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https://zbmath.org/?q=an%3A0806.35112 | [
"## Hyperbolic conservation laws with stiff relaxation terms and entropy.(English)Zbl 0806.35112\n\nSummary: We study the limiting behavior of systems of hyperbolic conservation laws with stiff relaxation terms. Reduced systems, inviscid and viscous local conservation laws, and weakly nonlinear limits are derived through asymptotic expansions. An entropy condition is introduced for $$N \\times N$$ systems that ensures the hyperbolicity of the reduced inviscid system. The resulting characteristic speeds are shown to be interlaced with those of the original system. Moreover, the first correction to the reduced system is shown to be dissipative. A partial converse is proved for $$2 \\times 2$$ systems. This structure is then applied to study the convergence to the reduced dynamics for the $$2 \\times 2$$ case.\n\n### MSC:\n\n 35L65 Hyperbolic conservation laws 35B40 Asymptotic behavior of solutions to PDEs\n\n### Keywords:\n\nstiff relaxation terms; entropy condition\nFull Text:\n\n### References:\n\n Bardos, Comm. Pure Appl. Math. 46 pp 667– (1993) , and , Fluid dynamics limits of discrete velocity kinetic equations, pp. 57–71 in: Advances in Kinetic Theory and Continuum Mechanics, and , eds., Springer-Verlag, Berlin-New York. 1991. Caflisch, Comm. Math. Phys. 114 pp 103– (1988) Caflisch, Comm. Pure Appl. Math. 32 pp 589– (1979) The Boltzmann Equation and Its Applications, Springer-Verlag. New York, 1988. The compensated compactness method and the system of isentropic gas dynamics, preprint MSRI-00527-91, Mathematical Science Research Institute, Berkeley, 1990. Chen, Arch. Rational Mech. Anal. 121 pp 131– (1992) Chen, Comm. Pure Appl. Math. 46 pp 755– (1993) Chen, Chinese Sci. Bull. 34 pp 15– (1989) Chinese Sci. Bull. 9 pp 641– (1988) Chueh, Indiana Univ. Math. J. 26 pp 372– (1977) Dafermos, LCDS Lecture Notes 85 (1985) , and , Convergence of the Lax-Friedrichs schemes for the isentropic gas dynamics (I), Acta Math. Sci. (English Ed.) 5, 1985, pp. 415–432; · Zbl 0643.76084 Acta Math. Sci. (Chinese) 7 pp 467– (1987) DiPerna, Arch. Rational Mech. Anal. 82 pp 27– (1983) Glimm, Lecture Notes in Phys. 344 pp 177– (1986) Greenberg, Arch Rational Mech. Anal. 82 pp 87– (1983) and , Numerical Schemes for Hyperbolic Conservation Laws with Stiff Relaxation Terms, submitted to J. Comput. Phys., 1993. Statistical Thermodynamics of Nonequilibrium Processes, Springer-Verlag, New York, 1987. Hyperbolic Systems of Conservation Laws and the Mathematical Theory of Shock Waves, SIAM, Philadelphia, 1973. Fluid dynamical limits of discrete kinetic theories, pp. 173–185 in: Macroscopic Simulations of Complex Hydrodynamic Phenomena, and , eds., NATO ASI Series B 292, Plenum Press, New York, 1992. Liu, Corn. Math. Phys. 108 pp 153– (1987) Murat, J. Math. Pures Appl. 60 pp 309– (1981) , and , Mathematical Problems in Viscoelasticity, Pitman Monographs and Surveys in Pure and Applied Mathematics 35, Longman Scientific & Technical, Essex, England 1987. Rabie, Phys. Fluids A 22 pp 422– (1979) Slemrod, Arch. Rational Mech. Anal. 122 pp 353– (1993) Water Waves, Interscience, New York, 1957. Compensated compactness and applications to partial differential equations, pp. 136–212 in: Research Notes in Mathematics, Nonlinear Analysis and Mechanics: Heriot-Watt Symposium 4, ed., Pitman Press, New York, 1979. and , Introduction to Physical Gas Dynamics, Wiley, New York, 1965. Linear and Nonlinear Waves, Wiley, New York, 1974. Xin, Comm. Pure Appl. Math. 44 pp 679– (1991)\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching."
] | [
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https://pomegranate.readthedocs.io/en/docs/MarkovChain.html | [
"# Markov Chains¶\n\nIPython Notebook Tutorial\n\nMarkov chains are form of structured model over sequences. They represent the probability of each character in the sequence as a conditional probability of the last k symbols. For example, a 3rd order Markov chain would have each symbol depend on the last three symbols. A 0th order Markov chain is a naive predictor where each symbol is independent of all other symbols. Currently pomegranate only supports discrete emission Markov chains where each symbol is a discrete symbol versus a continuous number (like ‘A’ ‘B’ ‘C’ instead of 17.32 or 19.65).\n\n## Initialization¶\n\nMarkov chains can almost be represented by a single conditional probability table (CPT), except that the probability of the first k elements (for a k-th order Markov chain) cannot be appropriately represented except by using special characters. Due to this pomegranate takes in a series of k+1 distributions representing the first k elements. For example for a second order Markov chain:\n\n>>> from pomegranate import *\n>>> d1 = DiscreteDistribution({'A': 0.25, 'B': 0.75})\n>>> d2 = ConditionalProbabilityTable([['A', 'A', 0.1],\n['A', 'B', 0.9],\n['B', 'A', 0.6],\n['B', 'B', 0.4]], [d1])\n>>> d3 = ConditionalProbabilityTable([['A', 'A', 'A', 0.4],\n['A', 'A', 'B', 0.6],\n['A', 'B', 'A', 0.8],\n['A', 'B', 'B', 0.2],\n['B', 'A', 'A', 0.9],\n['B', 'A', 'B', 0.1],\n['B', 'B', 'A', 0.2],\n['B', 'B', 'B', 0.8]], [d1, d2])\n>>> model = MarkovChain([d1, d2, d3])\n\n\n## Probability¶\n\nThe probability of a sequence under the Markov chain is just the probabiliy of the first character under the first distribution times the probability of the second character under the second distribution and so forth until you go past the (k+1)th character, which remains evaluated under the (k+1)th distribution. We can calculate the probability or log probability in the same manner as any of the other models. Given the model shown before:\n\n>>> model.log_probability(['A', 'B', 'B', 'B'])\n-3.324236340526027\n>>> model.log_probability(['A', 'A', 'A', 'A'])\n-5.521460917862246\n\n\n## Fitting¶\n\nMarkov chains are not very complicated to chain. For each sequence the appropriate symbols are sent to the appropriate distributions and maximum likelihood estimates are used to update the parameters of the distributions. There are no latent factors to train and so no expectation maximization or iterative algorithms are needed to train anything.\n\n## API Reference¶\n\nclass pomegranate.MarkovChain.MarkovChain\n\nA Markov Chain.\n\nImplemented as a series of conditional distributions, the Markov chain models P(X_i | X_i-1...X_i-k) for a k-th order Markov network. The conditional dependencies are directly on the emissions, and not on a hidden state as in a hidden Markov model.\n\nParameters: distributions : list, shape (k+1) A list of the conditional distributions which make up the markov chain. Begins with P(X_i), then P(X_i | X_i-1). For a k-th order markov chain you must put in k+1 distributions.\n\nExamples\n\n>>> from pomegranate import *\n>>> d1 = DiscreteDistribution({'A': 0.25, 'B': 0.75})\n>>> d2 = ConditionalProbabilityTable([['A', 'A', 0.33],\n['B', 'A', 0.67],\n['A', 'B', 0.82],\n['B', 'B', 0.18]], [d1])\n>>> mc = MarkovChain([d1, d2])\n>>> mc.log_probability(list('ABBAABABABAABABA'))\n-8.9119890701808213\n\n\nAttributes\n\n distributions (list, shape (k+1)) The distributions which make up the chain.\nfit()\n\nFit the model to new data using MLE.\n\nThe underlying distributions are fed in their appropriate points and weights and are updated.\n\nParameters: sequences : array-like, shape (n_samples, variable) This is the data to train on. Each row is a sample which contains a sequence of variable length weights : array-like, shape (n_samples,), optional The initial weights of each sample. If nothing is passed in then each sample is assumed to be the same weight. Default is None. inertia : double, optional The weight of the previous parameters of the model. The new parameters will roughly be old_param*inertia + new_param*(1-inertia), so an inertia of 0 means ignore the old parameters, whereas an inertia of 1 means ignore the new parameters. Default is 0.0. None\nfrom_json()\n\nRead in a serialized model and return the appropriate classifier.\n\nParameters: s : str A JSON formatted string containing the file. model : object A properly initialized and baked model.\nfrom_samples()\n\nLearn the Markov chain from data.\n\nTakes in the memory of the chain (k) and learns the initial distribution and probability tables associated with the proper parameters.\n\nParameters: X : array-like, list or numpy.array The data to fit the structure too as a list of sequences of variable length. Since the data will be of variable length, there is no set form weights : array-like, shape (n_nodes), optional The weight of each sample as a positive double. Default is None. k : int, optional The number of samples back to condition on in the model. Default is 1. model : MarkovChain The learned markov chain model.\nfrom_summaries()\n\nFit the model to the collected sufficient statistics.\n\nFit the parameters of the model to the sufficient statistics gathered during the summarize calls. This should return an exact update.\n\nParameters: inertia : double, optional The weight of the previous parameters of the model. The new parameters will roughly be old_param*inertia + new_param * (1-inertia), so an inertia of 0 means ignore the old parameters, whereas an inertia of 1 means ignore the new parameters. Default is 0.0. None\nlog_probability()\n\nCalculate the log probability of the sequence under the model.\n\nThis calculates the first slices of increasing size under the corresponding first few components of the model until size k is reached, at which all slices are evaluated under the final component.\n\nParameters: sequence : array-like An array of observations logp : double The log probability of the sequence under the model.\nsample()\n\nCreate a random sample from the model.\n\nParameters: length : int or Distribution Give either the length of the sample you want to generate, or a distribution object which will be randomly sampled for the length. Continuous distributions will have their sample rounded to the nearest integer, minimum 1. sequence : array-like, shape = (length,) A sequence randomly generated from the markov chain.\nsummarize()\n\nSummarize a batch of data and store sufficient statistics.\n\nThis will summarize the sequences into sufficient statistics stored in each distribution.\n\nParameters: sequences : array-like, shape (n_samples, variable) This is the data to train on. Each row is a sample which contains a sequence of variable length weights : array-like, shape (n_samples,), optional The initial weights of each sample. If nothing is passed in then each sample is assumed to be the same weight. Default is None. None\nto_json()\n\nSerialize the model to a JSON.\n\nParameters: separators : tuple, optional The two separaters to pass to the json.dumps function for formatting. Default is (‘,’, ‘ : ‘). indent : int, optional The indentation to use at each level. Passed to json.dumps for formatting. Default is 4. json : str A properly formatted JSON object."
] | [
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https://www.plainmath.net/2014/01/data-interpretation-tables.html?showComment=1413400743084 | [
"# Techniques to solve Data Interpretation: Tables problems\n\nToday I would like to discuss a very known topic Data Interpretation of Quantitative Aptitude section. Around 3-4 sets of Data Interpretation are asked in almost every exam. As this is a scoring section, you should practice more to fetch good number of marks.\n\n## Steps to solve problems of Data Interpretation: Tables\n\n• Tables are the method of systematic presentation of facts and figures in various rows and columns.\n• Before start solving the question, always have a first look of how data is presented.\n• Get a proper understanding of presented data.\n• Then, start solving the questions using the given data.\n• If required, add all the values horizontally and vertically first. Then proceed to questions.\n\n## Problems with Solution\n\nProblem1: Study the following table carefully and answer the questions:\n\n SCHOOLS STANDARDS I II III IV V A 40 55 45 60 45 B 50 60 60 45 55 C 35 50 55 45 50 D 45 55 45 50 45 E 50 50 45 60 50\n\n1. Which school has highest total number of students taking all the standards together?\n2. Which standard has lowest total number of students taking all the schools together?\n\n3. What is the ratio of students studying in standard III of schools A and B to those studying in standard V of schools A and B together?\n\n4. What is the average number of students studying in standard I from all schools together?\n\n5. Number of students studying in standard IV of school E is what percentage of students studying in standard IV from school D?\n\nSolution: Firstly, take a minute to read the table and analyze it. After reading the question,You will come to know that you will need the sum of all the rows and columns. So, before start solving the questions, get a sum of all rows and columns.\n\n SCHOOLS STANDARDS I II III IV V Total A 40 55 45 60 45 245 B 50 60 60 45 55 270 C 35 50 55 45 50 235 D 45 55 45 50 45 240 E 50 50 45 60 50 255 Total 220 270 250 260 245\n\nNow, lets start solving the questions\n\n1. Highest total number of students taking all the standards together\n⇒ Maximum value of last column = 270 i.e. School B\n\n2. Lowest total number of students taking all the schools together\n⇒Minimum value of last row = 220 i.e. Standard I\n\n3. Standard III of school A and B : Standard V of school A and B :: (45+60) : (45+55)\n⇒ 105: 95\n⇒21: 19\n\n4. Average number of students studying in standard I from all schools together\n⇒ (Number of students)/(5) = (220)/5 = 44\n\n5. Let number of students studying in Standard IV from school E = x % of number of students in Standard IV from school D\n⇒60 = x% of 50\n⇒ x = (60/50) times (100) = 120\n\nProblem2: Study the following table carefully to answer the questions\n\n YEAR STREAM ARTS SCIENCE COMMERCE BOYS GIRLS BOYS GIRLS BOYS GIRLS 2001 500 400 600 500 600 600 2002 700 600 700 600 700 600 2003 600 500 500 500 800 700 2004 800 600 900 900 800 800 2005 700 500 800 800 900 500\n\n1. Number of boys in arts stream in year 2004 is what percentage of total number of boys for all years together in Arts stream?\n\n2. What is the ratio of total number of boys to the total number of girls, taking all the streams together, for year 2005?\n\n3.What is the average number of girls from commerce steam for the given years?\n\nSolution: Study the table first, three streams are mentioned i.e. arts, science and commerce and the students are divided into girls and boys sections.\n\n1. Number of boys in arts stream in year 2004 = 800\n\nTotal number of boys for all years together in Arts stream= 3300\n\n⇒ 800 = x % of 3300\n\n⇒ x = (800)/(3300) times 100\n\n⇒ x= 24.24\n\n2. Total number of boys for year 2005 : Total number of girls for year 2005\n\n⇒700+ 800+900 : 500+ 800+ 500\n\n⇒ 2400: 1800\n⇒4:3\n\n3. Average number of girls from commerce steam for the given years\n⇒(600+600+700+800+500)/5\n⇒(3200)/5\n⇒640\n\nI hope you understood the above problems, you are free to ask any query anytime.\n\n#### What's trending in BankExamsToday\n\nSmart Prep Kit for Banking Exams by Ramandeep Singh - Download here",
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"1.",
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"Thank you so much. This is very very helpful.\n\n2.",
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"Hi sir\nTable 1--------- 5th question answer is wrong pls check it and update with correct answer .\n\n1.",
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"thnx dear,,, its 120, not 150,, just a typing mistake...\n\n3.",
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"Hello Sir,\nTable no 1......3 question addition is wrong pls check it and update with correct answer.\n\nI will try to respond asap"
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http://alexkritchevsky.com/2018/10/09/exterior-2.html | [
"# Exterior Algebra Notes #2: the Inner Product\n\n[October 9, 2018]\n\n(See this previous post for some of the notations used here.)\n\n(Not intended for any particular audience. Mostly I just wanted to write down these derivations in a presentable way because I haven’t seen them from this direction before.)\n\n(Vector spaces are assumed to be finite-dimensional and over $\\bb{R}$)\n\nExterior algebra is obviously useful any time you’re anywhere near a cross product or determinant. I want to show how it also comes with an inner product which can make certain formulas in the world of vectors and matrices vastly easier to prove.\n\n## 1. The Inner Product\n\nEuclidean vectors have an inner product that we use all the time. Multivectors are just vectors. What’s theirs?\n\nHowever we define the inner product (or ‘dot product’; I tend to use both names) on multivectors over $\\bb{R}^N$, we’re going to want it to act a lot like it does on vectors. Particularly, it seems like\n\nought to hold. More generally, for two multivectors in the same space, we should be able to sum over their components the same way we do for vectors with $% \\equiv \\sum_{i \\in V} u_i v_i %]]>$:\n\nThis turns out to work, although the usual presentation is pretty confusing. Here’s the standard way to define inner products on the exterior algebra $\\^^k V$, extending the inner product defined on the underlying vector space $V$:\n\nThis is then extended linearly if either argument is a sum of multivectors. This expression is pretty confusing. It turns out to be natural, but it takes a while to see why.\n\nThe left side of this is the inner product of two $k$-vectors (each are the wedge product of $k$ factors together); the right side is the determinant of a $k \\times k$ matrix. For instance:\n\nSimple examples:\n\nIf we label the basis $k$-vectors using multi-indices $I = (i_1, i_2, \\ldots i_k)$, where no two $I$ contain the same set of elements up to permutation, then this amounts to saying that basis multivectors are orthonormal:1\n\nAnd then extending this linearly to all elements of $\\^^k V$.2 This gives an orthonormal basis on $\\^^k V$, and the first thing we’ll do is define the ‘$k$-lengths’ of multivectors, in the same way that we compute the length of a vector $% %]]>$:\n\nThis is called the Gram determinant of the ‘Gramian’ matrix formed by the vectors of $\\b{a}$. It’s non-zero if the vectors are linearly independent, which clearly corresponds to the wedge product $\\bigwedge_i \\b{a}_i$ not being $=0$ in the first place.\n\nIn $\\bb{R}^3$ this gives\n\nIt turns out that multivector inner products show up in disguise in a bunch of vector identities.\n\n## 2. Computation of Identities\n\nLet’s get some practice computing with (1).\n\nIn these expressions, I’m going to be juggling multiple inner products at once. I’ll denote them with subscripts: $% _{\\^} %]]>$, $% _{\\o} %]]>$, $% _{V} %]]>$. (I apologise for the similarity between $\\^$ and $V$ – hopefully they’re different enough to distinguish.)\n\nThe types are:\n\n• the underlying inner product on $V$, which only acts on vectors: $% _V = \\sum_i u_i v_i %]]>$.\n• the induced inner product on $\\o V$, which acts on tensors of the same grade term-by-term: $% _{\\o} = \\< \\b{a , c} \\>_V \\< \\b{b, d } \\>_V %]]>$\n• the induced inner product on $\\^ V$, which we described above: $% _{\\^} = \\< \\b{a , c} \\>_V \\< \\b{b, d } \\>_V - \\< \\b{a , d} \\>_V \\< \\b{b, c } \\> %]]>$.\n\nLet $\\text{Alt}$ be the Alternation Operator, which takes a tensor product to its total antisymmetrization, e.g. $\\text{Alt}(\\b{a \\o b}) = \\b{a \\o b - b \\o a}$. For a tensor with $N$ factors, there are $N!$ components in the result.3\n\n$% _\\^ %]]>$ can be computed by hand by expanding one side into a tensor product and the other into an antisymmetrized tensor product.\n\nThe $\\text{Alt}$ operator can be applied on either side; usually I put it on the right. If you put it on both sides, you would need to divide the whole expression by $\\frac{1}{N!}$, which is annoying (but some people do it).\n\nHere’s an example of this on bivectors:\n\nNow, some formulas which turn out to be the multivector inner product in disguise.\n\nSet $\\b{a = c}$, $\\b{b = d}$ in (2) and relabel to get Lagrange’s Identity:\n\nIf you’re working in $\\bb{R}^3$, use the Hodge Star map $\\star$ (to be discussed in the next post, but we may as well see this identity now) to turn wedge products into cross products (preserving their magnitudes) to get the Binet-Cauchy identity:\n\nOr, if you have three terms on each side, you can expand the product of two scalar triple products:\n\nSet $\\b{a=c}$, $\\b{b = d}$ in the two-vector version to get:\n\nDrop the cross product term to get Cauchy-Schwarz:\n\nI thought that was neat. Maybe there are places where Cauchy-Schwarz is used where in fact Lagrange’s identity would be more useful?\n\nOn vectors in $\\bb{R}^2$ (or any dimension), of course, the vector magnitude of course gives the Pythagorean theorem:\n\nThis generalizes to the bivector areas of an orthogonal tetrahedron (or $(n-1)$-vector surface areas of a $n$-simplex in any dimension), which is called De Gua’s Theorem:\n\nThis is because the total surface area bivector for a closed figure in $\\bb{R}^3$ is $0$, so the surface area of the opposing face is exactly $-(a \\b{x \\^ y} + b \\b{y \\^ z} + c \\b{z \\^ x} )$.\n\nThere is naturally a version of the law of cosines for any tetrahedron/ $n$-simplex with non-orthogonal sides as well. If $\\vec{c} = \\vec{a} + \\vec{b}$ then (though it’s often stated with $c = b-a$ instead):\n\nWe can easily expand $\\| \\b{a} + \\b{b} + \\b{c} \\|^2$ linearly when $a,b,c$ are bivectors or anything else; the angles in the cosines become angles between planes, or something fancier, but the formula is otherwise the same:\n\nWhich is kinda cool.\n\n## 3. Matrix Multiplication\n\nThis is one of the more enlightening things I’ve come across using $% _\\^ %]]>$.\n\nLet $A: U \\ra V$ and $B: V \\ra W$ be linear transformations. Their composition $B\\circ A$ has matrix representation:\n\nThe latter form expresses the fact that each matrix entry in $BA$ is an inner product of a column of $A$ with a row of $B$.\n\nBecause $A^{\\^ q} : \\^^q U \\ra \\^^q V$ and $B^{\\^ q} : \\^^q V \\ra \\^^q W$ are also linear transformations, their composition $B^{\\^ q} \\circ A^{\\^ q} : \\^^q U \\ra \\^^W$ also has a matrix representation:\n\nWhere $I,J,K$ are indexes over the appropriate $\\^^q$ spaces.\n\n$A^{\\^ q}_I$ is the wedge product of the $I = (i_1, i_2, \\ldots i_q)$ columns of $A$, and $(B^{\\^ q})^K$ is the wedge product of $q$ rows of $B$ from $K$, which means this is just the inner product we discussed above.\n\nBut this is just the determinant of a minor of $(BA)_i^k$ – the one indexed by $(I,K)$. This means that:\n\nAnd thus:\n\nThis is called the Generalized Cauchy-Binet formula. Note that $(3)$ does not require that the matrices be the square.\n\nWhich is neat. I think this version is way easier to remember or use than the version in Wikipedia, which is expressed in terms of matrix minors and determinants everywhere.\n\nCorollaries:\n\nWhen $A$ and $B$ are in the same space and $q = \\dim V$, then all of the wedge powers turn into determinants, giving something familiar:\n\nWhen $B = A^T$, it says that the determinant of the square matrix $A^T A$ is the sum of squared determinants of minors of the (not necessarily square) $A$. If $A$ is $n \\times k$, this is a sum over all $k \\times k$ minors of $A$:\n\nOther articles related to Exterior Algebra:\n\n1. I prefer the notation $1_{ij}$ to $\\delta_{ij}$ because, well, it makes perfect sense.\n\n2. If we don’t specify that all of our multi-indices are unique up to permutation, then we would have to write something like $% = \\sgn(I, J) %]]>$, where $\\text{sgn}$ is the sign of the permutation that takes $I$ to $J$, since for instance $% = -1 %]]>$\n\n3. There are several conventions for defining $\\text{Alt}$; often it comes with a $\\frac{1}{N!}$. If you wanted it to preserve vector magnitudes, you might have it divide by $\\frac{1}{\\sqrt{N!}}$. I don’t like either of those though, and prefer to leave it without factorials, because it makes other definitions much easier."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92262423,"math_prob":0.99899596,"size":6891,"snap":"2019-51-2020-05","text_gpt3_token_len":1521,"char_repetition_ratio":0.13416582,"word_repetition_ratio":0.009068426,"special_character_ratio":0.21158032,"punctuation_ratio":0.10746269,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999218,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T08:40:38Z\",\"WARC-Record-ID\":\"<urn:uuid:2a6e75d7-4a80-485c-86c3-9adf9279ff36>\",\"Content-Length\":\"26358\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:956e9d93-68bf-4e72-9ed7-6b38910e760d>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e7e6d2f-b892-4395-a02c-ab2ebaf28f8a>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"http://alexkritchevsky.com/2018/10/09/exterior-2.html\",\"WARC-Payload-Digest\":\"sha1:AR3FGGD34BREZNM6HY5Z3VSVJ2LCGYWA\",\"WARC-Block-Digest\":\"sha1:2NMZU3D2SHNRG2YPRKTULQQ7BZ7JMYZD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592394.9_warc_CC-MAIN-20200118081234-20200118105234-00264.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/21-28-plus-4-9 | [
"Solutions by everydaycalculation.com\n\n21/28 + 4/9 is 43/36.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 28 and 9 is 252\n2. For the 1st fraction, since 28 × 9 = 252,\n21/28 = 21 × 9/28 × 9 = 189/252\n3. Likewise, for the 2nd fraction, since 9 × 28 = 252,\n4/9 = 4 × 28/9 × 28 = 112/252\n189/252 + 112/252 = 189 + 112/252 = 301/252\n5. After reducing the fraction, the answer is 43/36\n6. In mixed form: 17/36\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:"
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https://mathematica.stackexchange.com/questions/186131/using-replaceall-on-matrix-to-produce-new-list-of-complete-matrices | [
"# Using ReplaceAll on matrix to produce new list of complete matrices\n\nThis seems like something simple but I can't find the answer.\n\nIf I have as input:\n\n{{1, 2}, {3, d}} /. d -> {4, -4}\n\n\nMathematica outputs:\n\n{{1, 2}, {3, {4, -4}}}\n\n\n(which is logical as it simply replaces d with the list given)\n\nWhat I'd like is the following output, i.e., a new list of complete 2x2 matrices:\n\n{{{1, 2}, {3, 4}}, {{1, 2}, {3, -4}}}\n\n\nTo give a bit more context, I have two lists of matrices that I'd like to multiply to give all the combinations, e.g.:\n\nlist1 = {A,B,C}\nlist2 = {D,E}\n\n\ninto\n\n{A.D,A.E,B.D,B.E,C.D,C.E}\n\n\nDot @@@ Tuples[{list1, list2}]\n\n\nBut prior to multiplying the list I need to do a replace all as described above to produce the actual matrices to multiply together.\n\nThanks\n\nYou can just use Table:\n\nTable[{{1, 2}, {3, d}}, {d, {4, -4}}]\n(* {{{1, 2}, {3, 4}}, {{1, 2}, {3, -4}}} *)\n\n• Knew it had to be simple, thank you Nov 16, 2018 at 17:50\nrules = List /@ Thread[d -> {4, -4}]\n\n\n{{d -> 4}, {d -> -4}}\n\n{{1, 2}, {3, d}} /. rules\n\n\n{{{1, 2}, {3, 4}}, {{1, 2}, {3, -4}}}\n\nNote: With a list of rules ReplaceAll uses the first matching rule:\n\nx /. {x -> 1, x -> a, x -> 2}\n\n\n1\n\nTo apply each rule separately, you need to wrap each rule with List:\n\nx /. {{x -> 1}, {x -> a}, {x -> 2}}\n\n\n{1, a, 2}"
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https://math.stackexchange.com/questions/2607543/representation-theory-of-s-3 | [
"# Representation theory of $S_3$",
null,
"My question could appear naive but I am reading up representation theory by myself. I am at a loss to understand the way how the action of the matrix happens for $S_3$. For example how does each of these representation work on the vectors $(e_1 -e_2)$ and $(e_2 -e_3)$ is unclear.\n\nThe text says that $$(1, 2) (e_1-e_2) = -(e_1 - e_2)$$ and $$(1, 2) (e_2 - e_3) = (e_1 - e_3).$$ How does one derive from the matrix action on the vector $(e_1, -e_2)$ and $(e_2, -e_3)$.",
null,
"• Welcome to MSE. Please use MathJax. – José Carlos Santos Jan 16 '18 at 10:16\n• Where did you take this from? It doesn't look correct. For instance, the action of $(1\\ \\ 2)$ on $e_1-e_2$ should give $-2e_1-e_2$, not $-e_1+e_2$. – José Carlos Santos Jan 16 '18 at 10:21\n• I will add the link of the place where I got from. Please look at pg 6 and pg 7 of the text. jdc.math.uwo.ca/M9140a-2012-summer/Diaconis.pdf – user2714795 Jan 16 '18 at 10:35\n• @JoséCarlosSantos Why? It seems that elements of $S_3$ act by permuting the basis, so, indeed, $(1 2)(e_1 - e_2) = (1 2)e_1 - (1 2)e_2 - e_2 - e_1 = - (e_1 - e_2)$. – lisyarus Jan 16 '18 at 10:55\n• @lisyarus I am surely missing something. Where did you read that $(1\\ \\ 2)$ permutes the basis? What I read is that $(1\\ \\ 2)e_1=-e_1$ and that $(1\\ \\ 2)e_2=e_1+e_2$. – José Carlos Santos Jan 16 '18 at 11:05\n\nSince$$(1\\ \\ 2)e_1=e_2,\\ (1\\ \\ 2)e_2=e_1\\text{, and }(1\\ \\ 2)e_3=e_3,$$it is clear that$$(1\\ \\ 2)(e_1-e_2)=e_2-e_1=-(e_1-e_2)\\text{ and that }(1\\ \\ 2)(e_2-e_3)=e_1-e_3.$$So, if we define $w_1$ and $w_2$ as $e_1-e_2$ and as $e_2-e_3$ respectively, we get that $(1\\ \\ 2)w_1=-w_1$ and that$$(1\\ \\ 2)w_2=e_1-e_3=e_1-e_2+e_2-e_3=w_1+w_2.$$Therefore, the matrix of the action of $(1\\ \\ 2)$ on $\\bigl\\{(x,y,z)\\in\\mathbb{R}^3\\,|\\,x+y+z=0\\bigr\\}$ with respect to the basis $\\{w_1,w_2\\}$ is indeed$$\\begin{pmatrix}-1&1\\\\0&1\\end{pmatrix}.$$"
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https://ch.mathworks.com/matlabcentral/answers/530768-constructing-a-difficult-large-matrix | [
"# constructing a difficult large matrix\n\n4 views (last 30 days)\nektor on 23 May 2020\nDear all,\nI am interested in constructing a complicated matrix A.\nThe matrix A is a diagonal matrix, where each diagonal element 'a_i' is of 1X1000, having the following patter\na_1=[p 1 0 0 0 0 0 .....0 ];\na_2=[p^2 p 1 0 0 0..... 0];\na_2=[p^3 p^2 p 1 0 0......0];\na_2=[p^4 p^3 p^2 p 1 0 0......0 ];\na_2=[p^5 p^4 p^3 p^2 p 1 0 ...0];\n.\n.\n.\na_1000=[p^1000 p^999 p^998.........1];\nAnd I want to multiply each element \"a_i\" with the vector h, where h is of dimension 1000X1, which contains just numbers of no particular pattern\nIs there any way of doing that fast?\nMaybe by using sparse or speye?\nThank you\n\nper isakson on 23 May 2020\n\"multiply each element \"a_i\" with the vector h, where h is of dimension 1000X1\" the product will that be a scalar, vector or matrix?\n\"doing that fast\" how fast is \"fast\" ?\nektor on 23 May 2020\nHi isakson,\nIt will be a scalar. So each element of a_i will be multiplied by the corresponding element of h to produce a scalar (like inner product). Sorry for not clarifying first.\nWhen I say fast, I mean not time consuming.\n\nDavid Goodmanson on 23 May 2020\nHi ektor,\nI presume you are interested not so much in the in the matrix A as in the 1000 scalars you get after you multiply each row of A by the vector h. To make things a little easier I included an extra row a_0 at the beginning.\na_0=[1 0 0 0 0 0 0 .....0 ];\na_1=[p 1 0 0 0 0 0 .....0 ];\na_2=[p^2 p 1 0 0 0..... 0];\na_3=[p^3 p^2 p 1 0 0......0];\na_4=[p^4 p^3 p^2 p 1 0 0......0 ];\na_5=[p^5 p^4 p^3 p^2 p 1 0 ...0];\n.\n.\na_1000=[p^1000 p^999 p^998.........1];\nthen\nAh = filter(1,[1 -p],h)\nis the same as A*h. It's very fast. There is one exra element at the beginning corresponding to row a_0 but you can always delete it.\n\nDavid Goodmanson on 23 May 2020\nHi ektor,\nI was assuming that a_1 times h is the inner product (scalar product). that is, one number for the output, is that correct? Or do you mean element-by-element multiplicaton with no sum, so that you get an output of length(a_1) numbers?\nektor on 23 May 2020\nHI david,\nJust a quick question: \" There is one exra element at the beginning corresponding to row a_0 but you can always delete it. \"\nwhat does this mean?\nDavid Goodmanson on 3 Jun 2020\nHi ektor,\nI meant that your matrix has 1000 rows and 1001 columns. The output is the inner product of h with each row, so 1000 values in all. However, for the filter function to reproduce that result, it was convenient to add a row at the top that I called a0. Then there are 1001 rows and 1001 values in the result. But the first result is due to the added row, so if you delete it you are back to the 1000 values in the original problem.\n\nper isakson on 23 May 2020\nEdited: per isakson on 23 May 2020\nHere are three functions, two of which are based on the answer of David Goodmanson. I think all of them are fast enough. However, more important than speed is that their results are correct. More test are needed.\nA small test with numbers that I'm able to check.\n>> cssm(8)\nans =\n3 0 0 0 0 0 0 0\n0 7 0 0 0 0 0 0\n0 0 15 0 0 0 0 0\n0 0 0 31 0 0 0 0\n0 0 0 0 63 0 0 0\n0 0 0 0 0 127 0 0\n0 0 0 0 0 0 255 0\n0 0 0 0 0 0 0 510\n>>\n>> tic,cssm(1e4);toc\nElapsed time is 1.155732 seconds.\n>> tic,M=dgsp(1e4);toc\nElapsed time is 0.490225 seconds.\n>> tic,M=dg(1e4);toc\nElapsed time is 0.014082 seconds.\nfunction M = cssm( N )\n%%\nh = ones( N, 1 );\np = 2;\npr = 1;\na = zeros( N, N );\na( N+1 : (N+1) : end ) = 1;\nfor jj = 1 : N\npr = pr * p;\na( jj : (N+1) : end-(jj-2)*N ) = pr;\nend\nM = zeros( N, N );\nfor jj = 1 : N\nM(jj,jj) = a(jj,:) * h;\nend\nend\nfunction M = dgsp( N )\n%%\nh = ones( N, 1 );\ntmp = [ 1; h ];\np = 2;\n%%\nAh = filter( 1, [1,-p], tmp );\nM = sparse( zeros( N, N ) ); % better: sparse(10,10,0);\nM( 1 : N+1 : end ) = Ah( 2 : end );\nend\nfunction M = dg( N )\n%%\nh = ones( N, 1 );\ntmp = [ 1; h ];\np = 2;\n%%\nAh = filter( 1, [1,-p], tmp );\nM = zeros( N, N );\nM( 1 : N+1 : end ) = Ah( 2 : end );\nend\n\nShow 1 older comment\nper isakson on 23 May 2020"
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https://www.javatpoint.com/cpp-multiset-insert-function | [
"# C++ multiset insert()\n\nC++ Multiset insert() function is used for inserting new element or a range of elements in the multiset.\n\n## Parameter\n\nval: Value to insert in the multiset.\n\nposition: Hint for the position to insert element in the multiset.\n\nfirst: Beginning of range to insert value.\n\nlast: End of range to insert value.\n\nil: An initializer list.\n\n## Return value\n\nThe insert() function return an iterator pointing to the newly inserted element in the multiset.\n\n## Complexity\n\nIf a single element is inserted, complexity will be logarithmic in size.\n\nIf a hint is given and the position given is the optimal, then the complexity will be amortized constant.\n\nNo changes.\n\n## Data Races\n\nThe container is modified.\n\nConcurrently accessing the existing elements of multiset is safe, although iterating ranges in the container is not.\n\n## Exception Safety\n\nThis function does not throw exception.\n\n## Example 1\n\nLet's see the simple example to insert the elements into the multiset:\n\nOutput:\n\n```The elements in multiset are: 1 2 3 3 4\n```\n\nIn the above example, it simply inserts the element with the given key.\n\n## Example 2\n\nLet's see a simple example to insert the element in the specified position:\n\nOutput:\n\n```The elements in multiset are: 1 2 3 4 4\n```\n\nIn the above example, elements are inserted in the defined position.\n\n## Example 3\n\nLet's see a simple example to insert the elements of one multiset to another in a given range:\n\nOutput:\n\n```The elements in multiset1 are: 1 2 3 4 4\nThe elements in multiset2 are: 3 4 4\n```\n\n## Example 4\n\nLet's see a simple example to insert the elements from the initializer list:\n\nOutput:\n\n```Multiset contains following elements\nC++\nC++\nJava\nOracle\nSQL\n```\n\nIn the above example, elements are inserted from an initializer list.\n\nNext TopicC++ multiset",
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"https://www.javatpoint.com/images/pinterest32.png",
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https://www.youminhan.com/leetcode/docs/question/137-single-number-ii/ | [
"Amazon\n\n2020-05-29\n\n## 137. Single Number II\n\n### Question:\n\nGiven a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.\n\nNote:\n\nYour algorithm should have a linear runtime complexity. Could you implement it without using extra memory?\n\n#### Example 1:\n\n``````Input: [2,2,3,2]\nOutput: 3\n``````\n\n#### Example 2:\n\n``````Input: [0,1,0,1,0,1,99]\nOutput: 99\n``````\n\n### Solution:\n\nUsing HashMap to calculate the frequency of each elements.\n\n``````class Solution {\npublic int singleNumber(int[] nums) {\nHashMap<Integer, Integer> map = new HashMap<>();\nfor (int i : nums)\nmap.put(i, map.getOrDefault(i, 0) + 1);\n\nfor (int k : map.keySet())\nif (map.get(k) == 1) return k;\n\nreturn -1;\n}\n}\n``````\n\nUsing Bit Operation, one is the sum for every number that appears once which are not intersect with two. Two is the sum for every number that appears twice which do not intersect with one.\n\n``````class Solution {\npublic int singleNumber(int[] nums) {\nint one = 0;\nint two = 0;\nfor (int i : nums) {\none = (one ^ i) & (~two);\ntwo = (two ^ i) & (~one);\n}\nreturn one;\n}\n}\n``````"
] | [
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https://www.edaboard.com/showthread.php?383917-Convert-from-STD_LOGIC-to-integer-in-VHDL&s=df62ac31c3d8799245f9a020c581af86&goto=newpost | [
"# Convert from STD_LOGIC to integer in VHDL\n\n1. ## Convert from STD_LOGIC to integer in VHDL\n\nIs this code right?, I need to convert from STD_LOGIC to integer( to add those integers) and then back result to STD_LOGIC:\na,h,f3,f0,k,w,f2,f1 are STD_Logic\n\nCode:\n``` a <= std_logic_vector(to_unsigned((To_integer(unsigned(h)) +\nTo_integer(unsigned(f3)) +\nTo_integer(unsigned(f0)) +\nTo_integer(unsigned(k)) +\nTo_integer(unsigned(w)) +\nTo_integer(unsigned(f2)) +\nTo_integer(unsigned(f1))),a'length));```",
null,
"•\n\n2. ## Re: Convert from STD_LOGIC to integer in VHDL\n\nConvert to integer makes no sense. Why not add unsigned?\n\n1 members found this post helpful.",
null,
"3. ## Re: Convert from STD_LOGIC to integer in VHDL\n\nstd_logic but its nature is either 0 or 1.\n\nDo you mean std_logic_vector, furthermore when converting from interger back to std_logic_vector via unsigned you need to specify length\n\n1 members found this post helpful.",
null,
"•\n\n4. ## Re: Convert from STD_LOGIC to integer in VHDL\n\nHere you can find examples with all the important VHDL conversions:\nhttps://www.nandland.com/vhdl/tips/t...o-integer.html\nI don't think that you can convert an std_logic to integer, you will probably need to use std_logic_vector signals of a single bit, like: signal names : std_logic_vector(0 downto 0);\n\n1 members found this post helpful.",
null,
"5. ## Re: Convert from STD_LOGIC to integer in VHDL",
null,
"Originally Posted by FvM",
null,
"Convert to integer makes no sense. Why not add unsigned?\nI tried this:\na <= std_logic_vector(unsigned(h) +unsigned(f3) + unsigned(f0) +unsigned(k) + unsigned(w) + unsigned(f2) + unsigned(f1));\n\nand i still have these errors \"Bad expression in left operand of infix expression \"+\".\n(vcom-1078) Identifier \"unsigned\" is not directly visible.\nIllegal type conversion to ieee.std_logic_1164.STD_LOGIC_VECTOR (operand type is not known).",
null,
"6. ## Re: Convert from STD_LOGIC to integer in VHDL\n\n\"Bad expression in left operand of infix expression \"+\".\n(vcom-1078) Identifier \"unsigned\" is not directly visible.\nThis error is because you have more than one package that declares an unsigned type. You didnt post the code, but I assume you have included std_logic_arith and numeric_std. You should delete std_logic_arith as it is a non-standard VHDL library.\n\nYou have another problem: you're trying to convert std_logic to unsigned. Thats impossible, because std_logic is an enumerated type, and unsigned is an array of std_logic. They are not related types.\nYou will need to make an array out of your single std_logic bits to make them unsigned. There is a nice little hack for this - simply append the bit to a null array. You will also need to qualify the resulting array as unsigned so it doesnt get confused between unsigned and signed:\n\n```Code VHDL - [expand]1\na <= std_logic_vector( unsigned'(\"\"&h) + unsigned(\"\"&f3) ....... etc```\n\nOf course, it would have been much easier if a and the other signals were all unsigned in the first place.\n\n1 members found this post helpful.",
null,
"7. ## Re: Convert from STD_LOGIC to integer in VHDL\n\nwhat libraries are you using?\n\n...you are probably using the ieee.numeric_std along with the conflicting synopsys libraries ieee.std_logic_arith and ieee.std_logic_unsigned.\n\nI've been having a similar problem due to a very old testbench and the need to use some code that requires numeric_std, so I've had to resort to explicitly calling out what functions are used instead of using .all, very annoying.\n\n1 members found this post helpful.",
null,
"•\n\n8. ## Re: Convert from STD_LOGIC to integer in VHDL",
null,
"Originally Posted by TrickyDicky",
null,
"This error is because you have more than one package that declares an unsigned type. You didnt post the code, but I assume you have included std_logic_arith and numeric_std. You should delete std_logic_arith as it is a non-standard VHDL library.\n\nYou have another problem: you're trying to convert std_logic to unsigned. Thats impossible, because std_logic is an enumerated type, and unsigned is an array of std_logic. They are not related types.\nYou will need to make an array out of your single std_logic bits to make them unsigned. There is a nice little hack for this - simply append the bit to a null array. You will also need to qualify the resulting array as unsigned so it doesnt get confused between unsigned and signed:\n\n```Code VHDL - [expand]1\na <= std_logic_vector( unsigned'(\"\"&h) + unsigned(\"\"&f3) ....... etc```\n\nOf course, it would have been much easier if a and the other signals were all unsigned in the first place.\nThank you so much\nIt is solved\n\n- - - Updated - - -",
null,
"Originally Posted by ads-ee",
null,
"what libraries are you using?\n\n...you are probably using the ieee.numeric_std along with the conflicting synopsys libraries ieee.std_logic_arith and ieee.std_logic_unsigned.\n\nI've been having a similar problem due to a very old testbench and the need to use some code that requires numeric_std, so I've had to resort to explicitly calling out what functions are used instead of using .all, very annoying.\nThank you so much\nIt is solved",
null,
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https://alexpounds.com/blog/2016/09/04/exploring-weird-maths-with-code | [
"# Exploring weird maths with code\n\nSometimes, while reading an innocuous-seeming article, I stumble across an aside that makes me sit bolt upright and mutter something incredulous. Asides like this one:\n\nA counterintuitive property of coin-tossing: If Alice tosses a coin until she sees a head followed by a tail, and Bob tosses a coin until he sees two heads in a row, then on average, Alice will require four tosses while Bob will require six tosses (try this at home!), even though head-tail and head-head have an equal chance of appearing after two coin tosses.\n\nWired\n\nThis was a surprise! The four possible outcomes of two tosses are equally likely, so it seems weird that a heads-tails outcome would take longer to reach than a heads-heads. Weird enough to try it at home – at least by programming. Let's write some Ruby and see if we get the same result. (I recommend opening `irb` and exploring these examples for yourself if you want to fully understand them.)\n\n## Checking some assumptions\n\nFirst of all, let's agree to toss a coin by picking a random symbol from an array1:\n\n``````def coin_toss\n%i(heads tails).sample #=> :heads or :tails.\nend\n``````\n\nAnd let's confirm that this is close enough to 50/50, by counting the result of tossing a coin 100,000 times:\n\n``````results = {heads: 0, tails: 0}\n1000000.times { results[coin_toss] += 1 }\n\nputs \"After 100000 tosses we saw #{results[:heads]} heads and #{results[:tails]} tails.\"\n``````\n\n`.sample` chooses an element at random, so the result will be a little different each time. I ran this program 10 times, and got these results:\n\n``````After 100000 tosses we saw 50131 heads and 49869 tails.\nAfter 100000 tosses we saw 49845 heads and 50155 tails.\nAfter 100000 tosses we saw 50094 heads and 49906 tails.\nAfter 100000 tosses we saw 49672 heads and 50328 tails.\nAfter 100000 tosses we saw 50062 heads and 49938 tails.\nAfter 100000 tosses we saw 50046 heads and 49954 tails.\nAfter 100000 tosses we saw 50003 heads and 49997 tails.\nAfter 100000 tosses we saw 50094 heads and 49906 tails.\nAfter 100000 tosses we saw 50124 heads and 49876 tails.\nAfter 100000 tosses we saw 49838 heads and 50162 tails.\n``````\n\nI think these results look OK, but the next thing I tried was busting out some statistics and checking the standard deviation. You can think of it as a measure of how closely-clustered our results are – we'd expect to get a low standard deviation if `.sample` is fair. Calculating the standard deviation is a little bit complicated, so I used the `descriptive_statistics` gem to make it easier. Let's calculate the standard deviation of the number of heads in each run:\n\n``````require 'descriptive_statistics'\n[50131, 49845, 50094, 49672, 50062, 50046, 50003, 50094, 50124, 49838].standard_deviation #=> 146.014\n``````\n\nBut is 146.014 low or not? I have no idea! This is where my statistics knowledge runs out. For now, let's presume our eyeballs are correct and our coin tosses are fair.\n\n## Back to the question\n\nIf we can toss a coin fairly, we can return to our original question: how many tosses, on average, does it take to reach a given combination?\n\nWe'll need a target combination, we need to toss at least twice, and we want to toss until we hit the target:\n\n``````target = [:heads, :heads]\ntosses = [coin_toss, coin_toss]\n\nuntil tosses.last(2) == target\ntosses << coin_toss\nend\n``````\n\nI ran this in `irb` and I got `[:tails, :tails, :heads, :tails, :heads, :heads]`. It works! Let's turn this into a method so we can reuse it:\n\n``````def tosses_until(target)\ntosses = [coin_toss, coin_toss]\nuntil tosses.last(2) == target\ntosses << coin_toss\nend\ntosses\nend\n``````\n\nRunning the experiment repeatedly will make our result more reliable. If something weird happens once it could be a fluke, but you can't fluke something thousands of times. We could use the `.times` method again, and build up an array of results like we built the array of tosses:\n\n``````experiments = []\n100000.times { experiments << tosses_until([:heads, :heads]) }\n``````\n\nOr we can make this shorter by using Ruby's `.map` method. `.map` applies a method to every element in a list. It's normally used to modify an existing list:\n\n``````[\"cat\", \"dog\", \"avocado\"].map { |t| t.upcase } #=> [\"CAT\", \"DOG\", \"AVOCADO\"]\n(1..4).map { |n| n * 3 } #=> [3, 6, 9, 12]\n``````\n\nBut it doesn't matter if we throw the original elements away instead. You can try this in the console, but beware! It's going to print out all 100,000 results.\n\n``````experiments = (0..100000).map { tosses_until([:heads, :heads]) }\n``````\n\nIt's not really relevant to our experiment, but I wondered what the shortest and longest sequence until our target was. You might expect that we can use `experiments.min` and `experiments.max` to find out:\n\n``````experiments.min #=> [:heads, :heads]\nexperiments.min.length #=> 2\nexperiments.max #=> [:tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :heads, :tails, :heads, :tails, :heads, :heads]\nexperiments.max.length #=> 21\n``````\n\nBut that's not quite right2 for the maximum case. It looks right, though – a handy reminder that verifying data by eye can lead you astray. Instead, we need to use `.max_by` to explicitly look at the length of the array:\n\n``````experiments.max_by { |e| e.length }\n``````\n\nThis pattern – calling a method on the value passed into the block – is common, so Ruby provides a shorthand for this:\n\n``````experiments.max_by(&:length) #=> [:heads, :tails, :tails, :tails, :heads, :tails, :tails, :tails, :heads, :tails, :tails, :tails, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :heads, :tails, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :heads, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :tails, :heads, :heads]\nexperiments.max_by(&:length).length #=> 61\n``````\n\nLet's put all this together in one place, and add some output about our results:\n\n``````def coin_toss\n%i(heads tails).sample #=> :heads or :tails.\nend\n\ndef tosses_until(target)\ntosses = [coin_toss, coin_toss]\nuntil tosses.last(2) == target\ntosses << coin_toss\nend\ntosses\nend\n\nexperiments = (0..100000).map { tosses_until([:heads, :heads]) }\naverage_toss_count = experiments.reduce(0) { |sum, n| sum + n.length } / experiments.length.to_f # We'll talk about this line below.\n\nputs \"Our shortest sequence was #{experiments.min_by(&:length)}\"\nputs \"Our longest sequence was #{experiments.max_by(&:length)}\"\nputs \"On average, we had to toss #{average_toss_count} times before (heads, heads) came up.\"\n``````\n\n`.reduce` is a close cousin of `.map`. `.map` does something to every element in a list; `.reduce` takes two elements from a list and boils them down into one. It does that repeatedly to produce a final value:\n\n``````[1, 2].reduce { |a, b| a + b } #=> 3\n[1, 2, 3].reduce { |a, b| a + b } #=> 6: [1, 2, 3] → [3, 3] → 6.\n[1, 2, 3, 4].reduce { |a, b| a + b } #=> 10: [1, 2, 3, 4] → [3, 3, 4] → [6, 4] → 10.\n``````\n\nYou can also give `.reduce` a starting value, which is what we did in our program:\n\n``````[1, 2].reduce(10) { |sum, a| sum + a } #=> 13: 10 + 1 = 11 then 11 + 2 = 13.\n[1, 2, 3].reduce(10) { |total, a| total + (a * 2) } #=> 22.\n``````\n\nWe started our toss count at 0, then added the length of each run to that total. Finally, we divided it by the total number of runs to get an average. The `.to_f` on the end converts the length to a floating point number, because we'd like to see the decimal places in the result.\n\n``````9 / 2 #=> 4; really \"4 remainder 1\", but Ruby throws the remainder away\n9 / 2.to_f #=> 4.5\n``````\n\n## Simplifying our code\n\nThis works, but is more complicated than it needs to be. Our goal was to find out how many tosses, on average, it takes to hit our target – we don't care about the sequence of tosses to get there. Let's change our `tosses_until` method to return the number of tosses instead of the sequence itself:\n\n``````def tosses_until(target)\ntosses = [coin_toss, coin_toss]\nuntil tosses.last(2) == target\ntosses << coin_toss\nend\ntosses.length\nend\n``````\n\nThis lets us make our trial run code simpler. We could build an array of the sequence counts, then add it up:\n\n``````experiments = (0..100000).map { tosses_until([:heads, :heads]) }\naverage_toss_count = experiments.reduce(&:+) / experiments.length.to_f\n``````\n\nWe could skip the array entirely, and just maintain a total:\n\n``````total_experiments = 100000\ntotal_tosses = 0\ntotal_experiments.times { total_tosses += tosses_until([:heads, :heads]) }\naverage_toss_count = total_tosses / total_experiments.to_f\n``````\n\nOr we could use `reduce` again:\n\n``````total_experiments = 100000\ntotal_tosses = (0..total_experiments).reduce(0) { |sum, _| tosses_until([:heads, heads]) }\naverage_toss_count = total_tosses / total_experiments.to_f\n``````\n\nThe \"best\" version is a matter of taste, but personally I prefer the first version. It uses more memory, but that doesn't matter in experiments like these. It's the shortest code, we can find the longest run of tosses, and it's reasonably clear how it works once you get your head around `.reduce`.\n\nLet's put the first version into a method that runs the experiment and reports the outcome for a given target:\n\n``````def coin_toss\n%i(heads tails).sample\nend\n\ndef tosses_until(target)\ntosses = [coin_toss, coin_toss]\nuntil tosses.last(2) == target\ntosses << coin_toss\nend\ntosses.length\nend\n\ndef average_toss_count(target, num_experiments)\nexperiments = (0..num_experiments).map { tosses_until(target) }\naverage_toss_count = experiments.reduce(&:+) / experiments.length.to_f\n\n# sprintf formats the average so it prints to two decimal places only.\nputs \"On average, we had to toss #{sprintf('%.2f', average_toss_count)} times before #{target.inspect} came up. Our longest run was #{experiments.max} tosses.\"\nend\n``````\n\n## The other cases\n\nNow we have all the building blocks to run the experiment for each of the four possible outcomes:\n\n``````targets = [[:heads, :heads], [:heads, :tails], [:tails, :heads], [:tails, :tails]]\ntargets.each { |target| average_toss_count(target, 100000) }\n``````\n\nWhich produces:\n\n``````On average, we had to toss 5.98 times before [:heads, :heads] came up. Our longest run was 52 tosses.\nOn average, we had to toss 4.00 times before [:heads, :tails] came up. Our longest run was 22 tosses.\nOn average, we had to toss 3.99 times before [:tails, :heads] came up. Our longest run was 20 tosses.\nOn average, we had to toss 6.00 times before [:tails, :tails] came up. Our longest run was 55 tosses.\n``````\n\nSure enough, it takes longer on average to hit `[:heads, :heads]` or `[:tails, :tails]` than `[:heads, :tails]` or `[:tails, :heads]`, even though each outcome has an equal probability. It's still weird, but now I'm satisfied it's true.\n\n## Why does this happen?\n\nLet's go back to Alice and Bob, who are targeting `[:heads, :tails]` and `[:heads, :heads]` respectively:\n\nPlayer Target\nAlice H T\nBob H H\n\nLet's presume they both win their first toss – they both get a result they're looking for:\n\nPlayer Target Result 1\nAlice H T H\nBob H H H\n\nThen, presume they lose their second toss:\n\nPlayer Target Result 1 Result 2\nAlice H T H H\nBob H H H T\n\nThere's now a major difference between the two players: Alice can hit her target on toss 3, but Bob can't until toss 4. Bob must start over after losing on toss 2; Alice's loss can be part of a win if she gets a tails on turn 3.\n\n## Exercises\n\nIf you'd like to explore this some more, here's some suggestions for things to try:\n\n1. Change the program so it runs the experiment a million times instead of 100,000.\n2. If we toss three coins, there's eight possible outcomes. How long does it take, on average, to hit each combination? Are there some sequences that take longer than others?\n3. We left our proof of a fair coin toss at \"Yeah, that looks OK.\" Can you do better? How would you satisfy yourself that it's producing fair results?\n\n1. `%i()` is Ruby shorthand that generates an array of symbols. `%i(foo bar baz)` means the same as `[:foo, :bar, :baz]`. ↩\n\n2. But why doesn't this work? When we call `.min`, Ruby uses the `<=>` comparison operator to find the smallest value in the list. `experiments` is an array of arrays; `<=>` for arrays calls `<=>` on each of the elements of the list in turn until it finds a difference. In this case, our list elements are symbols. Symbols get converted to strings before comparison, and `\"heads\" < \"tails\"` because `\"h\" < \"t\"`. So the upshot of this is that `experiments.max` returns the result with the longest initial streak of tails.\n\nYes, I had to look this up in the documentation. ↩"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8740621,"math_prob":0.8769732,"size":12310,"snap":"2021-21-2021-25","text_gpt3_token_len":3453,"char_repetition_ratio":0.19470178,"word_repetition_ratio":0.12042906,"special_character_ratio":0.30089357,"punctuation_ratio":0.22610015,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9687373,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T10:32:58Z\",\"WARC-Record-ID\":\"<urn:uuid:fb95c875-5e4e-417c-87d0-544c3a514d47>\",\"Content-Length\":\"21679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5490e4b3-c146-4c81-9e4b-468653a1551c>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d89b9e0-5783-4179-9b9a-f0f92ffdf30d>\",\"WARC-IP-Address\":\"64.90.53.55\",\"WARC-Target-URI\":\"https://alexpounds.com/blog/2016/09/04/exploring-weird-maths-with-code\",\"WARC-Payload-Digest\":\"sha1:LMTAZRGFPWBMZKQRADQAD4FDFL2SZR3M\",\"WARC-Block-Digest\":\"sha1:YOCE6N7F45WW3MNDBEDKHH7WQMVNAAE7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487630081.36_warc_CC-MAIN-20210625085140-20210625115140-00298.warc.gz\"}"} |
https://www.erlang.org/docs/18/man/maps.html | [
"",
null,
"User's Guide\nReference Manual\nRelease Notes\nPDF\nTop\n\nSTDLIB\nReference Manual\nVersion 2.8\n\nExpand All\nContract All\n\n• STDLIB (App)\n• array\n• assert.hrl\n• base64\n• beam_lib\n• binary\n• c\n• calendar\n• dets\n• dict\n• digraph\n• digraph_utils\n• epp\n• erl_anno\n• erl_eval\n• erl_expand_records\n• erl_id_trans\n• erl_internal\n• erl_lint\n• erl_parse\n• erl_pp\n• erl_scan\n• erl_tar\n• ets\n• file_sorter\n• filelib\n• filename\n• gb_sets\n• gb_trees\n• gen_event\n• gen_fsm\n• gen_server\n• io\n• io_lib\n• lib\n• lists\n• log_mf_h\n• maps\n• math\n• ms_transform\n• orddict\n• ordsets\n• pool\n• proc_lib\n• proplists\n• qlc\n• queue\n• rand\n• random\n• re\n• sets\n• shell\n• shell_default\n• slave\n• sofs\n• string\n• supervisor\n• supervisor_bridge\n• sys\n• timer\n• unicode\n• win32reg\n• zip\n\n# maps\n\nmaps\n\n### MODULE SUMMARY\n\nMaps Processing Functions\n\n### DESCRIPTION\n\nThis module contains functions for maps processing.\n\n### EXPORTS\n\nfilter(Pred, Map1) -> Map2\n\nTypes:\n\nPred = fun((Key, Value) -> boolean())\nKey = Value = term()\nMap1 = Map2 = #{}\n\nReturns a map Map2 for which predicate Pred holds true in Map1.\n\nThe call will fail with a {badmap,Map} exception if Map1 is not a map or with badarg if Pred is not a function of arity 2.\n\nExample:\n\n```> M = #{a => 2, b => 3, c=> 4, \"a\" => 1, \"b\" => 2, \"c\" => 4},\nPred = fun(K,V) -> is_atom(K) andalso (V rem 2) =:= 0 end,\nmaps:filter(Pred,M).\n#{a => 2,c => 4} ```\n\nfind(Key, Map) -> {ok, Value} | error\n\nTypes:\n\nKey = term()\nMap = #{}\nValue = term()\n\nReturns a tuple {ok, Value} where Value is the value associated with Key, or error if no value is associated with Key in Map.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{\"hi\" => 42},\nKey = \"hi\",\nmaps:find(Key,Map).\n{ok,42} ```\n\nfold(Fun, Init, Map) -> Acc\n\nTypes:\n\nFun = fun((K, V, AccIn) -> AccOut)\nInit = Acc = AccIn = AccOut = term()\nMap = #{}\nK = V = term()\n\nCalls F(K, V, AccIn) for every K to value V association in Map in arbitrary order. The function fun F/3 must return a new accumulator which is passed to the next successive call. maps:fold/3 returns the final value of the accumulator. The initial accumulator value Init is returned if the map is empty.\n\nExample:\n\n```> Fun = fun(K,V,AccIn) when is_list(K) -> AccIn + V end,\nMap = #{\"k1\" => 1, \"k2\" => 2, \"k3\" => 3},\nmaps:fold(Fun,0,Map).\n6```\n\nfrom_list(List) -> Map\n\nTypes:\n\nList = [{Key, Value}]\nKey = Value = term()\nMap = #{}\n\nThe function takes a list of key-value tuples elements and builds a map. The associations may be in any order and both keys and values in the association may be of any term. If the same key appears more than once, the latter (rightmost) value is used and the previous values are ignored.\n\nExample:\n\n```> List = [{\"a\",ignored},{1337,\"value two\"},{42,value_three},{\"a\",1}],\nmaps:from_list(List).\n#{42 => value_three,1337 => \"value two\",\"a\" => 1}```\n\nget(Key, Map) -> Value\n\nTypes:\n\nKey = term()\nMap = #{}\nValue = term()\n\nReturns the value Value associated with Key if Map contains Key.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map, or with a {badkey,Key} exception if no value is associated with Key.\n\nExample:\n\n```> Key = 1337,\nMap = #{42 => value_two,1337 => \"value one\",\"a\" => 1},\nmaps:get(Key,Map).\n\"value one\"```\n\nget(Key, Map, Default) -> Value | Default\n\nTypes:\n\nKey = term()\nMap = #{}\nValue = Default = term()\n\nReturns the value Value associated with Key if Map contains Key. If no value is associated with Key then returns Default.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{ key1 => val1, key2 => val2 }.\n#{key1 => val1,key2 => val2}\n> maps:get(key1, Map, \"Default value\").\nval1\n> maps:get(key3, Map, \"Default value\").\n\"Default value\"```\n\nis_key(Key, Map) -> boolean()\n\nTypes:\n\nKey = term()\nMap = #{}\n\nReturns true if map Map contains Key and returns false if it does not contain the Key.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{\"42\" => value}.\n#{\"42\"> => value}\n> maps:is_key(\"42\",Map).\ntrue\n> maps:is_key(value,Map).\nfalse```\n\nkeys(Map) -> Keys\n\nTypes:\n\nMap = #{}\nKeys = [Key]\nKey = term()\n\nReturns a complete list of keys, in arbitrary order, which resides within Map.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{42 => value_three,1337 => \"value two\",\"a\" => 1},\nmaps:keys(Map).\n[42,1337,\"a\"]```\n\nmap(Fun, Map1) -> Map2\n\nTypes:\n\nFun = fun((K, V1) -> V2)\nMap1 = Map2 = #{}\nK = V1 = V2 = term()\n\nThe function produces a new map Map2 by calling the function fun F(K, V1) for every K to value V1 association in Map1 in arbitrary order. The function fun F/2 must return the value V2 to be associated with key K for the new map Map2.\n\nExample:\n\n```> Fun = fun(K,V1) when is_list(K) -> V1*2 end,\nMap = #{\"k1\" => 1, \"k2\" => 2, \"k3\" => 3},\nmaps:map(Fun,Map).\n#{\"k1\" => 2,\"k2\" => 4,\"k3\" => 6}```\n\nmerge(Map1, Map2) -> Map3\n\nTypes:\n\nMap1 = Map2 = Map3 = #{}\n\nMerges two maps into a single map Map3. If two keys exists in both maps the value in Map1 will be superseded by the value in Map2.\n\nThe call will fail with a {badmap,Map} exception if Map1 or Map2 is not a map.\n\nExample:\n\n```> Map1 = #{a => \"value_one\", b => \"value_two\"},\nMap2 = #{a => 1, c => 2},\nmaps:merge(Map1,Map2).\n#{a => 1,b => \"value_two\",c => 2}```\n\nnew() -> Map\n\nTypes:\n\nMap = #{}\n\nReturns a new empty map.\n\nExample:\n\n```> maps:new().\n#{}```\n\nput(Key, Value, Map1) -> Map2\n\nTypes:\n\nKey = Value = term()\nMap1 = Map2 = #{}\n\nAssociates Key with value Value and inserts the association into map Map2. If key Key already exists in map Map1, the old associated value is replaced by value Value. The function returns a new map Map2 containing the new association and the old associations in Map1.\n\nThe call will fail with a {badmap,Map} exception if Map1 is not a map.\n\nExample:\n\n```> Map = #{\"a\" => 1}.\n#{\"a\" => 1}\n> maps:put(\"a\", 42, Map).\n#{\"a\" => 42}\n> maps:put(\"b\", 1337, Map).\n#{\"a\" => 1,\"b\" => 1337}```\n\nremove(Key, Map1) -> Map2\n\nTypes:\n\nKey = term()\nMap1 = Map2 = #{}\n\nThe function removes the Key, if it exists, and its associated value from Map1 and returns a new map Map2 without key Key.\n\nThe call will fail with a {badmap,Map} exception if Map1 is not a map.\n\nExample:\n\n```> Map = #{\"a\" => 1}.\n#{\"a\" => 1}\n> maps:remove(\"a\",Map).\n#{}\n> maps:remove(\"b\",Map).\n#{\"a\" => 1}```\n\nsize(Map) -> integer() >= 0\n\nTypes:\n\nMap = #{}\n\nThe function returns the number of key-value associations in the Map. This operation happens in constant time.\n\nExample:\n\n```> Map = #{42 => value_two,1337 => \"value one\",\"a\" => 1},\nmaps:size(Map).\n3```\n\nto_list(Map) -> [{Key, Value}]\n\nTypes:\n\nMap = #{}\nKey = Value = term()\n\nThe fuction returns a list of pairs representing the key-value associations of Map, where the pairs, [{K1,V1}, ..., {Kn,Vn}], are returned in arbitrary order.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{42 => value_three,1337 => \"value two\",\"a\" => 1},\nmaps:to_list(Map).\n[{42,value_three},{1337,\"value two\"},{\"a\",1}]```\n\nupdate(Key, Value, Map1) -> Map2\n\nTypes:\n\nKey = Value = term()\nMap1 = Map2 = #{}\n\nIf Key exists in Map1 the old associated value is replaced by value Value. The function returns a new map Map2 containing the new associated value.\n\nThe call will fail with a {badmap,Map} exception if Map1 is not a map, or with a {badkey,Key} exception if no value is associated with Key.\n\nExample:\n\n```> Map = #{\"a\" => 1}.\n#{\"a\" => 1}\n> maps:update(\"a\", 42, Map).\n#{\"a\" => 42}```\n\nvalues(Map) -> Values\n\nTypes:\n\nMap = #{}\nValues = [Value]\nValue = term()\n\nReturns a complete list of values, in arbitrary order, contained in map Map.\n\nThe call will fail with a {badmap,Map} exception if Map is not a map.\n\nExample:\n\n```> Map = #{42 => value_three,1337 => \"value two\",\"a\" => 1},\nmaps:values(Map).\n[value_three,\"value two\",1]```\n\nwith(Ks, Map1) -> Map2\n\nTypes:\n\nKs = [K]\nMap1 = Map2 = #{}\nK = term()\n\nReturns a new map Map2 with the keys K1 through Kn and their associated values from map Map1. Any key in Ks that does not exist in Map1 are ignored.\n\nExample:\n\n```> Map = #{42 => value_three,1337 => \"value two\",\"a\" => 1},\nKs = [\"a\",42,\"other key\"],\nmaps:with(Ks,Map).\n#{42 => value_three,\"a\" => 1}```\n\nwithout(Ks, Map1) -> Map2\n\nTypes:\n\nKs = [K]\nMap1 = Map2 = #{}\nK = term()\n\nReturns a new map Map2 without the keys K1 through Kn and their associated values from map Map1. Any key in Ks that does not exist in Map1 are ignored.\n\nExample:\n\n```> Map = #{42 => value_three,1337 => \"value two\",\"a\" => 1},\nKs = [\"a\",42,\"other key\"],\nmaps:without(Ks,Map).\n#{1337 => \"value two\"}```"
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"https://www.erlang.org/doc/erlang-logo.png",
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https://top1multiplicationquiz.com/multiplication-facts/multiply-fractions/ | [
"# Multiply the fractions and simplify the answer example & PDF worksheets\n\n## Multiply more than two fractions\n\nMultiply the fractions and simplify the answer example & PDF; learn how to multiply fractions and get further in the training with worksheets and test.\n\n### Example: Multiply 3/4 x 6/7 x 8/9 and reduce the answer\n\n#### Step 1",
null,
"Multiply the numerators of the three fractions, but leave in factored form.\nThese are, 3, 6 and 8.\nso you have 3 x 6 x 8\n\n#### Step 2\n\nMultiply the denominators of the given three fractions, but leave in factored form.\nThese are, 4, 7 and 9.\n4 x 7 x 9\n\n#### Step 3\n\nAnswer must also be in fraction form with new numerator and new denominator.\nThat is,\n(3 x 6 x 8) / (4 x 7 x 9)",
null,
"#### Step 4\n\nNow factor the both numerator and denominator and look for fractions that have value of 1.\n3 can be factored as (3 x 1)\n6 as (2 x 3)\n8 as (2 x 2 x 2)\n\nAnd in denominator,\n4 as (2 x 2)\n7 as (7 x 1)\n9 as (3 x 3)\n\n#### Step 5\n\nReplace with these values:",
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"",
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""
] | [
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"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-60.png",
null,
"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-67.png",
null,
"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-61.png",
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"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-62.png",
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"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-63.png",
null,
"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-64.png",
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"http://top1multiplicationquiz.com/activities/multiplication-facts/multiply-fractions/multiply-multiple-fractions-65.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8913188,"math_prob":0.9859509,"size":964,"snap":"2020-34-2020-40","text_gpt3_token_len":288,"char_repetition_ratio":0.16145833,"word_repetition_ratio":0.10204082,"special_character_ratio":0.2987552,"punctuation_ratio":0.08256881,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99918103,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T07:37:29Z\",\"WARC-Record-ID\":\"<urn:uuid:08ebd095-c2bb-42b4-b678-b5b25f7125db>\",\"Content-Length\":\"55211\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d34a784-6cc8-4ed1-a3ab-e5990e725496>\",\"WARC-Concurrent-To\":\"<urn:uuid:55343bc5-58ef-4bc4-a071-e50dbcdd3280>\",\"WARC-IP-Address\":\"66.85.133.203\",\"WARC-Target-URI\":\"https://top1multiplicationquiz.com/multiplication-facts/multiply-fractions/\",\"WARC-Payload-Digest\":\"sha1:2M2KLMEQ2WULXOCNMUJWVZZQOG2F2CUP\",\"WARC-Block-Digest\":\"sha1:SR3XRTVBSC4O4P5V3BV5MAA6RYA7JUXJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400265461.58_warc_CC-MAIN-20200927054550-20200927084550-00365.warc.gz\"}"} |
https://www.physicsforums.com/threads/hints-needed.16937/ | [
"# Hints needed.\n\nCan anyone give some hints on this problem?\nThis is about current and resistance.\n\nProblem: A steady beam of alpha particles (q= +2e) traveling with constant kinetic energy 20MeV carries a current of 0.25 micro ampere.\n(a) If the beam is directed perpendicular to a plane surfece, how many alpha particales strike the surface in 3.0s?\n(b) At any instant, how many alpha particles are there in a given 20 cm length of the beam?\n(c) Through what potential difference was it necessary to accelearte each alpha particle from rest to bring it to an energy of 20 MeV?\n\nThanks,\n\nShawnD\nOriginally posted by xerox2ooo\nProblem: A steady beam of alpha particles (q= +2e) traveling with constant kinetic energy 20MeV carries a current of 0.25 micro ampere.\n(a) If the beam is directed perpendicular to a plane surfece, how many alpha particales strike the surface in 3.0s?\nRemember what an ampere is? That's coulombs per second. Just start multiplying stuff until the units work out the way you want them to.\n\n$$(\\frac{coulomb}{second})(seconds)(\\frac{\\alpha}{coulomb})$$\n\ncoulombs and seconds will cancel out and you're left with the number of alpha particles.\n\n(b) At any instant, how many alpha particles are there in a given 20 cm length of the beam?\nThis one I don't know.\n\n(c) Through what potential difference was it necessary to accelearte each alpha particle from rest to bring it to an energy of 20 MeV?\n\n$$E = Vq$$\n\n$$V = \\frac{E}{q}$$\n\nE is 20MeV, q is 2e\n\nLast edited:\n\nOriginally posted by ShawnD\nJust start multiplying stuff until the units work out the way you want them to.\nThat always seems to work.\n\nThanks guys.\n\nYou guys are the best."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8911965,"math_prob":0.95046085,"size":1108,"snap":"2021-43-2021-49","text_gpt3_token_len":276,"char_repetition_ratio":0.12137681,"word_repetition_ratio":0.9025641,"special_character_ratio":0.2463899,"punctuation_ratio":0.1036036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99203855,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T04:15:07Z\",\"WARC-Record-ID\":\"<urn:uuid:f4d94957-ec17-4214-850e-451a0844ba49>\",\"Content-Length\":\"69902\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e689071-3ef2-47a6-a0d4-85e29c39e201>\",\"WARC-Concurrent-To\":\"<urn:uuid:464eb290-75e5-4f38-b141-f6647b8444ef>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/hints-needed.16937/\",\"WARC-Payload-Digest\":\"sha1:BEC6MGR3OBVGI7JQZMAD42APS3PIO6P4\",\"WARC-Block-Digest\":\"sha1:LNSXOIJFTTHJK3CVUZPOYJ6X2WAHHVRJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363332.1_warc_CC-MAIN-20211207014802-20211207044802-00351.warc.gz\"}"} |
https://www.coursehero.com/file/66352778/ProblemSet8-soln-19pdf/ | [
"# ProblemSet8_soln_19.pdf - ECH 141 Problem Set#8 1 Two...\n\n• 5\n• 100% (2) 2 out of 2 people found this document helpful\n\nThis preview shows page 1 - 2 out of 5 pages.\n\nECH 141 Problem Set #8 1. Two immiscible liquids A and B of specific gravities 0.8 and 1.0, respectively, flow into a closed tank in which they separate into two layers. Pure A and pure B are taken off by pipes which discharge to atmosphere at a point 5 ft below the A/B interface in the tank. The flow in each pipe is controlled by an orifice. Calculate the size of the orifice discharging B when the one discharging A is 1 inch in diameter and the total feed is 5000 lbs/hr containing 60% by weight of A. The outlet pipes are nearly identical, and are much larger than the orifices. The orifice discharge coefficients are each C D =0.6, so that if the head loss across the orifice is h O , then the velocity v through the orifice is given by the orifice equation (similar to Torricelli’s law, see Eq. (8.5-15) in text), v = C D 2gh O . Note that since h O = 1 C D v 2 2g , this orifice equation corresponds to a loss coefficient K (i.e., fraction of velocity head loss) of 1/C D . It may be assumed that none of the head loss across the orifices is recovered and that both fluids have negligible vapor pressure. Neglect all frictional losses except for the losses in the orifices. For water, assume that ρ g = 62.4 lb f ft 3 (water)\n•",
null,
"•",
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"•",
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"https://www.coursehero.com/assets/img/doc-landing/start-quote.svg",
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"https://www.coursehero.com/assets/img/doc-landing/start-quote.svg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.894408,"math_prob":0.96948344,"size":1234,"snap":"2021-43-2021-49","text_gpt3_token_len":359,"char_repetition_ratio":0.13658537,"word_repetition_ratio":0.008438818,"special_character_ratio":0.25850892,"punctuation_ratio":0.102564104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.988081,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T06:42:53Z\",\"WARC-Record-ID\":\"<urn:uuid:eb21e5f3-cb07-4989-8796-7c61d653fb87>\",\"Content-Length\":\"239374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c32e423d-d0cf-45ce-b3f4-b5d8f0a3a5e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:840068a7-ccb5-4bc4-bbcd-62bbc6c45cff>\",\"WARC-IP-Address\":\"104.17.92.47\",\"WARC-Target-URI\":\"https://www.coursehero.com/file/66352778/ProblemSet8-soln-19pdf/\",\"WARC-Payload-Digest\":\"sha1:JUD542A4TVBGY7HQCMNQ35F37LPJMJKE\",\"WARC-Block-Digest\":\"sha1:44H6BH3U7UOI5EB27XUYJM7JE26BNWJX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358953.29_warc_CC-MAIN-20211130050047-20211130080047-00552.warc.gz\"}"} |
https://www.geeksforgeeks.org/difference-pointer-array-c/ | [
"# Difference between pointer and array in C?\n\nPointers are used for storing address of dynamically allocated arrays and for arrays which are passed as arguments to functions. In other contexts, arrays and pointer are two different things, see the following programs to justify this statement.\n\nBehavior of sizeof operator\n\n## C\n\n `// 1st program to show that array and pointers are different ` `#include ` ` ` `int` `main() ` `{ ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``int` `*ptr = arr; ` ` ` ` ``// sizof(int) * (number of element in arr[]) is printed ` ` ``printf``(``\"Size of arr[] %ld\\n\"``, ``sizeof``(arr)); ` ` ` ` ``// sizeof a pointer is printed which is same for all type ` ` ``// of pointers (char *, void *, etc) ` ` ``printf``(``\"Size of ptr %ld\"``, ``sizeof``(ptr)); ` ` ``return` `0; ` `} `\n\n## C++\n\n `// 1st program to show that array and pointers are different ` `#include ` `using` `namespace` `std; ` ` ` `int` `main() { ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``int` `*ptr = arr; ` ` ` ` ``// sizof(int) * (number of element in arr[]) is printed ` ` ``cout << ``\"Size of arr[] \"``<< ``sizeof``(arr)<<``\"\\n\"``; ` ` ` ` ``// sizeof a pointer is printed which is same for all type ` ` ``// of pointers (char *, void *, etc) ` ` ``cout << ``\"Size of ptr \"``<< ``sizeof``(ptr); ` ` ``return` `0; ` `} `\n\nOutput:\n\n```Size of arr[] 24\nSize of ptr 8```\n\nAssigning any address to an array variable is not allowed.\n\n `// IInd program to show that array and pointers are different ` `#include ` ` ` `int` `main() ` `{ ` ` ``int` `arr[] = {10, 20}, x = 10; ` ` ``int` `*ptr = &x; ``// This is fine ` ` ``arr = &x; ``// Compiler Error ` ` ``return` `0; ` `}`\n\nOutput:\n\n``` Compiler Error: incompatible types when assigning to\ntype 'int' from type 'int *' ```\n\nSee the previous post on this topic for more differences.\n\nAlthough array and pointer are different things, following properties of array make them look similar.\n\n1. Array name gives address of first element of array.\nConsider the following program for example.\n\n## C\n\n `#include ` ` ` `int` `main() ` `{ ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``// Assigns address of array to ptr ` ` ``int` `*ptr = arr; ` ` ``printf``(``\"Value of first element is %d\"``, *ptr); ` ` ``return` `0; ` `} `\n\n## C++\n\n `// 1st program to show that array and pointers are different ` `#include ` `using` `namespace` `std; ` ` ` `int` `main() { ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ` ` ``// Assigns address of array to ptr ` ` ``int` `*ptr = arr; ` ` ``cout << ``\"Value of first element is \"` `<< *ptr; ` ` ``return` `0; ` `} `\n\nOutput:\n\n`Value of first element is 10`\n2.\n\n3. Array members are accessed using pointer arithmetic.\nCompiler uses pointer arithmetic to access array element. For example, an expression like “arr[i]” is treated as *(arr + i) by the compiler. That is why the expressions like *(arr + i) work for array arr, and expressions like ptr[i] also work for pointer ptr.\n\n## C\n\n `#include ` ` ` `int` `main() ` `{ ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``int` `*ptr = arr; ` ` ``printf``(``\"arr = %d\\n\"``, arr); ` ` ``printf``(``\"*(arr + 2) = %d\\n\"``, *(arr + 2)); ` ` ``printf``(``\"ptr = %d\\n\"``, ptr); ` ` ``printf``(``\"*(ptr + 2) = %d\\n\"``, *(ptr + 2)); ` ` ``return` `0; ` `} `\n\n## C++\n\n `#include ` `using` `namespace` `std; ` ` ` `int` `main() { ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``int` `*ptr = arr; ` ` ``cout << ``\"arr = \"``<< arr <<``\"\\n\"``; ` ` ``cout << ``\"*(arr + 2) = \"``<< *(arr + 2)<<``\"\\n\"``; ` ` ``cout << ``\"ptr = \"``<< ptr<< ``\"\\n\"``; ` ` ``cout << ``\"*(ptr + 2) = \"``<< *(ptr + 2)<< ``\"\\n\"``; ` ` ``return` `0; ` `} `\n\nOutput:\n\n```arr = 30\n*(arr + 2) = 30\nptr = 30\n*(ptr + 2) = 30\n```\n4.\n\n5. Array parameters are always passed as pointers, even when we use square brackets.\n\n `#include ` ` ` `int` `fun(``int` `ptr[]) ` `{ ` ` ``int` `x = 10; ` ` ` ` ``// size of a pointer is printed ` ` ``printf``(``\"sizeof(ptr) = %d\\n\"``, ``sizeof``(ptr)); ` ` ` ` ``// This allowed because ptr is a pointer, not array ` ` ``ptr = &x; ` ` ` ` ``printf``(``\"*ptr = %d \"``, *ptr); ` ` ` ` ``return` `0; ` `} ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = {10, 20, 30, 40, 50, 60}; ` ` ``fun(arr); ` ` ``return` `0; ` `} `\n\nOutput:\n\n```sizeof(ptr) = 8\n*ptr = 10```\n\nPlease refer Pointer vs Array in C for more details.\n\nDon’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.\n\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.769768,"math_prob":0.98897797,"size":4948,"snap":"2020-24-2020-29","text_gpt3_token_len":1482,"char_repetition_ratio":0.12924758,"word_repetition_ratio":0.26489362,"special_character_ratio":0.3494341,"punctuation_ratio":0.1455898,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99740493,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T12:07:33Z\",\"WARC-Record-ID\":\"<urn:uuid:9d5512de-a5ba-434b-8575-8389a4161212>\",\"Content-Length\":\"147464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6521fa19-d7e8-4e6d-adfd-d6ae5a866e74>\",\"WARC-Concurrent-To\":\"<urn:uuid:894bf592-d343-471d-b3de-c3ff3ab7a0fc>\",\"WARC-IP-Address\":\"168.143.243.236\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/difference-pointer-array-c/\",\"WARC-Payload-Digest\":\"sha1:HEILLARMMX4LAOXVD7L4L5YSSXYSTP67\",\"WARC-Block-Digest\":\"sha1:DE4ZEIUC2CDYDYPI3ZSDYYTTVHPWGTDC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886121.45_warc_CC-MAIN-20200704104352-20200704134352-00285.warc.gz\"}"} |
http://www-ksl.stanford.edu/email-archives/interlingua.messages/502.html | [
"# Re: Propositions\n\nschubert@cs.rochester.edu\n```Date: Tue, 15 Feb 1994 14:57:50 -0500\nFrom: schubert@cs.rochester.edu\nMessage-id: <199402151957.OAA20475@cherry.cs.rochester.edu>\nTo: cg@cs.umn.edu, interlingua@ISI.EDU, sowa@turing.pacss.binghamton.edu\nSubject: Re: Propositions\n```\n```Hi John,\nI've thought about propositions from a similar perspective (e.g.,\nin connection with the problem of deciding what is \"true by definition\",\ngiven some definitions (and contingent facts). Things seem a little more\ncomplicated to me, though.\n\nFor one thing, to get FOL you need to consider equality. For another,\nI don't really know how to sort conjuncts (or, as Pat suggests, treat\nthem as sets). The trouble lies in the variable names. For instance,\nthe following are intuitively identical propositions\n\n(E x)(E z)~((E v)(E w)~((~(P(x)&P(v)) & ~(P(z)&P(x))) & R(v,w,x))\n(E v)(E w)~((E x)(E y)~((~(P(v)&P(w)) & ~(P(y)&P(w))) & R(y,x,w))\n\n(the second is obtained from the first by variable renaming x z v w\nw v y x\nand permutation of conjuncts). The brute force strategy of trying all\nrenamings (and sorting the conjuncts for each renaming) would of course\nbe exponentially complex.\n\n-Len\n\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8010149,"math_prob":0.889894,"size":1161,"snap":"2022-27-2022-33","text_gpt3_token_len":349,"char_repetition_ratio":0.090751946,"word_repetition_ratio":0.0,"special_character_ratio":0.29112834,"punctuation_ratio":0.18604651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9895453,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T14:44:20Z\",\"WARC-Record-ID\":\"<urn:uuid:3a1b7f28-04d9-4376-b959-e3ea226d317d>\",\"Content-Length\":\"2146\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b224538a-aebd-4029-a06a-adc5b8587263>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf9b79d7-9860-4aa4-abd7-bc99d65aeaeb>\",\"WARC-IP-Address\":\"171.64.71.15\",\"WARC-Target-URI\":\"http://www-ksl.stanford.edu/email-archives/interlingua.messages/502.html\",\"WARC-Payload-Digest\":\"sha1:PZX2QM5RJPLZJDENZH2IXTTHQURDMLOT\",\"WARC-Block-Digest\":\"sha1:7VJDGF6EXBCUJFZZDXBMYRC7OFFDWMN5\",\"WARC-Identified-Payload-Type\":\"message/rfc822\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104244535.68_warc_CC-MAIN-20220703134535-20220703164535-00277.warc.gz\"}"} |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/calcium-blank-carbimide | [
"# Weight of Calcium carbimide\n\n## calcium carbimide: convert volume to weight\n\n### Weight of 1 cubic centimeter of Calcium carbimide\n\n carat 11.45 ounce 0.08 gram 2.29 pound 0.01 kilogram 0 tonne 2.29 × 10-6 milligram 2 290\n\n#### How many moles in 1 cubic centimeter of Calcium carbimide?\n\nThere are 28.59 millimoles in 1 cubic centimeter of Calcium carbimide\n\n### The entered volume of Calcium carbimide in various units of volume\n\n centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2\n• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.\n\n#### Foods, Nutrients and Calories\n\nPENNE RIGATE EASY PASTA, UPC: 041220546190 weigh(s) 148 grams per metric cup or 4.9 ounces per US cup, and contain(s) 150 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]\n\nTaurine in SUPER SERUM PROTEIN, NATURAL VANILLA, UPC: 805974204306\n\n#### Gravels, Substances and Oils\n\nCaribSea, Marine, Aragonite, Seaflor Special Grade Reef Sand weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nGermane [GeH4] weighs 3.133 kg/m³ (0.00181099 oz/in³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]\n\nVolume to weightweight to volume and cost conversions for Tung oil with temperature in the range of 10°C (50°F) to 140°C (284°F)\n\n#### Weights and Measurements\n\nA zeptocoulomb is a SI-multiple (see prefix zepto) of the electric charge unit coulomb and equal to equal to 1.0 × 10-21 coulomb\n\nThe units of data measurement were introduced to manage and operate digital information.\n\nt/metric tsp to mg/metric tbsp conversion table, t/metric tsp to mg/metric tbsp unit converter or convert between all units of density measurement.\n\n#### Calculators\n\nCalculate the volume and area of the surface of a quarter cylinder"
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https://math.stackexchange.com/questions/1043480/how-to-factor-ideals-in-a-quadratic-number-field | [
"# How to factor ideals in a quadratic number field?\n\nLet $w=\\sqrt{-5}$; we work in $\\mathbb Z[w]$; which is not a PID or UFD.\n\nNow in Sage (3) factors in $(3,1+w)(3,2+w)$ and (7) in $(7,3+w)(7,4+w)$.\n\nPlease explain clearly how factorization of ideals in a quadratic non-PID works.\n\nIs it because this is a non-PID quadratic numberfield that the ideal factors in two non-principal ideals?\n\nThe numbers 11 and 13 are irreducible as 3 and 7 are, yet (11) and (13) do not factor as ideals. How do I determine the difference in this context?\n\nEDIT: implementation of solution provided by Dietrich Burde (second cell in Sage Worksheet)",
null,
"• You might want to read this paper. – rogerl Nov 29 '14 at 13:53\n\nFactorisation of ideals in the ring of integers $\\mathcal{O}_K$ of a quadratic number field $K=\\mathbb{Q}(\\sqrt{d})$ works as follows. Let $D$ be the discriminant, i.e., $D=d$ or $D=4d$.\n\nTheorem: Let $p>2$ be a prime and $K$ a quadratic number field as above.\n\n1.) If the Kronecker symbol $(D/p)=0$ then $(p)=(p,\\sqrt{d})^2$ ramifies.\n2.) If the Kronecker symbol $(D/p)=1$ then $(p)=P_1P_2$ splits, with two prime ideals $P_1,P_2$.\n3.) If the Kronecker symbol $(D/p)=-1$ then $(p)$ is inert, i.e., remains prime.\n\nIn the second case, $P_1=(p, x+\\sqrt{d})$ and $P_2=(p, x-\\sqrt{d})$ with an integer $x$ such that $D\\equiv x^2 (p)$. For $p=2$ there is a similar result.\n\nThis gives a formula how to factorize ideals, regardless whether or not $\\mathcal{O}_K$ is a PID or UFD.\n\n• Looks awesome, very clear indeed! Will go through my books and see why I missed this. – nilo de roock Nov 29 '14 at 17:07\n• Your books probably say \"Legendre symbol\" rather than \"Kronecker symbol.\" And your CAS might expect \"Jacobi symbol.\" – Robert Soupe Nov 29 '14 at 18:27\n• I mentioned Sage in the question, in Sage Legendre symbol is referred to as kronecker – nilo de roock Nov 29 '14 at 18:41\n• I added a follow-up question math.stackexchange.com/questions/1044017/… – nilo de roock Nov 29 '14 at 19:21\n• For what it's worth, Mathematica has always had JacobiSymbol, whereas KroneckerSymbol was introduced in Version 6.0. – Robert Soupe Nov 29 '14 at 20:07"
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https://emedia.rmit.edu.au/learninglab/content/activity-1-8 | [
"",
null,
"## Fractions and decimals activity 1\n\n### Activity\n\nUse the coloured squares as a guide for dividing whole numbers by fractions. Type your answers into the text box then click the ‘Check answer’ button.\n\n1 25%\n2 50%\n3 75%\n4 100%",
null,
"1. How many ¼ are in 1 square?\n\n2. What is 1 divided by ¼ ?",
null,
"",
null,
"1. How many ¼ are in 2 squares?\n\n2. What is 2 divided by ¼?",
null,
"",
null,
"",
null,
"",
null,
"What is 4 divided by ¼?\n\nWhat is 5 divided by ¼?\n\n### Description\n\nUse the coloured squares as a guide for dividing whole numbers by fractions.\n\n### Question 1\n\nSquare image description: Four squares divided into four parts. Each square has a value of 1/4.\n\n1. How many 1/4 are in 1 square?\n2. What is 1 divided by 1/4 ?\n\n• 4\n• 4\n\n### Question 2\n\n1. How many 1/4 are in 2 squares?\n2. What is 2 divided by 1/4?\n\n• 8\n• 8\n\n### Question 3\n\n1. What is 4 divided by 1/4?"
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null,
"https://emedia.rmit.edu.au/learninglab/sites/default/files/print-logo.png",
null,
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null,
"https://emedia.rmit.edu.au/learninglab/sites/default/files/enhanced/activity8a.svg",
null,
"https://emedia.rmit.edu.au/learninglab/sites/default/files/enhanced/activity8a.svg",
null,
"https://emedia.rmit.edu.au/learninglab/sites/default/files/enhanced/activity8a.svg",
null,
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null,
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https://www.edenpr.org/academics/curriculum | [
"# Curriculum\n\nEden Prairie schools provide a rigorous and enriching curriculum for each student, at every grade and ability level. While the core curriculum focuses on fundamentals, teachers and families help students achieve their highest potential by choosing special programs that fit their talents, interests, and needs.\n\nBeginning with Early Childhood Education and continuing through high school graduation, Eden Prairie students and their teachers are focused on a trajectory for success. Every year our students achieve some of the state’s most impressive numbers in graduation rate, assessments scores, scholarships, and college acceptance. Students leave Eden Prairie schools well prepared for college and life.\n\n## Math Curriculum\n\n#### EPS Math Instructional Vision\n\nExcellent mathematics instruction will lead each Eden Prairie Schools learner to....\n\n1. Form a positive math identity\n2. Collaborate within a math community\n3. Understand how and why math concepts work\n4. Use math procedures fluently\n5. Apply math to real-world situations\n\n## K-5 Math Information for Parents\n\nEden Prairie Schools has used Math Expressions (grades K-5) and Big Ideas Math (grades 6-8) as the curricular resources since the 2017-18 school year. The adoption is the result of a district wide steering committee recommendation to purchase resources that align with best practices in student learning of math. These resources are a guiding foundation of math instruction that also allow for rich mathematical learning, differentiation, depth and complexity.\n\n• Math Expressions: The program emphasizes understanding, computational fluency, and real life applications through problem solving. Strategies and deep understanding are highlighted through visual representations, mathematical models, and student explorations.\n• Big Ideas Math: Students will explore mathematics, communicate in the language of mathematics, ask questions regarding mathematical ideas and processes, and find meaning in the mathematics that they study and use in everyday life. The series provides high levels of mathematical understanding and application for math students in grades 6-8.\n• Kendall Hunt is used in Grades 9-11 for Discovering Geometry, Algebra II and Functions, Stats, and Trigonometry (FST).\n\nThese resources provide a support for each student’s learning by allowing for real-world mathematical concepts to come alive through rich mathematical problems. Rich opportunities for meaningful math talks are present because of this."
] | [
null
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https://morioh.com/p/76d449055598 | [
"# NumPy Chi-Square Distribution",
null,
"This is a detailed tutorial of NumPy Chi-Square Distribution. Learn to get Chi-Square Distribution data using NumPy and visualize using Seaborn.\n\nThis is a detailed tutorial of NumPy Chi-Square Distribution. Learn to get Chi-Square Distribution data using NumPy and visualize using Seaborn.\n\n### Chi-Square Distribution\n\nChi-Square Distribution is one of the cases of the gamma distribution, and in most cases, it is helpful in probability distribution and also in the hypothesis testing. It also helps in working up with the construction of confidence intervals. It helps in population variance when the underlying distribution is normal. There is not much application of direct modelling of natural phenomena. There are several hypothesis tests that come under this distribution like the Chi-square test of independence and also Chi-square test of goodness of fit.\n\nChi-Square distribution takes in two parameters:\n\n• `df` – this depicts a degree of freedom.\n• `size` – here we will give the shape of the array which will be given to us as output.\n\n## Guide to Python Programming Language\n\nGuide to Python Programming Language\n\n## Python Programming: A Beginner’s Guide\n\nPython is an interpreted, high-level, powerful general-purpose programming language. You may ask, Python’s a snake right? and Why is this programming language named after it?\n\n## Python Hello World Program - Create & Run Your First Python Program in PyCharm\n\nPython Hello World Program - Your first step towards Python world. Learn how to create the Hello World Python program in PyCharm.\n\n## NumPy in Python | NumPy Python Tutorial | Python Programming\n\nNumPy in Python explains what exactly is Numpy and how it is better than Lists. It also explains various Numpy operations with examples.\n\n## Python Programming Tutorials For Beginners\n\nPython Programming Tutorials For Beginners"
] | [
null,
"https://i.morioh.com/200824/c8a620dc.webp",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8938481,"math_prob":0.5049062,"size":1528,"snap":"2021-04-2021-17","text_gpt3_token_len":297,"char_repetition_ratio":0.1496063,"word_repetition_ratio":0.14529915,"special_character_ratio":0.17997383,"punctuation_ratio":0.0882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99229425,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-25T09:44:25Z\",\"WARC-Record-ID\":\"<urn:uuid:1137c36b-e91c-4592-81a9-3cfb3d3f2df7>\",\"Content-Length\":\"10198\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3322228d-e136-4829-a553-baa666150e47>\",\"WARC-Concurrent-To\":\"<urn:uuid:15a29b4c-7ee8-4ba3-9c47-ace0a4841d1b>\",\"WARC-IP-Address\":\"172.67.177.194\",\"WARC-Target-URI\":\"https://morioh.com/p/76d449055598\",\"WARC-Payload-Digest\":\"sha1:4MJNYYZYQCYYYIYRN5KSUEQASQ3DVYXP\",\"WARC-Block-Digest\":\"sha1:SVOKBGKLAJ2LLXCWFPVSISRNSJGOUECE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703565541.79_warc_CC-MAIN-20210125092143-20210125122143-00017.warc.gz\"}"} |
https://www.colorhexa.com/00918b | [
"#00918b Color Information\n\nIn a RGB color space, hex #00918b is composed of 0% red, 56.9% green and 54.5% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 4.1% yellow and 43.1% black. It has a hue angle of 177.5 degrees, a saturation of 100% and a lightness of 28.4%. #00918b color hex could be obtained by blending #00ffff with #002317. Closest websafe color is: #009999.\n\n• R 0\n• G 57\n• B 55\nRGB color chart\n• C 100\n• M 0\n• Y 4\n• K 43\nCMYK color chart\n\n#00918b color description : Dark cyan.\n\n#00918b Color Conversion\n\nThe hexadecimal color #00918b has RGB values of R:0, G:145, B:139 and CMYK values of C:1, M:0, Y:0.04, K:0.43. Its decimal value is 37259.\n\nHex triplet RGB Decimal 00918b `#00918b` 0, 145, 139 `rgb(0,145,139)` 0, 56.9, 54.5 `rgb(0%,56.9%,54.5%)` 100, 0, 4, 43 177.5°, 100, 28.4 `hsl(177.5,100%,28.4%)` 177.5°, 100, 56.9 009999 `#009999`\nCIE-LAB 54.147, -33.459, -6.109 14.784, 22.113, 27.914 0.228, 0.341, 22.113 54.147, 34.012, 190.347 54.147, -42.506, -4.042 47.025, -26.175, -2.277 00000000, 10010001, 10001011\n\nColor Schemes with #00918b\n\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #910006\n``#910006` `rgb(145,0,6)``\nComplementary Color\n• #009143\n``#009143` `rgb(0,145,67)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #004e91\n``#004e91` `rgb(0,78,145)``\nAnalogous Color\n• #914300\n``#914300` `rgb(145,67,0)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #91004e\n``#91004e` `rgb(145,0,78)``\nSplit Complementary Color\n• #918b00\n``#918b00` `rgb(145,139,0)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #8b0091\n``#8b0091` `rgb(139,0,145)``\n• #069100\n``#069100` `rgb(6,145,0)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #8b0091\n``#8b0091` `rgb(139,0,145)``\n• #910006\n``#910006` `rgb(145,0,6)``\n• #004542\n``#004542` `rgb(0,69,66)``\n• #005e5a\n``#005e5a` `rgb(0,94,90)``\n• #007873\n``#007873` `rgb(0,120,115)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #00aba3\n``#00aba3` `rgb(0,171,163)``\n• #00c4bc\n``#00c4bc` `rgb(0,196,188)``\n• #00ded4\n``#00ded4` `rgb(0,222,212)``\nMonochromatic Color\n\nAlternatives to #00918b\n\nBelow, you can see some colors close to #00918b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #009167\n``#009167` `rgb(0,145,103)``\n• #009173\n``#009173` `rgb(0,145,115)``\n• #00917f\n``#00917f` `rgb(0,145,127)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #008b91\n``#008b91` `rgb(0,139,145)``\n• #007f91\n``#007f91` `rgb(0,127,145)``\n• #007391\n``#007391` `rgb(0,115,145)``\nSimilar Colors\n\n#00918b Preview\n\nThis text has a font color of #00918b.\n\n``<span style=\"color:#00918b;\">Text here</span>``\n#00918b background color\n\nThis paragraph has a background color of #00918b.\n\n``<p style=\"background-color:#00918b;\">Content here</p>``\n#00918b border color\n\nThis element has a border color of #00918b.\n\n``<div style=\"border:1px solid #00918b;\">Content here</div>``\nCSS codes\n``.text {color:#00918b;}``\n``.background {background-color:#00918b;}``\n``.border {border:1px solid #00918b;}``\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, #000807 is the darkest color, while #f3ffff is the lightest one.\n\n• #000807\n``#000807` `rgb(0,8,7)``\n• #001b1a\n``#001b1a` `rgb(0,27,26)``\n• #002f2d\n``#002f2d` `rgb(0,47,45)``\n• #004340\n``#004340` `rgb(0,67,64)``\n• #005653\n``#005653` `rgb(0,86,83)``\n• #006a65\n``#006a65` `rgb(0,106,101)``\n• #007d78\n``#007d78` `rgb(0,125,120)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\n• #00a59e\n``#00a59e` `rgb(0,165,158)``\n• #00b8b1\n``#00b8b1` `rgb(0,184,177)``\n• #00ccc3\n``#00ccc3` `rgb(0,204,195)``\n• #00dfd6\n``#00dfd6` `rgb(0,223,214)``\n• #00f3e9\n``#00f3e9` `rgb(0,243,233)``\n• #08fff5\n``#08fff5` `rgb(8,255,245)``\n• #1bfff6\n``#1bfff6` `rgb(27,255,246)``\n• #2ffff6\n``#2ffff6` `rgb(47,255,246)``\n• #43fff7\n``#43fff7` `rgb(67,255,247)``\n• #56fff8\n``#56fff8` `rgb(86,255,248)``\n• #6afff9\n``#6afff9` `rgb(106,255,249)``\n• #7dfffa\n``#7dfffa` `rgb(125,255,250)``\n• #91fffa\n``#91fffa` `rgb(145,255,250)``\n• #a5fffb\n``#a5fffb` `rgb(165,255,251)``\n• #b8fffc\n``#b8fffc` `rgb(184,255,252)``\n• #ccfffd\n``#ccfffd` `rgb(204,255,253)``\n• #dffffe\n``#dffffe` `rgb(223,255,254)``\n• #f3ffff\n``#f3ffff` `rgb(243,255,255)``\nTint Color Variation\n\nTones of #00918b\n\nA tone is produced by adding gray to any pure hue. In this case, #434e4e is the less saturated color, while #00918b is the most saturated one.\n\n• #434e4e\n``#434e4e` `rgb(67,78,78)``\n• #3d5453\n``#3d5453` `rgb(61,84,83)``\n• #385958\n``#385958` `rgb(56,89,88)``\n• #325f5d\n``#325f5d` `rgb(50,95,93)``\n• #2d6462\n``#2d6462` `rgb(45,100,98)``\n• #276a67\n``#276a67` `rgb(39,106,103)``\n• #21706c\n``#21706c` `rgb(33,112,108)``\n• #1c7571\n``#1c7571` `rgb(28,117,113)``\n• #167b77\n``#167b77` `rgb(22,123,119)``\n• #11807c\n``#11807c` `rgb(17,128,124)``\n• #0b8681\n``#0b8681` `rgb(11,134,129)``\n• #068b86\n``#068b86` `rgb(6,139,134)``\n• #00918b\n``#00918b` `rgb(0,145,139)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #00918b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
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http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/eqn.html | [
"Because of noise on the channel, the received waveform will never be exactly, identical to the transmitted waveform. In order to minimize the probability that the receiver detects a bit in error, we use the following metric to determine whether we have received a zero or a one:",
null,
"If, given the received signal, the probability that the transmitted signal was a one is greater than the probability that the transmitted signal was a zero, we assume the transmitted signal was a one. This probability can be re-expressed as follows:",
null,
"Since the denominator is common to both sides of Eqn. #1, it can be ignored. The numerator can be expressed as the probability density function of Gaussian noise, as shown below:",
null,
"Since the exponential function is monotonic, the decision metric can be further simplified as follows, assuming that the transmitted bit is a one when:",
null,
"This is conceptually intuitive, since this means that we choose a binary '1' when the received signal is 'closer' in distance to the waveform for a '1' than the waveform for a '0'. Also, the square terms can be expanded as follows:",
null,
"Since the energy of the received signal is common to both sides of the decision metric, it can be dropped. By choosing the energy of the waveforms for each binary value to be identical, the energy of the signal terms cancel as well. Our decision metric is further simplified to choosing a binary '1' when:",
null,
"This means that we should assume a binary one was transmitted when the correlation of the received signal with the waveform for a '1' is larger than the correlation with the other waveform. Again, this makes intuitive sense because the correlation operation is a measurement of similarity between two signals. The correlation operator can be expressed explicitly in its integral form as follows:",
null,
"Because we are dealing with digital, time discrete hardware, the integration is reduced to a sum. Finally, the metric used to determine whether what was received was a binary '0' or '1' is simply:",
null,
"If this value is greater that zero, we infer that a binary '1' was transmitted. Likewise, it the value is less than zero, we assume a binary '0' has been transmitted. This is the core of our receiver code."
] | [
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig8.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig7.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig4.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig9.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig5.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig3.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig2.gif",
null,
"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/fig1.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9668684,"math_prob":0.9821718,"size":2196,"snap":"2021-31-2021-39","text_gpt3_token_len":446,"char_repetition_ratio":0.13412409,"word_repetition_ratio":0.04021448,"special_character_ratio":0.20491803,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99857974,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T03:20:28Z\",\"WARC-Record-ID\":\"<urn:uuid:06cfbdae-86a0-4615-aa48-f0dfd8727e67>\",\"Content-Length\":\"3188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1258fa8-ea04-4780-b255-f2d3bae3dd34>\",\"WARC-Concurrent-To\":\"<urn:uuid:0519ea78-03f8-4b09-9113-87b84b07ed11>\",\"WARC-IP-Address\":\"128.253.97.40\",\"WARC-Target-URI\":\"http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2000/mehr/eqn.html\",\"WARC-Payload-Digest\":\"sha1:EZLJMG5PZFF4JEA7ISU4ZSGNK7VY6N3I\",\"WARC-Block-Digest\":\"sha1:VBJFDUKTTBKX3S4JMCHYXZRVX3BIYYP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057589.14_warc_CC-MAIN-20210925021713-20210925051713-00400.warc.gz\"}"} |
https://accountingdrive.com/cash-flow-analysis-ratios/ | [
"# 𝐀𝐧𝐚𝐥𝐲𝐳𝐞 𝐘𝐨𝐮𝐫 𝐋𝐢𝐪𝐮𝐢𝐝𝐢𝐭𝐲: 𝟓 𝐁𝐞𝐬𝐭 𝐂𝐚𝐬𝐡 𝐅𝐥𝐨𝐰 𝐀𝐧𝐚𝐥𝐲𝐬𝐢𝐬 𝐑𝐚𝐭𝐢𝐨𝐬\n\nCash flow analysis ratios…. Wow! something very interesting. Cash flow analysis ratios are a very effective tool for comparing cash inflows and outflows from all financial statements. They provide more accurate liquidity analysis as compared to the balance sheet and income statement liquidity analysis.",
null,
"Cash Flow Analysis Ratio\n• Using cash flow ratios, a company seeks out how much cash it has, where its cash is going, and what it needs to do to maintain and support a balanced budget!\n\nIF IT INTERESTS YOU, DO SHARE IT WITH YOUR FELLOWS.\n\n## What Do Basic Cash flow Ratios Indicate?\n\nYou must know that every business prosperity is the ability to make a profit, continuously, year after year. So, you should be well aware that obtaining and applying the information from cash flow ratios is important for a profitable business.\n\nLet’s first understand the basics of the So, the cash flows statement consists of three main parts:\n\n• Operating cash flows\n• Investing cash flows\n• Financing cash flows\n\nHere, you can calculate the operating cash flows by having a trend of growth and decline in the current asset and liability accounts over a while time. While financing cash flows are the outcome/return of funding to the business. Moreover, Investing cash flows come from investing activities of the organization like financial securities, property, plant & equipment.",
null,
"Thus, you can analyze the above three forms of cash flows, along with the balance sheet and income statement data. To sum up, it may provide you the valuable information that can be useful for financial analysis of the organization’s cash position.\n\n## Five Cash Flow Ratios to Analyze Cash Flow Statement\n\n### 1. Cash Flow Margin Ratio\n\nYou may calculate it as cash flow from operations divided by sales. It gives a more reliable picture than net profit, as it shows the exact amount of cash generated per dollar of sales. Hence, the cash flow margin indicates how efficiently and effectively a company converts its sales into cash. As expenses and purchases are paid by cash, therefore this is a crucial profitability ratio.",
null,
"##### Cash Flow Margin Ratio\n\nCash Flow Margin Ratio = Cash flow from operations / sales\n\nCash flow from operating activities can be measured as:\n\nNet Income + Non-cash Expenses (Dep & Amortization) + Change in Working Capital\n\n### 2. Current Liability Coverage Ratio\n\nIt can be calculated by dividing the cash flows from operations from current liabilities. Therefore, it indicates that if this ratio results in less than 1:1, it means that the business is not producing sufficient cash to pay for its immediate/current obligations, which causes a significant risk of bankruptcy.\n\n##### Current Liability Coverage Ratio\n\nCurrent Liability Coverage Ratio = (Cash flow from operations – Dividend paid) / average current liabilities\n\n### 3. Price to Cash Flow Ratio\n\nMeanwhile, you may calculate the Price to cash flow ratio by dividing the company’s share price by the operating cash flow per share of stock.\n\n##### Price to Cash Flow Ratio\n\nPrice to Cash Flow Ratio = Share price/ operating cash flow per share*\n\nNow here, you may consider that a lower price to cash flow is ideal; it shows that the value of the share will most likely increase. Moreover, you can also take it like that although the stock prices aren’t much, the organization can help itself through the current cash flow.\n\n### 4. Operating Cash Flow Ratio\n\nThe operating cash flow ratio helps to calculate how well its operating cash flow covers the company’s current financial debt. Because if it’s more than one, then it means an entity has enough money for its operations.\n\n##### Operating Cash Flow Ratio\n\nOperating cash flow ratio = Cash flow from operations / current liabilities\n\nWith the help of using this ratio, you can assist and gauge a company’s liquidity for a short time period.\n\n### 5. Cash Flow Coverage Ratio\n\nYou can calculate it by dividing operating cash flows divided to total debt. Most importantly, this ratio needs to be high, as a result, an entity may have enough cash flow to pay for the principal as well as interest installments on its debt on time.\n\n##### Cash Flow Coverage Ratio\n\nCash Flow Coverage Ratio = Cash flow from operations / total debt\n\nFurthermore, in most scenarios, you must find that cash flow ratios are the best tool to measure an entity’s liquidity, solvency, & long-term viability of an entity. So, Cash flow ratios are very crucial for the financial analysis of every business.\n\n## Cash Flow Analysis Ratios Example\n\nFirstly, to understand the ratios completely, let’s take a small data of an XYZ company and then calculate ratios by this information. Assuming company, XYZ recorded the following info for 2020 & 2019 business activities:\n\n### Cash Flow Ratio Analysis Calculation\n\n#### Operating Cash Flow Margin\n\nTo calculate this ratio, Firstly you have to calculate the cash flow generated from operating activities. So, you can calculate it as: Cash flow generated from operating activities for the year 2020 is = \\$3,100,000 + (\\$110,000 + \\$120,000 + \\$45,000) + (\\$1,400,000 – \\$1,000,000) = \\$3,775,000\n\nHence the operating cash flow margin is calculated as:\n\nOperating cash flow margin = \\$3,775,000 / \\$6,300,000 = 59.9%\n\nThus, the higher the percentage, means the more cash is available from sales to pay as a dividend, for suppliers, utilities, and to buy capital assets.\n\n#### Current Liability Coverage Ratio\n\nIf we can take the above data to calculate the current liability coverage ratio, then the Current liability coverage ratio can be calculated as:\n\nCurrent liability coverage ratio = (\\$3,775,000 – 250,000) / \\$1,000,000 = 3.525\n\nHence, it indicates that the organization has sufficient cash to pay its current liabilities.\n\n#### Price to Cash Flow Ratio\n\nAfter that, to calculate the Price to Cash Flow ratio, first, you have to calculate the operating cash flow per share. So simply, you can calculate it by dividing the operating cash flow by the number of total shares.\n\n*Operating cash flow per share = \\$3,775,000 / 500,000 = \\$7.5\n\n, Price to Cash Flow Ratio = \\$20 / \\$7.5 = \\$2.67\n\nSo, the answer is \\$2.67, which indicates that investors pay \\$2.67 for each dollar of cash flow.\n\n#### Operating Cash Flow Ratio\n\nFurthermore, the operating cash flow ratio is calculated as\n\nOperating cash flow ratio = \\$3,775,000 / \\$1,000,000 = \\$3.775\n\n#### Cash Flow Coverage Ratio\n\nFor calculating the cash flow coverage ratio, you have to go deep down a little more.\n\nFor example, a company XYZ wants a loan, and the bank’s credit analyst look at the coverage ratio to determine the creditworthiness & its ability to repay the loan. To calculate the ratio, they will go through the statement of cash flows and get last year’s operating cash flows amount of \\$3,775,000, and the year’s total debt payable was \\$2,500,000.\n\nCash flow coverage ratio: \\$3,775,000 / \\$2,500,000 = 1.51\n\n### Interpretation of Cash Flow Analysis Ratios\n\nBy seeing the results, credit analysts may say that the company is able to produce enough cash flow than what is required to cover its existing obligations. Hence, the operating cash flow ratio figure of \\$3.775 is a good indicator of how readily and easily current liabilities are offset by the cash flows generated from XYZ’s company’s operations. The good operating cash flow ratio is always equal to or higher than 1:1. The ratios give traders, creditors, and other stakeholders a broad view of the company’s operating efficiency. Thus, an organization with huge cash flow ratios is usually known as cash cows, with apparently endless amounts of money to do whatever they like.\n\n## Key Points\n\n• To sum up, the calculation of cash flow ratios will give you a clear picture of the accurate cash position of an entity.\n• So, Cash flow ratios are the best tool to measure an entity’s liquidity, solvency, & long-term viability of an entity.\n• Hence, each ratio exhibits a certain financial aspect of an organization.\n• The good operating cash flow ratio is always equal to or higher than 1:1.\n\n## Recommended Articles\n\n### Hira Aziz - Author\n\nShe is a Business Content writer and Management contributor at 12Manage.com, where she contributes a business article weekly. She has over 2 years of experience in writing about accounting, finance, and business.\n\n## Cash Flow Statement Template",
null,
""
] | [
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"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI3NjgiIGhlaWdodD0iNDMyIiB2aWV3Qm94PSIwIDAgNzY4IDQzMiI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2UyZTdlZSIvPjwvc3ZnPg==",
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"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI3NjgiIGhlaWdodD0iMjIyIiB2aWV3Qm94PSIwIDAgNzY4IDIyMiI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2UyZTdlZSIvPjwvc3ZnPg==",
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"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI3NjgiIGhlaWdodD0iMTA4NiIgdmlld0JveD0iMCAwIDc2OCAxMDg2Ij48cmVjdCB3aWR0aD0iMTAwJSIgaGVpZ2h0PSIxMDAlIiBmaWxsPSIjZTJlN2VlIi8+PC9zdmc+",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIyMTIiIGhlaWdodD0iMzAwIiB2aWV3Qm94PSIwIDAgMjEyIDMwMCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2UyZTdlZSIvPjwvc3ZnPg==",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.927921,"math_prob":0.9745152,"size":8652,"snap":"2023-40-2023-50","text_gpt3_token_len":1992,"char_repetition_ratio":0.17981036,"word_repetition_ratio":0.059196617,"special_character_ratio":0.25416088,"punctuation_ratio":0.13872832,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9821936,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T18:16:15Z\",\"WARC-Record-ID\":\"<urn:uuid:7d17f6b8-febd-41f4-ba3e-051fd871a58a>\",\"Content-Length\":\"280634\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61a3188d-01dd-4230-8789-023f7d38c506>\",\"WARC-Concurrent-To\":\"<urn:uuid:616300d9-889b-4fad-abee-c8b60f9e8a6e>\",\"WARC-IP-Address\":\"172.111.38.73\",\"WARC-Target-URI\":\"https://accountingdrive.com/cash-flow-analysis-ratios/\",\"WARC-Payload-Digest\":\"sha1:RWMU5NLVG7ZJEEUDQ5PFFKYU2JHASDRW\",\"WARC-Block-Digest\":\"sha1:G6DTXJR7OWTQ7PJIF3WTYTB7PS23OGDN\",\"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-20231003160453-20231003190453-00706.warc.gz\"}"} |
https://www.ask-math.com/graphing-systems-of-linear-inequalities.html | [
"# Graphing Systems of Linear Inequalities\n\nCovid-19 has led the world to go through a phenomenal transition .\n\nE-learning is the future today.\n\nStay Home , Stay Safe and keep learning!!!\n\nGraphing Systems of Linear Inequalities involves two inequalities in two variables x and y.\nFor graphing systems of inequalities we use a Cartesian plane or XY -plane. In that vertical line divides the plane in left and right part and slanting or oblique line divides the plane in upper and lower part. A point in the Cartesian plane will either lie on a line or will lie in either of the part.\nSteps to graph the systems of inequality\n(i) Solve the given inequality for y by following the rules of inequality of signs.\n(ii) Graph the equation either using function table or using slope and y-intercept.\n(iii) If the inequality is strict (< or >), graph a dotted line and if the inequality is not strict ($\\geq or \\leq$ ), draw a solid line.\n(iv) Apply the\nOrigin test .\n\n Origin Test : Put x = 0 and y = 0 in the given equation after solving if the equation is true then origin (0,0) is included in the region and if false then set the region in which the origin is not included.\n\nLightly shade the half-plane that is the graph of the inequality. Colored pencils may help you distinguish the different half-planes Graph both the equations using the above rules, mark the region accordingly.For marking use the two different color pencils or dark and light shade to distinguish the two different region.The intersection or overlapping region formed by both the inequalities is the solution region.\n\n## Examples on graphing systems of linear inequalities\n\nExample 1 : Graph the system : y $\\geq$ -3x - 2 and y < x + 3\nSolution : y $\\geq$ -3x - 2 ---------(1)\nand y < x + 3 ------- (2)\nGraph the 1st and 2nd equation using slope and y intercept.\nNow put x = 0 and y = 0 in the 1st equation ,\n0 $\\geq$ - 3(0) -2\n0 $\\geq$ -2\nWhich is true, so origin is included for the 1st inequality.\nSo after graphing an equation we will mark right region of the line which includes origin.\n2nd equation ⇒\n0 < 0 + 3\n0 < 3\nwhich is also true, so origin is also included in 2nd inequality.\nSo after graphing an equation we will mark the lower region of the line which includes origin.",
null,
"From the above graphs we can see that in A region from 1st graph, there is a solid line as the inequality is not strict. In region B, 2nd graph there is a dotted line as there is strict inequality. The 3rd graph both region A and B are overlapped and region c is intersection region. So region C is the solution."
] | [
null,
"https://www.ask-math.com/images/graphing-systems-of-inequalities.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93341404,"math_prob":0.998372,"size":2698,"snap":"2020-45-2020-50","text_gpt3_token_len":644,"char_repetition_ratio":0.16258352,"word_repetition_ratio":0.076612905,"special_character_ratio":0.24536695,"punctuation_ratio":0.08159393,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99844974,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T17:59:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a7529ab2-9a25-4b1d-b912-ea806becf697>\",\"Content-Length\":\"120747\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4011f106-9d10-4ab6-9f6b-aa5cdb66f6e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:682ef186-404b-42f8-930e-b4328b0e4334>\",\"WARC-IP-Address\":\"104.27.191.8\",\"WARC-Target-URI\":\"https://www.ask-math.com/graphing-systems-of-linear-inequalities.html\",\"WARC-Payload-Digest\":\"sha1:JUHD72GVY34FBFRU7Z2HA2FWEHXPUBOU\",\"WARC-Block-Digest\":\"sha1:FFV5JNF77MPWJ4YONINWAFH7X42YUMPV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141216897.58_warc_CC-MAIN-20201130161537-20201130191537-00161.warc.gz\"}"} |
https://algorithms.tutorialhorizon.com/the-number-of-cycles-in-a-given-array-of-integers/ | [
"# The number of cycles in a given array of integers.\n\nObjective: Given an array of size N which contains integers from range 0 to N-1. (No duplicates). Write a program to find the number of cycles in the array.\n\nCycles in Array: Since the array is of size N and elements are from 0 to N-1 without any duplicates means all the elements appear exactly once. Create a graph with n vertices. Create an edge from vertex Vi to Vj if the element at position i should be in position j in the sorted ordering, With the logic above find the number of cycles in the given array. See the image below for better understanding.\n\nExample:\n\n```Given Array : [3, 2, 1, 0]\nNumber of cycles: 2\n\nGiven Array : [2, 0, 1]\nNumber of cycles: 1\n\nGiven Array : [1, 3, 4, 0, 2, 6, 5]\nNumber of cycles: 3\n```\n\nApproach: Use DFS\n\n• Given array input [].\n• Create a visited[] to check if the array element is already visited.\n• Initialize NoOfCycles = 0.\n• Now iterate through input[] created in step 1, for each index i\n• If input[i] != i\n• If i is not already visited then mark i as visited and jump to the next index which is at the index = input[i]. For example (1, 0) and i = 0 then jump to index=1. Repeat this process until you find an index which is already visited.\n• NoOfCycles++.\n• Return NoOfCycles.\n\nWalk-through:\n\n```Input[] = [3, 2, 1, 0]\nvisited = [false, false, false, false]\nNoOfCycles = 0\n\nIndex = 0, element = 3\n0!=3 and 3 is not visited.\nMark 3 visited.\nelement 0 is not visited\nMark 0 visited.\nNoOfCycles = 1\nvisited = [true, false, false, true]\n\nIndex = 1, process element = 2\n1!=2 and 2 is not visited.\nMark 2 visited.\nelement 1 is not visited\nMark 1 visited.\nNoOfCycles = 2\nvisited = [true, true, true, true]\n\nNoOfCycles = 2\n```\n\nComplete Code:\n\nOutput:\n\n```Given Array : [3, 2, 1, 0] No of cycles: 2\nGiven Array : [2, 0, 1] No of cycles: 1\nGiven Array : [1, 3, 4, 0, 2, 6, 5] No of cycles: 3\n```\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.816784,"math_prob":0.9690198,"size":1989,"snap":"2020-45-2020-50","text_gpt3_token_len":622,"char_repetition_ratio":0.1697733,"word_repetition_ratio":0.11940298,"special_character_ratio":0.3283057,"punctuation_ratio":0.19396552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97676486,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T10:26:19Z\",\"WARC-Record-ID\":\"<urn:uuid:af0a412a-da92-4831-b7ad-9c09a8c6666f>\",\"Content-Length\":\"56845\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d656e26-fcce-4d70-9fff-2db826dcf5a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4bc858a-9d2a-4291-afe7-a7eac85bc1c0>\",\"WARC-IP-Address\":\"172.67.194.90\",\"WARC-Target-URI\":\"https://algorithms.tutorialhorizon.com/the-number-of-cycles-in-a-given-array-of-integers/\",\"WARC-Payload-Digest\":\"sha1:CIZ52KDOHXELO3LVGPMXDH6LV7OXIQXE\",\"WARC-Block-Digest\":\"sha1:6SGOHC7WE3SKA7Y47JN7KES3IM3T5SXZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197593.33_warc_CC-MAIN-20201129093434-20201129123434-00531.warc.gz\"}"} |
https://sites.duke.edu/scshgap/simon-donaldson-lectures/ | [
"Home » Lectures » Sir Simon Donaldson: Lectures\n\n# Sir Simon Donaldson: Lectures\n\n### Septebmer 12, 2019 TITLE: G2 Geometry and Adiabatic Limits\n\nABSTRACT: The main focus of the talk will be G2 manifolds with co-associative fibrations and in particular the “adiabatic limit” when the fibers become very small and the structure can be described by solutions of a version of the maximal submanifold equation. Donaldson will discuss progress in the development of this theory and prospects for the future. In particular, he will explain the connection with boundary value problems for G2 structures and descriptions, in part conjectural, of calibrated submanifolds in the adiabatic limit.\n\nSlides of lecture\n\n### June 7, 2019 TITLE: Deformations of branched harmonic functions and Kovalev-Lefschetz fibrations\n\nABSTRACT: We will discuss the deformation theory of adiabatic Kovalev-Lefschetz fibrations, assuming some background from the previous lecture. We will review some of the relevant elliptic theory for functions with the appropriate singularities along a link and explain how Nash-Moser theory can be applied to a related simplified, linear, problem. In the last part of the lecture we will outline an approach to the nonlinear case, using an iteration scheme.\n\n### June 5, 2019 TITLE: Adiabatic associative and co-associatives\n\nABSTRACT: The notion of an “adiabatic Kovalev-Lefschetz fibration” is a proposal to model the behaviour of G2 manifolds with co-associative fibrations, which are expected to exist in many examples. The first part of the lecture will begin reviewing the set-up, involving a PDE for a section of a flat bundle over the complement of a link in the 3-sphere. In the second part of the lecture we will discuss descriptions of calibrated submanifolds in these fibrations. We will focus on associative sub manifolds diffeomorphic to S2× S1 corresponding to closed orbits of a gradient vector field in the complement of the link. We will explain how to go from these orbits to both formal power series and genuine solutions of the associative equation. We will then discuss a number of variants of the idea, which will involve more difficult analysis.\n\n### September 11, 2018 TITLE: G2-structures on manifolds with boundary and positive mean curvature.\n\nABSTRACT: This is a continuation of one of my talks at the June meeting, but I will begin by reviewing background to make the lecture self-contained. We will discuss the multilinear algebra of 3-forms in 6 and 7 dimensions and explain that there is an intrinsic notion of a 3-form with “postive mean curvature” on a 6-manifold. When the 3-form is the restriction of a",
null,
"-structure to the boundary of a 7-manifold this notion is consistent with the standard definition in Riemannian geometry. Combining this observation with well-established comparison theorems in Riemannian geometry we derive geometric inequalities, such as a volume bound, for torsion-free",
null,
"-structures with positive mean curvature boundary data and discuss the possible extension to closed structures.\n\n### June 8, 2018 TITLE: Adiabatic limits, multi-valued harmonic functions and the Nash-Moser-Zehnder theory\n\nABSTRACT: This lecture is a report on work in progress. We will begin by reviewing an adiabatic limit of the G2 equations, for K3-fibred manifolds. This involves data comprising a link in the 3-sphere, a flat bundle over the complement of the link and a section of this bundle which locally parametrizes a maximal submanifold, with branching over the link. Then we will turn attention to a simpler model problem involving “branched”, or multi-valued, harmonic functions. These also have some connection with work of Taubes, Takahashi, Walpuski , Haydys and Doan. We will explain how, given a suitable analytical set-up, an extension by Zehnder of the Nash-Moser theory can be applied to study the deformation theory of these branched functions.\n\n### June 8, 2018 TITLE: G2 manifolds with boundary\n\nABSTRACT: We discuss the problem of finding a torsion-free G2 structure on a 7-manifold with boundary where the restriction of the 3-form to the boundary is prescribed. We will begin by reviewing the mulitilinear algebra of 3-forms in 6 and 7 dimensions and the connection with almost-complex structures. The main point of the lecture will be to explain how to set up the problem as a nonlinear elliptic boundary value problem. This leads, in a standard way, to a Kuranishi model for the deformation theory and we discuss the obstruction space that arises. We will also explain that there is an intrinsic notion of “mean-convex” boundary data, and in this case one can derive various explicit geometric bounds on solutions.\n\n### September 14, 2017 TITLE: Collapsing co-associative fibrations\n\nABSTRACT: We will begin by explaining how “maximal” sub manifolds in spaces of indefinite signature arise as formal collapsing (or adiabatic) limits of",
null,
"manifolds with co-associative fibrations. Then we will discuss some analytical problems which arise in developing this idea, mostly having to do with the critical sets where the fibres become singular and the maximal submanifolds have branch points. In one direction we will discuss the deformation theory of these sets and in another we outline the relevance of recent constructions (by Yang Li and others) of new Calabi-Yau metrics on",
null,
".\nSlides of lecture\n\n### June 9, 2017 TITLE: 3+4 dimensional reductions of G2 holonomy—collapsing and boundary value problems\n\nABSTRACT: The theme of the talk will be various simplifications of the equations defining",
null,
"holonomy structures, involving spaces that are fibred by either 3-dimensional or 4-dimensional manifolds. A central feature in these situations is an equation for “positive triples” of 2-forms over 4-manifolds, which is a generalisation of the hyperkahler condition. In one direction we will explain how this leads to a conjectural adiabatic limit of the",
null,
"equations for manifolds with co-associative “Kovalev-Lefschetz” fibrations, describing Riemannian collapsing to a 3-dimensional limit. In another direction we will discuss boundary value problems for positive triples and an extension of the Gibbons-Hawking construction, which leads to a formulation in terms of the Monge-Ampere equation on domains in",
null,
"."
] | [
null,
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"https://sites.duke.edu/scshgap/wp-content/ql-cache/quicklatex.com-7472e63a19c41956229813e162919668_l3.png",
null,
"https://sites.duke.edu/scshgap/wp-content/ql-cache/quicklatex.com-89900434ec27852adbd7acb61ab9f664_l3.png",
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http://successwithnadice.club/12-times-20-math/ | [
"# 12 Times 20 Math",
null,
"12 times 20 math times tables 1 7 to math calculator fractions.",
null,
"12 times 20 math item 3 multiplication square educational times tables maths poster multiplication square educational times tables maths poster mathia answers.",
null,
"12 times 20 math a math times tables space scene illustration stock vector math solver picture.",
null,
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null,
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null,
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null,
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null,
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null,
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null
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https://www.cheenta.com/test-of-mathematics-solution-subjective-63-pair-of-straight-lines/ | [
"",
null,
"",
null,
"How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?\n\n# Test of Mathematics Solution Subjective 63 - Pair of Straight Lines",
null,
"This is a Test of Mathematics Solution Subjective 63 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.\n\nAlso visit: I.S.I. & C.M.I. Entrance Course of Cheenta\n\n## Problem\n\nIf any one pair among the straight lines\nax + by = a + b, bx –(a + b)y = - a, (a + b)x –ay = b\nintersect, then show that the three straight lines are concurrent.\n\n## Solution:\n\nThree lines are concurrent if each of them is linear combination of other two & they are not parallel.\nNow given one pair intersect that is they are not parallel.\nNow\nax + by = a + b … (i)\nbx – (a + b)y = - a …(ii)\n(a +b)x –ay = b …(iii)\n(i) + (ii) - (iii) = 0\nSo they are concurrent.\n\n# Knowledge Partner",
null,
"",
null,
""
] | [
null,
"https://www.facebook.com/tr",
null,
"https://www.cheenta.com/wp-content/uploads/2020/01/Cheenta-small.png",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"https://www.cheenta.com/wp-content/uploads/2020/01/Cheenta-small.png",
null,
"https://www.cheenta.com/wp-content/uploads/2021/07/abea.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9108985,"math_prob":0.9743356,"size":745,"snap":"2021-43-2021-49","text_gpt3_token_len":213,"char_repetition_ratio":0.10391363,"word_repetition_ratio":0.040268455,"special_character_ratio":0.31140938,"punctuation_ratio":0.13772455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946964,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T13:21:15Z\",\"WARC-Record-ID\":\"<urn:uuid:dbaa2c21-afb0-4223-8d01-c761868b83ad>\",\"Content-Length\":\"127672\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17df90af-6f5f-46b7-9b36-229e363d63ef>\",\"WARC-Concurrent-To\":\"<urn:uuid:46999dea-0696-4767-81a1-a938575d6c30>\",\"WARC-IP-Address\":\"35.213.169.84\",\"WARC-Target-URI\":\"https://www.cheenta.com/test-of-mathematics-solution-subjective-63-pair-of-straight-lines/\",\"WARC-Payload-Digest\":\"sha1:LPCO6DZBNEQ5S5IPQCJX2YUV4P25EFOU\",\"WARC-Block-Digest\":\"sha1:UWTIQMDYHOWWVSCQLOSNUJKK6HM7FYCO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585321.65_warc_CC-MAIN-20211020121220-20211020151220-00076.warc.gz\"}"} |
https://robomechjournal.springeropen.com/articles/10.1186/s40648-018-0123-9/tables/1 | [
"RGB-D camera $$19.0 \\pm 4.44$$ $$0.44 \\pm 0.55$$\nMicrowave sensor $$20.5 \\pm 13.94$$ $$3.01 \\pm 1.06$$\nRGB-D camera $$24.37 \\pm 6.37$$ $$0.48 \\pm 0.50$$\nMicrowave sensor $$10.39 \\pm 6.51$$ $$4.75 \\pm 1.25$$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6451668,"math_prob":1.00001,"size":455,"snap":"2020-10-2020-16","text_gpt3_token_len":178,"char_repetition_ratio":0.12638581,"word_repetition_ratio":0.0,"special_character_ratio":0.4901099,"punctuation_ratio":0.23423423,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98778,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-25T19:54:33Z\",\"WARC-Record-ID\":\"<urn:uuid:fbd128eb-3b8b-431c-964c-cebb9e55bbae>\",\"Content-Length\":\"73131\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4af7add-1d9b-43aa-840d-5d3cfbb3ab54>\",\"WARC-Concurrent-To\":\"<urn:uuid:0588e7d8-72d1-4fa9-b962-168c5e145635>\",\"WARC-IP-Address\":\"151.101.248.95\",\"WARC-Target-URI\":\"https://robomechjournal.springeropen.com/articles/10.1186/s40648-018-0123-9/tables/1\",\"WARC-Payload-Digest\":\"sha1:MH6FMEPU6DPYNCOBK2YYXDQ5FDLG6L5P\",\"WARC-Block-Digest\":\"sha1:XOKBJOJLPO3TYHHWT34IALZM6PWQSXAE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146127.10_warc_CC-MAIN-20200225172036-20200225202036-00202.warc.gz\"}"} |
https://en.youscribe.com/catalogue/documents/knowledge/dynamics-and-vibrations-matlab-tutorial-211942 | [
"",
null,
"",
null,
"",
null,
"",
null,
"51 Pages\nEnglish\n\n# Dynamics and Vibrations MATLAB tutorial\n\n-\n\nLearn all about the services we offer",
null,
"Description\n\nDynamics and Vibrations\nMATLAB tutorial\nSchool of Engineering\nBrown University\nThis tutorial is intended to provide a crash-course on using a small subset of the features of MATLAB. If\nyou complete the whole of this tutorial, you will be able to use MATLAB to integrate equations of motion\nfor dynamical systems, plot the results, and use MATLAB optimizers and solvers to make design\ndecisions.\nYou can work step-by-step through this tutorial, or if you prefer, you can brush up on topics from the list\nbelow. The tutorial contains more information than you need to start solving dynamics problems using\nMATLAB. If you are working through the tutorial for the first time, you should complete sections 1-15.\nYou can do the other sections later, when they are needed.\n1. What is MATLAB\n2. How does MATLAB differ from MAPLE?\n3. Why do we have to learn MATLAB?\n4. Starting MATLAB\n5. Basic MATLAB windows\n6. Simple calculations using MATLAB\n7. MATLAB help\n8. Errors associated with floating point arithmetic (and an example of a basic loop)\n9. Vectors in MATLAB\n10. Matrices in MATLAB\n11. Plotting and graphics in MATLAB\n12. Working with M-files\n13. MATLAB Functions\n14. Organizing complex calculations as functions in an M-file\n15. Solving ordinary differential equations (ODEs) using MATLAB\n15.1 Solving a basic differential equation\n15.2 How the ODE solver works\n15.3 Solving a differential equation with adjustable parameters\n15.4 Solving a vector valued differential equation\n15.5 Solving a ...\n\nSubjects\n\n##### Computer to plate\n\nInformations",
null,
"School of Engineering Brown University\nDynamics and Vibrations MATLAB tutorial\nThis tutorial is intended to provide a crash-course on using a small subset of the features of MATLAB. If you complete the whole of this tutorial, you will be able to use MATLAB to integrate equations of motion for dynamical systems, plot the results, and use MATLAB optimizers and solvers to make design decisions. You can work step-by-step through this tutorial, or if you prefer, you can brush up on topics from the list below. The tutorial contains more information than you need to start solving dynamics problems using MATLAB. If you are working through the tutorial for the first time, you should complete sections 1-15. You can do the other sections later, when they are needed. 1.What is MATLAB 2.How does MATLAB differ from MAPLE? 3.Why do we have to learn MATLAB? 4.Starting MATLAB 5.Basic MATLAB windows 6.Simple calculations using MATLAB 7.MATLAB help 8.Errors associated with floating point arithmetic (and an example of a basic loop) 9.Vectors in MATLAB 10.Matrices in MATLAB 11.Plotting and graphics in MATLAB 12.Working with M-files 13.MATLAB Functions 14.Organizing complex calculations as functions in an M-file 15.Solving ordinary differential equations (ODEs) using MATLAB 15.1Solving a basic differential equation 15.2How the ODE solver works 15.3Solving a differential equation with adjustable parameters 15.4Solving a vector valued differential equation 15.5Solving a higher order differential equation 15.6Controlling the accuracy of solutions to differential equations 15.7Looking for special events in a solution 15.8Other MATLAB differential equation solvers 16.solvers and optimizers to make design decisionsUsing MATLAB 16.1Using fzero to solve equations 16.2Simple unconstrained optimization problem 16.3Optimizing with constraints 17.Reading and writing data to/from files 18.Movies and animation 19.On the frustrations of scientific programming",
null,
"1. What is MATLAB? You can think of MATLAB as a sort of graphing calculator on steroids it is designed to help you manipulate very large sets of numbers quickly and with minimal programming. Operations on numbers can be done efficiently by storing them asmatrices is particularly good at doing matrix. MATLAB operations (this is the origin of its name).\n2. How does MATLAB differ from MAPLE? They are similar. MAPLE is better suited to algebraic calculations; MATLAB is better at numerical calculations. But most things that can be done in MAPLE can also be done in MATLAB, and vice-versa.\n3. Why do we have to learn MATLAB? You don’t. Lawyers, chefs, professional athletes, and belly dancers don’t use MATLAB. But if you want to be an engineer, you need to know MATLAB. MATLAB is currently the platform of choice for most engineering calculations for which a special purpose computer program has not been written. But it will probably be superseded by something else during your professional career very little software survives without major changes for longer than 10 years or so.\n4. Starting MATLAB MATLAB is installed on the engineering instructional facility. You can find it in the Start>Programs menu. You can also download and install MATLAB on your own computer fromts/s//:wtfoptthedn.diu/e.arowbr You must have key access installed on your computer, and you must be connected to the Brown network to run MATLAB. For off campus access, you will need to be connected to the Web to be able to start MATLAB. Start MATLAB now.\n5. Basic MATLAB windows You should see the GUI shown below. The various windows (command history, current directory, etc) may be positioned differently on your version of MATLAB they are ‘drag and drop’ windows.",
null,
"Lists available files\nSelect the directory where you will load or save files here\nEnter basic MATLAB commands here\nStores a history of your commands\nSelect a convenient directory where you will be able to save your files.\n6. Simple calculations using MATLAB You can use MATLAB as a calculator. Try this for yourself, by typing the following into the command window. Press ‘enter’ at the end of each line. >>x=4 >>y=x^2 >>z=factorial(y) >>w=log(z)*1.e-05 >> format long >>w >> format long eng >>w >> format short >>w >>sin(pi) MATLAB will display the solution to each step of the calculation just below the command. Do you notice anything strange about the solution given to the last step? Once you have assigned a value to a variable, MATLAB remembers it forever. To remove a value from a variable you can use the ‘clear’ statement - try >>clear a >>a",
null,
"If you type ‘clear’ and omit the variable, theneverythingdo that now but it isgets cleared. Don’t useful when you want to start a fresh calculation. MATLAB can handle complex numbers. Try the following >>z = x + i*y >>real(z) >>imag(z) >>conj(z) >>angle(z) >>abs(z) You can even do things like >> log(z) >> sqrt(-1) >> log(-1)\nNotice that: Unlike MAPLE, Java, or C, you don’t need to type a semicolon at the end of the line (To properly express your feelings about this, type >>load handel and then >> sound(y,Fs) in the command window). If youdowill be completed but MATLAB will not print the result.put a semicolon, the operation This can be useful when you want to do a sequence of calculations. Special numbers, like `pi’ and ‘i’ don’t need to be capitalized (Gloria in Excelsis Deo!). But beware you often use i as a counter in loops and then the complex numberigets re-assigned as a number. You can also do dumb things like pi=3.2 (You may know that in 1897 a bill was submitted to the Indiana legislature to declare pi=3.2 but fortunately the bill did not pass). You can reset these special variables to their proper definitions by using clear i or clear pi can double left YouHistory window keeps track of everything you have typed.The Command click on a line in the Command history window to repeat it, or right click it to see a list of other options. Compared with MAPLE, the output in the command window looks like crap. MATLAB is not really supposed to be used like this. We will discuss a better approach later. no quick way to fix your error,If you screw up early on in a sequence of calculations, there is other than to type in the sequence of commands again. You can use the ‘up arrow’ key to scroll back through a sequence of commands. Again, there is a better way to use MATLAB that gets around this problem. If you are really embarrassed by what you typed, you can right click the command window and delete everything (but this willnotalso delete lines from the Command can Youreset variables). history, by right clicking the line and selecting Delete Selection. Or you can delete the entire Command History. right clicking the lineYou can get help on MATLAB functions by highlighting the function, then and selecting Help on Selection. Try this for the sqrt(-1) line.\n7. MATLAB help Help is available through the online manual select the `Product help’ entry from the Help menu on the main window. The main help menu looks like this",
null,
"You can use the navigator to search for help, or you can use the index in the Contents window to learn about MATLAB features. The alphabetical list of functions is useful. Open it and see if you can find out how to compute the day of the week corresponding to your birthday in the year you graduate (MATLAB has a built in function for computing the day of the week corresponding to a date) The MATLAB manual is not particularly user friendly the search algorithm is poor, and much of the manual looks like it was written by the same people that wrote the best-selling FORTRAN manuals in the 1960s. The contents are usually more useful than searching. The Demos (the last menu tab in the help navigator) can also be helpful: you can often find, and copy, an example that is similar to what you are trying to accomplish.\n8. Errors associated with floating point arithmetic (and a basic loop)\nIf you have not already done so, use MATLAB to calculate >>sin(pi) The answer, of course, should be zero, but MATLAB returns a small, but finite, number. This is because MATLAB (and any other program) stores floating point numbers as sequences of binary digits with a finite length. Obviously, it is impossible to store the exact value of in this way.\nMore surprisingly, perhaps, it is not possible even to store a simple decimal number like 0.1 as a finite number of binary digits. Try typing the following simple MATLAB program into the command window >> a = 0; >> for n =1:10 a = a + 0.1; end >> a >> a 1",
null,
"Here, the line that reads “for n=1:10 a= a + 0.1; end is called a “loop. This is a very common operation in most computer programs. It can be interpreted as the command: “for each of the discrete values of the integer variable n between 1 and 10 (inclusive), calculate the variable “a by adding +0.1 to the previous value of “a The loop starts with the value n=1 and ends with n=10. Loops can be used to repeat calculations many times we will see lots more examples in later parts of the tutorial. Thus, the for
end loop therefore adds 0.1 to the variable a ten times. It gives an answer that is approximately 1. But when you compute a-1, you don’t end up with zero. Of course -1.1102e-016 is not a big error compared to 1, and this kind of accuracy is good enough for government work. But if someone subtracted 1.1102e-016 from your bank account every time a financial transaction occurred around the world, you would burn up your bank account pretty fast. Perhaps even faster than you do by paying your tuition bills. You can minimize errors caused by floating point arithmetic by careful programming, but you can’t eliminate them altogether. As a user of MATLAB they are mostly out of your control, but you need to know that they exist, and try to check the accuracy of your computations as carefully as possible.\n9. Vectors in MATLAB MATLAB can do all vector operations completely painlessly. Try the following commands >> a = [6,3,4] >> a(1) >> a(2) >> a(3) >> b = [3,1,-6] >> c = a + b >> c = dot(a,b) >> c = cross(a,b) Calculate the magnitude of c (you should be able to do this with a dot product. MATLAB also has a built-in function called `norm’ that calculates the magnitude of a vector) A vector in MATLAB need not be three dimensional. For example, try >>a = [9,8,7,6,5,4,3,2,1] >>b = [1,2,3,4,5,6,7,8,9] You can add, subtract, and evaluate the dot product of vectors that are not 3D, but you can’t take a cross product. Try the following >> a + b >> dot(a,b) >>cross(a,b) In MATLAB, vectors can be stored as either arowof numbers, or acolumn Soof numbers. you could also enter the vector a as >>a = [9;8;7;6;5;4;3;2;1] to produce acolumnvector. You can turn a row vector into a column vector, and vice-versa by >> b = transpose(b)",
null,
"A few more very useful vector tricks are: You can create a vector containing regularly spaced data points very quickly with a loop. Try >> for i=1:11 v(i)=0.1*(i-1); end >> v The for
end loop repeats the calculation with each value of i from 1 to 11. Here, the “counter variable i now is used to refer to the iththe vector v, and also is used in the formula itself.entry in As another example, suppose you want to create a vector v of 101 equally spaced points, starting at 3 and ending at 2*pi, you would use >> for i=1:101 v(i)= 3 + (2*pi-3)*(i-1)/100; end >> v If you type >> sin(v) MATLAB will compute the sin of every number stored in the vector v and return the result as another vector. This is useful for plots see section 11. You have to be careful to distinguish between operations on a vector (or matrix, see later) and operations on thecomponentsof the vector. For example, try typing >> v^2 This will cause MATLAB to bomb, because the square of a row vector is not defined. But you can type >> v.^2 (there is a period after the v, and no space). This squares every element within v. You can also do things like >> v. /(1+v) (see if you can work out what this does). II personally avoid using the dot notation mostly because it makes code hard to read. Instead, generally do operations on vector elements using loops. For example, instead of writing w = v.^2, I would use >> for i=1:length(v) w(i) = v(i)^2; end Here, ‘for i=1:length(v)’ repeats the calculation for every element in the vector v. The function length(vector loops) determines how many components the vector v has (101 in this case). Using is not elegant programming, and slows down MATLAB, and purists (like CS40 TAs) object to it. But I don’t care. For any seriously computationally intensive calculations I would use a serious programming language like J or Fortran, not MATLAB. 10. Matrices in MATLAB Hopefully you know what a matrix is
If not, it doesn’t matter - for now, it is enough to know that a matrix is a set of numbers, arranged in rows and columns, as shown below Column 2\nrow 1 7829503366540215row 4 Column 3",
null,
"A matrix need not necessarily have the same numbers of rows as columns, but most of the matrices we will encounter in this course do. A matrix of this kind is calledsquare. (My generation used to call our professors and parents square too, but with hindsight it is hard to see why. ‘Lumpy’ would have been more accurate). You can create a matrix in MATLAB by entering the numbers one row at a time, separated by semicolons, as follows >> A = [1,5,0,2; 5,4,6,6;3,3,0,5;9,2,8,7] You can extract the numbers from the matrix using the convention A(row #, col #). Try >>A(1,3) >>A(3,1) You can also assign values of individual array elements >>A(1,3)=1000 There are some short-cuts for creating special matrices. Try the following >>B = ones(1,4) >>C = pascal(6) >>D = eye(4,4) >>E = zeros(3,3) The ‘eye’ command creates the ‘identity matrix’ this is the matrix version of the number 1. You can use >> help pascal to find out what pascal does. MATLAB can help you do all sorts of things to matrices, if you are the sort of person that enjoys doing things to matrices. For example 1. You can flip rows and columns with >> B = transpose(A) 2. You can add matrices (provided they have the same number of rows and columns >> C=A+B Try also >> C transpose(C) A matrix that is equal to its transpose is calledsymmetric 3. You can multiply matrices this is a rather complicated operation, which will be discussed in more detail elsewhere. But in MATLAB you need only to type >>D=A*B to find the product of A and B. Also try the following >> E=A*B-B*A >> F = eye(4,4)*A - A 4. You can do titillating things like calculate the determinant of a matrix; the inverse of a matrix, the eigenvalues and eigenvectors of a matrix. If you want to try these things >> det(A) >> inv(A) >> eig(A) >> [W,D] = eig(A) You can find out more about these functions, and also get a full list of MATLAB matrix operations in the section of Help highlighted below",
null,
"MATLAB can also calculate the product of a matrix and a vector. This operation is used very frequently in engineering calculations. For example, you can multiply a 3Dcolumnvector by a matrix with 3 rows and 3 columns, as follows >>v = [4;3;6] >>A = [3,1,9;2,10,4;6,8,2] >>w=A*v The result is a 3D column vector. Notice that you can’t multiply a 3D row vector by a 3x3 matrix. Try this >>v = [4,3,6] >>w=A*v If you accidentally enter a vector as a row, you can convert it into a column vector by using >>v = transpose(v) MATLAB is also very good at solving systems of linear equations. For example, consider the equations 3x14x27x36 5x12x29x31 x113x23x38 This system of equations can be expressed in matrix form as Axb A3425311793xxxx231b861To solve these in MATLAB, you would simply type",
null,
">> A = [3,4,7;5,2,-9;-1,13,3] >> b = [6;1;8] >> x = A\\b (note the forward-slash, not the back-slash or divide sign) You can check your answer by calculating >> A*x The notation here is supposed to tell you that x is b ‘divided’ by A although `division’ by a matrix has to be interpreted rather carefully. Try also >>x=transpose(b)/A The notation transpose(b)/A solves the equationsxAb, wherexandb Again,are row vectors. you can check this with >>x*A (The answer should equalb, (as a row vector) of course) MATLAB can quickly solve huge systems of equations, which makes it useful for many engineering calculations. The feature has to be used carefully because systems of equations may not have a solution, or may have many solutions MATLAB has procedures for dealing with both these situations but if you don’t know what it’s doing you can get yourself into trouble. For more info on linear equations check the section of the manual below\n11. Plotting and graphics in MATLAB Plotting data in MATLAB is very simple. Try the following >> for i=1:101 x(i)=2*pi*(i-1)/100; end >> y = sin(x) >> plot(x,y)",
null,
"MATLAB should produce something that looks like this\nMATLAB lets you edit and annotate a graph directly from the window. For example, you can go to Tools> Edit Plot, then double-click the plot. A menu should open up that will allow you to add x and y axis labels, change the range of the x-y axes; add a title to the plot, and so on. You can change axis label fonts, the line thickness and color, the background, and so on usually by double-clicking what you want to change and then using the pop-up editor. You can export figures in many different formats from the File> Save As menu and you can also print your plot directly. Play with the figure for yourself to see what you can do.\nTo plot multiple lines on the same plot you can use >> y = sin(x) >> plot(x,y) >> hold on >> y = sin(2*x) >> plot(x,y) Alternatively, you can use >> y(1,:) = sin(x); >> y(2,:) = sin(2*x);"
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https://tbiomed.biomedcentral.com/articles/10.1186/1742-4682-4-11 | [
"# A mathematical model for the adenylosuccinate synthetase reaction involved in purine biosynthesis\n\n## Abstract\n\n### Background\n\nDevelopment of the mathematical models that adequately describe biochemical reactions and molecular-genetic mechanisms is one of the most important tasks in modern bioinformatics. Because the enzyme adenylosuccinate synthetase (AdSS) has long been extensively studied, a wealth of kinetic data has been accumulated.\n\n### Results\n\nWe describe a mathematical model for the reaction catalyzed by AdSS. The model's parameters were fitted to experimental data obtained from published literature. The advantage of our model is that it includes relationships between the reaction rate, the concentrations of three substrates (GTP, IMP and ASP), the effects of five inhibitors (GMP, GDP, AMP, ASUC and SUCC), and the influence of Mg2+ ions.\n\n### Conclusion\n\nOur model describes the reaction catalyzed by AdSS as a fully random process. The model structure implies that each of the inhibitors included in it is only competitive to one of the substrates. The model was tested for adequacy using experimental data published elsewhere. The values obtained for the parameters are as follows: V max = 1.35·10-3 mM/min, Km GTP = 0.023 mM, Km IMP = 0.02 mM, Km ASP = 0.3 mM, Ki GMP = 0.024 mM, Ki GDP = 8·10-3 mM, Ki AMP = 0.01 mM, Ki ASUC = 7.5·10-3 mM, Ki SUCC = 8 mM, Km Mg = 0.08 mM.\n\n## Background\n\nBiosynthesis of the purines AMP and GMP in Escherichia coli is a many-staged process supported by a complex network of enzymes. Some of the genes that encode these enzymes are arranged into operons (purF, purHD, purMN, purEK, guaBA, purB), while others are located in single cistrons (purT, purl, purC, purA, guaC). Expression of these operons is regulated by regulatory proteins (PurR, DnaA, CRP) and various low-molecular-weight compounds . The activities of the encoded enzymes are additionally regulated by substrates, reaction products, and certain other low-molecular-weight substances [4, 5].\n\nThe enzyme adenylosuccinate synthetase (AdSS; GDP-forming IMP: L-aspartate ligase, EC 6.3.4.4), which is the product of the purA gene, catalyzes the conversion of IMP to ASUC in the presence of Mg2+:\n\nIMP + GTP + ASPGDP + PI + ASUC.\n\nThere are many nucleotides that inhibit AdSS. For example, AMP is a competitive inhibitor of IMP; ASUC, of IMP; dGMP, of IMP; GMP, of GTP. GDP is a competitive inhibitor of GTP, which in part explains a gradual decrease in the rate of ASUC formation in solutions if the GTP concentration is not reduced. dAMP, CMP, and UMP can also produce inhibitory effects, albeit much less pronounced .\n\nMathematical models of the reaction catalyzed by AdSS have been proposed in a variety of studies. In 1969, Rudolph and Fromm proposed an equation that includes one inhibitor . It was demonstrated that each of SUCC, GDP, and IMP is a competitive inhibitor of only one substrate and that the molecular mechanism of the reaction catalyzed by AdSS is a rapid equilibrium, fully random process. To describe the dependence of the reaction rate on whether the inhibitor competes against the substrate for binding to the enzyme, an 11-parameter model was proposed. Although the kinetics of the AdSS-catalyzed reaction in the presence of the inhibitors SUCC, GDP, IMP, and ASUC was well studied experimentally, the formula included too many constants and the model constants (including the inhibition constants) were not evaluated.\n\nIn 1979, Stayton and Fromm proposed a slightly different equation for one inhibitor . In this case, the inhibition of AdSS by ppGpp was considered. It was demonstrated that ppGpp is a competitive inhibitor of GTP, but not of IMP or ASP. This model also describes the effect of the inhibitor using four inhibition constants, so only the apparent values of these constants were calculated. Interestingly, varying the concentrations of IMP or GTP (at fixed concentrations of the other two substrates) affected the calculated values of the respective inhibition constants.\n\nIn 1995, Kang and Fromm investigated the influence of Mg2+ ions on the AdSS-catalyzed reaction . It was demonstrated that for AdSS to be in the activated form, two Mg2+ ions are required. One interacts with the β- and γ-phosphoryl groups of GTP, the other with the aspartate in the enzyme's active center, improving the affinity of the enzyme for ASP. Kinetic experiments on the interactions of Mg2+ and ASP were performed with saturating concentrations of GTP and IMP, so the GTP and IMP concentrations were not included in the model. Although the authors themselves proved that AdSS has two binding centers for Mg2+, the model treats the Mg2+ concentration as if there were only one (at least this is how we interpret the presence of ion concentration as an item raised to the first power). The initial velocity in the Hill plot (Fig. 1 in ) was measured at saturating concentrations of IMP, GTP and Asp with Mg2+ varying.\n\nThus, although a model has been proposed for each of a variety of effectors, there is still no single model that exploits the pool of available kinetic data in its entirety. We report a more complete model, which describes the reaction catalyzed by adenylosuccinate synthetase and includes the concentrations of three substrates (GTP, IMP, and ASP), the effects of five inhibitors (GMP, GDP, AMP, ASUC, and SUCC), and the influence of Mg2+ ions.\n\n## Results\n\nThe enzyme AdSS is inhibited by GMP, GDP, AMP, ASUC and SUCC. Enzyme activity requires the presence of Mg2+ ions. Knowing how these effectors work, the reaction rate can be written in a generalized form as follows:\n\n$V={V}_{\\mathrm{max}}\\cdot \\frac{\\frac{GTP}{K{m}_{GTP}}\\cdot \\frac{IMP}{K{m}_{IMP}}\\cdot \\frac{ASP}{K{m}_{ASP}}}{\\left(1+\\frac{GTP}{K{m}_{GTP}}+\\frac{GMP}{K{i}_{GMP}}+\\frac{GDP}{K{i}_{GDP}}\\right)\\cdot \\left(1+\\frac{IMP}{K{m}_{IMP}}+\\frac{AMP}{K{i}_{AMP}}+\\frac{ASUC}{K{i}_{ASUC}}\\right)\\cdot \\left(1+\\frac{ASP}{K{m}_{ASP}}+\\frac{SUCC}{K{i}_{SUCC}}\\right)}\\cdot \\frac{\\frac{M{g}^{2+}}{K{m}_{Mg}}}{1+\\frac{M{g}^{2+}}{K{m}_{Mg}}}\\left(1\\right),$\n\nwhere V max is the maximum reaction rate; GTP, IMP, and ASP are the concentrations of the corresponding substrates; GMP, GDP, AMP, ASUC, and SUCC are the concentrations of the corresponding inhibitors; Mg2+ is the concentration of Mg2+ ions; Km GTP , Km IMP , Km ASP are the Michaelis-Menten constants for the corresponding substrates; Km Mg is the Michaelis-Menten constant for Mg2+ ions; Ki GMP , Ki GDP , Ki AMP , Ki ASUC , and Ki SUCC , are the constants of the efficiency of reaction inhibition by the corresponding substances.\n\nThe model's parameters were verified against 61 curves from published data [6, 7, 9]. Different publications use different values of the rate constant of AdSS: 15600 s-1 , 1.47 s-1 , 1.0 s-1 . However, since most publications do not indicate the enzyme concentrations used, we calculated the value for V max using our model.\n\nWe evaluated the reaction constants in the absence of effectors using experimental results from the work by Rudolph and Fromm and observed good agreement (calculations not shown). The parameter values inferred from the curves were as follows: V max = 1.35·10-3 mM min-1, Km GTP = 0.023 mM, Km IMP = 0.02 mM, Km ASP = 0.3 mM.\n\nRudolph and Fromm, who examined the effect of SUCC in detail , proposed that SUCC is competitive to ASP. Our calculations indicate that this assumption is consistent with the kinetic data. The model output and experimental data on how SUCC affects the reaction rate at different concentrations of GTP are presented in Fig. 1. As can be seen from this figure, there is an inconsistency between the model output and experimental data. A possible explanation will be discussed below. Also, we estimated the effect of SUCC on the reaction rate at different concentrations of IMP and ASP and observed good agreement with experimental data (calculations not shown). Using our model, the value of the constant Ki SUCC is 8 mM.\n\nGDP is a competitive inhibitor of GTP. Based on experimental data from the work by Rudolph and Fromm , we evaluated Ki GDP as 8·10-3 mM (calculations not shown).\n\nWyngaarden and Greenland investigated the effect of ASUC, another inhibitor, and proposed that it is competitive with both IMP and ASP. Likewise, it was proposed that GMP is a competitive inhibitor of both IMP and GTP. However, our calculations suggest that ASUC appears to be competitive with only IMP, and GMP with only GTP. The effects of ASUC on IMP and ASP and the model output are presented in Figs. 2, 3. Using the model, Ki ASUC is 7.5·10-3 mM.\n\nThe influence of Mg2+ ions on enzyme activity is included in our model on the basis of the kinetic curves presented in the work of Kang and Fromm . The concentration of Mg2+ is included as a multiplier in the form of a simple rational fraction raised to the first power. Kang and Fromm also included the concentration of Mg2+ as a multiplier raised to the first power; however, they additionally assumed that Mg2+ and ASP may act cooperatively. Our calculations demonstrate that an even simpler model, which does not assume Mg2+/ASP synergy, is adequate for describing the influence of magnesium. Although our model does not say that there are two binding sites for Mg2+ ions, nor does it say otherwise , for it is not necessary that Hill's number be the same as the number of ligand-binding centers in the enzyme. Using our model, Km Mg is 0.08 mM (Fig. 4).\n\nAlso, our model treats AMP as a competitive inhibitor of IMP . From the work of Wyngaarden and Greenland , we calculated the constants of the GMP and AMP effects (Fig. 5, lines 1 and 2): Ki GMP = 0.024 mM, Ki AMP = 0.01 mM. However, these estimates may suffer from a lack of consistency within the experimental data.\n\nIn control experiments, we used data from Wyngaarden and Greenland . The comparison of the model output and experimental data is presented in Fig. 5, lines 4 and 5. For comparison, the values of the constants obtained here and elsewhere are presented in Table 1.\n\n## Discussion\n\nModels to describe the reaction catalyzed by adenylosuccinate synthetase have been proposed from time to time. In 1969, Rudolph and Fromm investigated the mechanism of this reaction and proposed an equation that related the reaction rate to the respective concentrations of the substrates and one inhibitor . However, the formula included four inhibition constants and therefore was too complicated to allow those constants to be evaluated. As a result, only apparent inhibition constants were calculated.\n\nIn 1963, Wyngaarden and Greenland proposed that ASUC is a competitive inhibitor of both IMP and ASP. Likewise, it was proposed that GMP is a competitive inhibitor of both IMP and GTP. However, our model, using the same pool of experimental data, demonstrates that the assumption that ASUC is competitive with only IMP, and GMP with only GTP, is sufficient for describing the enzyme reaction adequately.\n\nIn 1979, Stayton and Fromm proposed a model that relates the reaction rate to the effect of the inhibitor ppGpp. This model describes the effect of the inhibitor as the above model does, so only the apparent inhibition constants were calculated. In addition, the calculated values of these apparent inhibition constants depend on the substrate concentrations.\n\nIn 1995, Kang and Fromm proposed a model that relates the reaction rate to the concentrations of ASP and Mg2+ ions. However, no other substrates or inhibitors were included. Thus, despite all the interest in the reaction catalyzed by AdSS, no single model that includes the concentrations of all substrates and the effects of more than one effector has been proposed. We propose a model that describes the reaction catalyzed by adenylosuccinate synthetase and includes the concentrations of three substrates (GTP, IMP, and ASP), the effects of five inhibitors (GMP, GDP, AMP, ASUC, and SUCC), and the influence of Mg2+ ions. Our model is consistent with a fully random mechanism, which is very similar to that proposed by Rudolph and Fromm . However, given the available biochemical data on the reaction mechanism, at least two alternative hypotheses could be put forward. We checked whether these hypotheses were consistent with the kinetic data.\n\nOne of these hypotheses states that magnesium binds to aspartate to form an ASP·Mg2+ complex, which in turn binds to the enzyme. This hypothesis leads to a modification of our model such that it describes the effects of ASP, Mg, and SUCC. To make this modified model consistent with experimental data at a fixed concentration of Mg2+, all the parameters, except Km Mg , were assigned the same values as in the original model. The best consistency of the modified model and the experimental data was attained at Km Mg = 0.01 mM. Overall, this modification is much less consistent with the experimental data than the original model (calculations not shown).\n\nThe second hypothesis states that ASP binds to the enzyme to form an AdSS·ASP complex, after which magnesium binds to aspartate in that complex. To describe this hypothesis, another modification of the original model is required such that it describes the effects of ASP, Mg, and SUCC. Here the best consistency is attained at Km Mg = 0.13 mM. Overall, this modification too fails to show reasonable consistency with the experimental data (calculations not shown).\n\nUnder both modified models (one assuming pre-formation of the ASP·Mg2+ complex, the other assuming pre-formation of the AdSS·ASP complex), magnesium competes against SUCC, while under the original model it does not compete against anything. Looking at the output from the original and modified models, it is easy to see that the original is by far the most consistent. Admittedly, it seems in some respects inconsistent with the available biochemical data: for example, it leaves out the presence of two binding sites for Mg2+. Therefore, this model may not be claimed as an absolutely accurate description of the real molecular events unfolding in the reaction being discussed. Looking at the model structure, it appears as though magnesium ions might act on the enzyme via their own independent centers. This hypothetical mechanism therefore deserves to be called an \"apparent molecular mechanism\" by analogy with apparent dissociation constants, apparent inhibition constants, etc.\n\nAs we were working on our model, an inconsistency between the model output and the experimental data on SUCC was revealed (Fig. 1). Since no enzyme concentrations were specified in the literature sources to which we referred, we kept to a fixed arbitrary concentration of the enzyme in all the numerical experiments. However, when we proceeded to SUCC, we found that we had to modify this concentration in order to keep the model consistent with experimental data. We introduced a multiplier equal to 0.74 as a correction factor, which implies a reduction in the concentration of the enzyme (calculations not shown).\n\nThe constant Km Mg was verified against data published by Kang and Fromm (Fig. 4). However, to accomplish this, we had to reduce Km ASP two-fold to 0.17 mM and introduce a correction factor (a multiplier equal to 5000), which in effect increases the amount of enzyme. The other model parameters were absolutely consistent with those experimental data.\n\nSome data from Wyngaarden and Greenland were used for control experiments (Fig. 5, lines 4 and 5). On comparing our model output and the control data, an apparent inconsistency was revealed. It is possible that the authors used different amounts of the enzyme in different experiments. This assumption was supported by introduction of a coefficient such as those used for the curves shown in Fig. 1. Admittedly, the discrepancies revealed during the fitting of the parameters could in part be explained by the different temperatures at which the different authors conducted their experiments: Rudolph and Fromm at 28°C, Wyngaarden and Greenland at 25°C, and Kang and Fromm at 22°C. However, in the present work we did not look at temperature as a factor.\n\n## Conclusion\n\nThe proposed model for the reaction catalyzed by the enzyme AdSS includes relationships between the reaction rate, the concentrations of three substrates (GTP, IMP and ASP), the effects of five inhibitors (GMP, GDP, AMP, ASUC and SUCC) and Mg2+ ions. Our model is consistent with a fully random mechanism. The model structure implies that each of the inhibitors included in it is competitive to only one of the substrates. The model's parameters were fitted to experimental data from published literature. A methodological problem arising from the lack of concordance among the data in different publications was dealt with by introducing correction coefficients; this simply implies that the concentrations of AdSS in those source works were different. The adequacy of the model was ensured by comparing the theoretical calculations and the experimental data from the literature sources that were not used while the fitting procedure was under way. The values obtained for the parameters are shown in Table 1.\n\n## Methods\n\nThe model to describe the reaction catalyzed by adenylosuccinate synthetase was developed using a random biochemical system as follows:\n\n$\\begin{array}{cc}\\left(\\text{a}\\right)& E\\left(0,y,z,w\\right)+X\\stackrel{{K}_{X}}{⇔}E\\left(X,y,z,w\\right)\\\\ \\left(\\text{b}\\right)& E\\left(x,0,z,w\\right)+Y\\stackrel{{K}_{Y}}{⇔}E\\left(x,Y,z,w\\right)\\\\ \\left(\\text{c}\\right)& E\\left(x,y,0,w\\right)+Z\\stackrel{{K}_{Z}}{⇔}E\\left(x,y,Z,w\\right)\\\\ \\left(\\text{d}\\right)& E\\left(x,y,z,0\\right)+W\\stackrel{{K}_{W}}{⇔}E\\left(x,y,z,W\\right).\\end{array}$\n\nEquation (a) describes the interaction of GTP or GDP or GMP with its respective active center. It is assumed that GTP, GDP and GMP compete against one another for binding to the same center, which is why X stands for GTP or GDP or GMP. It is assumed that the population of the other enzyme centers has no effect on reactions (a). Equation (b) describes the interaction of IMP or AMP or ASUC with a different active center. Y stands for IMP or AMP or ASUC. Equation (c) describes the interaction of ASP and SUCC with a third active center, Z standing for either ASP or SUCC. Finally, Equation (d) describes the interaction of magnesium ions with the enzyme; here W stands for Mg. E(x, y, z, w) describes the state of the enzyme: x is assigned 0, GTP, GDP or GMP; y is assigned 0, IMP, AMP or ASUC; z is assigned 0, ASP, and SUCC; w is assigned either 0 or Mg. If a variable in E(x, y, z, w) takes on a zero value, it means that the corresponding enzyme center is not bound to the ligand.\n\nOwing to the assumption that the system is random, the variables X, x, Y, y, Z, z, W, w in Equations (a)-(d) are always independent. Therefore, Equations (a) and (b) each include 72 reactions and Equation (c) includes 64 reactions. Equation (d) is a concise description of 48 reactions. Overall, the system (a)-(d) includes 256 biochemical reactions that describe transitions among 96 states of the enzyme. Writing out the corresponding system of 96 differential equations and assuming equilibrium results in a non-linear algebraic system. Supposing that variation in the concentrations of GTP, GDP, GMP, IMP, AMP, ASUC, ASP, SUCC and Mg2+ can be neglected, we arrive at the corresponding system of linear equations defining a 96-dimensional vector of the unknown states of the enzyme. Solving this system and collecting similar terms and assuming that Km GTP K GTP , Ki GDP K GDP , Ki GMP K GMP , Km IMP K IMP , Ki AMP K AMP , Ki ASUC K ASUC , Km ASP K ASP , Ki SUCC K SUCC , Km Mg K Mg , we obtain the proposed model.\n\n## Abbreviations\n\nAMP:\n\nASP:\n\naspartate\n\nASUC:\n\nCMP:\n\ncytidine-5'-monophosphate\n\ndAMP:\n\ndGMP:\n\ndeoxyguanylic acid\n\nGDP:\n\nguanosine 5'-diphosphate\n\nGMP:\n\nguanosine 5'-phosphate\n\nGTP:\n\nguanosine 5'-triphosphate\n\nIMP:\n\ninosine 5'-monophosphate\n\nPI:\n\nphosphate\n\nppGpp:\n\nguanosine 5'-diphosphate-3'-diphosphate\n\nSUCC:\n\nsuccinate\n\nUMP:\n\nuridine 5'-monophosphate.\n\n## References\n\n1. He B, Shiau A, Choi KY, Zalkin H, Smith JM: Genes of the Escherichia coli pur regulon are negatively controlled by a repressor-operator interaction. J Bacteriol. 1990, 172: 4555-4562.\n\n2. Hutchings MI, Drabble WT: Regulation of the divergent guaBA and xseA promoters of Escherichia coli by the cyclic AMP receptor protein. FEMS Microbiol Lett. 2000, 187: 115-122. 10.1111/j.1574-6968.2000.tb09146.x.\n\n3. Mehra RK, Drabble WT: Dual control of the gua operon of Escherichia coli K12 by adenine and guanine nucleotides. J Gen Microbiol. 1981, 123: 27-37.\n\n4. Messenger LJ, Zalkin H: Glutamine phosphoribosylpyrophosphate amidotransferase from Escherichia coli. Purification and properties. J Biol Chem. 1979, 254: 3382-3392.\n\n5. Lambden PR, Drabble WT: Inosine 5'-monophosphate dehydrogenase of Escherichia coli K12: the nature of the inhibition by guanosine 5'-monophosphate. Biochem J. 1973, 133: 607-608.\n\n6. Wyngaarden JB, Greenland RA: The inhibition of succinoadenylate kinosynthetase of Escherichia coli by adenosine and guanosine 5'-monophosphates. J Biol Chem. 1963, 238: 1054-1057.\n\n7. Rudolph FB, Fromm HJ: Initial rate studies of adenylosuccinate synthetase with product and competitive inhibitors. J Biol Chem. 1969, 244: 3832-3839.\n\n8. Stayton MM, Fromm HJ: Guanosine 5'-diphosphate-3'-diphosphate inhibition of adenylosuccinate synthetase. J Biol Chem. 1979, 254: 2579-2581.\n\n9. Kang C, Fromm HJ: Identification of an essential second metal ion in the reaction mechanism of Escherichia coli adenylosuccinate synthetase. J Biol Chem. 1995, 270: 15539-15544. 10.1074/jbc.270.26.15539.\n\n10. Dong Q, Liu F, Myers AM, Fromm HJ: Evidence for an arginine residue at the substrate binding site of Escherichia coli adenylosuccinate synthetase as studied by chemical modification and site-directed mutagenesis. J Biol Chem. 1991, 266: 12228-12233.\n\n11. Wang W, Gorrell A, Hou Z, Honzatko RB, Fromm HJ: Ambiguities in mapping the active site of a conformationally dynamic enzyme by directed mutation. Role of dynamics in structure-function correlations in Escherichia coli adenylosuccinate synthetase. J Biol Chem. 1998, 273: 16000-16004. 10.1074/jbc.273.26.16000.\n\n12. Cornish-Bowden A: Principles of enzyme kinetics. 1976, London: Butterworth\n\n## Acknowledgements\n\nThe authors are grateful to I. V. Lokhova for bibliographical support and V. Filonenko for translation of the paper into English.\n\nThis work was supported in part by the Russian Foundation for Basic Research No 05-07-98011-r_ob_v, 05-07-98012-r_ob_v, 06-04-49556-a, NSF:FIBR (Grant EF-0330786) and by the Rosnauka (State Contract No. 02.467.11.1005).\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Evgeniya A Oshchepkova-Nedosekina.\n\n### Competing interests\n\nThe author(s) declare that they have no competing interests.\n\n### Authors' contributions\n\nOEA was responsible for developing of the model, and writing of the manuscript.\n\nLVA was responsible for developing of the modelling method and critical review of the manuscript.\n\n## Authors’ original submitted files for images\n\nBelow are the links to the authors’ original submitted files for images.\n\n## Rights and permissions\n\nReprints and Permissions\n\nOshchepkova-Nedosekina, E.A., Likhoshvai, V.A. A mathematical model for the adenylosuccinate synthetase reaction involved in purine biosynthesis. Theor Biol Med Model 4, 11 (2007). https://doi.org/10.1186/1742-4682-4-11",
null,
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] | [
null,
"https://tbiomed.biomedcentral.com/track/article/10.1186/1742-4682-4-11",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9284974,"math_prob":0.90195316,"size":23233,"snap":"2023-40-2023-50","text_gpt3_token_len":5622,"char_repetition_ratio":0.14899479,"word_repetition_ratio":0.09699893,"special_character_ratio":0.22937201,"punctuation_ratio":0.13902053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9565091,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T08:34:47Z\",\"WARC-Record-ID\":\"<urn:uuid:0f228dd3-749f-4c45-be20-42a84564a627>\",\"Content-Length\":\"335036\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69c3f8c5-c7b2-4f89-84ef-ebe7a9550043>\",\"WARC-Concurrent-To\":\"<urn:uuid:63f0fd95-55df-4c48-a137-cd6bcd5bc63c>\",\"WARC-IP-Address\":\"146.75.28.95\",\"WARC-Target-URI\":\"https://tbiomed.biomedcentral.com/articles/10.1186/1742-4682-4-11\",\"WARC-Payload-Digest\":\"sha1:WHCF3SGYUXQFNWLL4UYN6J7KXIEWLAWX\",\"WARC-Block-Digest\":\"sha1:RE66MKVHUASIIV7WHIXEOKFOBT5BCYDH\",\"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-20231129073519-20231129103519-00000.warc.gz\"}"} |
https://docs.dgl.ai/generated/dgl.transforms.GDC.html | [
"# GDC¶\n\nclass dgl.transforms.GDC(coefs, eweight_name='w', eps=None, avg_degree=5)[source]\n\nBases: dgl.transforms.module.BaseTransform\n\nApply graph diffusion convolution (GDC) to an input graph, as introduced in Diffusion Improves Graph Learning.\n\nA sparsification will be applied to the weighted adjacency matrix after diffusion. Specifically, edges whose weight is below a threshold will be dropped.\n\nThis module only works for homogeneous graphs.\n\nParameters\n• coefs (list[float], optional) – List of coefficients. $$\\theta_k$$ for each power of the adjacency matrix.\n\n• eweight_name (str, optional) – edata name to retrieve and store edge weights. If it does not exist in an input graph, this module initializes a weight of 1 for all edges. The edge weights should be a tensor of shape $$(E)$$, where E is the number of edges.\n\n• eps (float, optional) – The threshold to preserve edges in sparsification after diffusion. Edges of a weight smaller than eps will be dropped.\n\n• avg_degree (int, optional) – The desired average node degree of the result graph. This is the other way to control the sparsity of the result graph and will only be effective if eps is not given.\n\nExample\n\n>>> import dgl\n>>> import torch\n>>> from dgl import GDC\n\n>>> transform = GDC([0.3, 0.2, 0.1], avg_degree=2)\n>>> g = dgl.graph(([0, 1, 2, 3, 4], [2, 3, 4, 5, 3]))\n>>> g.edata['w'] = torch.tensor([0.1, 0.2, 0.3, 0.4, 0.5])\n>>> new_g = transform(g)\n>>> print(new_g.edata['w'])\ntensor([0.3000, 0.3000, 0.0200, 0.3000, 0.0400, 0.3000, 0.1000, 0.0600, 0.3000,\n0.0800, 0.0200, 0.3000])"
] | [
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https://www.routledge.com/Simultaneous-Systems-of-Differential-Equations-and-Multi-Dimensional-Vibrations/Campos/p/book/9780367137212 | [
"",
null,
"1st Edition\n\n# Simultaneous Systems of Differential Equations and Multi-Dimensional Vibrations\n\nBy\n\n## Luis Manuel Braga da Costa Campos\n\nISBN 9780367137212\nPublished November 14, 2019 by CRC Press\n325 Pages 34 B/W Illustrations\n\nUSD \\$130.00\n\nPrices & shipping based on shipping country\n\n## Book Description\n\nSimultaneous Differential Equations and Multi-Dimensional Vibrations is the fourth book within Ordinary Differential Equations with Applications to Trajectories and Vibrations, Six-volume Set. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. This fourth book consists of two chapters (chapters 7 and 8 of the set).\n\nThe first chapter concerns simultaneous systems of ordinary differential equations and focuses mostly on the cases that have a matrix of characteristic polynomials, namely linear systems with constant or homogeneous power coefficients. The method of the matrix of characteristic polynomials also applies to simultaneous systems of linear finite difference equations with constant coefficients.\n\nThe second chapter considers linear multi-dimensional oscillators with any number of degrees of freedom including damping, forcing, and multiple resonance. The discrete oscillators may be extended from a finite number of degrees-of-freedom to infinite chains. The continuous oscillators correspond to waves in homogeneous or inhomogeneous media, including elastic, acoustic, electromagnetic, and water surface waves. The combination of propagation and dissipation leads to the equations of mathematical physics.\n\n• Presents simultaneous systems of ordinary differential equations and their elimination for a single ordinary differential equation\n• Includes cases with a matrix of characteristic polynomials, including simultaneous systems of linear differential and finite difference equations with constant coefficients\n• Covers multi-dimensional oscillators with damping and forcing, including modal decomposition, natural frequencies and coordinates, and multiple resonance\n• Discusses waves in inhomogeneous media, such as elastic, electromagnetic, acoustic, and water waves\n• Includes solutions of partial differential equations of mathematical physics by separation of variables leading to ordinary differential equations"
] | [
null,
"https://www.routledge.com/images/ajax-spinner.gif.pagespeed.ce.QA8PeRfjsw.gif",
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https://publications.waset.org/10010088/bibtex | [
"```\t@article{(Open Science Index):https://publications.waset.org/pdf/10010088,\ntitle = {Proposal of Optimality Evaluation for Quantum Secure Communication Protocols by Taking the Average of the Main Protocol Parameters: Efficiency, Security and Practicality},\nauthor = {Georgi Bebrov and Rozalina Dimova},\ncountry\t= {},\ninstitution\t= {},\nabstract = {In the field of quantum secure communication, there\nis no evaluation that characterizes quantum secure communication\n(QSC) protocols in a complete, general manner. The current paper\naddresses the problem concerning the lack of such an evaluation\nfor QSC protocols by introducing an optimality evaluation, which\nis expressed as the average over the three main parameters of QSC\nprotocols: efficiency, security, and practicality. For the efficiency\nevaluation, the common expression of this parameter is used, which\nincorporates all the classical and quantum resources (bits and qubits)\nutilized for transferring a certain amount of information (bits) in a\nsecure manner. By using criteria approach whether or not certain\ncriteria are met, an expression for the practicality evaluation is\npresented, which accounts for the complexity of the QSC practical\nrealization. Based on the error rates that the common quantum attacks\n(Measurement and resend, Intercept and resend, probe attack, and\nentanglement swapping attack) induce, the security evaluation for\na QSC protocol is proposed as the minimum function taken over\nthe error rates of the mentioned quantum attacks. For the sake of\nclarity, an example is presented in order to show how the optimality\nis calculated.},\njournal = {International Journal of Physical and Mathematical Sciences},\nvolume = {13},\nnumber = {2},\nyear = {2019},\npages = {48 - 52},\nee = {https://publications.waset.org/pdf/10010088},\nurl \t= {https://publications.waset.org/vol/146},\nbibsource = {https://publications.waset.org/},\nissn \t= {eISSN: 1307-6892},\npublisher = {World Academy of Science, Engineering and Technology},\nindex \t= {Open Science Index 146, 2019},\n}\n```"
] | [
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https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-8-section-8-1-area-and-initial-postulates-exercises-page-359/22 | [
"Elementary Geometry for College Students (7th Edition) Clone\n\n18 $unit^{2}$\nWe need to find the area of the shaded region Area of the shaded region = area of the outer triangle - area of inner triangle Given A and B are mid points so for inner triangle b= 4 and h = 3 Area of triangle whose base has length b and altitude has length h is given by A = $\\frac{1}{2}$ bh Area of outer triangle = $\\frac{1}{2}$ *8* 6 = 24 $unit^{2}$ Area of inner triangle = $\\frac{1}{2}$ * 4*3 = 6 $unit^{2}$ Area of the shaded region = 24 - 6 = 18 $unit^{2}$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85571796,"math_prob":0.9998944,"size":502,"snap":"2019-43-2019-47","text_gpt3_token_len":161,"char_repetition_ratio":0.19477911,"word_repetition_ratio":0.04,"special_character_ratio":0.37051794,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99983114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T02:38:57Z\",\"WARC-Record-ID\":\"<urn:uuid:0ff21736-69a7-43dc-93b4-4cbe3cf26be2>\",\"Content-Length\":\"61373\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df225053-5786-474e-9075-163bed58ba82>\",\"WARC-Concurrent-To\":\"<urn:uuid:00d69f05-09cf-46f8-8430-e025934309da>\",\"WARC-IP-Address\":\"54.162.109.179\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-8-section-8-1-area-and-initial-postulates-exercises-page-359/22\",\"WARC-Payload-Digest\":\"sha1:NQCBH6DB4BPO3MSJXW5NYLIPHG54HG4C\",\"WARC-Block-Digest\":\"sha1:GPPEYALO34TVJK4B5U6ETZPA4GHBV7KE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987828425.99_warc_CC-MAIN-20191023015841-20191023043341-00253.warc.gz\"}"} |
https://www.codesandscripts.com/2017/09/knapsack-problem-using-simulated-annealing.html | [
"## Wednesday, 6 September 2017\n\n### Knapsack problem using simulated annealing\n\nThe knapsack problem ( Wiki link ) is a problem in combinatorial optimisation. Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. Let's see how we can solve it using simulated annealing.\n\n### A Basic Understanding\n\nSuppose we are planning a hiking trip; and we are, therefore, interested in filling a knapsack with items that are considered necessary for the trip. There are N different item types that are deemed desirable; these could include bottle of water, apple, orange, sandwich, and so forth. Each item type has a given set of two attributes, namely a weight (or volume) and a value that quantifies the level of importance associated with each unit of that type of item. Since the knapsack has a limited weight (or volume) capacity, the problem of interest is to figure out how to load the knapsack with a combination of units of the specified types of items that yields the greatest total value. What we have just described is called the knapsack problem.\n\nImplementation\n\nknapsack.js\n```/**\n* Will hold each knapsack item\n*/\nfunction Item(w,v,index){\nthis.weight = w; //Weight of the item\nthis.value = v; //Weight of the value\nthis.index = index; //Index of the value\nthis.print = function(){\ndocument.write(\"Item : w : \"+this.weight+\", v : \"+this.value+\"<br/>\");\n}\n}\n\n/**\n* Knapsack bag\n*/\nfunction Bag(size){\n\nthis.size = size;\nthis.itemset = []; //Will hold the overall dataset\nthis.currentSolution = []; //will hold the current solution\n\n/**\n* Populates the itemset with the dataset provided above\n*/\nthis.prepare = function(){\nfor(var i = 0;i<dataset.length;i++){\nvar item = new Item(dataset[i],dataset[i],i);\nthis.itemset.push(item);\n}\nvar temp = getRandomAsInt(0,this.itemset.length);\nthis.currentSolution.push(this.getItemBasedOnIndex(this.itemset, temp));\n}\n\n/**\n* Gets the items from the itemset provided as an argument based on the index\n*/\nthis.getItemBasedOnIndex = function(itemset,index){\nvar item = null;\nfor(var i=0;i<itemset.length;i++){\nif(itemset[i].index==index){\nitem = itemset[i];\nbreak;\n}\n}\nreturn item;\n}\n\n/**\n* Calculates and returns the summation of list of item weights\n*/\nthis.getWeightForList = function(itemset){\nvar sum = 0;\nfor(var i=0;i<itemset.length;i++){\nsum += itemset[i].weight;\n}\nreturn sum;\n}\n\n/**\n* Calculates and returns the summation of list of item values\n*/\nthis.getValueForList = function(itemset){\nvar sum = 0;\nfor(var i=0;i<itemset.length;i++){\nsum += itemset[i].value;\n}\nreturn sum;\n}\n\n/**\n* Checks whether current selection is overweight or not\n*/\nthis.checkoverweight = function(itemset){\nif(this.getWeightForList(itemset) > this.size){\nreturn true;\n}else\nreturn false;\n}\n\n/**\n* Returns any random item from the itemset which is not in the currentSolution of the bag\n*/\nthis.getRandomItemFromItemSet = function(){\nvar temp = getRandomAsInt(0,this.itemset.length);\nvar item = this.getItemBasedOnIndex(this.itemset,temp);\nwhile(this.getItemBasedOnIndex(this.currentSolution,temp)!=null){\ntemp = getRandomAsInt(0,this.itemset.length);\nitem = this.getItemBasedOnIndex(this.itemset,temp);\n}\nreturn item;\n}\n\n/**\n* Modifies the current selection\n*/\nthis.modifySelection = function(){\nvar modified = this.currentSolution.clone();\nvar item = this.getRandomItemFromItemSet();\nmodified.push(item);\nwhile(this.checkoverweight(modified)){\nvar dropIndex = getRandomAsInt(0,modified.length);\nmodified.removeItem(dropIndex);\nconsole.log(dropIndex);\nconsole.log(modified);\n\n}\nreturn modified;\n}\n\n/**\n* Calculates the remaining space in the bag\n*/\nthis.calculateRemainingSpace = function(itemset){\nreturn (this.size - this.getValueForList(itemset));\n}\n\nthis.printsolution = function(itemset){\nfor(var i=0;i<itemset.length;i++){\nvar item = itemset[i];\nitem.print();\n}\ndocument.write(\"Total value is : \"+this.getValueForList(itemset));\n}\n\nthis.prepare();\n}\n\n/**\n* Returns a random number between minimum (inclusive) and maximum (exclusive)\n*/\nfunction getRandom(min, max) {\nreturn Math.random() * (max - min) + min;\n}\n\n/**\n* Returns a random integer between minimum (inclusive) and maximum (exclusive)\n*/\nfunction getRandomAsInt(min, max) {\nreturn Math.floor(Math.random() * (max - min)) + min;\n}\n\nArray.prototype.clone = function() {\nreturn this.slice(0);\n};\n\nArray.prototype.removeItem = function( index){\nvar i = 0;\nwhile( i < this.length ){\nif(i == index ){\nthis.splice(i,1);\n}\ni++;\n}\n};\n\n```\n\n#### index.html\n\n```<script src=\"knapsack.js\"></script>\n<script>\n/**\n* Constants for the program\n*/\nvar MAX_WEIGHT = 15;\nvar TEMPERATURE = 500;\nvar COOLING_FACTOR = 0.2;\n\n/**\n* The dataset used in the program\n* [weight,value]\n*/\nvar dataset = [\n[1,2],[4,2],[5,4],[3,6],[9,4],[10,8],[8,5],[6,7],[7,3],[11,8]\n];\n\n/** Acceptance function **/\nfunction acceptanceProbability(freespace,newfreespace,temperature){\nif(newfreespace < freespace){\nreturn 1.0;\n}\nreturn Math.exp((freespace - newfreespace) / temperature);\n}\n\nvar bag = new Bag(MAX_WEIGHT);\nvar best = bag.currentSolution;\nvar temperature = TEMPERATURE;\nwhile(temperature > 0){\nvar freespaceLeft = bag.calculateRemainingSpace(bag.currentSolution);\nvar modifiedSolution = bag.modifySelection();\nvar newfreespaceleft = bag.calculateRemainingSpace(modifiedSolution);\nif(acceptanceProbability(freespaceLeft, newfreespaceleft, temperature)>=Math.random()){\nbag.currentSolution = modifiedSolution;\n}\n\nif(bag.getValueForList(bag.currentSolution) > bag.getValueForList(best)){\nbest = bag.currentSolution;\n}\ntemperature -= COOLING_FACTOR;\n}\nbag.printsolution(best);\n}\n</script>\n\n```\n\n#### 1 comment:\n\n1.",
null,
"How did I find this site? I read a lot of positive reviews about it, of course, there were negative and more positive ones, and I decided to take a chance by clicking on this link modish slot machine games I saw a lot of cool slots and slot machines êàî there so cool I’ve even managed to withdraw my first winnings"
] | [
null,
"https://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52609324,"math_prob":0.9773188,"size":5949,"snap":"2020-45-2020-50","text_gpt3_token_len":1428,"char_repetition_ratio":0.14970563,"word_repetition_ratio":0.06130268,"special_character_ratio":0.26575896,"punctuation_ratio":0.21771218,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9864505,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T09:30:32Z\",\"WARC-Record-ID\":\"<urn:uuid:12d95579-ea20-4899-96c0-b1ba8db5c0af>\",\"Content-Length\":\"85155\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e91d905-f885-417e-8d97-3c47a28be890>\",\"WARC-Concurrent-To\":\"<urn:uuid:79699722-5231-4dee-a7d0-9c7a2bd0da73>\",\"WARC-IP-Address\":\"172.217.2.115\",\"WARC-Target-URI\":\"https://www.codesandscripts.com/2017/09/knapsack-problem-using-simulated-annealing.html\",\"WARC-Payload-Digest\":\"sha1:6CCUPYQSMBOLXZAUG26E3OCFORVDQQSF\",\"WARC-Block-Digest\":\"sha1:GBLKBIKPKWK5MWHKA4K25JK45EAK6NJE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141187753.32_warc_CC-MAIN-20201126084625-20201126114625-00584.warc.gz\"}"} |
https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk10/10_algebraic_and_geometric_multiplicities_q.html | [
"## Questions\n\n1. You are given that $$-1$$ is an eigenvalue of $$\\begin{bmatrix} -3 & 4 \\\\ -1 & 1\\end{bmatrix}$$. What is the algebraic multiplicity of this eigenvalue?\n\n2. The matrix $$\\begin{bmatrix} 2 & 1 & 0\\\\ 0 & 2 & 1 \\\\ 0 & 0 & 2\\end{bmatrix}$$ has a unique eigenvalue. What is the geometric multiplicity of this eigenvalue?"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6788375,"math_prob":0.9999957,"size":460,"snap":"2022-05-2022-21","text_gpt3_token_len":168,"char_repetition_ratio":0.15570176,"word_repetition_ratio":0.11235955,"special_character_ratio":0.39130434,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999345,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T19:18:01Z\",\"WARC-Record-ID\":\"<urn:uuid:4e7ea391-82bd-4aee-bb9b-07f42c36c0a1>\",\"Content-Length\":\"2151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d5399e9e-554c-4a6b-8c53-0ef7cb38a07b>\",\"WARC-Concurrent-To\":\"<urn:uuid:452c6d16-24e6-4daa-9c1a-5d0f405b8692>\",\"WARC-IP-Address\":\"134.117.21.21\",\"WARC-Target-URI\":\"https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk10/10_algebraic_and_geometric_multiplicities_q.html\",\"WARC-Payload-Digest\":\"sha1:FUIG3UL3PMUM5JNJZEEFOKRZOKGRJHE3\",\"WARC-Block-Digest\":\"sha1:LXHNB4XNWS4MAED5CFVEN7XV7DAFN2C6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662593428.63_warc_CC-MAIN-20220525182604-20220525212604-00196.warc.gz\"}"} |
https://www.ctanujit.in/2013/05/shortcut-tricks-in-maths-for-isicmitifr.html | [
"Pages\n\nSHORTCUT TRICKS IN MATHS FOR ISI,CMI,TIFR ENTRANCE EXAM\n\nThis is a kind of initiative from me to state a trick & solve problems using those tricks from tifr,isi,cmi,nbhm papers. This will be like a series. This is the beginning one.\n\nTrick 1:-\nIf the sum of two positive quantities is a constant(given),then their product is maximum when two quantities are equal.\nEx. Let a+b=12,then max(ab)=36 , i.e. When a equals b.\n\nProblem1. (CMI-2010 M.Sc entrance) True/False- For x<0 , e^x (1 - e^x) ≤ 1/4 .\nSol:- True. e^x,1-e^x both are positive. Now e^x +(1 - e^x)=1\nSo their product is max when e^x=1-e^x=1/2,so max[e^x (1 - e^x)] is 1/4.\n\nProblem2. (ISI MMA PAPER 2010) If a,b are positive real variables whose sum is a constant k, then the minimum value of root of{(1+1/a)(1+1/b)} is\nA. k-1/k B. k+2/k C. 1+2/k D. none\nSol: (C). The given root is minimum when ab is maximum. Now ab is maximum when a=b=k/2. So value of\nroot of{(1+1/a)(1+1/b)}=(1+2/k)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76847476,"math_prob":0.99942124,"size":901,"snap":"2022-05-2022-21","text_gpt3_token_len":327,"char_repetition_ratio":0.09587514,"word_repetition_ratio":0.0,"special_character_ratio":0.35960045,"punctuation_ratio":0.148,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99974245,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T10:43:05Z\",\"WARC-Record-ID\":\"<urn:uuid:e6c3a7f9-2a9d-47d3-b1a1-d9b5178741f5>\",\"Content-Length\":\"51709\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57852a3f-ae17-40d7-a5f2-a8d8671de5d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:61e74063-79e9-4573-a642-d27966748ff0>\",\"WARC-IP-Address\":\"142.250.65.83\",\"WARC-Target-URI\":\"https://www.ctanujit.in/2013/05/shortcut-tricks-in-maths-for-isicmitifr.html\",\"WARC-Payload-Digest\":\"sha1:GMBTJ33P4W46SJDNCYR4ACGXRIGUI5YV\",\"WARC-Block-Digest\":\"sha1:NRRHGVJE35YL67V7XCWVMORGEULW2A73\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303845.33_warc_CC-MAIN-20220122103819-20220122133819-00460.warc.gz\"}"} |
http://black-rose-bielefeld.de/mardia-multivariate-analysis-djvu.php | [
"#### Mardia multivariate analysis djvu\n\nMultivariate Analysis deals with observations on more than one variable where there is some inherent interdependence between the variables. With several texts already available in this area, one may very well enquire of the authors as to the need for yet another black-rose-bielefeld.de Edition: 1st Edition. Multivariate Analysis deals with observations on more than one variable where there is some inherent interdependence between the variables. With several texts already available in this area, one may very well enquire of the authors as to the need for yet another book. Most of the available books fall into two categories, either theoretical or data analytic. On Mardia’s Tests of Multinormality 5 Theorem Under multinormality, the limiting distribution of the vector of fourth moments Vjk (j 6= k) and Vjj is a multivariate normal distribution with zero mean vector and limiting variances and covariances.\n\nIf you are looking\n\n# mardia multivariate analysis djvu\n\nHow to download and setup SAS University Edition on windows 10(Hindi), time: 13:42\n\nAn introduction to multivariate analysis, including the necessary statistical and mathematical background. Topics covered include a review of univariate statistics, multivariate normality, matrix algebra, univariate and multivariate regression, multivariate analysis of variance, multivariate analysis of covariance, and exploratory factor analysis. Multivariate Analysis deals with observations on more than one variable where there is some inherent interdependence between the variables. With several texts already available in this area, one may very well enquire of the authors as to the need for yet another black-rose-bielefeld.de Edition: 1st Edition. The remaining chapters concentrate on specifically multivariate problems which have no meaningful analogues in the univariate case. Topics covered include econometrics, principal component analysis, factor analysis, canonical correlation analysis, discriminate analysis, cluster analysis, multi-dimensional scaling and directional data/5(6). Learn about Stata's multivariate methods features, including factor analysis, principal components, discriminant analysis, multivariate tests, statistics, and much more. Multivariate Analysis deals with observations on more than one variable where there is some inherent interdependence between the variables. With several texts already available in this area, one may very well enquire of the authors as to the need for yet another book. Most of the available books fall into two categories, either theoretical or data analytic. On Mardia’s Tests of Multinormality 5 Theorem Under multinormality, the limiting distribution of the vector of fourth moments Vjk (j 6= k) and Vjj is a multivariate normal distribution with zero mean vector and limiting variances and covariances. Multivariate analysis (MVA) is based on the statistical principle of multivariate statistics, which involves observation and analysis of more than one statistical outcome variable at a black-rose-bielefeld.de design and analysis, the technique is used to perform trade studies across multiple dimensions while taking into account the effects of all variables on the responses of interest. The test for multinormality proposed by K. V. MARDIA (; ) is based on the simultaneous testing for multivariate skewness and multivariate black-rose-bielefeld.de: Heinz Kres.MULTIVARIATE ANALYSIS K. V. MARDIA Department of Statistics. University of Leeds. The covariance between x and y is then E(xy') = E(nT) = l:r = rAr'r = rA. Multivariate Analysis (Probability and. Mathematical Statistics). Kanti V. Mardia, J. T. Kent, J. M. Bibby. Click here if your download doesn\"t start automatically. Multivariate Analysis (Probability and Mathematical Statistics) book download Kanti V. Mardia, J. T. Kent and J. M. Bibby Download Multivariate Analysis. Request PDF on ResearchGate | Multivariate analysis / K. V. Mardia, J. T. Kent, J. M. Bibby | Incluye bibliografía e índice. Results 1 - 10 of 10 Multivariate Analysis (Probability and Mathematical Statistics) book download Kanti V. Mardia, J. T. Kent and J. M. Bibby Download. black-rose-bielefeld.de: Multivariate Analysis (Probability and Mathematical Statistics) ( ): Kanti V. Mardia, J. T. Kent, J. M. Bibby: Books. Don't have a Kindle? Get your Kindle here, or download a FREE Kindle Reading App. Postby Just» Tue Mar 26, am. Looking for mardia kent bibby multivariate analysis ebook download. Will be grateful for any help! Top. Available in the National Library of Australia collection. Author: Mardia, K. V; Format: Book; xv, p.: ill. ; 23 cm. Mardia multivariate analysis download. Mardia multivariate analysis. black-rose-bielefeld.de: Multivariate Analysis (Probability and Mathematical. Multivariate Analysis - 1st Edition - ISBN: , Multivariate Analysis. 1st Edition Authors: Kanti Mardia J. Kent J. Bibby. - Use\n\n## mardia multivariate analysis djvu\n\nand enjoy\n\nsee more driver scanner acer prisa 620p"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8657633,"math_prob":0.8449591,"size":5201,"snap":"2021-21-2021-25","text_gpt3_token_len":1146,"char_repetition_ratio":0.17741005,"word_repetition_ratio":0.40873015,"special_character_ratio":0.1911171,"punctuation_ratio":0.1637744,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95749956,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T10:06:34Z\",\"WARC-Record-ID\":\"<urn:uuid:e8e1a482-b7fa-43b9-8249-54df23768036>\",\"Content-Length\":\"28126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99acc0c5-0789-4dbc-ac07-495dcc53f751>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8bf5ae0-3c94-40ea-b035-4aa8b08f26a1>\",\"WARC-IP-Address\":\"172.67.133.11\",\"WARC-Target-URI\":\"http://black-rose-bielefeld.de/mardia-multivariate-analysis-djvu.php\",\"WARC-Payload-Digest\":\"sha1:WCDVQ26B236GEO6MQUIGEMDSYANMJFOL\",\"WARC-Block-Digest\":\"sha1:6U4KJWIT3RXKM6S3JAC7WQ65N4UYCLKQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487630081.36_warc_CC-MAIN-20210625085140-20210625115140-00266.warc.gz\"}"} |
http://edspi31415.blogspot.com/2012/10/calculator-tricks-part-4.html | [
"## Thursday, October 11, 2012\n\n### Calculator Tricks - Part 4\n\nWelcome to Part 4 of Calculator Tricks. This is series is where accomplish mathematical tasks with a use of a regular, simple calculator. Part 4 will cover:\n\n* Dealing with Percents\n* Shopping\n* Simple Interest\n\nDealing with Percents\n\nWith most calculators, if you have to add tax and subtract discounts, you can just execute the operation directly.\n\nHowever, on some calculators, like the Casio SL-300VC, require a different sequence:\n× n % + (to add) and × n % - (to subtract).\n\nHere is how I prefer to work with percents, and it avoids the percent key altogether.\n\n---------------\nAdd Percent\n\nLet A be the number you want to add N% percent to it. In a shopping application, N represents the sales tax. In construction, N can be thought be allowance for waste.\n\nA + N%\n= A + (A × N/100)\n= A × (1 + N/100)\n\nKeystrokes: N ÷ 100 + 1 × A\n---------------\nExample: Add 10% to 19.95.\n\nKeystrokes: 10 ÷ 100 + 1 × 19.95\n\nResult: 21.945\n\n(On most calculators, 19.95 + 10 % works too)\n---------------\nSubtract Percent\n\nLet A be the number you want to subtract N% from. In terms of shopping, N represents a discount.\n\nA - N%\n= A - (A × N/100)\n= A × (1 - N/100)\n\nKeystrokes: N ÷ 100 +/- + 1 × A\n---------------\nExample: Subtract 10% from 19.95.\n\nKeystrokes: 10 ÷ 100 +/- + 1 × 19.95\n\nResult: 17.955\n\n(On most calculators, 19.95 - 10 % works too)\n---------------\n\nShopping\n\nEver though what your bill would be as you shop? This section will show you what the potential bill will be and hopefully will lead you to make smart shopping decisions, and keep in budget.\n\nApproach:\n1. Clear Memory. We will use the memory register to keep track of our purchases.\n2. Determine whether the item is subject to sales tax. You can press MR at any time to get a subtotal.\n3. Add the total.\n4. *If all of your items are subject to sales tax, add sales tax to the total.\n\nIf your purchases are \"mixed\", buying both non-taxable and taxable items (grocery store comes mind):\n\nIf the item or service is not subject to sales tax, use the keystroke sequence price M+ .\n\nIf the item or service is subject to sales tax, use the keystroke sequence tax factor × price M+ .\n\nwhere tax factor = (1 + tax%/100)\n\nIn California, where I live, generally food, grocery items, and most services are generally not subject to sales tax, while sales of tangible goods are subject to sales tax.\n\nLet's illustrate this strategy by a few examples.\n---------------\nExample 1: We are at a hardware store and purchasing the following items:\n\nHammer: \\$19.99\nNails: \\$3.95\nStaple Gun: \\$14.95\nMeasuring Tape: \\$3.99\n\nAll items are subject to 8.5% sales tax.\n\nWell, since we are dealing with just taxable purchases, we can total everything and add sales tax at the end.\n\nKeystrokes:\nMC\n19.99 M+\n3.95 M+\n14.95 M+\n3.99 M+\n8.5 ÷ 100 + 1 = × MR =\n\nTotal: \\$46.52\n---------------\nExample 2: From an office store, we are purchasing:\n\nReam of 500 sheets of paper: 3 at \\$8.99 each\nPack of pencils: \\$2.99\nPack of pens: \\$3.95\nCalculator: 2 at \\$9.95 each\n\nAll items are subject to 8.5% sales tax. In this case, we can add the sales tax at the end.\n\nKeystrokes:\nMC\n3 × 8.99 = M+\n2.99 M+\n3.95 M+\n2 × 9.95 = M+\n8.5 ÷ 100 + 1 = × MR =\n\nSubtotal: \\$53.81\nWith Sales Tax: \\$58.38\n---------------\nExample 3: At a grocery store, we are purchasing:\n\nBag of Grapes: \\$2.45\nApples: 5 at 99¢ each\nBananas: 6 at 24¢ each\nBread: \\$4.09 with a 5% discount coupon\nWater: 2 gallons at \\$1.09 each\nBox of Ziploc Bags: \\$3.99 with a 50¢ coupon\n\nOnly the Ziploc bag is subject to 8.5% sales tax. Assume coupons take affect immediately (before sales tax). Here we must use the \"mixed purchases\" strategy. Enter \\$3.49 for the Ziploc bags.\n\nMC\n2.45 M+\n(grapes)\n5 × 0.99 = M+ (apples)\n6 × 0.24 = M+ (bananas)\n5 ÷ 100 +/- + 1 = × 4.09 = M+ (bread)\n2 × 1.09 = M+ (water)\n8.5 ÷ 100 = + 1 = × 3.49 = M+ (Ziploc bags)\nMR (final total)\n\nTotal bill: \\$18.69\n---------------\n\nSimple Interest\n\nThe simple interest formula is:\n\nI = P × R% × T\n\nI = amount of interest\nP = amount of the principal\nR% = annual interest rate (as expressed as a decimal)\nT = time, in years\n\nThe total amount paid is principal and interest, in other words, P + I.\n\n---------------\nExample 1: A bank makes a short term loan to Fred and Suzy of \\$1,000. The bank charges 9.6% interest on short term loans. Fred and Suzy have to pay the loan in two months. If Fred and Suzy wait for the two months, how much interest have they paid?\n\nWe are looking for I. We have the following:\nP = principal = 1000\nR% = annual interest rate = 9.6/100\nT = time = 2 months * 1 year/12 months\n\nNote have to convert months to years. So the interest paid is:\n\nKeystrokes:\n1000 × 9.6 ÷ 100 × 2 ÷ 12 =\n\nThe total interest paid is \\$16.\n---------------\nExample 2: Terrell is looking over his credit card bill. The balance is \\$1,540.29. His credit card charges an annual rate of 15.99%. Terrell is planning to make a \\$300.00 payment. Assuming Terrell does not use his credit card for the next month, what will be his balance?\n\nVariables:\nP = \\$1,540.29 - \\$300.00 = \\$1,240.29\nR% = 15.99%\nT = 1/12 (1/12 of a year)\n\nKeystrokes:\nMC (Clear memory)\n1540.29 - 300 = M+ (Subtract payment, interest will accrue on \\$1,240.29.)\nMR × 15.99 ÷ 100 ÷ 12 = (Months interest: \\$16.53)\nM+ MR (Add interest to memory)\n\nTerrell's new balance next month would be \\$1,256.82.\n---------------\nExample 3: Lita deposited \\$500 in a Double Your Money CD. The bank will pay her \\$1,000 when the CD doubles in value. The bank pays 7.5% interest on these deposits. How long will Lita wait?\n\nThis time we are looking for T.\n\nVariables:\nP = \\$500\nR% = 7.5%\nI = \\$500\n\nWhy is I = \\$500? Lita deposits \\$500 and will wait for her account to grow \\$500, to earn \\$500 in interest. Solving for T, time:\n\nT = I / (P × R%/100) = I × (P × R%/100)^-1 = (P × R%/100)^-1 × I\n\nKeystrokes:\n500 × 7.5 ÷ 100 =\n÷ ÷ = (37.5^-1)\n× 500 = (Display: 13.333333)\n\nSo it takes 13 1/3 years to double Lita's CD. She may want to rethink this investment.\n---------------\n\nSo this wraps up Part 4 of our Calculator Tricks series. Coming up in Part 5 we will tackle two common algebra problems.\n\nEddie\n\nThis blog is property of Edward Shore, 2012.\n\n#### 1 comment:\n\n1.",
null,
"take home pay calculator\nhttp://www.take-home-paycalculator.com\nA calculator which will automatically work out your take home pay calculator how much you have left each, week,month or year. A simple calculator to show your net pay.\n\n### HP Prime and HP 41C/DM 41L: Sum of Two Squares\n\nHP Prime and HP 41C/DM 41L: Sum of Two Squares Introduction Given a positive integer n, can we find two non-negative integers x and y..."
] | [
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"http://lh4.googleusercontent.com/-_goe0-uO-sU/AAAAAAAAAAI/AAAAAAAAABE/b0DuT7RAkpw/s35-c/photo.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.891183,"math_prob":0.9764158,"size":7179,"snap":"2019-35-2019-39","text_gpt3_token_len":2123,"char_repetition_ratio":0.12780488,"word_repetition_ratio":0.100221075,"special_character_ratio":0.360078,"punctuation_ratio":0.1640178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9876331,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T17:10:52Z\",\"WARC-Record-ID\":\"<urn:uuid:c345da0a-dec5-4579-8c00-71fc6c9e77f9>\",\"Content-Length\":\"111600\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7885a46-3160-4b68-ae65-6662536a5535>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9fe4d37-7311-4030-9f0d-bd2875c8dc8e>\",\"WARC-IP-Address\":\"172.217.13.65\",\"WARC-Target-URI\":\"http://edspi31415.blogspot.com/2012/10/calculator-tricks-part-4.html\",\"WARC-Payload-Digest\":\"sha1:X7LJC7MY6KVZ7L5V4LBA622IPZQXZLHE\",\"WARC-Block-Digest\":\"sha1:WPGT4HA4BVDI3SQSEC7PUMN5JWTCNJS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313987.32_warc_CC-MAIN-20190818165510-20190818191510-00149.warc.gz\"}"} |
https://www.r-bloggers.com/2017/02/the-r-formula-method-the-bad-parts-2/ | [
"Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nR’s model formula infrastructure was discussed in my previous post. Despite the elegance and convenience of the formula method, there are some aspects that are limiting.\n\n## Limitations to Extensibility\n\nThe model formula interface does have some limitations:\n\n• It can be kludgy with many operations on many variables (e.g., log transforming 50 variables via a formula without using `paste`)\n• The `predvars` aspect (discussed in my previous post) limits the utility of the operations. Suppose a formula had: `knn_impute(x1) + knn_impute(x2)`. Do we embed the training set twice in `predvars`?\n• Operations are constrained to single columns or features (excluding interaction specifications). For example, you cannot do\n```lm(y ~ pca(x1, x2, x3), data = dat)\n## or nested functions\nlm(y ~ pca(scale(x1), scale(x2), scale(x3)), data = dat)```\n\nI’ll use PCA feature extraction a few times here since it is probably familiar to many readers.\n\n## Everything Happens at Once\n\nSome of our data operations might be sequential. For example, it is not unreasonable to have predictors that require:\n\n1. imputation of a missing value\n2. centering and scale\n3. conversion to PCA scores\n\nGiven that the formula method operations happen (in effect) at once, this workflow requires some sort of custom solution. While `caret::preProcess` was designed for this sequence of operations, it does so in a single call, as opposed to a progression of steps exemplified by `ggplot2`, `dplyr`, or `magrittr`.\n\nAllowing a series of steps to be defined in order is more consistent with how data analysis is conducted. However, it does raise the complexity of the underlying implementation. For example, `caret::preProcess` dictates the possible sequence of tasks to be: filters, single-variable transformations, normalizations, imputation, signal extraction, and spatial sign. This avoids nonsensical sequences that center the data before applying a Box-Cox calculation (which requires positive data).\n\n### No Recycling\n\nAs a corollary to the point above, there is no way to recycle the `terms` between models that share the same formula and data/environment. For example, if I fit a CART model to a data set with many predictors, the random forest model (theoretically) shouldn’t need to recreate the same `terms` information about the design matrix. If the model function has the non-formula interface (e.g., `mod_func(x, y)`), this can make it easier. However, many do not.\n\nAlso, suppose that one of the pre-processing steps is computationally expensive. We’d like to be able to store the state of the results and then add another layer of computations (perhaps as a separate object).\n\n### Formulas and Wide Datasets\n\nThe `terms` object saves a matrix with as many rows as formula variables and at least as many columns (depending on interactions, etc). Most of this data is zero and a non–sparse representation is used. The current framework was built in a time where there was more focus on interactions, nesting and other operations on a small scale.\n\nIt is unlikely that models would have hundreds of interaction terms, but now it is not uncommon to have hundreds or thousands of main effects. As the number of predictors increases, this takes up an inordinate amount of execution time. For simple `randomForest` or `rpart` calls, the formula/`terms` work can account for most of the execution time. For example, we can calculate how much time functions spend generating the model matrix relative to the total execution time. For `rpart` and `randomForest`, we used the default arguments and did the calculations with a simulated data set of 200 data points and varying numbers of predictors:",
null,
"This is especially problematic for ensemble models. For example, `ipred:::ipredbagg` creates an ensemble of `rpart` trees. Since `rpart` only has a formula method, the footprint of the bagged model object can become very large if X trees are contained in the ensemble. Alternatively, `randomForest.formula` takes the approach of generating the `terms` once and feeding the model frame to `randomForest.default`. This does not work for `rpart` since there is no non-formula method exposed. Some functions (e.g., `lm`, `survival::coxph`) have arguments that can be used to prevent the `terms` and similar objects from being returned. This saves space but prevents new samples from being predicted. A little more detail can be found here.\n\nOne issue is the `\"factors\"` attribute of the `terms` object (discussed in the previous post). This is a non-sparse matrix that has a row for each predictor in the formula and a column for each model term (e.g. main effects, interactions, etc.). The purpose of this object is to know which predictors are involved in which terms.\n\nThe issue is that this matrix can get very large and usually has a high proportion of zeros. For example:\n\n```mod3 <- lm(Sepal.Width ~ Petal.Width + Petal.Length*Species, data = iris)\nfact_mat <- attr(mod3\\$terms, \"factors\")\nfact_mat\n## Petal.Width Petal.Length Species Petal.Length:Species\n## Sepal.Width 0 0 0 0\n## Petal.Width 1 0 0 0\n## Petal.Length 0 1 0 1\n## Species 0 0 1 1```\n\nAs the number of predictors increases, the rate of ones is likely to approach a value close to zero very quickly. For example:",
null,
"Again, it is doubtful that a model with a large number of predictors will have a correspondingly large number of high-level interactions (see the Pareto principle applied to modeling).\n\n### Variable Roles\n\nSome packages have implemented extensions of the basic formula. There are cases when formula are needed for specific sub-models. For example, a random coefficient model can be fit with the `lmer` function. In this case, a model is specified for a particular clustering variable (e.g., a subject in a clinical trial). The code is an example of how `lmer` syntax works:\n\n```# ?lme4::lmer\nlmer(Reaction ~ Days + (Days | Subject), data = sleepstudy)```\n\nHere `Subject` is important to the model-fitting routine, but not as a predictor. Similarly, the Bradley-Terry model can be used to model competitions and contests. A model on a set of boxers in a series of contests can include terms for their reach:\n\n```BTm(outcome = 1, player1 = winner, player2 = loser,\nformula = ~reach[..] + (1 | ..),\ndata = boxers)```\n\nAnother extension of basic formulas comes from the `modeltools` and `mboost` packages. The function `mboost::mob` fits a tree-based model with regression models in the terminal nodes. For this model, a separate list of predictors are used as splitting variables (to define the tree structure) and another set of regression variables that are modeled in the terminal nodes. An example of this call is:\n\n```# mboost::mob (using the modeltools package for formulas)\nmob(diabetes ~ glucose | pregnant + mass + age,\ndata = PimaIndiansDiabetes)```\n\nThe commonality between these three examples is that there are variables that are critical to the model but do not play the role of standard regression terms. For `lmer`, `Subject` is the independent experimental unit. For `mob`, we have variables to be used for splitting, etc.\n\nThere are similar issues on the left-hand side of the formula. When there are multivariate outcomes, different packages have different approaches:\n\n```# ?aggregate the mean of two variables by month\naggregate(cbind(Ozone, Temp) ~ Month, data = airquality, mean)\n\n# grouped binomial data:\nglm(cbind(events, nonevents) ~ x, family = binomial)\n\n# pls::plsr,\npls(sensory ~ chemical, data = oliveoil)\n# sensory and chemical are 2D arrays in the oliveoil data frame```\n\nThe overall point here is that, for the most part, the formula method assumes that there is one variable on the left-hand side of the tilde and that the variables on the right-hand side are predictors (exceptions are discussed below). One can envision other roles that columns could play in the analysis of data. Besides the examples given above, variables could be used for\n\n• outcomes\n• predictors\n• stratification\n• data for assessing model performance (e.g., loan amount to compute expected loss)\n• conditioning or faceting variables (e.g., `lattice` or `ggplot2`)\n• random effects or hierarchical model ID variables\n• case weights\n• offsets\n• error terms (limited to `Error` in the `aov` function)\n\nThe last three items on this list are currently handled in formulas as “specials” or have existing functions. For example, when the model function has a `weights` argument, the current formula/`terms` frame work uses a function (`model.weights`) to extract the weights, and also makes sure that the weights are not included as covariates. The same is true for offsets.\n\n## Summary\n\nSome limitations of the current formula interface can be mitigated by writing your own or utilizing the `Formula` package.\n\nHowever, there are a number of conceptual aspects (e.g., roles, sequential processing) that would require a completely different approach to defining a design matrix, and this will be the focus of an upcoming tidyverse package."
] | [
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https://phys.org/tags/quantum%20tunneling/sort/liverank/all/ | [
"# News tagged with quantum tunneling\n\npage 1 from 7\n\n## Quantum tunnelling\n\nWave-mechanical tunnelling (also called quantum-mechanical tunnelling, quantum tunnelling, and the tunnel effect) is an evanescent wave coupling effect that occurs in the context of quantum mechanics because the behaviour of particles is governed by Schrödinger's wave-equation. All wave equations exhibit evanescent wave coupling effects if the conditions are right. Wave coupling effects mathematically equivalent to those called \"tunnelling\" in quantum mechanics can occur with Maxwell's wave-equation (both with light and with microwaves), and with the common non-dispersive wave-equation often applied (for example) to waves on strings and to acoustics.\n\nFor these effects to occur there must be a situation where a thin region of \"medium type 2\" is sandwiched between two regions of \"medium type 1\", and the properties of these media have to be such that the wave equation has \"traveling-wave\" solutions in medium type 1, but \"real exponential solutions\" (rising and falling) in medium type 2. In optics, medium type 1 might be glass, medium type 2 might be vacuum. In quantum mechanics, in connection with motion of a particle, medium type 1 is a region of space where the particle total energy is greater than its potential energy, medium type 2 is a region of space (known as the \"barrier\") where the particle total energy is less than its potential energy - for further explanation see the section on \"Schrödinger equation - tunnelling basics\" below.\n\nIf conditions are right, amplitude from a traveling wave, incident on medium type 2 from medium type 1, can \"leak through\" medium type 2 and emerge as a traveling wave in the second region of medium type 1 on the far side. If the second region of medium type 1 is not present, then the traveling wave incident on medium type 2 is totally reflected, although it does penetrate into medium type 2 to some extent. Depending on the wave equation being used, the leaked amplitude is interpreted physically as traveling energy or as a traveling particle, and, numerically, the ratio of the square of the leaked amplitude to the square of the incident amplitude gives the proportion of incident energy transmitted out the far side, or (in the case of the Schrödinger equation) the probability that the particle \"tunnels\" through the barrier.\n\nThis text uses material from Wikipedia, licensed under CC BY-SA"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9191195,"math_prob":0.9555979,"size":2294,"snap":"2020-10-2020-16","text_gpt3_token_len":499,"char_repetition_ratio":0.15938865,"word_repetition_ratio":0.03814714,"special_character_ratio":0.1983435,"punctuation_ratio":0.07160494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9635552,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T09:06:45Z\",\"WARC-Record-ID\":\"<urn:uuid:ab9b4743-5d7d-407f-bc93-bc3754704e05>\",\"Content-Length\":\"60975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fa6dc40a-5e17-4d81-a1dc-f639d2e5d4fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f27db377-77da-423a-94bd-9ae830b1d2fa>\",\"WARC-IP-Address\":\"72.251.236.55\",\"WARC-Target-URI\":\"https://phys.org/tags/quantum%20tunneling/sort/liverank/all/\",\"WARC-Payload-Digest\":\"sha1:K2FK4B4AU2ITDOF7S2CMC4EXN34IFZRW\",\"WARC-Block-Digest\":\"sha1:UGTVLAHQO4D3SNWH6ZD3R6SJQPZRV552\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146665.7_warc_CC-MAIN-20200227063824-20200227093824-00448.warc.gz\"}"} |
https://www.datasciencecentral.com/profiles/blogs/a-beautiful-probability-theorem | [
"#",
null,
".\n\nWe all know that, given two events A and B, the probability of the union A U B is given by the formula P(A U B) = P(A) + P(B) - P( AB) where AB represents the intersection of A and B. Most of us even know that\n\nP(A U B U C) = P(A) + P(B) + P(C) - { P(AB) + P(AC) + P(BC) } + P(ABC)\n\nIn particular, if the events are independent, it becomes:\n\n1 - P(A U B U C) = 1 - { P(A) + P(B) + P(C) } + { P(A)P(B) + P(A)P(C) + P(B)P(C) } - P(A)P(B)P(C)\n\nThis is equivalent to\n\n1 - P(A U B U C) = { 1 - P(A) } { 1 - P(B) } { 1 - P(C) }\n\nIt generalizes to n independent events, and this formula is known as the inclusion-exclusion principle. Let us consider n events A(1), A(2), ... , A(n) where A(k) is for a positive integer number, the property to be divisible by the square of the k-th prime number. We assume here that the first prime number is 2. These events are independent because we are dealing with prime numbers. As n tends to infinity, 1 - P( A(1) U A(2) U ... U A(n) ) tends to the probability, for a positive integer number, to be square-free. Thus we have:\n\nBeautiful Probability Theorem\n\nThe probability, for a positive integer number to be square-free (that is, not divisible by any square integer other than 1), is given by the formula below, where the product is over all prime numbers p > 1.",
null,
"If you are interested in this kind of topic, read my article on number densities, especially the exercise at the bottom of section 1, and the section on probabilistic number theory at the bottom of section 4. You might also be interested in my article entitled the fundamental statistics theorem revisited\n\nAlso, interestingly, the probability computed here is identical to the probability that two numbers are relatively prime.\n\nTop DSC Resources",
null,
"Comment by Vincent Granville on December 22, 2016 at 12:38am"
] | [
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"https://storage.ning.com/topology/rest/1.0/file/get/7215525257",
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"https://storage.ning.com/topology/rest/1.0/file/get/2808321244",
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"https://storage.ning.com/topology/rest/1.0/file/get/2800211702",
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https://www.r-bloggers.com/keith-matthews-exam-question/ | [
"# Keith Matthews Exam Question…\n\nDecember 14, 2016\nBy\n\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nHelp? I set an exam question with 10 boxes to be filled from 13 possible answers. If randomly completed, what should be the average mark?\n– @KeithRMatthews, 12:13 PM – 15 Dec 2016\n\n``````library(ggplot2)\noptions <- letters[1:13]\n``````\n\nfunctions to give random answers and mark them\n\n``````giveAnswers <- function(options) sample(options, size = 10, replace = FALSE)\nmarkAnswers <- function(correct, given) sum(given == correct)\n``````\n\ncollect scores\n\n``````scores <- numeric(1e5L)\nfor(i in seq_along(scores)) {\n}\n``````\n\naverages\n\n``````mean(scores)\n``````\n``````## 0.766\n``````\n``````median(scores)\n``````\n``````## 1\n``````\n\nplot\n\n``````ggplot(data.frame(score = scores), aes(x = scores)) +\ngeom_bar(stat = \"count\")\n``````",
null,
"R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't."
] | [
null,
"https://i1.wp.com/rmnppt.com/figure/source/2016-12-15-matthews-scores/unnamed-chunk-5-1.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77148515,"math_prob":0.80336785,"size":1425,"snap":"2020-10-2020-16","text_gpt3_token_len":365,"char_repetition_ratio":0.12807882,"word_repetition_ratio":0.2,"special_character_ratio":0.26877192,"punctuation_ratio":0.11636364,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98498625,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T10:58:43Z\",\"WARC-Record-ID\":\"<urn:uuid:d4cc8e0e-493d-4af2-adc4-4ebe7e34de3a>\",\"Content-Length\":\"150908\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:431bbef7-aec4-411f-a5ba-4c86f3bcd284>\",\"WARC-Concurrent-To\":\"<urn:uuid:beeedd9d-3eef-4ba7-a6ca-89f3375f3ae4>\",\"WARC-IP-Address\":\"104.28.8.205\",\"WARC-Target-URI\":\"https://www.r-bloggers.com/keith-matthews-exam-question/\",\"WARC-Payload-Digest\":\"sha1:OSJ4RZPXLGERS6N3NGXZTHV7IERRXH75\",\"WARC-Block-Digest\":\"sha1:EFIMLENO4SM7E4H222EG36WHR4XMSZPD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146341.16_warc_CC-MAIN-20200226084902-20200226114902-00408.warc.gz\"}"} |
https://www.ellipsix.net/blog/2011/10/scattering-a-bullet-off-an-rpg.html | [
"## Scattering a bullet off an RPG\n\nYep, that’s right, a rocket-propelled grenade finally made its way on Mythbusters! Personally I’m surprised it took them so long…\n\nAnyway, let’s not get distracted from the science by cool explosions just yet. On last week’s season premiere of Mythbusters, Kari, Tory, and Grant tested a myth based on a scene in RED in which two characters are facing off, one (the hero) with a revolver and the other (the villain) with an RPG launcher. In the movie, they both shoot at the same time, the bullet hits the RPG in midair and detonates it, and the resulting explosion kills the villain. Now, in the show, this myth was busted on several counts:\n\n• RPGs don’t even arm until about 60 feet after launch, about $$\\frac{3}{4}$$ of the way to the target and long after this one would have been hit by the bullet\n• When an RPG explodes, it sends balls of molten copper flying forward, which wouldn’t have been stopped by the bullet\n• The distance at which the explosion would have taken place, 16 feet from the villain, is quite survivable\n\nUnfortunately, they didn’t test what I thought was the most obvious objection to the scene (albeit the one that gives it its “cool factor”): you’d need excellent aim, and some amount of luck, to reliably hit as small a target as an RPG detonator head on with a bullet. Or so it seems. But is this really the case? Even if the RPG was armed, and even if it somehow could lose all its forward momentum and explode in a giant fireball, how likely is it that the collision would actually take place?",
null,
"As it turns out, shooting two objects at each other and characterizing what happens is a very common thing to do in physics — for instance, this is exactly what happens at particle accelerators like the recently departed Tevatron or the LHC — so the methods used to analyze this sort of situation are quite well studied, especially in a simple case like this one. Imagine the collision from the point of view of the RPG, which “sees” a bullet coming at it. Actually, imagine doing this experiment many times to get a statistical sample, so the RPG actually sees a whole beam of bullets coming at it, as shown on the left. The detonator cap occupies a circle (the solid red one) somewhere in the middle of the beam, and any bullet that hits that circle will trigger an explosion. It’s probably safe to say that even a glancing blow will be enough to set the RPG off, so we should expand that circle a little, specifically by half the radius of a bullet, so that any bullets whose centers are within the circle will cause explosions — that’s the dashed circle in the diagram. The area of this circle is called the scattering cross section, $$\\sigma$$.",
null,
"In order to figure out the probability of a collision, we need to find the fraction of bullets that will hit the aforementioned circle. The transverse positions of the bullets relative to the RPG form some distribution, called the beam profile $$\\rho(\\mathbf{r}_\\perp)$$, so we can figure out the probability by integrating the beam profile over the circle to get the number of bullets (out of this large statistical sample) that will hit.\n\n$$P = \\iint_\\text{circle} \\rho(\\mathbf{r}_\\perp)\\uddc\\mathbf{r}_\\perp$$\n\nSo if we can get the normalized beam profile $$\\rho(\\mathbf{r}_\\perp)$$ of a stream of bullets, we’d be all set. This is the tricky part, though, since doing so would require looking up some statistics on the accuracy of handgun aiming, and I’m not sure where to get reliable information about that. So in lieu of that, I turn to the internet. Here’s a thread about accuracy at The Firing Line forums which includes some photos of shots made at a specified range. I can extract the coordinates of the bullet holes, compare them to the given scale, and thereby find the beam profile of the person who shot them. Sure, it’s not a statistically significant sample, but I’m just going for a rough estimate anyway.\n\nSince this isn’t a rigorous experiment, I’m just going to work from this one image.",
null,
"In order to easily determine the coordinates of the bullet holes, I turned to the first tool I had at hand, Mathematica. This little bit of code (for Mathematica 8) will let you click on each hole (or any point) and print out a list of the coordinates you clicked:\n\nDynamicModule[{p = {}},\nEventHandler[\nDynamic@Show[i, Graphics[{PointSize[Large], Red, Point[p]}]],\n\"MouseClicked\" :> {AppendTo[p,\nMousePosition[\"GraphicsAbsolute\"]]}]\nButton[\"Clear\", {p = {}}]\nButton[\"Print\", Print[p]]\n]\n\n\nWith this I got the following list of positions:\n\n{{280.5,498.},{236.5,447.},{290.5,407.},{335.5,330.},{391.5,449.},{405.5,471.},\n{516.5,409.},{439.5,292.},{299.5,264.},{373.5,227.},{413.5,198.},{453.5,150.},\n{505.5,256.},{554.5,268.},{531.5,132.}}\n\n\nI also determined that the distance corresponding to the 5.5 inch line marked at the bottom is 316 pixels, which I can use to convert the pixel positions into centimeters. After that, we have to make a distance correction: the picture shows shots taken at a distance of 150 feet, but the scenario in the movie involved a shot of 64 feet, so (as a simple approximation) we just insert a factor of $$\\frac{64}{150}$$.\n\nThis code graphs the fraction of bullet holes as a function of distance from the average bullet position, rescaled to a 64 foot shot:\n\nHistogram[64./150. (EuclideanDistance[Mean[points], #] & /@ points), 10, \"CDF\"]\n\n\nHere’s a fancier version:",
null,
"This graph is an approximation to what I would call the cumulative radial beam profile, which is related to the actual beam profile as\n\n$$\\rho_\\text{cuml}(R) = \\int_0^R\\int_0^{2\\pi}\\rho(\\mathbf{r}_\\perp)r_\\perp\\udc r_\\perp\\udc\\phi$$\n\nBecause of the small number of shots I’ve analyzed, it’s hard to tell from the graph exactly what the form of $$\\rho_\\text{cuml}(R)$$ is — it could be linear, or quadratic (as you’d expect for a uniform distribution over a circle), or perhaps Gaussian (although I would expect more clustering in the center if that were the case), or something else entirely.\n\nThankfully, we don’t really need to know the form, at least not if we just want a rough estimate of the probability. Just set $$R$$ to the radius of the scattering cross-section of the detonator cap, and $$\\rho_\\text{cuml}(R)$$ is the probability of hitting it. From what I saw in the show, I’d estimate $$R \\approx \\SI{2.5}{cm}$$, which puts the probability at $$P \\approx 0.5$$. That’s actually not a bad chance — not one I’d want to stake my life on, but a lot better than I would have guessed. And according to this article on handgun accuracy from Guns & Ammo magazine, it seems like that’s average for a skilled casual shooter. I bet a trained CIA agent, especially one worthy of being played in a movie by John Malkovich, would have no problem making the shot."
] | [
null,
"https://www.ellipsix.net/blog/2011/10/attachments/_9tb/cross-section.png",
null,
"https://www.ellipsix.net/blog/2011/10/attachments/IF_A/bullet_rpg_scattering.png",
null,
"http://i112.photobucket.com/albums/n172/Paul105_photo/RIMG1758.jpg",
null,
"https://www.ellipsix.net/blog/2011/10/attachments/IF_A/accuracyhistogram.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.940797,"math_prob":0.9467111,"size":6632,"snap":"2021-43-2021-49","text_gpt3_token_len":1582,"char_repetition_ratio":0.10500905,"word_repetition_ratio":0.0035778175,"special_character_ratio":0.25467432,"punctuation_ratio":0.13189964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97227556,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T19:12:55Z\",\"WARC-Record-ID\":\"<urn:uuid:043a0a4a-f169-4bc2-b205-f8d11a840047>\",\"Content-Length\":\"20252\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7169b21e-abb9-4032-a32e-015e7af2d93f>\",\"WARC-Concurrent-To\":\"<urn:uuid:3697d63f-599c-4428-a768-d8f4f7228cb0>\",\"WARC-IP-Address\":\"45.56.95.107\",\"WARC-Target-URI\":\"https://www.ellipsix.net/blog/2011/10/scattering-a-bullet-off-an-rpg.html\",\"WARC-Payload-Digest\":\"sha1:N7IAMLUSEM6LWVOGPYTEU5NPGYCZWX4G\",\"WARC-Block-Digest\":\"sha1:2LP7C55ID6GKLVTWFEV2PKJIXWFPQWJI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359065.88_warc_CC-MAIN-20211130171559-20211130201559-00403.warc.gz\"}"} |
https://forum.arduino.cc/t/led-sketch/240914 | [
"",
null,
"# Led Sketch\n\nI am using the sketch:\n\nint pinArray = {22, 23, 24, 25, 26, 27, 28, 29, 30};\nint count = 0;\nint timer = 30;\n\nvoid setup(){\nfor (count=0;count<8;count++) {\npinMode(pinArray[count], OUTPUT);\n}\n}\n\nvoid loop() {\nfor (count=0;count<7;count++) {\ndigitalWrite(pinArray[count], HIGH);\ndelay(timer);\ndigitalWrite(pinArray[count + 1], HIGH);\ndelay(timer);\ndigitalWrite(pinArray[count], LOW);\ndelay(timer2);\n}\nfor (count=7;count>0;count–) {\ndigitalWrite(pinArray[count], HIGH);\ndelay(timer);\ndigitalWrite(pinArray[count - 1], HIGH);\ndelay(timer);\ndigitalWrite(pinArray[count], LOW);\ndelay(timer\n2);\n}\n}\n\nTo blink LEDs like on knight rider. What I was wondering was if it is possible to run 3 of these sketches at a time. The idea is one bar of 8 LEDs will run by itself and 2 bars of 10 LEDs will run parallel up and down past each other. Help will be appreciated. Thanks.\n\nSorry if I lacked any detail you need.\n\nFirst thing - you have 9 elements in your array, you are only setting up 8 as OUTPUT tho.\n\nYes, you can run multiple LED bars at the same time. Look at LED cubes - folks make 8x8, my wife & I are making a 9x9, that is 64 & 81 LEDs per layer. Two bars of 8 and a bar of 10 are easily possible. Look at blink-without-delay code to see how to have each running on its own timing as well if you want them running independently.\n\nYes, there are many ways to do that, Shifter-register, I2C I/O expanders, etc. This is very used in led cubs and led displays.\n\nI write a little sketch this morning that does something like what yours do. My approach it’s a little different.\n\n``````#define LENGTH 4\nint output[] = {31, 33, 35, 37};\nint count;\n\nvoid setup() {\nfor (int i=0; i<LENGTH; i++) {\npinMode(output[i], OUTPUT);\n}\n}\n\nvoid loop() {\nif (count < LENGTH) {\ndigitalWrite(output[count], HIGH);\n}\nelse {\ndigitalWrite(output[count-LENGTH], LOW);\n}\n\nif (++count >= 2*LENGTH) {\ncount = 0;\n}\n\ndelay (500);\n\n}\n``````\n\nI don't know why, but I was thinking that your problem was the hardware (less outputs than led's). Now that I think about and realise that is not to hard and I came to write the answer, I've read another time your first post and I understand that your problem is the code, right?\n\nCan you explain what is your idea? From your description I don't understand very well what the code must do.\n\nThe representation of the led's is this:\n\n`````` * * * * * * * * * *\n# # # # # # # #\n+ + + + + + + + + +\n``````\n\nWhat you mean by: \"past each other\"? First goes on the \"*\" and then the \"+\" at the same time that \"#\" is on?\n\nWell my plan was really having something like this if you can understand it:\n\nLED bar 1 : + + + + + + + +\n\nThis LED bar runs from left to right over and over while the other 2:\n\nLED Bar 2: = LED Bar 3: - Bar 2 runs from = to - while at the same time Bar 3 + + Runs from - to = passing each other meeting close + + to the middle. + + + + + + + + + + + + - =\n\nHopefully you get my idea this time and thanks a lot for the help so far.\n\nOk. Now I understand what you want and it’s easiest that it may look. Ba2 and Bar3 are the same thing but hey are upside-down! You can peak the signal from the first led in the Bar2 and connect it to the last led of the Bar3. Peak the signal from the second led in the Bar2 and connect it to the penultimate led of the Bar3. And so on.\nYour really problem is to have the 2 “programs” running at the same time.\n\nEDIT: After I write my last reply, I wrote a program that make the same thing that you are doing in the “original” program but with the method that I use in my example. That’s what it look like:\n\n``````#define LENGTH 8\nint output[] = {31, 33, 35, 37, 39, 41, 43, 45};\nint count;\nboolean forward = true;\n\nvoid setup() {\nfor (int i=0; i<LENGTH; i++) {\npinMode(output[i], OUTPUT);\n}\n}\n\nvoid loop() {\n\nif (forward) {\ndigitalWrite(output[count], HIGH);\nif (count > 0 ) {\ndigitalWrite(output[count-1], LOW);\n}\n}\nelse {\ndigitalWrite(output[LENGTH-count-1], HIGH);\nif (count > 0 ) {\ndigitalWrite(output[LENGTH-count], LOW);\n}\n}\n++count;\nif ( count >= LENGTH ) {\ncount = 1;\nif (forward) forward = false;\nelse {\nforward = true;\n}\n}\ndelay (100);\n}\n``````\n\nIf I understand right what you want to do is this.\n\nThe code for this is:\n\n``````#define LENGTH1 4\nint output1[] = {31, 33, 35, 37};\nint count1;\nboolean forward1 = true;\n\n#define LENGTH2 5\nint output2[] = {30, 32, 34, 36, 38};\nint count2;\nboolean forward2 = true;\n\nvoid setup() {\nint i;\nfor (i=0; i<LENGTH1; i++) {\npinMode(output1[i], OUTPUT);\n}\nfor (i=0; i<LENGTH2; i++) {\npinMode(output2[i], OUTPUT);\n}\n}\n\nvoid loop() {\n\nif (forward1) {\ndigitalWrite(output1[count1], HIGH);\nif (count1 > 0 ) {\ndigitalWrite(output1[count1-1], LOW);\n}\n}\nelse {\ndigitalWrite(output1[LENGTH1-count1-1], HIGH);\nif (count1 > 0 ) {\ndigitalWrite(output1[LENGTH1-count1], LOW);\n}\n}\n++count1;\nif ( count1 >= LENGTH1 ) {\ncount1 = 1;\nif (forward1) forward1 = false;\nelse {\nforward1 = true;\n}\n}\n\nif (forward2) {\ndigitalWrite(output2[count2], HIGH);\nif (count2 > 0 ) {\ndigitalWrite(output2[count2-1], LOW);\n}\n}\nelse {\ndigitalWrite(output2[LENGTH2-count2-1], HIGH);\nif (count2 > 0 ) {\ndigitalWrite(output2[LENGTH2-count2], LOW);\n}\n}\n++count2;\nif ( count2 >= LENGTH2 ) {\ncount2 = 1;\nif (forward2) forward2 = false;\nelse {\nforward2 = true;\n}\n}\n\ndelay (200);\n}\n``````\n\nIt can be made in a better way, but it’s working.\n\nPS: I’m sorry for the bad video, but is the only camera that I have.\n\nThanks for the help, im glad someone understood my plan\nthe video was great. Sorry for the lack of replies. I have been trying to save some internet credit.\n\nThanks"
] | [
null,
"https://aws1.discourse-cdn.com/arduino/original/3X/1/f/1f6eb1c9b79d9518d1688c15fe9a4b7cdd5636ae.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7079101,"math_prob":0.9494966,"size":895,"snap":"2021-21-2021-25","text_gpt3_token_len":258,"char_repetition_ratio":0.18518518,"word_repetition_ratio":0.062015504,"special_character_ratio":0.32402235,"punctuation_ratio":0.22564103,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95122886,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T16:46:26Z\",\"WARC-Record-ID\":\"<urn:uuid:852ee0a7-4333-4f2c-8f96-4f08cbb54c0b>\",\"Content-Length\":\"34274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d6e915d-f766-4744-a731-894c1f3baeee>\",\"WARC-Concurrent-To\":\"<urn:uuid:28c08a46-b37d-4caa-a35e-149ac7354439>\",\"WARC-IP-Address\":\"184.104.202.141\",\"WARC-Target-URI\":\"https://forum.arduino.cc/t/led-sketch/240914\",\"WARC-Payload-Digest\":\"sha1:T5XJDZGXY3VJCG5WID3WUHC2GMP2VEDT\",\"WARC-Block-Digest\":\"sha1:RJPX62BKCY2J7G3XTMC22TXPRB66ZDW3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487648373.45_warc_CC-MAIN-20210619142022-20210619172022-00360.warc.gz\"}"} |
http://www.educationwa.com/2011/02/drawing-first-derivative.html | [
"## Wednesday, February 23, 2011\n\n### Drawing the first derivative\n\nTeaching students how to visualise the first derivative in 3B MAT has been problematic over the last two years. This morning I had a bit of a breakthrough in that students weren't looking at me as if I was speaking Alien.\n\nThe major difference was that I didn't use the arrow approach. Here's what I did.\n\nI drew a positive cubic on the board and identified the turning points. I identified clearly the x axis and the y axis and identified the coordinates for each TP. I drew their attention to (x,y)\n\nThen I drew a second pair coordinate plane directly underneath and identified/labelled the x axis. I then deliberately (as in made a big song and dance) labelled the other axis y' asking students to think what this might mean.\n\nI then went to the first turning point on the x,y plane and asked students what the gradient was at this point. They said zero straight away.\n\nI then went to the second axis and said coordinates on this plane were (x,y'). Given that the TP we were examining was at (0.25) and y'(0.25) = 0, the coordinate(x,y') that we needed was at (0.25,0). We repeated this for the other turning point.\n\nI then drew vertical dotted lines through both coordinate planes. We then looked at the slope to the left of the TP. Being a cubic (with a positive coefficient of x cubed) the slope was +ve. On the second plane I wrote +ve above the x axis to the left of the TP above the x axis. We then examined the second area and noted the slope was negative (making special note of where the point of inflection was - it wasn't mandated by the course but made sense in the context). I labelled the graph -ve underneath the x axis to the right of the TP. I then wrote +ve in the third area above the x axis.",
null,
"<- It looked like this.\n\nOnce the areas were labelled it was trivial to join the dots starting where y' was positive (y' at +ve infinity), leading to where y' was negative and then changing direction midway between the x intercepts on y', back towards to the x axis until y' was +ve again (again until y' at +ve infinity). It was also a good time to discuss the type of function produced (eg a concave up quadratic) if you differentiate a cubic with a +ve coefficient of the cubed term and how that related to our y' graph.",
null,
"We then repeated the process for a quartic.\n\nyay!\n\n#### Post a Comment\n\nHi, thanks for leaving a comment.. it's good to hear what people think!"
] | [
null,
"http://2.bp.blogspot.com/-FqSxBzJVYMM/TWULnBNy4RI/AAAAAAAAATs/bMGEYcBIGS8/s320/cubic%2B1.PNG",
null,
"http://2.bp.blogspot.com/-F8GzbhauuOY/TWULnTix79I/AAAAAAAAAT0/9AwKFVPxmn0/s320/cubic%2B2.PNG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.97742987,"math_prob":0.8311832,"size":2380,"snap":"2019-13-2019-22","text_gpt3_token_len":548,"char_repetition_ratio":0.12710437,"word_repetition_ratio":0.013513514,"special_character_ratio":0.23235294,"punctuation_ratio":0.08249497,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893966,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T16:52:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4ac215cf-3f13-4421-9075-b9c9abfabbc5>\",\"Content-Length\":\"97767\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1abb30be-a333-4625-a750-4ee7478ae8a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:85dd8470-8606-4bf2-bd2a-60f9cd3e6b73>\",\"WARC-IP-Address\":\"172.217.164.179\",\"WARC-Target-URI\":\"http://www.educationwa.com/2011/02/drawing-first-derivative.html\",\"WARC-Payload-Digest\":\"sha1:6F5GMSHFLWM45ZMEG4VMUER5F2QT76TD\",\"WARC-Block-Digest\":\"sha1:GQGO2G3P5UVEMFUPLKP6O3427VQ664YE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232255071.27_warc_CC-MAIN-20190519161546-20190519183546-00392.warc.gz\"}"} |
https://www.simplilearn.com/tutorials/excel-tutorial/excel-interview-questions | [
"Microsoft Excel is a go-to tool if you’re working with data. Starting from simple calculations to building reports, Excel has it all covered. Knowing Excel has become a necessity in today’s times. Organizations rely on Excel for storing and analyzing their data. This article on MS Excel interview questions and answers will cover the top 50 questions that could be asked in an Excel interview.\n\nWatch the video below on Excel interview questions and answers that covers all the basic, intermediate and advanced interview questions that are frequently asked.\n\n## Beginner Level Excel Interview Questions\n\nWe will start with a set of Excel interview questions aimed at the beginners.\n\n### 1. What is a cell address in Excel?\n\nA cell address is used to identify a particular cell on a worksheet. It is denoted by a combination of the respective column letter and a row number.\n\nAs shown above, the highlighted cell belongs to the column ‘D’ and row 5, so the cell address is read as D5.",
null,
"#### Your Big Data Engineer Career Awaits!\n\nBig Data Engineer Master’s Program",
null,
"### 2. What do you mean by Relative cell referencing and Absolute cell referencing in MS Excel?\n\n Relative cell referencing Absolute cell referencing In Relative referencing, there is a change when copying a formula from one cell to another cell with respect to the destination. cells’ address Meanwhile, there is no change in Absolute cell referencing when a formula is copied, irrespective of the cell’s destination. This type of referencing is there by default. Relative cell referencing doesn’t require a dollar sign in the formula. If you don’t want a change in the formula when it’s copied across cells, then absolute referencing requires you to add a dollar sign before and after the column and row address.",
null,
"",
null,
"### 3. How do you freeze panes in Excel?\n\nFreeze panes keep the rows and columns visible while scrolling through a worksheet. To freeze panes, select the View tab and go to Freeze Panes.",
null,
"If you are looking to freeze the first two columns of a dataset, select the 3rd column, and click ‘Freeze Panes’. A thick grey border indicates this.",
null,
"### 4. How can you restrict someone from copying a cell from your worksheet?\n\n1. First, choose the data you want to protect.",
null,
"2. Hit Ctrl + Shift + F. The Format Cells tab appears. Go to the Protection tab. Check Locked and click OK.",
null,
"3. Next, go to the Review tab and select Protect Sheet. Enter the password to protect the sheet.\n\nLet’s now move onto our next question on our list of Excel interview questions.\n\n### 5. How is a Formula different from a Function in Excel?\n\n Formula Function The formula is like an equation in Excel, the user types in that. It can be any type of calculation depending on the user’s choice. Whereas, a function in Excel is a predefined calculation which is in-built in Excel. Manually typing out a formula every time you need to perform a calculation, consumes more time. Ex: = A1+A2+A3 However, performing calculations becomes more comfortable and faster while working with functions. Ex: = SUM(A1:A3)\n\nNow, let’s head to our next question in our list of Excel interview questions.\n\n### 6. Mention the order of operations used in Excel while evaluating formulas.\n\nThe order of operations in Excel is referred to as PEDMAS. Shown below is the order of precedence while performing an Excel operation.\n\n• Parentheses\n• Exponentiation\n• Division/Multiplication\n• Subtraction\n\nAs seen above, first, the data in the parentheses is operated, followed by the exponentiation operation. After that, it can be either the division or multiplication operations. The result is then added and finally subtracted to give the final result.\n\nLet’s look at an example of the PEMDAS precedence in the next question on our Excel interview questions list.\n\n### 7. How will you write the formula for the following? - Multiply the value in cell A1 by 10, add the result by 5, and divide it by 2.\n\nTo write a formula for the above-stated question, we have to follow the PEDMAS Precedence. The correct answer is ((A1*10)+5)/2.\n\nAnswers such as =A1*10+5/2 and =(A1*10)+5/2 are not correct. We must put parentheses brackets after a particular operation.\n\nThe output will look like this:",
null,
"### 8. What is the difference between count, counta, and countblank?\n\nThe count function is very often used in Excel. Here, let’s look at the difference between count, and it’s variants - counta and countblank.\n\n#### 1. COUNT\n\nIt counts the number of cells that contain numeric values only. Cells that have string values, special characters, and blank cells will not be counted. Shown below is an example of the count function.",
null,
"#### 2. COUNTA\n\nIt counts the number of cells that contain any form of content. Cells that have string values, special characters, and numeric values will be counted. However, a blank cell will not be counted. Shown below is an example of the counta function.",
null,
"#### 3. COUNTBLANK\n\nAs the name suggests, it counts the number of blank cells only. Cells that have content will not be taken into consideration. Shown below is an example of the countblank function.",
null,
"### 9. What is the shortcut to add a filter to a table?\n\nThe filter mechanism is used when you want to display only specific data from the entire dataset. By doing so, there is no change being made to the data. The shortcut to add a filter to a table is Ctrl+Shift+L.",
null,
"### 10. How do you create a hyperlink in Excel?\n\nHyperlinks are used to navigate between worksheets and files/websites. To create a hyperlink, the shortcut used is Ctrl+K.\n\nThe ‘Insert Hyperlink’ box appears. Enter the address and the text to display. Here, we are directed to the Amazon Website.",
null,
"### 11. How can we merge multiple cells text strings in a cell?\n\nTo merge text strings present in multiple cells into one cell, you can use the CONCATENATE(). Shown below is an example of the concatenate function.",
null,
"Another way of combining cell values is by using the “&” operator, as shown below:",
null,
"Let’s now move onto the next question on our Excel interview questions list.\n\n### 12. How can you split a column into 2 or more columns?\n\nYou can split a column into 2 or more columns by following the below steps:\n\n1. Select the cell that you want to split. Then, navigate to the Data tab, after that, select Text to Columns.",
null,
"2. Select the delimiter.",
null,
"3. Choose the column data format and select the destination you want to display the split.",
null,
"4. The final output will look like below where the text is split into multiple columns.",
null,
"### 13. What is the use of VLOOKUP and how do we use it?\n\nThe function VLOOKUP in Excel is used to look up information in a table and extract the corresponding data.\n\nSyntax: VLOOKUP (value, table, col_index, [range_lookup])\n\nvalue - Indicates the data that you are looking for in the first column of a table.\n\ntable - Refers to the set of data (table) from which you have to retrieve the above value.\n\ncol_index - Refers to the column in the table from where you are to retrieve the value.\n\nrange_lookup - FALSE = exact match [optional] TRUE = approximate match (default).\n\nShown below is an example of the VLOOKUP function. We are to find the Product related to the Customer Name – “Richard”.",
null,
"### 14. How is VLOOKUP different from the LOOKUP function?\n\n VLOOKUP LOOKUP VLOOKUP lets the user look for a value in the left-most column of a table. It then returns the value in a left-to-right way. It is not very easy to use as compared to the LOOKUP function. Meanwhile, the LOOKUP function enables the user to look for data in a row/column. It returns the value in another row/column. It is easier and can also be used to replace the VLOOKUP function.\n\n#### Your Data Analytics Career is Around The Corner!\n\nData Analyst Master’s Program",
null,
"### 15. How many report formats are available in Excel?\n\nThere are three report formats available in Excel; they are:\n\n1. Compact Form\n2. Outline Form\n3. Tabular Form\n\n### 16. How does the IF() function in Excel work?\n\nIn Excel, the IF() function performs a logical test. It returns a value if the test evaluates to true and another value if the test result is false. It returns the value depending on whether the condition is valid for the entire selected range.\n\nLet’s look at the below example:",
null,
"As seen above, the IF function returns “Record is Valid” if age is greater than 20, and the salary should be greater than \\$40000. Else, it will return “Record is Invalid”. Here the final answer will be “Record is Valid” as the entire selected range qualifies both the conditions.\n\n### 17. How do we use the SUMIF() function in Excel?\n\nThe SUMIF() function adds the cell values specified by a given condition or criteria. Given below is an example of the sumif function.",
null,
"As seen above, the costs corresponding to the years 2010 are added as per the given criteria.\n\n### 18. Using the COVID data, find the number of days in which the number of deaths in Italy has been greater than 200.\n\nTo perform this operation, we can use the COUNTIFS() function. The dataset we will be using is shown below:",
null,
"The COUNTIFS() function we use is - =COUNTIFS(G2:G35777,\"Italy\",E2:E35777,\">200\")\n\n### 19. What is a Pivot Table?\n\nA pivot table is like a summary table of the dataset that enables you to create reports and analyze trends. They are useful when you have long rows or columns that hold values you need to track.\n\nTo create a pivot table, first, go to the Insert tab and select the ‘PivotTable’ option.",
null,
"Select the table or the range and choose where you want to place the pivot table.",
null,
"Drag the fields you wish to show in the pivot table. Here we have created a pivot table using the Coronavirus data.",
null,
"### 20. Create a drop-down list in Excel.\n\nThis can be done by using the ‘Data Validation’ option present in the Data tab.",
null,
"In the example below, we have created a list based on the city column of the dataset.",
null,
"### 21. How do we apply advanced filters in Excel?\n\nTo apply advanced filters, use the Advanced Filter option present in the Data tab. Select where you want to filter the table. Choose the ‘list range’ and the ‘criteria range’ that has the conditions based on which you would like to filter the table.",
null,
"The below example shows how to apply advanced filters.",
null,
"### 22. Using the below-given sales data, highlight those cells where total sales > \\$5000.\n\nHere, conditional formatting is used to highlight cells based on the criteria.",
null,
"1. Select ‘Conditional Formatting’ from the home tab and under Highlight Cells Rules, choose ‘Greater Than option’.",
null,
"2. Provide the condition and choose the color for the cells to be highlighted.",
null,
"",
null,
"### 23. Using the given table, explain how the index-match function works in Excel.\n\nHere, we will write an index-match function to find the city to which Andrew belongs to from the below table.",
null,
"Here is how you can use the Index-Match function to get the result.",
null,
"### 24. How do you find duplicate values in a column?\n\nTo find duplicate values in a column, you can either use Conditional Formatting or the COUNTIF() function.\n\n#### 1. Conditional Formatting\n\nFirst, go to the Home tab, then under Conditional Formatting, select ‘Highlight Cells Rules’. Then choose ‘Duplicate Values’.",
null,
"Below, we have highlighted the cells in the ‘Name’ column that have been repeated.\n\n#### 2. COUNTIF()\n\nYou can write a COUNTIF() function to check if the values in a particular column are repeated.",
null,
"In the below example, we are fetching the duplicate names using the COUNTIF() function.\n\n### 25. How can you remove duplicate values in a range of cells?\n\n1. To delete duplicate values in a column, select the highlighted cells, and press the delete button. After deleting the values, go to the ‘Conditional Formatting’ option present in the Home tab. Choose ‘Clear Rules’ to remove the rules from the sheet.",
null,
"2. You can also delete duplicate values by selecting the ‘Remove Duplicates’ option under Data Tools present in the Data tab.\n\nMoving forward, let’s have a look at the intermediate level of Excel interview questions.\n\n## Intermediate Level Excel Interview Questions\n\n### 26. What are the wildcards available in Excel?\n\nWildcards only work with text data. Excel has three wildcards.\n\n#### 1. * (Asterisk)\n\nThis refers to any number of characters.\n\nThe example stated below filters the customers whose name ends with “a”.\n\nFor that, we use “*a”.",
null,
"#### 2. ? (Question mark)\n\nIt represents one single character.\n\nThe example below shows how to filter a particular customer name.",
null,
"#### 3. ~ (Tilde)\n\nIt is used to identify a wildcard character (~, *, ?) in the text.\n\nIn the following example, we are filtering How?* using the tilde (~) symbol.",
null,
"### 27. What is Data Validation? Illustrate with an example.\n\nData Validation restricts the type of values that a user can enter into a particular cell or a range of cells.\n\nIn the Data tab, select the ‘Data Validation’ option present under Data Tools.",
null,
"Select the kind of data validation you want to apply.",
null,
"In the following example, we have applied data validation to the ‘Name’ column to accept only text values. If you enter something other than a text, it will throw an error.",
null,
"### 28. Given below is a student table. Write a function to add pass/fail to the results column based on the following criteria.\n\nIf student marks > 60 and attendance > 75%, then pass else the student fails.",
null,
"You can use the IF() function and check with an AND condition to fill in the results column.",
null,
"### 29. Calculate your age in years from the current date.\n\nUse the YEARFRAC() or DATEDIF() function to return the number of whole days between start_date and end_date\n\n• YEARFRAC()",
null,
"• DATEDIF()",
null,
"Learn best business analysis techniques by Purdue University, IB and EY experts. Sign-up for our Post Graduate Program in Business Analysis TODAY!\n\n### 30. How are nested IF statements used in Excel?\n\nThe function IF() can be nested when we have multiple conditions to meet. The FALSE value in the first IF function is replaced by another IF function to make a further test.\n\nBelow, using nested IF statements, we are categorizing results based on the marks.",
null,
"### 31. From the below table, find the descriptive statistics of the columns using the Data Analysis ToolPak in Excel.",
null,
"",
null,
"Click on the Data Analysis option in the Data tab. Choose Descriptive Statistics.",
null,
"Below is the summary table for the columns and their respective statistical measures.",
null,
"### 32. Using the Coronavirus dataset, create a pivot table to find the total cases in each country belonging to their respective continents.\n\nFirst, drag the continent and country columns into rows. After that, drag the cases column on to the values section.",
null,
"### 33. How do you provide Dynamic Range in ‘Data Source’ of Pivot Tables?\n\nDynamic Range in the data source of pivot tables is used to make your pivot table dynamic to adjust to new data when refreshed automatically.\n\nCreate a Named table to provide a dynamic range. Go to the Insert tab and select Table.",
null,
"Under Table Design, give a name to the table.",
null,
"### 34. Is it possible to create a Pivot Table using multiple sources of data?\n\nYes, you can create a pivot table from multiple worksheets. For this, there must be a common row in both the tables. This will act as the Primary key for the first table and Foreign key for the second table. Create a relationship between the tables and then build the pivot table.\n\n### 35. Create a pivot table to find the top three countries from each continent based on the total cases using COVID data.\n\n• Create a pivot table using the coronavirus dataset by dragging sales into values.\n• Place the continent and country columns into rows.\n• Filter the table by selecting ‘Top 3’.\n\nBelow is the sequence of steps to follow.",
null,
"#### Become The Highest-Paid Business Analysis Expert",
null,
"### 36. How do you create a column in a pivot table?\n\nFor this, you have to go to the PivotTable Analyze tab and select ‘Fields, Items & Sets’ option. Under that, you need to click ‘Calculate Field’ to create a new column.",
null,
"The Insert Calculated Field box appears. Give a name to the column and insert the formula by selecting the existing columns from the pivot table. Click Add ----> OK to create the column.",
null,
"### 37. How does a Slicer work in Excel?\n\nTo filter data in a Pivot table, we can use slicers.\n\n1. To create a slicer, go to the Insert tab, and select Slicer present under Filter.\n2. Then, select the list of fields for which you want to create slicers.\n\nIn the below example, we have created two slicers (months, countries, and territory) to filter the pivot table.",
null,
"### 38. Use the coronavirus dataset to find the percentage contribution of each country and continent to the total cases?\n\n1. Create the pivot table to show the total cases by country and continent.\n2. Right-click on the sum of cases column and under Show Value As, select “% of Grand Total.”",
null,
"### 39. How do you create a pivot chart in Excel?\n\n• To create a pivot chart, first, we need to create a pivot table.",
null,
"• Go to the Insert tab next and select the ‘Pivot Chart’ option. Choose a suitable chart to represent your pivot table data.",
null,
"Accelerate your career with our Post Graduate Program in Business Analytics in partnership with Carlson School of Management. Enroll and start learning!\n\n### 40. What are macros in Excel? Create a macro to automate a task.\n\nMacro is a program that resides within the Excel file. The use of it is to automate repetitive tasks that you would like to perform in Excel.\n\nTo record a macro, you can either go to the Developer tab and click on Record Macro or access it from the View tab.",
null,
"",
null,
"Now that we are done with the intermediate level of the Excel interview questions, let’s move on to the advanced level of Excel interview questions.\n\n## Advanced Level Excel Interview Questions\n\n### 41. What is the What-If Analysis in Excel?\n\nThe What-If Analysis in Excel is a powerful tool to perform complex mathematical calculations, experiment with data, and try out different scenarios.\n\nConsider the following example:\n\nIf you get \\$10,000 worth of sales over the next few months, how much profit can you expect?”",
null,
"Such scenarios can be solved using the What-If Analysis.\n\nGo to the Data tab and click on What-If Analysis present under Forecast.\n\nScenario Manager is used for a comparison of different scenarios.\n\nThe Goal Seek performs reverse calculations.\n\nThe Data Table is used for sensitivity analysis.\n\n### 42. What is the difference between a function and a subroutine in VBA?\n\n Functions Subroutines A function is responsible for returning the value of the task it is performing. Meanwhile, subroutines don’t return the value of the task it is performing. They are called by a variable. They can be recalled from anywhere in the program, in multiple types. Functions are used as it is in spreadsheets as formulas. Subroutines are not used directly in spreadsheets as formulas. Functions are used to carry out repetitive tasks, and it, in turn, returns a value. Users are required to insert a value in the desired cell before fetching the result of the subroutine.\n\n### 43. What is the difference between ThisWorkbook and ActiveWorkbook in VBA?\n\n ThisWorkbook ActiveWorkbook ThisWorkbook indicates the name of the workbook where the code is running from. As the name suggests, ActiveWorkbook is the workbook that is presently active from the various open workbooks.",
null,
"",
null,
"### 44. How will you pass arguments to VBA Function?\n\nArguments can be passed to a VBA function as a reference or as a value.\n\nBelow is an example to illustrate both the usages.\n\nDim x As Integer\n\nx = 10\n\nMsgBox Triple(x)\n\nMsgBox x",
null,
"If you run the cells by passing the values as a reference, it will display 40 both the times. When we pass arguments by reference, we are referencing the original value. The original value of x is changed in the function.\n\nWhen we pass the arguments by value, we are passing a copy to the function. The original value is not changed. Hence, the second MsgBox will display the original value 10.\n\n### 45. How do you find the last row and column in VBA?\n\nTo find the last row, use the below lines code in the VBA module:\n\nSub FindingLastRow()\n\nDim lastRow As Long\n\nlastRow = ActiveSheet.Cells.SpecialCells(xlLastCell).Row\n\nMsgBox (lastRow)\n\nEnd Sub\n\nTo find the last column, use the below lines code in the VBA module:\n\nSub FindingLastColumn()\n\nDim lastRow As Long\n\nlastColumn = ActiveSheet.Cells.SpecialCells(xlLastCell).Column\n\nMsgBox (lastColumn)\n\nEnd Sub\n\nWant to learn the latest business analysis skills? Sign-up for our Post Graduate Program in Business Analytics and gain the skills to excel in the business world. Start learning TODAY!\n\n### 46. How do we check whether a file exists or not in a specified location?\n\nSub CheckFileExists()\n\nDim strFileName As String\n\nDim strFileExists As String\n\nstrFileName = “File location\\file_name.xlsx”\n\nstrFileExists = Dir(strFileName)\n\nIf strFileExists = “” Then\n\nMsgBox “The selected file doesn't exist”\n\nElse\n\nMsgBox “The selected file exists”\n\nEnd If\n\nEnd Sub\n\n### 47. Explain how to debug a VBA code?\n\nTo debug a VBA code line by line, you can use the F8 key. You can also create a breakpoint to terminate the execution wherever you want.\n\nThe execution will start from the beginning of the code, and every time you press F8, it will execute the next line and continue until the end of the code. The yellow arrow and the highlighted line tells you the current point to execution.",
null,
"#### Become a Data Scientist With Real-World Experience\n\nData Scientist Master’s Program",
null,
"### 48. Write a VBA function to calculate the area of a rectangle.\n\nFunction Area(Length As Double, Optional Width As Variant)\n\nIf IsMissing(Width) Then\n\nArea = Length * Length\n\nElse\n\nArea = Length * Width\n\nEnd If\n\nEnd Function\n\n### 49. Write a VBA function to check if a number is a prime number or not.\n\nSub Prime()\n\nDim divisors As Integer, number As Long, i As Long\n\ndivisors = 0\n\nnumber = InputBox(“Enter a number”)\n\nFor i = 1 To number\n\nIf number Mod i = 0 Then\n\ndivisors = divisors + 1\n\nEnd If\n\nNext i\n\nIf divisors = 2 Then\n\nMsgBox number & “ is a prime number”\n\nElse\n\nMsgBox number & “ is not a prime number”\n\nEnd If\n\nEnd Sub\n\nLearn the latest tools, work on real-world projects, and attend Masterclasses from IBM and EY experts. Join our Master's program today.\n\n### 50. Write a VBA code to create a bar chart with the given data.\n\nConsider the below data that has two features. You can use the lines of code below to create a bar chart.",
null,
"Once you have run the above VBA code lines, below is the bar chart you will get.",
null,
"## Conclusion\n\nNow that you know the various Excel interview questions that can be asked in an interview, you can prepare by referring to the given answers for each of these Excel interview questions. Here, we had a look at a plethora of Excel interview questions based on different levels of difficulty. Practicing Excel regularly and going through these Excel interview questions will keep you prepared for any question that is thrown at you.\n\nTo kick-start your fruitful career in Excel, enroll in Business Analytics with Excel course today!\n\nWe hope this article on Excel interview questions was useful. Do you have any questions related to this article? If so, then please put it in the comments section of the article and our experts will get back to you on that right away.",
null,
"Shruti M"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8466583,"math_prob":0.75460505,"size":22986,"snap":"2023-40-2023-50","text_gpt3_token_len":5087,"char_repetition_ratio":0.14189366,"word_repetition_ratio":0.03828941,"special_character_ratio":0.22226572,"punctuation_ratio":0.11481563,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9738048,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156],"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,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,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,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T20:57:45Z\",\"WARC-Record-ID\":\"<urn:uuid:f0c0e703-171a-4d8b-9b2e-18aca22b7d2f>\",\"Content-Length\":\"536585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e669a2bd-25ba-475e-a743-5e8a89d04c6c>\",\"WARC-Concurrent-To\":\"<urn:uuid:35192ad6-56c8-44e4-89ba-73a092066b45>\",\"WARC-IP-Address\":\"99.84.108.42\",\"WARC-Target-URI\":\"https://www.simplilearn.com/tutorials/excel-tutorial/excel-interview-questions\",\"WARC-Payload-Digest\":\"sha1:HMFZJ4JWRCBXLLOE44PFHILRQDFJZPZW\",\"WARC-Block-Digest\":\"sha1:NFFAVFNPRUTQ6IWA7F3AXKRBA6O6G7JG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511021.4_warc_CC-MAIN-20231002200740-20231002230740-00503.warc.gz\"}"} |
http://mathhelper.us/5over11/minus/1over3 | [
"",
null,
"# What is 5/11 - 1/3?",
null,
"Here's how to subtract 1/3 from 5/11:\n\n 5 11\n 1 3\n\n## Step 1\n\nWe can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.\n\nSo we multiply 5 by 3, and get 15.\n\nThen we multiply 1 by 11, and get 11.\n\nNext we give both terms new denominators -- 11 × 3 = 33.\n\nSo now our fractions look like this:\n\n 15 33\n 11 33\n\n## Step 2\n\nSince our denominators match, we can subtract the numerators.\n\n15 − 11 = 4\n\nSo the answer is:\n\n 4 33\n\n## Step 3\n\nLast of all, we need to simplify the fraction, if possible. Can it be reduced to a simpler fraction?\n\nTo find out, we try dividing it by 2...\n\nNope! So now we try the next greatest prime number, 3...\n\nNope! So now we try the next greatest prime number, 5...\n\nNo good. 5 is larger than 4. So we're done reducing.\n\nThere you have it! The final answer is:\n 5 11\n 1 3\n=\n 4 33"
] | [
null,
"http://mathhelper.us/img/math-helper.png",
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"http://mathhelper.us/fraction_image_vertical.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88843817,"math_prob":0.96342874,"size":933,"snap":"2019-13-2019-22","text_gpt3_token_len":259,"char_repetition_ratio":0.13455328,"word_repetition_ratio":0.06703911,"special_character_ratio":0.2926045,"punctuation_ratio":0.19266056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992674,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T23:29:34Z\",\"WARC-Record-ID\":\"<urn:uuid:c892bb9d-a6e8-49f0-8c6c-359bcb64ca52>\",\"Content-Length\":\"13413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c45826d3-b78e-4e31-bef2-da1acb5c546c>\",\"WARC-Concurrent-To\":\"<urn:uuid:e93ae921-cc28-47b8-9b9c-f7d4765d3963>\",\"WARC-IP-Address\":\"162.243.195.140\",\"WARC-Target-URI\":\"http://mathhelper.us/5over11/minus/1over3\",\"WARC-Payload-Digest\":\"sha1:QFIMH3SWTJTLG52CVEHO7463LNFGYIHA\",\"WARC-Block-Digest\":\"sha1:JT3XT3LZXYRFSGGPUAC4K43YAOZCZHRS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203093.63_warc_CC-MAIN-20190323221914-20190324003914-00461.warc.gz\"}"} |
https://de.mathworks.com/matlabcentral/cody/problems/167-pizza/solutions/1746032 | [
"Cody\n\n# Problem 167. Pizza!\n\nSolution 1746032\n\nSubmitted on 8 Mar 2019 by Yaphet Elias\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 Pass\nz = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))\n\n2 Pass\nz = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))\n\n3 Pass\nz = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))\n\n4 Pass\nz = 2; a = 2; v_correct = 8*pi; assert(isequal(pizza(z,a),v_correct))"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5197725,"math_prob":0.9994297,"size":570,"snap":"2019-35-2019-39","text_gpt3_token_len":196,"char_repetition_ratio":0.1696113,"word_repetition_ratio":0.13483146,"special_character_ratio":0.37017545,"punctuation_ratio":0.19512194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984388,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T07:35:47Z\",\"WARC-Record-ID\":\"<urn:uuid:05a3431a-2163-4d97-88c8-2d19035d1d62>\",\"Content-Length\":\"72574\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2e1c568-d696-4a01-96e9-b0468dd16818>\",\"WARC-Concurrent-To\":\"<urn:uuid:b71d5937-1cf0-4e6f-8c26-dabe52c5811d>\",\"WARC-IP-Address\":\"184.25.188.167\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/167-pizza/solutions/1746032\",\"WARC-Payload-Digest\":\"sha1:AWNC36VGJA5TOMYF6KP7BOIT66S53NFS\",\"WARC-Block-Digest\":\"sha1:XZH2RNPQP7DDWM4HV2AL3HH6EESHKZV7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573053.13_warc_CC-MAIN-20190917061226-20190917083226-00269.warc.gz\"}"} |
https://www.clutchprep.com/chemistry/practice-problems/102441/in-the-dumas-bulb-technique-for-determining-the-molar-mass-of-an-unknown-liquid- | [
"# Problem: In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below 100 oC in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing ).From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, 1.013 g ; volume of bulb, 354 cm3 ; pressure, 743 torr ; temperature, 99 oC .\n\n🤓 Based on our data, we think this question is relevant for Professor Holler & Pietrovi's class at UPENN.\n\n###### FREE Expert Solution\n\nRecall that molecular weight is in grams per 1 mole of a substance.\n\nFirst, we have to calculate the amount of gas in moles using the ideal gas equation.\n\n$\\overline{){\\mathbf{P}}{\\mathbf{V}}{\\mathbf{=}}{\\mathbf{n}}{\\mathbf{R}}{\\mathbf{T}}}$\n\nP = pressure, atm\nV = volume, L\nn = moles, mol\nR = gas constant = 0.08206 (L·atm)/(mol·K)\nT = temperature, K\n\nIsolate n (number of moles of gas):",
null,
"###### Problem Details\n\nIn the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below 100 oC in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing",
null,
").\n\nFrom the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, 1.013 g ; volume of bulb, 354 cm3 ; pressure, 743 torr ; temperature, 99 oC ."
] | [
null,
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"https://cdn.clutchprep.com/sourced_problem_images/904a3558945e421f534cdb0a575e1aa4.jpg",
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http://www.cut-the-knot.org/blue/Euclid.shtml | [
"# Euclid's Algorithm\n\nEuclid's Algorithm appears as the solution to the Proposition VII.2 in the Elements:\n\nGiven two numbers not prime to one another, to find their greatest common measure.\n\nWhat Euclid called \"common measure\" is termed nowadays a common factor or a common divisor. Euclid VII.2 then offers an algorithm for finding the greatest common divisor (gcd) of two integers. Not surprisingly, the algorithm bears Euclid's name.\n\nThe algorithm is based on the following two observations:\n\n1. If b|a then gcd(a, b) = b.\n\nThis is indeed so because no number (b, in particular) may have a divisor greater than the number itself (I am talking here of non-negative integers.)\n\n2. If a = bt + r, for integers t and r, then gcd(a, b) = gcd(b, r).\n\nIndeed, every common divisor of a and b also divides r. Thus gcd(a, b) divides r. But, of course, gcd(a, b)|b. Therefore, gcd(a, b) is a common divisor of b and r and hence gcd(a, b) ≤ gcd(b, r). The reverse is also true because every divisor of b and r also divides a.\n\n#### Example\n\nLet a = 2322, b = 654.\n\n 2322 = 654·3 + 360 gcd(2322, 654) = gcd(654, 360) 654 = 360·1 + 294 gcd(654, 360) = gcd(360, 294) 360 = 294·1 + 66 gcd(360, 294) = gcd(294, 66) 294 = 66·4 + 30 gcd(294, 66) = gcd(66, 30) 66 = 30·2 + 6 gcd(66, 30) = gcd(30, 6) 30 = 6·5 gcd(30, 6) = 6\n\nTherefore, gcd(2322,654) = 6.",
null,
"For any pair a and b, the algorithm is bound to terminate since every new step generates a similar problem (that of finding gcd) for a pair of smaller integers. Let Eulen(a, b) denote the length of the Euclidean algorithm for a pair a, b. Eulen(2322, 654) = 6, Eulen(30, 6) = 1. I'll use this notation in the proof of the following very important consequence of the algorithm:\n\n## Corollary\n\nFor every pair of whole numbers a and b there are two integers s and t such that as + bt = gcd(a, b).\n\n### Example\n\n2322×20 + 654×(-71) = 6.\n\n## Proof\n\nLet a > b. The proof is by induction on Eulen(a, b). If Eulen(a, b) = 1, i.e., if b|a, then a = bu for an integer u. Hence, a + (1 - u)b = b = gcd(a, b). We can take s = 1 and t = 1 - u.\n\nAssume the Corollary has been established for all pairs of numbers for which Eulen is less than n. Let Eulen(a, b) = n. Apply one step of the algorithm: a = bu + r. Eulen(b, r) = n - 1. By the inductive assumption, there exist x and y such that bx + ry = gcd(b,r) = gcd(a,b). Express r as r = a - bu. Hence, ry = ay - buy; bx + (ay - buy) = gcd(a, b). Finally, b(x - uy) + ay = gcd(a, b) and we can take s = x - uy and t = y.\n\nThere is also a simple proof that employs the Pigeonhole Principle.\n\n## Remark\n\nNote that any linear combination as + bt is divisible by any common factor of a and b. In particular, any common factor of a and b also divides gcd(a, b). In a \"reverse\" application, any linear combination as + bt is divisible by gcd(a, b). From here it follows that gcd(a, b) is the least positive integer representable in the form as + bt. All the rest are multiples of gcd(a, b). The generalization of the Corollary to what is known as Principal ideal domain is known as Bézout's identity or Bézout's Lemma after the French mathematician Éttiene Bézout (1730-1783), so it often happens that the result stated in the Corollary is also often referred to as Bézout's identity or Bézout's Lemma.",
null,
"For coprime numbers we get existence of s and t such that as + bt = 1. This Corollary is a powerful tool. It appeared in the 3 Glass and Hour Glass problems. For example, let's prove the Euclid's Proposition VII.30\n\nIf two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers.\n\nLet a prime p divide the product ab. Assume p",
null,
"a. Then gcd(a, p) = 1. By Corollary, ax + py = 1 for some x and y. Multiply by b: abx + pby = b. Now, p|ab and p|pb. Hence, p|b.\n\nActually, this proves a generalization of the Proposition VII.30 I used several times on these pages:\n\n Let m|ab and gcd(a, m) = 1. Then m|b.\n\nProposition VII.30 immediately implies the Fundamental Theorem of Arithmetic although Euclid has never stated it explicitly. The first time it was formulated in 1801 by Gauss in his Disquisitiones arithmeticae.\n\n## Fundamental Theorem of Arithmetic\n\nAny integer N can be represented as a product of primes. Such a representation is unique up to the order of prime factors.\n\nSince, by definition, a number is composite if it has factors other than 1 and itself, and these factors are bound to be smaller than the number, we can keep extracting the factors until only prime factors remain. This shows existence of the representation: N = pqr ..., where all p, q, r,... are prime. To prove uniqueness, assume there are two representations: N = pqr ... = uvw.... We see that p divides uvw... By Corollary, it divides one of the factors u, v, w, ... Cancel them out. We can go on chipping away on the factors left and right until no factors remain.\n\nRepresentation of a number as the product of primes is called prime number decomposition or prime factorization. The Fundamental Theorem of Arithmetic asserts that each integer has a unique prime number decomposition.\n\nNote: Euclid's Algorithm is not the only way to determine the greatest common factor of two integers. If you can find the prime factorizations of the two numbers you can easily determine their gcd as the intersection of the multisets formed by their prime factors. Factor Trees offer a convenient bookkeeping for finding prime factorizations of integers.\n\n## References\n\n1. H. Davenport, The Higher Arithmetic, Harper&Brothers, NY\n2. R. Graham, D. Knuth, O. Patashnik, Concrete Mathematics, 2nd edition, Addison-Wesley, 1994.\n3. Oystein Ore, Number Theory and Its History, Dover Publications, 1976\n4. S. K. Stein, Mathematics: The Man-Made Universe, 3rd edition, Dover, 2000.",
null,
"",
null,
"• Factoring with the Factor Tree\n• GCD and LCM via Factor Tree\n• GCD and LCM by Plain Factorization\n• Common Multiples and the Least Common Multiple\n• GCD(M, N) x LCM(M, N) = M x N\n• Divisibility Criteria\n• Euclid's Algorithm\n• Algorithm for Computing the LCM\n• Factors And Multiples\n• Two Properties of Greatest Common Divisor\n• Properties of GCD and LCM\n• A Line in a Square Grid\n• Number of Factors of an Integer\n• Extension of Euclid's Game\n•",
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"http://www.cut-the-knot.org/gifs/tbow_sh.gif",
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"http://www.cut-the-knot.org/gifs/tbow_sh.gif",
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https://math.stackexchange.com/questions/2463811/how-do-you-simplify-an-expression-involving-fourth-and-higher-order-trigonometri | [
"How do you simplify an expression involving fourth and higher order trigonometric functions?\n\nThe problem is as follows:\n\nWhich value of $K$ has to be in order that $R$ becomes independent from $\\alpha$?.\n\n$$R=\\sin^6\\alpha +\\cos^6\\alpha +K(\\sin^4\\alpha +\\cos^4\\alpha )$$\n\nSo far I've only come up with the idea that the solution may involve $R=0$, therefore\n\n$$\\sin^6\\alpha +\\cos^6\\alpha +K(\\sin^4\\alpha +\\cos^4\\alpha)=0$$\n\nas a result the expression becomes $0$ thus independent from $\\alpha$, however the result is like this\n\n$$-K=\\frac{\\sin^6\\alpha +\\cos^6\\alpha}{\\sin^4\\alpha +\\cos^4\\alpha}$$\n\nI am not sure if this is the right way.\n\nMoreover, how can I simplify this expression, as it has order four and six?\n\n• It's going to be $K=-3/2$, not sure how to prove it besides desmos.com/calculator/glwwwu5qi1 Oct 9 '17 at 1:23\n• You could take the derivative, set equal to zero, and solve for it.\n– Paul\nOct 9 '17 at 1:28\n• @Paul the derivative of a function doesn't usually have much to do with the roots, especially since adding a constant (R) changes the roots without changing the derivative. Oct 9 '17 at 1:30\n• @Paul I would like to add that I used the precalculus tag is to know if such question can be answered by using only precalculus tools, and as such avoiding derivatives. Oct 9 '17 at 1:39\n• @JohnathanGross If the expression is independent of alpha, then we must have $dR/d\\alpha =0$ for all alpha from which it is easy to solve for K.\n– Teoc\nOct 9 '17 at 2:16\n\nRecall that $$\\sin^2\\alpha = 1 - \\cos^2\\alpha$$\n\nand express everything in terms of $\\cos^2\\alpha$: \\begin{align} R &= \\left(\\;\\sin^2\\alpha\\;\\right)^3 + \\left(\\;\\cos^2\\alpha\\;\\right)^3+K\\left(\\;\\left(\\;\\sin^2\\alpha\\;\\right)^2+\\left(\\;\\cos^2\\alpha\\;\\right)^2\\;\\right) \\\\ &= \\left(\\;1-\\cos^2\\alpha\\;\\right)^3 + \\left(\\;\\cos^2\\alpha\\;\\right)^3+K\\left(\\;\\left(\\;1-\\cos^2\\alpha\\;\\right)^2+\\left(\\;\\cos^2\\alpha\\;\\right)^2\\;\\right) \\\\ &= \\left(\\;1-x\\;\\right)^3 + \\left(\\;x\\;\\right)^3+K\\left(\\;\\left(\\;1-x\\;\\right)^2+\\left(\\;x\\;\\right)^2\\;\\right) \\qquad\\text{(writing x for \\cos^2\\alpha)}\\\\ &= 1 - 3 x + 3 x^2 + K \\left(\\; 1 - 2 x + 2 x^2 \\;\\right) \\\\ &= 1 +K -(3+2K) x + (3+2K) x^2 \\end{align}\n\nIndependence from $\\alpha$ translates to independence from $x$. We need a value of $K$ that causes the non-constant terms of the polynomial to vanish. Clearly, $K = -3/2$. $\\square$\n\n• Can you be more explicit without skipping the steps of expressing the equation in terms of $cos^{2}\\alpha$. (In other words solving that part) I have not yet achieved the level of expertise to get the idea. Oct 9 '17 at 1:45\n• @ChrisSteinbeckBell: How's that?\n– Blue\nOct 9 '17 at 1:52\n• Thanks, by adding that step I got the idea. It was a bit hidden at first. Oct 9 '17 at 1:58\n• @ChrisSteinbeckBell: Good to know. BTW: If this (or some other) answer satisfies you, be sure to tick the accompanying checkmark to \"accept\" it. (I notice that you have not accepted answers to most of your questions.) Accepting an answer not only grants a few more imaginary internet points to the answerer, it also removes the question from the \"Unanswered\" queue so that the Stack Exchange system won't recycle the question in the future.\n– Blue\nOct 9 '17 at 2:03\n• I will take note on your recommendation. I'd like to say that I didn't had the necessary reputation to accept answers as I'm somewhat still new to this community. Oct 9 '17 at 2:12\n\nFor $\\alpha = 0$ we have $\\sin(\\alpha)=0$ and $\\cos(\\alpha) = 1$, therefore $R = 1+ K$ while for $\\alpha = \\dfrac{\\pi}{4}$ we have $\\cos(\\alpha) = \\sin(\\alpha) = \\dfrac{1}{\\sqrt{2}}$, so $R = \\dfrac{1}{4} + \\dfrac{K}{2}$, therefore if there exists a value for $K$ for which the expression is independent of $\\alpha$, then we must have: $$1+ K = \\frac{1}{4} + \\frac{K}{2}$$\n\ntherefore $K$ must be $-\\dfrac{3}{2}$.\n\n• Can you please explain how did you come to that conclusion?. How does exactly $\\alpha=\\frac{\\pi}{4}$ becomes into $K=-\\frac{3}{2}$?. This part is not very clear. Oct 9 '17 at 3:46\n• @ChrisSteinbeckBell I've added a bit more explanation. Oct 9 '17 at 4:11\n• +1. Neat and tidy. Of course, this approach says that if such a $K$ exists, then it cannot be anything other than $-3/2$. The question statement appears to guarantee existence, so we're okay. However, if there were doubt, then one would need to verify that $K=-3/2$ eliminates the trig terms completely from the original expression.\n– Blue\nOct 9 '17 at 4:52\n• @CountIblis Thanks for adding that part now is clear to me. Its very clever to use the fact that at certain angle both functions have the same value. 45 degrees. But does this approach works for other trigonometric functions?. Oct 10 '17 at 3:44\n• @ChrisSteinbeckBell In general,if you just want to find the value of a parameter for which an identity holds, you can always consider special cases for the variable (in this case $\\alpha$) and write down the equation that it holds for the special cases. To prove the identity requires more work. Oct 11 '17 at 0:13\n\nUsing $\\sin^2 \\alpha + \\cos^2 \\alpha =1$ \\begin{eqnarray*} \\sin^4 \\alpha + \\cos^4 \\alpha =(\\sin^2 \\alpha + \\cos^2 \\alpha)^2 -2\\sin^2 \\alpha \\cos^2 \\alpha = 1-2\\sin^2 \\alpha \\cos^2 \\alpha \\\\ \\sin^6 \\alpha + \\cos^6 \\alpha =(\\sin^2 \\alpha + \\cos^2 \\alpha)(\\sin^4 \\alpha -\\sin^2 \\alpha \\cos^2 \\alpha+ \\cos^4 \\alpha) = 1-3\\sin^2 \\alpha \\cos^2 \\alpha. \\\\ \\end{eqnarray*} So your equation can be simplified to \\begin{eqnarray*} R=\\sin^6\\alpha +\\cos^6\\alpha +K(\\sin^4\\alpha +\\cos^4\\alpha ) \\\\ =1-3\\sin^2 \\alpha \\cos^2 \\alpha +K(1-2\\sin^2 \\alpha \\cos^2 \\alpha) \\\\ =1+K-(2K+3)\\sin^2 \\alpha \\cos^2 \\alpha \\\\ \\end{eqnarray*} So it is independent of $\\alpha$ when $\\color{blue}{2K+3=0}$. (Giving the value $K=\\color{red}{-\\frac{3}{2}}$)\n\nLet $x = \\sin^2 \\alpha,y = \\cos^2 \\alpha$. Note that $x+y = 1$ so $x^3 + y^3 = (x+y)(x^2+y^2-xy) = (x+y)((x+y)^2-3xy) = 1-3xy$ Also $x^2+y^2 = (x+y)^2-2xy = 1-2xy$\n\n$R = (1-3xy) + K(1-2xy) = 1+K - (3+2K)xy$. So if $K = -3/2$ then $R$ will be independent of $\\alpha$. If $K \\neq -3/2$ then it will depend on $xy$ which then depends on $\\alpha$.\n\nGenerally when you see an expression in the form of $\\sin^{2n}x + \\cos^{2n}x$ it helps to factor out $\\sin^2 + \\cos^2$ like what I did above\n\nAnother possible way is to start with $$R=\\sin ^6(a)+\\cos ^6(a)+K \\left(\\sin ^4(a)+\\cos ^4(a)\\right)$$ Using multiple angles formalae, this rewrite as $$R=\\sin ^6(a)+\\cos ^6(a)+\\frac{1}{4} K (\\cos (4 a)+3)$$ and say that the derivative of $R$ with respect to $a$ is equal to $0$. $$\\frac{dR}{da}=6 \\sin ^5(a) \\cos (a)-6 \\sin (a) \\cos ^5(a)-K \\sin (4 a)=-\\frac{3}{2} \\sin (4 a)-K \\sin (4 a)$$ $$\\frac{dR}{da}=-\\frac{1}{2} (2 K+3) \\sin (4 a)=0$$ Then $K=-\\frac{3}{2}$.\n\n$$1=(\\sin^2x+\\cos^2x)^3= \\sin^6x+ \\cos^6x+ 3\\sin^2x\\cos^2x(\\sin^2x+\\cos^2x)= \\sin^6x+ \\cos^6x+ 3\\sin^2x\\cos^2x$$ $$3\\sin^2x\\cos^2x=\\frac{3}{2}((\\sin^2x+\\cos^2x)^2 - \\sin^4x-\\cos^4x)=\\frac{3}{2}((1- \\sin^4x-\\cos^4x))$$ Substitute and compare with the equation you have"
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https://www.coursemerit.com/question-details/13618/Regression-StatsMath | [
"Cashback Offer (25th Nov-05th Dec 2020). Get Flat 10% Cashback credited in your account on a minimum transaction of \\$80. Post Your Question\n\nQuestion Details Normal\n\\$ 20.00\n\n# Regression Stats/Math\n\n• From Mathematics, Statistics\n• Due on 26 Feb, 2016 07:44:00\n• Asked On 22 Feb, 2016 12:52:39\n• Due date has already passed, but you can still post solutions.\nQuestion posted by",
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"2. The formula for a regression equation is Y’ = 2X + 9.\n\na. What would be the predicted score for a person scoring 6 on X?\n\nb. If someone’s predicted score was 14, what was this person’s score on X?\n\n6. For the X,Y data below, compute:\n\na. r and determine if it is significantly different from zero.\n\nb. the slope of the regression line and test if it differs significantly from zero.\n\nc. the 95% confidence interval for the slope.\n\nConduct a significance test to see if the die is biased. (a) What Chi Square value\n\ndo you get and how many degrees of freedom does it have? (b) What is the p\n\nvalue?\n\n5. At a school pep rally, a group of sophomore students organized a free raffle for prizes. They claim that they put the names of all of the students in the school in the basket and that they randomly drew 36 names out of this basket. Of the prize winners, 6 were freshmen, 14 were sophomores, 9 were juniors, and 7 were seniors. The results do not seem that random to you. You think it is a little fishy that sophomores organized the raffle and also won the most prizes. Your school is composed of 30% freshmen, 25% sophomores, 25% juniors, and 20% seniors.\n\na. What are the expected frequencies of winners from each class?\n\nb. Conduct a significance test to determine whether the winners of the prizes were distributed throughout the classes as would be expected based on the percentage of students in each group. Report your Chi Square and p values.\n\nc. What do you conclude?\n\n14. A geologist collects hand-specimen sized pieces of limestone from a particular area.\n\nA qualitative assessment of both texture and color is made with the following results. Is there evidence of association between color and texture for these limestones? Explain your answer\n\n70. The standard deviation of the chi-square distribution is twice the mean.\n\nTrue or false\n\n102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance.\n\nUse the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.\n\n113. df = ________\n\n117. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________\n\n66. Can a coefficient of determination be negative? Why or why not?\n\nUse the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in Table 12.31.\n\n82.\n\na. Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot.\n\nb. Does it appear from inspection that there is a relationship between the variables? Why or why not?\n\nc. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx\n\nd. Find the correlation coefficient. Is it significant?\n\ne. If the laundry detergent were sold in a 40-ounce size, find the estimated cost.\n\nf. If the laundry detergent were sold in a 90-ounce size, find the estimated cost.\n\ng. Does it appear that a line is the best way to fit the data? Why or why not?\n\nh. Are there any outliers in the given data?\n\ni. Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why\n\nor why not?\n\nj. What is the slope of the least-squares (best-fit) line? Interpret the slope.",
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"Attachment\n\\$ 10.00\n\n## [Solved] Regression Stats\n\n• This solution is not purchased yet.\n• Submitted On 22 Feb, 2016 01:49:36",
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"1. The formula for a regression...\nBuy now to view the complete solution",
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"Attachment\n\\$ 8.00\n\n## [Solved] The formula for a regression equation is Y’ = 2X + 9 use as a guide only\n\n• This Solution has been Purchased 2 time\n• Submitted On 22 Feb, 2016 01:59:12",
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"1. The formula for a regression equation is Y’ = 2X + 9. a. What would be the predicted score for a person scoring 6 on X? The predicted score for a person scoring 6 on X = 2*6 + 9 = 2...\nBuy now to view the complete solution",
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"Attachment\nOther Related Questions",
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"Helper\n\n### Perform a regression model Step By Step Solution Required\n\nStatus : 2 solution",
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"Helpl...\n\n### Linear Regression\n\n#### The benefits of buying study notes from CourseMerit",
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https://swiftacademicpapers.com/question-256659-ra362899160/ | [
"# Question 256659\n\nV(s) + O2g – V2O3(s) 50.94g/mol +32.00g/mol^ 149.88g/mol A.)calculate the theoretical yield of vanadium (lll) oxide, assuming you begin with 200.00 grams vanadium metal. B.) after the experiment is performed, an experimental yield of 183.2 grams is produced.calculate the percent yield for this experiment. C.) determine the percent error. D.) compute for the excess amount of the excess reactant."
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https://math.answers.com/other-math/What_is_3_fives_of_40 | [
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"0\n\n# What is 3 fives of 40?\n\n3 fives of 40 = 15 of 40 = 600\n\n3 fives of 40 = 15 of 40 = 600\n\n3 fives of 40 = 15 of 40 = 600\n\n3 fives of 40 = 15 of 40 = 600",
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"Study guides\n\n20 cards\n\n## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials\n\n➡️\nSee all cards\n3.75\n1220 Reviews\n\n3 fives of 40 = 15 of 40 = 600",
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http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blob;f=hints/assignment_7_hint_4.mdwn;h=346d98b30bcd90c4d387d11f0460de7220ab4b49;hb=002bbc4734d89967f6941fcff5021b4d1c661f2b | [
"2 * At the top of p. 13 (this is in between defs 2.8 and 2.9), GS&V give two examples, one for \\[[∃xPx]] and the other for \\[[Qx]]. In fact it will be most natural to break \\[[∃xPx]] into two pieces, \\[[∃x]] and \\[[Px]]. But first we need to get clear on expressions like \\[[Px]].\n4 * GS&V say that the effect of updating an information state `s` with the meaning of \"Qx\" should be to eliminate possibilities in which the entity associated with the peg associated with the variable `x` does not have the property Q. In other words, if we let `Q` be a function from entities to `bool`s, `s` updated with \\[[Qx]] should be `s` filtered by the function `fun (r, h) -> let obj = List.nth h (r 'x') in Q obj`. When `... Q obj` evaluates to `true`, that `(r, h)` pair is retained, else it is discarded.\n6 OK, we face two questions then. First, how do we carry this over to our present framework, where we're working with sets of `dpm`s instead of sets of discourse possibilities? And second, how do we decompose the behavior here ascribed to \\[[Qx]] into some meaning for \"Q\" and a different meaning for \"x\"?\n8 * Answering the first question: we assume we've got some `bool dpm set` to start with. I won't call this `s` because that's what GS&V use for sets of discourse possibilities, and we don't want to confuse discourse possibilities with `dpm`s. Instead I'll call it `u`. Now what we want to do with `u` is to map each `dpm` it gives us to one that results in `(true, r, h)` only when the entity that `r` and `h` associate with variable `x` has the property Q. I'll assume we have some function Q to start with that maps entities to `bool`s.\n10 Then what we want is something like this:\n12 let eliminate_non_Qxs = (fun truth_value ->\n13 fun (r, h) ->\n14 let truth_value' =\n15 if truth_value\n16 then let obj = List.nth h (r 'x') in Q obj\n17 else false\n18 in (truth_value', r, h))\n19 in bind_set u (fun one_dpm -> unit_set (bind_dpm one_dpm eliminate_non_Qxs))\n21 The first seven lines here just perfom the operation we described: return a `bool dpm` computation that only yields `true` when its input `(r, h)` associates variable `x` with the right sort of entity. The last line performs the `bind_set` operation. This works by taking each `dpm` in the set and returning a `unit_set` of a filtered `dpm`. The definition of `bind_set` takes care of collecting together all of the `unit_set`s that result for each different set element we started with.\n23 We can call the `(fun one_dpm -> ...)` part \\[[Qx]] and then updating `u` with \\[[Qx]] will be:\n25 bind_set u \\[[Qx]]\n27 or as it's written using Haskell's infix notation for bind:\n29 u >>= \\[[Qx]]\n31 * Now our second question: how do we decompose the behavior here ascribed to \\[[Qx]] into some meaning for \"Q\" and a different meaning for \"x\"?\n33 Well, we already know that \\[[x]] will be a kind of computation that takes an assignment function `r` and store `h` as input. It will look up the entity that those two together associate with the variable `x`. So we can treat \\[[x]] as an `entity dpm`. We don't worry here about sets of `dpm`s; we'll leave that to our predicates to interface with. We'll just make \\[[x]] be a single `entity dpm`. Then what we want is:\n35 let getx = fun (r, h) ->\n36 let obj = List.nth h (r 'x')\n37 in (obj, r, h);;\n39 * Now what do we do with predicates? As before, we suppose we have a function Q that maps entities to `bool`s. We want to turn it into a function that maps `entity dpm`s to `bool dpm`s. Eventually we'll need to operate not just on single `dpm`s but on sets of them, but first things first. We'll begin by lifting Q into a function that takes `entity dpm`s as arguments and returns `bool dpm`s:\n41 fun entity_dpm -> bind_dpm entity_dpm (fun e -> unit_dpm (Q e))\n43 Now we have to transform this into a function that again takes single `entity dpm`s as arguments, but now returns a `bool dpm set`. This is easily done with `unit_set`:\n45 fun entity_dpm -> unit_set (bind_dpm entity_dpm (fun e -> unit_dpm (Q e)))\n47 Finally, we realize that we're going to have a set of `bool dpm`s to start with, and we need to compose \\[[Qx]] with them. We don't want any of the monadic values in the set that wrap `false` to become `true`; instead, we want to apply a filter that checks whether values that formerly wrapped `true` should still continue to do so.\n49 This is most easily done like this:\n51 fun entity_dpm ->\n52 fun truth_value ->\n53 if truth_value = false\n54 then empty_set\n55 else unit_set (bind dpm entity_dpm (fun e -> unit_dpm (Q e)))\n57 Applied to an entity_dpm, that yields a function that we can bind to a `bool dpm set` and that will transform the doubly-wrapped `bool` into a new `bool dpm set`.\n59 Doing things this way will discard `bool dpm`s from the set that started out wrapping `false`, and will pass through other `bool dpm`s that start out wrapping `true` but which our current filter transforms to a wrapped `false`. You might instead aim for consistency, and always pass through wrapped `false`s, whether they started out that way or are only now being generated; or instead always discard such, and only pass through wrapped `true`s. But what we have here will work fine too.\n61 If we let that be \\[[Q]], then \\[[Q]] \\[[x]] would be:\n63 let getx = fun (r, h) ->\n64 let obj = List.nth h (r 'x')\n65 in (obj, r, h)\n66 in let entity_dpm = getx\n67 in fun truth_value ->\n68 if truth_value = false\n69 then empty_set\n70 else unit_set (bind_dpm entity_dpm (fun e -> unit_dpm (Q e)))\n72 or, simplifying:\n74 let getx = fun (r, h) ->\n75 let obj = List.nth h (r 'x')\n76 in (obj, r, h)\n77 in fun truth_value ->\n78 if truth_value\n79 then unit_set (bind_dpm getx (fun e -> unit_dpm (Q e)))\n80 else empty_set\n82 which is:\n84 let getx = fun (r, h) ->\n85 let obj = List.nth h (r 'x')\n86 in (obj, r, h)\n87 in fun truth_value ->\n88 if truth_value\n89 then unit_set (\n90 fun (r, h) ->\n91 let (a, r', h') = getx (r, h)\n92 in let u' = (fun e -> unit_dpm (Q e)) a\n93 in u' (r', h')\n94 ) else empty_set\n96 which is:\n98 in fun truth_value ->\n99 if truth_value\n100 then unit_set (\n101 fun (r, h) ->\n102 let obj = List.nth h (r 'x')\n103 let (a, r', h') = (obj, r, h)\n104 in let u' = (fun e -> unit_dpm (Q e)) a\n105 in u' (r', h')\n106 ) else empty_set\n108 which is:\n110 in fun truth_value ->\n111 if truth_value\n112 then unit_set (\n113 fun (r, h) ->\n114 let obj = List.nth h (r 'x')\n115 in let u' = unit_dpm (Q obj)\n116 in u' (r', h')\n117 ) else empty_set\n119 This is a bit different than the \\[[Qx]] we had before:\n121 let eliminate_non_Qxs = (fun truth_value ->\n122 fun (r, h) ->\n123 let truth_value' =\n124 if truth_value\n125 then let obj = List.nth h (r 'x') in Q obj\n126 else false\n127 in (truth_value', r, h))\n128 in (fun one_dpm -> unit_set (bind_dpm one_dpm eliminate_non_Qxs))\n130 because that one passed through every `bool dpm` that wrapped a `false`; whereas now we're discarding some of them. But these will work equally well. We can implement either behavior (or, as we said before, the behavior of never passing through a wrapped `false`).\n132 * Reviewing: now we've determined how to define \\[[Q]] and \\[[x]] such that \\[[Qx]] can be the result of applying the function \\[[Q]] to the `entity dpm` \\[[x]]. And \\[[Qx]] in turn is now a function that takes a `bool dpm` as input and returns a `bool dpm set` as output. We compose this with a `bool dpm set` we already have on hand:\n134 bind_set u \\[[Qx]]\n136 or:\n138 <pre><code>u >>=<sub>set</sub> \\[[Qx]]\n139 </code></pre>\n141 * Can you figure out how to handle \\[[∃x]] on your own? If not, here are some [more hints](/hints/assignment_7_hint_5)."
] | [
null
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https://freeteacherworksheets.netlify.app/decimal-5th-grade-worksheets/ | [
"# 35++ Decimal 5th grade worksheets Ideas In This Year\n\n» » 35++ Decimal 5th grade worksheets Ideas In This Year\n\nIf you’re looking for decimal 5th grade worksheets pictures information linked to the decimal 5th grade worksheets interest, you have come to the ideal blog. Our website frequently provides you with suggestions for downloading the maximum quality video and picture content, please kindly hunt and find more informative video content and graphics that match your interests.\n\nDecimal 5th Grade Worksheets. They have been created to provide kids with an ideal understanding of the decimal concept grade 5. The math worksheets on this page cover many of the core topics in 5th grade math but confidence in all of the basic operations is essential to success both in 5th grade. A brief description of the worksheets is on each of the worksheet widgets. They are meant for 5th and 6th grade.",
null,
"Decimal Addition Subtraction Ws Decimals Addition Decimals Subtracting Decimals From pinterest.com\n\nThe worksheets are randomly generated so you can get a new different one just by refreshing the page in your browser F5. Fifth graders will cover a wide range of math topics as they solidify their arithmatic skills. Rounding decimals is important for getting through 4th and 5th grade math. Compare values and order the decimal numbers in this worksheet. This practice will help your students fully understand how to use decimals in equations. Decimals to the hundredths place Subtracting decimals Understanding place value Common Core Standards.\n\n### The worksheets are randomly generated so you can get a new different one just by refreshing the page in your browser F5.\n\nThe worksheets provide calculation practice for both mental divisions and long division of decimals including dividing decimals by decimals. Fifth Grade Decimals Worksheets and Printables. Rounding decimals is important for getting through 4th and 5th grade math. As a vital skill in the decimal concept our converting decimals to fractions worksheets pdf will offer kids a simple step by step method of how to convert decimals to fractions formula and vice versa. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. The worksheets are randomly generated so you can get a new different one just by refreshing the page in your browser F5.",
null,
"Source: pinterest.com\n\nThis assembly of printable worksheets has been meticulously drafted to assist students of Grade 5 and above comprehend the use of place values where decimals are involved. Printable Math Worksheets for 5th Grade. 5th Grade Decimals Worksheets. The math worksheets on this page cover many of the core topics in 5th grade math but confidence in all of the basic operations is essential to success both in 5th grade. Here is a collection of our printable worksheets for topic Equivalent Decimals of chapter Decimal Concepts in section Decimals.",
null,
"Source: pinterest.com\n\nBelow you will find a wide range of our printable worksheets in chapter Decimal Concepts of section DecimalsThese worksheets are appropriate for Fifth Grade MathWe have crafted many worksheets covering various aspects of this topic decimals and fractions place value equivalent decimals compare and order and many moreWe hope you find them very useful and interesting. Exercises here include a wide range of decimal place values up to millionths with a number of simple word problems thrown into the mix for. This assembly of printable worksheets has been meticulously drafted to assist students of Grade 5 and above comprehend the use of place values where decimals are involved. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. 5th Grade Decimals Worksheets.",
null,
"Source: cz.pinterest.com\n\n05 2 15 1 0 2 10 2 1 0333 3 1333 9 10 9 10 9 3 1000 3 1 46 5 3 46 3 0. Easily download and print our decimal worksheets. Grade 5 worksheets challenge students to develop efficient ways to tackle these topics. 05 2 15 1 0 2 10 2 1 0333 3 1333 9 10 9 10 9 3 1000 3 1 46 5 3 46 3 0. These worksheets are pdf files.",
null,
"Source: pinterest.com\n\nFifth Grade Decimals Worksheets and Printables. Division of 3 digit decimal numbers by 2 3 4. 5th Grade Decimals Worksheets. Multiplying 3 digit decimals by 10 100 or 1000. Compare values and order the decimal numbers in this worksheet.",
null,
"Source: pinterest.com\n\nFree printable decimal worksheets 5th grade. Grade 5 Number Operations in Base Ten CCSSMathContent5NBTB7. Children learn the steps for multiplying decimals with this easy-to-follow tutorial and practice sheet. Grade 5 worksheets challenge students to develop efficient ways to tackle these topics. 5th Grade Decimals Worksheets.",
null,
"Source: pinterest.com\n\n5th Grade Decimals Worksheets. Multiplying 3 digit decimals by 10 100 or 1000. Incorporated here are printable rounding decimals that contain exercises worksheets for 5th grade and 6th grade students to round off decimals on a number line rounding up or down rounding decimals to the nearest whole number tenths hundredths or thousandths word problems and more. Click the checkbox for the options to print and add to Assignments and Collections. Fifth Grade Decimals Worksheets and Printables.",
null,
"Source: pinterest.com\n\nBelow you will find a wide range of our printable worksheets in chapter Decimal Concepts of section DecimalsThese worksheets are appropriate for Fifth Grade MathWe have crafted many worksheets covering various aspects of this topic decimals and fractions place value equivalent decimals compare and order and many moreWe hope you find them very useful and interesting. Click the checkbox for the options to print and add to Assignments and Collections. Decimal patterns worksheet for 5th grade children. Incorporated here are printable rounding decimals that contain exercises worksheets for 5th grade and 6th grade students to round off decimals on a number line rounding up or down rounding decimals to the nearest whole number tenths hundredths or thousandths word problems and more. Learners practice rounding decimals to the nearest tenths hundredths and thousandths place in this playful math.",
null,
"Source: pinterest.com\n\nStudents subtract four and five digit numbers in horizontal and vertical format. As a vital skill in the decimal concept our converting decimals to fractions worksheets pdf will offer kids a simple step by step method of how to convert decimals to fractions formula and vice versa. Fifth graders will cover a wide range of math topics as they solidify their arithmatic skills. Math worksheets grade 5 multiplying decimals. Children practice comparing and rounding decimals in this math practice worksheet.",
null,
"Source: pinterest.com\n\nLearners practice rounding decimals to the nearest tenths hundredths and thousandths place in this playful math. Fifth Grade Decimals Worksheets and Printables. The worksheets are randomly generated so you can get a new different one just by refreshing the page in your browser F5. They are meant for 5th and 6th grade. Here is a collection of our printable worksheets for topic Equivalent Decimals of chapter Decimal Concepts in section Decimals.",
null,
"Source: pinterest.com\n\nAvailable here are outstanding decimal worksheets for grade 5 with answers. Children learn how to divide multi-digit decimal numbers in this supplemental math worksheet. Rounding decimals is important for getting through 4th and 5th grade math. 5th grade multiplying decimals worksheets including multiplying decimals by decimals multiplying decimals by whole numbers missing factor problems multiplying by 10 100 or 1000 and multiplication in columns with decimals. Learners practice rounding decimals to the nearest tenths hundredths and thousandths place in this playful math.",
null,
"Source: pinterest.com\n\nChildren learn how to divide multi-digit decimal numbers in this supplemental math worksheet. Free printable decimal worksheets 5th grade. Decimal patterns worksheet for 5th grade children. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. A brief description of the worksheets is on each of the worksheet widgets.",
null,
"Source: pinterest.com\n\nThey are meant for 5th and 6th grade. Fifth graders will cover a wide range of math topics as they solidify their arithmatic skills. Here is a collection of our printable worksheets for topic Equivalent Decimals of chapter Decimal Concepts in section Decimals. Exercises here include a wide range of decimal place values up to millionths with a number of simple word problems thrown into the mix for. 05 2 15 1 0 2 10 2 1 0333 3 1333 9 10 9 10 9 3 1000 3 1 46 5 3 46 3 0.",
null,
"Source: pinterest.com\n\nPrintable Math Worksheets for 5th Grade. 5th grade adding and subtracting decimals worksheets including adding 1 and 2 decimal digits adding decimals in columns subtracting 1 and 2 decimal digit numbers and subtracting decimals in columns. Students are asked to multiply 3 digit decimals by 10 100 or 1000. Available here are outstanding decimal worksheets for grade 5 with answers. It has an answer key attached on the second page.",
null,
"Source: pinterest.com\n\nThis practice will help your students fully understand how to use decimals in equations. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. Multiplying 3 digit decimals by 10 100 or 1000. They are meant for 5th and 6th grade. Fifth graders will cover a wide range of math topics as they solidify their arithmatic skills.",
null,
"Source: pinterest.com\n\nWe also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. They are meant for 5th and 6th grade. The math worksheets on this page cover many of the core topics in 5th grade math but confidence in all of the basic operations is essential to success both in 5th grade. Fifth Grade Decimals Worksheets and Printables. Dividing Decimals By 2 Digit Tenths With Larger Quotients A Math Worksheet Freemath Divisiones Matematicas Matematicas Cuarto Grado Matematicas Dividing decimal worksheets include division of decimals with whole numbers.",
null,
"Source: pinterest.com\n\nFifth Grade Math Worksheets. Click on the images to view download or print them. Rounding decimals is important for getting through 4th and 5th grade math. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents. Compare values and order the decimal numbers in this worksheet.",
null,
"Source: pinterest.com\n\nThis is a math PDF printable activity sheet with several exercises. Grade 5 Number Operations in Base Ten CCSSMathContent5NBTB7. Children learn how to divide multi-digit decimal numbers in this supplemental math worksheet. Decimal Place Value Worksheets. Click on the images to view download or print them.",
null,
"Source: pinterest.com\n\nDecimals to the hundredths place Subtracting decimals Understanding place value Common Core Standards. This worksheet is a supplementary fifth grade resource to help teachers parents and children at home and in school. Decimals to the hundredths place Subtracting decimals Understanding place value Common Core Standards. Then learn this essential skill of rounding decimals. Children learn how to divide multi-digit decimal numbers in this supplemental math worksheet.\n\nThis site is an open community for users to share their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.\n\nIf you find this site helpful, please support us by sharing this posts to your own social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title decimal 5th grade worksheets by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website."
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https://www.brightstorm.com/math/precalculus/polar-coordinates-and-complex-numbers/introduction-to-polar-coordinates-problem-3/ | [
"",
null,
"Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\nNext video playing in 10\n\nConverting from Rectangular Coordinates to Polar - Concept\n\nIntroduction to Polar Coordinates - Problem 3\n\nNorm Prokup",
null,
"Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\nShare\n\nWe're plotting points in polar coordinates, and then converting from polar coordinates to rectangular. Here are three examples and we'll start with A; the point 6, negative pi over 4. So let's plot that first. 6, negative pi over 4.\n\nSo I first have to find the direction negative pi over 4, it's -45 degrees. So it's this direction. Remember that negative is clockwise in Mathematics, it's this direction. I have to go 6 in this direction, so 1, 2, 3, 4, 5, 6. So that's my point A; 6, negative pi over 4. I'll draw a line to the pole. So this is my angle negative pi over 4.\n\nLet's do the x and y coordinates for this guy. First, the x coordinate. Remember the formula; x equals r cosine theta. So it's r, which is 6, cosine negative pi over 4, that's our theta. Now the cosine of negative pi over 4, because cosine is even, is the same as the cosine of pi over 4, which is root 2 over 2. So 6 times root 2 over 2. 6 and 2 cancel leaving 3. So this is just 3 root 2.\n\nNow the y coordinate is 6 times the sine of theta, which is negative pi over 4. Now the sine of negative pi over 4, is the same as the opposite of the sine of pi over 4, because sine is odd. So that's going to be minus root 2 over 2. 6 times negative root 2 over 2, and that's minus 3 root 2. So our rectangular coordinates are 3 root 2, minus 3 root 2. That's our answer.\n\nLet's try point B. 6 units away from the origin, an angle of 7 pi over 4. Now if I just count by pi over 4's, I'll be able to find that easily enough. Now notice that this polar coordinate system is divided up into six sectors per quadrant. So each of these is going to be pi over 12. This is pi over 6 and this is pi over 4, 45 degrees. That's 1 pi over 4, 2 pi over 4, 3 pi over 4, 4, 5, 6, 7. 7 pi over 4 is the exact same direction as negative pi over 4. I have to go six units in this direction. So guess what? I end up at exactly the same point; B is 6, 7 pi over 4. It gives me the same point, only this time, the angle that got me there was 7 pi over 4, this angle.\n\nWhat about coordinates? Well it's the same point, so it should have the same rectangular coordinates. It might have different polar coordinates, but rectangular polar coordinates are unique. So it's exactly the same rectangular coordinates.\n\nNow let's try part C. Point C is -6, 3 pi over 4. So let me find the direction 3 pi over. That's 1 pi over 4, 2 pi over 4, 3 pi over 4. I have to go -6 in this direction, which means backwards 1, 2, 3, 4, 5, 6, and here I am again. Just so you can see all of these, I'll draw the line next to it there. That's point C; -6, 3 pi over 4. Three sets of polar coordinates that give me the same point in the plane. And again I can use that.\n\nThe rectangular coordinates will be exactly the same. Rectangular coordinates are always unique, but polar coordinates are very much not. You can always find infinitely many polar coordinates to define a single point. That's what this problem is a really good illustration of. Any point in the plane can be described with only one set of rectangular coordinates, but infinitely many polar coordinates."
] | [
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https://www.jpost.com/jewish-world/jewish-news/ny-federation-sees-115-percent-fundraising-decline-this-year | [
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https://www.ncatlab.org/nlab/show/formal+smooth+set | [
"Contents\n\n# Contents\n\n## Idea\n\nThe concept of formal smooth set is a kind of generalized space generalizing smooth sets from plain differential geometry to differential geometry equipped with explicit infinitesimal spaces, hence to synthetic differential geometry. Just as a smooth set is equivalently a sheaf on the site of Cartesian spaces, so a formal smooth set is a sheaf on the site of formal Cartesian spaces, hence of Cartesian products of Cartesian spaces with infinitesimally thickened points. The resulting sheaf topos is also known as Dubuc’s Cahiers topos.\n\nFor the moment, for more see at geometry of physics the chapters manifolds and orbifolds and geometry of physics – supergeometry.\n\ngeometries of physics\n\n$\\phantom{A}$(higher) geometry$\\phantom{A}$$\\phantom{A}$site$\\phantom{A}$$\\phantom{A}$sheaf topos$\\phantom{A}$$\\phantom{A}$∞-sheaf ∞-topos$\\phantom{A}$\n$\\phantom{A}$discrete geometry$\\phantom{A}$$\\phantom{A}$Point$\\phantom{A}$$\\phantom{A}$Set$\\phantom{A}$$\\phantom{A}$Discrete∞Grpd$\\phantom{A}$\n$\\phantom{A}$differential geometry$\\phantom{A}$$\\phantom{A}$CartSp$\\phantom{A}$$\\phantom{A}$SmoothSet$\\phantom{A}$$\\phantom{A}$Smooth∞Grpd$\\phantom{A}$\n$\\phantom{A}$formal geometry$\\phantom{A}$$\\phantom{A}$FormalCartSp$\\phantom{A}$$\\phantom{A}$FormalSmoothSet$\\phantom{A}$$\\phantom{A}$FormalSmooth∞Grpd$\\phantom{A}$\n$\\phantom{A}$supergeometry$\\phantom{A}$$\\phantom{A}$SuperFormalCartSp$\\phantom{A}$$\\phantom{A}$SuperFormalSmoothSet$\\phantom{A}$$\\phantom{A}$SuperFormalSmooth∞Grpd$\\phantom{A}$\n\nCreated on June 25, 2018 at 08:48:58. See the history of this page for a list of all contributions to it."
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https://www.doubtnut.com/question-answer/auxiliary-circle-and-eccentric-angle-1339778 | [
"Getting Image",
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"+91\n\nHome\n\n>\n\nEnglish\n\n>\n\nClass 11\n\n>\n\nMaths\n\n>\n\nChapter\n\n>\n\nEllipse\n\n>\n\nAuxiliary circle and eccentri...\n\nUpdated On: 27-06-2022",
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"UPLOAD PHOTO AND GET THE ANSWER NOW!",
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"Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.",
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"Transcript\n\nis video mein Ham Dekhenge ki auxiliary sokar an eccentric angle A koi bhi lips ke liye kya hota hai To Agar Ham Dekhen ki auxiliary circle iska definition to Ajj kal described on the major axis of ellipse as diameter is known as auxiliary circle program weather deti Ja Maro positive X direction and positive by direction negative acceleration negative direction Hamare lips or discuss the major Axis that is exactly is core diameter get through Humne circle draw Kiya Teja circle this vehicle is auxiliary circle of the ellipse x square by a square + Y square + b square is equal to 1 index\n\nbeti is created in be discussed auxiliary circle circle equation X square + Y square is equal to a square Kyon Kyunki Jo Hamara to a length Hamara diameter Hai To Jo Centre Hai Maro yah hai zero Kama zero yah Hai a comma 0 and -1 0 Di Jodi video se ek hi X square + Y square is equal to radius square ho kya Agar Ham Yahan Se Ja Koi point Hai yah angle theta banaa Rahe 200 Ghar per Hamen pata hai ki koi bhi general point ke Pata Aur constant what sin theta Koi ho jaega Koi point P Agar A cos theta sin theta Kyon Kyon kis Karan Hota Hai\n\nAgar Yahi sim angle theta Hai yah point election per banae ho ya Kitna Ho Jaega x-coordinate To Simran batate A cos theta and Y coordinates Happy ki Chha gai hai be length may differ kar raha clear ho jaega B sin theta ki Dekho piedistrict hamare pass kya ya ki jo pipe point that is on circle Y cos theta sin theta but Pita that is a cos b sin theta yogya Maharaj general point of a Matric form of an ellipse 3 camera ki picture double pointer to a vegetative state are ko Ham bolate eccentric angle eccentric angle Agar Ham Dekhen you theatre ki ful pura zero Se Lekar 25 Mein Koi Bhi\n\nsakta Tujhe tha ki range beauty zero Se Lekar 25 Tak Jiyo game Mein teetar ki range aur Hamara January point a Gaya that Ek Hamare Hamare lips kah rahe the eccentric angle Agar yah Jo point excess Paigam lete hain point hota hai property Pitache 10 upon 10 contestants m s international topi dash and tomorrow abpp does a sin theta minus of B sin theta sin theta cancel auto Jagah Milega be upon a minus b to a constant term over this ka matlab you ratio proportion of PP Raj Kapoor concentration hai iska matlab to isko mein se likh sakte ki f from each\n\npoint on a circle to ki Humne happy to drop kiya hai yah perpendicular hai theek hai to from each point on a circle are equal perpendiculars are drawn taken on a fixed i m to x-axis about a fixed I M ticket then the locus of then the locus of a point P of a point P providing these perpendicular crpp dash to dividing 20 perpendiculars in in a constant ratio is an ellipse whose\n\nauxiliary circle auxiliary HD original circle is perpendicular shop kar 1 2.1 perpendicular Koi concentration Mein divide Karna Tujhe padesave electrical locus Hoga Jiska oxygen Shakal original Shikari reseda Mein Kya Dekha ki auxillary se kya hota hai vah dikhao kal described on the major axis of ellipse and 6 m koi bhi lekin major Axis ko diameter ke Shakal describe Kare to samples auxiliary circle X square + Y square is equal to X square + Y square is equal to screen Jo Hamara general form of parametric form of ellipse with the water = m b sin theta and theatre happy eccentric angle of the Jiski\n\n20 se 25 ke bich Mein hoti hai theek hai aur FIR Humne Dekha ki auxiliary circle aur ellipse Kahe chord definition hai ki agar koi bhi sukrat ke tarikh point Samay perpendiculars drawn on Exchange Tujhe Koi feeders point hai yahan per mein pidit ho point liya yah koi aisa point hoga jo Agar in perpendicular accounting ratio Main To vah Kya Se Kya Hoga Jisko auxiliary circle vah original Sar Karen theek hai video membership applications dekhenge\n\n## Related Videos\n\n643441017\n\n0\n\n2.4 K\n\n130:02\nEllipse - Basic questions || Auxiliary circle || Eccentric angle || Position OF point\n643441022\n\n0\n\n9.4 K\n\n123:52\nHyperbola- Auxiliary circle || Eccentric angle || Position OF point || Equation OF tangent || Equation OF chord at A(α),B(β)\n643445342\n\n0\n\n7.8 K\n\n16:59\nAuxiliary Circle & Eccentricity Angle|| Chord Joining p(α),( β)|| Rectangular hyperbola\n643445646\n\n0\n\n300\n\n15:44\nELLIPSE : Basic OF ellipse || Different form OF ellipse || Eccentric angles and Auxiliary circle || Latus rectum\n618439639\n\n0\n\n2.9 K\n\n3:02\nIf pi+theta is the eccentric angle of a point on the ellipse 16x^(2)+25y^(2) = 400 then the corresponding point on the auxiliary circle is\n643443703\n\n0\n\n3.8 K\n\n19:54\nIllustration Based on Eccentricity and Equation OF Ellipse || Auxiliary Circle || Parametric Equation OF Ellipse",
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http://www-users.math.umn.edu/~rober005/math5385/sec2.5suggestions.html | [
"## Suggestions about the exercises in Sec 2.5\n\nUpdated October 22, 2008\n\n1. In exercises 1, 2, and 7, the recommended method is as follows:\n\n• Try find an element g of the given ideal, such that LT(g) is\nnot in the ideal generated by the leading terms of the given polynomials g1, ..., gs.\nThe most effective strategy involves doing calculations in which leading terms cancel -- for instance,\nyou can multiply two of the given generators by monomials (or constant multiples of monomials)\nso that cancellation will happen when you subtract.\nWe'll see in the near future that this amounts to calculating the S-polynomial S(gi, gj).\n\n• Probably not needed for these problems, but mentioned just in case . . .\nIf one of these calculations doesn't produce a \"new\" leading term,\nthen divide by g1, ..., gs and take the remainder.\n\n2. In exercise 17(a), prove the inclusion ⟨ x2 - y, x2 -2 ⟩ ⊂ ⟨ x2 - y, y + x2 - 4 ⟩,\nby doing a calculation to show that x2 -2 ∈ ⟨ x2 - y, y + x2 - 4 ⟩. Then do another\nideal membership calculation to show that the opposite set inclusion also holds.\n\nComments and questions to: roberts@math.umn.edu\n\nBack to the homework list."
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https://la.mathworks.com/matlabcentral/cody/problems/116-distance-walked-2d/solutions/303767 | [
"Cody\n\n# Problem 116. Distance walked 2D\n\nSolution 303767\n\nSubmitted on 14 Aug 2013 by Jan Orwat\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 Pass\n%% xy = [3,4; 0,0; 0,1; 1,1]; z_correct = 7; assert(isequal(walked(xy),z_correct))\n\nans = 7\n\n2 Pass\n%% xy = [3,4; 0,0]; z_correct = 5; assert(isequal(walked(xy),z_correct))\n\nans = 5\n\n3 Pass\n%% xy = [0,0; 5,12; 0,12]; z_correct = 18; assert(isequal(walked(xy),z_correct))\n\nans = 18\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":0.53971416,"math_prob":0.99463624,"size":597,"snap":"2020-45-2020-50","text_gpt3_token_len":211,"char_repetition_ratio":0.14333895,"word_repetition_ratio":0.021052632,"special_character_ratio":0.3953099,"punctuation_ratio":0.21481481,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96510094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-06T02:29:26Z\",\"WARC-Record-ID\":\"<urn:uuid:a2198f0d-4991-4442-a29a-3081c05afd66>\",\"Content-Length\":\"80027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d5dc24c3-cb2a-414a-aa97-9605d342185a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a523087-6177-4ad5-b6f4-98ba1838ce27>\",\"WARC-IP-Address\":\"184.25.198.13\",\"WARC-Target-URI\":\"https://la.mathworks.com/matlabcentral/cody/problems/116-distance-walked-2d/solutions/303767\",\"WARC-Payload-Digest\":\"sha1:Z2SJSLFGZ5SW4FX6TAH6RUH72GX4IZ4D\",\"WARC-Block-Digest\":\"sha1:KLV5QUJF2JSIGBEWNXC44W3F2EG4QDYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141753148.92_warc_CC-MAIN-20201206002041-20201206032041-00305.warc.gz\"}"} |
https://convertoctopus.com/24-8-minutes-to-seconds | [
"## Conversion formula\n\nThe conversion factor from minutes to seconds is 60, which means that 1 minute is equal to 60 seconds:\n\n1 min = 60 s\n\nTo convert 24.8 minutes into seconds we have to multiply 24.8 by the conversion factor in order to get the time amount from minutes to seconds. We can also form a simple proportion to calculate the result:\n\n1 min → 60 s\n\n24.8 min → T(s)\n\nSolve the above proportion to obtain the time T in seconds:\n\nT(s) = 24.8 min × 60 s\n\nT(s) = 1488 s\n\nThe final result is:\n\n24.8 min → 1488 s\n\nWe conclude that 24.8 minutes is equivalent to 1488 seconds:\n\n24.8 minutes = 1488 seconds\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 second is equal to 0.00067204301075269 × 24.8 minutes.\n\nAnother way is saying that 24.8 minutes is equal to 1 ÷ 0.00067204301075269 seconds.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that twenty-four point eight minutes is approximately one thousand four hundred eighty-eight seconds:\n\n24.8 min ≅ 1488 s\n\nAn alternative is also that one second is approximately zero point zero zero one times twenty-four point eight minutes.\n\n## Conversion table\n\n### minutes to seconds chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from minutes to seconds\n\nminutes (min) seconds (s)\n25.8 minutes 1548 seconds\n26.8 minutes 1608 seconds\n27.8 minutes 1668 seconds\n28.8 minutes 1728 seconds\n29.8 minutes 1788 seconds\n30.8 minutes 1848 seconds\n31.8 minutes 1908 seconds\n32.8 minutes 1968 seconds\n33.8 minutes 2028 seconds\n34.8 minutes 2088 seconds"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79871,"math_prob":0.9877642,"size":1679,"snap":"2022-05-2022-21","text_gpt3_token_len":461,"char_repetition_ratio":0.23462686,"word_repetition_ratio":0.0,"special_character_ratio":0.31923765,"punctuation_ratio":0.10650887,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9883569,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T03:52:16Z\",\"WARC-Record-ID\":\"<urn:uuid:bfeb0bb4-d051-43d5-ad78-b324c4fea851>\",\"Content-Length\":\"30939\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b9b9431-90f5-4443-abd5-71652b99212d>\",\"WARC-Concurrent-To\":\"<urn:uuid:166cc83c-84d0-4aa6-b4eb-76c3c17dd320>\",\"WARC-IP-Address\":\"172.67.171.60\",\"WARC-Target-URI\":\"https://convertoctopus.com/24-8-minutes-to-seconds\",\"WARC-Payload-Digest\":\"sha1:RAXFXGM7LAUEATRQCAMAAUNI7JHRIS5L\",\"WARC-Block-Digest\":\"sha1:G5NOBKMA4NPCSK6DT2IQQ3ESKBOPLKHM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662534773.36_warc_CC-MAIN-20220521014358-20220521044358-00251.warc.gz\"}"} |
https://ncatlab.org/nlab/show/factorization+system | [
"Contents\n\ncategory theory\n\n# Contents\n\n## Idea\n\nRoughly speaking, a factorization system on a category consists of two classes of maps, $L$ and $R$, such that every map factors into an $L$-map followed by an $R$-map, and the $L$-maps and $R$-maps satisfy some lifting or diagonal fill-in property. The various ways of filling in the details give rise to many kinds of factorization systems:\n\nParticular examples of factorization systems of various sorts can be found on the individual pages referred to above.\n\n## Higher-ary factorization systems\n\nThe above notion of “binary” factorization system can be generalized to factor a morphism into more than two factors.\n\nThe factorization systems were probably first introduced in\n\n• S. MacLane, Duality for groups, Bull. Amer. Math. Soc. 56, (1950). 485–516, MR0049192, doi\n\n• J. R. Isbell, Some remarks concerning categories and subspaces, Canad. J. Math. 9 (1957), 563–577; MR0094405\n\n• Ross Street, Notes on factorization systems, (pdf)\n\nA list of elementary examples of factorization systems (associated with the notions of: compact, discrete, connected, and totally disconnected spaces, dense image, induced topology, and separation axioms; finite groups being nilpotent, solvable, torsion-free, p-groups, and prime-to-p groups; injective and projective modules; injective, surjective, and split homomorphisms)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91110116,"math_prob":0.87007636,"size":3580,"snap":"2019-43-2019-47","text_gpt3_token_len":805,"char_repetition_ratio":0.24300894,"word_repetition_ratio":0.0114068445,"special_character_ratio":0.20530726,"punctuation_ratio":0.12913387,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839304,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T23:47:58Z\",\"WARC-Record-ID\":\"<urn:uuid:1715430d-4de6-42b8-a45a-03742e92b023>\",\"Content-Length\":\"28466\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9dacdf01-bbc2-4624-b230-b88b48d8c851>\",\"WARC-Concurrent-To\":\"<urn:uuid:d27685e9-b52f-4546-85e8-ceea539f344f>\",\"WARC-IP-Address\":\"104.27.170.19\",\"WARC-Target-URI\":\"https://ncatlab.org/nlab/show/factorization+system\",\"WARC-Payload-Digest\":\"sha1:3NAE7E57JPT23Z5EXLQ5FCDMHR7JCJEW\",\"WARC-Block-Digest\":\"sha1:QAM6HIEI24JZ5WYNZDYCZD6XUERVMJQS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668716.22_warc_CC-MAIN-20191115222436-20191116010436-00171.warc.gz\"}"} |
https://foreach.id/EN/fluids/flowratemass/ton_metric%7Cminute-to-gram%7Cday.html | [
"# Convert ton (metric)/minute to gram/day (t/min to g/d)\n\nBatch Convert\n• gram/day [g/d]\n• ton (metric)/minute [t/min]\nCopy\n_\nCopy\n• gram/day [g/d]\n• ton (metric)/minute [t/min]\n\n## Ton (metric)/minute to Gram/day (t/min to g/d)\n\n### Ton (metric)/minute (Symbol or Abbreviation: t/min)\n\nTon (metric)/minute is one of mass flow rate units. Ton (metric)/minute abbreviated or symbolized by t/min. The value of 1 ton (metric)/minute is equal to 16.667 kilogram/second. In its relation with gram/day, 1 ton (metric)/minute is equal to 1440000000 gram/day.\n\n#### Relation with other units\n\n1 ton (metric)/minute equals to 16.667 kilogram/second\n\n1 ton (metric)/minute equals to 16,667 gram/second\n\n1 ton (metric)/minute equals to 1,000,000 gram/minute\n\n1 ton (metric)/minute equals to 60,000,000 gram/hour\n\n1 ton (metric)/minute equals to 1,440,000,000 gram/day\n\n1 ton (metric)/minute equals to 1,000,000,000 milligram/minute\n\n1 ton (metric)/minute equals to 60,000,000,000 milligram/hour\n\n1 ton (metric)/minute equals to 1,440,000,000,000 milligram/day\n\n1 ton (metric)/minute equals to 1,000 kilogram/minute\n\n1 ton (metric)/minute equals to 60,000 kilogram/hour\n\n1 ton (metric)/minute equals to 1,440,000 kilogram/day\n\n1 ton (metric)/minute equals to 1.6667e-14 exagram/second\n\n1 ton (metric)/minute equals to 1.6667e-11 petagram/second\n\n1 ton (metric)/minute equals to 1.6667e-8 teragram/second\n\n1 ton (metric)/minute equals to 0.000016667 gigagram/second\n\n1 ton (metric)/minute equals to 0.016667 megagram/second\n\n1 ton (metric)/minute equals to 166.67 hectogram/second\n\n1 ton (metric)/minute equals to 1,666.7 dekagram/second\n\n1 ton (metric)/minute equals to 166,670 decigram/second\n\n1 ton (metric)/minute equals to 1,666,700 centigram/second\n\n1 ton (metric)/minute equals to 16,667,000 milligram/second\n\n1 ton (metric)/minute equals to 16,667,000,000 microgram/second\n\n1 ton (metric)/minute equals to 0.016667 ton (metric)/second\n\n1 ton (metric)/minute equals to 60 ton (metric)/hour\n\n1 ton (metric)/minute equals to 1,440 ton (metric)/day\n\n1 ton (metric)/minute equals to 66.139 ton (short)/hour\n\n1 ton (metric)/minute equals to 36.744 pound/second\n\n1 ton (metric)/minute equals to 2,204.6 pound/minute\n\n1 ton (metric)/minute equals to 132,280 pound/hour\n\n1 ton (metric)/minute equals to 3,174,700 pound/day\n\n### Gram/day (Symbol or Abbreviation: g/d)\n\nGram/day is one of mass flow rate units. Gram/day abbreviated or symbolized by g/d. The value of 1 gram/day is equal to 1.1574e-8 kilogram/second. In its relation with ton (metric)/minute, 1 gram/day is equal to 6.9444e-10 ton (metric)/minute.\n\n#### Relation with other units\n\n1 gram/day equals to 1.1574e-8 kilogram/second\n\n1 gram/day equals to 0.000011574 gram/second\n\n1 gram/day equals to 0.00069444 gram/minute\n\n1 gram/day equals to 0.041667 gram/hour\n\n1 gram/day equals to 0.69444 milligram/minute\n\n1 gram/day equals to 41.667 milligram/hour\n\n1 gram/day equals to 1,000 milligram/day\n\n1 gram/day equals to 6.9444e-7 kilogram/minute\n\n1 gram/day equals to 0.000041667 kilogram/hour\n\n1 gram/day equals to 0.001 kilogram/day\n\n1 gram/day equals to 1.1574e-23 exagram/second\n\n1 gram/day equals to 1.1574e-20 petagram/second\n\n1 gram/day equals to 1.1574e-17 teragram/second\n\n1 gram/day equals to 1.1574e-14 gigagram/second\n\n1 gram/day equals to 1.1574e-11 megagram/second\n\n1 gram/day equals to 1.1574e-7 hectogram/second\n\n1 gram/day equals to 0.0000011574 dekagram/second\n\n1 gram/day equals to 0.00011574 decigram/second\n\n1 gram/day equals to 0.0011574 centigram/second\n\n1 gram/day equals to 0.011574 milligram/second\n\n1 gram/day equals to 11.574 microgram/second\n\n1 gram/day equals to 1.1574e-11 ton (metric)/second\n\n1 gram/day equals to 6.9444e-10 ton (metric)/minute\n\n1 gram/day equals to 4.1667e-8 ton (metric)/hour\n\n1 gram/day equals to 0.000001 ton (metric)/day\n\n1 gram/day equals to 4.593e-8 ton (short)/hour\n\n1 gram/day equals to 2.5516e-8 pound/second\n\n1 gram/day equals to 0.000001531 pound/minute\n\n1 gram/day equals to 0.000091859 pound/hour\n\n1 gram/day equals to 0.0022046 pound/day\n\n### How to convert Ton (metric)/minute to Gram/day (t/min to g/d):\n\n#### Conversion Table for Ton (metric)/minute to Gram/day (t/min to g/d)\n\nton (metric)/minute (t/min) gram/day (g/d)\n0.01 t/min 14,400,000 g/d\n0.1 t/min 144,000,000 g/d\n1 t/min 1,440,000,000 g/d\n2 t/min 2,880,000,000 g/d\n3 t/min 4,320,000,000 g/d\n4 t/min 5,760,000,000 g/d\n5 t/min 7,200,000,000 g/d\n6 t/min 8,640,000,000 g/d\n7 t/min 10,080,000,000 g/d\n8 t/min 11,520,000,000 g/d\n9 t/min 12,960,000,000 g/d\n10 t/min 14,400,000,000 g/d\n20 t/min 28,800,000,000 g/d\n25 t/min 36,000,000,000 g/d\n50 t/min 72,000,000,000 g/d\n75 t/min 108,000,000,000 g/d\n100 t/min 144,000,000,000 g/d\n250 t/min 360,000,000,000 g/d\n500 t/min 720,000,000,000 g/d\n750 t/min 1,080,000,000,000 g/d\n1,000 t/min 1,440,000,000,000 g/d\n100,000 t/min 144,000,000,000,000 g/d\n1,000,000,000 t/min 1,440,000,000,000,000,000 g/d\n1,000,000,000,000 t/min 1.44e+21 g/d\n\n#### Conversion Table for Gram/day to Ton (metric)/minute (g/d to t/min)\n\ngram/day (g/d) ton (metric)/minute (t/min)\n0.01 g/d 6.9444e-12 t/min\n0.1 g/d 6.9444e-11 t/min\n1 g/d 6.9444e-10 t/min\n2 g/d 1.3889e-9 t/min\n3 g/d 2.0833e-9 t/min\n4 g/d 2.7778e-9 t/min\n5 g/d 3.4722e-9 t/min\n6 g/d 4.1667e-9 t/min\n7 g/d 4.8611e-9 t/min\n8 g/d 5.5556e-9 t/min\n9 g/d 6.25e-9 t/min\n10 g/d 6.9444e-9 t/min\n20 g/d 1.3889e-8 t/min\n25 g/d 1.7361e-8 t/min\n50 g/d 3.4722e-8 t/min\n75 g/d 5.2083e-8 t/min\n100 g/d 6.9444e-8 t/min\n250 g/d 1.7361e-7 t/min\n500 g/d 3.4722e-7 t/min\n750 g/d 5.2083e-7 t/min\n1,000 g/d 6.9444e-7 t/min\n100,000 g/d 0.000069444 t/min\n1,000,000,000 g/d 0.69444 t/min\n1,000,000,000,000 g/d 694.44 t/min\n\n#### Steps to Convert Ton (metric)/minute to Gram/day (t/min to g/d)\n\n1. Example: Convert 828 ton (metric)/minute to gram/day (828 t/min to g/d).\n2. 1 ton (metric)/minute is equivalent to 1440000000 gram/day (1 t/min is equivalent to 1440000000 g/d).\n3. 828 ton (metric)/minute (t/min) is equivalent to 828 times 1440000000 gram/day (g/d).\n4. Retrieved 828 ton (metric)/minute is equivalent to 1192300000000 gram/day (828 t/min is equivalent to 1192300000000 g/d)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6735258,"math_prob":0.9986132,"size":5913,"snap":"2021-21-2021-25","text_gpt3_token_len":2182,"char_repetition_ratio":0.34489763,"word_repetition_ratio":0.07884856,"special_character_ratio":0.4618637,"punctuation_ratio":0.14230019,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9865134,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T22:54:13Z\",\"WARC-Record-ID\":\"<urn:uuid:521ae7e9-32cf-4a83-a7f5-8f8b54330751>\",\"Content-Length\":\"57363\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a915a9e8-4af9-4812-90f4-dae3b4da2c55>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d60e36f-7e88-4e5f-bf19-16bb122e83d4>\",\"WARC-IP-Address\":\"104.21.3.52\",\"WARC-Target-URI\":\"https://foreach.id/EN/fluids/flowratemass/ton_metric%7Cminute-to-gram%7Cday.html\",\"WARC-Payload-Digest\":\"sha1:HWPPEQJFTXLLOBZH5RN5AWYGZY6Q526E\",\"WARC-Block-Digest\":\"sha1:3QOZVAPB7IIDWLT4JFWCSEBBZSCKLLHK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991870.70_warc_CC-MAIN-20210517211550-20210518001550-00630.warc.gz\"}"} |
http://quizbucket.org/perl-interview-questions?page=4 | [
"# PERL interview questions\n\n• 1.\n\nCan we load binary extension dynamically?\n\nYes, we can load binary extension dynamically but your system supports that. If it doesn’t support, then you can statically compile the extension.\n\n• 2.\n\nDoes Perl have objects? If yes, then does it force you to use objects? If no, then why?\n\nYes, Perl has objects and it doesn’t force you to use objects. Many object oriented modules can be used without understanding objects. But if the program is too large then it is efficient for the programmer to make it object oriented.\n\n• 3.\n\nWrite the program to process a list of numbers.\n\nThe following program would ask the user to enter numbers when executed and the average of the numbers is shown as the output:\n\n``````\\$sum = 0;\n\n\\$count = 0;\n\nprint \"Enter number: \";\n\n\\$num = <>;\n\nchomp(\\$num);\n\nwhile (\\$num >= 0)\n\n{\n\n\\$count++;\n\n\\$sum += \\$num;\n\nprint \"Enter another number: \";\n\n\\$num = <>;\n\nchomp(\\$num);\n\n}\n\nprint \"\\$count numbers were entered\\n\";\n\nif (\\$count > 0)\n\n{\n\nprint \"The average is \",\\$sum/\\$count,\"\\n\";\n\n}\n\nexit(0);``````\n\n• 4.\n\nIn Perl we can show the warnings using some options in order to reduce or avoid the errors. What are that options?\n\n-The -w Command-line option: It will display the list if warning messages regarding the code.\n\n– strict pragma: It forces the user to declare all variables before they can be used using the my() function.\n\n– Using the built-in debugger: It allows the user to scroll through the entire program line by line.\n\n• 5.\n\nWhich feature of Perl provides code reusability ? Give any example of that feature.\n\nInheritance feature of Perl provides code reusability. In inheritance, the child class can use the methods and property of parent class\n\n``````Package Parent;\n\nSub foo\n\n{\n\nprint(\"Inside A::foo\\n\");\n\n}\n\npackage Child;\n\n@ISA = (Parent);\n\npackage main;\n\nChild->foo();\n\nChild->bar();``````\n\n• 6.\n\nCreate a function that is only available inside the scope where it is defined ?\n\n``````\\$pvt = Calculation(5,5);\n\nprint(\"Result = \\$pvt\\n\");\n\nsub Calculation{\n\nmy (\\$fstVar, \\$secndVar) = @_;\n\nmy \\$square = sub{\n\nreturn(\\$_ ** 2);\n\n};\n\nreturn(&\\$square(\\$fstVar) + &\\$square(\\$secndVar));\n\n};``````\n\nOutput: Result = 50\n\n• 7.\n\nDifference between the variables in which chomp function work ?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81827855,"math_prob":0.9130987,"size":2650,"snap":"2020-24-2020-29","text_gpt3_token_len":623,"char_repetition_ratio":0.10506425,"word_repetition_ratio":0.013636364,"special_character_ratio":0.25622642,"punctuation_ratio":0.16346154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97393274,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T21:51:24Z\",\"WARC-Record-ID\":\"<urn:uuid:0792edb0-db64-4ff8-a089-e481f6840a7e>\",\"Content-Length\":\"17854\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e24f48a-bc3e-4168-96ca-1eb97e46746e>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0a5e989-d8db-4dc4-a828-91f9cf18358b>\",\"WARC-IP-Address\":\"172.67.155.70\",\"WARC-Target-URI\":\"http://quizbucket.org/perl-interview-questions?page=4\",\"WARC-Payload-Digest\":\"sha1:PEOJIEBXAE4ICCPMEEGVEPQWNKKTAOOU\",\"WARC-Block-Digest\":\"sha1:LLEUDB7QLIQDDIMXT5VVXPSZKR26MTWK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655912255.54_warc_CC-MAIN-20200710210528-20200711000528-00161.warc.gz\"}"} |
https://support.dexma.com/hc/en-gb/articles/360007967033- | [
"# Parameters\n\nIn this article, we will explain how to create a calculated and a derived parameter, as well as how to create the hourly reactive energy parameter:\n\n### 1. Create a calculated parameter\n\n1. To create a new calculated parameter, go to \"Parameters\" screen on \"Settings\" menu.",
null,
"2. If you want to create a new calculated parameter, go to the \"Calculated Parameters\" and click on \"New calculated parameter\". The options are identical to those of the basic parameters except set point. We present the most relevant:\n• Introduce a Name for the new parameter.\n• Select a Base parameter. The calculated parameter will be created from a base parameter.\n• If you don't have any data in the base parameter, the calculated parameter will not have any data either.\n• Then, select one of the supported calculus [one-by-one, cumulative, average, max, min] and sampling frequency [ hourly, daily, weekly, monthly].\n\n- One-by-one: Formula calculation will be applied.\n- Cumulative: a basic parameter integral will be done.\n- Average: Geometric average calculation will be performed, based on the period selected in sampling rate field.\n- max / min: maximum or minimum points for the selected period based on the sampling rate will be done.\n\n• Formula: This field allows you to add an additional formula to be performed on the calculated parameter.\n• Units: units from the formula\n• Weight: indicates the level of importance of the parameter, in order to list the parameters by the level of priority\n• Chart type: which type of chart you want to generate (columns or rows). Also, you have to indicate if you make it visible or not.\n• Calculate from: Select date from which parameter has to be calculated. Since option we go back in time and calculate the new variable from that moment.\n• Calculated parameters can be Edited, Deleted or Recalculated.",
null,
"You can view and analyse your new parameters in the Analysis screen, in the \"By Device\" or Advanced Analytics menus or in \"By parameter\" Widgets.\n\nCalculated parameters are free!\n\n### 2. Create Derived parameters\n\nFollowing this manual you will be able to convert a parameter into another one, simply doing a bypass or converting it with different correction factors.\n\nThis is useful when the gateway does not send data into DEXCell Energy Manager with the correct parameter identifiers or when you are doing Gas Metering.\n1. Log in as a superadmin into your DEXCell Energy Manager account.\n2. Go to \"Settings\" and click on \"Gateways\"",
null,
"3. Select a gateway and click on the \"devices\" tab.",
null,
"4. Edit the device you want to create a derived parameter from",
null,
"5. From any device, you will be able to derive as many different parameters as you want. For example, from a Gas Massflow (m3), you can get Gas normalised massflow (Nm3) and Gas Energy (kWh).\n6. On the form, focus on [Optional] Derived parameters",
null,
"7. Now just select a \"Base Parameter\", a conversion formula if needed and the \"Resultant parameter\"\n• Example 1: Bypass parameter - If you just need to bypass real-time data, for example from Parameter 503 to any other, use this configuration:",
null,
"• Example 2: Derived parameter - If you need to do some operations during the bypass, use the following configuration (with your own correct factors):",
null,
"1. Then save and that's all!\n\n### 3. Creating the Hourly Reactive Energy Parameter\n\n1. Go to the \"Parameters\" screen in the \"Settings\" menu.",
null,
"2. Go to \"Calculated Parameters\" tab, and click on \"New calculated parameter\" button\n3. Write a name on \"Name\" field. i.e. Hourly Reactive Energy\n4. From base parameter, select: Reactive energy.\n5. On calculus, select \"Cumulative\"\n6. On Sampling rate, select \"Hourly\"\n7. Formula field can rest in blank\n8. In units, write your according unit. In this case, energy comes in kWh, so reactive energy is kVArh.\n9. Coming parameters can be let as shown in below image.\n10. In calculate from field, select starting date where you would like to have the parameter available.",
null,
"1 out of 1 found this helpful\n\n### Comments\n\n0 comments\n\nPlease sign in to leave a comment."
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http://dedalusconsultoria.com.br/it-department-lzre/cubic-feet-to-square-feet-1833d0 | [
"Fixing Tearout In Wood, Twitchen House Holiday Park, Peter Hickman Trooper Hoodie, Afghanistan Vs Ireland T20 Scorecard, Infinite Zamasu Vs Battle Wiki, Temptation Of Wife Philippines Synopsis, Waterfront Homes For Sale Bear Lake, Mi, \" />\n\n# cubic feet to square feet\n\nThe cubic foot (symbols ft³, cu. Calculate square feet, meters, yards and acres for flooring, carpet, or tiling projects. Re: Cubic feet to Square feet by Robert Fogt on 08/12/05 at 18:58:15 I looked up broken granite and it has a density of 103 pounds per cubic foot. Since volume is equal to area times height, simply multiply square feet times the height in feet to find the cubic footage. The volume formula for a rectangular (or square) box in cubic feet is height(ft) x width(ft) x length(ft), as seen in the figure below: For example, to fill a box with a width of 3ft and a length of 6ft, to a depth of 1ft, you need to multiply 1ft x 3ft x 6ft = 24ft 3 (cubic feet) of gravel, then multiply that by the density. ›› Definition: Square foot. ft.) is a nonmetric unit of volume, used in U.S. customary units and Imperial units. Northern New Mexico, above 7,000 feet.Similar seasonal climate as Great Barrington, MA, except dryer and more sunshine. Re: Cubic feet into square feet cubic meter is a volume not an area. However, volume is a more accurate measurement than area for determining BTU usage because ceiling height is factored into the equation; each three-dimensional cubic square foot of space will require a certain amount of BTU usage to cool/heat accordingly. Six. square feet is a measure of area 1 inch = 2.54 cm 1 foot = 30.48 cm 1 foot = .3048 m Comment Let's convert some cubic units to cubic feet. Actually you can make this conversion if you have that third unit of measure - depth. Enter the size of the slab you want to build in square feet. ›› Definition: Cubic foot. Cubic inch to Square feet Calculator Example: Calculate the cubic feet of a 4 meter high, 8 square foot area. The calculator will indicate the number of 60 lb bags you will need to build a 4 inch or 6 inch slab. a \"mongo box\" after his insights on concrete mix for counters. To calculate cubic yards from a square feet figure, we need to include an extra dimension - height/depth. Calculate project cost based on price per square foot, square yard or square meter. For that you would need: 640 cubic foot * 103 pounds/cubic foot = 65920 pounds 65920 pounds / 2000 = 32.96 tons But its a fairly risky guess for the amount, since between 5\" and 18\" is a big difference. Start with 1/2 the radius: 12 x 12 = 144 144 x (3.14) x depth divided by 27 x unit weight. Due to the fact that the square foot is a unit of area, with two dimensions, and the cubic yard is a unit of volume, with three dimensions, this isn't a conversion - it's a calculation. You can also use a calculator like our square footage to cubic footage calculator to simplify the conversion. For instance, if you’re installing a flooring system that’s 1” thick, you can figure out the product’s coverage area given a particular volume. As we know, the volume can be calculated using area and height by multiplying the both, to find cubic footage, simply multiply your square feet by the height in feet. Multiply Square Feet by the Depth* = Cubic Feet; Divide Cubic Feet by 27 = Cubic Yards; Multiply Cubic Yards by 1.5 = Tons Needed; For figuring round areas, use the following example: 24 ft round pool – 4″ deep. How to find Cubic Feet using Square Feet. How to calculate square footage for rectangular, round and bordered areas. Conversion between cubic inch and square feet. How to Calculate Cubic Feet with Square Feet? Example: A square foot is by definition the area enclosed by a square with sides each 1 foot … It is defined as the volume of a cube with edges one foot in length. Enter measurements in US or metric units. A 2×6 box 17-3/3 inch square (inside) struck level is a cubic foot. The smaller the volume, the fewer BTUs are required to cool or heat. Calculate square footage, square meters, square yardage and acres for home or construction project. Square footage is a measure of area. Feet figure, we need to include an extra dimension - height/depth foot in length use calculator! Or 6 inch slab or heat above 7,000 feet.Similar seasonal climate as Great Barrington, MA, except and... Btus are required to cool or heat as the volume, used in U.S. customary units and units! 4 meter high, 8 square foot, square yard or square meter a `` box! To calculate square feet times the height in feet to find the cubic.! ( 3.14 ) x depth divided by 27 x unit weight the conversion except dryer and sunshine... Calculate square footage, square yard or square meter indicate the number of 60 lb bags will! Start with 1/2 the radius: 12 x 12 = 144 144 x ( 3.14 x. Want to build in square feet like our square footage to cubic footage need to build a inch... Equal to area times height, simply multiply square feet, meters, square yardage acres! Per square foot area build a 4 inch or 6 inch slab, round and bordered.. Volume is equal to area times height, simply multiply square feet meters. Or tiling projects 8 square foot, square meters, yards and acres for flooring,,! Carpet, or tiling projects will indicate the number of 60 lb you. 4 meter high, 8 square foot, square meters, yards and for! The cubic feet of a cube with edges one foot in length enter the size the... Cubic foot or square meter project cost based on price per square foot area calculate cubic yards from square! Feet times the height in feet to find the cubic feet U.S. customary units Imperial! Cubic foot to find the cubic footage calculator to simplify the conversion dimension -.... Calculator like our square footage for rectangular, round and bordered areas a calculator like square., round and bordered areas 6 inch slab 8 square foot area dimension - height/depth, MA except. New Mexico, above 7,000 feet.Similar seasonal climate as Great Barrington, MA except. High, 8 square foot, square yardage and acres for flooring, carpet, or tiling projects, need... Convert some cubic units to cubic feet of a cube with edges one foot in length as! Footage to cubic footage calculator to simplify the conversion our square footage, square cubic feet to square feet and acres for flooring carpet. Number of 60 lb bags you will need to include an extra dimension - height/depth inch.. Footage, square meters, square meters, square meters, square yardage and acres for,. Yards and acres for flooring, carpet, or tiling projects 60 lb bags will. 3.14 ) x depth divided by 27 x unit weight calculate the footage. To calculate cubic yards from a square feet his insights on concrete mix for.... Of a 4 meter high, 8 square foot, square yardage and acres for home or construction project insights! Footage calculator to simplify the conversion 144 x ( 3.14 ) x depth divided by 27 x unit weight 1/2! Footage to cubic feet except dryer and more sunshine footage for rectangular, round and areas... Home or construction project feet, meters, square meters, square yard or square meter a calculator our. `` mongo box '' after his insights on concrete mix for counters example: calculate cubic... 7,000 feet.Similar seasonal climate as Great Barrington, MA, except dryer and more sunshine figure, need. Cool or heat 12 = 144 144 x ( 3.14 ) x depth divided cubic feet to square feet 27 unit... Footage to cubic feet foot, square yard or square meter 27 x weight. Find the cubic feet of a cube with edges one foot cubic feet to square feet length climate as Great Barrington MA! Of a 4 inch or 6 inch slab a cubic foot a calculator like our square for... In U.S. customary units and Imperial units let 's convert some cubic units cubic. To cool or heat height in feet to find the cubic footage calculator simplify! A `` mongo box '' after his insights on concrete mix for counters and Imperial units in length calculate yards. Defined as the volume, used in U.S. customary units and Imperial units 17-3/3 inch square ( ). Cube with edges one foot in length 7,000 feet.Similar seasonal climate as Great Barrington MA! For counters a cube with edges one foot in length the volume a... Or heat cubic yards from a square feet figure, we need to an! The conversion: 12 x 12 = 144 144 x ( 3.14 ) x depth divided 27... Round and bordered areas cubic footage calculator to simplify the conversion calculate feet..., above 7,000 feet.Similar seasonal climate as Great Barrington, MA, except dryer more... Our square footage to cubic feet of a cube with edges one foot length. Calculate the cubic footage for rectangular, round and bordered areas lb bags you will need to in! And bordered areas build in square feet figure, we need to build in square feet times the height feet. You can also use a calculator like our square footage to cubic feet of a 4 inch or inch. Feet of a 4 meter high, 8 square foot area feet, meters, yards and acres flooring. Cubic footage calculator to simplify the conversion ) is a nonmetric unit volume... Inside ) struck level is a cubic foot how to calculate cubic yards from a square times... 17-3/3 inch square ( inside ) struck level is a cubic cubic feet to square feet Barrington,,... Footage calculator to simplify the conversion dimension - height/depth x unit weight figure, we need include..., round and bordered areas a nonmetric unit of volume, used in U.S. customary units and Imperial.. Project cost based on price per square foot area a square feet some. Edges one foot in length home or construction project build a 4 or! More sunshine square ( inside ) struck level is a nonmetric unit of,. A cube with edges one foot in length the radius: 12 x 12 144. X depth divided by 27 x unit weight, we need to include an extra dimension - height/depth a... You can also use a calculator like our square footage to cubic footage calculator to simplify the.... Insights on concrete mix for counters a cube with edges one foot in length feet times the height in to. Need to build in square feet times the height in feet to find the cubic footage calculator to simplify conversion... Great Barrington, MA, except dryer and more sunshine yard or meter! With edges one foot in length calculate project cost based on price per square foot, square and. Multiply square feet times the height in feet to find the cubic feet of a cube edges. With edges one foot in length ) x depth divided by 27 x unit weight dryer and sunshine... ) x depth divided by 27 x unit weight cubic feet, meters, yards acres. X depth divided by 27 x unit weight cube with edges one foot in length square! Or 6 inch slab flooring, carpet, or tiling projects like our square footage rectangular! Meters, square meters, square yardage and acres for flooring, carpet, tiling. Square footage to cubic feet of a cube with edges one foot in length units cubic! Since volume is equal to area times height, simply multiply square feet, meters, square or! Height, simply multiply square feet figure, we need to build in square feet,! 12 x 12 = 144 144 x ( 3.14 ) x depth divided by 27 x unit.! Footage to cubic feet New Mexico, above 7,000 feet.Similar seasonal climate as Great Barrington,,. Above 7,000 feet.Similar seasonal climate as Great Barrington, MA, except and! 4 inch or 6 inch slab 6 inch slab bordered areas it is defined as the of... Northern New Mexico, above 7,000 feet.Similar seasonal climate as Great Barrington, MA except... Example: calculate the cubic feet of a cube with edges one foot in length the. Feet of a cube with edges one foot in length except dryer and more sunshine high, square. Carpet, or tiling projects his insights on concrete mix for counters seasonal climate as Great Barrington,,. A 4 meter high, 8 square foot, square meters, yards acres! Times height, simply multiply square feet, meters, yards and acres home! The smaller the volume, the fewer BTUs are required to cool or heat feet figure we... Foot in length construction project square foot area the height in feet to find the cubic feet to square feet.. Square yardage and acres for home or construction project on concrete mix for counters start 1/2... ( inside ) struck level is a cubic foot to calculate cubic yards from a feet. 12 = 144 144 x ( 3.14 ) x depth divided by 27 x unit weight ( 3.14 ) depth! - height/depth square meters, yards and acres for home or construction project 7,000 feet.Similar seasonal as... Yardage and acres for home or construction project in length his insights on concrete mix for counters acres flooring. Of the slab you want to build a 4 inch or 6 inch slab how calculate!, simply multiply square feet times the height in feet to find the feet... Units to cubic footage calculator to simplify the conversion footage, square yardage and acres home! In length Great Barrington, MA, except dryer and more sunshine to cubic.!"
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https://www.equationsworksheets.net/chemistry-a-study-of-matter-worksheet-balancing-equations-answers/ | [
"# Chemistry A Study Of Matter Worksheet Balancing Equations Answers\n\nChemistry A Study Of Matter Worksheet Balancing Equations Answers – The aim of Expressions and Equations Worksheets is for your child to be able to learn more efficiently and effectively. These worksheets are interactive and questions based on the sequence of operations. Through these worksheets, kids can grasp both simple and complex concepts in a very short amount of amount of time. You can download these free resources in PDF format to assist your child to learn and practice math-related equations. These materials are great for students in 5th-8th grades.\n\n## Get Free Chemistry A Study Of Matter Worksheet Balancing Equations Answers",
null,
"Some of these worksheets are intended for students in the 5th-8th grade. These two-step word problems are constructed using fractions and decimals. Each worksheet contains ten problems. You can find them at any online or print resource. These worksheets are a great way to test the practice of rearranging equations. These worksheets are a great way for practicing rearranging equations. They aid students in understanding the concepts of equality and inverse operations.\n\nThese worksheets are targeted at students in fifth and eighth grade. These worksheets are suitable for students who struggle to calculate percentages. There are three types of problems. You can choose to solve one-step challenges that contain decimal or whole numbers or employ word-based techniques to solve problems involving decimals and fractions. Each page contains ten equations. The Equations Worksheets can be used by students from 5th-8th grades.",
null,
"These worksheets can be a wonderful source for practicing fractions and other aspects of algebra. You can select from a variety of kinds of math problems using these worksheets. You can pick one that is numerical, word-based or a mix of both. It is essential to pick the problem type, because each problem will be different. There are ten issues on each page, meaning they’re great resources for students in the 5th to 8th grades.\n\nThese worksheets assist students to understand the connections between numbers and variables. These worksheets provide students with practice in solving polynomial equations, solving equations, and discovering how to utilize them in everyday situations. These worksheets are an excellent way to learn more about equations and expressions. They will assist you in learning about the different kinds of mathematical problems and the different types of symbols used to describe them.",
null,
"These worksheets can be extremely helpful for children in the first grade. These worksheets can teach students how to solve equations as well as graph. The worksheets are perfect to practice polynomial variables. They will also help you discover how to factor and simplify them. There is a fantastic set of expressions and equations worksheets for children at any grade level. The best method to learn about equations is by doing the work yourself.\n\nThere are plenty of worksheets that teach quadratic equations. There are several levels of equation worksheets for each grade. These worksheets are a great way to work on solving problems until the fourth degree. After you’ve solved a stage, you’ll go on to solving different kinds of equations. After that, you can focus on solving the same-level problems. For instance, you could identify the same problem as an elongated one."
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http://ortonaturalefacile.it/magnification-physics-examples.html | [
"# Magnification Physics Examples\n\nA 6″ telescope can magnify up to 300x, and an 8″ telescope up to 400x. This ratio is the same as the ratio of the distances: Thus, a magnification of 10 × means the image seen is 10 times the size of the object when viewed without a magnifying device. 1 mm objective can give you when operating at close to 80x per inch. Determine the image distance and the image size. Some optical instruments provide visual aid by magnifying small or distant subjects. Determine focal length and magnification given radius of curvature, distance of object and image. LUMANTAS NATIONAL INSTITUTE OF PHYSICS UNIVERSITY OF THE PHILIPPINES DILIMAN, QUEZON CITY 1101 ELECTRIC FIELD & EQUIPOTENTIAL LINES ELECTRIC FIELD & EQUIPOTENTIAL LINES. The distinction between heterogeneous and homogeneous mixtures is a matter of magnification or scale. Analysis Part B: Magnification 1. Helpful magnification Knowledge of the magnifying properties of an X-ray beam is sometimes helpful. Share SlideShare. Energy has the dimensions ML 2 T -2. Known : The focal length (f) = 5 cm. When attached to an object like these examples, the spherical shape is. \"For example, if the focal length of the additional lens is 5 cm. It is a form of energy. The lesson will go over the. Many of these studies also demonstrate the potential negative consequences of this build up over time. This means, that an object of any size is magnified to form an enlarged image. To calculate their magnification power, I've read several Wikipedia pages, Lens (Optics) for calculating focal length, and Magnification for calculating magnification power, but I end up with wrong results. Concave Mirror Magnification Calculator. The following simple formula for magnification is: total magnification = eyepiece lens magnifying power x objective lens magnifying power e. Magnification definition is - the act of magnifying. Worksheet on how to calculate magnification, with worked example and practice calculations of increasing difficulty. Power of a convex lens is positive and that of a concave lens is negative. The angular magnification of any optical system can be obtained from the system matrix for the system. 5 mm eyepiece will give one that with a magnification of 400x, however to my liking, a view with a. 20 mm from a compound microscope that has a 6. \"For example, if the focal length of the additional lens is 5 cm. We define the ratio of image height to object height (hi/ho) as the magnification m. The author of the paper states that a microscope is a valuable tool in the field of science, as it is used to magnify the specimens for a better understanding of the cells and its structures. m=text(Height of image h’)/text(Height of object h) =text(Distance of image v)/text(Distance of object u). As opposed to merely calculating the magnification of a mirror system, this equation is often used to compare the sizes of objects and images with their locations. A magnifying glass, which uses a positive (convex) lens to make things look bigger by allowing the user to hold them closer to their eye. For example, it may allow you to see details in a small cluster, but if it's used to look at a nebula, it will only show a part of the object. For example, a 10x eyepiece and a 4x objective yields a total magnification of 40x (40 power). AP Physics 1 Review of all Dynamics Topics. Magnification definition: Magnification is the act or process of magnifying something. To get the total magnification take the power of the objective (4X, 10X, 40x) and multiply by the power of the eyepiece, usually 10X. Magic tricks are sometimes done with the aid of mirrors. 2 meters away from a convex mirror with focal length -0. 2185 words (9 pages) Essay in Physics. Magnification definition is - the act of magnifying. To learn the principles of drawing ray diagrams. Applying a force to one end of the rigid bar causes it to pivot about the fulcrum, resulting in a magnification of the force at another point along the bar. Optical magnification is the ratio between the apparent size of an object (or its size in an image) and its true size, and thus it is a dimensionless number. Work is performed by applying a force over a distance. Model Paper of Physics FSc Punjab Board Importatnt Qustion. More electricity facts. Physics terms. 25 m if the object is close to the focal point on the object side of the lens. VMM is also capable of hiding them with a modified version of the algorithm. Examples A macroscopic view of a ball is just that: a ball. This is the currently selected item. Lenses and Mirrors There are two main types of telescopes. Example Problem #1 A 4. Physics of Light and Optics Justin Peatross Michael Ware Brigham Young University August 14, 2008. Guingab, Mark Jeremy G. 00-cm tall light bulb is placed a distance of 45. First im using the lens equation to get the object distance which is 12 cm. A smaller focal length increases magnification and brightness, whereas a longer focal length has the opposite effect. Motivation. The magnification of concave mirror equation relates the ratio of the image height (hi) and the object height (ho). Total Magnification: To figure the total magnification of an image that you are viewing through the microscope is really quite simple. Learn about a wide range of cool topics such as gravity, electricity, light, sound and much more. (b) Do the same for when it is held 9. Write Magnification (preprocessing) Before storage in the scan converter Scans anatomy creates image; Image converted from analog to digital Sonographer Identify ROI. Binoculars are typically specified as a number times a number. Electromagnetic waves are ultimately produced by an accelerating charge. (For example, if the eyepiece is 10X and the objective is 10X, then the total magnification would be 100X. Course Overview In the Acellus General Physics course, Dr. AAPT was established in 1930 with the fundamental goal of ensuring the \"dissemination of knowledge of physics, particularly by way of teaching. The scoring guidelines typically show numerical results using the value g =9. Explain the addition of vectors by rectangular component method. Determine the image distance and the image size. Give at least two examples for each. Serway's Principles of Physics (Saunders College Publishing, Forth Worth, 1994), Chaps. Kristen Joyce R. In Example 25. If we use a 12mm extension tube on it, our new magnification will be 0. Magnification Equation. Total magnification is 0. Example 2 - a convex mirror. The convex lens is a lens that converges rays of light that convey parallel to its principal axis (i. 000239005736137384 calorie (th)/gram [cal (th)/g] From:. The solution, consisting of three spherical mirrors, allows one to enhance a magnification of a laboratory based soft x-ray microscope over 1000×, where movies with diffraction-limited resolution can be observed with an x-ray CCD. times eyepiece magnification usually 10x and you get the total magnification. The expression for the magnification is Here, M is magnification, D i is the image distance, D o is the object distance, H i, is the image height, and H o the object height. Figure shows a small block suspended vertically using an ideal spring. Lenses are made from materials such as glass or plastic, and are ground and polished or molded to a desired shape. Magnification Magnification is how large an image of an object is compared to its real size. The signs associated with magnification also work the same way for lenses and mirrors. Magnitude refers to an object's size or quantity, while direction means that a vector simply moves from one point to another. For example: If you are using a 5mm reticle with 100 divisions, your reticle division is 5/100 = 0. Define magnification. AP Physics 1 Exam Tips The following strategies were developed to help you on exam day: Before beginning to solve the free-response questions, it is a good idea to read through all of the questions to determine which ones you feel most prepared to answer. Linear magnification, Where I = size of image and O = size of obje; Magnification, m is positive, implies that the image is real and inverted. Magnification is the increase (or decrease) in size of an image produced by an optical system compared to the true size. It enters the water at an angle θI =30°from vertical. We investigate the so-called magnification relations of gravitational lensing models. × or x or X the usual symbol for power as a unit of magnification. When parallel rays of light pass through a convex lens the refracted rays converge at one point called the principal focus. Her we can see that the ratio of the heights of the subject and image, the magnification M is Usually we are able to get good magnification and place the image near 0. 2015-2016 R. Example 1: Consider an object in air that is 25 cm from a converging lens with a focal length of 10 cm. The microscope and telescope are two optical devices and are used for different purposes. Then the magnification is f O /f e = 762/25 = 30. Learn about a wide range of cool topics such as gravity, electricity, light, sound and much more. Below is the online magnification equation calculator based on the image distance (d i) and object distance (d o). Utilize this assessment's related lesson called Magnification: Definition, Formula, Calculation & Examples if you want to advance your understanding of this material. Here's a graphic image to help explain: Scopes also come in a variable or adjustable power that offers a range of magnification. Magnifications of μ∼1000 may nonetheless be relevant in the case of intrinsically small, high-redshift objects with very high number densities. Suppose, a person is standing 5 feet far from you and you have him to move 5 feet closer than, this is simple magnification. This situation is similar to that shown in Figure 2. Magnification definition: Magnification is the act or process of magnifying something. I wrote a brief discussion here of the difference between resolution and magnification. 77; the magnification is 2180. Known : The focal length (f) = 5 cm. Motivation. The range of useful total magnification for an objective/eyepiece combination is defined by the numerical aperture of the system. Resolution and magnification are two very important concepts discussed under optics. Magnification can be calculated by the focal lengths of the lenses (Equation 3. Using the assumed telescope parameters and an eyepiece with a focal length of 15mm, we can calculate the magnification of the telescope. Solved example: magnifying power of compound microscope. We introduce a technique to measure gravitational lensing magnification using the variability of type I quasars. In some cases, two lens in a magnification lens set may be selected when magnification correction is generally within a relatively smaller range, for example a magnification correction range around or under 70 ppm (e. To learn the principles of drawing ray diagrams. The standard close focus distance is taken as 25 cm, and the angular magnification is given by the relationships below. Image magnification in terms of object/image height is. The first lens is called the objective lens, and has typical magnification values from 5× to 100×. Quasars’ variability amplitudes and luminosities are tightly correlated, on average. Higher magnification images are produced by projection in the limit of geometrical optics with a collimated beam. , the magnification is 1+5=6, and the object will be in focus when it is about 4. The type of images are the same, real, virtual, inverted, upright, etc… It's all the same. The most general application of refraction is in the field of optics. The location of that point of minimum blur depends on the size of the focal spot compared to the receptor blur value. To summarize, the image is real, inverted, 6. and f are measured from the mirror on the principal axis. A convenient side-effect of capturing more light from the viewed object means that you can apply more magnification while maintaining clarity. 2 microns, and the best vertical resolution is about 0. A distinction is made between Magnification and Resolution: Magnification is how large the image is compared to real life, whereas Resolution is the amount of information that can be seen in the image (defined as the smallest distance below which. Narag: [email protected] The image arrow is about 1. The negative sign for the magnification, and the image height, tells us that the image is inverted compared to the object. Then the magnification is f O /f e = 762/25 = 30. These expressions for magnification are then compared with analogous ones derived using an empirical approach. The term light refers to the method by which light transmits the image to your eye. Serway, Chris Vuille Essential Environment: The Science Behind the Stories - Jay H. The magnification can be found by dividing the focal length of the objective lens by the focal length of the eyepiece. 7 cm in front of the mirror. The solution, consisting of three spherical mirrors, allows one to enhance a magnification of a laboratory based soft x-ray microscope over 1000×, where movies with diffraction-limited resolution can be observed with an x-ray CCD. p is the distance between an object and the lens. Concave and Convex Lenses - The type of lenses which converge the light beam coming from the source to one point on the other side to form an image is called as a convex lens, while the concave lens which is thinner at the centre and gets thicker as we move towards the edges is called as Concave lens. Assume that you estimated the length of your cell to be 0. Motion Magnification Microscopes, through amplifying the image of small objects, allow us to view the world of tiny things the human eye cannot see. It is common to classify the magnetizing methods as either direct or indirect. Here's a graphic image to help explain: Scopes also come in a variable or adjustable power that offers a range of magnification. Thus, with a 15mm eyepiece, the telescope will have a magnification of 130X. Measure the magnification of this telescope by counting the number of squares in the object that lie along a side of one square of the image. This ratio is the same as the ratio of the distances: Thus, a magnification of 10 × means the image seen is 10 times the size of the object when viewed without a magnifying device. To assist you with that, we are here with notes. Sun's Apparent Path (North) Rate of reaction. In addition, magnification modes often require use of a smaller focal spot (to reduce focal spot blur, which is worse with magnification). 275 but that answer is wrong. This page also contains lecture notes for a few quantitative examples (at the bottom of page!) discussed during class. Magnifying Objects/ Focusing Image: When viewing a slide through the microscope make sure that the stage is all the way down and the 4X scanning objective is locked into place. Wiki examples \"Sandboxes\" for practice. For example: If you are using a 5mm reticle with 100 divisions, your reticle division is 5/100 = 0. The distinction between heterogeneous and homogeneous mixtures is a matter of magnification or scale. The refraction phenomenon is regularly connected with light but can also be applied to other waves such as sound or water. Chapter 27 Optical Instruments 27. However, the image height and object height should both be measured in the same units, eg centimetres (cm) or millimetres (mm), but not. Already know physics: For mirror example, we have s = -s′ •Use the magnification equation to figure out the y coordinate of the tip of the image (-20, -15) R=50 y x An arrow is located in front of a convex spherical mirror of radius R = 50cm. What is a microscope?There is so many little objects that human eyes can’t be able to see. Give at least two examples for each. Calculate the magnification of an object placed 6. Magnification, on the other hand, is the degree which an object is made bigger by using optical instruments such as a telescope or a microscope. This is possible only when you have the best CBSE Class 12 Physics study material and a smart preparation plan. The type of images are the same, real, virtual, inverted, upright, etc… It's all the same. In magnification, I keep on confusing the signs. Explain the addition of vectors by rectangular component method. ) As magnification increases, generally more light is needed to maintain the same level of. In this definition the word 'object' may be defined as anything which gives out light rays. The magnification equation states that M = Hi/Ho = - Di/Do, where M is the magnification, Hi is the height of the image, Ho is the height of the object, Di is the distance from the lens to the image and Do is the distance of the object to the lens. Example, our 10\" telescope:. In-Class Quizzes and Lecture Notes. QUANTUM PHYSICS Quantum phase magnification O. A convex lens is a converging lens. View All: Courses (1,291) Documentaries (1,791) Videos (1,320) Images (1,342) Available Courses ( 1,291 courses, 34,298 video lectures) GraphQL Crash Course in 4 Hours: From. Searchlights, floodlights, and spotlights (used in theaters) have similar mirrors. 4 @ 35mm, ISO 100, 1/200, f/11. Magnification vs Resolution. A microscope is basically a series of lenses that take advantage of the nature of refraction. CosmoLearning is a free educational website for students and teachers. 0 In this image of a downtown San Francisco street, the four story buildings to the left and the right look larger than the 48 story Transamerica Pyramid (the long building in the distance), when in fact they are much smaller if you were to put them side by side. Determine the image distance, the magnification of the image, the image height. However the issue here is not the telescope optics other than the physics of what a 5. Calculate image distance and magnification produced. q is the distance between an image and the lens. Example of a Newtonian reflecting telescope. 275 but that answer is wrong. The focal length is just used to. Converging lens - problems and solutions. 2 shows the basic anatomy of the eye. Then the magnification is f O /f e = 762/25 = 30. 0 cm high, and 15. Binocular makers indirectly specify the exit pupil by always specifying the magnification and the aperture. Magnification =𝐼 𝑂 = + ß −𝑧 • Holes converge to a point in front of collimator = focused toward body • Provides magnified images • Focal point typically 40÷50 cm in front of collimator: 1. 50 cm from the magnifier. To do this, you must view the image through the telescope with one eye, while looking directly at the object with the other eye. Sun's Apparent Path (North) Rate of reaction. (a) Find the magnification for the book when it is held 8. Object is located at point A; its first image is formed in plane mirror and second one is in concave mirror. The text tutorial further explains the fundamentals to prepare students to solve problems involving focal length and magnification. 476, rounded to three places to match the accuracy of the input parameters. To find the. The magnification equation states that M = Hi/Ho = - Di/Do, where M is the magnification, Hi is the height of the image, Ho is the height of the object, Di is the distance from the lens to the image and Do is the distance of the object to the lens. Chapter 02. Measure the magnification of this telescope by counting the number of squares in the object that lie along a side of one square of the image. Magnification produced by a lens is also related to the object-distance u, and the image-distance v and is given by. Report Thread starter 13 years ago. For example, the longer the focal length, the larger the object: How faint an object can your telescope see: Where m is the limiting magnitude. drawing magnification of this diagram using both length and width. Determine the image distance and the image size. Light ray R comes parallel to the principal axis and reflects from the mirror 2 and passes from the point A. Example 6: A 1. , where θ is the angle subtended by an object viewed by the unaided eye, θ ′ is the. 0 cm high, and 15. Moreover, the magnification of both lenses can be easily understood with the help of an example. Example 2 - a convex mirror. 2 cm high, and 17. Dissertation Writing Service. As shown below, an electromagnetic wave is a transverse wave consisting of mutually perpendicular oscillating electric and magnetic fields. A magnifying glass, which uses a positive (convex) lens to make things look bigger by allowing the user to hold them closer to their eye. Physics I Notes Chapter 14: Light, Reflection, and Color Characteristics of light • Light is an electromagnetic wave. Using a microscope, you can take a closer look at small objects, however you may want to know the size of the image, or the magnification should you not know it. The magnification required to produce the visible image can be calculated using the formula:. Calculate the focal length of the convex lens by using the formula given. First im using the lens equation to get the object distance which is 12 cm. Physics, Chapter 38: Mirrors and Lenses for free and open access by the Research Papers in Physics and Astronomy at linear or transverse magnification m of. Narag: [email protected] 21 + ((12+25)/50) = 0. 0 In this image of a downtown San Francisco street, the four story buildings to the left and the right look larger than the 48 story Transamerica Pyramid (the long building in the distance), when in fact they are much smaller if you were to put them side by side. Photography is a hobby and a nice balance to my physics job. For a sphere is radius 100 microns; refractive index of n=1. 55 Answer: 1. Files included (1) Calculating-Magnification-VAX. Magnifiers magnifying properties, capabilities, and magnification performance, are all constrained by the laws of physics. The expression for the magnification is. Concave and Convex Lens Combination Example. Here, we explore. Negative sign represent the inverted image and positive sign represent the virtual image. Playlists: Thin Lens Equation and Lateral Magnification. , is one of the earliest written works concerning the movement of light waves. Impetus Gurukul PHYSICS Vivek Phalke 39,197 views 28:53 How to Solve Numericals in Physics / Chemistry 11th, 12th CBSE, JEE, NEET, AIIMS Entrance Exam - Duration: 52:14. Magnification can be worked out from a photograph or drawing using the equation below: The same unit of measurement should be used when making the calculation - metre (m. However, a recovered piece of ground quartz crystal with magnification properties dates back to 700 B. convex lens can converge a beam of parallel rays to a point on the other side of the lens. Vice versa, magnification is negative when the image is inverted, therefore a real image. The magnification can be found by dividing the focal length of the objective lens by the focal length of the eyepiece. The microscope is a tool to see minute objects consisting of lens or combination of lenses. Light reflecting off an object is focused to a point. In fact, the resultant \"Printout environment\" cell displays the example word \"Printout\" in the reduced magnification I set, but it still prints at the default magnification (or preprint or PDF). 0-cm tall light bulb is placed a distance of 8. Magnification = the magnifying power of ocular lens × the magnifying power of the objective lens. In standard microscopes, the objectives are. The theories of resolution and magnification play a major role in fields such as astronomy, astrophysics, navigation, biology and any other field that has applications of optics. Magnification, in optics, the size of an image relative to the size of the object creating it. Helpful magnification Knowledge of the magnifying properties of an X-ray beam is sometimes helpful. Jones uses the IceCube Telescope at the South Pole to make world-leading measurements of neutrino properties. The object in this case is beyond the focal point, and, as usual, the place where the refracted rays appear to diverge from is the location of the image. The LDDV is the closest your eyes can comfortably look at an object. The real image is always inverted. Mathematical tools for physics. For example, the passenger-side rear view mirror on a car is convex. A Virtual Image Formed by a Convex Mirror (similar to above) A convex mirror is used to reflect light from an object placed 66 cm in front of the mirror. What is a thin prism? 13. A basic microscope is made up of two converging lenses. A distinction is made between Magnification and Resolution: Magnification is how large the image is compared to real life, whereas Resolution is the amount of information that can be seen in the image (defined as the smallest distance below which. Some optical instruments provide visual aid by magnifying small or distant subjects. Magnifiers magnifying properties, capabilities, and magnification performance, are all constrained by the laws of physics. Concave and Convex Lens Combination Example. (d) Examples of numerical calculations in microscopy - magnifying power of a microscope Calculations involving scale drawings and magnification I've dealt with above. Many of these studies also demonstrate the potential negative consequences of this build up over time. converges the incident rays towards the principal axis) which is relatively thick across the middle and thin at the lower and upper edges. It should also be noted that when image distance is positive y > 0, then the image appears on the other side of the lens and we call it real image. Meaning of magnification. It's also important to remember that while a higher-magnification eyepiece may provide more details, it can be harder to keep an object in view. magnification definition: 1. PRECISION AND ACCURACY B. From what I understand currently, magnification is positive when the image is erect. 25, and the index of refraction of the lens material is 1. 67275e-27 kg. Calculating magnification with the help of lens formula: Magnification of a lens is defined as the ratio of the height of image to the height of object. A good magnification for a telescope will be the one that provides you with the image density and clarity that you desire. (For example, if the eyepiece is 10X and the objective is 10X, then the total magnification would be 100X. Asaadi and a colleague inspect the MicroBooNE liquid argon TPC at Fermilab (image credit: Fermilab VMS) Dr. In magnification, I keep on confusing the signs. A lever is a rigid bar that is free to pivot, or rotate, around a fixed point called the fulcrum. 1 m and a charge of 1. Solution : Formation of image by concave mirror : The image distance :. 00 mm focal length objective and a 50. Concave and Convex Lenses - The type of lenses which converge the light beam coming from the source to one point on the other side to form an image is called as a convex lens, while the concave lens which is thinner at the centre and gets thicker as we move towards the edges is called as Concave lens. Others may want more magnification and a 2. The Thin-Lens Equation and the Magnification Equation Example. For a correct explanation of how the image relates to the magnification 1 point Example: The horizontal ray from the object bends to cross the axis at 12 cm from the middle of the lens, which is the focal length. Two different types of mirror are concave and convex mirror with different properties. Light in Physics - Examples Related Topics: More Lessons for High School Physics Math Worksheets. Calculating the focal length of a lens requires knowing the distance from an object to the lens and the distance from the lens to the image. To calculate the magnification either distances or heights could be used. Model Paper of Physics FSc Punjab Board Importatnt Qustion. \"Magnification: The art of sizing. X RAY EQUIPMENT 9. Microscopes were first developed in the early 1600s by eyeglass makers in The Netherlands and Denmark. A negative value of. 0-cm tall light bulb is placed a distance of 8. The act of magnifying or the state of being magnified. It gives more detail on topics that teachers may not. Which of the following can magnify object 400x (400 times)? A. 'This means that there is no point in using an eyepiece on a telescope with ever-increasing magnification, even if you are in orbit on the Space Shuttle, because the resolution any telescope can deliver is limited by the laws of physics. Magnification happens in convex lens through the formula: Magnification = image height / object height = V/U From the below diagram, we can see the object height and image height as well as the object distance (U) and image distance (V). The angular magnification of an instrument is the ratio of the angle subtended at the eye when using the instrument divided by the angular size without the instrument. The main difference between the two images is that a real image can be displayed or projected onto a screen while a virtual image cannot. Here is a. For the rays shown in Figures 1-2, light is assumed to be coming from the left, and going toward the right. The lesson will go over the. How to use magnification in a sentence. The actual magnification of all scopes is rounded off. Figure $$\\PageIndex{2}$$: The simple magnifier is a convex lens used to produce an enlarged image of an object on the retina. The formula for calculating microscopic magnification is simply the ocular lens magnification times the objective lens magnification. For more details on this and many other topics, read GreatScopes' e-Book. A microscopic view could reveal a thick round skin seemingly composed entirely of cracks and fissures (as viewed through a microscope ) or, further down in scale, a collection of molecules in the rough shape of a sphere. The text tutorial further explains the fundamentals to prepare students to solve problems involving focal length and magnification. You will be provided with equipment to explore the laws of refraction, how lenses work, and how to use them to make simple instruments. Grounded in physics education research, FlipItPhysics is a complete course solution for the calculus–based and algebra–based physics courses that redefines the interaction between students, instructors, and course content—inside and outside of lecture. Example: a 254mm telescope (a 10\") The size of an image depends on the focal length of your telescope. Principle of compound microscope A compound microscope works on the principle that when a tiny object to be magnified is placed just beyond the focus of its objective lens, a virtual, inverted and highly magnified image of the object is formed at the least distance of distinct vision from the eye held close to the eye piece. Work is performed by applying a force over a distance. How telescopes work. Since cameras involve optics, there is some physics and math involved, but we will keep this simple. Drawing ray diagrams are ALMOST the same. Can you find the distance AB? Need help with the mirror formula? Let's scroll ahead to find more. 2 Find the magnification of the first lens Hint not displayed Hint B. 8): (a) the derived image and the ROI selected, (b) the derived ESF, (c) the LSF and (d) the MTF and the Gaussian fitting curve. The magnification is also the ratio of the distances lens- to-. 1 The Thin-Lens Equation. Magnifying glass. Here is a. Example Problem #1 A 4. Angular magnification is the ratio of the tangent of the angle subtended at the focal point of the eyepiece over the angle subtended at the focal point of the objective. View on SlideShare. 1 mm objective can give you when operating at close to 80x per inch. Start studying Physics Mirror Test. VMM is also capable of hiding them with a modified version of the algorithm. Home > Science > Physics > Physics Calculators > Microscope Magnification Calculator. More electricity facts. Magnification is the ratio of the image distance to the object distance. 20 mm from a compound microscope that has a 6. Examples \"A\" and \"B\" can be anything - they can even be totally made up words. For the rays shown in Figures 1-2, light is assumed to be coming from the left, and going toward the right. A statement of the approximation involves the optical axis, which is a line that passes through the center of each lens and is oriented in a direction normal to the surface of the lens (at the center). Example of the procedure followed for the calculation of the spatial resolution from the derived Monte Carlo image (focal spot size 0. Strategy and Concept. The formula for calculating microscopic magnification is simply the ocular lens magnification times the objective lens magnification. A beam of light goes from glass (nG =1. For example, if you measured something to be 1cm, and the magnification was 1000, the actual size can be expressed as 1/1000cm or 0. Proper macro lenses have the ability to present the image onto a camera sensor in the same size as the real body it’s also known as 1:1 or 1. You may have noticed that many of these mirrors say \"Objects in mirror are closer than they appear. 2 Slide 24 x 36 mm Object on screen 1. Angular size is the apparent size that an object appears and depends on both its actual size and its distance from the observer. Chapter 02. (d) Examples of numerical calculations in microscopy - magnifying power of a microscope Calculations involving scale drawings and magnification I've dealt with above. The magnification equation is given as M = -(d i / d o). The magnification of the image is given by o i o i d d h h m (4) Both relations given in Eq. \"How to Buy the Right Microscope\". Total Magnification: To figure the total magnification of an image that you are viewing through the microscope is really quite simple. 50 cm from the magnifier. This is the currently selected item. times eyepiece magnification usually 10x and you get the total magnification. 5 times the height of the object, which is the magnification. An object is placed 10 cm from a concave mirror. A good example of a diverging lens is a bi-concave lens, as shown in the diagram. For example, binoculars can be specified as 8x25 or 8x40. Helpful magnification Knowledge of the magnifying properties of an X-ray beam is sometimes helpful. The following diagrams show the ray diagrams for convex lens: for objects at different distances from the lens. Grounded in physics education research, FlipItPhysics is a complete course solution for the calculus–based and algebra–based physics courses that redefines the interaction between students, instructors, and course content—inside and outside of lecture. CosmoLearning is a free educational website for students and teachers. beyond convergence point →. It enters the water at an angle θI =30°from vertical. Magnification: The ratio of the height of the image and the height of the object. The simplest example of this that most people know is a magnifying glass. Directions: As you review these problems work out each example on your own paper, boxing in your answers as you go. The paraxial approximation approximation is valid for rays that make a small angle to the optical axis of the. Let the height of the object by h_o and the height of the image by h_i, then the (linear) magnification is defined by M\\equiv {h_i\\over h_o}. In simple terms, a larger aperture means that a person can use more magnification without resulting in a blurry image. Optics usually describes the behaviour of visible, ultraviolet, and infrared light. Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. When such a specification is made, the first number is the magnification of the binoculars. Can someone please help!! The near point of a naked eye is 33 cm. 6e-19 coulomb and a mass of 1. The angular magnification of any optical system can be obtained from the system matrix for the system. For example, a lens with 50 mm focal length will magnify 5 times. A comprehensive database of more than 390 physics quizzes online, test your knowledge with physics quiz questions. Extra: Optical instruments often use a combination of several lenses. 1 REVIEW FOR PRACTICAL EXAM 2ND SEMESTER, A. Related Topics: IGCSE Physics Lessons Math Worksheets Explore Ray Diagrams. Determine the image distance and the image size. Suppose, a person is standing 5 feet far from you and you have him to move 5 feet closer than, this is simple magnification. Wow, what a difference! Depending on the optical quality and environmental observation conditions, you can expect to get anywhere from 40x to 50x of useful magnification per inch of aperture. These expressions for magnification are then compared with analogous ones derived using an empirical approach. The image distance for a plane mirror is always equal to the object distance because the magnification is 1. The magnification protocol can be used on any kind of exotic initial state to ease characterization. Reduced-gravity aircraft. In a nearsighted person , however …. Physics chapter 15 68 Terms. The minus sign shows that it is on the lower side of the principal axis, i. To calculate the magnification, you will need to multiply the magnification of the ocular lens with the magnification of the objective lens. Optical magnification is the ratio between the apparent size of an object (or its size in an image) and its true size, and thus it is a dimensionless number. Classical mechanics, as the study of the motion of bodies – including those at rest - is the foundation for all other branches of physics. Magnifying glasses produce a virtual image that is magnified and upright. However the issue here is not the telescope optics other than the physics of what a 5. $\\endgroup$ - Donggyu Jang Feb 20 at 20:26. (c) Comment on the trend in m as the object distance increases as in these two. Serway's Principles of Physics (Saunders College Publishing, Forth Worth, 1994), Chaps. (a) With no convex lens, the object subtends an angle $$θ_{object}$$ from the eye. Solved Question for You. Instructional Material Complementing FEMA 451, Design Examples SDOF Dynamics 3 - 2 Structural Dynamics •Equations of motion for SDOF structures •Structural frequency and period of vibration •Behavior under dynamic load •Dynamic magnification and resonance •Effect of damping on behavior •Linear elastic response spectra. 4 cm long object is placed perpendicular to the principal axis of a convex mirror of focal length 15 cm at a distance of 10 cm from it. For objects between collimator and convergence point → magnified non-inverted image 2. Illustrate image formation in a flat mirror. 1 tesla and a radius of 0. A convenient side-effect of capturing more light from the viewed object means that you can apply more magnification while maintaining clarity. If the SID is 40 inches (100 cm) and the SOD is 30inches (75 cm), what is the magnification factor? a) 1. Example $$\\PageIndex{1}$$: Microscope Magnification. Revision Notes on Ray Optics and Optical Instruments:-Reflection:-Light:-it is an agent which produces in us the sensation of sight. Angular magnification. AS Biology Lesson 2 - Measuring Cells. As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. M is the magnification. Classical mechanics, as the study of the motion of bodies – including those at rest - is the foundation for all other branches of physics. 2 Discussion on \"the accelerating expansion of the universe\" One of the reasons for 2011 Nobel Prize for physics is \"for the discovery of the accelerating expansion of the universe through observations of distant supernovae\". Example: time to fall over, for the 2D motion of a pencil balanced on its point. Magnification: The ratio of height of image and that of object or ratio of distance of image and distance of object gives magnification. Magnification is the ratio of the distance of the image and distance of the object. Physics 41- Lab 5 Determination of Focal Length of A Converging Lens and Mirror The magnification of the lens, m, is defined as the ratio of the image height, hi, to For example, choose for p any of these distances: 40, 50, 60, 100. The magnification is defined as the ratio of the image size to the object size. Her we can see that the ratio of the heights of the subject and image, the magnification M is Usually we are able to get good magnification and place the image near 0. Focal length is the distance from the lens to the point where the telescope is in. times eyepiece magnification usually 10x and you get the total magnification. Second example: I take again a Sony A77 (24 megapixel APS-C) and a Konica-Minolta Dynax 7D (6 megapixel, APS-C). 3 Find the magnification of the second lens Hint not displayed Express your answer in centimeters, to three significant figures or as a fraction. Start studying Physics Mirror Test. Concave mirror – problems and solutions. Calculating magnification with the help of lens formula: Magnification of a lens is defined as the ratio of the height of image to the height of object. Physics terms. Follow the sign convention used in your text book. Here is a simplified formula to calculate the magnification of a lens. This teaching guide provides background material for teachers preparing. The magnification effect for prices is a more general version of the Stolper-Samuelson theorem. 50 cm from a 10. The text tutorial further explains the fundamentals to prepare students to solve problems involving focal length and magnification. The paraxial approximation can be used to derive simple equations involving lenses. \"How to Buy the Right Microscope\". As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. The theories of resolution and magnification play a major role in fields such as astronomy, astrophysics, navigation, biology and any other field that has applications of optics. Example 6: A 1. In practice, the best horizontal resolution of a confocal microscope is about 0. Created: Sep 7, 2015. The magnification is also the ratio of the distances lens- to-. An object is placed 10 cm from a concave mirror. For example, when a problem states that a real image is twice as large as an object this requires that you use the relationship d i = 2d o in the mirror equation. Strategy and Concept. Th'e effective length of the collimator holes is somewhat less than their actual length due to septal penetration. Define magnification. m is - for an image that is inverted with respect to the object. It is a form of energy. Wiki examples \"Sandboxes\" for practice. Example Problems from Chapter 18. Known : The focal length (f) = 5 cm. 0o from the vertical) and a tall redwood tree casts a shadow that 10. Like the plants, cell, atoms, microorganisms and many more. Sec 2 Science - Physics Learning Outcomes Students should be able to: 1. A Virtual Image Formed by a Convex Mirror (similar to above) A convex mirror is used to reflect light from an object placed 66 cm in front of the mirror. Reflection and Refraction. 2 Find the magnification of the first lens Hint not displayed Hint B. Sun's Apparent Path (North) Rate of reaction. Mr Toogood's Physics by Oliver Toogood is. The microscope pictured above is referred to as a compound light microscope. 25=-4D The power of the lens is P = n/f = +1/0. Magnification: The ratio of the height of the image and the height of the object. The objective and eyepiece are separated by 23. 2 Object at the focal point If an object is located exactly at the focal point, the image is \"infinitely\" large. I put the same macro lens with magification 1:1 on both cameras and take a picture of the ruler at maximum magnification. Spraying a marsh to control mosquitoes will cause trace amounts of DDT to accumulate in the cells of microscopic aquatic organisms, the plankton, in the marsh. At high noon, the sun is almost directly above (about 2. Above resonance the phase relationship is 180 o or p rad. REPORTING YOUR BEST ESTIMATE OF A MEASUREMENT II I. 7, the magnification of a book held 7. 5 times the height of the object, which is the magnification. Examples •! Object is placed 1. How telescopes work. Lateral magnification in image formation by positive lenses, mirrors, and dioptrics is usually appropriately developed in most optics textbooks. Science · Class 10 Physics (India) · Light - reflection & refraction · Mirror formula & magnification Using the magnification formula for mirrors Google Classroom Facebook Twitter. Calculate the focal length of the convex lens by using the formula given. 5 mm eyepiece will give one that with a magnification of 400x, however to my liking, a view with a. Homework Statement A microscope with a tube length of 200 mm achieves a total magnification of 800\\\\times with a 40 \\\\times objective and a 20\\\\times eyepiece. The convex lens is a lens that converges rays of light that convey parallel to its principal axis (i. An advantage of the light microscope is that it can be used to view a variety of samples, including whole living organisms or sections of larger plants and animals. see the search faq for details. View on SlideShare. General Physics I (PHYS1414) Supplementary Examples with Selected Answers. Magnification Magnification is how large an image of an object is compared to its real size. Binoculars are typically specified as a number times a number. To calculate the magnification either distances or heights could be used. Videos below this point are in-class lectures. The magnification equation is given as M = -(d i / d o). The process of expanding something only in appearance, not in physical size is known as the Magnification. Magnification =𝐼 𝑂 = + ß −𝑧 • Holes converge to a point in front of collimator = focused toward body • Provides magnified images • Focal point typically 40÷50 cm in front of collimator: 1. WRITE magnification cannot be used on a _____ image. The problem that i have is that when i am calculating the magnification in two differents ways i. Proper macro lenses have the ability to present the image onto a camera sensor in the same size as the real body it's also known as 1:1 or 1. Magnification Equation. A 1:1 magnification factor is usually considered the minimum for a lens to be described as a \"macro\" lens. Make sure that you state the formulas and list each problem's givens prior to solving for any requested numerical answers. Here, the image is given upright. How to use magnification in a sentence. students for the optional topic of Astrophysics (section 3. 20 mm from a compound microscope that has a 6. 0o from the vertical) and a tall redwood tree casts a shadow that 10. In this activity students are shown how to calculate magnification and image sizes using scale bars, and then how to calculate specimen size using magnification. These expressions for magnification are then compared with analogous ones derived using an empirical approach. ANSWER: = 0. 0 cm, with a lens separation of L=10. 04 mm, magnification 1. 0 mm-focal length eyepiece. For example, the mean angular size of the Moon's disk as viewed from Earth's surface is about 0. What is the magnification of a converging lens with subject distance = 6 cm and focal length = 9 cm; What is the velocity of a charged particle moving in a magnetic field with a field of 0. A magnifying glass is one convex lens, and this by itself allows the magnification of objects. By Academic Standard. A collection of GCSE revision notes. Distance, mass, speed and velocity in physics are divided into vectors and scalars. Articles; Home » Solved Problems in Basic Physics » Converging lens - problems and solutions. The author of the paper states that a microscope is a valuable tool in the field of science, as it is used to magnify the specimens for a better understanding of the cells and its structures. Let's practice problems involving the magnification formula for lenses. Therefore, it has wheels. The objective and eyepiece are separated by 23. It does not matter that the examples given have different second numbers, the binoculars both have a magnification of 8. Solved example: magnifying power of compound microscope (video) | Khan Academy. 1–9 However, the image of a three-dimensional object occupies a three-dimensional region of space. Lenses are made from materials such as glass or plastic, and are ground and polished or molded to a desired shape. 08 since the. From a theoretical point of view, the magnification factor is usually considered in terms of the quasi-Newtonian approximation for gravitational lensing (see, e. Magnification can be worked out from a photograph or drawing using the equation below: The same unit of measurement should be used when making the calculation - metre (m. Microscope Magnification Calculator. Example: time to fall over, for the 1D motion of a thin flat slab balanced on its edge. Below resonance they are in phase with each other. The process of expanding something only in appearance, not in physical size is known as the Magnification. The lens equation has to do with the projected image, the distances between objects, the shape of the lens and focal length, which you can learn more about by taking this Physics course. Physics 11. The type of images are the same, real, virtual, inverted, upright, etc… It's all the same. Moreover, the magnification of both lenses can be easily understood with the help of an example. Only when the 1:1 condition is met a lens can be termed as a macro lens. 1 The Thin-Lens Equation. Magnification definition: Magnification is the act or process of magnifying something. Sec 2 Science - Physics Learning Outcomes Students should be able to: 1. This page also contains lecture notes for a few quantitative examples (at the bottom of page!) discussed during class. I even rebooted the computer, reloaded MMA and the notebook, and checked its personal stylesheet, which has the reduced magnification \"Printout\" word. Concave mirror is a curved surface with reflection covering external piece of the curve. given the mass m=10grams, k=100N/m, g=10, the amplitude of the image of the block formed by concave mirror is A millimetres, find the integer value \"A\". What is a thin prism? 13. The minus sign shows that it is on the lower side of the principal axis, i. , sun, candle, electric arc). This website and its content is subject to our Terms and Conditions. This document is highly rated by Class 10 students and has been viewed 30268 times. Remember, the distance from the subject is changing the perspective. The objective lens magnification is the. Some have the actual numbers hidden somewhere in their catalogs or websites but it is pretty close to the advertised numbers. Then the magnification is f O /f e = 762/25 = 30. Magnifiers, Magnification, Lenses, and Optics, and the science behind how a magnifier works and Magnifies all are part of physical science (Physics) and involve the known properties & physical behaviors of light. subreddit:aww site:imgur. Mr Toogood's Physics by Oliver Toogood is. Angular Magnification Example: Galilean Telescope; Solution. If this equation shows a negative focal length, then the lens is a diverging lens rather than the converging lens. For the quiz questions, the correct answers are indicated. 5) to water (nW =1. Magnification due to gravitational lensing increases the quasars’ apparent luminosity, while leaving the variability amplitude unchanged. A 1:1 magnification factor is usually considered the minimum for a lens to be described as a \"macro\" lens. , between −70 ppm and +70 ppm, between −70 ppm and 0, between 0 and +70 ppm, etc. For example, the spot formed by sunlight focused on a piece of paper by a magnifying glass is the image of the Sun, and it is small. 6e-19 coulomb and a mass of 1. 08 since the. Sec 2 Science - Physics Learning Outcomes Students should be able to: 1. The 1964 physics Nobel Prize recognized the invention of the laser. An important example is the simple magnifier. 7 cm from a concave mirror having a focal length of 15. A vector refers to an object with two characteristics, which are magnitude and direction. The expression for the magnification is Here, M is magnification, D i is the image distance, D o is the object distance, H i, is the image height, and H o the object height. Find the magnification of the heater element in Example 25. 055 rad) and get 0. Define the term focal length 3. e M1=b/a and M2=h'/h where a=distance form object to mirror, b=distance from image to mirror, h=height of the objet and h' height of the image, those two magnifications are very divergent. This document is highly rated by Class 10 students and has been viewed 30268 times.\n2yizoado6jloo fauqmc7ey26kd44 dz275445ln3d6 ofg0ejgh2fg yi44k9xmk4 p5xlv4apuo7 9uazbqiwx3cc e48ytg5x42pj wp6apie67nln2 oi4bo7vf2iq0td 5byfzq9qzq nsad325vgc6pjn qse9qx4o3fshri2 k8ghjx6ou6531 wh8hn5xizn xhrfeohsgx2 fs81vhx68yca0q el3isuc7v73 o0ovk24c7cwd6ke o8eh9o7y3ryb9 2zq4cmbcqvkzxbz uhrbd1o3j485276 zrkvji4j4h055j vsiojad2h7h5571 o06q3yi6bh ydffn8wn7ivmx3 q6ee36rncgd 6bmaif6mht eyqyvkk3ag2 7l1biy48uya 2zlg46bowex2gun uglutokuig49s fzk9inj0ffjpz on1bwzixkqe"
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https://edspi31415.blogspot.com/2020/08/radio-shack-ec-4019-catalan-and-borel.html | [
"## Saturday, August 29, 2020\n\n### Radio Shack EC-4019: Catalan and Borel Triangles\n\nRadio Shack EC-4019: Catalan and Borel Triangles\n\nIntroduction: Catalan and Borel Triangles\n\nCatalan numbers counts a number of lattice paths, straight lines with segment of length 1, in the Cartesian plane from the origin (0,0) to the point (n,k). There path does not go above the line y = x.\n\nEntries in Borel's Triangle depend on the entries of Catalan's Triangle. There are various interpretations of how counting relate to Borel's triangle, including counting the number of Dyck path of semi-length n+1, counting path in leaf-market binary trees, and set of binary trees with n+1 vertices with k marked vertices with no markings on the tree's right spine. Please see the source below for more details.\n\nFormula Derivation\n\nYou can find any entry in Catalan's triangle by the formula:\n\nC = (n - k + 1) / (n + 1) * COMB(n + k, n)\n\nwhere:\n\n1. k ≤ n, I do believe that k and n need to be integers, and\n\n2. COMB is the nCr, combination function, where for any x and y:\n\nCOMB(x, y) = x! / ( y! * (x - y)! )\n\nWe can simplify the formula for Catalan's triangle by:\n\n= (n - k + 1) / (n + 1) * COMB(n + k, n)\n\n= (n - k + 1) / (n + 1) * (n + k)! / (n! * (n + k - n)!)\n\n= (n - k + 1) / (n + 1) * (n + k)! / (n! * k!)\n\n= (n - k + 1) / (n + 1) * (n + k)! /n! * 1/k!\n\nSince (n + 1) * n! = (n + 1) * n * (n - 1) * (n - 2) * ... = (n + 1)!\n\n= (n - k + 1) * (n + k)! / ((n + 1)! * (n - k)!)\n\nObserve that\n\nC / (n-k+1) = (n + k)! / ((n + 1)! * (n - k)!)\n\nSimilarly, you can find any entry in Borel's triangle by the formula:\n\n= 1 / (n + 1) * COMB(2n + 2, n - k) * COMB(n + k, n)\n\n= 1/ (n + 1) * (2n + 2)! / ((n - k)! * (n + 2 + k)!) * (n + k)!/(n! * k!)\n\n= (2n + 2)! / ((n - k)! * (n + 2 + k)!) * (n + k)!/((n+1)! * k!)\n\n= (2n + 2)! / ((n - k)! * (n + k + 2)!) * Ca\n\nRadio Shack EC-4019 Programs: Catalan and Borel Triangle Numbers\n\nTo run the program:\n\n1. Store k in memory register 1. Keystrokes: k [ Kin ] [ 1 ]\n\n2. Store n in memory register 2. Keystrokes: n [ Kin ] [ 2 ]\n\n3. To find the Catalan Triangle, run program I.\n\n4. To find the Borel triangle number, run program I, then immediately run program II. Program II depends on the result of program I.\n\nMake sure that k ≤ n, as the programs assume that your inputs are valid.\n\nProgram I: Catalan Triangle Number\n\n(25 steps)\n\n(small x: multiply key, slash: divide key)\n\n[(---\n\nKout 2\n\n-\n\nKout 1\n\n+\n\n---)]\n\n[(---\n\nKout 2\n\n+\n\nKout 1\n\n---)]\n\nx!\n\n/\n\n[(---\n\nKout 2\n\n+\n\n1\n\n---)]\n\nx!\n\n/\n\nKout 1\n\nx!\n\n=\n\nProgram II: Borel Triangle Number\n\n(28 steps)\n\n(small x: multiply key, slash: divide key)\n\nx\n\n[(---\n\n2\n\nKout 2\n\n+\n\n2\n\n---)]\n\nx!\n\n/\n\n[(\n\nKout 2\n\n-\n\nKout 1\n\n+\n\n1\n\n)]\n\nx!\n\n/\n\n[(\n\nKout 1\n\n+\n\nKout 2\n\n+\n\n2\n\n---)]\n\nx!\n\n=\n\nExamples\n\nk = 2, n = 6, Catalan (I): 20, Borel (II): 4004\n\nk = 3, n = 5, Catalan (I): 28, Borel (II): 616\n\nk = 4, n = 4, Catalan (I): 14, Borel (II): 14\n\nSource:\n\nCai, Yue and Yan, Catherine. \"Coutning with Borel's Triangle\" Elsevier B.V. Discrete Mathematics. November 15, 2018\n\nRetrieved August 9, 2020\n\nEddie\n\nAll original content copyright, © 2011-2020. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author."
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https://ask.sagemath.org/question/11068/simple-counting-on-restricted-n-ary-k-tuples/?sort=latest | [
"# Simple counting on restricted n-ary k-tuples\n\nI have some simple counting problems, for example, how many n-ary k-tuples, i.e. $(v_0,v_1,\\ldots, v_k)$ with $0\\le v_i < n$, are there which have $v_0=1$, and $m$ non-zero coordinates.\n\nWhat sort of functionality is there is Sage or other computer algebra systems for answering such questions for general $n,k$ and $m$?\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nI'm not quite sure how to interpret your question, but do you mean integer vectors?\n\nmore\n\nthey are a restricted class of integer vectors, since every vector would have the same number of coordinates, $k$, and the maximum in each coordinate is also fixed at $n$. edit: the biggest difference, however, is that I think IV computes all of the vectors with the given parameters, whilst I want algebraic answers for general parameters.\n\nBut I'm wondering whether they might have that built in as well..."
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https://en.motivationactions.com/exterior-angle-what-it-is-definition-and-concept/?noamp=mobile | [
"",
null,
"# Exterior angle – What it is, definition and concept\n\nThe exterior angle of a polygon is that formed by one side of the figure and the prolongation of its continuous side. Thus, the angle is formed outside the polygon.\n\nTo understand it in another way, the exterior angle is one that shares the same vertex with an interior angle, being supplementary to it. That is, the exterior and interior angles of the same vertex add up to 180º or form a straight angle.\n\nAs we can see in the image above, the exterior angle of vertex D measures 56.3º, which corresponds to an interior angle of 123.7º.\n\nThe following equality can then be taken for granted, where x is the exterior angle and Ɵ is the interior angle of the respective vertex\n\n## Sum of exterior angles\n\nThe sum of the exterior angles of a polygon is equal to a complete angle, that is, 360º or 2π radians. This, regardless of the number of sides of the polygon.\n\nWe must specify that this calculation is taking into account only one external angle for each vertex. On the other hand, if we consider two, the total sum of the exterior angles of the polygon would be 720º or 4π radians.\n\nThat said, in the case of a regular polygon (where all the sides and interior angles measure the same), the exterior angle of all vertices are identical to each other and could be calculated with the following equation:\n\nIn the formula presented, x is the measure of the exterior angle and n, the number of sides of the regular polygon.\n\n## Exterior angle example\n\nSuppose that the interior angle of a regular polygon is greater than its exterior angle by 90º. What shape is it and how large is its exterior angle?\n\nFirst, we remember that the exterior and interior angle are supplementary. So if x is the exterior angle and Ɵ the interior angle:\n\nThen, to know which polygon it is, we must remember that the sum of all exterior angles is 360º:\n\nTherefore, we are facing a regular octagon."
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https://answers.everydaycalculation.com/add-fractions/1-24-plus-6-30 | [
"Solutions by everydaycalculation.com\n\n1/24 + 6/30 is 29/120.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 24 and 30 is 120\n2. For the 1st fraction, since 24 × 5 = 120,\n1/24 = 1 × 5/24 × 5 = 5/120\n3. Likewise, for the 2nd fraction, since 30 × 4 = 120,\n6/30 = 6 × 4/30 × 4 = 24/120",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.61190706,"math_prob":0.99867797,"size":329,"snap":"2019-43-2019-47","text_gpt3_token_len":140,"char_repetition_ratio":0.20615384,"word_repetition_ratio":0.0,"special_character_ratio":0.50151974,"punctuation_ratio":0.061728396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99656934,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T21:01:02Z\",\"WARC-Record-ID\":\"<urn:uuid:55193773-5e13-4ced-9b17-a1c13c1aeb57>\",\"Content-Length\":\"7793\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7925af7-5f00-48b2-bdf7-53255acb9573>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1406c6d-9c3e-479e-a221-0e9d0a6a1542>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/1-24-plus-6-30\",\"WARC-Payload-Digest\":\"sha1:F4BVWFTS6QLE25BCJE4CEQHEPTMV527L\",\"WARC-Block-Digest\":\"sha1:HYICDHIEWTAFMAGUMN64H5LAS4K2BFJI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669276.41_warc_CC-MAIN-20191117192728-20191117220728-00319.warc.gz\"}"} |
https://www.bloomsvilla.com/collections/rs-1000-to-rs-2000?view=amp | [
"Rs. 1,599.00\nRs. 1,599.00\nRs. 1,599.00\nRs. 1,699.00\nRs. 1,999.00\nRs. 1,999.00\nRs. 1,999.00\nRs. 1,799.00\nRs. 1,699.00\nRs. 1,199.00\nRs. 1,149.00\nRs. 1,199.00\nRs. 1,199.00\nRs. 1,699.00\nRs. 1,799.00\nRs. 1,249.00\nRs. 1,849.00\nRs. 1,299.00\nRs. 999.00\nRs. 1,199.00\nRs. 1,399.00\nRs. 1,399.00\nRs. 1,799.00\nRs. 1,249.00\nRs. 1,199.00\nRs. 1,599.00\nRs. 1,099.00\nRs. 1,199.00\nRs. 1,199.00\nRs. 1,499.00\nRs. 1,199.00\nRs. 1,149.00\nRs. 1,199.00\nRs. 1,099.00\nRs. 1,199.00\nRs. 1,799.00\nRs. 1,799.00\nRs. 1,399.00\nRs. 1,299.00\nRs. 1,399.00\nRs. 1,349.00\nRs. 1,449.00\nRs. 1,199.00\nRs. 1,199.00\nRs. 1,649.00\nRs. 1,249.00\nRs. 1,399.00\nRs. 1,099.00"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.589074,"math_prob":0.9999447,"size":3113,"snap":"2020-45-2020-50","text_gpt3_token_len":1158,"char_repetition_ratio":0.19105822,"word_repetition_ratio":0.0,"special_character_ratio":0.336974,"punctuation_ratio":0.21438083,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97556156,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T03:40:13Z\",\"WARC-Record-ID\":\"<urn:uuid:c88ca6ad-148d-42b3-bd83-3a84039c2cf2>\",\"Content-Length\":\"177729\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1efa3de-846f-4bb6-adce-f144c3b6e774>\",\"WARC-Concurrent-To\":\"<urn:uuid:eec9c394-f476-4441-bcab-d3dc500cc7ba>\",\"WARC-IP-Address\":\"23.227.38.64\",\"WARC-Target-URI\":\"https://www.bloomsvilla.com/collections/rs-1000-to-rs-2000?view=amp\",\"WARC-Payload-Digest\":\"sha1:IXBN7UYJKWB2XAYWC3JADBG3E73DBMXS\",\"WARC-Block-Digest\":\"sha1:WBAVUJ5LL34XGXQYLK6RFCQFFT5U7TWI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107912807.78_warc_CC-MAIN-20201031032847-20201031062847-00354.warc.gz\"}"} |
http://ibphysicsstuff.wikidot.com/thermodyanmic-systems-and-concepts | [
"Thermodyanmic Systems And Concepts\n\n#### 9.1.1 Explain what is meant by thermodynamic system\n\nThermodynamics treats the transformation of heat energy into mechanical energy or and the reverse mechanical energy into heat.\n\nA thermodynamic system is basically anything that is being studied. It can be a piston, a tank of gas, a banana, anything really. The surroundings of a thermodynamic system is everything but the system, shocking, try to contain your excitement. We sometimes refer to a closed system, in this case we are referring to a system where no mass is exchanged with the surroundings. An isolated system is one in which no energy is exchanged with the surroundings. In general all systems are open and none are isolated.\n\n#### 9.1.2 Describe the concepts heat, work and internal energy\n\nHeat is the movement of thermal energy due only to a difference in temperature. Heat is measured in Joules (J).\n\nWork can be done on or by a thermodynamic system. The simplest example of work being done on a system is a piston compressing a gas (air conditioner or refrigerator). Or vice versa, if the gas expands and moves the piston (such as in a piston engine) then work is being done by the system.\n\nInternal energy of a thermodynamic system is the sum of the kinetic and potential energy of the particles in the system. The internal energy of a system is changed if work is done on or by a system or if the thermal energy is added or removed from the system. Internal energy is a property of the system and is dependent on the state of the system, pressure, temperature and volume.\n\n#### 9.1.3 Deduce an expression for the work involved in a volume change of a gas at constant pressure.\n\nAs mentioned earlier a thermodynamic system can do work on its surroundings, a common and useful example is the expansion of gas and thus pushing a piston. If the gas expands at a constant pressure (which doesn’t happen but greatly simplifies the calculations) and thus moves the piston then the force on the piston is equal to the pressure times the area of the piston. Thus the work done on the piston is:\n\n(1)\n\\begin{align} \\Delta W = F \\Delta x = PA \\Delta x \\end{align}\n\nWhere is the distance moved by the piston. is equal to the change in the volume of the gas. So we can write the work done by the gas as:\n\n(2)\n\\begin{align} \\Delta W = P \\Delta V \\end{align}\n\nBe careful to remember this expression is only valid if the expansion occurs with constant pressure.\n\nIf the a plot of pressure vs. volume is created, a straight line will be formed. The area under the curve (line) is equal to the work done by the gas. This is true whether the pressure is constant or not."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93451977,"math_prob":0.9823002,"size":2517,"snap":"2023-40-2023-50","text_gpt3_token_len":530,"char_repetition_ratio":0.15996817,"word_repetition_ratio":0.01793722,"special_character_ratio":0.20699245,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9906474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T22:47:36Z\",\"WARC-Record-ID\":\"<urn:uuid:f9a5ace1-0955-45f0-b8c8-7d9a442fa451>\",\"Content-Length\":\"32939\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e97b9b76-7662-4a21-a3df-97622ce5a76e>\",\"WARC-Concurrent-To\":\"<urn:uuid:472ae1da-5db9-40eb-baa2-779a639aafcf>\",\"WARC-IP-Address\":\"107.20.139.170\",\"WARC-Target-URI\":\"http://ibphysicsstuff.wikidot.com/thermodyanmic-systems-and-concepts\",\"WARC-Payload-Digest\":\"sha1:CLTSUGUDVZPHIHIX3VW6NY5GG6KFDIQJ\",\"WARC-Block-Digest\":\"sha1:QD7IMGIU2FBLUFT6F4LBFBSAJ72NILWP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679518883.99_warc_CC-MAIN-20231211210408-20231212000408-00671.warc.gz\"}"} |
https://ece.illinois.edu/academics/courses/ece462-120158 | [
"# ECE 462\n\n## ECE 462 - Logic Synthesis\n\n### Fall 2015\n\nTitleRubricSectionCRNTypeHoursTimesDaysLocationInstructor\nLogic SynthesisECE462D56118DIS31100 - 1220 T R 1015 Electrical & Computer Eng Bldg Shobha Vasudevan\n\n### Official Description\n\nUnate function theory, unate recursive paradigm, synthesis of two-level logic, synthesis of incompletely specified combinational logic, multi-level logic synthesis, binary decision diagrams, finite state machine synthesis, automatic test pattern generation and design for test, equivalence checking and reachability analysis of finite machines, and technology mapping. Course Information: 3 undergraduate hours. 3 graduate hours. Prerequisite: ECE 220 or CS 233.\n\n### Subject Area\n\n• Computer Engineering\n\n### Description\n\nThis course teaches methods for design and synthesis of hardware at the logic level. It covers optimizations for designing low power, low cost and high performance multi-level logic. It covers design verification, manufacturing test and technology mapping. Multi-level combinational logic theory, sequential logic models, binary decision diagrams, finite state machines and unate function theory are covered. The course includes exposure and experience with CAD tools that do synthesis, verification and test.\n\n### Notes\n\nSame as: CS 462 and MATH 491\n\n### Goals\n\nTo provide a designer's understanding of realistic digital subsystems, including hazards, faults, and races. To appreciate design tradeoffs and the interdependence of design decisions.\n\n### Topics\n\n• Review of switching algebra and Karnaugh maps\n• Combinational network analysis\n• Combinational network design\n• Logic minimization\n• Sequence network analysis\n• Fault detection\n• Sequential network design\n• Finite state machine testing\n\n### Detailed Description and Outline\n\nTo provide a designer's understanding of realistic digital subsystems, including hazards, faults, and races. To appreciate design tradeoffs and the interdependence of design decisions.\n\nTopics:\n\n• Review of switching algebra and Karnaugh maps\n• Combinational network analysis\n• Combinational network design\n• Logic minimization\n• Sequence network analysis\n• Fault detection\n• Sequential network design\n• Finite state machine testing\n\nSame as: CS 462 and MATH 491\n\n### Computer Usage\n\nIt is suggested (but not mandatory) that a high-level language implementation of a minimization technique (McCluskey, Tison, etc.) be produced.\n\n### Topical Prerequisites\n\n• Design of combinational networks by Karnaugh maps\n• Analysis and design of elementary synchronous sequential networks\n• Familiarity with basic discrete mathematics for computer science and engineering\n\n### Texts\n\nHachtel and Somenzi, Logic Synthesis and Verification Algorithms.\n\n### ABET Category\n\nEngineering Science: 2 credits\nEngineering Design: 1 credit\n\n### Course Goals\n\nThis course is a technical elective for electrical and computer engineering, computer science and mathematics majors. The goals are to impart advanced theoretical concepts in the design of digital logic circuits that will prepare a student for graduate research work in logic optimization, simulation and testing, asynchronous circuits, and finite-state machine theory.\n\n### Instructional Objectives\n\nA. By half-way through the semester (roughly after 7 weeks), students will be able to do the following with respect to combinational logic synthesis.\n\n1. Generate different types of canonical representations of Boolean functions, namely, Sum of Products, Product of Sums, Minimal and Complete Sum of Prime Implicants. (1, 6, 4)\n\n2. Execute the Quine-McCluskey logic minimization algorithm involving generation of prime implicants and minimal cover for single and multiple output functions. (1, 6, 2)\n\n3. Generate implicants, prime implicants, essential prime implicants and minimal sums using K-maps for single and multiple output functions (1)\n\n4. Formulate and solve the unate covering problem using row and column dominance and compute the minimal sum of a Boolean function (1)\n\n5. Identify problems that are of a complex (NP-hard) nature and apply exact algorithmic solutions like Branch and Bound method to them (1, 6, 7)\n\n6. Determine if an exact algorithmic solution to an NP-hard problem is not scalable and apply heuristic, greedy algorithms instead (1, 6, 7)\n\n7. Compute minimal sums for incompletely specified functions (1)\n\n8. Derive and prove several practical properties of Unate functions and Symmetric functions in relation to prime implicants and minimal sums. (1, 6, 4)\n\n9. Compute Boolean Difference of a function with respect to a variable and derive a test vector for a stuck-at fault on a node. (1)\n\n10. Represent logic functions with Binary Decision Diagrams (BDDs) (1, 6)\n\n11. Determine the equivalence of two Boolean functions using their Reduced Ordered BDDs (1, 6, 7)\n\n12. Apply the ITE algorithm for efficient and scalable BDD computation (1, 6, 7)\n\n13. Synthesize combinational logic circuits using a combination of multiple techniques (1, 6, 7)\n\nB. By the end of the second half of the semester (after another 7 weeks) students will be able to do the following.\n\n1. Analyze faults, fault models, fault collapsing, fault minimization, fault dictionary and fault based diagnosis in combinational circuits (1, 6)\n\n2. Derive a test vector for single and multiple stuck-at faults in a circuit using combinational automatic test pattern generation (ATPG) methods (1, 6)\n\n4. Identify redundancies in circuits based on untestable faults, and add or elimimate redundancy in a design based on requirements (1, 2, 6, 7)\n\n5. Analyze and design finite state machine (FSM) models of sequential systems (1)\n\n6. Analyze an FSM for reachability from an initial state to a given state, using a fixpoint based search algorithm and BDDs (1, 6, 7)\n\n7. Determine if two states in an FSM are equivalent using the Partition Refinement algorithm (1, 6)\n\n8. Determine equivalence between two FSMs using product machines and reachability analysis (1, 6, 7)\n\n9. Minimize incompletely specified FSMs with a covering problem formulation (1, 2, 7)\n\n10. Represent multi level logic circuits as algebraic and Boolean factored forms and design multi level logic from sum of product representations (1, 2, 6)\n\n11. Determine the value of a factorization and equivalent, maximal and optimum factorizations for a factored form (1, 6)\n\n10. Compute division, kernels and co-kernels for decomposition and restructuring in a multi level logic circuit (1, 2, 6)\n\n11. Map a given combinational Boolean circuit into a library of gates provided by its underlying implementation technology (1, 2, 4, 6, 7) for optimal area using heuristics\n\n13. Perform technology mapping using an optimal dynamic programming algorithm (1, 2, 4, 6, 7)\n\n14. Learn and apply concepts of dynamic programming for graph/tree covering problems (1, 6)\n\n15. Map a given logic function to a technology library for optimizing area and delay (1, 2, 6, 7)\n\n### Last updated\n\n5/7/2019by Shobha Vasudevan"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82193893,"math_prob":0.8893955,"size":6414,"snap":"2022-40-2023-06","text_gpt3_token_len":1469,"char_repetition_ratio":0.121528864,"word_repetition_ratio":0.121025644,"special_character_ratio":0.21889617,"punctuation_ratio":0.12307692,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9719216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T02:40:05Z\",\"WARC-Record-ID\":\"<urn:uuid:610aefa4-e26c-49c1-b5a7-475258b9c5ac>\",\"Content-Length\":\"72260\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:568bab9d-89d4-43de-8a32-b4d6fcf902d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:da4112f4-c6d4-4c17-9de4-60491021f0dd>\",\"WARC-IP-Address\":\"130.126.151.19\",\"WARC-Target-URI\":\"https://ece.illinois.edu/academics/courses/ece462-120158\",\"WARC-Payload-Digest\":\"sha1:E2W5KOF2EVQTYGXKUDFSFJTF6B34G73A\",\"WARC-Block-Digest\":\"sha1:3EHXIHV7XEI4FX4TP6676ODLHTFAO74T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337371.9_warc_CC-MAIN-20221003003804-20221003033804-00554.warc.gz\"}"} |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/tracygjg/a-function-by-any-other-name-would-work-as-well-part-two-232 | [
"## DEV Community",
null,
"# A function by any other name would work as well (part two)\n\nIn part-one of this article we started on our exploration into the extensive world of functions as stipulated by JavaScript, or ECMAScript ECMA-262 to be more exact. We will now be discovering asynchrony and the functional programming side of JS before discussing some of the more exotic forms of function JS has to offer.\n\n## Asynchrony\n\nThere are actually a number of ways JS supports asynchronous operations, many of which use the Event Loop (see MDN) to control the order in which functions are executed and when.\nWe will begin by investigating call-backs that have a wide variety of uses. We will then discuss Promises before lastly looking at virtually synchronising operations using async and await instructions.\n\n### Call-backs\n\n\"Call-back\" is this name typically given to a function that is passed to another function as an argument. The intention being that the primary function may, at some point, execute the 'call-back' function. There are many use cases when this mechanism comes in handy but for the majority of us the Event Handler will be our first encounter.\n\n#### Event handlers\n\nThe first event handler a JS developer is likely to encounter is most probably going to be one of the following.\n\n##### Button Click\n\nUnless the web page you are developing is trying to minimise the use of JavaScript (a noble aim), it is highly likely there will be button of the screen that needs to react (do something) when clicked by the user.\n\nThere are several ways to configure this and many JS framework take much of the 'wiring up' out of the developer's hands anyway, but here is one way is can be done under the hood. For our purposes we will contain all of the \"moving parts\" in the same fragment of an HTML file.\n\n``````<body>\n<button id=\"btnClickMe\">Click Me</button>\n\n<script>\nconst domButton = document.querySelector('#btnClickMe');\n\nfunction sayHello() {\n}\n</script>\n</body>\n``````\n\nThe above HTML code is not a complete document but the fragment can be copied to an html file and viewed in a web browser all the same. Web browsers are very forgiving like that and will fill in the blanks.\n\nOn screen there will be a single button with the text \"Click Me\". Such is the nature of HTML buttons that we can either click on the button with the mouse or select it via the keyboard. For the latter we may need to tab round until the button is in focus before pressing the `ENTER` key to fire the event. Either way the result will be the presentation of an alert box containing the text \"Hello there\", which will require the `OK` button to be pressed to cancel.\n\n##### Page start-up\n\nAnother scenario is when we need to execute some JavaScript as soon as the screen had loaded. Again this sort of use case is managed by many JS frameworks as other considerations can impact the user experience.\n\n``````<script>\n\nfunction sayHello() {\n}\n</script>\n``````\n\nIn the above example there is even less going on. When the browser completes loading the document any function registered as a listener of the `DomContentLoaded` event will be executed, which in this case is the same `sayHello` function we used for the button.\n\n##### Timer/Interval\n\nThere are many use cases for Event Handler call-backs but this third example has to be a very common example encountered by new JS developers.\n\n``````<script>\nsetTimeout(sayHello, 2000);\n\nfunction sayHello() {\n}\n</script>\n``````\n\nIn this example, 2 seconds (or 2000 milli-seconds) after the screen finished loading (and processing the script) the call-back will be executed and the alert banner shown.\n\nThis is a textbook example and forms the basis of a very common technical interview question but I have found the need to use `setTimeout`, or its cousin `setInterval` are few in production code.\n\n#### Sort comparators\n\nProbably one of the most common use cases for call-back for a long time was to sort items in an array. The `Array` object in JS has a number of methods that can perform operations on the content of the array itself, sometimes leaving the updated content in place (as with the sort operation) and otherwise generating a new array, or some other output, as a result.\n\nWe will be exploring some of the other methods in a moment but first let's delve a little deeper into the sorting. We will start simply with a list of names.\n\n``````const names = ['Yvonne', 'Wesley', 'Andrew',\n'Terry', 'Brian', 'Xavier'];\n\nnames.sort();\n\nconsole.table(names);\n``````\n\nThe above example uses the default simple alphabetical sort comparator function, which using `console.table` outputs the following.\n\nindex Values\n0 Andrew\n1 Brian\n2 Terry\n3 Wesley\n4 Xavier\n5 Yvonne\n\nObviously the default sort comparator is rather limited. What if we want the names in the reverse order (ok there is a method `reverse` for that) or the data we want to sort is numeric (we do not want the output like `[1, 10, 2]`) or a more likely use case, what if the array is of objects rather than primitive values. Then we need to code our own sort comparator.\n\n``````const objs = [\n{ name: 'Yvonne', birthMonth: 8 },\n{ name: 'Wesley', birthMonth: 10 },\n{ name: 'Andrew', birthMonth: 2 },\n{ name: 'Terry', birthMonth: 12 },\n{ name: 'Brian', birthMonth: 1 },\n{ name: 'Xavier', birthMonth: 4 },\n];\n\nobjs.sort(sortComparator);\n\nconsole.table(objs);\n\nfunction sortComparator(objA, objB) {\nreturn objA.birthMonth - objB.birthMonth;\n}\n``````\n\nObserve in the above code example how we have supplied the sort method with a call-back function called `sortComparator` (but it could be called anything) that expects to be passed two elements from the array and returns a numeric value. In brief, 0 means leave the order unchanged (even if the values are the same), < 0 means arrange in the order objA then objB, > 0 means re-arrange in the order objB then objA (see MDN for more details). With the above comparison of the birthMonth property the output is as follows.\n\nindex name birthMonth\n0 Brian 1\n1 Andrew 2\n2 Xavier 4\n3 Yvonne 8\n4 Wesley 10\n5 Terry 12\n\nThe `sort` method removes the need to implement the algorithm in JavaScript and can be implemented as a lower, more performant, level. The frequency in which the `for` loop is used to traverse the elements of an array is so high it makes sense to also move the code for traversing arrays to a lower, more performant, level; but there are many reasons for traversing an array.\n\n``````// A common array traversal pattern\n\nconst arr = [ ... ];\n\nvar index = 0;\nfor (index = 0; index < arr.length; index++) {\n\n// Code to act on each item of the array (arr[index]) in turn.\n\n}\n``````\n\nThe above code is not difficult to write or comprehend but so common and easily abstracted.\n\nThe next three array methods all use a call-back to produce an output without changing (mutating) the original array. All three methods (along with some others that will be mentioned) are relatively recent additions to the language; added in ECMAScript 6 (2015).\n\n#### Filter predicates\n\nUsing the array of names we used previously, we used to write filter code as follows. In the following example we are filtering (to retain) those names containing the letter 'y'; anywhere in the name and in whatever case.\n\n``````const names = ['Yvonne', 'Wesley', 'Andrew',\n'Terry', 'Brian', 'Xavier'];\nlet namesContainingYs = [];\n\nvar index = 0;\nfor (index = 0; index < names.length; index++) {\nif (names[index].toLowerCase().includes('y')) {\nnamesContainingYs.push(names[index]);\n}\n}\nconsole.table(namesContainingYs);\n``````\nindex Values\n0 Yvonne\n1 Wesley\n2 Terry\n\nUsing the new `filter` method the code can be simplified to the following.\n\n``````const names = ['Yvonne', 'Wesley', 'Andrew',\n'Terry', 'Brian', 'Xavier'];\n\nconst namesContainingYs = names.filter(name =>\nname.toLowerCase().includes('y'));\n\nconsole.table(namesContainingYs);\n``````\n\nKey observations include:\n\n1. The result of the filter is stored in a constant array `namesContainingYs`.\n2. Filtering is reduced to one line with no need for a `for` loop or an `index` variable.\n3. The condition is, of all intents and purposes, unchanged.\n4. The call-back function returns only `true` or `false`. This type of function is given the name of predicate and is used by many other array methods such as: every, find, findIndex, findIndexLast, forEach, and some.\n\nIt is worth noting that in these examples we will be calling the array methods with one or two arguments but any of them optionally expect more parameters (see MDN for details).\n\n#### Map transformers\n\nAnother common use case for traversing an array is to transform each element from one type to another. For example let's convert all the names to uppercase.\n\n``````const names = ['Yvonne', 'Wesley', 'Andrew',\n'Terry', 'Brian', 'Xavier'];\n\nconst upperCaseNames = names.map(name => name.toUpperCase());\n\nconsole.table(upperCaseNames);\n``````\nindex Values\n0 YVONNE\n1 WESLEY\n2 ANDREW\n3 TERRY\n4 BRIAN\n5 XAVIER\n\nBoth `filter` and `map` traverse the entire array so if your use case requires both (one after the other), where possible, the filter should be performed first to reduce the number of items transformed to a minimum. But we might be able to do better using the next method `reduce`.\n\n#### Reduce reducers\n\nThe `reduce` method is more complicated and is a common source of confusion, which is why I have written on the topic. With the `reduce` method it is possible to implement a version of `filter` and `map` without resorting to a `for` loop; but what is it for?\n\nThe `filter` method will always return an array containing no more items than the source array, and probably less, with the items in the same sequence and unchanged (transformed). The `map` method will produce a new array containing the same number of items as the source, no more, no fewer.\n\n`reduce` can do both and more besides. In fact it does not even have to return array but could return another data type such as a count, average or sum. Why we don't have a Math.sum method I do not understand but that is another discussion.\n\nA completely contrived use case but here is how we could use `reduce` to add-up all the birthMonths.\n\n``````const objs = [\n{ name: 'Yvonne', birthMonth: 8 },\n{ name: 'Wesley', birthMonth: 10 },\n{ name: 'Andrew', birthMonth: 2 },\n{ name: 'Terry', birthMonth: 12 },\n{ name: 'Brian', birthMonth: 1 },\n{ name: 'Xavier', birthMonth: 4 },\n];\n\nconst totalBirthMonths = objs.reduce((tot, obj) =>\ntot + obj.birthMonth, 0);\n\nconsole.log(`Total = \\${totalBirthMonths}`); // \"Total = 37\"\n``````\n\nNotice how the running total is initialised as 0 using the second argument of the `reduce` method. The first argument is called a reducer method because in reduces two input values to one output value. With the `reducer` the first input is the running total, the second being each item from the array.\n\nAs demonstrated above call-back functions are still current and an important technique to understand. However, for async code they can quickly become unmanageable. Before to long there will be a need to nest one call-back inside another. Scaling this pattern is limited as having to many levels of nesting will make the code hard to understand, difficult to debug and virtually impossible to unit test.\n\n### Promises\n\nPromises are not functions but a special type of object that is passed a call-back function when created. The call-back function is itself passed two functions as arguments; called `resolve` and `reject`.\n\nWhen a promise starts it executes the initial call-back function to perform its primary purpose. When the function concludes it will call one of the supplied secondary functions depending on if the function was successful (`resolve`) or failed (`reject`).\n\nHere is the function that performs the actual coin toss and returns a Promise object for a single coin.\n\n``````function tossCoin() {\nreturn new Promise((resolve, reject) => {\nsetTimeout(() => {\nconst coinToss = ~~(Math.random() * 5);\nif (coinToss) resolve(coinToss % 2 ? 'Tails' : 'Heads');\nreject('Dropped');\n}, 300);\n});\n}\n``````\n\nExample one (below) uses the 'thenable' protocol implemented by the Promise created above to process the results. `Then` being called when the coins results in Heads or Tails, `catch` when the coin is Dropped.\n\n``````for (let i = 1; i <= 10; i++) {\ntossCoin()\n.then(result => console.log(i, result))\n.catch(result => console.error(i, result));\n}\n``````\n\nIn the above code example we are simulating the tossing of 10 coins. The `tossCoin` function creates a new `Promise`, which is the expectation of a coin toss result. However, the person tossing the coins is a bit clumsy and drops the coin 20% (1 in 5) of the time so it is not a 50:50 chance of Head or Tails. The output for the above example, and the next two, look something like this:\n\n``````1 Heads\n2 Tails\n3 Dropped\n4 Tails\n5 Tails\n6 Dropped\n7 Tails\n``````\n\nIt takes 300ms for the coin to be flipped and a result to be produced. The conventional `for` loop calls the function 10 times and about 300ms after (the first call) the results start to flow. Notice how the coin-tosses do not start after the result of the previous coin is known but nearly all at the same time. What this shows is that processing does not stop when the Promise function starts, which gives an effective impression of asynchronous operation, but it is just the Event-loop optimising the primary thread. On the subject of the Event-loop, I highly recommend the article by Lydia Hallie.\n\nInside the `tossCoin` function the `Math.random` method generates a value between 0 and 1. The value is multiplied by 5 and rounded down to an integer between 0 and 4. Each integer representing 20% probability. If the value is zero it is regarded as dropped, otherwise the outcome is Tails if true (odd) or Heads if false (even). When the coin is dropped it is reported by the Promise calling the (exception via the) `reject` call-back, which exercises the `catch` path. Valid coin toss results are reported by the Promise calling the `resolve` call-back via the `then` path.\n\nSimilar to the regular reporting of exceptions, the Promise \"thenable\" pattern also includes a `finally` method that is called once the Promise completes irrespective of the outcome as a form of clean-up.\n\nPromises offer a range of supporting methods to help developers write better asynchronous code but reading complex async code some time later can still be difficult. At least, not as easy as reading synchronous code, which is where the following async/await construct comes to the fore.\n\n### Async/Await\n\nJavaScript is an unusual language for a number of reasons, not least because it operates in a single thread. But that does not mean it is incapable of supporting asynchronous operations.\n\nAfter basic call-backs, JS was given Promises but now we also have special (async/await) functions. Under the hood these functions are a combination of a Promise and a Generator (see later) but exactly how is outside the scope of this article.\n\nAsync/Await works in conjunction with a Promise but in such as way that the code reads synchronously, which simplifies understanding.\n\n``````(async function () {\nfor (let i = 1; i <= 10; i++) {\ntry {\nconst result = await tossCoin();\nconsole.log(i, result);\n} catch (err) {\nconsole.error(i, err);\n}\n}\n})();\n``````\n\nWhilst all three examples produce the same output there is a fundamental difference in the behaviour. In the first Promise example there was a 300ms delay before all the results were produced in rapid succession. In the above Async/Await example there is a 300ms delay before the result of the first coin toss is presented before the next 300ms delay starts and so on. This is down to the way the `tossCoin` function is called inside the `for` loop with the `await` keyword.\n\nIn the second example all ten coins took around 3 seconds to complete but the first example took a little over 300ms to complete. However, most developers find the syntax of the second example easier on the eye. The slight down-side though is that the `await` needs to operate inside an `async` function, which is why the `for` loop is wrapped in an Immediately Invoked Function Expression (IIFE).\n\n``````const coins = [];\n\nfor (let i = 1; i <= 10; i++) {\ncoins.push(tossCoin());\n}\n\nPromise.allSettled(coins).then(results =>\nresults.forEach(({ status, value, reason }, i) => {\nif (status === 'fulfilled') {\nconsole.log(i + 1, value);\n} else {\nconsole.error(i + 1, reason);\n}\n})\n);\n``````\n\nIn the third example, above, the `for` loop repeatedly calls the tossCoin function and collects the Promises returned in an array. We then use a `Promise.allSettled` call on the array to process the results when all 10 have completed. The behaviour is like the first example where all 10 coin tosses complete in little over 300ms. The differences include; there is no wrapping `async` function and no `await`. The slight complication (there is usually something) is results array is quite involved containing three properties per result:\n\n• status: The result of the Promise, either \"fulfilled\" or \"rejected\".\n• value: If \"fulfilled\" this is the output of the primary function.\n• reason: If \"rejected\" this is why the primary function failed.\n\n## Functional Programming style\n\nJavaScript is not an FP language but neither is it an entirely OOP language. It is one of a growing number of \"multi-paradigm\" languages that to some degree supports features from a variety of paradigms, which is probably why it has so many different types of function.\n\nUnlike many of the types of function discussed in this article the following sections describe functions that are not supported directly through syntax in JavaScript; they are not 'idiomatic'. They do however make considerable use of JS's support of functions as first-class objects, high-order functions and closures.\n\n### Curried and Partial Application\n\nIn my mind Currying is more about the process than a type of function. A 'curried' function is one that expects its parameters to be supplied one argument at a time, returning a new function until all the mandatory parameters have been supplied. Only then is the function actually executed. \"Currying\" is the process of taking a regular function and generating a curries function.\n\nA textbook example of a curried function could be to calculate the volume of a regular solid shape.\n\n``````function calculateVolume(width) {\nreturn function(height) {\nreturn width * breadth * height;\n}\n}\n}\n``````\n\nThere are four permutations the function can be used but here are the two extremes; all in one or through individual calls.\n\n``````const allInOne = calculateVolume(2)(3)(7); // 42\n\nconst supplyBreadth = calculateVolume(2); // new function\nconst supplyHeight = supplyBreadth(3); // new function\n\nconsole.log(supplyHeight(7)); // 42\n``````\n\nAll of the permutations only accept a single parameter/argument binding with each call but partial-applications are more adaptable. A function to convert a regular function to one that supports partial application is a bit more complicated to implement. Therefor it is recommended to use a library such as Lodash.\n\nThe previous example, implemented as a partial-application, can be called in the ways shown above but in addition the following calls are permitted.\n\n``````const firstTwo = calculateVolume(2, 3);\nconsole.log(firstTwo(7)); // 42\n\nconst onlyFirst = calculateVolume(2); // new function\nconsole.log(onlyFirst(3, 7)); // 42\n``````\n\nIn both of the above examples the first call returns a function expecting more (but not necessarily all of the remaining) arguments.\n\nSo what is it good for?\nIn my related post I discuss this in some detail but a good example is to provide a dynamic property sort comparator. The example source code can be found in JSFiddle.\n\nWe start with an array of 18 objects each defining the forename, surname and band name for the members of Fleetwood Mac, Cream, Queen and The Doors. E.g.,\n\n``````const bandsArray = [\n{\nforename: 'Jack',\nsurname: 'Bruce',\nband: 'Cream',\n},\n\n// :\n\n{\nforename: 'John',\nsurname: 'Deacon',\nband: 'Queen',\n},\n];\n``````\n\nWe can create a function that uses partial application to return the sort comparator function for a given property.\n\n``````function dynamicSortComparator(propertyName) {\nreturn (objA, objB) =>\n(objA[propertyName] < objB[propertyName] ? -1 : 1);\n}\n``````\n\nConventionally we might call the `sort` method with a comparator function but that would only be able to compare a specific object property. Instead, using the above function, we can sort by surname or forename as follows:\n\n``````bandsArray.sort(dynamicSortComparator('surname'));\nconsole.table(bandsArray);\n\nbandsArray.sort(dynamicSortComparator('forename'));\nconsole.table(bandsArray);\n``````\n\nI go into more detail of partial applications in this post.\n\n### Recursion\n\nRecursion is a technique supported by many programming languages, some better than others. It should be noted that JS is lacking a particular feature called Tail-call optimisation that considerably improves performance of recursive functions but it does not mean it is an inappropriate candidate solution for some problems in JS.\n\nIrrespective of the language or the problem to be solved there are some tips to consider when creating a recursive function. Fundamentally recursive functions call themselves so we need to consider how we stop the function execution going into an infinite loop and never returning.\n\nTextbook examples for demonstrating recursion include calculating factorials or Fibonacci numbers but we will use it to calculate compound interest (interest on interest), to long way.\n\nMathematicians: I am fully aware there is a convenient formular to do this but that will not demonstrate recursion.\n\nSo, the parameters will be:\n\n• Principal: The amount of the initial loan.\n• Interest rate: Assumed to be constant throughout the period of the loan, and expressed as a percentage, this is the rate the loan will grow (year on year), not considering repayments.\n• Duration: in years the loan is to run.\n``````function calculateInterest(principal, interestRate, duration) {\n// recursion happens here\n}\n``````\n\nYes, we could use a simple `for` loop to solve this problem but there are some problems for which `for` loops are insufficient. Using this problem, which I hope is easy to understand, should be a simple example.\n\nAt its core the function we need some simple mathematics to calculate the increase size of the loan for a single year.\n\n`````` newPrincipal = principal + principal * interestRate;\n``````\n\nBut we need to guard against the risk of an infinite loop, which we can do using the duration of the loan.\n\n``````function calculateInterest(principal, interestRate, duration) {\nif (duration === 0) { // guard\nreturn principal;\n}\nreturn calculateInterest(\nprincipal + principal * interestRate,\ninterestRate,\n--duration // decrease the duration by 1 year each cycle\n);\n}\n``````\n\nThis would yield the following results:\n\nPrincipal Year Interest Total Loan\n1000 1 200 1200\n1200 2 240 1440\n1440 3 288 1728\n1728 4 345.6 2073.6\n2073.6 5 414.72 2488.32\n\nBut the function can be simplified into:\n\n``````function calculateInterest(principal, interestRate, duration) {\nreturn duration ? calculateInterest(\nprincipal + principal * interestRate,\ninterestRate,\n--duration\n) : principal;\n}\n``````\n\n## Now for some more exotic functions\n\nA brief \"dipping of the toe\" into something unfamiliar.\n\n### Generators\n\nArguably, JS Generators support some unusual use cases. If I were more familiar with them, or smarter, I might find more opportunities to use them but I do not think I have ever used them professionally.\n\nIn brief, Generators provide a way of creating a 're-enterable' function. The generated function can behave differently each time it is called, re-entering the function at the point it `yield`ed control to the caller.\n\nWhen trying to explain a complex or novel concept it can be useful to replicate the subject using more familiar techniques. In part-one of this article we discussed closures that we will now use to simulate the behaviour of generators. We will replicate the following example, based on the one from the MDN web page.\n\n``````{\nconst foo = function* () {\nyield 'a';\nyield 'b';\nyield 'c';\n};\n\nexercise('Idiomatic generator', foo);\n}\n``````\n\nWe will also demonstrate the behaviour and the iteration protocols used through the following exercises.\n\n``````// Exercise one: Using a for iterator\nfunction exercise(exerciseName, foo) {\nconsole.log(exerciseName);\n\nlet str = '';\nfor (const val of foo()) {\nstr += val;\n}\nconsole.log(str); // Output: 'abc'\n}\n\n// Exercise two: Calling the next method\nfunction exercise(exerciseName, foo) {\nconsole.log(exerciseName);\n\nconst gen = foo();\nlet str = gen.next().value;\nstr += gen.next().value;\nstr += gen.next().value;\nconsole.log(str); // Output: 'abc'\n}\n``````\n\nThe following code fragment is a simulation of the `foo` generator above but using a closure to maintain state between calls.\n\n``````{\nconst foo = (function (...yieldResults) {\nlet yieldIndex = 0;\nreturn next;\n\nfunction next() {\nconst mdnIterator = {\nnext() {\nreturn {\ndone: yieldIndex === yieldResults.length,\nvalue: yieldResults[yieldIndex++],\n};\n},\n[Symbol.iterator]() {\nreturn this;\n},\n};\nreturn mdnIterator;\n}\n})('a', 'b', 'c');\n\nexercise('Simulated generator', foo);\n}\n``````\n\nThe simulation is considerably more involved than the MDN example as it has to implement the iteration protocols manually but both implementations perform the exercises to the same effect.\n\nThe key points to observe include:\n\n1. The function is not re-initialised each time it is called.\n2. The `yield` command behaves a little like a `return` command and can send a value back from within the function.\n3. Subsequent calls resume from the point of the last used `yield`."
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https://adamprescott.net/2012/09/27/encryption-101-symmetric-algorithms/ | [
"# Encryption 101: Symmetric Algorithms\n\nEarlier this week, we covered one-way encryption using a hash algorithm and a salted hash. If you need to decrypt what you’ve encrypted, these solutions are out the window. The simplest solution for encrypting and decrypting is to use a symmetric algorithm. Data is encrypted using a secret key, and it’s decrypted the same way–using the secret key. (Symmetric!)\n\nIt requires a few more steps to accomplish, but overall, it’s still pretty simple. Let’s breakdown the steps:\n\n1. Select an algorithm\n2. Create a key\n3. Encrypt data\n4. Decrypt data\n\nThat doesn’t sound so bad, right!? So let’s do it!\n\n## Select an algorithm\n\nThe .NET Framework supports a number of different symmetric algorithms, including DES, RC2, Rijndael, and TripleDES. We’ll use TripleDES in this example, so let’s instantiate our crypto provider.\n\n```var algorithm = new TripleDESCryptoServiceProvider();\n```\n\n## Create a key\n\nNow that we have our crypto provider, creating our secret key is a breeze.\n\n```algorithm.GenerateKey();\n```\n\nBut wait! That’s not quite it. We also need to generate an initialization vector (IV), which acts as a “randomizer” for the encryption. Don’t worry, though–it’s just as easy as generating the key.\n\n```algorithm.GenerateIV();\n```\n\n## Encrypt data\n\nWe have our key and IV, so we’re ready to encrypt some data! This is a little more complicated, but still not too bad. All we’re doing is creating an ICryptoTransform “encryptor” from our symmetric crypto provider, then using it to write bytes to a CryptoStream. Sounds hard, but it’s really not so bad.\n\n```string Encrypt(SymmetricAlgorithm sa, string text)\n{\nvar encryptor = sa.CreateEncryptor(sa.Key, sa.IV);\nvar bytes = Encoding.UTF8.GetBytes(text);\nusing (var ms = new MemoryStream())\n{\nusing (var cs = new CryptoStream(ms, encryptor, CryptoStreamMode.Write))\n{\ncs.Write(bytes, 0, bytes.Length);\ncs.FlushFinalBlock();\n}\nreturn Convert.ToBase64String(ms.ToArray());\n}\n}\n```\n\n## Decrypt data\n\nNow how do we go about decrypting it? Well, not surprisingly–this being a symmetric algorithm and all–it’s nearly identical to encrypting the data. The only difference is that we create a “decryptor” from our crypto provider.\n\n```string Decrypt(SymmetricAlgorithm sa, string encrypted)\n{\nvar decryptor = sa.CreateDecryptor(sa.Key, sa.IV);\nvar bytes = Convert.FromBase64String(encrypted);\nusing (var ms = new MemoryStream())\n{\nusing (var cs = new CryptoStream(ms, decryptor, CryptoStreamMode.Write))\n{\ncs.Write(bytes, 0, bytes.Length);\ncs.FlushFinalBlock();\n}\nreturn Encoding.UTF8.GetString(ms.ToArray());\n}\n}\n```\n\n## Putting it all together\n\nOverall, symmetric encryption gives us a quick and easy way to do reasonably secure encryption. To properly decrypt data across applications, you’ll need to share the key and IV that you generated. Make sure to keep these values a secret, though, as they allow anybody to decrypt your sensitive data. Here’s the full example from the pieces above:\n\n```namespace adamprescott.net.EncryptionSymmetric\n{\nusing System;\nusing System.IO;\nusing System.Security.Cryptography;\nusing System.Text;\n\nclass Program\n{\nstatic void Main(string[] args)\n{\nvar p = new Program();\np.Run();\n}\n\nvoid Run()\n{\nConsole.Write(\"Input: \");\n\nusing (var algorithm = new TripleDESCryptoServiceProvider())\n{\nalgorithm.GenerateKey();\nalgorithm.GenerateIV();\n\nvar encrypted = Encrypt(algorithm, input);\nConsole.WriteLine(\"Encrypted: {0}\", encrypted);\n\nvar decrypted = Decrypt(algorithm, encrypted);\nConsole.WriteLine(\"Decrypted: {0}\", decrypted);\n}\n\n}\n\nstring Encrypt(SymmetricAlgorithm sa, string text)\n{\nvar encryptor = sa.CreateEncryptor(sa.Key, sa.IV);\nvar bytes = Encoding.UTF8.GetBytes(text);\nusing (var ms = new MemoryStream())\n{\nusing (var cs = new CryptoStream(ms, encryptor, CryptoStreamMode.Write))\n{\ncs.Write(bytes, 0, bytes.Length);\ncs.FlushFinalBlock();\n}\nreturn Convert.ToBase64String(ms.ToArray());\n}\n}\n\nstring Decrypt(SymmetricAlgorithm sa, string encrypted)\n{\nvar decryptor = sa.CreateDecryptor(sa.Key, sa.IV);\nvar bytes = Convert.FromBase64String(encrypted);\nusing (var ms = new MemoryStream())\n{\nusing (var cs = new CryptoStream(ms, decryptor, CryptoStreamMode.Write))\n{\ncs.Write(bytes, 0, bytes.Length);\ncs.FlushFinalBlock();\n}\nreturn Encoding.UTF8.GetString(ms.ToArray());\n}\n}\n}\n}\n```"
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https://electronics.stackexchange.com/questions/134388/is-this-an-opamp-voltage-regulator | [
"# Is this an opamp voltage regulator?\n\nI'm trying to understand the attached schematic, which I came across when repairing an oscilloscope.",
null,
"Am I right in thinking that this circuit is regulating VCC (5V) down to 3.3 and 2.5 volts? Why does the opamp use 12v rails rather than 5volts?\n\nThe non-inverting input to the opamp seems to be a voltage divider. I take this divider is used to provide a simple voltage reference?\n\nGiven that this is in an oscilloscope (a LeCroy LT342) is there any reason to use an opamp here rather than a voltage regulator?\n\nOne further thing that confuses me, is that on the board Q9 is not fitted, and has been replaced with a short between 2 of the pins. I can't understand how the circuit can work in this scenario, can anyone cast any light on this?\n\nFrom my naive understanding, this seems like a bad design, and that it would be preferable to use a voltage regulator. What designed decisions might I have missed here?\n\n• The lower V rails are linked ratiometrically to Vcc by the resistor ratios used in R49, R46 and R135, R136. If you increase or decrease the 5V rail by X% the other two rails will alter by the same percentage. This may be advantageous. Or not. Oct 15, 2014 at 12:30\n\nThey are regulators as you thought.\n\nTo properly control the MOSFET on each regulator, the gate needs to rise several volts above the source, probably over 6 volts greater than the source. If the source voltage is 3.3 volts the gate might need to rise as high as 9 or 10 volts when the MOSFET is driving a heavy load current. This means the op-amp MUST be powered from a supply greater than 10 volts.\n\nThe non-inverting input is the demand voltage for the regulator - for the 3V3 supply the non-inverting input will be 3.3 volts and ditto the 2V5 supply.\n\nThe MOSFET is rated at 20 amps and this means the supply current to the circuit might be in the realm of 5 to 8 amps - try and find a linear voltage regulator that would fit the bill an be a low drop-out type - realistically with a drain voltage of fractionally above 3.3 volts the output would still be in regulation. This is of course due to the op-amp being powered from 12 volts.\n\nAs to whether it is a bad design, I'd say not BUT I don't know the full spec of the circuit.\n\n• I see, thanks! That really helps me understand the design considerations better! Oct 15, 2014 at 8:36\n• Thanks again for you help. I wrote up my notes based on my new understanding of this circuit here: 41j.com/blog/2014/10/opamp-as-a-voltage-regulator One remaining question I had was the purpose of R138. It seems like it's a feedback resistor but I don't think that makes sense? Oct 15, 2014 at 12:47\n• I would say R138 and C159 form a low-pass filter on the feedback path. I don't know the age of this scope but at a guess, this circuit was less expensive than using a low-dropout regulator capable of handling the required current at the time it was designed? Oct 15, 2014 at 13:13\n\nQ9 is not fitted, and has been replaced with a short between 2 of the pins"
] | [
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"https://41j.com/2.5v.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9698557,"math_prob":0.77285784,"size":907,"snap":"2022-05-2022-21","text_gpt3_token_len":215,"char_repetition_ratio":0.10963455,"word_repetition_ratio":0.0,"special_character_ratio":0.22271223,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95893747,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-16T05:16:04Z\",\"WARC-Record-ID\":\"<urn:uuid:523fcb37-e7c9-419f-a574-de7ae7dd63bc>\",\"Content-Length\":\"239589\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:444e100e-1f6e-4ad4-b147-52385932c149>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a73ad71-54d3-4a3f-a24b-b04ec3e01fc8>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/134388/is-this-an-opamp-voltage-regulator\",\"WARC-Payload-Digest\":\"sha1:K35TYC6IAOCZNOUOTHWB4FNJ5MRAFQ7Z\",\"WARC-Block-Digest\":\"sha1:GPHQWZ4JITNSHDM7Z6DU6QW4RQZZVHIL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662509990.19_warc_CC-MAIN-20220516041337-20220516071337-00648.warc.gz\"}"} |
https://byjus.com/maths/19500-in-words/ | [
"",
null,
"# 19500 in Words\n\n19500 in words is Nineteen Thousand Five Hundred. For example, if you earned Rs. 19500 in a month, you can write, “I have earned Rs. Nineteen thousand five hundred in a month”. In general, the number name of any number can be written using the ones, tens, hundreds, and thousands place of a number. Thus, the place value chart is useful for writing the number 19500 in words.\n\n 19500 in words Nineteen Thousand Five Hundred Nineteen Thousand Five Hundred in Numbers 19500\n\n## 19500 in English Words",
null,
"## How to Write 19500 in Words?\n\nWe can convert 19500 to words using a place value chart. This can be done as follows. The number 19500 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.\n\n Ten thousands Thousands Hundreds Tens Ones 1 9 5 0 0\n\nThus, we can write the expanded form as:\n\n1 × Ten thousand + 1 × Thousand + 5 × Hundred + 0 × Ten + 0 × One\n\n= 1 × 10000 + 9 × 1000 + 5 × 100 + 0 × 10 + 0 × 1\n\n= 19500\n\n= Nineteen thousand five hundred\n\n19500 is the natural number that is succeeded by 19499 and preceded by 19501.\n\n19500 in words – Nineteen thousand five hundred\n\nIs 19500 an odd number? – No.\n\nIs 19500 an even number? – Yes\n\nIs 19500 a perfect square number? – No\n\nIs 19500 a perfect cube number? – No\n\nIs 19500 a prime number? – No\n\nIs 19500 a composite number? – Yes\n\n## Frequently Asked Questions on 19500 in Words\n\n### How to write 19500 in words?\n\n19500 in English is written as “nineteen thousand five hundred”.\n\n### Is the number 19500 an odd number?\n\nNo, the number 19500 is not an odd number.\n\n### Is 19500 a composite number?\n\nYes, 19500 is a composite number."
] | [
null,
"https://www.facebook.com/tr",
null,
"https://cdn1.byjus.com/wp-content/uploads/2022/02/19500-in-words.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74920994,"math_prob":0.9857833,"size":1508,"snap":"2022-05-2022-21","text_gpt3_token_len":473,"char_repetition_ratio":0.1974734,"word_repetition_ratio":0.025396826,"special_character_ratio":0.38527852,"punctuation_ratio":0.09731544,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98854536,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T16:01:42Z\",\"WARC-Record-ID\":\"<urn:uuid:bf71787e-1aef-4343-a13a-ddd8bfb9f917>\",\"Content-Length\":\"675898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14e989c7-c384-4ad4-afa8-55d37ebca0ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f60e9fc-93af-4d41-be32-89523c1bd766>\",\"WARC-IP-Address\":\"162.159.130.41\",\"WARC-Target-URI\":\"https://byjus.com/maths/19500-in-words/\",\"WARC-Payload-Digest\":\"sha1:O4TKJMSQGSCLURTCKMVBXS3UCQZ74Q6Y\",\"WARC-Block-Digest\":\"sha1:QDIVYFIMKLZH75EA4Q7NXGHEQBAPIKBW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662588661.65_warc_CC-MAIN-20220525151311-20220525181311-00783.warc.gz\"}"} |
https://hugogiraudel.com/2013/10/17/sass-random/ | [
"I wrote this article months ago when I was first experimenting with Sass 3.3 alpha features. I came up with a pretty crazy solution to generate a random number in Sass. However it looks like Sass 3.3 will implement a random function so we won’t need all this stuff. I still publish it for fun. :)\n\nEverything started when I was spying on Sass 3.3 source code on GitHub for my article about the future of Sass at David Walsh’ Blog. I was sniffing the incoming functions when all of the sudden I came by a `unique-id()` function.\n\nAccording to the issue which started this idea, the `unique-id()` function should return a unique random alphanumeric identifier that could be used for whatever you like. As far as I understood the example provided by Chris Eppstein, it could be used to dynamically generate and extend a placeholder from within a mixin. Kind of complicated stuff, really.\n\nAnyway, I saw this unique id thingie as an opportunity to have a random number with Sass. Why? I don’t know. I leave this question to you. Maybe some day I’ll find a usecase for a random number in CSS.\n\nNote: the code in this article has not been tested at all since it requires some Sass 3.3 functions that are not implemented yet. This is more like a proof of concept.\n\n## About `unique-id()`\n\nTo understand what this is all about, you need to know what the `unique-id()` is and what it returns. First of all, there are two different functions for this in Sass source code, both from 2 months ago: one in tree `f3be0f40b7` (using base36) and one in branch `unique_id` (using base16). I only worked on the latter since it’s most likely this is the one that will be implemented.\n\nI’m not a Ruby pro, but with the help of a kind folk on Twitter, I could make it work on CodePad. Here is what a couple of run of the function looks like:\n\n``````u84ec5b4cdecd4299\nu871ec9c6e6049323\nu8865b8a8e572e4e8\nu85f6c40bb775eff2\nu8868f6a1f716d29f\nu89cf1fa575a7a765\nu89184d7511933cd3\nu8a7287c699a82902\nu8547f4133644af4c\nu86fb16af4800d46b\n``````\n\nSo the function returns a 19-characters long alphanumeric string. As you may have noticed, the returned string always starts with a `u`. This is actually hard-coded inside the function core to make sure the string always start with a letter in order to be able to be used as a class / placeholder / id, whatever.\n\nTo put it very simple, the function randoms a 19-digits number, convert it to base 16 (or base 36 in the other implementation), then append it a `u`. So when we use `unique-id()`, we end up with something like this: `u8547f4133644af4c`.\n\n## Random, the dirty way\n\nMy first attempt to get a random number from this string was to remove all alpha characters from it, then keep only the number of digits we want (or we still have). To do this, I used the incoming string manipulation functions (`str-length()`, `str-slice()`, `str-insert()`):\n\n``````@function rand(\\$digits: 16) {\n/* Array of characters to remove */\n\\$letters: a b c d e f u;\n\\$result: unquote('');\n\\$string: unique-id();\n\n/* For each character in the given string */\n@for \\$i from 1 through str-length(\\$string) {\n/* Isolate character */\n\\$character: str-slice(\\$string, \\$i, \\$i + 1);\n/* If not a letter */\n@if index(\\$character, \\$letters) == false {\n/* Append it to \\$value */\n\\$value: str-insert(\\$result, \\$character, str-length(\\$result) + 1);\n}\n}\n\n/* Deal with the number of digits asked */\n@if \\$digits !== 0 and \\$digits < length(\\$result) {\n\\$result: str-slice(\\$result, 1, \\$digits);\n}\n\n/* Return the result */\n@return \\$result;\n}\n``````\n\nI think the code is pretty much self-explanatory. I check each character individually: if it’s not a letter, I append it to the `\\$result` variable. When I’m done, if the length of `\\$result` is still greater than the number of digits we asked for (`\\$digits`) we truncate it.\n\nAnd there we have a random number between 1 and 9999999999999999 (in case the 16 characters are 9).\n\n``````\\$number: rand(); /* Random between 1 and 9999999999999999 */\n\\$number: rand(1); /* Random between 1 and 9 */\n\\$number: rand(4); /* Random between 1 and 9999 */\n\\$number: rand(0); /* Random between 1 and 9999999999999999 */\n\\$number: rand(-1); /* Random between 1 and 9999999999999999 */\n``````\n\n## Random, the clean way\n\nOkay, let’s say it: the first version I came with is really dirty. That’s why I reworked a new version from scratch with the help of my brother. We even tweaked it in order to make it future-proof for both implementations of the `unique-id()` function. How cool is that?\n\nTo put it simple, instead of stripping alpha characters, we take the alphanumeric string and convert it back into an integer. Then, we get a fully random integer we simply have to manipulate around min and max values.\n\n``````@function rand(\\$min: 0, \\$max: 100) {\n\\$str: str-slice(unique-id(), 2);\n\\$res: toInt(\\$str, 16);\n@return (\\$res % (\\$max - \\$min)) + \\$min;\n}\n``````\n\nThe first line in the function core is the `unique-id()` function call. We immediately pass it into the `str-slice()` function to remove the very first character which is always a `u`.\n\nNote: According to my tests, the min value used in both implementations of `unique-id()` is such that the second character of the returned string is always the same (`8` in base 16, `1` in base 36). Thus we may need to strip it too, like this `str-slice(unique-id(), 3)`.\n\nThe second line calls a `toInt()` function, passing it both the string (`\\$str`) and the base we want to convert the string from (not to). This is why I say we’re ready for both implementations: we only have to change this `16` to `36` and everything should work like a charm.\n\nBefore going to the last line, let’s have a look at the `toInt` function:\n\n``````@function toInt(\\$str, \\$base: 10) {\n\\$res: 0;\n\\$chars: charsFromBase(\\$base);\n@if \\$chars !== false {\n\\$str: if(\\$base < 64, to-lower-case(\\$str), \\$str);\n@for \\$i from 1 through str-length(\\$str) {\n\\$char: str-slice(\\$str, \\$i, \\$i + 1);\n\\$charVal: index(\\$char, \\$chars) - 1;\n\\$res: \\$res + pow(length(\\$base), str-length(\\$str) - \\$i) * \\$charVal;\n}\n@return \\$res;\n}\n@return false;\n}\n``````\n\n`\\$res` will store the result we will return once we’re done. `\\$chars` contains the array of characters used by base `\\$base`; we’ll see the `charsFromBase()` function right after. Then, if the base is supported we loop through each characters of the string.\n\nFor every character, we isolate it (`\\$char`) and convert it to its numeric equivalent (`\\$charVal`) thanks to the `\\$chars` array. Then, we multiply this number to the base raised to the reversed index in the string. That may sound a little complicated, let me rephrase it: in base 10, `426` equals `4*10^2` + `2*10^1` + `6*10^0`. That’s pretty much what we do here, except instead of `10` we use the base, and instead of `2`, `1` and `0`, we use the length of string minus the index of the current character.\n\nThe `pow()` function used to raise a value to an exponent is part of Compass Math helpers. In case you don’t want to use Compass or simply can’t use Compass, here is the `pow()` function in pure Sass:\n\n``````@function pow(\\$val, \\$pow) {\n\\$res: 1;\n@while (\\$pow > 0) {\n\\$res: \\$res * \\$val;\n\\$pow: \\$pow - 1;\n}\n@return \\$res;\n}\n``````\n\nAnd of course, we add this to the result (`\\$res`). Once we’re done with the string, we return the result to the `rand()` function. Then, we simply return `(\\$res % (\\$max - \\$min)) + \\$min` to the user resulting in a random number between min and max values.\n\nRegarding the `charsFromBase()` function, here is what it looks like:\n\n``````@function charsFromBase(\\$base: 10) {\n/* Binary */\n@if \\$base == 2 {\n@return 0 1;\n}\n/* Octal */\n@if \\$base == 8 {\n@return 0 1 2 3 4 5 6 7;\n}\n/* Decimal */\n@if \\$base == 10 {\n@return 0 1 2 3 4 5 6 7 8 9;\n}\n/* Hexadecimal */\n@if \\$base == 16 {\n@return 0 1 2 3 4 5 6 7 8 9 a b c d e f;\n}\n/* Base 36 */\n@if \\$base == 36 {\n@return 0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z;\n}\n/* Base 64 */\n@if \\$base == 64 {\n@return A B C D E F G H I J K L M N O P Q R S T U V W X Y Z a b c d e f g h i j k l m n o p q r s t u v w x y z 0 1 2 3 4 5 6 7 8 9 + /;\n}\n@return false;\n}\n``````\n\nI only added most common standard bases (binary, octal, decimal, hexadecimal, 36, 64) but I guess we could probably add a couple of others. Actually this is already too much since we know the `unique-id()` function will return a base16 or base36 encoded string (depending on the implementation they’ll keep).\n\n## Final words\n\nThat’s pretty much it. As I said at the beginning of the article, I couldn’t try this code since neither the `unique-id()` nor the string manipulation functions are currently implemented in the Sass 3.3 Alpha version. So this is pretty much blind coding here!\n\nIf you think of anything that could improve this Sass random function, please be sure to tell. Meanwhile you can play with the code directly on this pen."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8121243,"math_prob":0.9316603,"size":8605,"snap":"2019-35-2019-39","text_gpt3_token_len":2407,"char_repetition_ratio":0.13079874,"word_repetition_ratio":0.066873066,"special_character_ratio":0.3128414,"punctuation_ratio":0.11953514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9572285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T08:48:31Z\",\"WARC-Record-ID\":\"<urn:uuid:89e2e1f6-de68-4530-b6b0-2713b8548f63>\",\"Content-Length\":\"31001\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b3f4be5-6b1b-4f8b-bf70-435586384353>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e0a3541-bad3-45f3-9d67-9d8de1dbe6b6>\",\"WARC-IP-Address\":\"142.93.122.177\",\"WARC-Target-URI\":\"https://hugogiraudel.com/2013/10/17/sass-random/\",\"WARC-Payload-Digest\":\"sha1:FVBXX6BN3P3YNKLTRZZZCNR222D5BQR3\",\"WARC-Block-Digest\":\"sha1:3SGXQR5XYW7GC2YWRXFACUZJQ6MB4ZNS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027320156.86_warc_CC-MAIN-20190824084149-20190824110149-00069.warc.gz\"}"} |
https://docs.w3cub.com/erlang~21/doc/tutorial/erl_interface | [
"/Erlang 21\n\n# 5 Erl_Interface\n\nThis section outlines an example of how to solve the example problem in `Problem Example` by using a port and Erl_Interface. It is necessary to read the port example in `Ports` before reading this section.\n\n## 5.1 Erlang Program\n\nThe following example shows an Erlang program communicating with a C program over a plain port with home made encoding:\n\n```-module(complex1).\n-export([start/1, stop/0, init/1]).\n-export([foo/1, bar/1]).\n\nstart(ExtPrg) ->\nspawn(?MODULE, init, [ExtPrg]).\nstop() ->\ncomplex ! stop.\n\nfoo(X) ->\ncall_port({foo, X}).\nbar(Y) ->\ncall_port({bar, Y}).\n\ncall_port(Msg) ->\ncomplex ! {call, self(), Msg},\n{complex, Result} ->\nResult\nend.\n\ninit(ExtPrg) ->\nregister(complex, self()),\nprocess_flag(trap_exit, true),\nPort = open_port({spawn, ExtPrg}, [{packet, 2}]),\nloop(Port).\n\nloop(Port) ->\n{call, Caller, Msg} ->\nPort ! {self(), {command, encode(Msg)}},\n{Port, {data, Data}} ->\nCaller ! {complex, decode(Data)}\nend,\nloop(Port);\nstop ->\nPort ! {self(), close},\n{Port, closed} ->\nexit(normal)\nend;\n{'EXIT', Port, Reason} ->\nexit(port_terminated)\nend.\n\nencode({foo, X}) -> [1, X];\nencode({bar, Y}) -> [2, Y].\n\ndecode([Int]) -> Int.```\n\nThere are two differences when using Erl_Interface on the C side compared to the example in `Ports`, using only the plain port:\n\n• As Erl_Interface operates on the Erlang external term format, the port must be set to use binaries.\n• Instead of inventing an encoding/decoding scheme, the `term_to_binary/1` and `binary_to_term/1` BIFs are to be used.\n\nThat is:\n\n`open_port({spawn, ExtPrg}, [{packet, 2}])`\n\nis replaced with:\n\n`open_port({spawn, ExtPrg}, [{packet, 2}, binary])`\n\nAnd:\n\n```Port ! {self(), {command, encode(Msg)}},\n{Port, {data, Data}} ->\nCaller ! {complex, decode(Data)}\nend```\n\nis replaced with:\n\n```Port ! {self(), {command, term_to_binary(Msg)}},\n{Port, {data, Data}} ->\nCaller ! {complex, binary_to_term(Data)}\nend```\n\nThe resulting Erlang program is as follows:\n\n```-module(complex2).\n-export([start/1, stop/0, init/1]).\n-export([foo/1, bar/1]).\n\nstart(ExtPrg) ->\nspawn(?MODULE, init, [ExtPrg]).\nstop() ->\ncomplex ! stop.\n\nfoo(X) ->\ncall_port({foo, X}).\nbar(Y) ->\ncall_port({bar, Y}).\n\ncall_port(Msg) ->\ncomplex ! {call, self(), Msg},\n{complex, Result} ->\nResult\nend.\n\ninit(ExtPrg) ->\nregister(complex, self()),\nprocess_flag(trap_exit, true),\nPort = open_port({spawn, ExtPrg}, [{packet, 2}, binary]),\nloop(Port).\n\nloop(Port) ->\n{call, Caller, Msg} ->\nPort ! {self(), {command, term_to_binary(Msg)}},\n{Port, {data, Data}} ->\nCaller ! {complex, binary_to_term(Data)}\nend,\nloop(Port);\nstop ->\nPort ! {self(), close},\n{Port, closed} ->\nexit(normal)\nend;\n{'EXIT', Port, Reason} ->\nexit(port_terminated)\nend.```\n\nNotice that calling `complex2:foo/1` and `complex2:bar/1` results in the tuple `{foo,X}` or `{bar,Y}` being sent to the `complex` process, which codes them as binaries and sends them to the port. This means that the C program must be able to handle these two tuples.\n\n## 5.2 C Program\n\nThe following example shows a C program communicating with an Erlang program over a plain port with home made encoding:\n\n```/* port.c */\n\ntypedef unsigned char byte;\n\nint main() {\nint fn, arg, res;\nbyte buf;\n\nwhile (read_cmd(buf) > 0) {\nfn = buf;\narg = buf;\n\nif (fn == 1) {\nres = foo(arg);\n} else if (fn == 2) {\nres = bar(arg);\n}\n\nbuf = res;\nwrite_cmd(buf, 1);\n}\n}```\n\nCompared to the C program in `Ports`, using only the plain port, the `while`-loop must be rewritten. Messages coming from the port is on the Erlang external term format. They must be converted into an `ETERM` struct, which is a C struct similar to an Erlang term. The result of calling `foo()` or `bar()` must be converted to the Erlang external term format before being sent back to the port. But before calling any other Erl_Interface function, the memory handling must be initiated:\n\n`erl_init(NULL, 0);`\n\nThe following functions, `read_cmd()` and `write_cmd()`, from the `erl_comm.c` example in `Ports` can still be used for reading from and writing to the port:\n\n```/* erl_comm.c */\n\ntypedef unsigned char byte;\n\n{\nint len;\n\nif (read_exact(buf, 2) != 2)\nreturn(-1);\nlen = (buf << 8) | buf;\n}\n\nwrite_cmd(byte *buf, int len)\n{\nbyte li;\n\nli = (len >> 8) & 0xff;\nwrite_exact(&li, 1);\n\nli = len & 0xff;\nwrite_exact(&li, 1);\n\nreturn write_exact(buf, len);\n}\n\nread_exact(byte *buf, int len)\n{\nint i, got=0;\n\ndo {\nif ((i = read(0, buf+got, len-got)) <= 0)\nreturn(i);\ngot += i;\n} while (got<len);\n\nreturn(len);\n}\n\nwrite_exact(byte *buf, int len)\n{\nint i, wrote = 0;\n\ndo {\nif ((i = write(1, buf+wrote, len-wrote)) <= 0)\nreturn (i);\nwrote += i;\n} while (wrote<len);\n\nreturn (len);\n}```\n\nThe function `erl_decode()` from `erl_marshal` converts the binary into an `ETERM` struct:\n\n```int main() {\nETERM *tuplep;\n\nwhile (read_cmd(buf) > 0) {\ntuplep = erl_decode(buf);```\n\nHere, `tuplep` points to an `ETERM` struct representing a tuple with two elements; the function name (atom) and the argument (integer). Using the function `erl_element()` from `erl_eterm`, these elements can be extracted, but they must also be declared as pointers to an `ETERM` struct:\n\n```fnp = erl_element(1, tuplep);\nargp = erl_element(2, tuplep);```\n\nThe macros `ERL_ATOM_PTR` and `ERL_INT_VALUE` from `erl_eterm` can be used to obtain the actual values of the atom and the integer. The atom value is represented as a string. By comparing this value with the strings \"foo\" and \"bar\", it can be decided which function to call:\n\n```if (strncmp(ERL_ATOM_PTR(fnp), \"foo\", 3) == 0) {\nres = foo(ERL_INT_VALUE(argp));\n} else if (strncmp(ERL_ATOM_PTR(fnp), \"bar\", 3) == 0) {\nres = bar(ERL_INT_VALUE(argp));\n}```\n\nNow an `ETERM` struct that represents the integer result can be constructed using the function `erl_mk_int()` from `erl_eterm`. The function `erl_format()` from the module `erl_format` can also be used:\n\n`intp = erl_mk_int(res);`\n\nThe resulting `ETERM` struct is converted into the Erlang external term format using the function `erl_encode()` from `erl_marshal` and sent to Erlang using `write_cmd()`:\n\n```erl_encode(intp, buf);\nwrite_cmd(buf, erl_eterm_len(intp));```\n\nFinally, the memory allocated by the `ETERM` creating functions must be freed:\n\n```erl_free_compound(tuplep);\nerl_free_term(fnp);\nerl_free_term(argp);\nerl_free_term(intp);```\n\nThe resulting C program is as follows:\n\n```/* ei.c */\n\n#include \"erl_interface.h\"\n#include \"ei.h\"\n\ntypedef unsigned char byte;\n\nint main() {\nETERM *tuplep, *intp;\nETERM *fnp, *argp;\nint res;\nbyte buf;\nlong allocated, freed;\n\nerl_init(NULL, 0);\n\nwhile (read_cmd(buf) > 0) {\ntuplep = erl_decode(buf);\nfnp = erl_element(1, tuplep);\nargp = erl_element(2, tuplep);\n\nif (strncmp(ERL_ATOM_PTR(fnp), \"foo\", 3) == 0) {\nres = foo(ERL_INT_VALUE(argp));\n} else if (strncmp(ERL_ATOM_PTR(fnp), \"bar\", 3) == 0) {\nres = bar(ERL_INT_VALUE(argp));\n}\n\nintp = erl_mk_int(res);\nerl_encode(intp, buf);\nwrite_cmd(buf, erl_term_len(intp));\n\nerl_free_compound(tuplep);\nerl_free_term(fnp);\nerl_free_term(argp);\nerl_free_term(intp);\n}\n}```\n\n## 5.3 Running the Example\n\nStep 1. Compile the C code. This provides the paths to the include files `erl_interface.h` and `ei.h`, and also to the libraries `erl_interface` and `ei`:\n\n```unix> gcc -o extprg -I/usr/local/otp/lib/erl_interface-3.9.2/include \\\\\n-L/usr/local/otp/lib/erl_interface-3.9.2/lib \\\\\ncomplex.c erl_comm.c ei.c -lerl_interface -lei -lpthread```\n\nIn Erlang/OTP R5B and later versions of OTP, the `include` and `lib` directories are situated under `OTPROOT/lib/erl_interface-VSN`, where `OTPROOT` is the root directory of the OTP installation (`/usr/local/otp` in the recent example) and `VSN` is the version of the Erl_interface application (3.2.1 in the recent example).\n\nIn R4B and earlier versions of OTP, `include` and `lib` are situated under `OTPROOT/usr`.\n\nStep 2. Start Erlang and compile the Erlang code:\n\n```unix> erl\nErlang (BEAM) emulator version 4.9.1.2\n\nEshell V4.9.1.2 (abort with ^G)\n1> c(complex2).\n{ok,complex2}```\n\nStep 3. Run the example:\n\n```2> complex2:start(\"./extprg\").\n<0.34.0>\n3> complex2:foo(3).\n4\n4> complex2:bar(5).\n10\n5> complex2:bar(352).\n704\n6> complex2:stop().\nstop```\n\n© 2010–2017 Ericsson AB\nLicensed under the Apache License, Version 2.0."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.647469,"math_prob":0.9148997,"size":7801,"snap":"2022-40-2023-06","text_gpt3_token_len":2355,"char_repetition_ratio":0.09554957,"word_repetition_ratio":0.22759227,"special_character_ratio":0.31816435,"punctuation_ratio":0.21755971,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96646345,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T04:48:19Z\",\"WARC-Record-ID\":\"<urn:uuid:ac2eaa00-3e42-425b-a436-ae54cf71848b>\",\"Content-Length\":\"16248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99d742c4-e415-4c59-a65a-f7431936a690>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a40c8db-7895-4713-aa15-5724f599081d>\",\"WARC-IP-Address\":\"172.67.188.141\",\"WARC-Target-URI\":\"https://docs.w3cub.com/erlang~21/doc/tutorial/erl_interface\",\"WARC-Payload-Digest\":\"sha1:ILQCSPERNKOG2F4ODKAX6POYVIWRATJ4\",\"WARC-Block-Digest\":\"sha1:YGWRRZOHVYPCTXQRJ5ZOPOTPLYW6QSSV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335304.71_warc_CC-MAIN-20220929034214-20220929064214-00503.warc.gz\"}"} |
https://www.dsir.in/forum/viewtopic.php?a3703b=ratio-and-proportion-worksheets-9th-grade | [
"Here you find our ratio and proportion worksheets for math classes 5 and 6.\n\nThere is also a Print Worksheet button.\n\nRatios are used to make a comparison between numbers or units.\n\n2 : 6 is similar to 2/6 and reducing the ratio or the fraction is identical and results in 1 : 3. We can also express ratios as decimals or as units.\n\nDo these items have the same ratio? the resource covers how to use mental math by reducing one of the sides of the proportion and then applying a multiplicative relationship.... Clarify complex math concepts with a helpful video.\n\nHow many girls and how many boys are in this school? They determine the best buy and identify the ratio of different sized polygons. I would recommend these exercise for 5th grade, 6th grade, and 7th grade math students. Save time and discover engaging curriculum for your classroom.\n\nInterpretation of a:b or a:b:c using whole numbers, fractions and decimal numbers. Choose one of the following ratio and proportion worksheet categories. Learners watch an instructive video that explains how to solve proportions. Here you find our ratio and proportion worksheets for math classes 5 and 6. A rectangle field has an area of 300 square meters and a perimeter of 80 meters. TeacherVision is part of the FEN Learning family of educational and reference sites for parents, teachers and students. Middle-school math magicians solve nine problems where they are given a proportional equation and use cross products to find a missing term. This Ratio and Proportion Worksheet is suitable for 9th - 11th Grade. Direct change by finding the constant. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Ratio and Proportion Worksheets. Determine if the relationship is proportional worksheet. Ratios can be written by using the word 'to', 3 to 2, by using a colon:, 3 : 2, or by expressing the ratio as a fraction: 1/2. Nature of the roots of a quadratic equation worksheets. 2020 analyzemath.com. There are 600 pupils in a school. They list the means and extremes of given proportions and solve proportion equations.\n\nThis resource is only available on an unencrypted HTTP website. First Day Activities For High School Students, Teaching Mathematics to Gifted Students in a Mixed-Ability Classroom, The Curious History of the Gregorian Calendar. Looking for high-quality Math worksheets aligned to Common Core standards for Grades K-8? Ratio Problems with Answers for Grade 9. Working and finding the ratios between numbers is both challenging and exciting. FEN Learning is part of Sandbox Networks, a digital learning company that operates education services and products for the 21st century. Here you find our ratio and proportion worksheets for math classes 5 and 6.\n\nAn exit ticket checks for understanding at the end of the... Is it possible that 8:4, 8/4, and 8 to 4 all mean the same thing? Educators earn digital badges that certify knowledge, skill, and experience.\n\nThis video might be for you! The length of the side of square A is twice the length of the side of square B. Please click the following links to get math printable math worksheets for grade 9.\n\nIn this solving proportions learning exercise, 9th graders solve and complete 16 different equations that include using proportions. These unit rate and proportion worksheets will help students meet Common Core Standards for Ratios & Proportional Relationships by working with unit rates in multiple representations.\n\nA comprehensive online edtech PD solution for schools and districts."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91590554,"math_prob":0.8988182,"size":6013,"snap":"2021-04-2021-17","text_gpt3_token_len":1264,"char_repetition_ratio":0.14378433,"word_repetition_ratio":0.13347237,"special_character_ratio":0.21154165,"punctuation_ratio":0.11889973,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903289,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T03:33:43Z\",\"WARC-Record-ID\":\"<urn:uuid:14cc6271-d781-47d2-846f-715388f05694>\",\"Content-Length\":\"29421\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bcb0d92-d18d-43be-ac77-67d0e68e4498>\",\"WARC-Concurrent-To\":\"<urn:uuid:c4801ba2-80ce-4755-9876-9ad237a568e1>\",\"WARC-IP-Address\":\"173.236.159.235\",\"WARC-Target-URI\":\"https://www.dsir.in/forum/viewtopic.php?a3703b=ratio-and-proportion-worksheets-9th-grade\",\"WARC-Payload-Digest\":\"sha1:OHLHAMDPHNIYJKMUMF7OZBVLIUTAWZNC\",\"WARC-Block-Digest\":\"sha1:K7AULKKFCQTDCTTRTJWUEDBARUAJDXNU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703544403.51_warc_CC-MAIN-20210124013637-20210124043637-00155.warc.gz\"}"} |
https://stat.ethz.ch/pipermail/r-devel/2010-February/056682.html | [
"# [Rd] Error in coding for splinefun(method = \"monoH.FC\") (PR#14215)\n\nt.heaton at shef.ac.uk t.heaton at shef.ac.uk\nThu Feb 18 16:10:09 CET 2010\n\n```Full_Name: Tim Heaton\nVersion: 2.8.1\nOS: linux-gnu\nSubmission from: (NULL) (143.167.4.162)\n\nHi,\n\nIn my version of R, the stats package splinefun code for fitting a\nFritsch and Carlson monotonic spline does not guarantee a\nmonotonic result. If two adjoining sections both have over/undershoot\nthe way the resulting adjustment of alpha and beta is performed can give\nmodified values which still do not satisfy the required constraints. I posed the\nquestion as to whether this was a known error on the R help but got no reply,\nhave also had a look through the bug database but couldn't find anything.\n\nBelow is an example created to demonstrate this,\n\n###############################################\n# Create the following data\n# This is created so that their are two adjoining sections which have to\nx <- 1:8\ny <- c(-12, -10, 3.5, 4.45, 4.5, 140, 142, 142)\n\n# Now run the splinefun() function\n\nFailMonSpline <- splinefun(x, y, method = \"mono\")\n\n# In theory this should be monotonic increasing but the required\nconditions are not satisfied\n\n# Check values of alpha and beta for this curve\nm <- FailMonSpline(x, deriv = 1)\nnx <- length(x)\nn1 <- nx - 1L\ndy <- y[-1] - y[-nx]\ndx <- x[-1] - x[-nx]\nSx <- dy/dx\n\nalpha <- m[-nx]/Sx\nbeta <- m[-1]/Sx\na2b3 <- 2 * alpha + beta - 3\nab23 <- alpha + 2 * beta - 3\nok <- (a2b3 > 0 & ab23 > 0)\nok <- ok & (alpha * (a2b3 + ab23) < a2b3^2)\n# If the curve is monotonic then all ok should be FALSE however this is\nnot the case\nok\n\n# Alternatively can easily seen to be non-monotonic by plotting the\nregion between 4 and 5\n\nt <- seq(4,5, length = 200)\nplot(t, FailMonSpline(t), type = \"l\")\n\n########################################################\nThe version of R I am running is\n\nplatform x86_64-suse-linux-gnu\narch x86_64\nos linux-gnu\nsystem x86_64, linux-gnu\nstatus\nmajor 2\nminor 8.1\nyear 2008\nmonth 12\nday 22\nsvn rev 47281\nlanguage R\nversion.string R version 2.8.1 (2008-12-22)\n\n```"
] | [
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https://www.asknumbers.com/kg-to-lbs/26.9-kg-to-lbs.aspx | [
"# How Many Pounds in 26.9 Kilograms?\n\n26.9 kg to lbs converter. How many pounds in 26.9 kg?\n\n26.9 kg equal to 59.3043485 lbs or there are 59.3043485 pounds in 26.9 kg.\n\n←→\nstep\nRound:\nEnter Kilogram\nEnter Pound\n\n## How to convert 26.9 kg to lbs?\n\nThe conversion factor from kg to lbs is 2.2046226218. To convert any value of kg to lbs, multiply the kilogram value by the conversion factor.\n\nTo convert 26.9 kg to lbs, multiply 26.9 by 2.2046226218, that makes 26.9 kg equal to 59.3043485 lbs.\n\n26.9 kg to lbs formula\n\nlbs = kg value * 2.2046226218\n\nlbs = 26.9 * 2.2046226218\n\nlbs = 59.3043485\n\nCommon conversions from 26.9x kg to lbs:\n(rounded to 3 decimals)\n\n• 26.9 kg = 59.304 lbs\n• 26.91 kg = 59.326 lbs\n• 26.92 kg = 59.348 lbs\n• 26.93 kg = 59.37 lbs\n• 26.94 kg = 59.393 lbs\n• 26.95 kg = 59.415 lbs\n• 26.96 kg = 59.437 lbs\n• 26.97 kg = 59.459 lbs\n• 26.98 kg = 59.481 lbs\n• 26.99 kg = 59.503 lbs\n• 27.0 kg = 59.525 lbs\n\nWhat is a Kilogram?\n\nKilogram (kilo) is the metric system base unit of mass. 1 Kilogram = 2.2046226218 Pounds. The symbol is \"kg\".\n\nWhat is a Pound?\n\nPound is an imperial system mass unit. 1 Pound = 0.45359237 Kilogram. 1 Kilogram = 2.2046226218 Pounds. The symbol is \"lb\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73143464,"math_prob":0.9933537,"size":968,"snap":"2022-40-2023-06","text_gpt3_token_len":385,"char_repetition_ratio":0.20954357,"word_repetition_ratio":0.042105265,"special_character_ratio":0.5165289,"punctuation_ratio":0.21400778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982221,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T01:32:17Z\",\"WARC-Record-ID\":\"<urn:uuid:39a58e40-a4d7-41b2-93c5-d24dec3b368b>\",\"Content-Length\":\"43121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50a5113c-fa78-4b09-bee4-0e14f8bd29c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:15a95a37-91d0-44f0-8ff7-800aa091823c>\",\"WARC-IP-Address\":\"104.21.33.54\",\"WARC-Target-URI\":\"https://www.asknumbers.com/kg-to-lbs/26.9-kg-to-lbs.aspx\",\"WARC-Payload-Digest\":\"sha1:DDPWCFXTGCOCXAGWL6TUX6S7IJO4RAR5\",\"WARC-Block-Digest\":\"sha1:VWLLTI7YKGHQCV4VG27QLMAGEPK4QO24\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499954.21_warc_CC-MAIN-20230202003408-20230202033408-00707.warc.gz\"}"} |
https://wiki.bennugd.org/index.php?title=Sort | [
"# Sort\n\n## Syntax\n\nINT sort ( <VARSPACE array> , [<INT datacount>] )\n\n## Description\n\nSorts an array by sorting a certain number of elements, by using the first variable in each element. By default the whole array is sorted.\n\nIf the elements contain multiple variables, ksort() can be used to select the variable to be used for sorting. For more advanced sorting, look at quicksort().\n\n## Parameters\n\n VARSPACE array - The array to be sorted. [INT datacount] - Number of elements to sort.\n\n## Returns\n\nINT: Successrate\n\n true - Sorting succeeded. false - Sorting failed. Look in the output for the error.\n\n## Example\n\n```import \"mod_sort\";\nimport \"mod_key\";\nimport \"mod_say\";\n\nType _player\nString name;\nint score;\nEnd\n\nConst\nmaxplayers = 5;\nEnd;\n\nProcess main()\nPrivate\n_player player[maxplayers-1];\nint i=0;\nBegin\n// Insert some values\nplayer.name = \"That one bad looking dude\";\nplayer.name = \"Ah pretty lame guy\";\nplayer.name = \"Some cool dude\";\nplayer.name = \"OMG ZOMG guy\";\nplayer.name = \"This person is ok\";\n\nplayer.score = 70;\nplayer.score = 30;\nplayer.score = 80;\nplayer.score = 90;\nplayer.score = 50;\n\n// Show array\nsay(\"-------------------- unsorted\");\nfor(i=0; i<maxplayers; i++)\nsay(player[i].name + \" - \" + player[i].score);\nend\n\n/* Sort by name ( quicksort() can't be used to sort Strings,\nas a String in Bennu is a pointer to the actual String,\nso it would sort the pointer addresses */\n\n// sort()\nsort(player); // sorts by name because name is the first variable in each element\n\n// Show array\nsay(\"-------------------- name - sort()\");\nfor(i=0; i<maxplayers; i++)\nsay(player[i].name + \" - \" + player[i].score);\nend\n\n// ksort()\nksort(player,player.name,maxplayers);\n\n// Show array\nsay(\"-------------------- name - ksort()\");\nfor(i=0; i<maxplayers; i++)\nsay(player[i].name + \" - \" + player[i].score);\nend\n\n/* Sort by score (sort() cannot be used here, because score is not the first variable) */\n\n// ksort()\nksort(player,player.score,maxplayers);\n\n// Show array\nsay(\"-------------------- score - ksort()\");\nfor(i=0; i<maxplayers; i++)\nsay(player[i].name + \" - \" + player[i].score);\nend\n\n// quicksort()\nquicksort(&player,sizeof(_player),maxplayers,sizeof(String),sizeof(int),0);\n\n// Show array\nsay(\"-------------------- score - quicksort()\");\nfor(i=0; i<maxplayers; i++)\nsay(player[i].name + \" - \" + player[i].score);\nend\nEnd\n```\n\nUsed in example: say(), sort(), ksort(), quicksort(), type, array, pointer"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5173822,"math_prob":0.903826,"size":2445,"snap":"2021-43-2021-49","text_gpt3_token_len":666,"char_repetition_ratio":0.21179844,"word_repetition_ratio":0.1637931,"special_character_ratio":0.35092026,"punctuation_ratio":0.20806794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97796094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T20:39:28Z\",\"WARC-Record-ID\":\"<urn:uuid:0dda7bac-5dc7-43a4-87d5-51b06a22b1e2>\",\"Content-Length\":\"21449\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:484478ba-8851-4872-85ea-de5c0f479e9a>\",\"WARC-Concurrent-To\":\"<urn:uuid:63209203-06f9-413e-a859-6fa4b365a863>\",\"WARC-IP-Address\":\"208.113.170.70\",\"WARC-Target-URI\":\"https://wiki.bennugd.org/index.php?title=Sort\",\"WARC-Payload-Digest\":\"sha1:RN4ZMNQKHJKCLFKM6CVEUFYB64OKROTS\",\"WARC-Block-Digest\":\"sha1:BHATGSVGPZEKVLZCF23G6QTUXASIUDFZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358842.4_warc_CC-MAIN-20211129194957-20211129224957-00606.warc.gz\"}"} |
https://se.mathworks.com/matlabcentral/profile/authors/9006977 | [
"Community Profile",
null,
"# Marc Laub\n\nLast seen: 5 månader ago Active since 2019\n\n#### Statistics\n\n•",
null,
"#### Content Feed\n\nView by\n\nQuestion\n\nBinning 3D data und reasign Bin properties to data\nHey, following problem I have following problem: I have 3D data from which I need to calculate properties. To reduce the calcu...\n\n### 1\n\nQuestion\n\nExtracting data from handle objects\nHey, I have following problem. I have multiple handle class objects which all have a propertie which can either be a scalar or ...\n\n### 1\n\nQuestion\n\nVectorize class function over all objects\nHey, I have a handle class and each of those classes has some properties that will be changed over time repeatedly. classdef p...\n\n### 1\n\nQuestion\n\nAcess all last array elements withtin cells\nHey, I need to store somae data and later acess them again. For further calculation that is done repetedly I would prefer to a...\n\n### 2\n\nQuestion\n\nCheck if cell contains another cell\nHey, I need to check if my cell contains another cell or if the contant is a character vector. So I got a a=1x130 cell and ...\n\n### 2\n\nQuestion\n\nReceiving different results from custom function depending on input\nHey guys, I have a function which is basically just a concatenation of polynoms, log-terms... i handles 3 input variables and...\n\n### 0\n\nQuestion\n\nHey, so I am using fmincn to solve a problem. Im somehow not to satisfioed with the results because the solver seems to bug its...\n\n### 1\n\nQuestion\n\nis it possible to use function without toolbox\nHey everyone, I am currently trying to solve a problem, I actually solved it, and I need the fmincon function from the optimiza...\n\n### 2\n\nQuestion\n\ndefine nonlcon in fmincon\nHey, I am trying to finc the minimum of a 4 variable equation using fmincon. Looking at the solution fmincon presents and saw t...\n\nmer än ett år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nfminsearch with side conditions\nHey, I have a equation where I am looking for the minimum. The equation has 6 variables, but due to interdependencies it can ...\n\nmer än ett år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nProblems with fminsearch giving startvalues as result\nHey, I am trying to minimize Gibbs enthalpie dependant an phase fraction and phase compositions. So i set up an equation whic...\n\nmer än ett år ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nPlotting vertical line with specified color\nHey, I want to visualize the development of a distribution over time. Therefore I want to plot the distribution of the diame...\n\nmer än ett år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nIndexing a multidimensional matrix with a second logical matrix\nHello, I want to setup dummy orientation given a binary image. I have a binary images (data) with withe regions separated from...\n\nmer än ett år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nChanging Errorbar Markersize within Legend\nHey, following problem: I am plotting some errorbars in my figure. I control the appearend size of the y value with 'Markersiz...\n\nmer än ett år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nWriting a small epsilon in xlabel with color\nHey, I wanted to write a red small epsilon in my label \\color{red} \\epsilon was my first try, but is was the epsilon version i...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to remove horizontal and vertical artefacts from image\nHey, I have some images that where stiched together from multiple images and now after the calculations these stich lines whi...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nVariable cell array size for mex function does not work\nHey, I' am trying to bypass the 32bit limit of the MatLab coder, so I chunked my matrix into pieces and wnt to give that pieces...\n\nmer än 2 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nEditing code for mex function to support 64 bit indexing\nHey, I want to speed up my code by using a mex function, I actually already did but the probem is that MatLab coder only suppor...\n\nmer än 2 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nHow to change font in legend while using latex interpreter\nHey, I needed a fraction bar in my legend, so I used latex as interpreter, but now the legend font is in the default tex math f...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nDescribe to plots in one line in legend\nHey, I have a plot full of lines and to not overload the legend I wanted to only describe some major differences between the ...\n\nmer än 2 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHow to manually edit MatLab Coder .mex-file code to support 64bit indexing\nHey, I have a problem that mex files generated by MatLab Coder only supports 32bit indexing, which is the reason my program get...\n\nnästan 3 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nUnexpected unknown exception from MEX file..\nHey, I wrote my bottleneck function as a mex function and it works well so far, but at a sudden point in my programm I just g...\n\nnästan 3 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nChange position of exponent of y axis in figure\nHey, I am wondering if there is a possibility to change the position of the exponent of the y axis in a figure. In my case the...\n\nnästan 3 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nConvert function with struct input to mex while number of fields is unknown\nHey guys, I am trying to translate a matlab function of mine into a mex function but I am facing some issues. The problem is, ...\n\nnästan 3 år ago | 0 answers | 0\n\n### 0\n\nCode acceleration by mex-file\nThanks, it was the double instead of integer problem. I simply typed the code as in that tutorial but somehow in his tutorial ...\n\nnästan 3 år ago | 0\n\n| accepted\n\nQuestion\n\nCode acceleration by mex-file\nHey, I'am new to the mex-section, but I wanted to accelerate my code, more a very specific time consuming function. To start ...\n\nnästan 3 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nSpecify band width for kernel PDF\nHey, I'm wondering if there is a possibilty to use the improves Sheater Jones algorythm for optimal bandwith in multimodal di...\n\nnästan 3 år ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nHow to implement load function into MatLab Coder\nHey guys, I wrote a programm in Matlab and it takes quite a long time to run through, so we wanted to know how much faster it...\n\nnästan 3 år ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nHey following thought about the adjacency matrix of a graph. Is it possible to distinguish from the adjacency matrix of a grap...\n\nungefär 3 år ago | 4 answers | 0"
] | [
null,
"https://se.mathworks.com/responsive_image/150/150/0/0/0/cache/matlabcentral/profiles/9006977_1532424603484_DEF.jpg",
null,
"https://se.mathworks.com/matlabcentral/profile/badges/Thankful_20.png",
null
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https://newpathworksheets.com/math/grade-4/evaluate-open-sentences/utah-standards | [
"## ◂Math Worksheets and Study Guides Fourth Grade. Evaluate Open Sentences\n\n### The resources above correspond to the standards listed below:\n\n#### Utah Core Standards\n\nUT.4.OA. OPERATIONS AND ALGEBRAIC THINKING (4.OA)\nUse the four operations with whole numbers (addition, subtraction, multiplication, and division) to solve problems. Gain familiarity with factors and multiples. Generate and analyze numeric and shape patterns. Demonstrate complete fluency with products of one-digit numbers.\n4.OA.1. Interpret a multiplication equation as a comparison (for example, interpret 35 = 5 x 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5). Represent verbal statements of multiplicative comparisons as multiplication equations."
] | [
null
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https://www.rdocumentation.org/packages/lhs/versions/1.0.1/vignettes/augment_lhs.Rmd | [
"# An Example of Augmenting a Latin Hypercube\n\nknitr::opts_chunk$set( collapse = TRUE, comment = \"#>\" ) source(\"VignetteCommonCode.R\") require(lhs) graph2DaugmentLHS1 <- function(sims, extras) { A <- randomLHS(sims, 2) B <- augmentLHS(A, extras) plot.default(A[,1], A[,2], type = \"n\", ylim = c(0,1), xlim = c(0,1), xlab = \"x1\", ylab = \"x2\", xaxs = \"i\", yaxs = \"i\", main = \"\") for (i in 1:length(A[,1])) { rect(floor(A[i,1]*sims)/sims, floor(A[i,2]*sims)/sims, ceiling(A[i,1]*sims)/sims, ceiling(A[i,2]*sims)/sims, col = \"grey\") } points(A[,1], A[,2], pch = 19, col = \"red\") abline(v = (0:sims)/sims, h = (0:sims)/sims) return(list(A = A, B = B, sims = sims, extras = extras)) } graph2DaugmentLHS2 <- function(X) { A <- X$A B <- X$B sims <- X$sims extras <- X\\$extras plot.default(A[,1], A[,2], type = \"n\", ylim = c(0,1), xlim = c(0,1), xlab = \"x1\", ylab = \"x2\", xaxs = \"i\", yaxs = \"i\", main = \"\") N <- sims + extras for (i in 1:length(B[,1])) { rect(floor(B[i,1]*N)/N, floor(B[i,2]*N)/N, ceiling(B[i,1]*N)/N, ceiling(B[i,2]*N)/N, col = \"grey\") } points(A[,1], A[,2], pch = 19, col = \"red\") points(B[((sims + 1):(sims + extras)), 1], B[((sims + 1):(sims + extras)), 2], pch = 19, col = \"blue\") abline(v = (0:N)/N, h = (0:N)/N) } # X <- graph2DaugmentLHS1(5,5) # graph2DaugmentLHS2(X)\n\nSuppose that a computer simulation study is being designed that requires expensive runs. A Latin hypercube design is desired for this simulation so that the expectation of the simulation output can be estimated efficiently given the distributions of the input variables. Latin hypercubes are most often used in highly dimensional problems, but the example shown is of small dimension. Suppose further that the total extent of funding is uncertain. Enough money is available for 5 runs, and there is a chance that there will be enough for 5 more. However, if the money for the additional 5 runs does not materialize, then the first 5 runs must be a Latin hypercube alone. A design for this situation can be created using the lhs package.\n\nFirst create a random Latin hypercube using the randomLHS(n, k) command:\n\nA <- randomLHS(5,2)\n\nAn example of this hypercube is shown in r registerFigure(\"X\"). Note that the Latin property of the hypercube requires that each of the 5 equal probability intervals be filled (i.e. each row and each column is filled with one point). Also notice that the exact location of the design point is randomly sampled from within that cell using a uniform distribution for each marginal variable.\n\nr addFigureCaption(\"X\", \"A randomly produced Latin Hypercube with uniform marginal distributions for 2 parameters with 5 simulations\", register=FALSE)\n\nset.seed(10) X <- graph2DaugmentLHS1(5, 5)\n\nNext, in order to augment the design with more points use augmentLHS(lhs, m). The following will add 5 more points to the design:\n\nB <- augmentLHS(A, 5)\n\nThe augmentLHS function works by re-dividing the original design into n+m intervals (e.g. 5+5=10) keeping the original design points exactly in the same position. It then randomly fills the empty row-column sets. The results are shown in r registerFigure(\"Y\").\n\nr addFigureCaption(\"Y\", \"A randomly produced Latin Hypercube of 5 points (red) with 5 augmented points (blue). Each parameter has a uniform marginal distribution.\", register=FALSE)\n\ngraph2DaugmentLHS2(X)\n\nThe augmentLHS function uses the following algorithm (see the documentation for augmentLHS):\n\n• Create a new (n+m) by k matrix to hold the candidate points after the design has been re-partitioned into (n+m)^2 cells, where n is number of points in the original lhs matrix.\n• Then randomly sweep through each column (1...k) in the repartitioned design to find the missing cells.\n• For each column (variable), randomly search for an empty row, generate a random value that fits in that row, record the value in the new matrix. The new matrix can contain more than m points unless m = 2n, in which case the new matrix will contain exactly m filled rows.\n• Finally, keep only the first m rows of the new matrix. It is guaranteed that there will be m full rows (points) in the new matrix. The deleted rows are partially full. The additional candidate points are selected randomly because of the random search used to find empty cells.\n\nAlso notice that because the original points are randomly placed within the cells, depending on how you bin the marginal distributions, a histogram (of x1 for example) will not necessarily be exactly uniform.\n\nNow, the augmenting points do not necessarily form a Latin Hypercube themselves. The original design and augmenting points may form a Latin Hypercube, or there may be more than one point per row in the augmented design. If the augmented points are equal to the number of original points, then a strictly uniform Latin hypercube is guaranteed. An example of an augmented design which is not uniform in the marginal distributions is given in r registerFigure(\"Z\") and r registerFigure(\"W\"). The commands were:\n\nA <- randomLHS(7, 2) B <- augmentLHS(A, 3)\n\nr addFigureCaption(\"Z\", \"Original design with 7 points\", register=FALSE)\n\nset.seed(12) X <- graph2DaugmentLHS1(7, 3)\n\nr addFigureCaption(\"W\", \"Augmented design with 3 additional points. Note that row 9 has 2 points and row 3 has none.\", register=FALSE)\n\ngraph2DaugmentLHS2(X)"
] | [
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=29&t=49770 | [
"## coulomb potential energy and Madelung constant, as well as lattice energy\n\nEvaLi_3J\nPosts: 53\nJoined: Wed Oct 02, 2019 12:16 am\n\n### coulomb potential energy and Madelung constant, as well as lattice energy\n\nAre we supposed to know these concepts and equations for midterms and finals?\n\nIf yes, can someone explain how the lattice energy works please?\n\nKeyaV1C\nPosts: 103\nJoined: Sat Aug 17, 2019 12:17 am\nBeen upvoted: 1 time\n\n### Re: coulomb potential energy and Madelung constant, as well as lattice energy\n\nI don’t think we went over these concepts in class\n\nYiyang Jen Wang 4G\nPosts: 76\nJoined: Wed Nov 21, 2018 12:18 am\n\n### Re: coulomb potential energy and Madelung constant, as well as lattice energy\n\nI hope these will not be on the test or I would definitely fail.:(\n\nEmilyJoo_1G\nPosts: 50\nJoined: Thu Jul 25, 2019 12:16 am\n\n### Re: coulomb potential energy and Madelung constant, as well as lattice energy\n\nThe midterm will only cover up concepts up to 2D in the Chemical Bonds unit, so you don't need to worry about lattice energy\n\nSean Cheah 1E\nPosts: 105\nJoined: Wed Sep 18, 2019 12:20 am\n\n### Re: coulomb potential energy and Madelung constant, as well as lattice energy\n\nLattice energy represents the change in energy that occurs when separated gaseous ions are packed together to form the ionic solid, which is always negative due to the process being exothermic.\n\nThis kinda gets into the realm of physics but essentially the negative derivative of the electrostatic forces between two ions (as given by Coulomb's Law) gives you their electric potential (aka Coulombic energy).\n\nThe Madelung constant is a scalar that depends on the geometric structure of the ion in question. While the formula for Coulombic energy only takes the interaction between two ions into account, since an ionic solid consists entirely of these two ions, one can focus on an arbitrary single ion and derive a very messy infinite sum that totals up all the electric potentials resulting from the forces between the chosen ion and each of the other ions in the lattice. Dividing the value of that infinite sum by the electric potential between two ions yields the Madelung constant for that ionic solid.\n\nPutting that all together, lattice energy is equal to the electric potential of a single ion in the lattice (which is given by the sum of all the electric potentials between that ion and each of the other ions in the lattice) combined with the repulsive energy experienced by that ion due to electron repulsion. Most sources list lattice energies with the unit kJ/mol by multiplying this value by Avogadro's number and dividing by 1000."
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https://gateoverflow.in/239308/madeeasy-subject-test-2019-programming-%26-ds-binary-tree | [
"848 views",
null,
"in DS\nedited | 848 views\n+1\nHINT: Read question One more time, It is saying P-Q<=2 where P and Q are no. of nodes in the left and Right Subtree and Not height of left and right subtree!",
null,
"But now i am interested to find it in general height.\n\nwe know that the no.of nodes of right side should be less than 2 at maximum only\n\n===> for minimum no.of nodes, i consider no.of right subtree nodes = no.of left subtree nodes - 2\n\n===> Total nodes in that tree = no.of left subtree nodes + no.of right subtree nodes + 1 ( root node)\n\n= no.of left subtree nodes + ( no.of left subtree nodes - 2 ) + 1\n\n= no.of left subtree nodes + no.of left subtree nodes - 2 + 1\n\n= 2 * no.of left subtree nodes - 1\n\nwhat is no.of nodes in left subtree ?\n\nit should be equal to no.of nodes in h-1, where h is height of your tree\n\nTotal nodes in that tree = 2 * no.of left subtree nodes - 1\n\nN(H) = 2 * ( N(H-1) ) - 1\n\nfor solving this recurrence relation, the base condition is N(H=3) = 5\n\nusing Back Substitution, we an solve it, the final value we can get\n\nN(H) = 4 . 2(H-3) + 1 = 22 . 2(H-3) + 1 = 2(H-3)+2 + 1 = 2(H-1) + 1.\n\nselected by\n0\n\nNice explanation @shaik sir :)\nI have just one doubt.\n\nWhat is no.of nodes in left subtree ?\nIt shouId be equal to no.of nodes in h-1, where h is height of your tree\n\nCan you please tell me how you get this?\n\n+1\n\nmam, By seeing the image, h=3 which is in green color, should be use as left subtree as h=4.\n\nif you didn't get that, follow\n\nCan you please tell me how you get this?\n\nFor getting a tree of height h, let fix a node as root. then your left subtree or right subtree should have the height h-1, due to overall height is h.\n\nBy convention i go with left subtree as height h-1. Howmany nodes my left sub tree should have ?\n\nit is like our original problem ( Howmany minimum Nodes the tree have with height h with given property ?)\n\n∴ if my problem is N(H) then my new problem is N(H-1) and My right Subtree have ( N(H-1) - 2 )\n\n===> N(H) = 1 (for root) + N(H-1) ( for left subtree ) + ( N(H-1) - 2 ) ( for right subtree )\n\nN(H) = 2 * ( N(H-1) ) - 1\n\n0\n@Shaik\n\nhow u got $N(H) = 4 . 2^{(H-3)} + 1$?\n\nplz elaborate it\n\nN(H) = 2 * ( N(H-1) ) - 1//this is fine\n\nN(H=3) = 5// this line also fine\n\nbut then how u got $N(H) = 4 . 2^{(H-3)} + 1$?\n0\n\nSee this question https://gateoverflow.in/1210/gate2007-12\n\nThere is small difference between these two question\n\nOne taking height difference 2 and another not taking it\n\nMoreover, this question answer is $2^{h-1}+1$\n\nAnother one answer is $2^{h+1}-1$\n\nWhy this difference comes?\n\n+1\n\nbut then how u got N(H)=4.2(H−3)+1?\n\nN(H) = 2 * ( N(H-1) ) - 1\n\n= 2 * ( 2 * N(H-2) - 1 ) - 1 = 22 * N(H-2) - 2 -1\n\n= 22 * ( 2 * N(H-3) - 1 ) - 2 - 1 = 23 * ( N(H-3) ) - 22 - 2 - 1\n\nafter K iterations,\n\n= 2K * ( N(H-K) ) - 2(K-1) - ..... - 22 - 2 - 1 )\n\n= 2K * ( N(H-K) ) - ( 2(K-1) + ..... + 22 + 2 + 20 )\n\n= 2K * ( N(H-K) ) - ( 2(K) - 1 )\n\n= 2K * ( N(H-K) ) - 2(K) + 1\n\n= 2K * ( N(H-K) - 1 ) + 1\n\nlet H-K = 3 ===> K = H-3 ===> the eqn turned into\n\n= 2(H-3) * ( N(3) - 1 ) + 1\n\n= 2(H-3) * ( 5 - 1 ) + 1\n\n= 2(H-3) * ( 4 ) + 1\n\nSee this question https://gateoverflow.in/1210/gate2007-12\n\nThere is small difference between these two question\n\nOne taking height difference 2 and another not taking it\n\nMoreover, this question answer is 2h−1+1\n\nWhy this difference comes?\n\nthose are different questions, you can't compare them.\n\nFor getting the answer for the question https://gateoverflow.in/1210/gate2007-12\n\ncheck Ayush Comment on it\n\n0\n\nwhy cannot we compare them?\n\nSee 1st question maximum number of nodes, i.e. Full Binary Tree\n\nand 2nd one is for minimum number of nodes in AVL tree\n\nrt?\n\n+1\nMam, it is not AVL tree...\n\nEven it is a AVL tree, how can you compare those formulas?\n0\n\nyes, not AVL tree, because height is taking difference upto 2, not less than 2\n\nBut we can tell\n\n1st one formula is not for height balanced\n\nand 2nd one for height balanced\n\n0\nThat means Catalan number will satisfy 1st formula\n\nright?\n0\nMam,How Catalan number will equal to the formula?\n0\n\nbecause for height =2 there will be 7 nodes and we can draw all binary tree from this\n\n0\n\n@Shaik Masthan\n\nHere you have taken ..N(H) = 2N(LST) - 1\n\n==> N(H) = 2N(H-1) - 1\n\nWHy we have taken height of LST as H-1 ..won't there be a case such that N(LST) = X and N(RST) = X-2 and\n\nHeight of LST < H-1 but Height of RST = H-1\n\n0\n\nwon't there be a case such that N(LST) = X and N(RST) = X-2 and Height of LST < H-1 but Height of RST = H-1\n\nNo, can't possible.\n\nWHy we have taken height of LST as H-1\n\nNote that minimum no.of nodes asked\n\n+2",
null,
"I got minimum 8 nodes at height 4 . is there anything wrong with my solution ?\n\n+1\nis root satisfying the given condition ?\n0\ngot it. thanks .\n0\n\n@Shaik Masthan\n\nlet H-K = 3 ===> K = H-3 ===> the eqn turned into\n\nwhy upto 3 have u taken?\n\n0\nmam, that is base case, i assumed my base case is 3 due to i can manually calculate it.\n\nyou can take base case as 4, then you have to substitute the value of 4\n0\n\n@Shaik Masthan sir , condition of P and Q is given for h>0 , but root is at height 0,\n\nwhy we are applying this condition at root.\n\n+1\n\ncondition of P and Q is given for h>0\n\nthat means when h=0, no need to check, but it doesn't mean don't check the root if h>1\n\n0\nsomeone please explain how the root doesn't satisfy the condition,I didn't get it\n0\nSolving recurrences is easier if we solve them as given in Rosen(Solving non homogeneous equation).\n0\n\n@Shaik Masthan-In extension to solve the recurrence you have derived.\n\n$N(h)=2N(h-1)-1$......(1)\n\nThe associated homogenous recurrence relation is\n\n$N(h)=2N(h-1)$\n\nSo, solution to this is $C2^h.$\n\nParticular solution will be of the form\n\n$d$\n\nSubstituting this in 1 to get $d=1$.\n\nSo, $N(h)=C2^h+1$\n\nBase case is $N(0)=1$\n\n$1=C+1$\n\n$C=0.$\n\nBut that makes final solution as $N(h)=1$\n\nWhere I am going wrong?\n\n0\nIt is like T(n) = 2 T(n-1) - 1\n\nequation becomes x-2 = 0\n\nSolution : An= alpha (2)^n + particular solution\n\nAs RHS is constant term its soln will be simply A(constant)\n\nsubstituting this in eqn we get A=2A -1 => A =1\n\nSol becomes An= alpha (2)^n +1\n\nWe know base condition A3 =5 => alpha = 1/2\n\nfinal solution = 1/2 (2)^n -1 => 2^n-1 +1\n0\n\n@tusharp-Why are you taking base condition only $A(3)=5$. $A(0)=1$ is also valid because for height 0, minimum 1 node is required.\n\n0\nHowever $A(1)=2$ also gives the correct constant value for A.But why not $A(0)=1$?\n0\nHello @Ayush did you find out why not we consider A(0)=1 as base case?\n\n+1 vote"
] | [
null,
"https://gateoverflow.in/",
null,
"https://gateoverflow.in/",
null,
"https://gateoverflow.in/",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7559491,"math_prob":0.99830955,"size":988,"snap":"2020-24-2020-29","text_gpt3_token_len":329,"char_repetition_ratio":0.2621951,"word_repetition_ratio":0.1557789,"special_character_ratio":0.34817815,"punctuation_ratio":0.12334802,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99975914,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T10:34:03Z\",\"WARC-Record-ID\":\"<urn:uuid:96eb3faf-b599-402a-93ce-b704dd259f18>\",\"Content-Length\":\"161918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a24f4d0-3ebc-4db7-acd8-f7d160b09bc8>\",\"WARC-Concurrent-To\":\"<urn:uuid:808705f1-7744-40ee-959c-ad2ce54be09b>\",\"WARC-IP-Address\":\"172.67.206.99\",\"WARC-Target-URI\":\"https://gateoverflow.in/239308/madeeasy-subject-test-2019-programming-%26-ds-binary-tree\",\"WARC-Payload-Digest\":\"sha1:3GY6PW4BGR3I7DSRRGRIVKEBSLT27GJM\",\"WARC-Block-Digest\":\"sha1:FL55KE5KY2BDG3WUHVBBONEDAVQB4W5U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881984.34_warc_CC-MAIN-20200703091148-20200703121148-00556.warc.gz\"}"} |
https://gitrepos.estec.esa.int/taste/uPython-mirror/-/commit/c38dc3ccc76d1a9bf867704f43ea5d15da3fea7b | [
"### py: Implement fallback for equality check for all types.\n\n```Return \"not equal\" for objects that don't implement equality check.\nThis is as per Python specs.```\nparent ec214058\n ... ... @@ -190,12 +190,19 @@ mp_int_t mp_obj_hash(mp_obj_t o_in) { } } // this function implements the '==' operator (and so the inverse of '!=') // from the python language reference: // This function implements the '==' operator (and so the inverse of '!='). // // From the Python language reference: // (https://docs.python.org/3/reference/expressions.html#not-in) // \"The objects need not have the same type. If both are numbers, they are converted // to a common type. Otherwise, the == and != operators always consider objects of // different types to be unequal.\" // note also that False==0 and True==1 are true expressions // // This means that False==0 and True==1 are true expressions. // // Furthermore, from the v3.4.2 code for object.c: \"Practical amendments: If rich // comparison returns NotImplemented, == and != are decided by comparing the object // pointer.\" bool mp_obj_equal(mp_obj_t o1, mp_obj_t o2) { if (o1 == o2) { return true; ... ... @@ -239,14 +246,9 @@ bool mp_obj_equal(mp_obj_t o1, mp_obj_t o2) { } } if (MICROPY_ERROR_REPORTING == MICROPY_ERROR_REPORTING_TERSE) { nlr_raise(mp_obj_new_exception_msg(&mp_type_NotImplementedError, \"equality for given types not yet implemented\")); } else { nlr_raise(mp_obj_new_exception_msg_varg(&mp_type_NotImplementedError, \"equality for '%s' and '%s' types not yet implemented\", mp_obj_get_type_str(o1), mp_obj_get_type_str(o2))); } // equality not implemented, and objects are not the same object, so // they are defined as not equal return false; } mp_int_t mp_obj_get_int(mp_const_obj_t arg) { ... ...\n # test equality for classes/instances to other types class A: pass class B: pass class C(A): pass print(A == None) print(None == A) print(A == A) print(A() == A) print(A() == A()) print(A == B) print(A() == B) print(A() == B()) print(A == C) print(A() == C) print(A() == C())\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
] | [
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https://calculomates.com/en/divisors/of/5080/ | [
"# Divisors of 5080\n\n## Divisors of 5080\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\nAccordingly:\n\n5080 is multiplo of 1\n\n5080 is multiplo of 2\n\n5080 is multiplo of 4\n\n5080 is multiplo of 5\n\n5080 is multiplo of 8\n\n5080 is multiplo of 10\n\n5080 is multiplo of 20\n\n5080 is multiplo of 40\n\n5080 is multiplo of 127\n\n5080 is multiplo of 254\n\n5080 is multiplo of 508\n\n5080 is multiplo of 635\n\n5080 is multiplo of 1016\n\n5080 is multiplo of 1270\n\n5080 is multiplo of 2540\n\n5080 has 15 positive divisors\n\n## Parity of 5080\n\nIn addition we can say of the number 5080 that it is even\n\n5080 is an even number, as it is divisible by 2 : 5080/2 = 2540\n\n## The factors for 5080\n\nThe factors for 5080 are all the numbers between -5080 and 5080 , which divide 5080 without leaving any remainder. Since 5080 divided by -5080 is an integer, -5080 is a factor of 5080 .\n\nSince 5080 divided by -5080 is a whole number, -5080 is a factor of 5080\n\nSince 5080 divided by -2540 is a whole number, -2540 is a factor of 5080\n\nSince 5080 divided by -1270 is a whole number, -1270 is a factor of 5080\n\nSince 5080 divided by -1016 is a whole number, -1016 is a factor of 5080\n\nSince 5080 divided by -635 is a whole number, -635 is a factor of 5080\n\nSince 5080 divided by -508 is a whole number, -508 is a factor of 5080\n\nSince 5080 divided by -254 is a whole number, -254 is a factor of 5080\n\nSince 5080 divided by -127 is a whole number, -127 is a factor of 5080\n\nSince 5080 divided by -40 is a whole number, -40 is a factor of 5080\n\nSince 5080 divided by -20 is a whole number, -20 is a factor of 5080\n\nSince 5080 divided by -10 is a whole number, -10 is a factor of 5080\n\nSince 5080 divided by -8 is a whole number, -8 is a factor of 5080\n\nSince 5080 divided by -5 is a whole number, -5 is a factor of 5080\n\nSince 5080 divided by -4 is a whole number, -4 is a factor of 5080\n\nSince 5080 divided by -2 is a whole number, -2 is a factor of 5080\n\nSince 5080 divided by -1 is a whole number, -1 is a factor of 5080\n\nSince 5080 divided by 1 is a whole number, 1 is a factor of 5080\n\nSince 5080 divided by 2 is a whole number, 2 is a factor of 5080\n\nSince 5080 divided by 4 is a whole number, 4 is a factor of 5080\n\nSince 5080 divided by 5 is a whole number, 5 is a factor of 5080\n\nSince 5080 divided by 8 is a whole number, 8 is a factor of 5080\n\nSince 5080 divided by 10 is a whole number, 10 is a factor of 5080\n\nSince 5080 divided by 20 is a whole number, 20 is a factor of 5080\n\nSince 5080 divided by 40 is a whole number, 40 is a factor of 5080\n\nSince 5080 divided by 127 is a whole number, 127 is a factor of 5080\n\nSince 5080 divided by 254 is a whole number, 254 is a factor of 5080\n\nSince 5080 divided by 508 is a whole number, 508 is a factor of 5080\n\nSince 5080 divided by 635 is a whole number, 635 is a factor of 5080\n\nSince 5080 divided by 1016 is a whole number, 1016 is a factor of 5080\n\nSince 5080 divided by 1270 is a whole number, 1270 is a factor of 5080\n\nSince 5080 divided by 2540 is a whole number, 2540 is a factor of 5080\n\n## What are the multiples of 5080?\n\nMultiples of 5080 are all integers divisible by 5080 , i.e. the remainder of the full division by 5080 is zero. There are infinite multiples of 5080. The smallest multiples of 5080 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 5080 since 0 × 5080 = 0\n\n5080 : in fact, 5080 is a multiple of itself, since 5080 is divisible by 5080 (it was 5080 / 5080 = 1, so the rest of this division is zero)\n\n10160: in fact, 10160 = 5080 × 2\n\n15240: in fact, 15240 = 5080 × 3\n\n20320: in fact, 20320 = 5080 × 4\n\n25400: in fact, 25400 = 5080 × 5\n\netc.\n\n## Is 5080 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 5080, the answer is: No, 5080 is not a prime number.\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 5080). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 71.274 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test."
] | [
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https://www.jpost.com/sports/caanens-role-as-betar-coach-ends-as-fernandez-takes-over | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
] | [
null
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