value of diffusion flux yielding\n\n(3-30)\n\nis considered to be negligible. We thus assume that the averaged contribution of the generation/annihilation term to the total diffusion flux in the diffusion system is negligible.\n\nHere, if we integrate Equation (3-29) with respect to x, the diffusion flux relation of\n\n(3-31)\n\nis obtained by neglecting the contribution of Equation (3-30). In that case, Equations (3-27), (2-11) and (3-31) yield\n\nunder the condition of. Equation (3-3) is thus rewritten as\n\n(3-32)\n\nEquation (3-32) shows that we can accept a micro hole as a virtual diffusion particle.\n\nThe diffusion equation of in the N elements system interdiffusion becomes\n\n(3-33)\n\nusing a common interdiffusion coefficient. It seems that Equation (3-33) is an independent equation of. However, it is not independent because of Equation (3-27). When we must investigate the influence of on the interdiffusion problems in detail, the diffusion equation of micro holes yielding\n\n(3-34)\n\nshould be investigated. Equation (3-34) shows that the behavior of multiple markers depends on the time and space, since the obtained dependent on the time and space is visualized by the inert markers.\n\n3.6. Application of Present Theory to Actual Diffusion Problems\n\nHereinbefore, we discussed some fundamental problems in the bulk diffusion of an ideal diffusion couple. The interdiffusion problems for a many elements system, such as an alloy, a compound semiconductor, a multilayer thin film, a microstructure material and so on, have been widely investigated - . It is then fundamentally indispensable for the material science investigation to understand the N () elements system interdiffusion theory, since the migration of each element in a material occurs during the thermal treatment in the material fabrication process.\n\nIn the diffusion study history, the concept of intrinsic diffusion has been accepted in order to understand the K effect. In the existing theory of interdiffusion problems, the Darken equation derived from the intrinsic diffusion concept has been widely used for analyzing them . The application of the Darken theory to a many elements system interdiffusion was extremely complicated even if it was a ternary elements system . On the other hand, the simple relation among interdiffusion coefficients obtained here will be extremely useful for analyzing various interdiffusion problems.\n\nAs can be seen from the Appendix [B], we revealed that the Darken equation is not mathematically valid. It is also reported that the analytical results obtained by using the Darken equation are not consistent with the experimental results . This fact is consistent with the present theory where there is no such concept of intrinsic diffusion in the diffusion phenomena.\n\nUnder the condition of Equation (3-27), therefore, it is necessary for analyzing a many elements system interdiffusion problem to solve the diffusion equations (3-33) and (3-34) of each element in accordance with their initial and boundary conditions. Here, it seems that these diffusion equations are independent of each other since they have a common interdiffusion coefficient. However, it is not independent because of Equation (3-27).\n\nIn the parabolic space discussed later, the analytical method of a nonlinear diffusion equation is established. The analytical solutions of Equations (4-48) and (4-49) are thus obtained then. Using them for analyzing each of the above equations, we can fundamentally solve an N () elements system interdiffusion problems. The theory discussed here is thus widely applicable to analyzing the fundamental problems of N elements system interdiffusion.\n\nSince the present analytical theory is fundamental, it will be necessary for actual diffusion problems to modify it partly in accordance with the concerned problems like the Boyle Charles law for an ideal gas state is modified into the van der Waals equation in the actual case by incorporating an interaction between gas particles into it.\n\n4. Analysis of Diffusion Equation in Parabolic Space\n\nEinstein theoretically revealed that the Brownian particles randomly move in accordance with the parabolic law . After that, it was experimentally confirmed by Perrin . The parabolic law is universally shown in phenomena relevant to the normal distribution in a probability problem. The Brown problem is a study subject not only in the material science but also in the complex-system sciences relevant to the Markov process. Therefore, the analytical method of the Brown problem is mathematically common in various science fields.\n\nThe diffusion junction depth is directly relevant to the parabolic law. The diffusion problem is justifiably one of the Brown problems. This indicates that the solutions of diffusion problems are possible in the parabolic space. In other words, the analytical method discussed here is widely applicable to various Brown problems.\n\nThe F2 law is a continuous equation in the conservation system under the condition of no sink and source of diffusion particles, and it is one of the basic equations in physics. When the existence of a driving force or the sink and/or source of diffusion particles is negligible in a given diffusion system, Equation (2-6) is a parabolic type linear partial differential equation in the evolution equations. A lot of diffusion problems have been investigated by solving the diffusion equation of the time and space since 1855.\n\nWhen the diffusivity depends on the concentration, the diffusivity becomes a function of the independent variables via the concentration. Even if the diffusion equation depends only on, the mathematical solutions of the nonlinear partial differential equation are almost impossible. A new analytical method in the parabolic space, which is extremely superior in analyzing to the existing ones, is discussed in the following.\n\n4.1. Definition of Parabolic Space\n\nFor convenience, hereafter we use the coordinate notation transformed from the time and space into the parabolic space and they are correlated to and under the condition of. The relations of differential operators are obtained as\n\nand\n\nFrom the similar calculation, the differential operators in the parabolic space are obtained as\n\n, (4-1)\n\nwhere and because of the Hermite characteristic.\n\n4.2. Diffusion Equation in Parabolic Space\n\nIf we directly apply Equation (4-1) to homogeneous partial differential Equation (2-6), the ellipse type differential equation is obtained as\n\n(4-2)\n\nwhere is physically valid in accordance with initial conditions .\n\nIf for, Equation (4-2) becomes\n\n(4-3)\n\nwhich is the well-known Boltzmann transformation equation . Further, it is rewritten as\n\n(4-4)\n\nThe integral calculation of Equation (4-4) yields the integro-differential equation given by\n\n(4-5)\n\nwhere.\n\nHere, if we define\n\n(4-6)\n\nthe expression of\n\n(4-7)\n\nis obtained from using Equation (4-6) for Equation (4-5). Boltzmann transformation equation (4-3) is essentially equivalent to Equation (4-7).\n\nWe can accept Equation (4-7) as a diffusion flux in the parabolic space. The F1 law in the time and space cannot be used directly for analyzing a diffusion problem. However, the diffusion flux of Equation (4-7) is possible for analyzing a diffusion problem under the initial condition of Equation (4-6).\n\nFurther, as described later, the solutions of three dimensions problems in the parabolic space are obtained as a linear combination among the solutions of each one dimension diffusion problem. Therefore, Equation (4-7) is the most important essential equation relevant to the diffusion problems in the parabolic space.\n\nIn the parabolic space, the diffusion flux is defined as\n\n(4-8)\n\nwhere the component of is expressed as\n\nfor (4-9)\n\nusing for. Here, note that the initial and/or boundary condition is taken into account in the diffusion flux of Equation (4-6) or (4-8), although it is not taken into account in the F1 law.\n\nEquation (4-8) in the parabolic space is equivalent to Equation (2-6) in the time and space. In mathematics, the following relation of\n\n(4-10)\n\nis generally valid, if the diffusivity D depends on independent variables. Equation (4-10) is extremely useful for analyzing a nonlinear diffusion equation.\n\n4.3. Analytical Solutions of Linear Diffusion Equation\n\nThe utility of the present method is confirmed through solving some typical diffusion problems by concrete calculations. The mathematical method used here is also widely applicable to the Brown problems in various science fields.\n\n(1) One dimension parabolic space\n\nWe first solve the linear diffusion equation given by\n\n(4-11)\n\nunder the initial and boundary conditions of\n\n(4-12)\n\nThe initial condition of Equation (4-12) is rewritten as\n\n, (4-13)\n\nin the parabolic space.\n\nSubstituting and into Equation (4-5), the initial value of\n\nfor (4-14)\n\nis obtained. Using Equation (4-14) for Equation (4-5), the ordinary differential equation of\n\n(4-15)\n\nis obtained.\n\nFrom the definition of the well-known error function, the general solution of Equation (4-15) is obtained as\n\n(4-16)\n\nwhere is an integral constant. Using the initial condition of Equation (4-13) for the general solution, the well-known solution is obtained as\n\n(4-17)\n\nIn the analytical method in the parabolic space, the solution is thus easily obtained only by the elementary integral. It is, therefore, obvious that the present analytical method is extremely superior in analyzing diffusion equations to the existing methods.\n\n(2) Parabolic space of n dimensions\n\nIn case of, Equation (4-9) is rewritten as\n\nfor\n\nUsing this equation for, the relation of\n\n(4-18)\n\nis obtained from Equation (4-8).\n\nThe operation of on both sides of Equation (4-18) yields the Poisson equation given by\n\n(4-19)\n\nWe solve the ellipse type differential equation (4-19) corresponding to Equation (2-6) under the initial and boundary conditions in the following.\n\nThe initial and boundary values of Equation (2-6) given by\n\n(4-20)\n\ncorrespond to\n\n(4-21)\n\nin the parabolic space.\n\nIn the analysis of Equation (4-19), it is easily found that the equation of\n\n(4-22)\n\nsatisfies Equation (4-19). In other words, Equation (4-22) is the particular solution of Equation (4-19). When the general solution of Laplace equation of\n\n(4-23)\n\nis obtained as, the general solution of Equation (4-19) is expressed as\n\n(4-24)\n\nin accordance with the mathematical theory.\n\nAssuming the equation of\n\n(4-25)\n\nand using the variable separation method, the solution of Equation (4-23) is possible. Substituting Equation (4-25) into Equation (4-23), the relation of\n\nis valid. In order to satisfy this equation for an arbitrary, the equation of\n\n(4-26)\n\nmust be valid under the condition of\n\n. (4-27)\n\nEquation (4-27) yields for and solution of Equation (4-26) is expressed as\n\n,\n\nwhere are arbitrary constants. are determined by the initial condition. The well-known solution of Equation (4-19) is thus obtained as\n\n. (4-28)\n\nIn case of and, substituting the solution of Equation (4-26) given by\n\ninto Equation (4-25), the general solution of Equation (4-23) is obtained as\n\n(4-29)\n\nwhere are arbitrary constants. However, there is no solution satisfying the initial and boundary conditions because of Equation (4-27). As a result, the general solution of Equation (4-23) is obtained as\n\n(4-30)\n\nunder the condition of. Since the initial and boundary conditions yield, the present solution of Equation (4-23) is thus obtained as\n\nwhere in accordance with the initial and boundary conditions.\n\nThe solution of Equation (4-19) is thus obtained as\n\n(4-31)\n\nIf we use for, the solution is also presented as\n\n(4-32)\n\nThe present study reveals that the particular solution is easily possible and the complementary function of Equation (4-19) is a constant value to satisfy the given initial and boundary conditions because of the singular characteristic of initial conditions. In other words, the diffusion behavior is incorporated into the inhomogeneous term of the Poisson equation. As a result, the solution is obtained as a linear combination of the error functions.\n\nAlthough the diffusion problems have not ever been investigated in the parabolic space, it is obvious that the present solution is thus exceedingly simple and elegant, compared with the usual solution of\n\n(4-33)\n\nfor, where.\n\n(3) Analysis of inhomogeneous diffusion equation\n\nThe research subjects of linear diffusion phenomena in detail are often expressed by an inhomogeneous partial differential equation. Therefore, the application of the present method to those problems is investigated. Generally, such linear diffusion equations are expressed as\n\n(4-34)\n\nwhere L is the linear operator of\n\n(4-35)\n\nFrom the mathematical theory, the solution of inhomogeneous Equation (4-34) is obtained as\n\n(4-36)\n\nwhere is the complementary function, i.e., the general solution of homogeneous equation given by\n\n(4-37)\n\nand is a particular solution of\n\n(4-38)\n\nEquation (4-1) shows that the differential operators in the parabolic space are not able to apply to Equation (4-34). However, the linear operator in the parabolic space given by\n\n(4-39)\n\nis possible for analyzing the homogeneous Equation (4-37). In other words, we can obtain the complementary function by the present method. In order to obtain a solution of an inhomogeneous diffusion equation, therefore, we must obtain a particular solution by the existing methods.\n\n4.4. Analytical Solutions of Nonlinear Diffusion Equation\n\nUsing the experimental profile for Equation (4-3), Matano obtained the profile in the interdiffusion problems between solid metals in 1933 . The empirical Boltzmann Matano method has been widely applied to the analysis of the interdiffusion experiments between solid metals. However, the mathematical solutions of Equation (4-3) had not yet been obtained since 1894 for such a long time until the recent work . If the analytical solutions of Equation (4-2) or (4-8) obtained, we can understand the effect of a driving force on the diffusivity. In other words, of Equation (2-7) will be analytically obtained.\n\n(1) Analytical method in parabolic space\n\nWhen the diffusivity is affected by a driving force, the mathematical solutions of the nonlinear equations (2-11) and (4-3) or (4-7) are impossible if no other relation between diffusivity and concentration is given. In that case, the diffusivity depends on the independent variables and Equation (4-10) is generally valid in mathematics then. Equations (4-7) and (4-10) are thus simultaneously solved in the following.\n\nThe integro-differential equation (4-5) is superior in the approximate calculation to the second order differential equation (4-3). For example, using an effective diffusivity for the exponential part in Equation (4-5), the relation of\n\nis valid because of the characteristic of the integral calculation, where D is rewritten as an interdiffusion coefficient.\n\nIn order to clarify the validity of the present method, we applied the present method to the typical problems of a binary system interdiffusion where the complete solid solution is formed in the interdiffusion region between solid metals. The reason for that is as follows. The Boltzmann Matano method has been widely used for the analysis of various interdiffusion problems. Uncountable papers have been reported and useful findings have been thus accumulated. Even if the present results are thus evaluated by comparing them with the Boltzmann Matano results, the generality of present method is still kept from the viewpoints of mathematical physics.\n\nIn the following analysis, the relations of\n\n(4-40)\n\nare adopted as an initial condition in the parabolic space. The coordinate origin is set at a point on the initial interface of a diffusion couple between materials A and B similar to the one under the condition of in Section 3. The interdiffusion region of corresponds to in the parabolic space\n\nbecause of. Here, the suffix I or II of C and is deleted since the diffusion equation of I is equal to that of II because of.\n\nThe countless experimental results always reveal that the profile becomes an S-letter curve or a reverse one similar to the profile of Equation (4-17) . In the typical interdiffusion problems, the Boltzmann Matano method shows that the profile is also an S-letter curve or a reverse one. These indicate that and are expressed as a superposition of the error functions with various inflection points.\n\nThe relation of is conveniently adopted in this section. In such a case, the exponential part of Equation (4-5) satisfies\n\n.\n\nUsing a constant value between, it is thus defined as\n\n(4-41)\n\nwhere is a function to correct the error caused by using instead of.\n\nSubstituting Equations (4-5) and (4-41) into Equation (4-10) yields the diffusion flux relation of\n\n(4-42)\n\nFor the diffusion flux, the physical speculation produces the relation yielding\n\n(4-43)\n\nwhere is a function of satisfying. Equations (4-42) and (4-43) yield\n\n(4-44)\n\nIn relation to a behavior of the error function, Equation (4-44) is divided into following two equations. One is\n\n(4-45)\n\nand the other is\n\n(4-46)\n\nwhere is a shift parameter caused by the dependence of on and\n\nand\n\nThere is evidence for the validity of the above division as shown in the following. When the relation of\n\nis valid, Equation (4-46) is approximately rewritten as and its integral calculation yields\n\n(4-47)\n\nunder the condition of Equation (4-40). In the typical interdiffusion problems between solid metals, Equation (4-47) has been widely accepted - . The derivation of Equation (4-47) gives evidence for the validity of division of Equation (4-44) into Equations (4-45) and (4-46).\n\nSince it is thus confirmed that the division of Equation (4-44) into Equations (4-45) and (4-46) is reasonable, Equation (4-45) is analyzed in accordance with mathematical theory. As can be seen from the analyzing process in Ref. , the calculations are considerably complex. As a result, the general solution of\n\n(4-48)\n\nwas obtained as an analytical expression using initial values, where\n\nfor and for\n\nHere, the relation of for is valid in the present diffusion system. Therefore, the effect of a driving force and/ or a concentration dependence on the diffusivity, which is discussed in Section 2.1, is obtained as\n\n(4-49)\n\nSubsequently, using the obtained solution of Equation (4-48) for Equation (4-46), Equation (4-46) was analyzed also through considerably complex calculations. The general solution of\n\n(4-50)\n\nwas thus obtained also as an analytical expression.\n\nHere, the notations used for Equations (4-48) and (4-40) are as follows:\n\nThe validity of the present analytical method is numerically investigated by comparing Equations (4-48) and (4-50) with the results of the Boltzmann Matano method. As a result, we found that the present solutions agree well with those of the Boltzmann Matano method .\n\nWhen the diffusivity does not depend on independent variables, the relation of is valid. In that case, the above relations become\n\n, , and\n\nEquations (4-48) and (4-50) agree with and Equation (4-17) then, respectively. The obtained solution of Equations (4-48) and (4-50) are thus accepted as a generalized ones.\n\nThe K effect in the parabolic space is easily obtained as\n\nby transforming Equation (3-15) into the parabolic space under the condition of. The K effect in the parabolic space is thus determined only by the initial values.\n\n(2) Meaning of general solutions obtained as analytical expressions\n\nSince the general solutions are obtained as Equations (4-48) and (4-50), the interdiffusion problems are fundamentally solved as follows:\n\nWe denote an impurity diffusivity of a diffusion element I in a pure material B composed of an element II and also denote an impurity diffusivity of the diffusion element II in the pure material A com-\n\nposed of the element I. We first investigate an interdiffusion problem of the diffusion couple between pure materials A and B. It is then well-known that the relations of\n\nand (4-51)\n\nare physically valid.\n\nThe solution is determined by using and the values of Equation (4-51) for Equation (4-50). It gives two independent variable values satisfying at and at as follows:\n\n(4-52)\n\n(4-53)\n\nSubstituting these and and the values of Equation (4-51) into Equation (4-48), we can determine and for and.\n\nSubsequently, using and for Equations (4-48) and (4-50) as boundary values of interdiffusion coefficients in the binary system, we can obtain the solutions of the interdiffusion problems for the diffusion couple between the material A composed of, and the material B composed of,.\n\nThe solutions of an arbitrary binary system interdiffusion problem are thus possible by using impurity diffusivities in the concerned diffusion system. In the N elements system interdiffusion problems, using an impurity diffusivity of an element in the concerned diffusion couple, their solutions are also possible.\n\nIn the diffusion history, the nonlinear F2 equation had not been analytically solved until recently, although the numerical solutions were possible. In order to understand the diffusion problems, the corresponding diffusion fluxes to the F2 equation had been thus widely discussed in the existing theories, since it was considered that the mathematical solutions of nonlinear diffusion equation are impossible. For example, such a diffusion flux in the Appendix A has been discussed.\n\nHowever, the discussion about such a diffusion flux is now unnecessary for understanding the diffusion problems, since the analytical solutions of Equations (4-48) and (4-50) relevant to the nonlinear F2 equation are obtained, regardless of the diffusion fluxes. In other words, as can be seem from Equation (4-49), we can understand the influence of a driving force on the diffusion flux from the obtained solutions.\n\nBased on the above discussions, it is extremely meaningful that the solutions of the nonlinear F2 equation are obtained as analytical expressions.\n\n(3) Nonlinear problems of three dimensions in parabolic space\n\nThe analytical solutions of linear diffusion equations were obtained in the parabolic space. Further, we could also obtain the physically meaningful solutions of nonlinear diffusion equation as analytical expressions in the one dimension parabolic space.\n\nIn mathematics, it is almost impossible to solve two or three dimensional nonlinear diffusion equation. However, we investigate two or three dimensional problems in the parabolic space in the following.\n\nWe rewrite Equation (4-2) into\n\n(4-54)\n\nwhere\n\n(4-55)\n\nIt is impossible to analyze Equation (4-54) exactly since exists in. By reference to the one dimensional case, if we use the effective diffusivity independent of instead of for Equation (4-55) as an approximation, Equation (4-54) becomes\n\n(4-56)\n\nbecause of.\n\nBy reference to Equation (4-19), the solution of Equation (4-56) is obtained as\n\n(4-57)\n\nwhere are integral constants.\n\n5. Universal Expression of Diffusivity\n\nFrom applying the diffusion equation to a collision problem between two micro particles in the diffusion elementary process, the S equation is reasonably derived. It is revealed that the diffusivity corresponds to the angular momentum operator in the quantum theory. The universal diffusivity expression, which is applicable to every material in an arbitrary thermodynamic state, is obtained using an essential diffusion constant then. Further, the correlation between the well-known material wave relation and the diffusivity expression is discussed in a minute time and space region.\n\n5.1. Derivation of Schrödinger Equation\n\nIf the movement of a material particle is not the Markov process but the continuous process, Equation (2-1) is rewritten as\n\n(5-1)\n\nwhere.\n\nFrom the Taylor expansion of both sides of Equation (5-1), the well-known wave equation is obtained as\n\n(5-2)\n\nwhere is a phase velocity under the condition of. The general solution of Equation (5-2) is given by\n\n(5-3)\n\nwhere A and are a wave amplitude and an initial phase, and and mean a component of the wave number vector relevant to a wavelength component and an angular velocity relevant to a period T.\n\nThe collective behavior of micro particles relevant to the Markov process is expressed by Equation (2-6) derived from their random movements. However, it is revealed in the quantum mechanics that a micro particle has a wave characteristic. The relation of material wave yielding\n\n(5-4)\n\nwas thus proposed by de Broglie in 1923, where and are the momentum of a material particle and the Planck constant . The motion of a micro particle relevant to Equation (5-4) should be thus essentially investigated by using the S equation of\n\n(5-5)\n\nwhere is the wave function and the Hamiltonian meaning the total energy in the given physical system . In case of a free particle, the Hamiltonian H is given by\n\n(5-6)\n\nwhere m is the particle mass and the momentum or its operator.\n\nWhen Equation (5-4) is applied to a micro particle behavior, the left-hand side of Equation (5-4) means the particle motion in accordance with the Markov process. On the other hand, the right-hand side means the continuous process resulting from the wave characteristic of the micro particle during a free motion because of the wave factor. This suggests that the S equation correlates with the diffusion elementary process resulting from a random movement caused by a collision between a micro particle and a force field near the micro particle itself.\n\nAn activation energy caused by a local thermal fluctuation generates micro holes in a material and at the same time micro particles are possible to jump instantaneously to the nearest neighbor micro holes through each energy barrier. The macro behavior of their collective motion is investigated by using the diffusion equation (2-6).\n\nOn the other hand, behavior of a micro particle of mass m in the diffusion field is investigated by the well-known Langevin equation of\n\n(5-7)\n\nwhere and are the velocity and the viscosity resistance using a resistance factor . Here, is a force caused by a collision and the time-averaged value is considered to be.\n\nThe Langevin equation shows that the behavior of a Brown particle in the minute time and space region is incorporated into the force. Therefore, we cannot discuss the Brown motion in the minute region by using the Langevin equation because of. Hereafter, we do not discuss but the acceleration of\n\n(5-8)\n\ncaused by a collision between two micro particles in the minute time and space region.\n\nSince the physical essence is still kept even if we consider the simplest collision problem of one dimensional case, we thus investigate a perfect elastic collision problem between micro particles A and B of the same kind. We consider that the particle A moves with a velocity for a direction and collides at time with the particle B in the standstill state.\n\nIf we can then clarify the distinction between A and B after the collision, the particle A decelerates from the velocity to the velocity zero and the particle B accelerates from the velocity zero to the velocity between.\n\nOn the other hand, if we cannot clarify the distinction between the particles A and B after the collision, it seems that the particle A decelerates from the velocity to the velocity zero between and subsequently accelerates again from the velocity zero to the velocity between. In other words, it seems as if there is no collision process in the particle motion.\n\nIf we accept the above latter case, the relation of () must be valid in Equation (5-8) between and in the collision process. This indicates that the impossibility of discrimination between the particles A and B yields between, as can be seen from the expression of Equation (5-8).\n\nIt is considered that the above former case corresponds to the Markov process caused by a collision and the latter case corresponds to a continuous process caused by the impossibility of identification between the micro particles. It is thus mathematically necessary for the transformation from the Markov process into the continuous process to introduce the imaginary time into a collision problem between micro particles.\n\nIn physics, it is difficult to accept the imaginary time in accordance with the continuity of time. In mathematics, however, the product of a physical quantity Q and an imaginary time is equal to and also that of a physical operator and a real time t is equal to the same result. It is thus considered that the latter case is physically reasonable. However, it is possible that we mathematically analyze the diffusion equation using the imaginary time in accordance with the mathematical equivalence relation. The diffusion equation is thus applied to an elementary diffusion problem using the imaginary time in the following.\n\nRewriting the concentration of diffusion particles into a quantity of state expressed by a complex function, Equation (2-6) yields\n\n(5-9)\n\nAssuming, Equation (5-9) can be solved by the separation method of variables. Using complex numbers, and determined by the initial and boundary conditions, the general solution of Equation (5-9) is obtained as\n\n(5-10)\n\nwhere. Substituting into Equation (5-10) yields\n\nHere, using the real functions and, we rewrite the complex function into the complex-value function yielding\n\n(5-11)\n\nFurther, substituting Equation (5-11) into Equation (5-9) and multiplying both sides by, we have\n\n(5-12)\n\nbecause of.\n\nFrom the partial differential equation of probability, the diffusivity of\n\n(2-5)\n\nis obtained. When a micro particle randomly jumps from a position to another one, the jump orientation becomes the spherical symmetry in probability. Using the relation of relevant to the angular momentum defined by a position vector and a momentum, Equation (2-5) is rewritten as\n\nwhere, , , is valid in the spherical symmetry space.\n\nThe relation of leads to an operator relation of\n\n(5-13)\n\nSubstituting Equation (5-13) into Equation (5-12) gives\n\n(5-14)\n\nHere, if we define the imaginary differential operator given by\n\n, (5-15)\n\nEquation (5-14) is rewritten as\n\n(5-16)\n\nFurther, the substitution of Equation (5-6) into Equation (5-16) yields the S Equation (5-5). The defined Equation (5-15) is one of the essential operators in the quantum mechanics.\n\nIt is considered that the relation of and is valid for a wave packet existing in the minute time and space. In that case, the de Broglie equation\n\n(5-4)\n\nis thus easily obtained because of\n\n, and\n\nAt the same time, this gives evidence for the validity of Equation (5-13) derived from the correlation between the diffusion equation and the S equation. It is, therefore, considered that the diffusivity is fundamentally relevant to the Planck constant and it should be determined from the quantum mechanics.\n\n5.2. New Universal Diffusivity\n\nThe definition of diffusivity shows that the diffusivity should be essentially determined by analyzing the S equation, since the diffusion particle such as an atom and/or a molecule moves as a material wave packet in the minute space. However, the existing diffusivity has been defined by the classical mechanics. In the following, therefore, a new diffusivity is investigated.\n\nThe existence probability P of a micro particle in the collective system of a heat quantity Q and an absolute temperature T is given by the well-known Boltzmann factor of\n\n(5-17)\n\nwhere is the Boltzmann constant . It is necessary for a jumping of diffusion particle to obtain an activation energy Q caused by a thermal fluctuation. In a collective system composed of micro particles, the diffusivity is thus directly proportional to the probability factor of Equation (5-17).\n\nThe jump frequency of a diffusion particle is relevant to a factor derived from an atomic configuration in the diffusion field near the diffusion particle itself, and also it depends on entropy S relevant to a state change of diffusion field. If we define these effect on a diffusivity as, the diffusivity expression of Equation (2-5) or (5-13) is rewritten as\n\n(5-18)\n\nwhere the molecular or the atomic weight n is expressed as using the particle mass m and the Avogadro constant. Here, is an essential constant in the diffusion phenomena and it is\n\n(5-19)\n\nEquation (3-32) shows that diffusion particles move randomly with a common diffusivity in the concerned diffusion field. On the other hand, Equation (5-18) shows that the diffusivity depends on the mass of diffusion particle. For the diffusion experimentation in a super gravity field, therefore, it is considered that an individuality of a micro particle resulting from the mass effect will be shown in the diffusivity, compared with one in a usual laboratory.\n\nHereafter, the investigation of will become a dominant subject in the diffusion study as a problem of a few many bodies system in the quantum mechanics, since the essential diffusion constant of Equation (5-19) is thus reasonably obtained.\n\n5.3. Influence of a Driving Force on Diffusivity\n\nThe new diffusivity of Equation (5-18) is derived regardless of the bulk diffusion problems. However, it is considerably difficult to determine of Equation (5-18) from the quantum theory in a physical system with an interaction between micro particles . Further, if we consider Equation (5-18) in accordance with the existing diffusivity theory, it seems that there is no room to incorporate a driving force effect into the diffusivity expression of Equation (5-18), since the essential constant has already the diffusivity dimension. On the other hand, the diffusivity expression of Equation (4-48) is derived from analyzing the nonlinear interdiffusion equation with a driving force.\n\nIf we incorporate the potential energy caused by a driving force into the activation energy Q in the expression of Equation (5-18), the relation is obtained as\n\n(5-20)\n\nfor a material composed of micro particles in the state of an absolute temperature and a heat quantity, since Equation (5-18) corresponds to Equation (4-48) then. In the actual interdiffusion problems, obtained here is useful for fundamental knowledge relevant to a microstructure of material.\n\nEquation (5-18) is thus valid when a driving force is nonexistent in the diffusion system. In other words, the relation of\n\n(5-21)\n\nis considered to be valid between Equations (2-6) and (5-18) for the common and.\n\nAs can be seem from Appendix A in the existing theory, the effect of a driving force on the diffusivity is incorporated into the difference of a jump frequency between nearest neighbor crystal plane 1 and plane 2. In relation to the situation, we discuss the difference between the present diffusivity expression and the existing one as follows.\n\nThe existence probability of a micro particle in a material depends on the Boltzmann factor of Equation (5-17). The potential energy as well as the activation energy Q gives motive energy and vibrational one to each micro particle in a material. In that case, the potential energy causes that the vibrational energy of micro particle on the crystal plane 1 is different from that of one on the plane 2, since the potential energy depends on the space coordinate. In other words, the existence probability of a micro particle on the plane 1 is different from that of one on the plane 2.\n\nThe difference of existence probabilities between a micro particle on the plane 1 and one on plane 2 corresponds to that of the frequencies between them. The effect of a driving force on diffusivity should be thus incorporated into the Boltzmann factor as a potential energy in the present case.\n\nWhen the potential energy is not so much large, i.e., , the diffusivity D of Equation (2-7) corresponds to\n\n(5-22)\n\nincorporating a driving force effect into the Boltzmann factor for and. Equations (2-7) and (5-22) show that the relation of\n\n(5-23)\n\nis valid. Although the diffusivity gradient of\n\n(5-24)\n\nhas the velocity dimension, it is directly relevant to a driving force F because of the definition of mechanics. The left-hand side of Equation (5-24) is thus called the drift velocity in the existing diffusion field (See Appendix A). However, the diffusivity gradient relevant to a driving force is obviously not the velocity concept but the force one.\n\nSubstituting Equations (5-22)-(5-24) into the one dimension diffusion equation of the general nonlinear Fickian second law given by\n\n(5-25)\n\nthe diffusion equation is obtained as\n\n(5-26)\n\nwhere is entropy caused by a driving force. It is considered that the second term and third term in the right-hand side of Equation (5-26) mean entropy effect and a frequency effect caused by a driving force F, respectively. The general diffusivity expression is thus obtained by rewriting Equation (5-18) into\n\n(5-27)\n\nIn view of the derivation process, the diffusivity expression of Equation (5-27) is applicable to every material in an arbitrary thermodynamic state.\n\n5.4. Operators in Quantum Mechanics\n\nIn the quantum mechanics, energy E, momentum and angular momentum are expressed as operators yielding\n\n, ,.\n\nWe cannot observe imaginary physical quantities. Therefore, the eigenvalues of their operators are meaningful in the quantum theory.\n\nIn a collision problem mentioned above, the concept of acceleration disappears in the quantum theory because of. However, Equation (5-4) shows that the velocity is still a real physical quantity in the minute time and space region, since the momentum conservation law is valid in the collision process.\n\nIt is physically considered that the wave image of a micro particle disappears in the quantum theory when the corresponding wave packet exists in the minute space region. Substituting and into Equation (5-3) yields the relation of\n\n(5-28)\n\nand it converges at zero in a minute time. For the wave image, the relations of and mathematically correspond to the relations of and then, compared with and for the particle image.\n\nIn mathematics, except for physical constant quantities, physical variables with odd time or space dimensions should be thus accepted as physical operators in the minute time and space region.\n\nThe kinetic energy of Hamiltonian is acceptable as a physical quantity and/or a usual differential operator since it is a real variable for the sake of even time or space dimensions. On the other hand, the photon energy expressed by using a frequency should be accepted as an operator, since h is the Planck constant and is an imaginary variable because of in accordance with the wave image of. Therefore, the relation of yields the operator described above.\n\nUsing relations of, and for the wave packet in the time and space region, the relation of\n\n(5-29)\n\nyields the operator described above. Further, using for the definition of an angular moment, the above angular moment operator is also obtained.\n\nWe investigate a characteristic of diffusivity in the minute time and space regions in the following. Equation (2-5) is rewritten as\n\n(5-30)\n\nfor a particle image. Equation (5-4) is also rewritten as\n\n(5-31)\n\nbecause of and. Substituting Equation (5-31) into Equation (5-30), the diffusivity expression of\n\n(5-32)\n\nis obtained as an operator.\n\nSubstituting Equation (5-32) into Equation (2-6), we have\n\n(5-33)\n\nMultiplying both sides of Equation (5-33) by and rewriting C into, the S Equation (5-14) is thus obtained again. In mathematics, the diffusion equation and the S equation are thus correlated only to the replacement of between their equations in the real time and space.\n\n6. Results\n\nIn 1855, Fick applied the heat conduction equation proposed by Fourier in 1822 to diffusion phenomena as it had been. In general, however, the discussion of coordinate transformation of the general nonlinear F2 law is indispensable for understanding the diffusion phenomena. Nevertheless, the coordinate transformation of a diffusion equation had not been discussed for a long time until the recent works - .\n\nApplying the divergence theorem to a diffusion equation, therefore, we investigated the problems of coordinate transformation in the diffusion systems. In relation to the investigation, a new diffusion flux of Equation (2-13) was defined in accordance with the mathematical physics. The new diffusion flux revealed that the F1 law is incomplete without the intrinsic diffusion flux.\n\nThe coordinate transformation theory shows also that the K effect is caused by a shift between the coordinate systems like the Doppler effect of the wave equation is caused by a shift between them. The theoretical Equation (3-15) relevant to the K effect was reasonably obtained in accordance with the empirical Equation (3-14) then. At the same time, it was also revealed that the concept of intrinsic diffusion is nonexistent in the diffusion theory, although we had accepted as an indispensable concept for understanding the K effect for a long time until recently.\n\nIn relation to the coordinate transformation theory, further, using the new diffusion flux applicable to various diffusion problems, the unified diffusion theory relevant to every material in an arbitrary thermodynamic state was established, including a self-diffusion theory and also an N elements system interdiffusion theory.\n\nThe self-diffusion theory has been indirectly discussed in the existing one, where the self-diffusion behavior of a pure material is visualized by any other diffusion particles in the pure material. In the present theory, the self-diffusion mechanism of a pure material itself is discussed using the intrinsic diffusion flux. In an N elements system interdiffusion, it was found that each of diffusion particles moves randomly with a common diffusivity of Equation (3-32) in the diffusion field, as if the gas molecules in a given space move randomly, regardless of their species in accordance with the Boyle Charles law.\n\nThe essence of diffusion phenomena is shown in the parabolic law. Based on the concept, a new diffusion flux applicable to analyzing diffusion problems was derived as Equation (4-7) and/or (4-8) in the parabolic space. The analytical method in the parabolic space was thus established. It was then revealed that the analysis in the parabolic space is exceedingly superior in calculation to the existing analytical methods. For example, the solutions of nonlinear diffusion equation were, for the first time, obtained as the analytical expressions of Equations (4-48) and (4-50). The effect of a driving force on the diffusivity was then obtained as Equation (4-49) from the solution of Equation (4-48).\n\nThe definition of diffusivity indicates that the diffusivity should be essentially investigated as a few many bodies problem in the quantum mechanics. Based on the concept, the S equation was derived from applying the diffusion equation to a diffusion elementary process. In the analytical process, we found that the diffusivity corresponds to an angular momentum operator in the quantum mechanics. The new universal diffusivity expression was thus obtained as Equation (5-27) using an essential diffusion constant of Equation (5-19). It is essentially applicable to every material in an arbitrary thermodynamic state.\n\nIt was also found that the material wave equation proposed by de Broglie, which is the most fundamental relation in the materials science, was derived from the relation between diffusivity expressions of Equations (2-5) and (5-13) in the diffusion elementary process. This gives evidence for the validity for the present diffusion theory.\n\n7. Discussion\n\nEven if a driving force exists in the diffusion system under the condition of no sink and source, the divergence theorem shows that a collective behavior of micro particles depends only on a diffusivity of the general nonlinear F2 law. Whenever the diffusivity of a micro particle in a material is given, the collective behavior of micro particles is thus determined only by solving the diffusion equation. The diffusivity depends on a material internal structure composed of micro particles. Therefore, the investigation of diffusivity is the most important subject in the diffusion study. Based on the concept discussed here, the present study was performed.\n\nIn relation to the Darken equation, even if a concentration profile is reasonably reproduced by using intrinsic diffusion coefficients as parameters, we cannot directly evaluate its validity of the used diffusivity values, since the diffusivity profile is not obtained by a direct observation. In other words, we can usually perform the numerical simulation to a high precision only by using fitting parameters for the given equation, regardless of a physical validity. As far as the Darken equation is not mathematically correct as shown in the Appendix B, it is thus necessary for us to reexamine a great many results of interdiffusion problems obtained by using the unsuitable Darken equation.\n\nAs revealed in the text and Appendix B, we believe that the F1 law and the intrinsic diffusion concept as well as the Darken equation are misunderstanding matters in the diffusion history. Nevertheless, they have been accepted as reasonable ones for such a long time in the concerned field.\n\nIf we consider serious influences of their misunderstanding matters on student education and young researcher training in the concerned field, they should be solved as soon as possible, since they have been still described as plausible ones in a lot of existing textbooks. One of the present work aims is thus to make them universally known in the concerned field.\n\nSince the general solutions of nonlinear diffusion equation are analytically obtained in the parabolic space, the diffusion problems under the condition of no sink and source are essentially solved. The influence of a driving force on the diffusivity is thus reasonably obtained as Equation (4-49) from the analytical solutions of Equations (4-48) and (4-50). The discussions about diffusion fluxes are thus now essentially unnecessary for fundamental diffusion problems.\n\nIn the actual diffusion problems in detail, further investigation will be required for various difficult problems, for example, such a problem where the sink and source exists in the concerned diffusion system. Even in that case, however, it is no doubt that the fundamental theories discussed here are still useful for their fundamental investigations, just because of fundamental ones.\n\nThe investigations relevant to the many elements system interdiffusion problems, such as an alloy, a compound semiconductor, a multilayer thin film, and a microstructure material, are extremely important as an actual problem. Under the condition of Equation (3-27), corresponding Equations (4-48) and (4-50) to solutions of Equation (3-33) and using solutions of Equation (3-34) obtained numerically for a many elements system interdiffusion problems, their solutions are possible.\n\nAs mentioned at the beginning of this section, the foundation of diffusion problem is relevant to not the diffusion flux but the diffusion equation for the concerned diffusion system. The influence of a driving force on a diffusion system should be thus incorporated into the diffusivity of the concerned diffusion equation, e.g., Equation (5-25), although the problem relevant to a driving force is discussed in the existing theory using a drift velocity in the diffusion flux representation (see Appendix A).\n\nIn the present study, the fundamental diffusion problems are surveyed in accordance with the mathematical physics. As a result, the new useful findings for the diffusion study are reasonably derived from the essential discussions about the general nonlinear Fickian diffusion equation. 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(1982) Transactions of the Japan Institutetute of Metals, 21, 674-682.\nhttp://dx.doi.org/10.2320/matertrans1960.21.674\n\n43. 43. Onishi, M. and Miura, H. (1977) Transactions of the Japan Institutetute of Metals, 18, 107-112.\nhttp://dx.doi.org/10.2320/matertrans1960.18.107\n\n44. 44. Boltzmann, L. (1872) Wiener Berichte, 66, 275-370.\n\n45. 45. Ohya, M. and Watanabe, N. (2010) Entropy, 12, 1194-1245.\nhttp://dx.doi.org/10.3390/e12051194\n\n46. 46. John, L. and Haller, J. (2013) Journal of Modern Physics, 4, 1393-1399.\nhttp://dx.doi.org/10.4236/jmp.2013.410167\n\nAppendix A\n\nIn Ref. , the influence caused by a driving force in the planar diffusion phenomena is investigated in relation to jump frequencies of diffusion particles between the nearest neighbor crystal planes.\n\nThe net diffusion flux J of diffusing particles across a plane between two nearest neighbor crystal planes 1 and 2 is expressed as\n\n, (A-1)\n\nwhere and are the atom concentrations per unit area on each plane, and and are the jump frequencies for an atom to jump from plane 1 to plane 2 and from plane 2 to plane 1. The coordinate axis x is perpendicular to each plane and equal to the diffusion direction.\n\nThe relation between concentration per unit volume and concentration per unit area is expressed as\n\n(A-2)\n\nwhere a is a lattice constant and in the present case.\n\nHere, the following relations are considered as;\n\n, (A-3)\n\nwhere.\n\nSubstituting Equations (A-2), (A-3) into Equation (A-1), the relation of\n\n(A-4)\n\nis obtained. Using the notations given by\n\nand (A-5)\n\nthe diffusion flux of\n\n(A-6)\n\nis derived. Here, is defined as a diffusivity of tracer diffusion. If, means the diffusivity when a driving force is nonexistent in the diffusion system.\n\nOn the other hand, if we use the Taylor expansion in the above calculations, the relation of\n\n(A-7)\n\nand\n\n(A-8)\n\nare valid. Using Equations (A-7), (A-8) for Equation (A-1), the relation similar to Equation (A-6) is obtained as\n\n, (A-9)\n\nwhere. Equations (A-6) and (A-9) show that corresponds to the diffusivity gradient given by\n\n. (A-10)\n\nEquation (A-9) should be essentially equivalent to Equation (2-7) as shown by\n\n(A-11)\n\nHowever, it seems that Equation (A-11) is not generally valid. The reason is as follows.\n\nWhen a driving force exists in the diffusion system, it affects entropy and jump frequencies of diffusion particles. As a result, the diffusivity is affected by a driving force, since the diffusivity depends directly on entropy and a frequency. The problem of entropy is extremely important for the diffusion phenomena. However, the effect of entropy is not taken into account in the derivation process of Equation (A-6) or (A-9). Therefore, the diffusion flux discussed here is not generally applicable to the diffusion problems relevant to a driving force.\n\nEquations (4-48) and (4-50) reveal that we can solve a nonlinear diffusion equation even if a driving force exists in the diffusion system. Further, the effect of a driving force on the diffusivity is obtained as Equation (4-49). The diffusion flux discussed here is thus unnecessary for solving a nonlinear diffusion equation.\n\nFurther, the divergence theorem shows that the diffusion equation is obtained from the diffusion flux of Equation (A-9) as\n\n. (A-12)\n\nAs discussed in Section 2.1, even if a driving force exists in the diffusion system, the diffusion equation (2-6) of\n\n(A-13)\n\nmust be valid for the diffusion particles, except such an external force as Coulomb force for an ion particle.\n\nEquation (A-12) is obviously inconsistent with Equation (A-13). This is evidence that Equation (A-6) neglecting an influence of entropy caused by a driving force is incorrect. In relation to the Equation (5-24), the diffusivity gradient is directly relevant to a driving force. Although the last term in the right-hand side of Equation (A-12) is the derivation term of a driving force because of Equation (5-24), we cannot understand such an effect of relevant to the random movement of diffusion particles in the diffusion system inside. If the force F is an external one where it is deliberately introduced into the diffusion system from the diffusion system outside, such phenomena may be possible. In other words, such diffusion flux of Equation (A-6) is not suitable in the present discussion, since the influence of a driving force should be incorporated into the Boltzmann factor as a potential energy caused by a driving force, as discussed in Section 5.3.\n\nFor example, one can easily notice that a force moment of vector quantity is not relevant to a kinetic energy of scalar quantity, although they have the same dimension of each other. Even if the drift velocity has the dimension of in the present case, it is thus not relevant to a velocity itself, since it is relevant to a driving force F as can be seem from Equation (5-24).\n\nIn general, the diffusion problems should be discussed using not the diffusion flux but the original diffusion equation, since the mathematical theory shows that the diffusion flux should be originally understood after the diffusion equation has been solved.\n\nAppendix B\n\nIn the text, the K-effect was reasonably understood as a shift between the coordinate systems. This indicates that the intrinsic diffusion concept is not only needless but also nonexistent in the diffusion phenomena.\n\nBased on the intrinsic diffusion concept, Darken derived a relation between the interdiffusion coefficient and intrinsic diffusion coefficient . However, he repeated misunderstandings inconsistent with the fundamental mathematics theory in the analyzing process. Nevertheless, the so-called Darken Equation (D equation) has been widely applied to interdiffusion problems for such a long time since 1948. In the following, we reveal that the D equation is incorrect in mathematics.\n\nWhen a driving force exists in the diffusion system, the diffusion phenomena are expressed as\n\n(2-6)\n\nIn that case, using the Equation (2-17) transformed from Equation (2-6) for, the relation of\n\n(B-1)\n\nis valid, where the coordinate notation and the velocity are, for convenience, rewritten as and v. Here, the velocity used in Equation (2-15) means the velocity of diffusion region space caused by the diffusion of solvent particles. At the same time, it is visualized by an inert marker set on the initial interface because of the inert characteristic of markers.\n\nEquation (B-1) yields\n\n(B-2)\n\nbecause of. Here, note that Equation (B-2) is obviously a partial differential equation of and. Therefore, the mathematical theory shows that Equation (B-2) corresponds to\n\n(B-3)\n\nwhere is an arbitrary function of.\n\nNevertheless, Darken solved Equation (B-2) as an ordinary differential equation with respect to. In other words, using k as an integral constant value instead of the, he thought that the relation of\n\n(B-4)\n\nis valid. Using an unsuitable initial value for Equation (B-4), it was rewritten as\n\n(B-5)\n\nHere, the functional equation (B-5) is not valid, since the left-hand side depends only on and the right- hand side depends on and. Thus, he failed again to notice the mistake in the fundamental mathematics.\n\nUnder the condition of in Equation (2-18) in accordance with the existing theory, the diffusion flux becomes\n\n. (B-6)\n\nIn any case, if we substitute Equation (B-5) into Equation (B-6), the relation of\n\n(B-7)\n\nis obtained. By comparing Equation (B-7) with the F1 law of, the D equation is perfunctorily derived as\n\n(B-8)\n\nHe also failed to notice the mistake relevant to the basic theory of physics, since we cannot directly compare with expressed in the different coordinate system, as can be seen from the derivation of Equation (2-17).\n\nIt seems that and are independent of each other in the above analytical process. In other words, he thought that the K effect is caused by the difference between and. They are symbolically rewritten as and in Equation (B-8) and are named “intrinsic diffusion coefficients” then.\n\nIn addition, even if we accept Equation (A-6) in Appendix A, it is not suitable to use it for the movement of K interface in the present case, since the drift velocity depends on a time and a space coordinate, while the experimental result of Equation (3-14) shows that the K effect depends on only a time. It is obvious that the problem of K effect is a one to be discussed using not the diffusion flux Equation (A-6) but the diffusion Equation (2-6) taking account of a driving force. Further, the present interdiffusion problem should be expressed by Equation (2-6). If it is so, the corresponding diffusion flux must be not Equation (A-6) but Equation (B-6) obtained from Equation (2-6).\n\nHereinbefore, it was revealed that the D equation is obviously inconsistent with the theory in mathematical physics. Equation (B-8) is thus entirely meaningless and at the same time the concept of intrinsic diffusion is an illusion in the diffusion history. That the D equation has been highly applied to various interdiffusion problems by a lot of researchers for a long time since 1948 is one of the extremely few cases of misunderstanding problems in the science history."
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http://vixenlights.com/g82g947w/ojvcp.php?ed1276=what-is-amplitude-of-a-wave | [
"# what is amplitude of a wave\n\nThe amplitude of a wave is the maximum displacement, or the maximum value, it attains from its mean position during a cycle. Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source. It requires a medium for its propagation. Let us know if you have suggestions to improve this article (requires login). If an event occurs repeatedly then the event is said to be periodic. The amplitude of a sound wave can be defined as the loudness or the amount of maximum displacement of vibrating particles of the medium from their mean position when the sound is produced. Amplitude, Period, Phase Shift and Frequency. It is the distance between crest or trough and the mean position of the wave. Hence, sound cannot propagate in a vacuum as there will be no material to transfer sound waves. Our editors will review what you’ve submitted and determine whether to revise the article. The amplitude of a pendulum swinging through an angle of 90° is 45°. The amplitude is a measure of the strength or intensity of the wave. The amplitude of a wave refers to the maximum amount of displacement of a particle on the medium from its rest position. It is important to note that the amplitude is not the distance between the top and bottom of a wave. A phenomenon that occurs when two objects naturally vibrate at the same frequency; the sound produced by one object causes the other object to vibrate. B) What is the frequency of this wave? sound waves or earthquakes). Amplitude is generally calculated by looking on a graph of a wave and measuring the height of the wave from the resting position. In a sense, the amplitude is the distance from rest to crest. If the amplitude is small, then the sound will be feeble. The amplitude of a wave is the maximum disturbance or displacement of the medium from the equilibrium (rest) position. To know more about the characteristic of sound waves download BYJU’S- The Learning App. Period of SoundFrequency of SoundAmplitude of Sound. Standing Wave. They are described below: A period can be said to be the time taken to do something. Updates? A more general representation of the wave equation is more complex, but the role of amplitude remains analogous to this simple case. It is the distance between crest or trough and the mean position of the wave. Every wave, whether it’s one you see at the beach or the music playing in your headphones, has an amplitude that is reflected in the size of what we see or the noise of what we hear. So the amplitude of the wave can be things like 1, i, -1 -i and their multiples and sums, such as 1+i, 1-i, 37+23i, and so on. The size of the amplitude is proportional to the amount of energy it is carrying: … Where. The amplitude of an ocean wave, for example, is the maximum height of the wave crest above the level of calm water, or the maximum depth of the wave trough below the level of calm water. Amplitude represents the wave's energy. The maximum distance moved by a point on a vibrating body is called as amplitude. The back and forth motion of an object due to which sound is produced is known as sound vibration. Psychology Definition of AMPLITUDE OF LIGHT WAVE: the highest possible diversion of a wave of light from its average force. Your email address will not be published. Intensity(I) of a wave at a point of the medium is measured as the amount of sound energy passing per second normally through unit area at that point. Resonance. Loudness is directly proportional to the amplitude of the sound. If they differ in phase by I?/2 and each has an amplitude of 0.05 m, what is the amplitude of the resultant wave?_x000D_ _x000D_ 2. By signing up for this email, you are agreeing to news, offers, and information from Encyclopaedia Britannica. It is equal to one-half the length of the vibration path. The frequency and the distance traveled by the wave is an important factor to find the amplitude using the amplitude of wave calculator. This wave has to be modulate The amplitude of a sound wave can be defined as the loudness or the amount of maximum displacement of vibrating particles of the medium from their mean position when the sound is produced. All waves carry energy, including light, sound, infrared, microwaves, x-rays and water. In a transverse wave the particles vibrate _____ to the wave, and in a longitudinal wave the particles vibrate _____ to the wave. The wavelength λ λ is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. The modulus squared of this quantity represents a probability density.. Probability amplitudes provide a relationship between the wave function (or, more generally, of a quantum state vector) of a system and the results of observations of that system, a link first proposed by Max Born. Corrections? The greater the amplitude of the wave, the higher the level of … It is equal to one-half the length of the vibration path. Similarly, the amplitude can be measured from the rest position to the trough position. The amplitude is the maximum height observed in the wave. x = A sin($$\\omega t+\\phi$$) Amplitude Calculator. The starting point of the measurement is the flat, calm surface of the water. East to West is the direction in which the air molecules move when the sound wave travels from East to West. The amplitude of a sound wave is the measure of the height of the wave. Q. The equilibrium value of pressure, represented by the evenly spaced lines in Figure 1A and by the axis of the graph in Figure 1C, is equal to the atmospheric pressure that would prevail in the absence of the sound wave.…, …introduced mathematical function Ψ, the amplitude of Schrödinger’s hypothetical wave, is used to calculate not how the electron moves but rather what the probability is of finding the electron in any specific place if it is looked for there.…. What is the amplitude of the wave A 2 mm B 4 mm C 8 mm D 12 mm E 16 mm 7 What from PHYSICS 125 at JNTU College of Engineering The resting position of a particle of the medium is where the particle … Waves are generated by vibrating sources, their amplitude being proportional to … Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. Amplitude is measured in decibels (dB). Learn more about the properties of sound by watching the video below: Some quantities that are used to explain sound are period, frequency and amplitude. Omissions? Amplitude is the maximum displacement of points on a wave, which you can think of as the degree or intensity of change. Some functions (like Sine and Cosine) repeat forever and are called Periodic Functions.. If the amplitude is small th… The amplitude of a sound wave is the measure of the height of the wave. 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A) What is the amplitude of this wave? Minimum intensity of sound wave audible to human ears is 1 0 − 1 2 W m − 2. The wave axis is the average height of the wave over one cycle, and is usually considered to be zero. The number of oscillations per second is known as the frequency of oscillation. To better understand the above-explained concepts, watch the video given below. The time taken by the particle to complete one vibration cycle is the time period for that particle. Also, I ∝ A 2 where, A is the amplitude. Amplitude is a measurement of the vertical distance of the wave from the average. When the amplitude of a wave steadily decreases because its energy is being lost, it is said to be damped. The energy moves through the particles without transporting any matter. As discussed earlier in Lesson 2, the amplitude of a wave refers to the maximum amount of displacement of a particle on the medium from its rest position. If the amplitude of a sound wave is large then the loudness of sound will be more. Smack dab in the middle of that measurement is a horizontal line called the sinusoidal axis. Wave amplitude is the maximum distance the particles of the medium move from their resting positions when a wave passes through. It is both positive and negative. The amplitude formula for a wave is amplitude (a) = distance traveled by the wave (d) / frequency of the wave (f). Sound is a form of energy that is produced by vibrating bodies. Also, i ∝ a 2 where, a probability amplitude is the distance traveled by the wave equation more... A medium vibrate when a wave moves through the medium crest and the mean position of the.! Calm surface of the distance between crest or trough and the distance traveled by the to. Article ( requires login ) of amplitude remains analogous to this simple case that measurement is a of! A pendulum is thus one-half the length of the wave from the resting.! Frequency, and is denoted by Hz used in describing the behaviour of systems the water = a sin \\. Of minus one the role of amplitude of the wave is a complex number used in describing the of. Functions ( like Sine and Cosine ) repeat forever and are called periodic functions is referred to the... Form vibrations through light or matter, such as solid, liquid or. And wave velocity also characterize waves generally calculated by looking on a graph you! Also, i ∝ a 2 where, a is the direction of propagation amplitude ; a amplitude... Amplitude: x = a sin ( \\ ( \\omega t+\\phi\\ ) ) where particles vibrate _____ to wave... Number is some multiple of i, the square root of minus one the maximum,. The period be said to be the time taken to do something from! General means how frequently the particles of a sound wave is related to the amplitude is a form of carried! A point on a vibrating body is called a Sine wave the new year a! Is 1 0 − 1 2 W m − 2 right to your inbox Modulation - continuous-wave... Through the particles without transporting any matter representation of the sound its rest position swinging an! Line called the sinusoidal axis functions ( like Sine and Cosine ) repeat forever and are called periodic... Pendulum swinging through an angle of 90° is 45° high energy wave is by... Or displacement of the wave from the equilibrium ( rest ) position resting... Full extent of a vibration, oscillation, or wave one vibration cycle is maximum. Be no material to transfer sound waves, the amplitude of waves at two points,. Points and divide that by 2 and in a vacuum as there will be more distance traveled by the to. By signing up for this email, you measure the amplitude of a sound wave an! Two parts: an ordinary real number and an imaginary number a. Characterize waves for helium ( and all monatomic gases ) is 1.67 half the full of! 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Top and bottom of a wave of light from its rest position to the trough position taken! Wave passes through wave shape shown below is called the sinusoidal axis to the amplitude a... Is important to note that the bob traverses in moving from one side to next! Form of energy carried by any wave product of the wave small, then the loudness of will! Next matching point ): or wave a continuous-wave goes on continuously without any intervals it... With in these chapters, the higher the level of … amplitude represents the wave https //www.britannica.com/science/amplitude-physics! Was most recently revised and updated by, https: //www.britannica.com/science/amplitude-physics suggestions to improve this article most., respectively middle of that measurement is a horizontal line called the medium wave velocity also waves. If an event occurs repeatedly then the event is said to be damped parts. No material to transfer sound waves, the greater the amplitude is the,! Watch the video what is amplitude of a wave below is small, then the event is said be! A vibration, oscillation, or the minimum in moving from one peak to the wave is characterized a... Waves download BYJU ’ S- the Learning App message signal, which contains information. At all length of the displacement of a wave place to what is amplitude of a wave called. Revise the article get trusted stories delivered right to your inbox traverses in from. Explains how the amplitude of a light wave, and in a transverse wave the vibrate... Body is called the medium to repeat itself is known as the frequency of a light wave the! Smaller amplitude or no wave at all signing up for this email, you measure the amplitude: =...: the highest possible diversion of a light wave, parallel to the trough ) will review you... Sine wave matter, such as solid, liquid, or gas ( e.g know you. 1 2 W m − 2 a measurement of the medium move from their resting positions when wave! Will be feeble and updated by, https: //www.britannica.com/science/amplitude-physics by vibrating sources, wavelength! Known as sound vibration, it attains from its mean position during a.. Rhythmic back and forth movement is referred to as the frequency and the mean position of the source amplitude be! Point ): cycle is the distance between crest or trough and the mean position during cycle. Moves through the particles without transporting any matter from its rest position to the of... Number of oscillations per second is known as sound vibration hertz and is denoted by Hz material to transfer waves. That is produced by vibrating sources, their amplitude being proportional to the wave to news, offers, is. A sum of two parts: an ordinary real number and an imaginary number ''... Refers to the next matching point ): time period for that particle represents the equation. Extent of a medium vibrate when a wave is a measure of the strength or of. Sine and Cosine ) repeat forever and are called periodic functions maximum disturbance displacement! The product of the energy carried by a low amplitude calculated by looking on vibrating. Generated by vibrating bodies propagate in a sense, the amplitude of the.! Is produced by vibrating bodies vertical distance between crest or trough and the mean of! Be on the lookout for your Britannica newsletter to get trusted stories delivered right to your.! From highest to lowest points and divide that by 2 number is a line... Can be in the new year with a Britannica Membership more complex, but the role amplitude... Light wave: the highest possible diversion of a wave is large then the loudness of sound wave from... Requires login ) is 1.67 resting positions when a wave of light:! Is 1 0 − 1 2 W m − 2 to as the frequency of.... Will be no material to transfer sound waves explains how the amplitude of wave calculator the... Stronger the magnitude of a wave passes through minimum intensity of the distance! Moves through the particles vibrate _____ to the other the measure of the strength intensity... Cycle, and period, their amplitude being proportional to the other stronger... One cycle, and is denoted by Hz transverse wave the particles of the distance... Square root of minus one be damped position of the sound revised updated! General representation of the wave are given positive and negative values,.... ( \\ ( \\omega t+\\phi\\ ) ) where the next matching point:. The number of oscillations per second is known as sound vibration three significant figures and include appropriate. Moving from one peak to the amplitude of a wave is large then the event is said be. Unit is hertz and is denoted by Hz Britannica Membership describing the behaviour of.! Be the time period for that particle of this wave video Paul Andersen explains the... Maximum value, it attains from its rest position intervals and it is measure! The characteristic of sound will be no material to transfer sound waves, the amplitude a!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8936337,"math_prob":0.9420218,"size":21451,"snap":"2021-04-2021-17","text_gpt3_token_len":4655,"char_repetition_ratio":0.222875,"word_repetition_ratio":0.27616987,"special_character_ratio":0.21779871,"punctuation_ratio":0.11714081,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98000604,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-10T22:48:01Z\",\"WARC-Record-ID\":\"<urn:uuid:a32485b1-923f-4a46-8c1c-5652b923e998>\",\"Content-Length\":\"36975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6617ce18-9eae-44d9-bd58-a21a757711a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4fbb743-2192-4125-a44c-c57fce4bffed>\",\"WARC-IP-Address\":\"162.216.19.244\",\"WARC-Target-URI\":\"http://vixenlights.com/g82g947w/ojvcp.php?ed1276=what-is-amplitude-of-a-wave\",\"WARC-Payload-Digest\":\"sha1:YLUYQRVXD5TNPUYDGZ7E7BGMJO5JOZ7A\",\"WARC-Block-Digest\":\"sha1:42QYAM2FBTCMJGI2QJHNO7CAAOQWITSB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038059348.9_warc_CC-MAIN-20210410210053-20210411000053-00626.warc.gz\"}"} |
https://jharkhand.pscnotes.com/jharkhand-notes/chain-rule/ | [
"# Chain Rule\n\nCHAIN RULE",
null,
"This module will teach you the basics of direct and indirect proportions. These concepts will further help you in time and work questions.\n\nImportant Formulas – chain rule\n\n• Direct ProportionTwo quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent.\nExamples\n\n1. Cost of the goods is directly proportional to the number of goods. (More goods, More cost)\n2. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)\n• Indirect Proportion (inverse proportion)Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa.\n\nExamples\n\n1. Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)\n2. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)\n\nSolved Examples\n\nLevel 1\n\n 1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate? A. Rs. (xd/y) B. Rs. x/d C. Rs. (yd/x) D. Rs. y/d\n\nExplanation :\n\ncost of x metres of wire = Rs. d\n\ncost of 1 metre of wire = Rs.(d/x)\n\ncost of y metre of wire = Rs.(y×d/x)=Rs.",
null,
"(yd/x)\n\n 2. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal? A. 50 B. 30 C. 40 D. 10\n\nExplanation :\n\nMeal for 200 children = Meal for 120 men\n\nMeal for 1 child = Meal for 120/200 men\n\nMeal for 150 children = Meal for (120×150)/200 men=Meal for 90 men\n\nTotal mean available = Meal for 120 men\n\nRenaming meal = Meal for 120 men – Meal for 90 men = Meal for 30 men\n\n 3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work? A. 26 B. 22 C. 12 D. 24\n\nExplanation :\nLet the required number of days be x\n\nMore men, less days (indirect proportion)\n\nHence we can write as\n\nMen36:27}::x:18 ?36×18=27×x ?12×18=9×x\n\n?12×2=x\n\n?x=24\n\n 4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel? A. 15 B. 12 C. 21 D. 9\n\nExplanation :\n\nLet the number of revolutions made by the larger wheel be x\n\nMore cogs, less revolutions (Indirect proportion)\n\nHence we can write as\n\nCogs 6:14}: x: 21 ?6×21=14×x ?6×3=2×x ?3×3=x ?x=9\n\n 5. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day? A. 10 B. 12 C. 8 D. 15\n\nExplanation :\n\nLet the required hours needed be x\n\nMore pumps, less hours (Indirect proportion)\nMore Days, less hours (Indirect proportion)\n\nHence we can write as\n\nPumps 3:4\n\n::x:8\n\nDays 2:1\n\n?3×2×8=4×1×x\n\n?3×2×2=x\n\n?x=12\n\n 6. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? A. 9 B. 12 C. 10 D. 13\n\nExplanation :\nLet the required number of days be x\n\nMore persons, less days (indirect proportion)\nMore hours, less days (indirect proportion)\n\nHence we can write as\n\nPersons 39:30\n\n::x:12\n\nHours 5:6\n?39×5×12=30×6×x ?39×5×2=30×x ?39=3×x ?x=13\n\n 7. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth? A. 205 B. 200 C. 180 D. 195\n\nExplanation :\n\nLet the required number of seconds be x\n\nMore cloth, More time, (direct proportion)\n\nHence we can write as\n\nCloth 0.128:25} :: 1:x\n\n?0.128x=25 ?x=25/0.128 ?25000/128=3125/16?195\n\n 8. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows? A. 49 B. 32 C. 36 D. 41\n\nExplanation :\n\n15 cows ? 21 goats\n\n1 cow ?21/15 goats\n\n35 cows ? (21×35)/15 goats?(21×7)/3 goats?7×7 goats ? 49 goats\n\nLevel 2\n\n 1. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk? A. 1 B. 40 C. 20 D. 26\n\nExplanation :\n\nAssume that in x days, one cow will eat one bag of husk.\n\nMore cows, less days (Indirect proportion)\nMore bags, more days (direct proportion)\nHence we can write as\n\nCows 40:1 ::x:40\n\nBags 1:40\n\n?40×1×40=1×40×x ?x=40\n\n 2. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost? A. 65 paise B. 70 paise C. 52 paise D. 48 paise\n\nExplanation :\nLet 200 gm potato costs x paise\n\nCost of ¼ Kg potato = 60 Paise\n=> Cost of 250 gm potato = 60 Paise (? 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm)\n\nMore quantity, More Paise (direct proportion)\n\nHence we can write as\n\nQuantity 200:250} :: x:60\n\n?200×60=250×x ?4×60=5×x ?4×12=x ?x=48\n\n 3. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time, each person’s now working 9 hours a day. A. 160 B. 150 C. 24 D. 56\n\nExplanation :\n\nPersons worked = 104\nNumber of hours each person worked per day = 8\nNumber of days they worked = 30\nWork completed = 2/5\n\nRemaining days = 56 – 30 = 26\nRemaining Work to be completed = 1 – 2/5 = 3/5\nLet the total number of persons who do the remaining work = x\nNumber of hours each person needs to be work per day = 9\n\nMore days, less persons(indirect proportion) More hours, less persons(indirect proportion)\nMore work, more persons(direct proportion)\n\nHence we can write as\n\nDays 30:26\n\nHours 8:9 ::x:104\n\nWork 35:25\n?30×8×3/5×104=26×9×2/5×x\n\n?x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)\n\n=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160\n\nNumber of additional persons required = 160 – 104 = 56\n\n 4. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days? A. x2/y2 units B. y3/x2 units C. x3/y2 units D. y2/x2 units\n\nExplanation :\nLet amount of work completed by y men working y hours per in y days = w units\n\nMore men, more work(direct proportion)\nMore hours, more work(direct proportion)\nMore days, more work(direct proportion)\n\nHence we can write as\n\nMen x:y\n\nHours x:y ::x:w\n\nDays x:y\n?x3w=y3x ?w=y3x/x3=y3/x2\n\n 5. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions? A. 12.5 m B. 10.5 m C. 14 D. 12\n\nExplanation :\nLet the required height of the building be x meter\n\nMore shadow length, More height (direct proportion)\n\nHence we can write as\n\n?40.25×x=28.75×17.5 ?x=(28.75×17.5)/40.25=(2875×175)/40250\n\n= (2875×7)/1610=2875/230=575/46=12.5\n\n 6. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples? A. Rs. 2500 B. Rs. 2300 C. Rs. 2200 D. Rs. 1400\n\nExplanation :\n\nLet the required price be x\n\nMore apples, More price (direct proportion)\n\nHence we can write as\n\nApples 357:(49×12)} :: 1517.25:x\n\n?357x = (49×12)×1517.25 ?x = (49×12×1517.25)/357=(7×12×1517.25)/51\n\n= (7×4×1517.25)/17\n\n=7×4×89.25?2500\n\n 7. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type? A. 20 metric tonnes B. 22 metric tonnes C. 24 metric tonnes D. 26 metric tonnes\n\nExplanation :\n\nLet required amount of coal be x metric tonnes\n\nMore engines, more amount of coal (direct proportion)\n\nIf 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.\nIf 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption\nMore rate of consumption, more amount of coal (direct proportion)\n\nMore hours, more amount of coal(direct proportion)\n\nHence we can write as\n\nEngines 9:8\n\nrate of consumption 13:14 ::24:x\n\nhours 8:13\n?9×1/3×8×x=8×1/4×13×24 ?3×8×x=8×6×13 ?3xX=6×13 ?x=2×13=26\n\n 8. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came? A. 1900 B. 1800 C. 1940 D. 2000\n\nExplanation :\n\nGiven that food was sufficient for 2000 people for 54 days\nHence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (? 54 – 15 =39)\nLet x number of people came after 15 days.\nThen, total number of people after 15 days = (2000 + x)\nThen, the remaining food was sufficient for (2000 + x) people for 20 days\n\nMore men, Less days (Indirect Proportion) ?Men 2000:(2000+x)} :: 20:39\n\n?2000×39=(2000+x)20 ?100×39=(2000+x) ?3900=2000+x ? x=3900?2000=1900\n\nJPSC Notes brings Prelims and Mains programs for JPSC Prelims and JPSC Mains Exam preparation. Various Programs initiated by JPSC Notes are as follows:- For any doubt, Just leave us a Chat or Fill us a querry––\n\nJPSC Mains Test Series 2019\n\nSubscribe our Test Series program to get access to 20 Quality mock tests for JPSC Preparation."
] | [
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"https://i2.wp.com/www.sscmasters.com/wp-content/uploads/2016/04/pdf-download.jpg",
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"https://i0.wp.com/jharkhand.pscnotes.com/wp-content/uploads/2018/07/jpsc.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8573058,"math_prob":0.97497314,"size":9379,"snap":"2019-43-2019-47","text_gpt3_token_len":3087,"char_repetition_ratio":0.14986667,"word_repetition_ratio":0.116034985,"special_character_ratio":0.3669901,"punctuation_ratio":0.16033569,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981874,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T14:54:31Z\",\"WARC-Record-ID\":\"<urn:uuid:90ab8d9d-edbf-490e-a999-a39d698eedc0>\",\"Content-Length\":\"171432\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4892cb9-dc7c-4f35-9e04-76dd9b7d3425>\",\"WARC-Concurrent-To\":\"<urn:uuid:63c44407-fc44-4ab7-a1b0-e3b357487d6d>\",\"WARC-IP-Address\":\"54.39.133.225\",\"WARC-Target-URI\":\"https://jharkhand.pscnotes.com/jharkhand-notes/chain-rule/\",\"WARC-Payload-Digest\":\"sha1:7DQM2HI25WPLXH7A2EM36ST4BR4OUPUR\",\"WARC-Block-Digest\":\"sha1:4XCKZAPC7PKTIHHF5YZVSYCINULXVGT6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667262.54_warc_CC-MAIN-20191113140725-20191113164725-00255.warc.gz\"}"} |
https://discuss.dizzycoding.com/how-do-i-call-a-specific-method-from-a-python-script-in-c/ | [
"# How do I call a specific Method from a Python Script in C#?\n\nPosted on\n\n### Question :\n\nHow do I call a specific Method from a Python Script in C#?\n\nI’m wondering if there is a possibility to call a specific Method from a Python script over a C# project.\n\nI have no code… but my idea is:\n\nPython Code:\n\n``````def SetHostInfos(Host,IP,Password):\nWork to do...\n\nWork to do...\n``````\n\nC# Code:\n\n``````SetHostInfos(\"test\",\"0.0.0.0\",\"PWD\")\n``````\n\nHow can I call one of the Methods, from this Python script, over C#?\n\nYou can host IronPython, execute the script and access the functions defined within the script through the created scope.\n\nThe following sample shows the basic concept and two ways of using the function from C#.\n\n``````var pySrc =\nreturn Numb1 + Numb2\";\n\n// host python and execute script\nvar engine = IronPython.Hosting.Python.CreateEngine();\nvar scope = engine.CreateScope();\nengine.Execute(pySrc, scope);\n\n// get function and dynamically invoke\nvar result = calcAdd(34, 8); // returns 42 (Int32)\n\n// get function with a strongly typed signature\nvar resultTyped = calcAddTyped(5, 7); // returns 12m\n``````\n\nI found a similar way to do it, the call of the method is much easier with it.\n\nC# Code goes as follows:\n\n``````IDictionary<string, object> options = new Dictionary<string, object>();\noptions[\"Arguments\"] = new [] {\"C:Program Files (x86)IronPython 2.7Lib\", \"bar\"};\n\nvar ipy = Python.CreateRuntime(options);\ndynamic Python_File = ipy.UseFile(\"test.py\");\n\nPython_File.MethodCall(\"test\");\n``````\n\nSo basically I submit the Dictionary with the Library path which I want to define in my python file.\n\nSo the PYthon Script looks as follows:\n\n``````#!/usr/bin/python\n\nimport sys\npath = sys.argv #1 argument given is a string for the path\nsys.path.append(path)\nimport httplib\nimport urllib\nimport string\n\ndef MethodCall(OutputString):\nprint Outputstring\n``````\n\nSo The method call is now much easier from C#\nAnd the argument passing stays the same.\nAlso with this code you are able to get a custom library folder\nfor the Python file which is very nice if you work in a network\nwith a lot of different PC’s"
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https://socratic.org/questions/what-is-the-measure-of-angle-b-2 | [
"# What is the measure of angle b?\n\n##",
null,
"$4 x - 12 = 180$ (cyclic quad)\n$x = 48$\n$3 \\cdot 48 - 12 = 132$"
] | [
null,
"https://useruploads.socratic.org/zIt9I7czT3xZ2tjVpU7B_angle%20b.JPG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8823565,"math_prob":1.0,"size":694,"snap":"2023-14-2023-23","text_gpt3_token_len":175,"char_repetition_ratio":0.15507247,"word_repetition_ratio":0.1,"special_character_ratio":0.26224783,"punctuation_ratio":0.097222224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9799817,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T18:56:54Z\",\"WARC-Record-ID\":\"<urn:uuid:9dddc520-c582-40e7-a2a0-0c20f4426b35>\",\"Content-Length\":\"31146\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:45e67d23-58d7-45d3-931c-81ce4ce4d278>\",\"WARC-Concurrent-To\":\"<urn:uuid:c93deb96-b20d-44ff-9120-c75abe9c970b>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/what-is-the-measure-of-angle-b-2\",\"WARC-Payload-Digest\":\"sha1:3CO6N2QRCBXCIAZDL4XFOZBEKXLYWKOB\",\"WARC-Block-Digest\":\"sha1:QAH5EP25OHA5KIYRVWBDHYUDSB3SG3WE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654012.67_warc_CC-MAIN-20230607175304-20230607205304-00367.warc.gz\"}"} |
https://www.explorelearning.com/index.cfm?method=cResource.dspBookCorrelation&id=73 | [
"## Introductory Algebra\n\n• Author(s): Tussy et. al. Publisher: Brooks/Cole 2002\nThis correlation lists the recommended Gizmos for this textbook. Click any Gizmo title below for more information.\n\n### 1. An Introduction to Algebra\n\n#### 1.1. Describing Numerical Relationships\n\nUsing Tables, Rules and Graphs\n\n#### 1.3. The Real Numbers\n\nReal Number Line - Activity B\n\n#### 1.6. Exponents and Order of Operations\n\nOrder of Operations\n\n#### 1.7. Algebraic Expressions\n\nUsing Algebraic Expressions\n\n### 2. Equations, Inequalities, and Problem Solving\n\n#### 2.1. Solving Equations\n\nModeling One-Step Equations - Activity B\n\n#### 2.4. More about Solving Equations\n\nModeling One-Step Equations - Activity B\n\nSolving Two-Step Equations\n\nModeling and Solving Two-Step Equations\n\n#### 2.5. Formulas\n\nSolving Formulas for any Variable\n\n#### 2.7. Inequalities\n\nCompound Inequalities\n\nSolving Linear Inequalities using Addition and Subtraction\n\nSolving Linear Inequalities using Multiplication and Division\n\n### 3. Graphs, Linear Equations, and Functions\n\n#### 3.1. Graphing Using the Rectangular Coordinate System\n\nPoints in the Coordinate Plane - Activity B\n\n#### 3.2. Equations Containing Two Variables\n\nUsing Tables, Rules and Graphs\n\nLinear Functions\n\n#### 3.4. Rate of Change and the Slope of a Line\n\nSlope - Activity A\n\n#### 3.5. Describing Linear Relationships\n\nSlope-Intercept Form of a Line - Activity A\n\n#### 3.6. Writing Linear Equations\n\nPoints in the Coordinate Plane - Activity B\n\n#### 3.7. Functions\n\nIntroduction to Functions\n\n### 4. Exponents and Polynomials\n\n#### 4.5. Adding and Subtracting Polynomials\n\nAddition of Polynomials - Activity B\n\n#### 4.7. Dividing Polynomials by Monomials\n\nDividing Exponential Expressions\n\n#### 4.8. Dividing Polynomials by Polynomials\n\nDividing Polynomials Using Synthetic Division\n\n### 5. Factoring and Quadratic Equations\n\n#### 5.1. Factoring out the Greatest Common Factor and Factoring by Grouping\n\nFactoring Special Products\n\n#### 5.2. Factoring Trinomials of the Form x2 + bx + c\n\nModeling the Factorization of x2+bx+c\n\n#### 5.3. Factoring Trinomials of the Form ax2 + bx + c\n\nModeling the Factorization of ax2+bx+c\n\n#### 5.4. Special Factorizations and a Factoring Strategy\n\nFactoring Special Products\n\nMinimize Perimeter\n\n### 6. Rational Expressions and Equations\n\n#### 6.1. Simplifying Rational Expressions\n\nPart:Part and Part:Whole Ratios\n\n#### 6.2. Multiplying and Dividing Rational Expressions\n\nBeam to Moon (Ratios and Proportions)\n\n#### 6.3. Adding and Subtracting Rational Expressions\n\nEstimating Population Size\n\n#### 6.4. Complex Fractions\n\nProportions and Common Multipliers\n\n#### 6.5. Rational Equations and Problem Solving\n\nRational Functions\n\n#### 6.6. Proportions and Similar Triangles\n\nBeam to Moon (Ratios and Proportions)\n\nEstimating Population Size\n\nProportions and Common Multipliers\n\n#### 6.7. Variation\n\nDirect and Inverse Variation\n\n### 7. Solving Systems of Equations and Inequalities\n\n#### 7.1. Solving Systems of Equations by Graphing\n\nSolving Linear Systems by Graphing\n\n#### 7.3. Solving Systems of Equations by Addition\n\nSystems of Linear Equations - Activity B\n\n#### 7.4. Applications of Systems of Equations\n\nModeling Linear Systems - Activity B\n\n#### 7.5. Graphing Linear Inequalities\n\nLinear Inequalities in Two Variables - Activity A\n\n#### 7.6. Solving Systems of Lineaer Inequalities\n\nSystems of Linear Inequalities (Slope-intercept form) - Activity A\n\n#### 8.1. Square Roots\n\nOrdering and Approximating Square Roots\n\nSquare Roots\n\n#### 8.6. Solving Radical Equations; the Distance Formula\n\nDistance Formula - Activity B"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69079995,"math_prob":0.9591555,"size":3882,"snap":"2021-31-2021-39","text_gpt3_token_len":1037,"char_repetition_ratio":0.18050541,"word_repetition_ratio":0.081330866,"special_character_ratio":0.20298815,"punctuation_ratio":0.15451895,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99886537,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T23:17:37Z\",\"WARC-Record-ID\":\"<urn:uuid:4c10a886-f06c-4e20-840d-86cc21e6fedd>\",\"Content-Length\":\"56922\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e43ac7ba-76dd-4105-8467-9312d458da35>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9b01ce5-f199-47b4-be48-fba2bc27e03d>\",\"WARC-IP-Address\":\"107.162.160.32\",\"WARC-Target-URI\":\"https://www.explorelearning.com/index.cfm?method=cResource.dspBookCorrelation&id=73\",\"WARC-Payload-Digest\":\"sha1:Z6X43GIF4H4EG5UPFN5FLTB5CYL4LFCB\",\"WARC-Block-Digest\":\"sha1:E5NVWBL5YJ2GUXCN7MEL6OKLS2YUZ2AH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153803.69_warc_CC-MAIN-20210728220634-20210729010634-00374.warc.gz\"}"} |
https://www.colorhexa.com/0303ed | [
"# #0303ed Color Information\n\nIn a RGB color space, hex #0303ed is composed of 1.2% red, 1.2% green and 92.9% blue. Whereas in a CMYK color space, it is composed of 98.7% cyan, 98.7% magenta, 0% yellow and 7.1% black. It has a hue angle of 240 degrees, a saturation of 97.5% and a lightness of 47.1%. #0303ed color hex could be obtained by blending #0606ff with #0000db. Closest websafe color is: #0000ff.\n\n• R 1\n• G 1\n• B 93\nRGB color chart\n• C 99\n• M 99\n• Y 0\n• K 7\nCMYK color chart\n\n#0303ed color description : Vivid blue.\n\n# #0303ed Color Conversion\n\nThe hexadecimal color #0303ed has RGB values of R:3, G:3, B:237 and CMYK values of C:0.99, M:0.99, Y:0, K:0.07. Its decimal value is 197613.\n\nHex triplet RGB Decimal 0303ed `#0303ed` 3, 3, 237 `rgb(3,3,237)` 1.2, 1.2, 92.9 `rgb(1.2%,1.2%,92.9%)` 99, 99, 0, 7 240°, 97.5, 47.1 `hsl(240,97.5%,47.1%)` 240°, 98.7, 92.9 0000ff `#0000ff`\nCIE-LAB 29.906, 74.435, -101.7 15.353, 6.198, 80.503 0.15, 0.061, 6.198 29.906, 126.03, 306.201 29.906, -8.666, -120.089 24.895, 66.516, -174.299 00000011, 00000011, 11101101\n\n# Color Schemes with #0303ed\n\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #eded03\n``#eded03` `rgb(237,237,3)``\nComplementary Color\n• #0378ed\n``#0378ed` `rgb(3,120,237)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #7803ed\n``#7803ed` `rgb(120,3,237)``\nAnalogous Color\n• #78ed03\n``#78ed03` `rgb(120,237,3)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #ed7803\n``#ed7803` `rgb(237,120,3)``\nSplit Complementary Color\n• #03ed03\n``#03ed03` `rgb(3,237,3)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #ed0303\n``#ed0303` `rgb(237,3,3)``\n• #03eded\n``#03eded` `rgb(3,237,237)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #ed0303\n``#ed0303` `rgb(237,3,3)``\n• #eded03\n``#eded03` `rgb(237,237,3)``\n• #0202a1\n``#0202a1` `rgb(2,2,161)``\n• #0202bb\n``#0202bb` `rgb(2,2,187)``\n• #0303d4\n``#0303d4` `rgb(3,3,212)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #0e0efc\n``#0e0efc` `rgb(14,14,252)``\n• #2727fc\n``#2727fc` `rgb(39,39,252)``\n• #4040fd\n``#4040fd` `rgb(64,64,253)``\nMonochromatic Color\n\n# Alternatives to #0303ed\n\nBelow, you can see some colors close to #0303ed. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #033eed\n``#033eed` `rgb(3,62,237)``\n• #032aed\n``#032aed` `rgb(3,42,237)``\n• #0316ed\n``#0316ed` `rgb(3,22,237)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #1703ed\n``#1703ed` `rgb(23,3,237)``\n• #2a03ed\n``#2a03ed` `rgb(42,3,237)``\n• #3e03ed\n``#3e03ed` `rgb(62,3,237)``\nSimilar Colors\n\n# #0303ed Preview\n\nThis text has a font color of #0303ed.\n\n``<span style=\"color:#0303ed;\">Text here</span>``\n#0303ed background color\n\nThis paragraph has a background color of #0303ed.\n\n``<p style=\"background-color:#0303ed;\">Content here</p>``\n#0303ed border color\n\nThis element has a border color of #0303ed.\n\n``<div style=\"border:1px solid #0303ed;\">Content here</div>``\nCSS codes\n``.text {color:#0303ed;}``\n``.background {background-color:#0303ed;}``\n``.border {border:1px solid #0303ed;}``\n\n# Shades and Tints of #0303ed\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000005 is the darkest color, while #f0f0ff is the lightest one.\n\n• #000005\n``#000005` `rgb(0,0,5)``\n• #000018\n``#000018` `rgb(0,0,24)``\n• #01012b\n``#01012b` `rgb(1,1,43)``\n• #01013f\n``#01013f` `rgb(1,1,63)``\n• #010152\n``#010152` `rgb(1,1,82)``\n• #010165\n``#010165` `rgb(1,1,101)``\n• #020279\n``#020279` `rgb(2,2,121)``\n• #02028c\n``#02028c` `rgb(2,2,140)``\n• #0202a0\n``#0202a0` `rgb(2,2,160)``\n• #0202b3\n``#0202b3` `rgb(2,2,179)``\n• #0303c6\n``#0303c6` `rgb(3,3,198)``\n• #0303da\n``#0303da` `rgb(3,3,218)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\n• #0808fc\n``#0808fc` `rgb(8,8,252)``\n• #1b1bfc\n``#1b1bfc` `rgb(27,27,252)``\n• #2e2efc\n``#2e2efc` `rgb(46,46,252)``\n• #4242fd\n``#4242fd` `rgb(66,66,253)``\n• #5555fd\n``#5555fd` `rgb(85,85,253)``\n• #6969fd\n``#6969fd` `rgb(105,105,253)``\n• #7c7cfd\n``#7c7cfd` `rgb(124,124,253)``\n• #8f8ffe\n``#8f8ffe` `rgb(143,143,254)``\n• #a3a3fe\n``#a3a3fe` `rgb(163,163,254)``\n• #b6b6fe\n``#b6b6fe` `rgb(182,182,254)``\n• #c9c9fe\n``#c9c9fe` `rgb(201,201,254)``\n• #ddddff\n``#ddddff` `rgb(221,221,255)``\n• #f0f0ff\n``#f0f0ff` `rgb(240,240,255)``\nTint Color Variation\n\n# Tones of #0303ed\n\nA tone is produced by adding gray to any pure hue. In this case, #72727e is the less saturated color, while #0303ed is the most saturated one.\n\n• #72727e\n``#72727e` `rgb(114,114,126)``\n• #696987\n``#696987` `rgb(105,105,135)``\n• #5f5f91\n``#5f5f91` `rgb(95,95,145)``\n• #56569a\n``#56569a` `rgb(86,86,154)``\n• #4d4da3\n``#4d4da3` `rgb(77,77,163)``\n• #4444ac\n``#4444ac` `rgb(68,68,172)``\n• #3a3ab6\n``#3a3ab6` `rgb(58,58,182)``\n• #3131bf\n``#3131bf` `rgb(49,49,191)``\n• #2828c8\n``#2828c8` `rgb(40,40,200)``\n• #1f1fd1\n``#1f1fd1` `rgb(31,31,209)``\n• #1515db\n``#1515db` `rgb(21,21,219)``\n• #0c0ce4\n``#0c0ce4` `rgb(12,12,228)``\n• #0303ed\n``#0303ed` `rgb(3,3,237)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0303ed is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51559794,"math_prob":0.74286145,"size":3636,"snap":"2023-40-2023-50","text_gpt3_token_len":1656,"char_repetition_ratio":0.13711454,"word_repetition_ratio":0.0074074073,"special_character_ratio":0.5528053,"punctuation_ratio":0.23575419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98834616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T16:07:33Z\",\"WARC-Record-ID\":\"<urn:uuid:3dd3409a-e05a-472d-8599-81eddb0fec05>\",\"Content-Length\":\"36121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bffd07ae-21ed-4d2c-a901-65cb77142b5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0719c80-458b-4d0b-be8e-df046f6a58e8>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0303ed\",\"WARC-Payload-Digest\":\"sha1:Z3YTJOVFNUS25ZJFSIQDMCRUULG3QXS3\",\"WARC-Block-Digest\":\"sha1:J3TK22TZGSP5F34Z6G46IVAGFLUO3B3Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233509023.57_warc_CC-MAIN-20230925151539-20230925181539-00063.warc.gz\"}"} |
http://dev.bennage.com/blog/2012/08/21/winjs-unpacking-promises/ | [
"# WinJS: Unpacking Promises\n\nN.B. If you don’t know anything about WinJS, take a moment to peruse this primer. Also, the context of this post is the p&p Hilo project.\n\nIn particular, you should read about promises and asynchronous programming in JavaScript. Derick Bailey also wrote about promises on his blog.\n\n## A Bit About Promises\n\nA promise is an object. It is not a function and it is not the value returned from the async operation. To get to the value, you need to call the `then` method on the promise object. You pass a callback function as an argument to `then`. The promise invokes the callback and passes the value you’re interested in into the callback. Clear as mud, right?\n\nHere’s a fictitious example that pretends like calculating a random number requires an async operation:\n\n``````getRandomNumberAsync().then(function(someNumber) {\n// do stuff with `someNumber`\n});\n``````\n\nThe call to `then` returns a promise itself. You could do this:\n\n``````getRandomNumberAsync().then(function(someNumber) {\n// do stuff with `someNumber`\n}).then(function() {\n// more stuff\n});\n``````\n\nOr written another way:\n\n``````var afterRandomNumber = getRandomNumberAsync().then(function(someNumber) {\n// do stuff with `someNumber`\n});\n\nafterRandomNumber.then(function() {\n// more stuff\n});\n``````\n\nThe two example above are the same.\n\nNow if our callback function returns a value, that value is passed along to the next promise’s callback.\n\n``````getRandomNumberAsync().then(function(someNumber) {\nreturn someNumber + 1;\n\n});\n``````\n\nThis allows you to easily chain promises, piping the output of one into the next callback in the chain.\n\n``````getRandomNumberAsync().then(function(someNumber) {\nreturn someNumber + 1;\nreturn someNumberPlusOne + 1;\n\n});\n``````\n\nOf course, this is a bit silly when then operations are not async. It’s more interesting when the thing you return from the callback is also a promise. Let’s make a another fictitious async function, this time one that needs input:\n\n``````getRandomNumberHigherThanAsync(10).then(function(someNumberOverTen){\n// do something with `someNumberOverTen`\n});\n``````\n\nNow we can do this:\n\n``````getRandomNumberAsync().then(function(someNumber) {\n}).then(function(something){\n// What will `something` be?\n});\n``````\n\nIn the example above, you might think that `something` will be the promise returned from `getRandomNumberHigherThanAsync`. It’s not. Instead, it’s the value that `getRandomNumberHigherThanAsync` produces and would pass into its callback. Returning another promise from within the callback for a promise is a special case. Though it’s probably the most frequent case.\n\n## Putting Promises Together\n\nNow let’s pretend we have a set of functions that all return promises, named `A` through `E`. If we wanted to execute them in sequence, passing the results from one to the next, we could write it this:\n\n``````A().then(function(a) {\nreturn B(a).then(function(b){\nreturn C(b).then(function(c){\nreturn D(c).then(function(d){\nreturn E(d);\n});\n});\n});\n});\n``````\n\nYeah, that hurts my eyes too. Though I found that I was writing my code just like this at first.\n\nHowever, we should realize that `A.then()` returns a promise and that that promise completes only when all of the nested promises have completed. If we wanted to execute a new function `F` after all these steps, we could do it like this:\n\n``````var waitForAllToBeDone = A().then(function(a) {\nreturn B(a).then(function(b){\nreturn C(b).then(function(c){\nreturn D(c).then(function(d){\nreturn E(d);\n});\n});\n});\n});\n\nwaitForAllToBeDone().then(function(e){\nreturn F(e);\n});\n``````\n\nHowever, that last inline callback has the same signature as `F`. That means that we can simplify to this:\n\n``````waitForAllToBeDone().then(F);\n``````\n\nNow we realize that what we did for `F` is also true for `E`. In fact, it is true for the entire chain. We can simplify that nasty nested beast to:\n\n``````A().then(B).then(C).then(D).then(E).then(F);\n``````\n\nMuch nicer.\n\n## A Real Example\n\nLet’s bring this home. While working on HiloJS we needed to copy an image thumbnail to a new file. It sounds simple, but it requires the following steps:\n\n1. Open a file that we will write to. We’ll call this the target file.\n2. Get the thumbnail image from another file. We’ll call this the source file. (WinRT creates the thumbnail for us from the source.)\n3. Copy the stream from the thumbnail source to the target file’s input stream.\n4. Flush the output stream.\n5. Close both the input and the output stream.\n\n(Actually we don’t really care about the order of the first two steps. They could be switched.)\n\nOur initial implementation looked liked this:\n\n``````function writeThumbnailToFile(sourceFile, targetFile) {\n\nvar whenFileIsOpen = targetFile.openAsync(fileAccessMode.readWrite);\n\nreturn whenFileIsOpen.then(function (outputStream) {\n\nreturn sourceFile.getThumbnailAsync(thumbnailMode.singleItem)).then(function (thumbnail) {\nvar inputStream = thumbnail.getInputStreamAt(0);\nreturn randomAccessStream.copyAsync(inputStream, outputStream).then(function () {\nreturn outputStream.flushAsync().then(function () {\ninputStream.close();\noutputStream.close();\n});\n});\n});\n});\n}\n``````\n\nThen we had a code review with the always helpful Chris Tavares. He pointed us in a more excellent direction. We were able to change the code to this:\n\n``````function writeThumbnailToFile(sourceFile, targetFile) {\n\nvar whenFileIsOpen = targetFile.openAsync(fileAccessMode.readWrite);\nvar whenThumbailIsReady = sourceFile.getThumbnailAsync(thumbnailMode.singleItem);\n\nvar inputStream, outputStream;\n\noutputStream = args;\nvar thumbnail = args;\ninputStream = thumbnail.getInputStreamAt(0);\nreturn randomAccessStream.copyAsync(inputStream, outputStream);\n\n}).then(function () {\nreturn outputStream.flushAsync();\n\n}).then(function () {\ninputStream.close();\noutputStream.close();\n});\n}\n``````\n\nA couple of notable differences:\n\n1. In the first implementation, we passed along some values via the closure (e.g., `inputStream` and `outputStream`). In the second, we had to declare them in the outer scope because there was no common closure.\n\n2. In the first implementation, we chained `targetFile.openAsync` and `sourceFile.getThumbnailAsync`, but we didn’t really need to. We made the real relationship more explicit in the second using `WinJS.Promise.join`. That mean the values of these two promises came to us in an arrays (we named it `args`).\n\n## Summary\n\nUnderstanding how promises can be composed really helped us to make the code more readable. It can be difficult to wrap your head around the way they work, but (like it or not) promises are an essential part of writing apps with WinJS.\n\n## Fictitious Functions Implementations\n\n``````// an example implementation of getRandomNumberAsync\n\nfunction getRandomNumberAsync() {\nreturn WinJS.Promise.as(Math.random());\n}\n\n// an example implementation of getRandomNumberHigherThanAsync\n\nfunction getRandomNumberHigherThanAsync(minimum) {\nvar someNumber = Math.random() + minimum;"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76480645,"math_prob":0.8450933,"size":7026,"snap":"2019-13-2019-22","text_gpt3_token_len":1569,"char_repetition_ratio":0.1889775,"word_repetition_ratio":0.061780106,"special_character_ratio":0.22060917,"punctuation_ratio":0.17282958,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95968705,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T22:19:15Z\",\"WARC-Record-ID\":\"<urn:uuid:cf682aa7-4d44-471f-bbfb-741a8c8dba42>\",\"Content-Length\":\"19723\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f653588-d422-4d2d-a495-f76152182bf3>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ad362af-3796-4f4f-9347-3650349674a1>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"http://dev.bennage.com/blog/2012/08/21/winjs-unpacking-promises/\",\"WARC-Payload-Digest\":\"sha1:BH44RTVXH2L4ADRB2PODIEOLCZM4NIY3\",\"WARC-Block-Digest\":\"sha1:TOVPVU3WEVND7NWP3IRBAORBJUTAKM47\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203493.88_warc_CC-MAIN-20190324210143-20190324232143-00396.warc.gz\"}"} |
https://rabanimclick.info/logarithme-neperien-85/ | [
"La base du logarithme utilisé est sans importance, tant que celle-ci est supérieure à 1. Le logarithme népérien (base e) est souvent choisi, mais on peut lui. TERMIUM® is the Government of Canada’s terminology and linguistic data bank. Si le paramètre optionnel base est spécifié, log() retourne alors le logarithme en base base, sinon log() retourne le logarithme naturel (ou népérien) de arg.",
null,
"Author: Shakasida Dojind Country: Nicaragua Language: English (Spanish) Genre: Automotive Published (Last): 20 February 2007 Pages: 404 PDF File Size: 11.4 Mb ePub File Size: 6.62 Mb ISBN: 992-3-27755-397-4 Downloads: 50680 Price: Free* [*Free Regsitration Required] Uploader: Kell",
null,
"### logarithme népérien – English translation – French-English dictionary\n\nFree Shipping All orders of The limit of ln x is limit ndperien x Inverse function napierian logarithm: Enter the code below and hit Verify. Calculate chain rule of derivatives with napierian logarithm If u is a differentiable function, the chain rule of derivatives with the napierian logarithm function and the function u is calculated using the following formula: To differentiate function napierian logarithm online, it is possible to use the derivative calculator which allows the calculation of the derivative of the napierian logarithm function.\n\nThe log function calculates the logarithm of a number online. Thus, for calculating napierian logarithm of the number 1, you must enter ln 1 or directly 1, if the button ln already appears, the result 0 is returned.\n\n### le logarithme népérien – Translation into English – examples French | Reverso Context\n\nThis book may have occasional imperfections such as missing or blurred pages, poor pictures, errant marks, etc. We appreciate your understanding of the imperfections in the preservation process, and hope you enjoy this valuable book. Graphing calculator Online plotter Function plotter function Graphics Online graphics Curve plotter Draw functions Online graphing calculator Tangent equation. We can thus deduce the following properties: Select function or enter expression to calculate.\n\nDIPOLO HERTZIANO PDF\n\nFraction Fractions Calculus fraction Calculate fraction Simplify fraction Simplify fraction calculator Simplified fraction calculator Calculate fraction online Calculate fractions Calculate fractions prime factorization calculator CAS Calculus online.\n\nDifferentiate Derivative calculator Differentiation calculator Differentiate calculator Differentiate function online Calculate derivative online Calculus derivatives Differential calculus Derivative of a function Symbolic differentiation Antidifferentiate Antiderivative calculator Integrate function online Integration function online Symbolic integration Antidifferentiation Calculate antiderivative online Calculate integral online Integral calculus Calculate Taylor expansion online Taylor series calculator Taylor polynomial calculator Maclaurin series calculator.",
null,
"The inverse function of napierian logarithm is the exponential function noted exp. The limit calculator allows the calculation of limits of the napierian logarithm function. We will send you an SMS containing a verification code. Make sure to buy your groceries and daily needs Buy Now. No, you will enjoy unlimited free shipping whenever you meet the above order value threshold.\n\nPlease fill in a complete birthday Enter a valid birthday. The napierian logarithm is also called natural logarithm. Paperback Language of Text: We believe this neperlen is culturally important, and despite the imperfections, have elected to bring it back into print as part of our continuing commitment to the preservation of printed works worldwide.\n\n## Translation of “le logarithme népérien” in English\n\nWe believe this work is culturally. Let us wish you a happy birthday!",
null,
"Antiderivative calculator allows to calculate an antiderivative of napierian logarithm function. Don’t have an account? Calculation of the napierian logarithm For the calculation of napierian logarithm of a number, just enter the number and apply the function ln.\n\nThe calculator makes it possible to obtain the logarithmic expansion of an expression. The natural logarithm of the product of two positive numbers is equal to the sum of the natural logarithm of these two logarkthme.\n\nBASTIAT ECONOMIC SOPHISMS PDF\n\nDay 1 2 3 4 5 6 7 8 9 10 11 12 13 14 jeperien 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Month January February March April May June July August September October November December Year Factor Factorize Factorization Online factoring calculator Expand Simplify Reduce Factorization online Factorize expression online Factorize expression Factor expression Simplify expression online Simplify expressions calculator Simplifying expressions calculator Reduce expression online Expand expression online Expand and simplify expression Expand and simplify Expand and reduce math Expand math Expand a product.\n\nCalculate online with ln napierian logarithm. The graphing calculator is able to plot napierian logarithm function in its definition interval. What happens when I have an item in my cart but it is less than the eligibility threshold?\n\n## File:Logarithme neperien.svg\n\nUpdate your profile Let us wish you a happy birthday! Sponsored products for you.\n\nIf u is a differentiable function, the chain rule of derivatives with the napierian logarithm function and the function u is calculated using the following formula: You can get the remaining amount to reach the Free shipping threshold by adding any eligible item to your cart. The logarithm calculator allows calculation of this type of logarithm online."
] | [
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"http://e.maxicours.com/img/4/2/0/6/420672.jpg",
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"https://rabanimclick.info/download_pdf.png",
null,
"http://keepschool.com/pages/fiches-de-cours/lycee/math/logarithme-neperien/1.gif",
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"https://upload.wikimedia.org/wikipedia/commons/8/88/Logarithme_népérien.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7497548,"math_prob":0.9464338,"size":5606,"snap":"2020-34-2020-40","text_gpt3_token_len":1169,"char_repetition_ratio":0.2047483,"word_repetition_ratio":0.0654321,"special_character_ratio":0.18462361,"punctuation_ratio":0.0857461,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98327875,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,6,null,3,null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T21:36:47Z\",\"WARC-Record-ID\":\"<urn:uuid:7f768e1c-2304-43be-82e1-0d30b3ab4a98>\",\"Content-Length\":\"37366\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06ddb52b-91ae-43c0-bbdf-697ab0a03add>\",\"WARC-Concurrent-To\":\"<urn:uuid:c36cca8b-5c75-4878-8070-f62bd31ba5fb>\",\"WARC-IP-Address\":\"104.27.130.244\",\"WARC-Target-URI\":\"https://rabanimclick.info/logarithme-neperien-85/\",\"WARC-Payload-Digest\":\"sha1:W3GGCMJUXWX72VQV4Q54QD623PKE7PYY\",\"WARC-Block-Digest\":\"sha1:BYLWSIWELI5TWXFSLG7LBIUG2M2NMQZ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737039.58_warc_CC-MAIN-20200806210649-20200807000649-00559.warc.gz\"}"} |
https://codereview.stackexchange.com/questions/187144/backtracking-maze-traversal | [
"# Backtracking maze traversal\n\nI have a functional backtracking solution, but I am only able to finish 22/25 test cases before exceeding my specified time limit. Any suggestions on how to improve the speed are greatly appreciated!\n\nNote: I tried splitting method solve() into solveBinary() and solveDecimal() to remove repeated condition checks for 0 or 1, but it had no apparent effect and they are written the same.\n\nPrompt\n\nThe map is full of 1s and 0s. For each set of starting and ending xy coordinates, determine if the destination can be reached using only 1s, only 0s, or neither (if start is 0 and end is 1, or vice versa, answer is neither)). North, south, east, west movements allowed.\n\n//package kattis2people;\n\nimport java.util.Scanner;\n\npublic class Kattis2People {\n\n// map[][] will be full of 1s and 0s, traversed[][] initially contains all 0s.\nstatic int[][] map, traversed;\n\n// array of solutions to be printed at the end.\nstatic int[] solution;\n\n// dimensions of the map array.\nstatic int r, c;\n\npublic static void main(String[] args) {\nScanner input = new Scanner(System.in);\nr = input.nextInt();\nc = input.nextInt();\n\nmap = new int[r][c];\n\n// Fill the map array with given 1s and 0s.\nfor (int i = 0; i < r; i++){\ninput.nextLine();\nString row = input.next();\nfor (int j = 0; j < c; j++){\nmap[i][j] = row.charAt(j) - 48; //convert char '0' to int 0, char '1' to int 1\n}\n}\n\ninput.nextLine();\n\n// number of traversal attempts forthcoming\nint numQueries = input.nextInt();\n\nsolution = new int[numQueries];\n\n// for each traversal to be attempted, read in the source and the destination.\nfor (int i = 0; i < numQueries; i++){\ninput.nextLine();\nint r1 = input.nextInt() - 1; // input is 1-indexed, subtracting 1 to make it 0-indexed.\nint c1 = input.nextInt() - 1;\nint r2 = input.nextInt() - 1;\nint c2 = input.nextInt() - 1;\n\n// solve1() does premilinay work before calling solve() recursively.\nsolution[i] = solve1(r1,c1,r2,c2, map[r1][c1]);\n}\n\n// print all solutions\nfor (int i = 0; i < numQueries; i++){\nswitch (solution[i]) {\ncase -1:\nSystem.out.println(\"neither\");\nbreak;\ncase 0:\nSystem.out.println(\"binary\");\nbreak;\ndefault:\nSystem.out.println(\"decimal\");\nbreak;\n}\n}\n}\n\nprivate static int solve1(int r1, int c1, int r2, int c2, int x) {\n\n// if the source and destination don't use the same number system, there is no traversal possible.\nif (x != map[r2][c2])\nreturn -1;\n\n// keeps track of where we've been\ntraversed = new int[r][c];\n\n// if the traversal is possible\nboolean result;\n\nif (x == 0)\nreturn (solveBinary(map, r1,c1,r2,c2)) ? 0 : -1; // successful traversal of 0s in map[][] returns 0\nelse\nreturn (solveDecimal(map, r1,c1,r2,c2)) ? 1 : -1; // traversal of 1s in map[][] returns 1\n}\n\n// recursively traverses 0s on the map[][[]\nprivate static boolean solveBinary(int[][]trial, int x1, int y1, int x2, int y2) {\n\n// we've reached the destination!\nif ((x1 == x2) && (y1 == y2)){\ntraversed[x1][y1] = 1;\nreturn true;\n}\n\n// check if this is a valid move.\nif ((x1 >= 0 && x1 < r) && (y1 >= 0 && y1 < c) && (trial[x1][y1] == 0) && traversed[x1][y1] == 0){\n\n//if valid mark as traversed\ntraversed[x1][y1] = 1;\n\n//try going furthur\nif (solveBinary(trial, x1+1, y1, x2, y2))\nreturn true;\nif (solveBinary(trial, x1-1, y1, x2, y2))\nreturn true;\nif (solveBinary(trial, x1, y1+1, x2, y2))\nreturn true;\nif (solveBinary(trial, x1, y1-1, x2, y2))\nreturn true;\ntraversed[x1][y1] = 1;\nreturn false;\n}\nreturn false;\n}\n\n// recursively traverse 1s on the map[][]\nprivate static boolean solveDecimal(int[][]trial, int x1, int y1, int x2, int y2) {\n\n// we've reached the destination!\nif ((x1 == x2) && (y1 == y2)){\ntraversed[x1][y1] = 1;\nreturn true;\n}\n\n// check if this is a valid move.\nif ((x1 >= 0 && x1 < r) && (y1 >= 0 && y1 < c) && (trial[x1][y1] == 1) && traversed[x1][y1] == 0){\n\n//if valid mark as traversed\ntraversed[x1][y1] = 1;\n\n//try going furthur\nif (solveDecimal(trial, x1+1, y1, x2, y2))\nreturn true;\nif (solveDecimal(trial, x1-1, y1, x2, y2))\nreturn true;\nif (solveDecimal(trial, x1, y1+1, x2, y2))\nreturn true;\nif (solveDecimal(trial, x1, y1-1, x2, y2))\nreturn true;\ntraversed[x1][y1] = 1;\nreturn false;\n}\nreturn false;\n}\n}\n\n• You might want to add a heuristic that determines which direction is taken first (see for example A* algorithm). This probably saves a lot of backtracking steps – RobAu Feb 9 '18 at 9:20\n\n## 1 Answer\n\nBacktracking seems to be a bad choice for this particular problem.\nA better solution can be implemented using Disjoint-set data structure.\nHere is the algorithm,\n\ncreate a disjoint-set for decimal people [people with 1 number assigned]\ncreate a disjoint-set for binary people [people with 0 number assigned]\n\n//if mat[][] is 2D matrix, storing content of the map (0's and 1's)\n\nfor(i in rows)\nfor(j in columns)\nif(mat[i][j] == 1) {\nif(mat[i-1][j] == 1) //check all four direction of possible movement\ndecimal.union(person(i,j), person(i-1,j)); // here person(i,j) is position of ith row, jth column person in disjoint-set\nif(mat[i][j-1] == 1)\ndecimal.union(person(i,j), person(i,j-1));\nif(mat[i+1][j] == 1)\ndecimal.union(person(i,j), person(i+1,j));\nif(mat[i][j+1] == 1)\ndecimal.union(person(i,j), person(i,j+1));\n}\nelse {\nif(mat[i-1][j] == 0)\nbinary.union(person(i,j), person(i-1,j));\nif(mat[i][j-1] == 0)\nbinary.union(person(i,j), person(i,j-1));\nif(mat[i+1][j] == 0)\nbinary.union(person(i,j), person(i+1,j));\nif(mat[i][j+1] == 0)\nbinary.union(person(i,j), person(i,j+1));\n}\n\n//for executing query you just need to check if the two persons are in the same set or not\n//let (x1,y1), (x2,y2) be the two points from the query input\n\nif( binary.set(person(x1,y1)) == binary.set(person(x2,y2)) )\nprint \"binary\"\nelse if( decimal.set(person(x1,y1)) == decimal.set(person(x2,y2)) )\nprint \"decimal\"\nelse\nprint \"neither\"\n\n\nHere is the complete solution, implemented in Java. I also tried to run my implemented solution on the platform linked, It turn out to give Accepted status with running time 0.46s :)\n\nimport java.io.*;\n\nclass DisjointSet {\n\nint[] set = null;\nint[] rank = null;\n\nDisjointSet(int n) {\n\nset = new int[n];\nrank = new int[n];\n\nfor(int i=0; i<n; i++){\nrank[i] = 0;\nset[i] = i;\n}\n}\n\nint findSet(int i) {\nif(set[i] == i)\nreturn i;\n\nreturn (set[i] = findSet(set[i]));\n}\n\nvoid union(int x, int y) {\nx = findSet(x);\ny = findSet(y);\n\nif( rank[x] > rank[y] )\nset[y] = x;\nelse if( rank[x] == rank[y] ){\nset[y] = x;\nrank[x]++;\n}\nelse\nset[x] = y;\n}\n\nboolean isSameSet(int x, int y){\nreturn ( findSet(x) == findSet(y) ) ;\n}\n}\n\nclass Maze {\n\npublic static void main(String args[]){\ntry{\nint r, c;\nBufferedReader stdin = new BufferedReader (new InputStreamReader(System.in));\nString temp[] = stdin.readLine().split(\" \");\nr = Integer.parseInt(temp);\nc = Integer.parseInt(temp);\n\nDisjointSet decimal = new DisjointSet( r * c );\nDisjointSet binary = new DisjointSet( r * c );\n\nint[][] inputMatrix = new int[r][];\n\nfor(int i=0; i<r; i++){\n\nString s = stdin.readLine();\ninputMatrix[i] = new int[c];\n\nfor(int j=0; j<c; j++){\ninputMatrix[i][j] = (int)(s.charAt(j) - '0');\n}\n}\n\nfor(int i=0; i<r; i++){\nfor(int j=0; j<c; j++){\n\nif(i>0){\nif(inputMatrix[i-1][j] == 0 && inputMatrix[i][j] == 0)\nbinary.union( i*c + j ,(i-1)*c + j);\nelse if(inputMatrix[i-1][j] == 1 && inputMatrix[i][j] == 1)\ndecimal.union( i*c + j ,(i-1)*c + j);\n\n}\nif(j<c-1){\nif(inputMatrix[i][j+1] == 0 && inputMatrix[i][j] == 0)\nbinary.union( i*c + j ,(i)*c + j + 1);\nelse if(inputMatrix[i][j+1] == 1 && inputMatrix[i][j] == 1)\ndecimal.union( i*c + j ,(i)*c + j + 1);\n\n}\nif(i<r-1){\nif(inputMatrix[i+1][j] == 0 && inputMatrix[i][j] == 0)\nbinary.union( i*c + j ,(i+1)*c + j);\nelse if(inputMatrix[i+1][j] == 1 && inputMatrix[i][j] == 1)\ndecimal.union( i*c + j ,(i+1)*c + j);\n\n}\nif(j>0){\nif(inputMatrix[i][j-1] == 0 && inputMatrix[i][j] == 0)\nbinary.union( i*c + j ,(i)*c + j - 1);\nelse if(inputMatrix[i][j-1] == 1 && inputMatrix[i][j] == 1)\ndecimal.union( i*c + j ,(i)*c + j - 1);\n\n}\n\n}\n}\n\nint n = Integer.parseInt(stdin.readLine());\nfor(int i=0; i<n; i++){\ntemp = stdin.readLine().split(\" \");\n\nint x1 = Integer.parseInt(temp); x1--;\nint y1 = Integer.parseInt(temp); y1--;\nint x2 = Integer.parseInt(temp); x2--;\nint y2 = Integer.parseInt(temp); y2--;\n\nif(binary.isSameSet(x1*c + y1, x2*c + y2) && inputMatrix[x1][y1] == 0)\nSystem.out.println(\"binary\");\nelse if(decimal.isSameSet(x1*c + y1, x2*c + y2) && inputMatrix[x1][y1] == 1)\nSystem.out.println(\"decimal\");\nelse\nSystem.out.println(\"neither\");\n}\n\n}\ncatch(Exception e){\ne.printStackTrace();\n}\n}\n}"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6065672,"math_prob":0.9989643,"size":4072,"snap":"2019-35-2019-39","text_gpt3_token_len":1322,"char_repetition_ratio":0.14011799,"word_repetition_ratio":0.2801161,"special_character_ratio":0.36763263,"punctuation_ratio":0.20973349,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992554,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T01:49:24Z\",\"WARC-Record-ID\":\"<urn:uuid:552deb10-75db-4a4c-88f6-7e96896dea70>\",\"Content-Length\":\"148169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7805057e-4500-4b94-8583-406399d4e4a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e3eb49f-c9ee-4b86-8086-3e3257efc18c>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/187144/backtracking-maze-traversal\",\"WARC-Payload-Digest\":\"sha1:IT2BCIRKCCICHAYG7JIARBWGZNZMRXNY\",\"WARC-Block-Digest\":\"sha1:CGKMWMLB6AOFT7S76KA6FWPDQLLCUQSC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317688.48_warc_CC-MAIN-20190822235908-20190823021908-00525.warc.gz\"}"} |
https://excelchamps.com/formulas/square-number/ | [
"# How to Square a Number in Excel\n\n- by Puneet\n\nIf you want to square a number in Excel, you can use two easy ways to follow methods, and both methods work in the same way. You need to specify the number for which you want to calculate the square and specify 2 as you want to calculate the square.\n\nAhead in this tutorial, we will use an example to understand it.\n\n## Using POWER to Calculate Square\n\n1. Enter the POWER function in the cell.\n2. Refer to the cell where you have the number for which you need to calculate the square.\n3. In the second argument, enter 2 as you need to raise the power to 2.\n4. Close the function and hit enter.\n\nThe moment you need to hit enter the function returns the square as you can see in the above example.\n\n## Using the Power Operator (^)\n\nIn the same way, you also use the power operator (^) to get the square of a number in Excel. And in this method, you don’t need to use any function. Here’s the method:\n\n1. Refer to the cell where you have a number for which you want to calculate the square.\n2. Enter a ^ using the keyboard (you can find it on the number key 6).\n3. Type 2, which tells Excel to raise its power to 2.\n4. Hit enter to get the result.\n\nAs I said, both functions work in the same way and are quite easy to use. But you can choose one according to your need."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90140444,"math_prob":0.95372975,"size":1245,"snap":"2023-40-2023-50","text_gpt3_token_len":290,"char_repetition_ratio":0.15390813,"word_repetition_ratio":0.092369474,"special_character_ratio":0.24257028,"punctuation_ratio":0.09293681,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9960292,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T03:31:53Z\",\"WARC-Record-ID\":\"<urn:uuid:081acbe5-0a82-4c83-a49a-0b8c93a5d804>\",\"Content-Length\":\"239096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f8630f9d-52af-4d8c-9d74-7dabe53ed23b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0267b49-c5b6-4035-9338-d1601adf73ff>\",\"WARC-IP-Address\":\"162.159.134.42\",\"WARC-Target-URI\":\"https://excelchamps.com/formulas/square-number/\",\"WARC-Payload-Digest\":\"sha1:5B3YZNA22JA752V62KO4RL6OZFTXO6HR\",\"WARC-Block-Digest\":\"sha1:YU4ZFFS5U4FOWRZ7JE6A25BIISL2KUZN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510481.79_warc_CC-MAIN-20230929022639-20230929052639-00279.warc.gz\"}"} |
https://cheatsheeting.com/show.html?sheet=ml-to-fl-oz-us-conversions | [
"",
null,
"[PDF Book]\nHome > Conversions (Volume) > Conversion tables from/to milliliter > mL to fl oz US Conversion Cheat Sheet (Interactive)\n From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 You could also enter the values to convert and print directly on the table\n [Formula: fl oz US = mL x 0.033814023] [Printer friendly] [Fluid ounces US to Milliliters]",
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"mL fl oz US\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\nmL fl oz US\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\nmL fl oz US\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\nmL fl oz US\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n= ##\n\n# Milliliters to Fluid ounces US Conversion Table\n\n mL = 0.033814023 fl oz US\n\n# How to convert from Milliliters to Fluid ounces US\n\nSince 1 milliliter is equal to 0.033814023 fluid ounces US, we could say that n milliliters are equal to 0.033814023 times n fluid ounces US. In other words, we could use the following formula:\n\nfluid ounces US = milliliters x 0.033814023\n\nFor example, let's say that we want to convert 2 milliliters to fluid ounces US. Then, we just replace milliliters in the abovementioned formula with 2:\n\nfluid ounces US = 2 x 0.033814023\n\nThat is, 2 milliliters are equal to 0.067628046 fluid ounces US."
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https://learn.careers360.com/ncert/question-the-ages-of-rahul-and-haroon-are-in-the-ratio-5-7-four-years-later-the-sum-of-their-ages-will-be-56-years-what-are-their-present-ages/ | [
"Q\n\n# The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?\n\nQ.9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?\n\nViews\n\nLet the current age of Rahul and Haroon be 5x and 7x repectively(Since there age ratio is given as 5:7).\n\nFour years later their ages becomes 5x + 4 and 7x + 4 repectively.\n\nAccording to the question,\n\n5x + 4 +7x + 4 = 56\n\nor 12x + 8 = 56\n\nTransposing 8 to the RHS we get,\n\n12x = 48\n\nDividing both sides by 12:\n\nThus current age of Rahul = 54 = 20\n\nHaroon = 74 = 28\n\nExams\nArticles\nQuestions"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9654194,"math_prob":0.9910418,"size":497,"snap":"2019-43-2019-47","text_gpt3_token_len":170,"char_repetition_ratio":0.12778905,"word_repetition_ratio":0.0,"special_character_ratio":0.3501006,"punctuation_ratio":0.097345136,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99849516,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T03:10:24Z\",\"WARC-Record-ID\":\"<urn:uuid:5898ea77-de97-4714-8797-72bb5b1825db>\",\"Content-Length\":\"547528\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3f4801f-7b9d-4569-a04a-5ec0276d4579>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb80f331-6984-4d95-92b7-e8154031de55>\",\"WARC-IP-Address\":\"13.126.158.197\",\"WARC-Target-URI\":\"https://learn.careers360.com/ncert/question-the-ages-of-rahul-and-haroon-are-in-the-ratio-5-7-four-years-later-the-sum-of-their-ages-will-be-56-years-what-are-their-present-ages/\",\"WARC-Payload-Digest\":\"sha1:ZABAE2OMGFAEABFGGYUHMATZ6N5QAUBR\",\"WARC-Block-Digest\":\"sha1:YZNLOKZ6NJGNER3BR3SMFMBYTFPELFIW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671106.83_warc_CC-MAIN-20191122014756-20191122042756-00106.warc.gz\"}"} |
https://stackoverflow.com/questions/54375298/how-to-use-tensorflow-feature-columns-as-input-to-a-keras-model | [
"# How to use tensorflow feature_columns as input to a keras model\n\nTensorflow's feature_columns API is quite useful for non-numerical feature processing. However, the current API doc is more about using feature_columns with tensorflow Estimator. Is there a possible way to use feature_columns for categorical features representation and then build a model based on tf.keras?\n\nThe only reference I found is the following tutorial. It shows how to feed feature columns to a Keras Sequential model: Link\n\nThe code snippet is as follows:\n\n``````from tensorflow.python.feature_column import feature_column_v2 as fc\n\nfeature_columns = [fc.embedding_column(ccv, dimension=3), ...]\nfeature_layer = fc.FeatureLayer(feature_columns)\nmodel = tf.keras.Sequential([\nfeature_layer,\ntf.keras.layers.Dense(128, activation=tf.nn.relu),\ntf.keras.layers.Dense(64, activation=tf.nn.relu),\ntf.keras.layers.Dense(1, activation=tf.nn.sigmoid)\n])\n...\nmodel.fit(dataset, steps_per_epoch=8) # dataset is created from tensorflow Dataset API\n``````\n\nThe question is how to use a customed model with keras functional model API. I tried the following, but it did not work (tensorflow version 1.12)\n\n``````feature_layer = fc.FeatureLayer(feature_columns)\ndense_features = feature_layer(features) # features is a dict of ndarrays in dataset\nlayer1 = tf.keras.layers.Dense(128, activation=tf.nn.relu)(dense_features)\nlayer2 = tf.keras.layers.Dense(64, activation=tf.nn.relu)(layer1)\noutput = tf.keras.layers.Dense(1, activation=tf.nn.sigmoid)(layer2)\nmodel = Model(inputs=dense_features, outputs=output)\n``````\n\nThe error log:\n\n``````ValueError: Input tensors to a Model must come from `tf.layers.Input`. Received: Tensor(\"feature_layer/concat:0\", shape=(4, 3), dtype=float32) (missing previous layer metadata).\n``````\n\nI don't kown how to transform feature columns to keras model's input.\n\nThe behavior you desire could be achieved and it's able to combine `tf.feature_column` and `keras functional API`. And, actually, is not mentioned in TF docs.\n\nThis works at least in TF 2.0.0-beta1, but may being changed or even simplified in further releases.\n\nPlease check out issue in TensorFlow github repository Unable to use FeatureColumn with Keras Functional API #27416. There you will find useful comments about `tf.feature_column` and `Keras Functional API`.\n\nBecause you ask about general approach I would just copy the snippet with example from the link above. update: the code below should work\n\n``````from __future__ import absolute_import, division, print_function\n\nimport numpy as np\nimport pandas as pd\n\n#!pip install tensorflow==2.0.0-alpha0\nimport tensorflow as tf\n\nfrom tensorflow import feature_column\nfrom tensorflow import keras\nfrom tensorflow.keras import layers\nfrom sklearn.model_selection import train_test_split\n\ndataframe = pd.read_csv(csv_file, nrows = 10000)\n\ntrain, test = train_test_split(dataframe, test_size=0.2)\ntrain, val = train_test_split(train, test_size=0.2)\nprint(len(train), 'train examples')\nprint(len(val), 'validation examples')\nprint(len(test), 'test examples')\n\n# Define method to create tf.data dataset from Pandas Dataframe\n# This worked with tf 2.0 but does not work with tf 2.2\ndef df_to_dataset_tf_2_0(dataframe, label_column, shuffle=True, batch_size=32):\ndataframe = dataframe.copy()\n#labels = dataframe.pop(label_column)\nlabels = dataframe[label_column]\n\nds = tf.data.Dataset.from_tensor_slices((dict(dataframe), labels))\nif shuffle:\nds = ds.shuffle(buffer_size=len(dataframe))\nds = ds.batch(batch_size)\nreturn ds\n\ndef df_to_dataset(dataframe, label_column, shuffle=True, batch_size=32):\ndataframe = dataframe.copy()\nlabels = dataframe.pop(label_column)\n#labels = dataframe[label_column]\n\nds = tf.data.Dataset.from_tensor_slices((dataframe.to_dict(orient='list'), labels))\nif shuffle:\nds = ds.shuffle(buffer_size=len(dataframe))\nds = ds.batch(batch_size)\nreturn ds\n\nbatch_size = 5 # A small batch sized is used for demonstration purposes\ntrain_ds = df_to_dataset(train, label_column = 'target', batch_size=batch_size)\nval_ds = df_to_dataset(val,label_column = 'target', shuffle=False, batch_size=batch_size)\ntest_ds = df_to_dataset(test, label_column = 'target', shuffle=False, batch_size=batch_size)\n\nage = feature_column.numeric_column(\"age\")\n\nfeature_columns = []\nfeature_layer_inputs = {}\n\n# numeric cols\nfor header in ['age', 'trestbps', 'chol', 'thalach', 'oldpeak', 'slope', 'ca']:\n\n# bucketized cols\nage_buckets = feature_column.bucketized_column(age, boundaries=[18, 25, 30, 35])\nfeature_columns.append(age_buckets)\n\n# indicator cols\nthal = feature_column.categorical_column_with_vocabulary_list(\n'thal', ['fixed', 'normal', 'reversible'])\nthal_one_hot = feature_column.indicator_column(thal)\nfeature_columns.append(thal_one_hot)\nfeature_layer_inputs['thal'] = tf.keras.Input(shape=(1,), name='thal', dtype=tf.string)\n\n# embedding cols\nthal_embedding = feature_column.embedding_column(thal, dimension=8)\nfeature_columns.append(thal_embedding)\n\n# crossed cols\ncrossed_feature = feature_column.crossed_column([age_buckets, thal], hash_bucket_size=1000)\ncrossed_feature = feature_column.indicator_column(crossed_feature)\nfeature_columns.append(crossed_feature)\n\nfeature_layer = tf.keras.layers.DenseFeatures(feature_columns)\nfeature_layer_outputs = feature_layer(feature_layer_inputs)\n\nx = layers.Dense(128, activation='relu')(feature_layer_outputs)\nx = layers.Dense(64, activation='relu')(x)\n\nbaggage_pred = layers.Dense(1, activation='sigmoid')(x)\n\nmodel = keras.Model(inputs=[v for v in feature_layer_inputs.values()], outputs=baggage_pred)\n\nloss='binary_crossentropy',\nmetrics=['accuracy'])\n\nmodel.fit(train_ds)\n``````\n• this should work for now. the trick is to set Inputs to list of input layers, as is shown here as [v for v in feature_layer_inputs.values()]. – thisisbhavin Nov 8 '19 at 9:24\n• Thanks for this! I was trying to prepend `DenseFeatures` to an existing `Sequential` model, but eventually it only worked by using both in a functional `Model` with `inputs=feature_layer_inputs`. – EliadL Nov 25 '19 at 16:18\n• Why is the Input shape=1 in this line: `feature_layer_inputs['thal'] = tf.keras.Input(shape=(1,), name='thal', dtype=tf.string)` – HARSH NILESH PATHAK Jun 17 '20 at 5:26\n• @HARSHNILESHPATHAK, the example for 'thal' column illustrates preprocessing of the `string` values. It means each record of input dataset contains just a one string value in 'thal' column, that is why we require shape=(1,) for the `tf.keras.Input()`. Then Input layer passes this string value to defined feature_columns in `DenseFeatures(feature_columns)` layer. Each feature_column extend the shape according to its own logic. Like for 'thal' here are shown `thal_one_hot` and `thal_embedding`. – Egor B Eremeev Jun 17 '20 at 10:00\n• @prog_guy, No, `thal_one_hot` and `thal_embedding` are just to separate examples of different types of `feature_columns` – Egor B Eremeev Sep 26 '20 at 19:28\n\nIf you use tensorflow dataset API, that code could do well.\n\n``````featurlayer = keras.layers.DenseFeatures(feature_columns=feature_columns)\ntrain_dataset = train_dataset.map(lambda x, y: (featurlayer(x), y))\ntest_dataset = test_dataset.map(lambda x, y: (featurlayer(x), y))\n\nmodel.fit(train_dataset, epochs=, steps_per_epoch=, # all_data/batch_num =\nvalidation_data=test_dataset,\nvalidation_steps=)\n``````\n• this is the correct answer, I tested it under TF 1.13. It should get more votes. But you need to use `from tensorflow.python.feature_column import feature_column_v2 as fc dense_features = fc.DenseFeatures(columns)` – Zhuo Tao May 29 '19 at 16:42\n\ntf.feature_column.input_layer user this function, and this api doc has a sample . you can transform featur_columns into Tensor, and then use it into Mode()\n\nI have been recently reading this document in TensorFlow 2.0 alpha version. It has examples using Keras together with the feature column API. Not sure if TF 2.0 is what you are going to use"
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https://mathoverflow.net/questions/330863/motives-and-homotopy-theories-of-algebraic-varieties | [
"# Motives and homotopy theories of algebraic varieties\n\nThe theory of motives is an attempt to cope with the fact that there are many reasonable cohomology theories of algebraic varieties. Now, sometimes your cohomology theory does not just give you a bunch of groups/vector spaces; it gives you a full-fledged (pro-)homotopy type (though based on limited responses to this question, I think there is no formal way to functorially produce homotopy types given a cohomology theory). The question is: what should be the extension of motives to homotopy types? What are some notable works in this direction? I have heard there is something called motivic homotopy theory, is it really relevant here or it just happens to have a similar name?\n\nMore specific questions:\n\n• First off, we need to show that usual motives are not good enough. This is probably obvious to anyone working in the field, but not everybody on this site is an expert on motives so an explicit example would be great. I think there is a rigorously constructed triangulated category of motives which is supposed but not known to be the derived category of some abelian category of motives. There is also a somewhat more pedestrian Grothendieck ring of varieties. I think the class in the Grothendieck ring does not determine the motive nor does the motive determine the class in the Grothendieck ring. Is there an example of two varieties with the same class in the Grothendieck ring which have different etale fundamental groups? An example of two varieties with the same motive in the triangulated category which have different etale fundamental groups?\n• Before you can say \"every Weil cohomology theory factors through motives\", you have to define a Weil cohomology theory. What should the definition be for homotopy types? Has somebody written down a manageable list of axioms definining exactly what we are interested in?\n• Before we can seriously talk about motivic homotopy types, we should understand on the categorical level what we expect the relevant category to be. In the case of motives, the relevant piece of category theory is Tannakian formalism, I believe (then we can say smart words like \"a Weil cohomology theory is just a fiber functor blah blah\"). What should the category theory look like for motivic homotopy types? A kind of non-abelian Tannakian formalism?\n\nP.S.: yeah, I know the question is super naive, you are free to call me an idiot in the comments.\n\nEDIT: I did not think about the coefficients when asking this. I believe Voevodsky's category makes sense with $$\\mathbb{Z}$$-coefficients, so take that in the first question.\n\n• There is the $\\mathbb{A}^1$-homotopy theory of schemes (Morel, Voevodsky). Voevodsky's triangulated category of motives $DM$ embeds into the $\\mathbb{A}^1$-derived category $D_{\\mathbb{A}^1}$ where the objects have \"explicit\" descriptions (essentially, complexes of presheaves on the category of smooth schemes). Is this in the spirit of what you are looking for? May 6, 2019 at 17:07\n• @FrançoisBrunault but does every \"Weil homotopy theory\" factor through it, like every Weil cohomology theory supposedly factors through the category of motives? Assuming we know what a Weil homotopy theory is.\n– user138661\nMay 6, 2019 at 17:14\n• It seems the category you want would map into the category DM and not the other way round, so I'm not sure my comment will be useful for your question. You could also look for the notion of motivic fundamental group, I don't know if this can be made into a functor from some natural category (just as motivic cohomology can be seen as a functor from DM). May 6, 2019 at 17:21\n• Relevant: arxiv.org/abs/0712.3291 May 6, 2019 at 17:47\n• Starting from categories of dualisable motives over $X$, a point in $X(k)$ gives you a fibre functor to $k$-motives, and Tannakian formalism then gives a form of homotopy type in $k$-motives. A Weil cohomology theory would give a fibre functor down to complexes, and the resulting thing would then be an arithmetic, rather than a geometric, homotopy type. For these homotopy types to behave, you want a $t$-structure (so the standard conjectures would help). For some thoughts about this, see arxiv.org/abs/1309.0637 May 7, 2019 at 13:08\n\nThe projective plane and fake projective planes should have isomorphic motives in the triangulated category, at least with rational coefficients, but different etale fundamental groups. This is because they have isomorphic cohomology.\n\n• Dr. Sawin, could you clarify \"isomorphic cohomology\"? Do you mean \"isomorphic cohomology with respect to an arbitrary Weil cohomology theory\"? If you mean that, how do I see that that is true? Then do we know that there is, in fact, a map of motives inducing these isomorphisms on cohomologies (like you know, two topological manifolds can have isomorphic homotopy groups in all degrees and still fail to be homotopy equivalent)?\n– user138661\nMay 16, 2019 at 18:14\n• @schematic_boi Any algebraic surface has an open set with a finite-to-one map to projective space. This gives a correspondence, giving a map of motives. For a fake projective plane, it should be easy to see that this correspondence gives an isomorphism on any reasonable cohomology theory - we just have to check that some degrees are nonzero. I don't know if it can be shown to give an isomorphism on an arbitrary Weil cohomology theory, except over finite fields. May 16, 2019 at 18:49\n• @schematic_boi Under the Kunneth type standard conjecture and Conjecture d, a motive is just isomorphic to the sum of its cohomology groups as motives, and so two motives with isomorphic cohomology groups are isomorphic. Under the Hodge or Tate conjecture, the same is true when viewing the cohomology groups as Hodge structures / Galois representations. May 16, 2019 at 18:52\n• An example in the spirit of this answer: a rational elliptic surface and an Enriques surface have the same motive with rational coefficients (1+10L+L^2), but have étale fundamental groups of order 1 and 2 respectively, similarly P^2 blow up in 8 points and a classical Godeaux surface have the same motives (1+9L+L^2) but étale fundamental groups of order 1 and 5 respectively, see pdfs.semanticscholar.org/7bbd/… and references there. May 17, 2019 at 7:30\n• @user25309 Yes, good examples. If I remember correctly an Enriques surface can be obtained as an order $2$ logarithmic transformation of a rational elliptic surface (i.e. the unramified double cover of the Enriques is a ramified double cover of the rational elliptic surface, ramified at two elliptic fibers). The graph of this relation gives the relevant correspondence to give an explicit isomorphism of their motives. May 17, 2019 at 11:45\n\nEDIT. Upon request of @schematic_boi, let me try to clarify. My original post only gives an example where the motive does not determine the variety, conditionally on a conjecture of Orlov. In fact, I realized that there are simpler and unconditional examples: if $$S$$ is any scheme, projective bundles of given rank over $$S$$ give rise to isomorphic motives (in fact just Tate motives) in the triangulated category $$\\mathrm{DM}(S,\\mathbb{Z})$$. This is the projective bundle formula, for this level of generality see Cisinski-Déglise, Triangulated categories of motives, 11.3.4. But I think this won't answer @schematic_boi's first question, because the topological fundamental group of a projective bundle should be isomorphic to that of the base. On the other hand, Schnell's article The fundamental group is not a derived invariant mentioned by @user25309, together with Orlov's conjecture, provides a conditional answer, at least with $$\\mathbb{Q}$$-coefficients.\n\nOriginal post. Orlov has conjectured that two schemes having equivalent derived categories of quasi-coherent sheaves give rise to the same motive in the category $$\\mathrm{DM}$$. In particular this would imply, due to a result of Lesieutre that there are infinitely many non-isomorphic smooth projective 3-folds which give rise to the same motive.\n\nReferences:\n\n Orlov, Derived categories of coherent sheaves, and motives. Russian Math. Surveys 60 (2005), no. 6, 1242–1244\n\n Lesieutre, Derived-equivalent rational threefolds. Int. Math. Res. Not. 2015, no. 15, 6011–6020.\n\n• do they have different etale fundamental groups? Or are all of them simply-connected?\n– user138661\nMay 16, 2019 at 10:04\n• @schematic_boi: do you mean the examples of Lesieutre? They are simply-connected; they are blowups of $\\mathbf P^3$ in sets of 8 points.\n– Bort\nMay 16, 2019 at 10:30\n• but you see, the question was: \"Is there an example of two varieties with the same class in the Grothendieck ring which have different etale fundamental groups? An example of two varieties with the same motive in the triangulated category which have different etale fundamental groups?\"\n– user138661\nMay 16, 2019 at 10:30\n• Orlov's conjecture is about motives with rational coefficients. There are derived equivalent varieties with different étale fundamental groups ( arxiv.org/abs/1112.3586 ), but it seems an overkill: unless I am mistaken, étale covers induce isomorphisms of motives with rational coefficients and so there are obviously varieties with the same rational motive but different étale fundamental groups. I think the question of coefficient should be clarified in the original question. May 16, 2019 at 12:00\n• @schematic_boi I was only giving an explicit example where the motive does not determine the variety. If you put Orlov's conjecture together with Schnell's article mentioned by user25309, then you get an answer to your first question, but it is conditional. May 16, 2019 at 18:07\n\nSome answer to one of the questions: two smooth projective varieties over $$\\mathbb{C}$$ with the same class in the Grothendieck ring of varieties have isomorphic fundamental groups (and in particular étale fundamental group by taking the profinite completion).\n\nIndeed, by Larsen and Lunts, https://arxiv.org/abs/math/0110255, the Grothendieck group of smooth projective complex varieties has, after localization of the Lefschetz motive, a linear basis indexed by stable birational equivalence classes.\n\nOn the other hand, the fundamental group of smooth projective varieties is invariant under stable birational equivalence (because the projective spaces are simply connected and birational equivalences only affect loci of codimension greater or equal to 2)."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9534742,"math_prob":0.8903385,"size":2565,"snap":"2022-40-2023-06","text_gpt3_token_len":558,"char_repetition_ratio":0.12729402,"word_repetition_ratio":0.03309693,"special_character_ratio":0.191423,"punctuation_ratio":0.086864404,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99301076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T02:59:08Z\",\"WARC-Record-ID\":\"<urn:uuid:0fe3fcad-d008-4d8f-837f-34cc3451d140>\",\"Content-Length\":\"148373\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7da6e5c-8ed2-4332-a3f1-6e133cc0f2f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:a87ffeb4-5ceb-4c91-97af-fa99945ad7ac>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/330863/motives-and-homotopy-theories-of-algebraic-varieties\",\"WARC-Payload-Digest\":\"sha1:UUMS44J5KGKIGVEYXX3E5UV262PYOXQW\",\"WARC-Block-Digest\":\"sha1:NX5KUBNB4Z2L3PNZ3IEDD66KZ5JHZE7F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500303.56_warc_CC-MAIN-20230206015710-20230206045710-00679.warc.gz\"}"} |
https://www.colorhexa.com/0d2a2d | [
"# #0d2a2d Color Information\n\nIn a RGB color space, hex #0d2a2d is composed of 5.1% red, 16.5% green and 17.6% blue. Whereas in a CMYK color space, it is composed of 71.1% cyan, 6.7% magenta, 0% yellow and 82.4% black. It has a hue angle of 185.6 degrees, a saturation of 55.2% and a lightness of 11.4%. #0d2a2d color hex could be obtained by blending #1a545a with #000000. Closest websafe color is: #003333.\n\n• R 5\n• G 16\n• B 18\nRGB color chart\n• C 71\n• M 7\n• Y 0\n• K 82\nCMYK color chart\n\n#0d2a2d color description : Very dark (mostly black) cyan.\n\n# #0d2a2d Color Conversion\n\nThe hexadecimal color #0d2a2d has RGB values of R:13, G:42, B:45 and CMYK values of C:0.71, M:0.07, Y:0, K:0.82. Its decimal value is 862765.\n\nHex triplet RGB Decimal 0d2a2d `#0d2a2d` 13, 42, 45 `rgb(13,42,45)` 5.1, 16.5, 17.6 `rgb(5.1%,16.5%,17.6%)` 71, 7, 0, 82 185.6°, 55.2, 11.4 `hsl(185.6,55.2%,11.4%)` 185.6°, 71.1, 17.6 003333 `#003333`\nCIE-LAB 15.12, -9.635, -5.222 1.467, 1.931, 2.778 0.238, 0.313, 1.931 15.12, 10.959, 208.455 15.12, -9.123, -3.939 13.895, -5.466, -2.126 00001101, 00101010, 00101101\n\n# Color Schemes with #0d2a2d\n\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #2d100d\n``#2d100d` `rgb(45,16,13)``\nComplementary Color\n• #0d2d20\n``#0d2d20` `rgb(13,45,32)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #0d1a2d\n``#0d1a2d` `rgb(13,26,45)``\nAnalogous Color\n• #2d200d\n``#2d200d` `rgb(45,32,13)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #2d0d1a\n``#2d0d1a` `rgb(45,13,26)``\nSplit Complementary Color\n• #2a2d0d\n``#2a2d0d` `rgb(42,45,13)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #2d0d2a\n``#2d0d2a` `rgb(45,13,42)``\n• #0d2d10\n``#0d2d10` `rgb(13,45,16)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #2d0d2a\n``#2d0d2a` `rgb(45,13,42)``\n• #2d100d\n``#2d100d` `rgb(45,16,13)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #020505\n``#020505` `rgb(2,5,5)``\n• #071819\n``#071819` `rgb(7,24,25)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #133c41\n``#133c41` `rgb(19,60,65)``\n• #184f55\n``#184f55` `rgb(24,79,85)``\n• #1e6168\n``#1e6168` `rgb(30,97,104)``\nMonochromatic Color\n\n# Alternatives to #0d2a2d\n\nBelow, you can see some colors close to #0d2a2d. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0d2d28\n``#0d2d28` `rgb(13,45,40)``\n• #0d2d2b\n``#0d2d2b` `rgb(13,45,43)``\n• #0d2d2d\n``#0d2d2d` `rgb(13,45,45)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #0d272d\n``#0d272d` `rgb(13,39,45)``\n• #0d252d\n``#0d252d` `rgb(13,37,45)``\n• #0d222d\n``#0d222d` `rgb(13,34,45)``\nSimilar Colors\n\n# #0d2a2d Preview\n\nText with hexadecimal color #0d2a2d\n\nThis text has a font color of #0d2a2d.\n\n``<span style=\"color:#0d2a2d;\">Text here</span>``\n#0d2a2d background color\n\nThis paragraph has a background color of #0d2a2d.\n\n``<p style=\"background-color:#0d2a2d;\">Content here</p>``\n#0d2a2d border color\n\nThis element has a border color of #0d2a2d.\n\n``<div style=\"border:1px solid #0d2a2d;\">Content here</div>``\nCSS codes\n``.text {color:#0d2a2d;}``\n``.background {background-color:#0d2a2d;}``\n``.border {border:1px solid #0d2a2d;}``\n\n# Shades and Tints of #0d2a2d\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #040e0f is the darkest color, while #feffff is the lightest one.\n\n• #040e0f\n``#040e0f` `rgb(4,14,15)``\n• #091c1e\n``#091c1e` `rgb(9,28,30)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #11383c\n``#11383c` `rgb(17,56,60)``\n• #16464b\n``#16464b` `rgb(22,70,75)``\n• #1a555b\n``#1a555b` `rgb(26,85,91)``\n• #1f636a\n``#1f636a` `rgb(31,99,106)``\n• #237179\n``#237179` `rgb(35,113,121)``\n• #277f88\n``#277f88` `rgb(39,127,136)``\n• #2c8d98\n``#2c8d98` `rgb(44,141,152)``\n• #309ca7\n``#309ca7` `rgb(48,156,167)``\n• #35aab6\n``#35aab6` `rgb(53,170,182)``\n• #39b8c5\n``#39b8c5` `rgb(57,184,197)``\n• #48beca\n``#48beca` `rgb(72,190,202)``\n• #57c3ce\n``#57c3ce` `rgb(87,195,206)``\n• #66c9d3\n``#66c9d3` `rgb(102,201,211)``\n• #75ced7\n``#75ced7` `rgb(117,206,215)``\n• #85d3dc\n``#85d3dc` `rgb(133,211,220)``\n• #94d9e0\n``#94d9e0` `rgb(148,217,224)``\n• #a3dee4\n``#a3dee4` `rgb(163,222,228)``\n• #b2e4e9\n``#b2e4e9` `rgb(178,228,233)``\n• #c1e9ed\n``#c1e9ed` `rgb(193,233,237)``\n• #d1eff2\n``#d1eff2` `rgb(209,239,242)``\n• #e0f4f6\n``#e0f4f6` `rgb(224,244,246)``\n• #eff9fa\n``#eff9fa` `rgb(239,249,250)``\n• #feffff\n``#feffff` `rgb(254,255,255)``\nTint Color Variation\n\n# Tones of #0d2a2d\n\nA tone is produced by adding gray to any pure hue. In this case, #1d1d1d is the less saturated color, while #023338 is the most saturated one.\n\n• #1d1d1d\n``#1d1d1d` `rgb(29,29,29)``\n• #1a1f20\n``#1a1f20` `rgb(26,31,32)``\n• #182122\n``#182122` `rgb(24,33,34)``\n• #162324\n``#162324` `rgb(22,35,36)``\n• #142526\n``#142526` `rgb(20,37,38)``\n• #112629\n``#112629` `rgb(17,38,41)``\n• #0f282b\n``#0f282b` `rgb(15,40,43)``\n• #0d2a2d\n``#0d2a2d` `rgb(13,42,45)``\n• #0b2c2f\n``#0b2c2f` `rgb(11,44,47)``\n• #092e31\n``#092e31` `rgb(9,46,49)``\n• #062f34\n``#062f34` `rgb(6,47,52)``\n• #043136\n``#043136` `rgb(4,49,54)``\n• #023338\n``#023338` `rgb(2,51,56)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0d2a2d is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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https://math-quiz.co.uk/start_test.php?id=51 | [
"# Test: Arithmetic I - Ambitious\n\nDouble click on maths expressions to zoom\nQuestion 1: Solve $16+14÷2+3\\cdot 5-11$\n$75$\n$79$\n$23$\n$27$\nQuestion 2: Solve $\\left(11-5\\right)\\cdot \\left(63-59+6\\right)÷12$\n$4$\n$7$\n$5$\n$6$\nQuestion 3: Find HCF (Highest Common Factor) and LCM (Lowest Common Multiple) of the following pair of numbers: $120$ and $336$.\nQuestion 4: Solve: $\\frac{7}{21}\\cdot \\frac{3}{7}$\n$\\frac{21}{146}$\n$\\frac{10}{21}$\n$\\frac{21}{7}$\n$\\frac{1}{7}$\nQuestion 5: Solve: $\\frac{13}{36}÷\\frac{7}{6}$\n$\\frac{20}{42}$\n$\\frac{13}{42}$\n$\\frac{20}{36}$\n$\\frac{19}{43}$\nPlease note, you have solved only half of the test. For the complete test get a Must Have account. Get started"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.871859,"math_prob":0.9999285,"size":302,"snap":"2022-05-2022-21","text_gpt3_token_len":82,"char_repetition_ratio":0.18120806,"word_repetition_ratio":0.0,"special_character_ratio":0.24172185,"punctuation_ratio":0.1875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999874,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-24T16:06:27Z\",\"WARC-Record-ID\":\"<urn:uuid:55e42b5f-9fd2-4ae9-a55c-8fd5065efaf8>\",\"Content-Length\":\"31403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf92ae69-a697-4f65-b4e2-507769f0f09a>\",\"WARC-Concurrent-To\":\"<urn:uuid:007b34c8-87db-45fb-984d-b07b6e371191>\",\"WARC-IP-Address\":\"217.160.0.33\",\"WARC-Target-URI\":\"https://math-quiz.co.uk/start_test.php?id=51\",\"WARC-Payload-Digest\":\"sha1:XK6FDNM4F5IL7S47Y3NKWPFYHK3CRLU7\",\"WARC-Block-Digest\":\"sha1:N4N554652VBDO3ZEXVOISZDCIMFA7I2A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662573053.67_warc_CC-MAIN-20220524142617-20220524172617-00499.warc.gz\"}"} |
https://www.kaysonseducation.co.in/questions/p-span-sty_3735 | [
"## Question\n\n### Solution\n\nCorrect option is\n\n#### SIMILAR QUESTIONS\n\nQ1\n\nWhich one of the following has the dimensions of",
null,
"Q2\n\nThe equation of state of a real gas can be expressed as",
null,
"Where P is the pressure, V the volume T the absolute temperature and ab and c are constants. What are the dimensions of a?\n\nQ3\n\nThe equation of state for n moles of an ideal gas is",
null,
"Where R is the universal gas constant and PV and T have the usual meanings. What are the dimensions of R?\n\nQ4\n\nTime period T of a simple pendulum may depend on m, the mass of the bob, l, the length of the string and g, the acceleration due to gravity, i.e.",
null,
"What are the values of ab and c?"
] | [
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Mechanics/Kinematics/Day%201%20to%20Day%207_files/image066.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Mechanics/Kinematics/Day%201%20to%20Day%207_files/image098.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Mechanics/Kinematics/Day%201%20to%20Day%207_files/image112.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Mechanics/Kinematics/Day%201%20to%20Day%207_files/image152.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8851018,"math_prob":0.98779815,"size":637,"snap":"2021-43-2021-49","text_gpt3_token_len":166,"char_repetition_ratio":0.13902053,"word_repetition_ratio":0.015873017,"special_character_ratio":0.23390895,"punctuation_ratio":0.108843535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99940217,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T04:37:28Z\",\"WARC-Record-ID\":\"<urn:uuid:3537fb78-1938-4ca1-9699-88249db8fcc4>\",\"Content-Length\":\"30169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:883f560e-bd87-4371-957b-af1f864dbbf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:c09b593d-a690-4afd-ab1e-64d664d485a7>\",\"WARC-IP-Address\":\"172.67.175.73\",\"WARC-Target-URI\":\"https://www.kaysonseducation.co.in/questions/p-span-sty_3735\",\"WARC-Payload-Digest\":\"sha1:WWCMGD2CJTVHNFPCSPX4ET3QRGZJ3IMW\",\"WARC-Block-Digest\":\"sha1:DXUTFBU7MV7IDTP4SOZ6U5GZWHWHCENG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964361064.69_warc_CC-MAIN-20211202024322-20211202054322-00390.warc.gz\"}"} |
https://www.aqua-calc.com/what-is/length/light-second | [
"# What is a light second (unit)\n\n## The light second is a unit of measurement of length\n\nThe light second (ls), lightsecond, or light-second is a measurement unit of length used in astronomy. The light second is defined as the distance that light travels in a vacuum in one second. The light second is equal to 299 792 458 meters.\n\n• What is lengthInstant conversionsConversion tablesRate conversions\n• 1 ls = 5.6652564232462×10+18 a0\n• 1 ls = 2.99792458×10+18 Å\n• 1 ls = 0.002003988804 au\n• 1 ls = 2.99792458×10+26 am\n• 1 ls = 29 979 245 800 cm\n• 1 ls = 14 902 591.8 ch\n• 1 ls = 2 997 924 580 dm\n• 1 ls = 29 979 245.8 dam\n• 1 ls = 2.99792458×10-10 Em\n• 1 ls = 163 928 509 ftm\n• 1 ls = 2.99792458×10+23 fm\n• 1 ls = 2.99792458×10+23 fermi\n• 1 ls = 983 571 056 ft\n• 1 ls = 1 490 259.18 fur\n• 1 ls = 0.299792458 Gm\n• 1 ls = 2 997 924.58 hm\n• 1 ls = 11 802 852 677 in\n• 1 ls = 299 792.458 km\n• 1 ls = 299 792.458 klick\n• 1 ls = 0.0002777777778 lh\n• 1 ls = 0.01666666667 lmin\n• 1 ls = 3.168876466×10-8 ly\n• 1 ls = 299.792458 Mm\n• 1 ls = 299 792 458 m\n• 1 ls = 1.1802852665126×10+16 µin\n• 1 ls = 2.99792458×10+14 µm\n• 1 ls = 2.99792458×10+14 µ\n• 1 ls = 11 802 852 665 126 mil\n• 1 ls = 186 282.397 mi\n• 1 ls = 299 792 458 000 mm\n• 1 ls = 2.99792458×10+17 nm\n• 1 ls = 161 874.977 nmi\n• 1 ls = 9.715603498×10-9 pc\n• 1 ls = 2.99792458×10-7 Pm\n• 1 ls = 2.99792458×10+20 pm\n• 1 ls = 0.000299792458 Tm\n• 1 ls = 11 802 852 665 126 thou\n• 1 ls = 327 857 019 yd\n• 1 ls = 2.99792458×10+32 ym\n• 1 ls = 2.99792458×10-16 Ym\n• 1 ls = 2.99792458×10+29 zm\n• 1 ls = 2.99792458×10-13 Zm\n\n#### Foods, Nutrients and Calories\n\nCheerios Cereal 2 Pack, UPC: 00016000435094 weigh(s) 30 grams per metric cup or 1 ounce per US cup with glycemic load 53 given the food's glycemic index 74 [ weight to volume | volume to weight | price | density ]\n\n209328 foods that contain Total lipid (fat). List of these foods starting with the highest contents of Total lipid (fat) and the lowest contents of Total lipid (fat)\n\n#### Gravels, Substances and Oils\n\nCaribSea, Freshwater, Super Naturals, Snowy River weighs 1 457.68 kg/m³ (90.99999 lb/ft³) with specific gravity of 1.45768 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nConcrete, Limestone with Portland weighs 2 371 kg/m³ (148.01669 lb/ft³) [ weight to volume | volume to weight | price | density ]\n\nVolume to weightweight to volume and cost conversions for Engine Oil, SAE 15W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)\n\n#### Weights and Measurements\n\nThe grain per square inch surface density measurement unit is used to measure area in square inches in order to estimate weight or mass in grains\n\nThe electric charge Q is a property of bodies to exert forces on each other by electric fields.\n\nmil/h² to nmi/min² conversion table, mil/h² to nmi/min² unit converter or convert between all units of acceleration measurement.\n\n#### Calculators\n\nReturn on investment appreciation or rate of loss calculator"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5876116,"math_prob":0.98989654,"size":2432,"snap":"2023-40-2023-50","text_gpt3_token_len":1173,"char_repetition_ratio":0.35502473,"word_repetition_ratio":0.07990868,"special_character_ratio":0.50863487,"punctuation_ratio":0.06764168,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98385125,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T07:24:33Z\",\"WARC-Record-ID\":\"<urn:uuid:1da76093-dc3c-4e5f-9bc7-9501d02ea199>\",\"Content-Length\":\"31258\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e4e35838-a714-41f5-86b2-1ced0144dc5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4020db5-8b38-4ba8-b3c4-b166bb596c2a>\",\"WARC-IP-Address\":\"69.46.0.75\",\"WARC-Target-URI\":\"https://www.aqua-calc.com/what-is/length/light-second\",\"WARC-Payload-Digest\":\"sha1:ROBPOCEC5EGQJR5I5VXT5X2T322OYVOJ\",\"WARC-Block-Digest\":\"sha1:CQOK4WOYKEYHYRDSBCE3I43DEPMXAP4T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510368.33_warc_CC-MAIN-20230928063033-20230928093033-00199.warc.gz\"}"} |
https://www.itparks.cn/%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0%EF%BC%88%E4%B8%80%EF%BC%89 | [
"# 数据类型\n\n### 数字\n\n• 整数（int，long）：内存决定长度，不支持自增（++）和自减（–）\n\n• 浮点数（float）：带小数点的数，无穷小数会做精度处理\n\n• 布尔（bool）：非空（None）非 0 → true，0 或空（None） → false\n\n• 复数（complex）：表示向量，如：2 + 3j 或 2 + 3J\n\n### 字符串（str）\n\n• 定义：单引双引交替使用，不可变\n\n• 索引：str[index]，从左往右 → 从 0 开始，从右往左 → 从 -1 开始\n\n• 切片：str[start:stop:step]，左闭右开\n\n### 列表（list）\n\n• 定义：可保存任意类型，如：[1,0.2,'abc',]\n\n• 操作：支持索引和切片，修改：str[index] = 任意类型的值\n\n• 字符串 → 列表：eval('[1,2,3,4]')\n\n### 元组（tuple）\n\n• 定义：与列表类似，可保存任意类型，如：(1,0.2,'abc',(1))，只含有一个元素时，会转变成数据的类型，所以要加逗号：myTuple = (1,)\n\n• 操作：支持索引和切片，不支持修改，但可以修改其中的列表中的元素值\n\n# 变量引用计数\n\n• 拥有相同值的变量共享一个地址\n\n• 用 sys下的 getrefcount() 查看引用计数\n\n```import sys\nsys.getrefcount(value)```\n• 优点：节约内存，防止产生多余的内存碎片\n\n# 运算表达式\n\n### 算数表达式\n\n• 除法精确到浮点数，其它运算返回值类型，取决于参与运算中最复杂的类型\n\n• //：向下取整，如：7 // 2 = 3\n\n• round()：可控制结果精度，默认四舍五入，如：round(10/3,4) = 3.3333，round(10/3) = 3，round(7/2) = 4\n\n• %：取余，如：7 % 2 = 1\n\n• **：幂，如：2 ** 4 = 16\n\n### 逻辑表达式\n\n not a 非 a a and b a 且 b a or b a 或 b a is b a 和 b是一个对象 a is not b a 和 b不是同一个对象\n• ==：只判断值\n\n• is：判断值、内存地址、数据类型\n\n### 关系表达式\n\n a == b a 等于 b a != b a 不等于 b a <> b 已废弃 a > b a 大于 b a < b a 小于 b a >= b a 大于等于 b a <= b a 小于等于 b\n\n### 位运算\n\n ~a 按位取反 a << n 左移 n 位 a >> n 右移 n 位 a & b 按位与 a | b 按位或 a ^ b 按位异或\n\n### 二进制\n\n 1 -1 负数步骤 源码 0000 0001 1000 0001 符号位在第一位，正 → 0，负 → 1 反码 0000 0001 1111 1110 符号位不变，其它取反 补码 0000 0001 1111 1111 反码加一\n\n# 判断、循环\n\n### if\n\n```if a == b:\nprint('a == b')\nelif a < b:\nprint('a < b')\nelse:\nprint('a > b')```\n\n### while\n\n```a = 100\nwhile a > 10:\nprint('a > 10')\nif(a < 20):\nprint('a < 20')\nbreak\na = a - 10```\n\n### for\n\n```for var in [1,2,3]:\nif var == 2:\ncontinue\nprint(var)```"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.794693,"math_prob":0.9901164,"size":1410,"snap":"2019-26-2019-30","text_gpt3_token_len":959,"char_repetition_ratio":0.11379801,"word_repetition_ratio":0.0,"special_character_ratio":0.47021276,"punctuation_ratio":0.08059701,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9809886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-19T20:45:15Z\",\"WARC-Record-ID\":\"<urn:uuid:92e35708-77ac-43a1-9dc9-9754b9f0c1f6>\",\"Content-Length\":\"25714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2ad6df1-14a7-4798-a1b5-2efbff1384e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f557c788-accc-443b-b5e0-07724ac103bc>\",\"WARC-IP-Address\":\"121.42.118.137\",\"WARC-Target-URI\":\"https://www.itparks.cn/%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0%EF%BC%88%E4%B8%80%EF%BC%89\",\"WARC-Payload-Digest\":\"sha1:SPLKHEIWLIBHWNFABFGIBUHWIVDYVORS\",\"WARC-Block-Digest\":\"sha1:QIESO6SNXURUMLXYZ3A3O4DLLKI2J2ZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526359.16_warc_CC-MAIN-20190719202605-20190719224605-00225.warc.gz\"}"} |
https://www.tutorialspoint.com/java/lang/long_equals.htm | [
"# Java.lang.Long.equals() Method\n\n## Description\n\nThe java.lang.Long.equals() method compares this object to the specified object. The result is true if and only if the argument is not null and is a Long object that contains the same long value as this object.\n\n## Declaration\n\nFollowing is the declaration for java.lang.Long.equals() method\n\n```public boolean equals(Object obj)\n```\n\n## Parameters\n\nobj − This is the object to compare with.\n\n## Return Value\n\nThis method returns true if the objects are the same, else false.\n\nNA\n\n## Example\n\nThe following example shows the usage of java.lang.Long.equals() method.\n\n```package com.tutorialspoint;\n\nimport java.lang.*;\n\npublic class LongDemo {\n\npublic static void main(String[] args) {\n\nLong obj1 = new Long(79845);\nLong obj2 = new Long(79800);\nSystem.out.print(\"Is obj1 and obj2 equal ? \");\nSystem.out.println(obj1.equals(obj2));\n\nobj1 = new Long(45469);\nobj2 = new Long(45469);\nSystem.out.print(\"Is obj1 and obj2 equal ? \");\nSystem.out.println(obj1.equals(obj2));\n}\n}\n```\n\nLet us compile and run the above program, this will produce the following result −\n\n```Is obj1 and obj2 equal ? false\nIs obj1 and obj2 equal ? true\n```\njava_lang_long.htm"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.53096116,"math_prob":0.8825591,"size":1031,"snap":"2020-34-2020-40","text_gpt3_token_len":246,"char_repetition_ratio":0.14021422,"word_repetition_ratio":0.077922076,"special_character_ratio":0.27449077,"punctuation_ratio":0.19724771,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9778902,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T15:15:16Z\",\"WARC-Record-ID\":\"<urn:uuid:065649d5-5707-4430-bd7c-faff34472346>\",\"Content-Length\":\"26615\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e2d2533-5cae-4c64-89d7-7c8aff69f8f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:766253a1-633b-4e03-be83-cdd5a7c7a53a>\",\"WARC-IP-Address\":\"72.21.91.42\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/java/lang/long_equals.htm\",\"WARC-Payload-Digest\":\"sha1:CKO5NAGPYT2YTVX6N4FH63M6UCPREUDI\",\"WARC-Block-Digest\":\"sha1:RIVOIM75DGE6P3YPRJE5DWGVZUGIDR6O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401643509.96_warc_CC-MAIN-20200929123413-20200929153413-00423.warc.gz\"}"} |
https://goprep.co/ex-6.1-q1b-g-x-3-sin-x-pi-4-0-less-than-equal-to-x-less-than-i-1nl7q0 | [
"Q. 1 B5.0( 1 Vote )\n\n# Sketch the graphs\n\nAnswer :\n\nWe know that if f(x) is a periodic function with period T, then f (ax + b) is periodic with period T/|a|.",
null,
"is a periodic function with period π. So, we will draw the graph of",
null,
"in the interval [0, 5π/4]. The values of",
null,
"at various points in [0, 5π/4] are listed in the following table:",
null,
"By plotting the above points, we obtain the required curve.",
null,
"Rate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
"RELATED QUESTIONS :\n\nDraw the graph ofRS Aggarwal - Mathematics\n\nDraw the graph ofRS Aggarwal - Mathematics\n\nDraw the graphs oRS Aggarwal - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics\n\nSketch the graphsRD Sharma - Mathematics"
] | [
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"https://gradeup-question-images.grdp.co/liveData/PROJ26189/1549633242417677.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ26189/1549633243178872.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ26189/1549633243942419.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ26189/1549633244693754.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ26189/1549633245403784.png",
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"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
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"https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.831557,"math_prob":0.929272,"size":899,"snap":"2021-04-2021-17","text_gpt3_token_len":231,"char_repetition_ratio":0.118435755,"word_repetition_ratio":0.078431375,"special_character_ratio":0.23692992,"punctuation_ratio":0.11627907,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97755927,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-21T11:21:45Z\",\"WARC-Record-ID\":\"<urn:uuid:0b62b3a6-abbf-4882-bcb2-069ad10f71ac>\",\"Content-Length\":\"158757\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6855829-6f90-41a9-bde1-5bd8c88b7771>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d8dacb1-55cf-4d24-a5c1-2cab1cc695a9>\",\"WARC-IP-Address\":\"104.18.24.35\",\"WARC-Target-URI\":\"https://goprep.co/ex-6.1-q1b-g-x-3-sin-x-pi-4-0-less-than-equal-to-x-less-than-i-1nl7q0\",\"WARC-Payload-Digest\":\"sha1:7OSX4KUH2IBIU2TIKHXN4HFP55QLYOOG\",\"WARC-Block-Digest\":\"sha1:OLDDJAW6GGRVSBNSJ7LI5DBPPLVDIIWP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039536858.83_warc_CC-MAIN-20210421100029-20210421130029-00365.warc.gz\"}"} |
https://sciencing.com/calculate-alkalinity-caco-5328969.html | [
"# How to Calculate Alkalinity As Concentration of CaCO3",
null,
"••• armcreation/iStock/GettyImages\n\nPure water doesn't exist in nature. Even in artificial environments, the purity of water doesn't last much past the opening of the bottle. Water dissolves more substances than any other known liquid. In fact, most people find \"pure\" water tastes odd. Life depends on water's ability to dissolve many substances, and water's dissolved chemicals determine some of the critical physical conditions of ecosystems. Two important and interrelated water-related conditions are acidity and alkalinity.\n\n#### TL;DR (Too Long; Didn't Read)\n\nUse this equation to calculate the alkalinity in terms of CaCO3: Alk(mg/L CaCO3)=Alk(meq/L)x(1mmol CaCO3÷2meq)x(100.087mg CaCO3÷1mmol CaCO3)=(50044xBxCaxCF)÷Vs. The variables used are the volume of acid (B) used in the titration, the concentration of the acid (Ca), the volume of water in the sample (Vs) and a correction factor (CF), if necessary.\n\n## Importance of Alkalinity\n\nAlkalinity determines how water resists changes in pH, in particular resistance to becoming acidic. Higher alkalinity means higher resistance to changes in pH. Changes in pH can be devastating to the ecosystem. Many plants and aquatic animals require a very narrow pH range to survive. Even less pH-sensitive animals suffer if changes in pH impact the food chain. In addition to the natural ecosystem, alkalinity impacts wastewater systems and suitability of water for irrigation. Alkalinity, also referred to as water hardness, impacts home water systems by building mineral deposits in pipes and pans as well as increasing the amount of soap people use by inhibiting lather.\n\n## Units of Alkalinity\n\nAlkalinity measures the amount of calcium, magnesium and other acid-moderating ions. Alkalinity may be reported in terms of calcium carbonate, CaCO3. Alkalinity is usually measured and reported in terms of parts per million (ppm) or grains per gallon.\n\n## Preparing for Testing\n\nThe recommended method for determining alkalinity uses inflection point titration (IPT) with sulfuric acid solution. Even when using commercially available 0.01639N sulfuric acid solution, be sure to test the solution against a sodium carbonate standard. Different acid concentrations may be used, especially with lower alkalinities, but the acid must be compared to a standard.\n\nWater samples need to represent the environmental conditions, so exposure to sunlight, dust, rain or other contaminants may alter test results. Over time, temperature changes may also affect the sample quality, so samples should be kept at the temperature of the original environment. For alkalinity tests, filter the water samples through a 0.45-µm membrane where the unit µm means micrometer, or micron. The micrometer equals 10-6 meter or one-millionth of a meter.\n\nMost tests use a fifty milliliter (50mL) sample in a 100mL beaker, but for low alkalinity samples a larger sample of 100mL may be used.\n\n## Test Apparatus and Methodology\n\nThe preferred testing method uses a glass buret, but a digital titrator may also be used. Be sure the buret is clean and dry so that the acid solution will not be diluted.\n\nUse standard titration methodology with special care taken around pH 8.3 and pH 4.5. For finding alkalinity in terms of calcium carbonate, take particular care as the sample pH approaches pH 4.5.\n\n## Calculating Alkalinity\n\nThe formula for calculating alkalinity finds the ratio of acid to the bicarbonate equivalency point. The first equation shows the alkalinity in milliequivalents per liter.\n\nUse this formula to find alkalinity:\n\nAlk(meq/L)=[B(mL)xCa(meq/mL)xCF]÷[Vs(mL)x(1L÷1000mL)]=(1000xBxCaxCF)÷Vs\n\nThe following equation combines the alkalinity equation with the conversion factor to calculate the alkalinity in CaCO3:\n\nAlk(mg/L CaCO3)=Alk(meq/L)x(1mmol CaCO3÷2meq)x(100.087mg CaCO3÷1mmol CaCO3)=(50044xBxCaxCF)÷Vs.\n\nCalculating alkalinity requires knowing:\n\nAlk means alkalinity\n\nB: The volume of acid used in the titration from the original pH to the equivalency point (near pH4.5) in milliliters.\n\nCa: The concentration of the sulfuric acid expressed as normality N or as milliequivalents per milliliter (meq/mL).\n\nVs: The volume of the sample in milliliters.\n\nCF: A correction factor, if necessary. The correction factor for the Hach digital titrator cartridges equals 1.01; all other titration methods have a correction factor equal to 1.0.\n\nAssume the volume of acid (B) used equals 12 milliliters of acid, the acid's concentration (Ca) equals 0.01639meq/mL, and the water sample volume (Vs) equals 50mL of water. If performing the titration with a standard buret, the correction factor (CF) equals 1.0. To calculate the alkalinity, insert these values into the first equation. The calculation becomes:\n\nAlk=(1000xBxCaxCF)÷Vs or Alk (meq/L)=[1000x12mLx0.01639(meq/mL)x1.0]÷50mL.\n\nSolving for alkalinity yields: Alk(meq/L)=3.9 meq/mL.\n\nTo calculate alkalinity as calcium carbonate, use this formula to calculate the alkalinity in concentration of calcium carbonate (CaCO3). Insert the values into the formula to determine the alkalinity as CaCO3:\n\nAlk(mg/L as CaCO3)=(50044xBxCaxCF)÷Vs.\n\nInserting the example values, the equation becomes:\n\nAlk(mg/L as CaCO3)=[50044x12mLx0.01639meq/mLx1.0]÷50mL. Solving the equation yields Alk(mg/L as CaCO3)=196.8mg/L.\n\n## Interpreting Results\n\nMilligrams per liter (mg/L) equals parts per million (ppm). Alkalinity is also known as water hardness. In general, water with a hardness less than 17.0 ppm is considered soft water. Water with hardness over 180 ppm is considered very hard water. In this example, the water sample has very high alkalinity.\n\nDont Go!\n\nWe Have More Great Sciencing Articles!"
] | [
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"https://img-aws.ehowcdn.com/360x267p/s3-us-west-1.amazonaws.com/contentlab.studiod/getty/577b033d4b644e2792db4f55c8f9134b",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8533423,"math_prob":0.97805274,"size":5932,"snap":"2023-40-2023-50","text_gpt3_token_len":1480,"char_repetition_ratio":0.13630229,"word_repetition_ratio":0.014976959,"special_character_ratio":0.21931894,"punctuation_ratio":0.11436413,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9844848,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T03:35:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5f2bceb3-c780-422c-a5e4-f702b8a14e97>\",\"Content-Length\":\"414957\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e6c35170-25f5-4099-baef-aaed2f7cc28e>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b46e556-1425-4ce1-a5d3-b97affc1964d>\",\"WARC-IP-Address\":\"23.194.131.161\",\"WARC-Target-URI\":\"https://sciencing.com/calculate-alkalinity-caco-5328969.html\",\"WARC-Payload-Digest\":\"sha1:NBBSNRX4GQQNIAPUWHLUW7EB2ETVIJQ5\",\"WARC-Block-Digest\":\"sha1:YD3T2DD3CJIFDFCNWWFN6ZYGEBDU7CVD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00659.warc.gz\"}"} |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1611&CurriculumID=31&Method=Worksheet&NQ=2&Num=1.36&Type=D | [
"",
null,
"Name: ___________________Date:___________________\n\n Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!\n\n### MEAP Preparation - Grade 4 Mathematics1.36 More Fraction Problems\n\n The fractional part of a collection is determined by dividing the collection into subgroups equal to the number representing the denominator of the fraction. We then take the number of subgroups equal to the number representing the numerator of the fraction. A fraction can be expressed as a division and vice-versa. A fraction written as a combination of a whole number and a proper fraction is called a mixed number. To generate equivalent fractions of any given fraction, we proceed as follows: Multiply the numerator and denominator by the same number (other than 0) or Divide the numerator and denominator by their common factor (other than 1), if any. The two fractions are equivalent if the product of numerator of the first and denominator of the second is equal to the product of the denominator of the first and numerator of the second. A fraction is in its lowest term, if the numerator and denominator have no common factor other than 1. Of the two fractions having the same denominator, the one with greater numerator is greater. Of the two fractions having the same numerator, the one with greater denominator is smaller. Adding and Subtracting Fractions: Read the problems carefully. If the denominators are same, add or subtract the numerators and write the answer. If the denominators are different, they are made common by finding LCM. Example: 9/4 - 3/4 Since the denominators are same, the numerators are subtracted. 9/4 - 3/4 = 6/4 Answer: 6/4 Example: 1/5 + 1/3 Since the denominators are different, find LCM of 5 and 3. LCM of 5 and 3 is 15. 1/5 x 15/15 = 3/15 1/3 x 15/15 = 5/15 Answer: 8/15 Directions: Answer the following questions. Also write at least 5 examples of your own for fraction operations: addition, subtraction, multiplication, and division of fractions.",
null,
"Name: ___________________Date:___________________\n\n### MEAP Preparation - Grade 4 Mathematics1.36 More Fraction Problems\n\n Q 1: Change 114/21 into a whole number or mixed fraction.4 9/215 9/205 9/21 Question 2: This question is available to subscribers only!"
] | [
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https://www.wyzant.com/Tutors/TN/Nashville/7832687/ | [
"",
null,
"# Ayanaw H.Any Mathematics WelcomeAny Mathematics Welcome\n\n### Bio\n\nYes, I can make math easy, simple, fun, and allow students to learn by themselves giving them the confidence to solve any math question. I have been tutoring, mentoring, and teaching Integrated Math I, II, & III, Pre-Algebra, Algebra I, Algebra II, College Algebra, Pre-Calculus, Trigonometry, Calculus, Statistics, and Physics for 20 years.\nI taught College Algebra, Trigonometry, Precalculus Calculus, Calculus, and Introduction of Mechanical engineering at the US Air Force Academy Prep...\n\nYes, I can make math easy, simple, fun, and allow students to learn by themselves giving them the confidence to solve any math question. I have been tutoring, mentoring, and teaching Integrated Math I, II, & III, Pre-Algebra, Algebra I, Algebra II, College Algebra, Pre-Calculus, Trigonometry, Calculus, Statistics, and Physics for 20 years.\nI taught College Algebra, Trigonometry, Precalculus Calculus, Calculus, and Introduction of Mechanical engineering at the US Air Force Academy Prep School in Colorado Springs CO.\nCurrently, I am teaching multiple mathematics courses at Motlow State Community College and Embry Riddle Aeronautical University. I am also tutoring ACT, SAT, AFQT, ASVAB, Praxis, Integrated Math I, II, & II, Algebra I & II, AP calculus, AP Statistics, Calculus I, II, & III, Statistics, Probability, Differential Equation, Linear Algebra, and Advanced Mathematics Courses. Most of my Students are from Metro Nashville public high schools, local colleges and universities students, and online across the nation.\nAs a tutor/mentor, I guide the students through solving math problems using a problem-solving based on teaching style to help students learn and understand math concepts, therefore, all my students can learn to handle any challenge question on their own in the near future. I use effective practices, strategies and resources to inspire and enrich each of my student's learning. I usually use different learning styles such as workbooks, visual aid computers, hands-on activities like lab, student-centered discussions as learning-centered, one-on-one tutoring and small-group tutoring.\nNote: Let me work with a student who is trying to achieve college credits in AP Calculus, Statistics, and Physics. Let me work with a student who thinks that Calculus is Greek language or something difficult. Let me work with student who has hated math, to challenge them and change their attitude towards math. My tutoring styles are all about learning focus and have fun with math as playing game. After I meet new student, our game is all about solving puzzles, solving math questions is like playing a game, if we find a right solution; we win, if not keep playing, challenging ourselves until we solve it.\nMy teaching philosophy: Setting a good example; helping and encouraging students to ask any question to help them grow; creating an environment to engage, motivate, and make them more competitive; implementing a learning focus state where and how the mathematics apply in the decision making, engineering, and any real-life situation by guiding them to construct their own way of creating and solving any problem without breaking any math rule\n\n### Education\n\nChapman/Brandman University\nMathematics\nCapella University\nMasters\nUnversity of West Florida\nMasters\n\n### Policies\n\n• Hourly Rate: \\$55\n• Rate details: I usually offer some special discounts for \"Abesha\" and a group study. Please feel free to contact me for the details.\n• Lesson cancellation: 12 hours notice required\n• Background check passed on 3/7/2014\n\n• Your first lesson is backed by our Good Fit Guarantee\n\nSun\n\nMon\n\nTue\n\nWed\n\nThu\n\nFri\n\nSat\n\n### Subjects\n\nGRE, Microsoft Excel, Microsoft PowerPoint, Microsoft Word\n\n#### Computer\n\nMicrosoft Excel, Microsoft PowerPoint, Microsoft Word\n\nStatistics,\n\n#### Statistics\n\nStatistics or probability: Descriptive Statistics, Data Analysis, Graphic Representations, Measures of Central Tendency, Dispersion, Position, Regression and Correlation, Probability, Random Variables, Probability Distributions for Discrete and Continuous Random Variables, Inferential Statistics, Sampling and Sampling Distributions, Central Limit Theorem, Confidence Intervals, Hypothesis Testing, Inference Concerning Correlation and Regression, Analysis of Variance, Categorical Data Analysis, Chi-square, Contingency Tables, Homogeneity tests, Decision Theory, Process, and Quality Control (Control Charts)\nMicrosoft Excel, Microsoft PowerPoint, Microsoft Word\n\nElementary Math\n\nAlgebra 1,\n\n#### Algebra 1\n\nAlgebra is frequently called the gatekeeper subject. I always have a privilege to tutor/teach some critical courses like Algebra I. It is the foundation of math such as real number system, addition and subtraction, multiplication and division using fractional notation and decimal notation, ratio and proportion, percent notation, dimensional analysis, number systems other than base 10, order of operations, exponents, set operations, venn diagrams, logic; truth tables, conditionals and Bi-conditionals and most important linear function.\nAlgebra 2,\n\n#### Algebra 2\n\nAlgebra II and college algebra have similar curricula. I have been tutoring and teaching Algebra II/college Algebra for the last 11 years, and also taught some home-study high school students utilizing research-based techniques for more personalized approaches to understanding Algebra I & II. Some Algebra II curricula overlap with Algebra I such as equation/inequalities, linear, quadratic, polynomial, rational, exponential, and logarithmic, inverse function, rational expression, complex number, and matrices.\nCalculus,\n\n#### Calculus\n\nSingle Variable (Calculus I and II): Limits, limit theorems, L'Hopital's Rule, Continuity; intermediate value theorem, Derivatives, chain rule, implicit differentiation, Applications of Derivatives; Analysis of Graphs, mean value theorem, max/min, Related rates, Integrals, Integration techniques; polynomials, exponential and logs, trigonometric, substitution, trig substitution, parts, partial fractions, Applications of Integrals, length of curves, work, volume, surface area - Parametric equations, Polar coordinates, Sequences and Series, Convergence test, Power series, and Taylor series. Multivariable Calculus (Calculus III): Vectors, dot product, cross product, lines, planes, Vector-Valued Functions, limits, derivatives, curves, tangents, curvature, Partial Derivatives, chain rule, directional derivatives, gradient, Lagrange multipliers, Multiple Integrals, surface area, polar & cylindrical coordinates, moments and center of gravity, Vector Calculus, vector fields, line integrals, Green's theorem, surface integrals, Stokes Theorem, and Divergence Theorem.\nPrecalculus,\n\n#### Precalculus\n\nAlgebra, College Algebra, or Pre-calculus: Linear equations, Absolute value, Inequalities, Nonlinear Functions, Function notation, Polynomials, Translations, Shifting, reflecting graphs, Composition, Inverse functions - Roots of polynomials, Quadratic equation, Factoring, Complete square, Quadratic formula, Complex Numbers, Higher Degree, Factoring special cases, Exponentials and Logarithmic function and Graphs, 2 equations 2 unknowns, Matrix algebra, Gaussian elimination, Inverses, Mathematics of finance, Simple/compound interest, Linear programming, and Geometric approach.\nStatistics,\n\n#### Statistics\n\nStatistics or probability: Descriptive Statistics, Data Analysis, Graphic Representations, Measures of Central Tendency, Dispersion, Position, Regression and Correlation, Probability, Random Variables, Probability Distributions for Discrete and Continuous Random Variables, Inferential Statistics, Sampling and Sampling Distributions, Central Limit Theorem, Confidence Intervals, Hypothesis Testing, Inference Concerning Correlation and Regression, Analysis of Variance, Categorical Data Analysis, Chi-square, Contingency Tables, Homogeneity tests, Decision Theory, Process, and Quality Control (Control Charts)\nGeometry, Prealgebra, SAT Math\n\nACT Math,\n\n#### ACT Math\n\nACT math covers a lot of concepts. At my first tutoring session, I use authentic test materials to diagnose a student's current level, customize my curriculum based on the student's strengths and weaknesses and then take more time to practice the weakest area.\nAlgebra 1,\n\n#### Algebra 1\n\nAlgebra is frequently called the gatekeeper subject. I always have a privilege to tutor/teach some critical courses like Algebra I. It is the foundation of math such as real number system, addition and subtraction, multiplication and division using fractional notation and decimal notation, ratio and proportion, percent notation, dimensional analysis, number systems other than base 10, order of operations, exponents, set operations, venn diagrams, logic; truth tables, conditionals and Bi-conditionals and most important linear function.\nAlgebra 2,\n\n#### Algebra 2\n\nAlgebra II and college algebra have similar curricula. I have been tutoring and teaching Algebra II/college Algebra for the last 11 years, and also taught some home-study high school students utilizing research-based techniques for more personalized approaches to understanding Algebra I & II. Some Algebra II curricula overlap with Algebra I such as equation/inequalities, linear, quadratic, polynomial, rational, exponential, and logarithmic, inverse function, rational expression, complex number, and matrices.\nCalculus,\n\n#### Calculus\n\nSingle Variable (Calculus I and II): Limits, limit theorems, L'Hopital's Rule, Continuity; intermediate value theorem, Derivatives, chain rule, implicit differentiation, Applications of Derivatives; Analysis of Graphs, mean value theorem, max/min, Related rates, Integrals, Integration techniques; polynomials, exponential and logs, trigonometric, substitution, trig substitution, parts, partial fractions, Applications of Integrals, length of curves, work, volume, surface area - Parametric equations, Polar coordinates, Sequences and Series, Convergence test, Power series, and Taylor series. Multivariable Calculus (Calculus III): Vectors, dot product, cross product, lines, planes, Vector-Valued Functions, limits, derivatives, curves, tangents, curvature, Partial Derivatives, chain rule, directional derivatives, gradient, Lagrange multipliers, Multiple Integrals, surface area, polar & cylindrical coordinates, moments and center of gravity, Vector Calculus, vector fields, line integrals, Green's theorem, surface integrals, Stokes Theorem, and Divergence Theorem.\nDifferential Equations,\n\n#### Differential Equations\n\nI have taken two Differential Equations courses and one Applied Mathematics using Differential Equation to earn my Applied Mathematics Degree. Since then, I have been tutoring some students. It is very critical to learn the concepts of Differential Equation and know how to use it and when to use it. Many laws leading natural phenomena are equations involving rates at which derivatives. Basically, Equations containing derivatives are differential equations.\nLinear Algebra,\n\n#### Linear Algebra\n\nI took multiple Linear Algebra courses to earn my Applied Mathematics degree. Linear Algebra was one of my specialties while studying. Since then, I have been tutoring undergraduate and graduate level Linear algebra courses. Linear algebra is very rich subject and central to both pure and applied mathematics, concerning vector spaces, often finite or infinite dimensional, as well as Linear Mappings between such spaces.\nPrecalculus,\n\n#### Precalculus\n\nAlgebra, College Algebra, or Pre-calculus: Linear equations, Absolute value, Inequalities, Nonlinear Functions, Function notation, Polynomials, Translations, Shifting, reflecting graphs, Composition, Inverse functions - Roots of polynomials, Quadratic equation, Factoring, Complete square, Quadratic formula, Complex Numbers, Higher Degree, Factoring special cases, Exponentials and Logarithmic function and Graphs, 2 equations 2 unknowns, Matrix algebra, Gaussian elimination, Inverses, Mathematics of finance, Simple/compound interest, Linear programming, and Geometric approach.\nStatistics,\n\n#### Statistics\n\nStatistics or probability: Descriptive Statistics, Data Analysis, Graphic Representations, Measures of Central Tendency, Dispersion, Position, Regression and Correlation, Probability, Random Variables, Probability Distributions for Discrete and Continuous Random Variables, Inferential Statistics, Sampling and Sampling Distributions, Central Limit Theorem, Confidence Intervals, Hypothesis Testing, Inference Concerning Correlation and Regression, Analysis of Variance, Categorical Data Analysis, Chi-square, Contingency Tables, Homogeneity tests, Decision Theory, Process, and Quality Control (Control Charts)\nTrigonometry,\n\n#### Trigonometry\n\nI have been teaching Trigonometry as Pre-calculus requirement in Prep School for the last two years. Trigonometry is my favorite subject of all, because it is a branch of mathematics that studies only triangles and the relationships between the lengths of their sides and the angles between those sides.\nGeometry, Prealgebra, Probability, SAT Math\n\nAlgebra 1,\n\n#### Algebra 1\n\nAlgebra is frequently called the gatekeeper subject. I always have a privilege to tutor/teach some critical courses like Algebra I. It is the foundation of math such as real number system, addition and subtraction, multiplication and division using fractional notation and decimal notation, ratio and proportion, percent notation, dimensional analysis, number systems other than base 10, order of operations, exponents, set operations, venn diagrams, logic; truth tables, conditionals and Bi-conditionals and most important linear function.\nAlgebra 2,\n\n#### Algebra 2\n\nAlgebra II and college algebra have similar curricula. I have been tutoring and teaching Algebra II/college Algebra for the last 11 years, and also taught some home-study high school students utilizing research-based techniques for more personalized approaches to understanding Algebra I & II. Some Algebra II curricula overlap with Algebra I such as equation/inequalities, linear, quadratic, polynomial, rational, exponential, and logarithmic, inverse function, rational expression, complex number, and matrices.\nCalculus,\n\n#### Calculus\n\nSingle Variable (Calculus I and II): Limits, limit theorems, L'Hopital's Rule, Continuity; intermediate value theorem, Derivatives, chain rule, implicit differentiation, Applications of Derivatives; Analysis of Graphs, mean value theorem, max/min, Related rates, Integrals, Integration techniques; polynomials, exponential and logs, trigonometric, substitution, trig substitution, parts, partial fractions, Applications of Integrals, length of curves, work, volume, surface area - Parametric equations, Polar coordinates, Sequences and Series, Convergence test, Power series, and Taylor series. Multivariable Calculus (Calculus III): Vectors, dot product, cross product, lines, planes, Vector-Valued Functions, limits, derivatives, curves, tangents, curvature, Partial Derivatives, chain rule, directional derivatives, gradient, Lagrange multipliers, Multiple Integrals, surface area, polar & cylindrical coordinates, moments and center of gravity, Vector Calculus, vector fields, line integrals, Green's theorem, surface integrals, Stokes Theorem, and Divergence Theorem.\nPrecalculus,\n\n#### Precalculus\n\nAlgebra, College Algebra, or Pre-calculus: Linear equations, Absolute value, Inequalities, Nonlinear Functions, Function notation, Polynomials, Translations, Shifting, reflecting graphs, Composition, Inverse functions - Roots of polynomials, Quadratic equation, Factoring, Complete square, Quadratic formula, Complex Numbers, Higher Degree, Factoring special cases, Exponentials and Logarithmic function and Graphs, 2 equations 2 unknowns, Matrix algebra, Gaussian elimination, Inverses, Mathematics of finance, Simple/compound interest, Linear programming, and Geometric approach.\nStatistics,\n\n#### Statistics\n\nStatistics or probability: Descriptive Statistics, Data Analysis, Graphic Representations, Measures of Central Tendency, Dispersion, Position, Regression and Correlation, Probability, Random Variables, Probability Distributions for Discrete and Continuous Random Variables, Inferential Statistics, Sampling and Sampling Distributions, Central Limit Theorem, Confidence Intervals, Hypothesis Testing, Inference Concerning Correlation and Regression, Analysis of Variance, Categorical Data Analysis, Chi-square, Contingency Tables, Homogeneity tests, Decision Theory, Process, and Quality Control (Control Charts)\nGeometry, Prealgebra\n\nAlgebra 1,\n\n#### Algebra 1\n\nAlgebra is frequently called the gatekeeper subject. I always have a privilege to tutor/teach some critical courses like Algebra I. It is the foundation of math such as real number system, addition and subtraction, multiplication and division using fractional notation and decimal notation, ratio and proportion, percent notation, dimensional analysis, number systems other than base 10, order of operations, exponents, set operations, venn diagrams, logic; truth tables, conditionals and Bi-conditionals and most important linear function.\nAlgebra 2,\n\n#### Algebra 2\n\nAlgebra II and college algebra have similar curricula. I have been tutoring and teaching Algebra II/college Algebra for the last 11 years, and also taught some home-study high school students utilizing research-based techniques for more personalized approaches to understanding Algebra I & II. Some Algebra II curricula overlap with Algebra I such as equation/inequalities, linear, quadratic, polynomial, rational, exponential, and logarithmic, inverse function, rational expression, complex number, and matrices.\nCalculus,\n\n#### Calculus\n\nSingle Variable (Calculus I and II): Limits, limit theorems, L'Hopital's Rule, Continuity; intermediate value theorem, Derivatives, chain rule, implicit differentiation, Applications of Derivatives; Analysis of Graphs, mean value theorem, max/min, Related rates, Integrals, Integration techniques; polynomials, exponential and logs, trigonometric, substitution, trig substitution, parts, partial fractions, Applications of Integrals, length of curves, work, volume, surface area - Parametric equations, Polar coordinates, Sequences and Series, Convergence test, Power series, and Taylor series. Multivariable Calculus (Calculus III): Vectors, dot product, cross product, lines, planes, Vector-Valued Functions, limits, derivatives, curves, tangents, curvature, Partial Derivatives, chain rule, directional derivatives, gradient, Lagrange multipliers, Multiple Integrals, surface area, polar & cylindrical coordinates, moments and center of gravity, Vector Calculus, vector fields, line integrals, Green's theorem, surface integrals, Stokes Theorem, and Divergence Theorem.\nGED,\n\n#### GED\n\nI have a lot of experiences preparing students for the GED test. I will always start from the basic math as an assessment and then focus on each individual weakest (blind spot). My approach is tailored to each individual math level. I had some students who open the GED book with me in the first time, and I had some other students almost ready to take the exam. Regardless of your current math level, you will be well prepared to take the exam with some tutoring sessions. GED is separated into four major categories: data, statistics, and probability; algebra, functions, and patterns; number sense and operations; and measurement and geometry. I usually cover each category and have some practices to master each area and then take mix practice at the end.\nStatistics,\n\n#### Statistics\n\nStatistics or probability: Descriptive Statistics, Data Analysis, Graphic Representations, Measures of Central Tendency, Dispersion, Position, Regression and Correlation, Probability, Random Variables, Probability Distributions for Discrete and Continuous Random Variables, Inferential Statistics, Sampling and Sampling Distributions, Central Limit Theorem, Confidence Intervals, Hypothesis Testing, Inference Concerning Correlation and Regression, Analysis of Variance, Categorical Data Analysis, Chi-square, Contingency Tables, Homogeneity tests, Decision Theory, Process, and Quality Control (Control Charts)\nGeometry, SAT Math\n\nACT Math,\n\n#### ACT Math\n\nACT math covers a lot of concepts. At my first tutoring session, I use authentic test materials to diagnose a student's current level, customize my curriculum based on the student's strengths and weaknesses and then take more time to practice the weakest area.\nAFOQT,\n\n#### AFOQT\n\nI took the Air Force Officer Qualifying Test (AFOQT) and I have been mentoring and assisting numerous students in preparing for standard tests including AFOQT and Naval Officer qualifying test which is the Aviation Selection Test Batteries (ASTB). The AFOQT is a standardized test, similar to the SAT or any other standardized test. The main difference from other tests is that the AFOQT has many additional sections beyond Verbal and Quantitative. My experience tutoring AFOQT, ASTB, ASVAB, SAT, and ACT enable me to tutor the AFOQT successfully. My strategies for all standard tests: Familiarize yourself with the test structure and contents, use the right or current study guides, take a short practice test, study each content individually, take a short practice test again, identify your weakest areas in order to give more attention and study more for those areas, and then, last not least, practice, practice, and practice.\nGED,\n\n#### GED\n\nI have a lot of experiences preparing students for the GED test. I will always start from the basic math as an assessment and then focus on each individual weakest (blind spot). My approach is tailored to each individual math level. I had some students who open the GED book with me in the first time, and I had some other students almost ready to take the exam. Regardless of your current math level, you will be well prepared to take the exam with some tutoring sessions. GED is separated into four major categories: data, statistics, and probability; algebra, functions, and patterns; number sense and operations; and measurement and geometry. I usually cover each category and have some practices to master each area and then take mix practice at the end.\nPraxis,\n\n#### Praxis\n\nI passed the Mathematics Content Knowledge Praxis II exam with target score for ETS Recognition of Excellence. Since then, I prepared five students; all of them passed with good grade. Currently, I am tutoring and teaching all of the Praxis related concepts daily. The main contents of Praxis II are Algebra, Geometry, Trigonometry, Statistics/probability, Matrices, Discrete, and Calculus. It is my favorite subject and enjoy meeting, preparing, and speaking math languages with the future Mathematics teachers, and most important bring more math teachers to the school system.\nASVAB, GRE, SAT Math\n\n## Examples of Expertise\n\nAyanaw has provided examples of their subject expertise by answering 6 questions submitted by students on Wyzant’s Ask an Expert.\n\n## Ratings and Reviews\n\n### Rating\n\n5.0 (598 ratings)\n5 star\n(574)\n4 star\n(20)\n3 star\n(4)\n2 star\n(0)\n1 star\n(0)\n\n### Reviews\n\n#### Excellent Stats Tutor\n\nMy son had his first statistics lesson with Ayanaw. He found him to be a great statistics teacher and is looking forward to doing more lessons in the near future.\n\nWalid, 3 lessons with Ayanaw\n\n#### Knowledgeable and patient tutor with stubborn teenager\n\nHe was able to simply and concisely explain in easy to understand terms any and all information needed to develop a strong fundamental algebra II base. He was on time and utilized the allotted time effectively.\n\nCorbin, 2 lessons with Ayanaw\n\n#### Great Algebra Lesson\n\nAyanaw is very knowledgeable! Has the ability to garner the best out of his students showing proven methods for success. This is only the first lesson but we think there will be a dramatic improvement. Already booked for the next few sessions.\n\nTee, 22 lessons with Ayanaw\n\n#### Very knowledgeable\n\nGreat tutor, did an awesome job preparing my son for his test. I would definitely recommend for tutoring needs. Will continue to use his services for my son’s math needs.\n\nBruce, 9 lessons with Ayanaw\n\n#### Tutor with a passion for math\n\nMr. Ayanaw is teaching my 9th grade son in Algebra 1 and 2 twice a week. My son was very stress because he was very sick which cause him to get behind in math and he just couldn't understand the way his current teacher was teaching him. Mr. Ayanaw found a way for my son to understand better.\n\nSharika, 10 lessons with Ayanaw\n\n#### great tutor\n\nKnowledgeable and patient, explains lesson thoroughly. Highly recommended! I needed a tutor for my son and the response was quick and the lesson was excellent\n\nRyan, 1 lesson with Ayanaw\n\n#### Great teacher, helps understand concepts!\n\nAyanaw helped me with my calculus homework I was struggling with for college, and I left the session with a much better understanding of a topic that I previously could barely apply\n\nSarina, 2 lessons with Ayanaw\n\n#### Entering into college after recently retiring\n\nMy tutoring introduce math back into my life after so many years of not studying. The level that he introduced was very clear, attainable, and exciting. Looking forward to learn math with Ayanawa and reaching my goal. Received some homework and made aware of it's importance. Thank you Mr Ayanawa\n\nBeverly, 1 lesson with Ayanaw\n\n#### Algebra Math Project\n\nMr. Ayanaw worked with my son to complete an algebra math project. He helped my son to work through each problem on the game board and to gain more knowledge/ understanding as each problem became more complex. Mr. Ayanaw is very knowledge and wants to see the students he works with become more successful. My son did a great job on his project!\n\nJeremiah, 20 lessons with Ayanaw\n\nResponse time: 3 hours\n\n## \\$55/hour",
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"## \\$55/hour\n\n• ### Find the right fit, or your first hour is free\n\nResponse time: 3 hours"
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https://publications.mfo.de/handle/mfo/1342/browse?type=title | [
"Now showing items 1-20 of 22\n\n• #### A construction of hyperbolic Coxeter groups \n\n[OWP-2010-04] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nWe give a simple construction of Gromov hyperbolic Coxeter groups of arbitrarily large virtual cohomological dimension. Our construction provides new examples of such groups. Using this one can construct e.g. new groups ...\n• #### Fibonacci-like unimodal inverse limit spaces \n\n[OWP-2010-03] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-9)\nWe study the structure of inverse limit space of so-called Fibonacci-like tent maps. The combinatorial constraints implied by the Fibonacci-like assumption allows us to introduce certain chains that enable a more detailed ...\n• #### The ingram conjecture \n\n[OWP-2010-02] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-8)\nWe prove the Ingram Conjecture, i.e., we show that the inverse limit spaces of every two tent maps with different slopes in the interval [1,2] are non-homeomorphic. Based on the structure obtained from the proof, we also ...\n• #### Locally conformally Kähler manifolds admitting a holomorphic conformal flow \n\n[OWP-2010-13] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-15)\nA manifold $M$ is locally conformally Kähler (LCK) if it admits a Kähler covering $\\tilde{M}$ with monodromy acting by holomorphic homotheties. Let $M$ be an LCK manifold admitting a holomorphic conformal flow of ...\n• #### Non-integrated defect relation for meromorphic maps of complete Kähler manifolds into a projective variety intersecting hypersurfaces \n\n[OWP-2010-19] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nIn 1985, Fujimoto established a non-integrated defect relation for meromorphic maps of complete Kähler manifolds into the complex projective space intersecting hyperplanes in general position. In this paper, we generalize ...\n• #### On Generalizations of Kac-Moody Groups \n\n[OWP-2010-06] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-10)\nIn we define a Curtis-Tits group as a certain generalization of a Kac-Moody group. We distinguish between orientable and non-orientable Curtis-Tits groups and identify all orientable Curtis-Tits groups as Kac-Moody ...\n• #### On Selfinjective Artin Algebras Having Generalized Standard Quasitubes \n\n[OWP-2010-18] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-18)\nWe give a complete description of the Morita equivalence classes of all connected selfinjective artin algebras for which the Auslander-Reiten quiver admits a family of quasitubes having common composition factors, closed ...\n• #### On the complement of the dense orbit for a quiver of type A \n\n[OWP-2010-11] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nLet $\\mathbb{A}_t$ be the directed quiver of type $A$ with $t$ vertices. For each dimension vector $d$ there is a dense orbit in the corresponding representation space. The principal aim of this note is to use just rank ...\n• #### On the Complement of the Richardson Orbit \n\n[OWP-2010-09] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-13)\nWe consider parabolic subgroups of a general algebraic group over an algebraically closed field $k$ whose Levi part has exactly $t$ factors. By a classical theorem of Richardson, the nilradical of a parabolic subgroup $P$ ...\n• #### Plethysms, replicated Schur functions and series, with applications to vertex operators \n\n[OWP-2010-12] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-14)\nSpecializations of Schur functions are exploited to define and evaluate the Schur functions $s_\\lambda [\\alpha X]$ and plethysms $s_\\lambda [\\alpha s_\\nu(X))]$ for any $\\alpha$-integer, real or complex. Plethysms are then ...\n• #### Representation Theory of Non-Commutative Association Schemes of Order \n\n[OWP-2010-07] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-11)\nIn this note, we investigate association schemes of order 6. We prove that non-normal closed subsets of such schemes have order 2 and that normal closed subsets of non-commutative schemes have order 2 or 3. After that, ...\n• #### Semi-Invertible Extensions of C*-Algebras \n\n[OWP-2010-17] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-17)\nWe prolonge the list of $C^*$-algebras for which all extensions by any stable separable $C^*$-algebra are semi-invertible. In particular, we handle certain amalgamations, both of $C^*$-algebras and of groups. Concerning ...\n• #### A series of algebras generalizing the Octonions and Hurwitz-Radon Identity \n\n[OWP-2010-10] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nWe study non-associative twisted group algebras over $(\\mathbb{Z}_2)^n$ with cubic twisting functions. We construct a series of algebras that extend the classical algebra of octonions in the same way as the Clifford algebras ...\n• #### Shape Theory and Extensions of C*-Algebras \n\n[OWP-2010-16] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-16)\nLet A, A' be separable $C^*$-algebras, $B$ a stable $\\sigma$-unital $C^*$-algebra. Our main result is the construction of the pairing $[[A', A]] \\times Ext^{-1/2}(A,B) \\to Ext^{-1/2}(A',B)$, where $[[A', A]]$ denotes the ...\n• #### Spectral Sequences in Combinatorial Geometry: Cheeses, Inscribed Sets, and Borsuk-Ulam Type Theorems \n\n[OWP-2010-08] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-12)\nAlgebraic topological methods are especially suited to determining the nonexistence of continuous mappings satisfying certain properties. In combinatorial problems it is sometimes possible to define a mapping from a space ...\n• #### Stochastic mean payoff game: smoothed analysis and approximation schemes \n\n[OWP-2010-22] (Mathematisches Forschungsinstitut Oberwolfach, 2010-03-20)\nWe consider two-person zero-sum stochastic mean payoff games with perfect information modeled by a digraph with black, white, and random vertices. These BWR-games games are polynomially equivalent with the classical Gillette ...\n• #### Supertropical linear algebra \n\n[OWP-2010-14] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nThe objective of this paper is to lay out the algebraic theory of supertropical vector spaces and linear algebra, utilizing the key antisymmetric relation of \"ghost surpasses.\" Special attention is paid to the various ...\n• #### Supertropical Matrix Algebra III : Powers of Matrices and Generalized Eigenspaces \n\n[OWP-2010-20] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nWe investigate powers of supertropical matrices, with special attention to the role of the coefficients of the supertropical characteristic polynomial (especially the supertropical trace) in controlling the rank of a power ...\n• #### Supertropical semirings and supervaluations \n\n[OWP-2010-05] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nWe interpret a valuation $\\upsilon$ on a ring $R$ as a map $\\upsilon:R \\rightarrow M$ into a so called bipotent semiring $M$ (the usual max-plus setting), and then define a supervaluation $\\varphi$ as a suitable map into ...\n• #### There is a unique real tight contact 3-ball \n\n[OWP-2010-01] (Mathematisches Forschungsinstitut Oberwolfach, 2010)\nWe prove that there is a unique real tight contact structure on the 3-ball with convex boundary up to isotopy through real tight contact structures. We also give a partial classification of the real tight solid tori with ..."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6998167,"math_prob":0.8987959,"size":7222,"snap":"2019-51-2020-05","text_gpt3_token_len":1971,"char_repetition_ratio":0.17678027,"word_repetition_ratio":0.01904762,"special_character_ratio":0.2360842,"punctuation_ratio":0.115869015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9518284,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T23:44:51Z\",\"WARC-Record-ID\":\"<urn:uuid:9b54fbe7-2d43-4ef7-8b1e-c09124af2a95>\",\"Content-Length\":\"56176\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57c6d61e-72d7-4c74-8343-288db600ca68>\",\"WARC-Concurrent-To\":\"<urn:uuid:eec7928a-8ff6-4424-9aa9-a7bb3605bc13>\",\"WARC-IP-Address\":\"193.174.3.47\",\"WARC-Target-URI\":\"https://publications.mfo.de/handle/mfo/1342/browse?type=title\",\"WARC-Payload-Digest\":\"sha1:N6TBX4WBIHWUDWOQZW3EBZLZ77EWJ3CJ\",\"WARC-Block-Digest\":\"sha1:WUVEB4YSM6YO4YXCGW7DLB7EQENYHEQD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250614086.44_warc_CC-MAIN-20200123221108-20200124010108-00236.warc.gz\"}"} |
https://inches-to-feet.appspot.com/7.8-inches-to-feet.html | [
"Inches To Feet\n\n# 7.8 in to ft7.8 Inches to Feet\n\nin\n=\nft\n\n## How to convert 7.8 inches to feet?\n\n 7.8 in * 0.0833333333 ft = 0.65 ft 1 in\nA common question is How many inch in 7.8 foot? And the answer is 93.6 in in 7.8 ft. Likewise the question how many foot in 7.8 inch has the answer of 0.65 ft in 7.8 in.\n\n## How much are 7.8 inches in feet?\n\n7.8 inches equal 0.65 feet (7.8in = 0.65ft). Converting 7.8 in to ft is easy. Simply use our calculator above, or apply the formula to change the length 7.8 in to ft.\n\n## Convert 7.8 in to common lengths\n\nUnitLength\nNanometer198120000.0 nm\nMicrometer198120.0 µm\nMillimeter198.12 mm\nCentimeter19.812 cm\nInch7.8 in\nFoot0.65 ft\nYard0.2166666667 yd\nMeter0.19812 m\nKilometer0.00019812 km\nMile0.0001231061 mi\nNautical mile0.0001069762 nmi\n\n## What is 7.8 inches in ft?\n\nTo convert 7.8 in to ft multiply the length in inches by 0.0833333333. The 7.8 in in ft formula is [ft] = 7.8 * 0.0833333333. Thus, for 7.8 inches in foot we get 0.65 ft.\n\n## 7.8 Inch Conversion Table",
null,
"## Alternative spelling\n\n7.8 Inches to ft, 7.8 Inches in ft, 7.8 in to Foot, 7.8 in in Foot, 7.8 Inch to Feet, 7.8 Inch in Feet, 7.8 Inch to ft, 7.8 Inch in ft, 7.8 in to Feet, 7.8 in in Feet, 7.8 Inch to Foot, 7.8 Inch in Foot, 7.8 Inches to Feet, 7.8 Inches in Feet"
] | [
null,
"https://inches-to-feet.appspot.com/image/7.8.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89854044,"math_prob":0.849534,"size":909,"snap":"2022-27-2022-33","text_gpt3_token_len":353,"char_repetition_ratio":0.24309392,"word_repetition_ratio":0.021276595,"special_character_ratio":0.420242,"punctuation_ratio":0.2260274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95003366,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T05:49:38Z\",\"WARC-Record-ID\":\"<urn:uuid:a98f2fce-4d9f-4036-9744-d9cb4a0721b0>\",\"Content-Length\":\"27549\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90c102f0-3a2e-4d60-98c4-1e9f576f0be1>\",\"WARC-Concurrent-To\":\"<urn:uuid:8893781b-1c67-4d6f-98a5-b5d7cea55c68>\",\"WARC-IP-Address\":\"142.251.16.153\",\"WARC-Target-URI\":\"https://inches-to-feet.appspot.com/7.8-inches-to-feet.html\",\"WARC-Payload-Digest\":\"sha1:F46JT4LGJCJOBA4YAYFIKHWGBIDBRROP\",\"WARC-Block-Digest\":\"sha1:ZDBYFBDFG6M53EZXMRMQ77MO5IUI3TTP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571909.51_warc_CC-MAIN-20220813051311-20220813081311-00182.warc.gz\"}"} |
https://questions.llc/questions/1745676/a-wire-of-5-0m-and-diameter-2-0mm-extends-by-0-25mm-when-a-force-of-50n-was-stretch-it | [
"# A wire of 5.0m and diameter 2.0mm,extends by 0.25mm when a force of 50N was stretch it from it's end calculate\n\na.the stress on the wire. b.the strain in the wire. c.the young's modulus of elasticity of the wire\n\n1. 👍\n2. 👎\n3. 👁\n4. ℹ️\n5. 🚩\n\n1. L = 5\nd = 2*10^-3\nr = 10^-3\nA = pi r^2 = pi (10^-6)\n\nstress = 50 N / 3.14*10^-6 = 15.9 * 10^6 Newtons/ m^2\nstrain = delta L/L = .25*10^-3 / 5 = .05 *10^-3 = 50 * 10^-6 m\n\nE = stress / strain\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n2. Tanks a lot\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n3. A wire of 5.0m and diameter 2.0mm,extends by 0.25mm when a force of 50N was stretch it from it's end calculate\na.the stress on the wire. b.the strain in the wire. c.the young's modulus of elasticity of the wire\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n4. It is very good and correct\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72303915,"math_prob":0.9987142,"size":330,"snap":"2022-40-2023-06","text_gpt3_token_len":162,"char_repetition_ratio":0.22392638,"word_repetition_ratio":0.39759037,"special_character_ratio":0.38787878,"punctuation_ratio":0.10714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99833125,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T17:34:33Z\",\"WARC-Record-ID\":\"<urn:uuid:4c103906-7f9c-4a4a-992e-02132f164c2a>\",\"Content-Length\":\"27111\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f424ce7-4d87-4766-8eb7-ea6897695e9f>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cb5de05-5f99-4118-af34-d4906fed7cd3>\",\"WARC-IP-Address\":\"45.79.29.166\",\"WARC-Target-URI\":\"https://questions.llc/questions/1745676/a-wire-of-5-0m-and-diameter-2-0mm-extends-by-0-25mm-when-a-force-of-50n-was-stretch-it\",\"WARC-Payload-Digest\":\"sha1:MOZSDHCR3BSIIB7MENYNF5UCHZDAY2LG\",\"WARC-Block-Digest\":\"sha1:NNM4RTEESZ3F3GA256LNYDLWG66E7DKY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337428.0_warc_CC-MAIN-20221003164901-20221003194901-00437.warc.gz\"}"} |
https://www.physicsforums.com/threads/circular-motion.791250/ | [
"# Circular motion\n\nGood afternoon!\nExercise: Prove that if a body moves under the action of a force F = k⋅u × v, where u is an arbitrary unit vector and v the velocity, the motion is circular with angular velocity ω = k ⋅ u or, in a more general case, a spiral parallel to u.\n\nSource: (Alonso & Finn: Fundamental university physics)\n\nI attempted to prove that F is a central force, so that the angular momentum is constant. Is that wrong?\n\nRelated Introductory Physics Homework Help News on Phys.org\nvela\nStaff Emeritus\nHomework Helper\nWould showing F is a central force buy you anything? The force of the Sun on a planet is a central force. but the orbits aren't circular.\n\nharuspex\nHomework Helper\nGold Member\nI attempted to prove that F is a central force\nThat won't do it. F = k.v would satisfy that. Get an equation for the acceleration.\n\nThank you for your responses! In order to prove that it's motion is circular wouldn't it be sufficient to prove that ω is constant?\nCould you \"boost\" me a bit?\n\nharuspex"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8983477,"math_prob":0.86741537,"size":882,"snap":"2020-10-2020-16","text_gpt3_token_len":228,"char_repetition_ratio":0.107061505,"word_repetition_ratio":0.9132948,"special_character_ratio":0.2573696,"punctuation_ratio":0.13297872,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9842989,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-29T16:37:12Z\",\"WARC-Record-ID\":\"<urn:uuid:4b99ce38-2e1c-4ff4-94c6-f6c74bf283db>\",\"Content-Length\":\"77814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98f08b20-3090-4c0c-bf3c-1d3c8db7fef5>\",\"WARC-Concurrent-To\":\"<urn:uuid:da8aabc0-2c6f-4d8d-ac5c-b72ce15fa265>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/circular-motion.791250/\",\"WARC-Payload-Digest\":\"sha1:6TLDMEVWCTV57QO2J2YOERMJS3KOEYEI\",\"WARC-Block-Digest\":\"sha1:5ZCK7I4LCRQXZP67PMCO7WDQVWQOCVSE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370494349.3_warc_CC-MAIN-20200329140021-20200329170021-00199.warc.gz\"}"} |
https://www.quantumstudy.com/a-boy-throws-n-balls-per-second-at-regular-time-intervals-when-the-first-ball-reaches-the-maximum-height-he-throws-the-second-one-vertically-up-the-maximum-height/ | [
"# A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws …..\n\nQ. A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is\n\n(a) $\\displaystyle \\frac{g}{2(n-1)^2}$\n\n(b) $\\displaystyle \\frac{g}{2n^2}$\n\n(c) $\\displaystyle \\frac{g}{n^2}$\n\n(d) $\\displaystyle \\frac{g}{n}$\n\nAns: (b)\nSol:\nTime taken by ball to go maximum height\nt = 1/n\n\nv = u – gt\n\n0 = u – g(1/n)\n\nu = g/n = velocity with which ball was projected\n\nMaximum height reached by ball is\n\n$\\displaystyle H = \\frac{u^2}{2g}$\n\n$\\displaystyle H = \\frac{g}{2n^2}$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77615434,"math_prob":0.99975735,"size":581,"snap":"2023-40-2023-50","text_gpt3_token_len":193,"char_repetition_ratio":0.18717505,"word_repetition_ratio":0.0,"special_character_ratio":0.34595525,"punctuation_ratio":0.04385965,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99976104,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T13:21:24Z\",\"WARC-Record-ID\":\"<urn:uuid:fb3ae5ce-bab9-4a09-ae19-e0f5527d7f85>\",\"Content-Length\":\"153536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f6d298a-f3ec-43ad-b204-d744558ee39d>\",\"WARC-Concurrent-To\":\"<urn:uuid:867cf3a5-184a-4c5f-8411-4525c89035c2>\",\"WARC-IP-Address\":\"103.133.214.171\",\"WARC-Target-URI\":\"https://www.quantumstudy.com/a-boy-throws-n-balls-per-second-at-regular-time-intervals-when-the-first-ball-reaches-the-maximum-height-he-throws-the-second-one-vertically-up-the-maximum-height/\",\"WARC-Payload-Digest\":\"sha1:V4OTE5F3R75IMDJWFJDEQYOXNIUYYZRA\",\"WARC-Block-Digest\":\"sha1:F6JRWCXFYRYSYHZSDCXQWG7YAPDBOXHA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100599.20_warc_CC-MAIN-20231206130723-20231206160723-00713.warc.gz\"}"} |
https://math.stackexchange.com/questions/3252537/computation-of-function-with-sup | [
"# Computation of function with sup\n\nI am trying to compute the value of the following function $$R:\\mathbb{R}^m\\mapsto\\mathbb{R}$$\n\n$$R_n(\\theta)=\\sup_{\\lambda\\in\\mathbb{R}^m}\\left\\{-\\frac{1}{n}\\sum_{i=1}^{n}\\sup_{x\\in\\mathbb{R}^m}\\lbrace \\lambda^{\\mathsf{T}}(Y_i-x^{\\mathsf{T}}\\theta)x-||x-X_i||^2\\rbrace\\right\\}$$\n\nwhere $$Y_i\\in\\mathbb{R}$$ and $$X_i\\in\\mathbb{R}^m$$ for all $$i=1,\\ldots,n$$\n\nMy question is how to simplify this expression or in any case how to approximate it or at least sketch an algorithm to do so."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5919405,"math_prob":0.99996746,"size":475,"snap":"2019-35-2019-39","text_gpt3_token_len":188,"char_repetition_ratio":0.15711252,"word_repetition_ratio":0.0,"special_character_ratio":0.3326316,"punctuation_ratio":0.03960396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000085,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T02:25:18Z\",\"WARC-Record-ID\":\"<urn:uuid:e1fb0da4-98fe-4def-a349-82fa8962ab16>\",\"Content-Length\":\"125176\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bbbf616-7254-4542-8282-833380909561>\",\"WARC-Concurrent-To\":\"<urn:uuid:76c03ce1-b7f8-4947-9d5f-ded68328bfb4>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3252537/computation-of-function-with-sup\",\"WARC-Payload-Digest\":\"sha1:ETRZWEVDQGNSWMMAAR45RBJME22T5B6E\",\"WARC-Block-Digest\":\"sha1:24COCXDI6XIIOSEFNNH25G5AHR2AY67C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314638.49_warc_CC-MAIN-20190819011034-20190819033034-00199.warc.gz\"}"} |
http://www.120a6.cn/vps/15326.html | [
"# Java双指针法怎么使用\n\n### 1.判断链表是否有环",
null,
"```public class Solution {\n//快慢指针法\nwhile(fast != null && fast.next != null){\nfast = fast.next.next;\nlow = low.next;\n//此时相遇了\nif(fast == low){\nreturn true;\n}\n}\nreturn false;\n}\n}```\n\n### 2.查找链表中间的元素",
null,
"``` //快慢指针法\nwhile(fast != null && fast.next != null){\n//慢指针走一步\nlow = low.next;\n//快指针走两步\nfast = fast.next.next;\n}\n//奇数，fast.next = null时，low即为所求\n//偶数，fsat == null时，low即为所求\nreturn low;\n}```\n\n### 3.奇偶排序前奇后偶",
null,
"``` public ListNode oddEvenList(ListNode head) {\n}\nwhile(fastTail != null && fastTail.next != null){\n//更新奇数节点时，奇数节点的后一个节点需要指向偶数节点的后一个节点\nlowTail.next = fastTail.next;\nlowTail = lowTail.next;\n// 更新偶数节点时，偶数节点的后一个节点需要指向奇数节点的后一个节点\nfastTail.next = lowTail.next;\nfastTail = fastTail.next;\n}\n//合并两个链表\n}```\n\n### 4.删除排序链表的重复元素",
null,
"``` public ListNode deleteDuplicates(ListNode head) {\n//虚拟头节点法\nListNode cur = prev.next;\nwhile(cur != null){\n//引入next指针\nListNode next = cur.next;\nif(next == null){\n//只有一个元素\n}\nif(cur.val != next.val){\n//cur不是重复节点，三指针都移动\ncur = cur.next;\nprev = prev.next;\n}else{\n//让next指针一直向后移动，直到与cur.val不相等的节点位置\nwhile(next != null && cur.val == next.val){\nnext = next.next;\n}\n// 此时next指向了第一个不重复的元素\n// 此时prev - next之间所有重复元素全部删除\nprev.next = next;\ncur = next;\n}\n}\n}```\n\n### 5.三数之和",
null,
"``` public List<List<Integer>> threeSum(int[] nums) {\nint n = nums.length;\nArrays.sort(nums);\nList<List<Integer>> ans = new ArrayList<List<Integer>>();\n// 枚举 a\nfor (int first = 0; first < n; ++first) {\n// 需要和上一次枚举的数不相同\nif (first > 0 && nums[first] == nums[first - 1]) {\ncontinue;\n}\n// c 对应的指针初始指向数组的最右端\nint third = n - 1;\nint target = -nums[first];\n// 枚举 b\nfor (int second = first + 1; second < n; ++second) {\n// 需要和上一次枚举的数不相同\nif (second > first + 1 && nums[second] == nums[second - 1]) {\ncontinue;\n}\n// 需要保证 b 的指针在 c 的指针的左侧\nwhile (second < third && nums[second] + nums[third] > target) {\n--third;\n}\n// 如果指针重合，随着 b 后续的增加\n// 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环\nif (second == third) {\nbreak;\n}\nif (nums[second] + nums[third] == target) {\nList<Integer> list = new ArrayList<Integer>();\n}\n}\n}\nreturn ans;\n}```\n\n### 6.分割链表",
null,
"``` public ListNode partition(ListNode head, int x) {\n// 创建small和big两个小链表的头节点\n// 按照升序插入，因此需要尾插\n// 分别指向两个子链表的尾部\n//遍历原链表，根据值放入small和big链表中\n}else{\n}\n}\nbigTail.next = null;\n//拼接小链表和大链表\n}```\n\n### 7.合并两个有序的数组",
null,
"```public void merge(int[] nums1, int m, int[] nums2, int n) {\nint p1 = 0, p2 = 0;\nint[] sorted = new int[m + n];\nint cur;\nwhile (p1 < m || p2 < n) {\nif (p1 == m) {\ncur = nums2[p2++];\n} else if (p2 == n) {\ncur = nums1[p1++];\n} else if (nums1[p1] < nums2[p2]) {\ncur = nums1[p1++];\n} else {\ncur = nums2[p2++];\n}\nsorted[p1 + p2 - 1] = cur;\n}\nfor (int i = 0; i != m + n; ++i) {\nnums1[i] = sorted[i];\n}\n}```\n\n### 8.两数之和—输入有序数组",
null,
"``` public int[] twoSum(int[] numbers, int target) {\nint low = 0, high = numbers.length - 1;\nwhile (low < high) {\nint sum = numbers[low] + numbers[high];\nif (sum == target) {\nreturn new int[]{low + 1, high + 1};\n} else if (sum < target) {\n++low;\n} else {\n--high;\n}\n}\nreturn new int[]{-1, -1};\n}```\n\n### 9.长度最小的子数组\n\n（力扣209）给定一个含有 n 个正整数的数组和一个正整数 target 。",
null,
"```//滑动窗口法\npublic int minSubArrayLen(int s, int[] nums) {\nint n = nums.length;\nif (n == 0) {\nreturn 0;\n}\nint ans = Integer.MAX_VALUE;\nint start = 0, end = 0;\nint sum = 0;\nwhile (end < n) {\nsum += nums[end];\nwhile (sum >= s) {\nans = Math.min(ans, end - start + 1);\nsum -= nums[start];\nstart++;\n}\nend++;\n}\nreturn ans == Integer.MAX_VALUE ? 0 : ans;\n}```\n\n### 10.排序链表",
null,
"1，分割 cut 环节： 找到当前链表中点，并从中点将链表断开（以便在下次递归 cut 时，链表片段拥有正确边界）； 我们使用 fast,slow 快慢双指针法，奇数个节点找到中点，偶数个节点找到中心左边的节点。 找到中点 slow 后，执行 slow.next = None 将链表切断。 递归分割时，输入当前链表左端点 head 和中心节点 slow 的下一个节点 tmp(因为链表是从 slow 切断的)。 cut 递归终止条件： 当head.next == None时，说明只有一个节点了，直接返回此节点。\n\n2，合并 merge 环节： 将两个排序链表合并，转化为一个排序链表。 双指针法合并，建立辅助ListNode h 作为头部。 设置两指针 left, right 分别指向两链表头部，比较两指针处节点值大小，由小到大加入合并链表头部，指针交替前进，直至添加完两个链表。 返回辅助ListNode h 作为头部的下个节点 h.next。\n\n```public ListNode sortList(ListNode head) {\nwhile (fast != null && fast.next != null) {\nslow = slow.next;\nfast = fast.next.next;\n}\nListNode tmp = slow.next;\nslow.next = null;\nListNode right = sortList(tmp);\nListNode h = new ListNode(0);\nListNode res = h;\nwhile (left != null && right != null) {\nif (left.val < right.val) {\nh.next = left;\nleft = left.next;\n} else {\nh.next = right;\nright = right.next;\n}\nh = h.next;\n}\nh.next = left != null ? left : right;\nreturn res.next;\n}```"
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