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https://au.mathworks.com/help/dsp/ug/minimax-fir-filter-design.html	[ "# Minimax FIR Filter Design\n\nThis example shows how to use some of the key features of the generalized Remez FIR filter design function. This function provides all the functionality included in `firpm` plus many additional features showcased here.\n\n### Weighted-Chebyshev Design\n\nThe following is an illustration of the weighted-Chebyshev design. This example shows the compatibility of `firgr` with `firpm`.\n\n```N = 22; % Filter order F = [0 0.4 0.5 1]; % Frequency vector A = [1 1 0 0]; % Magnitude vector W = [1 5]; % Weight vector b = firgr(N,F,A,W); fvtool(b,1)```", null, "The following is a weighted-Chebyshev design where a type 4 filter (odd-order, asymmetric) has been explicitly specified.\n\n```N = 21; % Filter order F = [0 0.4 0.5 1]; % Frequency vector A = [0 0 1 1]; % Magnitude vector W = [2 1]; % Weight vector b = firgr(N,F,A,W,'4'); fvtool(b,1)```", null, "### \"Least-Squares-Like\" Design\n\nThe following illustrates a \"least-squares-like\" design. A user-supplied frequency-response function (`taperedresp.m`) is used to perform the error weighting.\n\n```N = 53; % Filter order F = [0 0.3 0.33 0.77 0.8 1]; % Frequency vector fresp = {@taperedresp, [0 0 1 1 0 0]}; % Frequency response function W = [2 2 1]; % Weight vector b = firgr(N,F,fresp,W); fvtool(b,1)```", null, "### Filter Designed for Specific Single-Point Bands\n\nThis is an illustration of a filter designed for specified single-point bands. The frequency points f = 0.25 and f = 0.55 are single-band points. These points have a gain that approaches zero.\n\nThe other band edges are normal.\n\n```N = 42; % Filter order F = [0 0.2 0.25 0.3 0.5 0.55 0.6 1]; % Frequency vector A = [1 1 0 1 1 0 1 1]; % Magnitude vector S = {'n' 'n' 's' 'n' 'n' 's' 'n' 'n'}; b = firgr(N,F,A,S); fvtool(b,1)```", null, "### Filter Designed for Specific In-Band Value\n\nHere is an illustration of a filter designed for an exactly specified in-band value. The value is forced to be exactly the specified value of 0.0 at f = 0.06.\n\nThis could be used for 60 Hz rejection (with Fs = 2 kHz). The band edge at 0.055 is indeterminate since it should abut the next band.\n\n```N = 82; % Filter order F = [0 0.055 0.06 0.1 0.15 1]; % Frequency vector A = [0 0 0 0 1 1]; % Magnitude vector S = {'n' 'i' 'f' 'n' 'n' 'n'}; b = firgr(N,F,A,S); zerophase(b,1)```", null, "### Filter Design with Specific Multiple Independent Approximation Errors\n\nHere is an example of designing a filter using multiple independent approximation errors. This technique is used to directly design extra-ripple and maximal ripple filters. One of the interesting properties that these filters have is a transition region width that is locally minimal. Further, these designs converge very quickly in general.\n\n```N = 12; % Filter order F = [0 0.4 0.5 1]; % Frequency vector A = [1 1 0 0]; % Magnitude vector W = [1 1]; % Weight vector E = {'e1' 'e2'}; % Approximation errors b = firgr(N,F,A,W,E); fvtool(b,1)```", null, "### Extra-Ripple Bandpass Filter\n\nHere is an illustration of an extra-ripple bandpass filter having two independent approximation errors: one shared by the two passbands and the other for the stopband (in blue). For comparison, a standard weighted-Chebyshev design is also plotted (in green).\n\n```N = 28; % Filter order F = [0 0.4 0.5 0.7 0.8 1]; % Frequency vector A = [1 1 0 0 1 1]; % Magnitude vector W = [1 1 2]; % Weight vector E = {'e1','e2','e1'}; % Approximation errors b1 = firgr(N,F,A,W,E); b2 = firgr(N,F,A,W); fvtool(b1,1,b2,1)```", null, "### Designing an In-Band-Zero Filter Using Three Independent Errors\n\nWe'll now re-do our in-band-zero example using three independent errors.\n\nNote: It is sometimes necessary to use independent approximation errors to get designs with forced in-band values to converge. This is because the approximating polynomial could otherwise be come very underdetermined. The former design is displayed in green.\n\n```N = 82; % Filter order F = [0 0.055 0.06 0.1 0.15 1]; % Frequency vector A = [0 0 0 0 1 1]; % Magnitude vector S = {'n' 'i' 'f' 'n' 'n' 'n'}; W = [10 1 1]; % Weight vector E = {'e1' 'e2' 'e3'}; % Approximation errors b1 = firgr(N,F,A,S,W,E); b2 = firgr(N,F,A,S); fvtool(b1,1,b2,1)```", null, "### Checking for Transition-Region Anomalies\n\nWith the `'check'` option, one is made aware of possible transition region anomalies in the filter that is being designed. Here is an example of a filter with an anomaly. The `'check'` option warns one of this anomaly: One also get a results vector `res.edgeCheck`. Any zero-valued elements in this vector indicate the locations of probable anomalies. The \"-1\" entries are for edges that were not checked (there can't be an anomaly at f = 0 or f = 1).\n\n```N = 44; % Filter order F = [0 0.3 0.4 0.6 0.8 1]; % Frequency vector A = [1 1 0 0 1 1]; % Magnitude vector b = firgr(N,F,A,'check');```\n```Warning: Probable transition-region anomalies. Verify with freqz. ```\n`fvtool(b,1)`", null, "### Determination of the Minimum Filter Order\n\nThe `firpm` algorithm repeatedly designs filters until the first iteration wherein the specifications are met. The specifications are met when all of the required constraints are met. By specifying `'minorder'`, `firpmord` is used to get an initial estimate. There is also `'mineven'` and `'minodd'` to get the minimum-order even-order or odd-order filter designs.\n\n```F = [0 0.4 0.5 1]; % Frequency vector A = [1 1 0 0]; % Magnitude vector R = [0.1 0.02]; % Deviation (ripple) vector b = firgr('minorder',F,A,R); zerophase(b,1)```", null, "### Differentiators and Hilbert Transformers\n\nWhile using the minimum-order feature, an initial estimate of the filter order can be made. If this is the case, then `firpmord` will not be used. This is necessary for filters that `firpmord` does not support, such as differentiators and Hilbert transformers as well as user-supplied frequency-response functions.\n\n```N = {'mineven',18}; % Minimum even-order, start order estimate at 18 F = [0.1 0.9]; % Frequency vector A = [1 1]; % Magnitude vector R = 0.1; % Deviation (ripple) b = firgr(N,F,A,R,'hilbert'); freqz(b,1,'whole')```", null, "### Design of an Interpolation Filter\n\nThis section illustrates the use of an interpolation filter for upsampling band-limited signals by an integer factor. Typically one would use `intfilt(r,l,alpha)` from the Signal Processing Toolbox™ to do this. However, `intfilt` does not give one as much flexibility in the design as does `firgr`.\n\n```N = 30; % Filter order F = [0 0.1 0.4 0.6 0.9 1]; % Frequency vector A = [4 4 0 0 0 0]; % Magnitude vector W = [1 100 100]; % Weight vector b = firgr(N,F,A,W); fvtool(b,1)```", null, "### A Comparison Between `firpm` and `intfilt`\n\nHere is a comparison made between a filter designed using `firpm` (blue) and a 30-th order filter designed using `intfilt` (green).\n\nNotice that by using the weighting function in `firpm`, one can improve the minimum stopband attenuation by almost 20 dB.\n\n```b2 = intfilt(4, 4, 0.4); fvtool(b,1,b2,1)```", null, "Notice that the equiripple attenuation throughout the second stopband is larger than the minimum stopband attenuation of the filter designed with `intfilt` by about 6 dB. Notice also that the passband ripple, although larger than that of the filter designed with `intfilt`, is still very small.\n\n### Design of a Minimum-Phase Lowpass Filter\n\nHere is an illustration of a minimum-phase lowpass filter.\n\n```N = 42; % Filter order F = [0 0.4 0.5 1]; % Frequency vector A = [1 1 0 0]; % Magnitude vector W = [1 10]; % Weight-constraint vector b = firgr(N,F,A,W, {64},'minphase');```\n\nThe pole/zero plot shows that there are no roots outside of the unit circle.\n\n`zplane(b,1)`", null, "" ]	[ null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_01.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_02.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_03.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_04.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_05.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_06.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_07.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_08.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_09.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_10.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_11.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_12.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_13.png", null, "https://au.mathworks.com/help/examples/dsp/win64/MinimaxFIRFilterDesignExample_14.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81551766,"math_prob":0.9928677,"size":7311,"snap":"2023-14-2023-23","text_gpt3_token_len":2174,"char_repetition_ratio":0.15368824,"word_repetition_ratio":0.20906201,"special_character_ratio":0.31103817,"punctuation_ratio":0.16801493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99572784,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T22:48:27Z\",\"WARC-Record-ID\":\"<urn:uuid:d261ac65-27ec-4799-a6b1-238e635216ef>\",\"Content-Length\":\"92266\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c74f576-2996-4726-987b-a552a9d85286>\",\"WARC-Concurrent-To\":\"<urn:uuid:a110b005-7a21-4d9a-a770-dd4fbf987491>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://au.mathworks.com/help/dsp/ug/minimax-fir-filter-design.html\",\"WARC-Payload-Digest\":\"sha1:NENR2LYYK5JK7FP6FA6P2JQIAGWPCRA5\",\"WARC-Block-Digest\":\"sha1:D44VROXNBP6W2DJUVSIQ4IPTYKMLFI5H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00401.warc.gz\"}"}
https://encyclopedia2.thefreedictionary.com/absolute+retract	[ "absolute retract\n\nabsolute retract\n\n[¦ab·sə‚lüt ri′trakt]\n(mathematics)\nA topological space, A, such that, if B is a closed subset of another topological space, C, and if A is homeomorphic to B, then B is a retract of C.\nMcGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.\nReferences in periodicals archive ?\nNote that any absolute retract (AR-space) is an AMR-space and any absolute neighbourhood retract (ANR-space) is an ANMR-space.\nA space X is called an [R.sub.[delta]]-set if there exists a decreasing sequence {[X.sub.n]} of compact absolute retracts such that X = [[intersection].sub.n] [X.sub.n].\nsets which are homologically equivalent to one point spaces) contains, besides standard convex sets, more general absolute retracts, R-sets and contractible sets (i.e.\nSite: Follow: Share:\nOpen / Close" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7955953,"math_prob":0.92920244,"size":330,"snap":"2022-05-2022-21","text_gpt3_token_len":96,"char_repetition_ratio":0.11349693,"word_repetition_ratio":0.0,"special_character_ratio":0.24545455,"punctuation_ratio":0.16176471,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96768254,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T20:43:49Z\",\"WARC-Record-ID\":\"<urn:uuid:4e358b3d-048d-4cb6-b878-b766f7429380>\",\"Content-Length\":\"41469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:025ca5ff-968d-437f-b5e3-e6a5e49025e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:09ee550c-1910-4b99-bef7-d41c2e7fb6b4>\",\"WARC-IP-Address\":\"91.204.210.225\",\"WARC-Target-URI\":\"https://encyclopedia2.thefreedictionary.com/absolute+retract\",\"WARC-Payload-Digest\":\"sha1:DV7SVIFXQ7Q7T7PKEJFUXOQERT2HH2YI\",\"WARC-Block-Digest\":\"sha1:6LADLZWYBTJCNQKYBM2LCQT4GP33ZJYU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306335.77_warc_CC-MAIN-20220128182552-20220128212552-00498.warc.gz\"}"}
https://www.onlinemath4all.com/fractions-in-mathematics.html	[ "# FRACTIONS IN MATHEMATICS\n\nIn simple words, a fraction is the part of a whole.\n\nIn fractions (for example 17/23), the number above the line (17) is called numerator and the number below the line (23) is called denominator.\n\nThe numerator represents a number of equal parts, and the denominator, which cannot be zero, indicates how many of those parts make up a unit or a whole.\n\nFor example, in the fraction 3/4, the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole.\n\nThe picture given below illustrates the fraction 3/4.", null, "## Different Types of Fractions\n\nWithin the world of fractions, we do have several types and ways of writing them. Let's discuss these now.\n\nProper fraction :\n\nA fraction is called a proper fraction if its\n\nDenominator > Numerator.\n\nExample : 3/4, 1/2, 9/10, 5/6\n\nImproper fraction :\n\nA fraction is called an improper fraction if its\n\nNumerator > Denominator.\n\nExample : 5/4, 6/5, 41/30, 51/25\n\nMixed fraction :\n\nA fraction consisting of a natural number and a proper fraction is called a mixed fractions.\n\nExample : 2 3/4, 1 4/5, 5 1/7\n\nLike Fractions:\n\nIn two or more fractions, the denominators (bottom numbers) are same, they are called as like fractions.\n\nExamples : 3/5 , 6/5, 2/5, 7/5\n\nIn the above fractions, all the denominators are same. That is 5.\n\nUnlike Fractions :\n\nIn two or more fractions, the denominators (bottom numbers) are different, they are called as unlike fractions.\n\nExamples : 3/5 , 6/7, 2/9, 7/2\n\nIn the above fractions, all the denominators are different. They are 5, 7, 9 and 2.\n\n## Converting Improper Fraction to Mixed Fraction\n\nThe picture shown below illustrates how to convert an improper fraction to mixed fraction.", null, "## Converting Mixed Fraction to Improper Fraction\n\nThe picture shown below illustrates how to convert a mixed fraction to improper fraction.", null, "## Addition and Subtraction of Fractions with Same Denominator\n\nExample 1 :\n\nSimplify : 2/5 + 3/5\n\nSolution :\n\nHere, for both the fractions, we have the same denominator, we have to take only one denominator and add the numerators.\n\nThen, we get\n\n2/5 + 3/5 = (2 + 3) / 5\n\n2/5 + 3/5 = 5/5\n\n2/5 + 3/5 = 1\n\nExample 2 :\n\nSimplify : 7/5 - 3/5\n\nSolution :\n\nHere, for both the fractions, we have the same denominator, we have to take only one denominator and subtract the numerators.\n\nThen, we get\n\n7/5 - 3/5 = (7-3) / 5\n\n7/5 - 3/5 = 4/5\n\n## Addition and Subtraction of Fractions with Different Denominators\n\nHere, we explain two methods to add two fractions with different denominators.\n\n1) Cross-multiplication method\n\n2) L.C.M method\n\nCross-Multiplication Method :\n\nIf the denominators of the fractions are co-prime or relatively prime, we have to apply this method.\n\nFro example, let us consider the two fractions 1/8, 1/3.\n\nIn the above two fractions, denominators are 8 and 3.\n\nFor 8 and 3, there is no common divisor other than 1. So 8 and 3 are co-prime.\n\nHere we have to apply cross-multiplication method to add the two fractions 1/8 and 1/3 as given below.", null, "L.C.M Method :\n\nIf the denominators of the fractions are not co-prime (there is a common divisor other than 1), we have to apply this method.\n\nFro example, let us consider the two fractions 5/12, 1/20.\n\nIn the above two fractions, denominators are 12 and 20.\n\nFor 12 and 20, if there is at least one common divisor other than 1, then 12 and 20 are not co-prime.\n\nFor 12 & 20, we have the following common divisors other than 1.\n\n2 and 4\n\nSo 12 and 20 are not co-prime.\n\nIn the next step, we have to find the L.C.M (Least common multiple) of 12 and 20.\n\n12 = 22 3\n\n20 = 22 5\n\nWhen we decompose 12 and 20 in to prime numbers, we find 2, 3 and 5 as prime factors for 12 and 20.\n\nTo get L.C.M of 12 and 20, we have to take 2, 3 and 5 with maximum powers found above.\n\nSo, L.C.M of 12 and 20 is\n\n= 22 3 5\n\n= 4 3 5\n\n= 60\n\nNow we have to make the denominators of both the fractions to be 60 and add the two fractions 5/12 and 1/20 as given below.", null, "Note :\n\nWe have to do the same process for subtraction of two fractions with different denominators.\n\n## Multiply a Fraction by a Whole Number\n\nTo multiply a proper or improper fraction and a whole number,first, we have to multiply the whole number and numerator of the fraction, keeping the denominator same.\n\nFor example,\n\n3/5 = 6/5\n\n7/11 = 21/11\n\nTo multiply a mixed fraction by a whole number, first convert the mixed fraction to an improper fraction and proceed as explained above.\n\nFor example,\n\n3 4/7 = 4 25/7\n\n⋅ 3 4/7 = (4 ⋅ 25)/7\n\n⋅ 3 4/7 = 100/7\n\n⋅ 3 4/7 = 14 2/7\n\n## Multiply a Fraction by a Fraction\n\nTo multiply a proper or improper fraction by another proper or improper fraction, we have to multiply the numerators and denominators.\n\nFor example,\n\n2/3 4/5 = 8/15\n\n1/3 7/11 = 7/33\n\n## Divide a Whole Number by a Fraction\n\nTo divide a whole number by any fraction, multiply the whole number by the reciprocal of the fraction.\n\nFor example,\n\n6 ÷ 2/5 = 6 5/2\n\n÷ 2/5 = (6 ⋅ 5)/2\n\n÷ 2/5 = (3 ⋅ 5)/1\n\n÷ 2/5 = 15\n\n## Divide a Fraction by a Whole Number\n\nTo divide a fraction by a whole number, we have to multiply the denominator of the fraction by the whole number and simplify, if possible.\n\nFor example,\n\n2/5 ÷ 6 = 2/(5 6)\n\n2/5 ÷ 6 = 1/(5 3)\n\n2/5 ÷ 6 = 1/15\n\n## Divide a Whole Number by a Mixed Fraction\n\nWhile dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and proceed.\n\nFor example,\n\n6 ÷ 3 4/5 = 6 ÷ 19/5\n\n6 ÷ 3 4/5 = 6 ⋅ 5/19\n\n6 ÷ 3 4/5 = (6 ⋅ 5)/19\n\n6 ÷ 3 4/5 = 30/19\n\n## Divide a Fraction by a Fraction\n\nTo divide a fraction by another fraction, multiply the first fraction by the reciprocal of the second fraction.\n\nFor example,\n\n1/5 ÷ 3/7 = 1/5 7/3\n\n1/5 ÷ 3/7 = (1 ⋅ 7) / (5 ⋅ 3)\n\n1/5 ÷ 3/7 = 7/15", null, "Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.\n\nKindly mail your feedback to [email protected]\n\n## Recent Articles", null, "1. ### Worksheet on Speed Distance and Time\n\nDec 10, 23 10:09 PM\n\nWorksheet on Speed Distance and Time\n\n2. ### Time Speed and Distance Problems\n\nDec 10, 23 07:42 PM\n\nTime Speed and Distance Problems - Step by Step Solutions" ]	[ null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 279 279'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 518 228'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 466 263.713636363636'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 660 282'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 662 489'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 510 99'%3E%3C/svg%3E", null, "https://www.onlinemath4all.com/objects/xrss.png.pagespeed.ic.nVmUyyfqP7.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8909383,"math_prob":0.986602,"size":5473,"snap":"2023-40-2023-50","text_gpt3_token_len":1800,"char_repetition_ratio":0.19637959,"word_repetition_ratio":0.13302326,"special_character_ratio":0.34058103,"punctuation_ratio":0.13195549,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99968624,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T06:23:07Z\",\"WARC-Record-ID\":\"<urn:uuid:377ce44c-9c90-4b75-a4e6-c7ab9849b27c>\",\"Content-Length\":\"51824\",\"Content-Type\":\"application/http; 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https://search.r-project.org/CRAN/refmans/EFA.dimensions/html/CONGRUENCE.html	[ "CONGRUENCE {EFA.dimensions} R Documentation\n\n## Factor solution congruence\n\n### Description\n\nAligns two factor loading matrices and computes the factor solution congruence and the root mean square residual.\n\n### Usage\n\nCONGRUENCE(target, loadings, verbose)\n\n### Arguments\n\n target The target loading matrix. loadings The loading matrix that will be aligned with the target. verbose Should detailed results be displayed in console? TRUE (default) or FALSE\n\n### Details\n\nThe function first searches for the alignment of the factors from the two loading matrices that has the highest factor solution congruence. It then aligns the factors in \"loadings\" with the factors in \"target\" without changing the loadings. The alignment is based solely on the positions and directions of the factors. The function then produces the Tucker-Wrigley-Neuhaus factor solution congruence coefficient as an index of the degree of similarity between between the aligned loading matrices (see Guadagnoli & Velicer, 1991; and ten Berge, 1986, for reviews).\n\n### Value\n\nA list with the following elements:\n\n rcBefore The factor solution congruence before factor alignment rcAfter The factor solution congruence after factor alignment rcFactors The congruence for each factor rmsr The root mean square residual residmat The residual matrix loadingsNew The aligned loading matrix\n\n### Author(s)\n\nBrian P. O'Connor\n\n### References\n\nGuadagnoli, E., & Velicer, W. (1991). A comparison of pattern matching indices. Multivariate Behavior Research, 26, 323-343.\n\nten Berge, J. M. F. (1986). Some relationships between descriptive comparisons of components from different studies. Multivariate Behavioral Research, 21, 29-40.\n\n### Examples\n\n\n# Rosenberg Self-Esteem scale items\nrotate='VARIMAX', verbose=FALSE)\n\ntarget <- PCA(data_RSE[151:300,], corkind='pearson', Nfactors = 3,\nrotate='VARIMAX', verbose=FALSE)\nCONGRUENCE(target = target$loadingsV, loadings = loadings$loadingsV, verbose=TRUE)\n\n# NEO-PI-R scales\nCONGRUENCE(target$loadingsV, loadings$loadingsV, verbose=TRUE)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6653441,"math_prob":0.7821465,"size":2197,"snap":"2022-05-2022-21","text_gpt3_token_len":584,"char_repetition_ratio":0.15686275,"word_repetition_ratio":0.048275862,"special_character_ratio":0.24214838,"punctuation_ratio":0.17438692,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700685,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T05:29:37Z\",\"WARC-Record-ID\":\"<urn:uuid:fa2b8b67-95fa-4d40-a0de-a22cf50910cb>\",\"Content-Length\":\"4823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:168a409e-e804-4cff-a09a-830087dc6189>\",\"WARC-Concurrent-To\":\"<urn:uuid:511d8446-242a-4f64-a79f-0add921bbffa>\",\"WARC-IP-Address\":\"137.208.57.46\",\"WARC-Target-URI\":\"https://search.r-project.org/CRAN/refmans/EFA.dimensions/html/CONGRUENCE.html\",\"WARC-Payload-Digest\":\"sha1:A6MWFPBYMWYIQVPEHM3GE4UR4DF3SFE4\",\"WARC-Block-Digest\":\"sha1:BPJADJTKNQXKSTKW7A6O3UARX5O3NONB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663039492.94_warc_CC-MAIN-20220529041832-20220529071832-00369.warc.gz\"}"}
https://www.brightstorm.com/math/calculus/the-definite-integral/the-definite-integral-problem-2/	[ "", 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\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# The Definite Integral - Problem 2\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\nWhen a graph is a curve, find the definite integral of the function to find the area under the curve. So, antidifferentiate the function. The definite integral from points a to b is the antiderivative at b minus the antiderivative at a. For example, if looking at the function is f(x)=x2 from x=1 to x=4, the antiderivative of f(x) is x3/3. The area under the curve is 43/3 - 13/3 = 64/3 - 1/3 = 63/3 = 21.\n\nLet's do another definite integral problem. This one is going to require a fact from Geometry. What's the area under a parabolic arch?\n\nSo imagine I construct a parabolic arch which has a shape of a parabola, and a flat base. It's kind of like a triangle except that the sides are curved. If this were a triangle, the area would be 1/2 base times height. Archimedes found that for a parabolic arch, the area is 2/3 of the base times height. So we're going to use this in the next problem.\n\nBefore I take you to the next problem, we need to review one fact about definite integrals. Now I've made the point of saying that, when the function that we're integrating is greater than or equal to 0, the definite integral gives me the exact area under the curve. However, when the function is less than or equal to 0, when the graph is below the x axis, the definite integral doesn't give me the area. It gives me the opposite of the area. Now you know that area is always positive. So this means that definite integrals are going to have negative values when you're integrating a curve that's below the x axis. So it's something to remember.\n\nSo here is the problem. The graph of y equals the quantity x minus 3 squared is shown below, this graph here. Evaluate first, the integral from 0 to 6 of that function. Well, let's do that.\n\nFirst let's observe what area that represents. From 0 to 6, the interval's from here to here. Since its function is non-negative on this interval, the integral is just going to equal the area under the curve. So all we have to do is figure out what this area is. Of course that's not straightforward, because this is not a parabolic arch. However, this is a parabolic arch in here. So what we can do is find the area of this rectangle, and subtract the area of the parabolic arch. So let me write that down; equals area of rectangle minus area of I'll write para; parabolic arch.\n\nNow the area of the rectangle is going to be the base times height. The base is 6, the height (this is not to the same scale), but the height is going to be 9. So 6 times 9 minus the area of the parabolic arch. Remember that formula; 2/3 base times height. So 2/3 (and this is the base, the flat part) the base has a length of 6. So 2/3, 6 times, (and the height is actually the same as the height of the rectangle was) it's going to be 9, the y value here. So 6 times 9. So this is 54 minus 2/3 of 54, is going to be 1/3 of 54, which is 18. This will be 54 minus 36, so that's 18. Very easy if you use subtractions. You can use the fact that the area under a parabolic arch is 2/3 base height.\n\nLet's take a look at this slightly harder problem. The integral from 0 to 6 of this function. It's our original function minus 9. So imagine taking this red curve, and subtracting 9. If we subtract 9, these two points are going to be down here. This point will be 9 units lower, so let me translate it downwards. These two points go here. This point goes nine units down, and the graph will look something like this. So this would be the graph of this function. Let's call it g(x). So this is y equals g(x).\n\nNow the integral from 0 to 6 of g(x), where would that be? Well, because this function lies entirely below the x axis, the value of the definite integral is going to e the opposite of the area. This is the fact that we had before. So if I calculate the area, and put a minus in front of it, that's the answer.\n\nWell, what would this area be? Equals 18 area. What would this area be? Well, it's the same parabolic arch we had before. It's going to be 2/3 the base of 6 times the height of 9. I put a minus in front of that. So minus 2/3 base of 6 height of 9. This is -36\n\nSo the lesson to learn from this is, when you're integrating a function that's below the x axis, expect to get an answer that's negative. When you integrate a function that's entirely above the x axis, expect to get an answer that's positive.\n\nIn the first case, the integral exactly equals the area under the curve; the area between the curve, and the x axis. In the second case, the integral equals the opposite. Put a minus in front of the area between the curve, and the x axis." ]	[ null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-3.png", null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9458386,"math_prob":0.98181,"size":4943,"snap":"2022-27-2022-33","text_gpt3_token_len":1263,"char_repetition_ratio":0.15853411,"word_repetition_ratio":0.09225875,"special_character_ratio":0.25288287,"punctuation_ratio":0.12079927,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99947566,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T23:42:08Z\",\"WARC-Record-ID\":\"<urn:uuid:82198626-6387-4bc4-8f25-30a8602585fc>\",\"Content-Length\":\"91934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81cd21e6-eedd-47db-a382-6ad5256e7d5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:340f5e9c-0594-420f-b7eb-015a8acc3a6f>\",\"WARC-IP-Address\":\"54.83.15.47\",\"WARC-Target-URI\":\"https://www.brightstorm.com/math/calculus/the-definite-integral/the-definite-integral-problem-2/\",\"WARC-Payload-Digest\":\"sha1:NHAFNN7UZIQ35SPLRIDZZ6DHG7KPMSHL\",\"WARC-Block-Digest\":\"sha1:TULHNJXCAL4UFEHKDIAWURA7UB2OQCNE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103947269.55_warc_CC-MAIN-20220701220150-20220702010150-00095.warc.gz\"}"}
https://www.cut-the-knot.org/Generalization/MatrixGroups.shtml	[ "# Matrix Groups\n\nThe problem below was proposed in the Mathematics Magazine (September, 1959) and discussed in The American Mathematics Monthly (vol. 70, n. 4, Apr 1963, p. 427) by R. A. Rosenbaum:\n\n (I) The set of nonsingular n×n matrices such that the sum of the elements in each row of each matrix is 1 forms a group under multiplication.", null, "R. A. Rosenbaum points out that a generalization of that statement is of the kind that gets closer \"to the heart of the matter\" than the statement itself, by stripping away nonessentials and exposing the significant relationships. As he notes, the significant hypothesis is that the row-sums be constant - not necessarily 1, and suggests to consider another problem instead.\n\n (II) Let A = [aij ] be an m×n matrix, B = [bij ] be an n×p matrix, and C = A×B. If the row-sums of A are all equal to a and the row-sums of B are all equal to b, then the row-sums of C are also constant and are equal to ab. If the row-sums of B are all equal to b ≠ 0, and the row-sums of C are all equal to c, then the row-sums of A are also constant and are equal to c/b.\n\nThe reformulation has a\n\n### Corollary\n\nThe set of all nonsingular n×n matrices (over any field) such that, for any one matrix, the sum of the elements of each row is constant (but perhaps not the same constant for different matrices) forms a group under multiplication.\n\nIf in the generalization a, b, c are all taken to be 1, then, with m = n = p, the original problem comes out as another corollary.", null, "The proof of the generalization is indeed straightforward and, were a, b, c to be replaced by 1, would not differ of that for the original statement.\n\n ∑j cij = ∑j ∑k aik bkj = ∑k ∑j aik bkj = ∑k aik ∑j bkj = ∑k aik b = b ∑k aik = b × a = a × b.", null, "Professor W. McWorter has remarked that, for square n×n matrices, having constant row-sums means exactly having an eigenvector 1 = (1 1 ... 1)T:", null, "a11 ... a1n a21 ... a2n ... an1 ... ann", null, "", null, "1 1 ... 1", null, "=", null, "∑j a1j ∑j a2j ... ∑j anj", null, "=a", null, "1 1 ... 1", null, "Thus we can write A1 = a1, B1 = b1, and C1 = c1, implying\n\n c1 = C1 = AB1 = A(b1) = b A1 = b×a1 = a×b 1.\n\nSo, again, c = ab, as expected.\n\nAs Professor McWorter has observed, the latter derivation works just fine when vector 1 is replaced with any other vector v. More accurately, the following statement holds:\n\n (III) Let A = [aij ] and B = [bij ] be an n×n square matrices, and C = A×B. If A and B have a common eigenvector v with the eigenvalues a and b respectively, then v is also an eigenvector of C with the eigenvalue c = ab. If B and C have a common eigenvector v with the eigenvalues c and b (so that b ≠ 0 by definition), then v is also the eigenvector of A with the eigenvalue a = c/b.\n\nFurthermore, similar claims for the addition of matrices and the multiplication by a scalar also hold true.", null, "• Matrices Help Relationships\n• Eigenvalues of an incidence matrix\n• Addition of Vectors and Matrices\n• Multiplication of a Vector by a Matrix\n• Multiplication of Matrices\n• Eigenvectors by Inspection\n• Vandermonde matrix and determinant\n• When the Counting Gets Tough, the Tough Count on Mathematics\n• Merlin's Magic Squares\n•", null, "" ]	[ null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/SYMB/LP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/RP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/LP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/RP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/LP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/RP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/LP.GIF", null, "https://www.cut-the-knot.org/gifs/SYMB/RP.GIF", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90013325,"math_prob":0.9978195,"size":2138,"snap":"2020-34-2020-40","text_gpt3_token_len":624,"char_repetition_ratio":0.1049672,"word_repetition_ratio":0.02020202,"special_character_ratio":0.28531337,"punctuation_ratio":0.12790698,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99966836,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T03:09:48Z\",\"WARC-Record-ID\":\"<urn:uuid:5add374a-7270-4548-aa5b-204c8b7c3294>\",\"Content-Length\":\"18522\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1968d2cf-781f-40e3-8fe8-929992c8e693>\",\"WARC-Concurrent-To\":\"<urn:uuid:81471442-109c-46f9-8c9e-bf3fe5d835d5>\",\"WARC-IP-Address\":\"107.180.50.227\",\"WARC-Target-URI\":\"https://www.cut-the-knot.org/Generalization/MatrixGroups.shtml\",\"WARC-Payload-Digest\":\"sha1:EJNRZVVRWP3PSS4MBR2ZIWNMTUX7HROX\",\"WARC-Block-Digest\":\"sha1:ZJB2ZB4GPFNDDNHUN3NQXVBYDFDJRDOX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738950.61_warc_CC-MAIN-20200813014639-20200813044639-00269.warc.gz\"}"}
https://www.colorhexa.com/00e3fe	[ "# #00e3fe Color Information\n\nIn a RGB color space, hex #00e3fe is composed of 0% red, 89% green and 99.6% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 10.6% magenta, 0% yellow and 0.4% black. It has a hue angle of 186.4 degrees, a saturation of 100% and a lightness of 49.8%. #00e3fe color hex could be obtained by blending #00ffff with #00c7fd. Closest websafe color is: #00ccff.\n\n• R 0\n• G 89\n• B 100\nRGB color chart\n• C 100\n• M 11\n• Y 0\n• K 0\nCMYK color chart\n\n#00e3fe color description : Pure (or mostly pure) cyan.\n\n# #00e3fe Color Conversion\n\nThe hexadecimal color #00e3fe has RGB values of R:0, G:227, B:254 and CMYK values of C:1, M:0.11, Y:0, K:0. Its decimal value is 58366.\n\nHex triplet RGB Decimal 00e3fe `#00e3fe` 0, 227, 254 `rgb(0,227,254)` 0, 89, 99.6 `rgb(0%,89%,99.6%)` 100, 11, 0, 0 186.4°, 100, 49.8 `hsl(186.4,100%,49.8%)` 186.4°, 100, 99.6 00ccff `#00ccff`\nCIE-LAB 82.961, -35.84, -25.934 45.353, 62.089, 103.355 0.215, 0.295, 62.089 82.961, 44.239, 215.889 82.961, -61.318, -36.737 78.797, -35.154, -22.611 00000000, 11100011, 11111110\n\n# Color Schemes with #00e3fe\n\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #fe1b00\n``#fe1b00` `rgb(254,27,0)``\nComplementary Color\n• #00fe9a\n``#00fe9a` `rgb(0,254,154)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #0064fe\n``#0064fe` `rgb(0,100,254)``\nAnalogous Color\n• #fe9a00\n``#fe9a00` `rgb(254,154,0)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #fe0064\n``#fe0064` `rgb(254,0,100)``\nSplit Complementary Color\n• #e3fe00\n``#e3fe00` `rgb(227,254,0)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #fe00e3\n``#fe00e3` `rgb(254,0,227)``\n• #00fe1b\n``#00fe1b` `rgb(0,254,27)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #fe00e3\n``#fe00e3` `rgb(254,0,227)``\n• #fe1b00\n``#fe1b00` `rgb(254,27,0)``\n• #009fb2\n``#009fb2` `rgb(0,159,178)``\n• #00b5cb\n``#00b5cb` `rgb(0,181,203)``\n• #00cce5\n``#00cce5` `rgb(0,204,229)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #19e6ff\n``#19e6ff` `rgb(25,230,255)``\n• #32e9ff\n``#32e9ff` `rgb(50,233,255)``\n• #4cecff\n``#4cecff` `rgb(76,236,255)``\nMonochromatic Color\n\n# Alternatives to #00e3fe\n\nBelow, you can see some colors close to #00e3fe. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00feda\n``#00feda` `rgb(0,254,218)``\n• #00feef\n``#00feef` `rgb(0,254,239)``\n• #00f8fe\n``#00f8fe` `rgb(0,248,254)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #00cefe\n``#00cefe` `rgb(0,206,254)``\n• #00b9fe\n``#00b9fe` `rgb(0,185,254)``\n• #00a4fe\n``#00a4fe` `rgb(0,164,254)``\nSimilar Colors\n\n# #00e3fe Preview\n\nThis text has a font color of #00e3fe.\n\n``<span style=\"color:#00e3fe;\">Text here</span>``\n#00e3fe background color\n\nThis paragraph has a background color of #00e3fe.\n\n``<p style=\"background-color:#00e3fe;\">Content here</p>``\n#00e3fe border color\n\nThis element has a border color of #00e3fe.\n\n``<div style=\"border:1px solid #00e3fe;\">Content here</div>``\nCSS codes\n``.text {color:#00e3fe;}``\n``.background {background-color:#00e3fe;}``\n``.border {border:1px solid #00e3fe;}``\n\n# Shades and Tints of #00e3fe\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, #001113 is the darkest color, while #feffff is the lightest one.\n\n• #001113\n``#001113` `rgb(0,17,19)``\n• #002226\n``#002226` `rgb(0,34,38)``\n• #00343a\n``#00343a` `rgb(0,52,58)``\n• #00454d\n``#00454d` `rgb(0,69,77)``\n• #005761\n``#005761` `rgb(0,87,97)``\n• #006875\n``#006875` `rgb(0,104,117)``\n• #007a88\n``#007a88` `rgb(0,122,136)``\n• #008b9c\n``#008b9c` `rgb(0,139,156)``\n• #009db0\n``#009db0` `rgb(0,157,176)``\n• #00aec3\n``#00aec3` `rgb(0,174,195)``\n• #00c0d7\n``#00c0d7` `rgb(0,192,215)``\n• #00d1ea\n``#00d1ea` `rgb(0,209,234)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\n• #13e6ff\n``#13e6ff` `rgb(19,230,255)``\n• #26e8ff\n``#26e8ff` `rgb(38,232,255)``\n• #3aeaff\n``#3aeaff` `rgb(58,234,255)``\n• #4decff\n``#4decff` `rgb(77,236,255)``\n• #61eeff\n``#61eeff` `rgb(97,238,255)``\n• #75f0ff\n``#75f0ff` `rgb(117,240,255)``\n• #88f2ff\n``#88f2ff` `rgb(136,242,255)``\n• #9cf4ff\n``#9cf4ff` `rgb(156,244,255)``\n• #b0f7ff\n``#b0f7ff` `rgb(176,247,255)``\n• #c3f9ff\n``#c3f9ff` `rgb(195,249,255)``\n• #d7fbff\n``#d7fbff` `rgb(215,251,255)``\n• #eafdff\n``#eafdff` `rgb(234,253,255)``\n• #feffff\n``#feffff` `rgb(254,255,255)``\nTint Color Variation\n\n# Tones of #00e3fe\n\nA tone is produced by adding gray to any pure hue. In this case, #758789 is the less saturated color, while #00e3fe is the most saturated one.\n\n• #758789\n``#758789` `rgb(117,135,137)``\n• #6b8e93\n``#6b8e93` `rgb(107,142,147)``\n• #62969c\n``#62969c` `rgb(98,150,156)``\n• #589ea6\n``#589ea6` `rgb(88,158,166)``\n• #4ea5b0\n``#4ea5b0` `rgb(78,165,176)``\n``#44adba` `rgb(68,173,186)``\n• #3bb5c3\n``#3bb5c3` `rgb(59,181,195)``\n• #31bdcd\n``#31bdcd` `rgb(49,189,205)``\n• #27c4d7\n``#27c4d7` `rgb(39,196,215)``\n• #1dcce1\n``#1dcce1` `rgb(29,204,225)``\n• #14d4ea\n``#14d4ea` `rgb(20,212,234)``\n``#0adbf4` `rgb(10,219,244)``\n• #00e3fe\n``#00e3fe` `rgb(0,227,254)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00e3fe is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5113109,"math_prob":0.8109928,"size":3695,"snap":"2023-40-2023-50","text_gpt3_token_len":1663,"char_repetition_ratio":0.14738554,"word_repetition_ratio":0.0073664826,"special_character_ratio":0.5328823,"punctuation_ratio":0.22954546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98474735,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T14:26:00Z\",\"WARC-Record-ID\":\"<urn:uuid:3998a498-17f5-4f25-bc7b-61f73efc9b0a>\",\"Content-Length\":\"36215\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d6bda87-93f8-428e-823f-9579c20d19b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a9a7edb-ea83-4624-bb36-5927059a281c>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00e3fe\",\"WARC-Payload-Digest\":\"sha1:JVICQVSEOOIWYEOTSF4JJQPZ4O5SI64C\",\"WARC-Block-Digest\":\"sha1:HKAOIEUTKDWDFODZYHCRCWFSBEYJIZNX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.23_warc_CC-MAIN-20231203125921-20231203155921-00055.warc.gz\"}"}
https://www.hackmath.net/en/word-math-problems/reason?page_num=17	[ "Reason - math word problems - page 17\n\n1. Engine pulley", null, "The engine has a 1460 rev / min (RPM). Disc diameter is 350 mm. What will be the disc peripheral speed in RPM? Pulleys on the engine has diameter 80mm, on a disc has diameter 160mm.\n2. Big factorial", null, "How many zeros end number 116! ?\n3. The farmer", null, "The farmer harvested 840 tons of grain in 2006, which was 44% less than in 2005, and one-fifth more than in 2004. How many tons of grain harvested in years 2005 and 2004?\n4. Grandmother's clocks", null, "Grandmother's clock is late every hour by half a minute. Grandmother it by set exactly at 8.00 am. How many hours will show after 24 hours.?\n5. Book", null, "To number pages of thick book was used 4201 digits. How many pages has this book?\n6. Cars speeds", null, "Distance from city A to city B is 108 km. Two cars were simultaneously started from both places. The speed of a car coming from city A was 2 km/h higher than the second car. What was the speed of each car if they met in 54 minutes?\n7. Three workers", null, "Three workers were rewarded CZK 9200 and the money divided by the work they have done. First worker to get twice than the second, the second three times more than the third. How much money each worker received?\n8. Fewer than 500 sheep,", null, "There are fewer than 500 sheep, but if they stand in a double, triple, quadruple, five and sixth order, one sheep will remain. But they can stand in the seventh order. How many are sheep?\n9. Eva", null, "When Eva buys 8 packages of cookies 4 CZK left her. If she wanted to buy 10 packages, she would have to borrow 20 CZK. How much money has Eva in her wallet?\n10. Rings - intersect", null, "There are 15 pupils on the sporting ring. 10 pupils go to football, 8 pupils go to floorball. How many pupils go to both rings at the same time?\n11. Ratio of two unknown numbers", null, "Two numbers are given. Their sum is 30. We calculate one-sixth of a larger number and add to both numbers. So we get new numbers whose ratio is 5:7. Which two numbers were given?\n12. Extreme temperatures", null, "USA hit extreme heat wave. Calculate percentage of the air temperature change from normal summer temperature 23 °C to 40 °C.\n13. Unknown number", null, "Unknown number is divisible by exactly three different primes. When we compare these primes in ascending order, the following applies: • Difference first and second prime number is half the difference between the third and second prime numbers. • The produ\n14. Have solution", null, "The sum of four consecutive even numbers is 96. Determine these numbers.\n15. Pipe", null, "Steel pipe has a length 2.5 meters. About how many decimetres is 1/3 less than 4/8 of this steel pipe?\n16. Travel by bus", null, "Five girls traveling by bus. Each holds in each hand two baskets. Each basket contains 5 cats. Each cat has 5 kittens. 3 girls getting off the bus and another girl without baskets go on board. The bus control driver. How many legs are in the bus?\n17. Hexagon rotation", null, "A regular hexagon of side 6 cm is rotated through 60° along a line passing through its longest diagonal. What is the volume of the figure thus generated?\n18. Four sides of trapezoid", null, "In the trapezoid ABCD is \|AB\| = 73.6 mm; \|BC\| = 57 mm; \|CD\| = 60 mm; \|AD\| = 58.6 mm. Calculate the size of its interior angles.\n19. Obtuse angle", null, "The line OH is the height of the triangle DOM, line MN is the bisector of angle DMO. obtuse angle between the lines MN and OH is four times larger than the angle DMN. What size is the angle DMO? (see attached image)\n20. Velocipede", null, "The front wheel of velocipede from year 1880 had a diameter 1.8 m. If the front wheel turned again one then rear wheel 6 times. What was the diameter of the rear wheel?\n\nDo you have an interesting mathematical word problem that you can't solve it? Enter it, and we can try to solve it.\n\nTo this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox." ]	[ null, "https://www.hackmath.net/thumb/60/t_3260.jpg", null, "https://www.hackmath.net/thumb/6/t_106.jpg", null, "https://www.hackmath.net/thumb/72/t_1572.jpg", null, "https://www.hackmath.net/thumb/23/t_2223.jpg", null, "https://www.hackmath.net/thumb/99/t_399.jpg", null, "https://www.hackmath.net/thumb/9/t_5209.jpg", null, "https://www.hackmath.net/thumb/28/t_1428.jpg", null, "https://www.hackmath.net/thumb/98/t_4998.jpg", null, "https://www.hackmath.net/thumb/96/t_1496.jpg", null, "https://www.hackmath.net/thumb/33/t_4933.jpg", null, "https://www.hackmath.net/thumb/13/t_5213.jpg", null, "https://www.hackmath.net/thumb/71/t_171.jpg", null, "https://www.hackmath.net/thumb/31/t_1331.jpg", null, "https://www.hackmath.net/thumb/64/t_4764.jpg", null, "https://www.hackmath.net/thumb/13/t_213.jpg", null, "https://www.hackmath.net/thumb/15/t_1915.jpg", null, "https://www.hackmath.net/thumb/39/t_3739.jpg", null, "https://www.hackmath.net/thumb/2/t_4902.jpg", null, "https://www.hackmath.net/thumb/22/t_1422.jpg", null, "https://www.hackmath.net/thumb/19/t_1519.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9543072,"math_prob":0.93414074,"size":3684,"snap":"2019-43-2019-47","text_gpt3_token_len":940,"char_repetition_ratio":0.099728264,"word_repetition_ratio":0.0,"special_character_ratio":0.26058632,"punctuation_ratio":0.11298701,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98471695,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T22:43:24Z\",\"WARC-Record-ID\":\"<urn:uuid:778bbf5f-972f-491d-8249-aaece0adbe7e>\",\"Content-Length\":\"25977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa12bf88-ad6e-432b-bd2c-443127e97125>\",\"WARC-Concurrent-To\":\"<urn:uuid:25bceda6-951b-4a7f-9b67-da87713f7c8d>\",\"WARC-IP-Address\":\"104.24.105.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/word-math-problems/reason?page_num=17\",\"WARC-Payload-Digest\":\"sha1:RERT74YVD3LDDFKVXZYU5KY4DB43TFG5\",\"WARC-Block-Digest\":\"sha1:QN7AGMKZV3XHJOLTAYQ27PCKATNHVW2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986700435.69_warc_CC-MAIN-20191019214624-20191020002124-00341.warc.gz\"}"}
https://tools.carboncollective.co/compound-interest/43721-at-9-percent-in-10-years/	[ "# What is the compound interest on $43721 at 9% over 10 years? If you want to invest$43,721 over 10 years, and you expect it will earn 9.00% in annual interest, your investment will have grown to become $103,503.51. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$43,721 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 10 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 9.00% into a decimal, which would be 0.09.\n\n$$A = 43721(1 + \\dfrac{ 0.09 }{1})^{ 10}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 10 because our example is for annual compounding, or one period per year, so 10 × 1 = 10.\n\n## How the compound interest on $43,721 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 10 years it is compounding: Start Balance Interest End Balance 1$43,721.00 $3,934.89$47,655.89\n2 $47,655.89$4,289.03 $51,944.92 3$51,944.92 $4,675.04$56,619.96\n4 $56,619.96$5,095.80 $61,715.76 5$61,715.76 $5,554.42$67,270.18\n6 $67,270.18$6,054.32 $73,324.49 7$73,324.49 $6,599.20$79,923.70\n8 $79,923.70$7,193.13 $87,116.83 9$87,116.83 $7,840.51$94,957.35\n10 $94,957.35$8,546.16 $103,503.51 We can also display this data on a chart to show you how the compounding increases with each compounding period. In this example we have 10 years of compounding, but to truly see the power of compound interest, it might be better to look at a larger number of compounding periods to see how much$43,721 can grow.\n\nIf you want an example with more compounding years, click here to view the compounding interest of $43,721 at 9.00% over 30 years. As you can see if you view the compounding chart for$43,721 at 9.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.\n\n## How long would it take to double $43,721 at 9% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$43,721 assuming an interest rate of 9.00%.\n\nWe can calculate this very approximately using the Rule of 72.\n\nThe formula for this is very simple:\n\n$$Years = \\dfrac{72}{Interest\\: Rate}$$\n\nBy dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:\n\n$$Years = \\dfrac{72}{ 9 } = 8$$\n\nUsing this, we know that any amount we invest at 9.00% would double itself in approximately 8 years. So $43,721 would be worth$87,442 in ~8 years.\n\nWe can also calculate the exact length of time it will take to double an amount at 9.00% using a slightly more complex formula:\n\n$$Years = \\dfrac{log(2)}{log(1 + 0.09)} = 8.04\\; years$$\n\nHere, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.\n\nAs you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.\n\nHopefully, this article has helped you to understand the compound interest you might achieve from investing \\$43,721 at 9.00% over a 10 year investment period." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93269277,"math_prob":0.997305,"size":4046,"snap":"2022-40-2023-06","text_gpt3_token_len":1148,"char_repetition_ratio":0.16006927,"word_repetition_ratio":0.014450867,"special_character_ratio":0.33811173,"punctuation_ratio":0.15409836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998548,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T01:50:28Z\",\"WARC-Record-ID\":\"<urn:uuid:fef03273-7d44-4e46-a3fd-c827d01d64fd>\",\"Content-Length\":\"26278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:deba2c86-f529-4fb2-a8b7-5d85c5a30773>\",\"WARC-Concurrent-To\":\"<urn:uuid:32b84f0d-0883-4624-b1bb-62737c0b4f25>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/compound-interest/43721-at-9-percent-in-10-years/\",\"WARC-Payload-Digest\":\"sha1:SXAQDUKN6XBU6YENL7G5NMBEZFUEMH44\",\"WARC-Block-Digest\":\"sha1:IPQKDWKFVILNZAMDWY7MNWQNLTFAQJUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500158.5_warc_CC-MAIN-20230205000727-20230205030727-00457.warc.gz\"}"}
https://ww2.mathworks.cn/help/econ/jcitest.html	[ "# jcitest\n\nJohansen cointegration test\n\n## Syntax\n\n``h = jcitest(Y)``\n``h = jcitest(Tbl)``\n``h = jcitest(___,Name=Value)``\n``````[h,pValue,stat,cValue] = jcitest(___)``````\n``````[h,pValue,stat,cValue,mles] = jcitest(___)``````\n\n## Description\n\nexample\n\n````h = jcitest(Y)` returns the rejection decisions `h` from conducting the Johansen test, which assesses each null hypothesis H(r) of cointegration rank less than or equal to r among the `numDims`-dimensional multivariate time series `Y` against the alternatives H(`numDims`) (`trace` test) or H(r + 1) (`maxeig` test). The tests produce maximum likelihood estimates of the parameters in a vector error-correction (VEC) model of the cointegrated series.```\n\nexample\n\n````h = jcitest(Tbl)` returns rejection decisions from conducting the Johansen test on the variables of the table or timetable `Tbl`.To select a subset of variables in `Tbl` to test, use the `DataVariables` name-value argument.```\n\nexample\n\n````h = jcitest(___,Name=Value)` uses additional options specified by one or more name-value arguments, using any input-argument combination in the previous syntaxes.Some options control the number of tests to conduct. The following conditions apply when `jcitest` conducts multiple tests: `jcitest` treats each test as separate from all other tests.Each row of all outputs contains the results of the corresponding test. For example, `jcitest(Tbl,Model=\"H2\",DataVariables=1:5)` tests the first 5 variables in the input table `Tbl` using the Johansen model that excludes all deterministic terms.```\n\nexample\n\n``````[h,pValue,stat,cValue] = jcitest(___)``` displays, at the command window, the results of the Johansen test and returns the p-values `pValue`, test statistics `stat`, and critical values `cValue` of the test. The results display includes the ranks r, corresponding rejection decisions, p-values, decision statistics, and specified options.```\n\nexample\n\n``````[h,pValue,stat,cValue,mles] = jcitest(___)``` also returns a structure of maximum likelihood estimates associated with the VEC(q) model of the multivariate time series yt.```\n\n## Examples\n\ncollapse all\n\nTest a multivariate time series for cointegration using the default values of the Johansen cointegration test. Input the time series data as a numeric matrix.\n\nLoad data of Canadian inflation and interest rates `Data_Canada.mat`, which contains the series in the matrix `Data`.\n\n```load Data_Canada series'```\n```ans = 5x1 cell {'(INF_C) Inflation rate (CPI-based)' } {'(INF_G) Inflation rate (GDP deflator-based)'} {'(INT_S) Interest rate (short-term)' } {'(INT_M) Interest rate (medium-term)' } {'(INT_L) Interest rate (long-term)' } ```\n\nTest the interest rate series for cointegration by using the Johansen cointegration test. Use default options and return the rejection decision.\n\n`h = jcitest(Data(:,3:end))`\n```h=1×7 table r0 r1 r2 Model Lags Test Alpha _____ _____ _____ ______ ____ _________ _____ t1 true true false {'H1'} 0 {'trace'} 0.05 ```\n\nBy default, `jcitest` conducts the `trace` test and uses the `H1` Johansen form by default. The test fails to reject the null hypothesis of rank 2 cointegration in the series.\n\nConduct the Johansen cointegration test on a multivariate time series using default options, which tests all table variables.\n\nLoad data of Canadian inflation and interest rates `Data_Canada.mat`. Convert the table `DataTable` to a timetable.\n\n```load Data_Canada dates = datetime(dates,12,31); TT = table2timetable(DataTable,RowTimes=dates); TT.Observations = [];```\n\nConduct the Johansen cointegration test by passing the timetable to `jcitest` and using default options. `jcitest` tests for cointegration among all table variables by default.\n\n`h = jcitest(TT)`\n```h=1×9 table r0 r1 r2 r3 r4 Model Lags Test Alpha _____ _____ _____ _____ _____ ______ ____ _________ _____ t1 true true false false true {'H1'} 0 {'trace'} 0.05 ```\n\nThe test fails to reject the null hypotheses of rank 2 and 3 cointegration among the series.\n\nBy default, `jcitest` includes all input table variables in the cointegration test. To select a subset of variables to test, set the `DataVariables` option.\n\n`jcitest` supports two types Johansen tests. Conduct a test for each type.\n\nLoad data of Canadian inflation and interest rates `Data_Canada.mat`. Convert the table `DataTable` to a timetable. Identify the interest rate series.\n\n```load Data_Canada dates = datetime(dates,12,31); TT = table2timetable(DataTable,RowTimes=dates); TT.Observations = []; idxINT = contains(TT.Properties.VariableNames,\"INT\");```\n\nConduct the Johansen cointegration test to assess cointegration among the interest rate series. Specify both test types `trace` and `maxeig`, and set the level of significance to 2.5%.\n\n`h = jcitest(TT,DataVariables=idxINT,Test=[\"trace\" \"maxeig\"],Alpha=0.025)`\n```h=2×7 table r0 r1 r2 Model Lags Test Alpha _____ _____ _____ ______ ____ __________ _____ t1 true false false {'H1'} 0 {'trace' } 0.025 t2 false false false {'H1'} 0 {'maxeig'} 0.025 ```\n\nh is a 2-row table; rows contain results of separate tests. At 2.5% level of significance:\n\n• The `trace` test fails to reject the null hypotheses of ranks 1 and 2 cointegration among the series.\n\n• The `maxeig` test fails to reject the null hypotheses for each cointegration rank.\n\nLoad data of Canadian inflation and interest rates `Data_Canada.mat`. Convert the table `DataTable` to a timetable. Identify the interest rate series.\n\n```load Data_Canada dates = datetime(dates,12,31); TT = table2timetable(DataTable,RowTimes=dates); TT.Observations = []; idxINT = contains(TT.Properties.VariableNames,\"INT\");```\n\nConduct the Johansen cointegration test to assess cointegration among the interest rate series. Specify both test types `trace` and `maxeig`.\n\n`[h,pValue,stat,cValue] = jcitest(TT,DataVariables=idxINT,Test=[\"trace\" \"maxeig\"])`\n```********************** Results Summary (Test 1) Data: TT Effective sample size: 40 Model: H1 Lags: 0 Statistic: trace Significance level: 0.05 r h stat cValue pValue eigVal ---------------------------------------- 0 1 37.6886 29.7976 0.0050 0.4101 1 1 16.5770 15.4948 0.0343 0.2842 2 0 3.2003 3.8415 0.0737 0.0769 ******************** Results Summary (Test 2) Data: TT Effective sample size: 40 Model: H1 Lags: 0 Statistic: maxeig Significance level: 0.05 r h stat cValue pValue eigVal ---------------------------------------- 0 0 21.1116 21.1323 0.0503 0.4101 1 0 13.3767 14.2644 0.0687 0.2842 2 0 3.2003 3.8415 0.0737 0.0769 ```\n```h=2×7 table r0 r1 r2 Model Lags Test Alpha _____ _____ _____ ______ ____ __________ _____ t1 true true false {'H1'} 0 {'trace' } 0.05 t2 false false false {'H1'} 0 {'maxeig'} 0.05 ```\n```pValue=2×7 table r0 r1 r2 Model Lags Test Alpha _________ ________ ________ ______ ____ __________ _____ t1 0.0050497 0.034294 0.073661 {'H1'} 0 {'trace' } 0.05 t2 0.050346 0.06874 0.073661 {'H1'} 0 {'maxeig'} 0.05 ```\n```stat=2×7 table r0 r1 r2 Model Lags Test Alpha ______ ______ ______ ______ ____ __________ _____ t1 37.689 16.577 3.2003 {'H1'} 0 {'trace' } 0.05 t2 21.112 13.377 3.2003 {'H1'} 0 {'maxeig'} 0.05 ```\n```cValue=2×7 table r0 r1 r2 Model Lags Test Alpha ______ ______ ______ ______ ____ __________ _____ t1 29.798 15.495 3.8415 {'H1'} 0 {'trace' } 0.05 t2 21.132 14.264 3.8415 {'H1'} 0 {'maxeig'} 0.05 ```\n\n`jcitest` prints a results display for each test to the command window. All outputs are tables containing the corresponding statistics and test options.\n\nLoad data of Canadian inflation and interest rates `Data_Canada.mat`. Convert the table `DataTable` to a timetable.\n\n```load Data_Canada dates = datetime(dates,12,31); TT = table2timetable(DataTable,RowTimes=dates); TT.Observations = []; idxINT = contains(TT.Properties.VariableNames,\"INT\");```\n\nPlot the interest series.\n\n```plot(TT.Time,TT{:,idxINT}) legend(series(idxINT),Location=\"northwest\") grid on```", null, "Test the interest rate series for cointegration; use the default Johansen form `H1`. Return all outputs.\n\n`[h,pValue,stat,cValue,mles] = jcitest(TT,DataVariables=idxINT);`\n```********************** Results Summary (Test 1) Data: TT Effective sample size: 40 Model: H1 Lags: 0 Statistic: trace Significance level: 0.05 r h stat cValue pValue eigVal ---------------------------------------- 0 1 37.6886 29.7976 0.0050 0.4101 1 1 16.5770 15.4948 0.0343 0.2842 2 0 3.2003 3.8415 0.0737 0.0769 ```\n`h`\n```h=1×7 table r0 r1 r2 Model Lags Test Alpha _____ _____ _____ ______ ____ _________ _____ t1 true true false {'H1'} 0 {'trace'} 0.05 ```\n`pValue`\n```pValue=1×7 table r0 r1 r2 Model Lags Test Alpha _________ ________ ________ ______ ____ _________ _____ t1 0.0050497 0.034294 0.073661 {'H1'} 0 {'trace'} 0.05 ```\n\nThe test fails to reject the null hypothesis of rank 2 cointegration in the series.\n\nPlot the estimated cointegrating relations ${B}^{\\prime }{y}_{t-1}+{c}_{0}$:\n\n```TTLag = lagmatrix(TT,1); T = height(TTLag); B = mles.r2.paramVals.B; c0 = mles.r2.paramVals.c0; plot(TTLag.Time,TTLag{:,idxINT}B+repmat(c0',T,1)) grid on```", null, "## Input Arguments\n\ncollapse all\n\nData representing observations of a multivariate time series yt, specified as a `numObs`-by-`numDims` numeric matrix. Each column of `Y` corresponds to a variable, and each row corresponds to an observation.\n\nData Types: `double`\n\nData representing observations of a multivariate time series yt, specified as a table or timetable with `numObs` rows. Each row of `Tbl` is an observation.\n\nTo select a subset of variables in `Tbl` to test, use the `DataVariables` name-value argument.\n\nNote\n\n`jcitest` removes the following observations from the specified data:\n\n• All rows containing at least one missing observation, represented by a `NaN` value\n\n• From the beginning of the data, initial values required to initialize lagged variables\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nBefore R2021a, use commas to separate each name and value, and enclose `Name` in quotes.\n\nExample: `jcitest(Tbl,Model=\"H2\",DataVariables=1:5)` tests the first 5 variables in the input table `Tbl` using the Johansen model that excludes all deterministic terms.\n\nJohansen form of the VEC(q) model deterministic terms , specified as a Johansen form name in the table, or a string vector or cell vector of character vectors of such values (for model parameter definitions, see Vector Error-Correction (VEC) Model).\n\nValueError-Correction TermDescription\n`\"H2\"`\n\nAB´yt − 1\n\nNo intercepts or trends are present in the cointegrating relations, and no deterministic trends are present in the levels of the data.\n\nSpecify this model only when all response series have a mean of zero.\n\n`\"H1\"`\n\nA(B´yt−1+c0)\n\nIntercepts are present in the cointegrating relations, and no deterministic trends are present in the levels of the data.\n\n`\"H1\"`\n\nA(B´yt−1+c0)+c1\n\nIntercepts are present in the cointegrating relations, and deterministic linear trends are present in the levels of the data.\n\n`\"H\"`A(B´yt−1+c0+d0t)+c1\n\nIntercepts and linear trends are present in the cointegrating relations, and deterministic linear trends are present in the levels of the data.\n\n`\"H\"`A(B´yt−1+c0+d0t)+c1+d1t\n\nIntercepts and linear trends are present in the cointegrating relations, and deterministic quadratic trends are present in the levels of the data.\n\nIf quadratic trends are not present in the data, this model can produce good in-sample fits but poor out-of-sample forecasts.\n\n`jcitest` conducts a separate test for each value in `Model`.\n\nExample: `Model=\"H1\"` uses the Johansen form `H1` for all tests.\n\nExample: `Model=[\"H1\" \"H1\"]` uses Johansen form `H1` for the first test, and then uses Johansen form `H1` for the second test.\n\nData Types: `string` \| `char` \| `cell`\n\nNumber of lagged differences q in the VEC(q) model, specified as a nonnegative integer or vector of nonnegative integers.\n\n`jcitest` conducts a separate test for each value in `Lags`.\n\nExample: `Lags=1` includes Δyt – 1 in the model for all tests.\n\nExample: `Lags=[0 1]` includes no lags in the model for the first test, and then includes Δyt – 1 in the model for the second test.\n\nData Types: `double`\n\nTest to perform, specified as a value in this table, or a string vector or cell vector of character vectors of such values.\n\nValueDescription\n`\"trace\"`\n\nThe alternative hypothesis is H(`numDims`), and the test statistics are\n\n`$-T\\left[\\mathrm{log}\\left(1-{\\lambda }_{r+1}\\right)+\\dots +\\mathrm{log}\\left(1-{\\lambda }_{m}\\right)\\right].$`\n\n`\"maxieig\"`\n\nThe alternative hypothesis is H(r + 1), and the test statistics are\n\n`$-T\\mathrm{log}\\left(1-{\\lambda }_{r+1}\\right).$`\n\nBoth tests assess the null hypothesis H(r) of cointegration rank less than or equal to r. `jcitest` computes statistics using the effective sample size T ≤ n`numObs` and ordered estimates of the eigenvalues of C = AB′, λ1 > ... > λm, where m = `numDims`.\n\n`jcitest` conducts a separate test for each value in `Test`.\n\nExample: `Test=\"maxeig\"` conducts the `maxeig` test for all tests.\n\nExample: `Test=[\"maxeig\" \"trace\"]` conducts the `maxeig` test for the first test, and then conducts the `trace` test for the second test.\n\nData Types: `char` \| `string` \| `cell`\n\nNominal significance level for the hypothesis test, specified as a numeric scalar between `0.001` and `0.999` or a numeric vector of such values.\n\n`jcitest` conducts a separate test for each value in `Alpha`.\n\nExample: `Alpha=[0.01 0.05]` uses a level of significance of `0.01` for the first test, and then uses a level of significance of `0.05` for the second test.\n\nData Types: `double`\n\nCommand window display control, specified as a value in this table.\n\nValueDescription\n`\"off\"``jcitest` does not display the results to the command window. If `jcitest` returns `h` or no outputs, this display is the default.\n`\"summary\"`\n\n`jcitest` displays a tabular summary of test results. The tabular display includes null ranks r = `0:(numDims − 1)` in the first column of each summary. `jcitest` displays multiple test results in separate summaries.\n\nWhen `jcitest` returns any other output than `h` (for example, `pValue`), this display is the default. You cannot set this display when `jcitest` returns `h` or no outputs.\n\n`\"params\"``jcitest` displays maximum likelihood estimates of the parameter values associated with the reduced-rank VEC(q) model of yt. You can set this display only when `jcitest` returns `mles`. `jcitest` returns the displayed parameter estimates in the field `mles.rn(j).paramVals` for null rank r = n and test `j`.\n`\"full\"``jcitest` displays both `\"summary\"` and `\"params\"`.\n\nExample: `Display=\"off\"`\n\nData Types: `char` \| `string`\n\nVariables in `Tbl` for which `jcitest` conducts the test, specified as a string vector or cell vector of character vectors containing variable names in `Tbl.Properties.VariableNames`, or an integer or logical vector representing the indices of names. The selected variables must be numeric.\n\nExample: `DataVariables=[\"GDP\" \"CPI\"]`\n\nExample: `DataVariables=[true true false false]` or `DataVariables=[1 2]` selects the first and second table variables.\n\nData Types: `double` \| `logical` \| `char` \| `cell` \| `string`\n\nNote\n\n• When `jcitest` conducts multiple tests, the function applies all single settings (scalars or character vectors) to each test.\n\n• All vector-valued specifications that control the number of tests must have equal length.\n\n• A lagged and differenced time series has a reduced sample size. Absent presample values, if the test series yt is defined for t = 1,…,T, the lagged series yt– k is defined for t = k+1,…,T. The first difference applied to the lagged series yt– k further reduces the time base to k+2,…,T. With p lagged differences, the common time base is p+2,…,T and the effective sample size is T–(p+1).\n\n## Output Arguments\n\ncollapse all\n\nTest rejection decisions, returned as a `numTests`-by-(`numDims` + 3) table, where `numTests` is the number of tests, which is determined by specified options.\n\nRow `j` of `h` corresponds to test `j` with options.\n\nRows of `h` correspond to tests specified by the values of the last three variables `Model`, `Test`, and `Alpha`. Row labels are `t1`, `t2`, …, `tu`, where `u` = `numTests`.\n\nVariables of `h` correspond to different, maintained cointegration ranks r = 0, 1, …, `numDims` – 1 and specified name-value arguments that control the number of tests. Variable labels are `r0`, `r1`, …, `rR`, where `R` = `numDims` – 1, and `Model`, `Test`, and `Alpha`.\n\nTo access results, for example, the result for test `j` of null rank `k`, use `h.rk(j)`.\n\nVariable k, labeled `rk`, is logical vector whose entries have the following interpretations:\n\n• `1` (`true`) indicates rejection of the null hypothesis of cointegration rank k in favor of the alternative hypothesis.\n\n• `0` (`false`) indicates failure to reject the null hypothesis of cointegration rank k.\n\nTest statistic p-values, returned as a table with the same dimensions and labels as `h`. Variable k, labeled `rk`, is a numeric vector of p-values for the corresponding tests. The p-values are right-tailed probabilities.\n\nWhen test statistics are outside tabulated critical values, `jcitest` returns maximum (`0.999`) or minimum (`0.001`) p-values.\n\nTest statistics, returned as a table with the same dimensions and labels as `h`.\n\nThe `Test` setting of a particular test determines the test statistic.\n\nCritical values, returned as a table with the same dimensions and labels as `h`. Variable k, labeled `rk`, is a numeric vector of critical values for the corresponding tests. The critical values are for right-tailed probabilities determined by `Alpha`.\n\n`jcitest` loads tables of critical values from the file `Data_JCITest.mat`, and then linearly interpolates test critical values from the tables. Critical values in the tables derive from methods described in .\n\nMaximum likelihood estimates associated with the VEC(q) model of yt, returned as a table with the same dimensions and labels as `h`. Variable k, labeled `rk`, is a structure array of MLEs with elements for the corresponding tests.\n\nEach element of `mles.rk` has the fields in this table. You can access a field using dot notation, for example, `mles.r2(3).paramVals` contains the parameter estimates of the third test corresponding to the null hypothesis of rank 2 cointegration.\n\nFieldDescription\n`paramNames`\n\nCell vector of parameter names, of the form:\n\n{`A`, `B`, `B1`, … `Bq`, `c0`, `d0`, `c1`, `d1`}\n\nElements depend on the values of the `Lags` and `Model` name-value arguments.\n\n`paramVals`Structure of parameter estimates with field names corresponding to the parameter names in `paramNames`.\n`res` T-by-`numDims` matrix of residuals, where T is the effective sample size, obtained by fitting the VEC(q) model of yt to the input data.\n`EstCov`Estimated covariance Q of the innovations process εt.\n`eigVal`Eigenvalue associated with H(r).\n`eigVec`Eigenvector associated with the eigenvalue in `eigVal`. Eigenvectors v are normalized so that vS11v = 1, where S11 is defined as in .\n`rLL` Restricted loglikelihood of yt under the null.\n`uLL`Unrestricted loglikelihood of yt under the alternative.\n\ncollapse all\n\n### Vector Error-Correction (VEC) Model\n\nA vector error-correction (VEC) model is a multivariate, stochastic time series model consisting of a system of m = `numDims` equations of m distinct, differenced response variables. Equations in the system can include an error-correction term, which is a linear function of the responses in levels used to stabilize the system. The cointegrating rank r is the number of cointegrating relations that exist in the system.\n\nEach response equation can include a degree q autoregressive polynomial composed of first differences of the response series, a constant, a time trend, and a constant and time trend in the error-correction term.\n\nExpressed in lag operator notation, a VEC(q) model for a multivariate time series yt is\n\n`$\\begin{array}{c}\\Phi \\left(L\\right)\\left(1-L\\right){y}_{t}=A\\left(B\\prime {y}_{t-1}+{c}_{0}+{d}_{0}t\\right)+{c}_{1}+{d}_{1}t+{\\epsilon }_{t}\\\\ =c+dt+C{y}_{t-1}+{\\epsilon }_{t},\\end{array}$`\n\nwhere\n\n• yt is an m = `numDims` dimensional time series corresponding to m response variables at time t, t = 1,...,T.\n\n• $\\Phi \\left(L\\right)=I-{\\Phi }_{1}-{\\Phi }_{2}-...-{\\Phi }_{q}$, I is the m-by-m identity matrix, and Lyt = yt – 1.\n\n• The cointegrating relations are B'yt – 1 + c0 + d0t and the error-correction term is A(B'yt – 1 + c0 + d0t).\n\n• r is the number of cointegrating relations and, in general, 0 ≤ rm.\n\n• A is an m-by-r matrix of adjustment speeds.\n\n• B is an m-by-r cointegration matrix.\n\n• C = AB′ is an m-by-m impact matrix with a rank of r.\n\n• c0 is an r-by-1 vector of constants (intercepts) in the cointegrating relations.\n\n• d0 is an r-by-1 vector of linear time trends in the cointegrating relations.\n\n• c1 is an m-by-1 vector of constants (deterministic linear trends in yt).\n\n• d1 is an m-by-1 vector of linear time-trend values (deterministic quadratic trends in yt).\n\n• c = Ac0 + c1 and is the overall constant.\n\n• d = Ad0 + d1 and is the overall time-trend coefficient.\n\n• Φj is an m-by-m matrix of short-run coefficients, where j = 1,...,q and Φq is not a matrix containing only zeros.\n\n• εt is an m-by-1 vector of random Gaussian innovations, each with a mean of 0 and collectively an m-by-m covariance matrix Σ. For ts, εt and εs are independent.\n\nIf m = r, then the VEC model is a stable VAR(q + 1) model in the levels of the responses. If r = 0, then the error-correction term is a matrix of zeros, and the VEC(q) model is a stable VAR(q) model in the first differences of the responses.\n\n## Algorithms\n\n• `jcitest` identifies deterministic terms that are outside of the cointegrating relations, c1 and d1, by projecting constant and linear regression coefficients, respectively, onto the orthogonal complement of A.\n\n• If `jcitest` fails to reject the null hypothesis of cointegration rank r = 0, the inference is that the error-correction coefficient C is zero, and the VEC(q) model reduces to a standard VAR(q) model in first differences. If `jcitest` rejects all cointegration ranks r less than `numDims`, the inference is that C has full rank, and yt is stationary in levels.\n\n• The parameters A and B in the reduced-rank VEC(q) model are not identifiable, but their product C = AB is identifiable. `jcitest` constructs `B` = `V(:,1:r)` using the orthonormal eigenvectors `V` returned by `eig`, and then renormalizes so that `V'S11*V = I` .\n\n• The time series in the specified input data can be stationary in levels or first differences (that is, I(0) or I(1)). Rather than pretesting series for unit roots (using, e.g., `adftest`, `pptest`, `kpsstest`, or `lmctest`), the Johansen procedure formulates the question within the model. An I(0) series is associated with a standard unit vector in the space of cointegrating relations, and the `jcontest` can test for its presence.\n\n• Deterministic cointegration, where cointegrating relations, perhaps with an intercept, produce stationary series, is the traditional sense of cointegration introduced by Engle and Granger (see `egcitest`). Stochastic cointegration, where cointegrating relations produce trend-stationary series (that is, `d0` is nonzero), extends the definition of cointegration to accommodate a greater variety of economic series.\n\n• Unless higher-order trends are present in the data, models with fewer restrictions can produce good in-sample fits, but poor out-of-sample forecasts.\n\n## Alternative Functionality\n\n### App\n\nThe Econometric Modeler app enables you to conduct the Johansen cointegration test.\n\n Engle, R. F. and C. W. J. Granger. \"Co-Integration and Error-Correction: Representation, Estimation, and Testing.\" Econometrica. Vol. 55, 1987, pp. 251–276.\n\n Hamilton, James D. Time Series Analysis. Princeton, NJ: Princeton University Press, 1994.\n\n Johansen, S. Likelihood-Based Inference in Cointegrated Vector Autoregressive Models. Oxford: Oxford University Press, 1995.\n\n MacKinnon, J. G. \"Numerical Distribution Functions for Unit Root and Cointegration Tests.\" Journal of Applied Econometrics. Vol. 11, 1996, pp. 601–618.\n\n Turner, P. M. \"Testing for Cointegration Using the Johansen Approach: Are We Using the Correct Critical Values?\" Journal of Applied Econometrics. v. 24, 2009, pp. 825–831." ]	[ null, "https://ww2.mathworks.cn/help/examples/econ/win64/TestMultipleSeriesForCointegrationUsingJcitestExample_01.png", null, "https://ww2.mathworks.cn/help/examples/econ/win64/TestMultipleSeriesForCointegrationUsingJcitestExample_02.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6390669,"math_prob":0.9683932,"size":16719,"snap":"2023-40-2023-50","text_gpt3_token_len":4541,"char_repetition_ratio":0.15925816,"word_repetition_ratio":0.22509076,"special_character_ratio":0.28189486,"punctuation_ratio":0.14526382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961979,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T19:57:37Z\",\"WARC-Record-ID\":\"<urn:uuid:be02f53c-e3ce-40f0-bc83-32c13633d74d>\",\"Content-Length\":\"178569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f90237f-8f14-498e-a649-70b77ea1a081>\",\"WARC-Concurrent-To\":\"<urn:uuid:c49a82f0-02b5-4a44-be1c-eaee05f67d29>\",\"WARC-IP-Address\":\"104.99.53.76\",\"WARC-Target-URI\":\"https://ww2.mathworks.cn/help/econ/jcitest.html\",\"WARC-Payload-Digest\":\"sha1:6RR23LZTU4VNIAVLKCUD5OTE3SLMXNMM\",\"WARC-Block-Digest\":\"sha1:OZ7PBSN645XSKP4A3I4HUA5BDLJE5FSX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510924.74_warc_CC-MAIN-20231001173415-20231001203415-00539.warc.gz\"}"}
https://lessonplanet.com/search?keywords=quartic+function	[ "### We found17 reviewed resources for quartic function\n\nLesson Planet\n\nFor Teachers 11th - 12th\nStudents explore cubic and quartic functions. In this Pre-calculus/Calculus lesson, students examine the properties of third and fourth degree functions. Students examine the intercepts and inflection points and examine the conditions...\nLesson Planet\n\n#### Polynomial Functions\n\nFor Students 11th\nIn this polynomial functions worksheet, 11th graders solve and complete 8 different problems that include various polynomial functions. First, they define a polynomial function. Then, students write the polynomial part in standard form...\nLesson Planet\n\n#### Saxon Math: Algebra 2 (Section 11)\n\nFor Teachers 9th - 12th Standards\nAs we approach the end of the Saxon math sections, the 11th of 12 units begins the process of summarizing and combining material from the previous lessons. The relationships between functions and their transformations are investigated,...\nLesson Planet\n\n#### Quartic Regions\n\nFor Teachers 9th - 12th Standards\nYoung scholars explore quartic functions in this calculus lesson. They investigate an application of derivatives and definite integrals, then explore the question of which of the \"three bumps\" of a quartic is the largest.\nLesson Planet\n\n#### Polynomials\n\nFor Teachers 11th\nEleventh graders explore the graphs of polynomial functions. In this Algebra II instructional activity, 11th graders investigate the graphs of first through fourth degree equations as they identify x and y intercepts, points of...\nLesson Planet\n\n#### Polynomial Functions\n\nFor Teachers 11th - 12th\nLearners explore polynomial functions. In this Algebra II lesson, students explore graphs of polynomial functions as classify the functions as linear, quadratic, cubic, or quartic. Learners determine the regression equation for each...\nLesson Planet\n\n#### Quartic Regions\n\nFor Teachers 11th - 12th\nStudents explore the concept of quartic regions. In this quartic regions instructional activity, students find the area under a curve using their Ti-89 calculator. Students find definite integrals of two curves. Students determine the...\nLesson Planet\n\n#### Cubic Splines\n\nFor Teachers 10th - Higher Ed\nStudents generate graphs on the calculator. In this math lesson, the students graph parabolas and other functions on the calculator with the intention of analyzing the graph. Students discover how to alter and modify each \"spine\" on the...\nEngageNY\n\n#### Modeling Riverbeds with Polynomials (part 2)\n\nFor Students 10th - 12th Standards\nExamine the power of technology while modeling with polynomial functions. Using the website wolfram alpha, learners develop a polynomial function to model the shape of a riverbed. Ultimately, they determine the flow rate through the river.\nLesson Planet\n\n#### Quartic Inflection\n\nFor Teachers 12th\nTwelfth graders investigate an application of derivatives. In this calculus instructional activity students explore quartic polynomials and the midpoint of the line segment between the two inflection points.\nLesson Planet\n\n#### Eco-Tourism: Whale Watching ID: 8307\n\nFor Teachers 11th\nEleventh graders explore a quartic function. In this Algebra II lesson, 11th graders examine data regarding the number of whale sightings for a given period. Students use a quartic regression to model the data with a function and use...\nLesson Planet\n\n#### Polynomial Functions\n\nFor Students 11th\nIn this Algebra II instructional activity, 11th graders use a graphing calculator to determine the regression equation of the given data, graph polynomial functions, factor polynomials, and determine the quotient and remainder in...\nLesson Planet\n\n#### Curve Fitting\n\nFor Students 6th - Higher Ed Standards\nBeing a statistician means never having to say you are certain. An interactive simulation allows scholars to plot numerous data points to see the line of best-fit on a graph. They may also choose linear, quadratic, cubic, or quartic and...\nLesson Planet\n\n#### Building Curves\n\nFor Teachers 9th - 12th\nThis activity looks at polynomial operations from a grahing perspective. Using the basic operations of addition, subtraction, multiplication, and division on polynomials, learners investigate graphs. In addition, regression modeling is...\nLesson Planet\n\n#### Building Curves\n\nFor Teachers 10th - 12th\nExplore operations on polynomials through graphing on the TI-Nspire. Learners investigate addition, subtraction, multiplication and division of polynomials through graphing. They graph polynomials and identify key concepts.\nLesson Planet\n\n#### Stacking Bricks\n\nFor Teachers 9th - 12th\nStudents stack bricks and find a pattern that will tell the number of bricks they would need to make a large stack of 50 rows high. Students first build a stack of bricks and look for a pattern, they use the detailed directions to...\nLesson Planet\n\n#### The Slope of a Curve\n\nFor Teachers 8th - 10th\nStudents identify the slope of a curve. In this algebra activity, students identify the maximum and minimum of a parabola. They use technology and tables to create graphs." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86521333,"math_prob":0.94280064,"size":4574,"snap":"2020-24-2020-29","text_gpt3_token_len":904,"char_repetition_ratio":0.18183808,"word_repetition_ratio":0.027149322,"special_character_ratio":0.17818102,"punctuation_ratio":0.13810742,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9958102,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-12T07:34:56Z\",\"WARC-Record-ID\":\"<urn:uuid:fc1df3e8-84eb-4d38-a0ca-3a3dbff831af>\",\"Content-Length\":\"103039\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09c7bceb-491b-4657-a37d-adea0fb9a6d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:a24ddf43-075e-4d03-9cb0-dd58157c73a7>\",\"WARC-IP-Address\":\"18.216.209.228\",\"WARC-Target-URI\":\"https://lessonplanet.com/search?keywords=quartic+function\",\"WARC-Payload-Digest\":\"sha1:PTXKOYVHVANKZQ5XMNDP5QB47ZW4PRTL\",\"WARC-Block-Digest\":\"sha1:IXLMS5FLYEKULDKXHNDHDYPT6DH7SPLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657131734.89_warc_CC-MAIN-20200712051058-20200712081058-00194.warc.gz\"}"}
https://personalpages.manchester.ac.uk/staff/theodore.voronov/volumes.html	[ "Volumes of classical supermanifolds\n\nThis work started from a question by E. Witten, who asked me (in fall 2012), whether the volume of every compact (even) symplectic supermanifold with respect to the super analog of the Liouville measure should vanish (except for the case of ordinary symplectic manifolds). He had some arguments in favor of such a conjecture. I presented a counterexample, which is the volume of the complex projective superspace, the simplest nontrivial case being CP1\|1 where the volume is 2π. The general formula for the volume of CPn\|m of radius R (with respect to the super analog of the Fubini-Study metric) is R2(n-m)πn 2m / Γ(n-m+1).\n\nTo put Witten's question into context, it was long known (Berezin's theorem, 1970s) that the volume of the super analog of the unitary group, the unitary supergroup U(n\|m), vanishes unless m=0 or n =0 (which is the case of an ordinary group, either U(n) or U(m)).\n\nThe above formula led me to the investigation of other formulas for \"super\" volumes. As it turns out, - an \"experimental fact\" - the volumes of various classical supermanifolds (upon some universal normalization) can be obtained by formulas that are analytic continuation of the formulas for the volumes of the corresponding ordinary manifolds. This looks as a mystery if one thinks about the formal algebraic nature of the Berezin integral in comparison with the ordinary measure theory. Also, this fact prompts for the investigation of a meaning of these \"normalized volumes\" in non-integer dimensions and in infinite dimension. There is another challenging connection with the \"universal Lie algebra\" program of Vogel-Deligne (see the works by R. Mkrtchyan, Mkrtchyan-Veselov and Mkrtchyan-Khudaverdian).\n\nReference:\n\nThere has been some new development in this area very recently, see a preprint by Stanford and Witten: arXiv:1907.03363. Among other things, they have deduced a formula for the Liouville volume of a symplectic supermanifold in topological terms (via the Chern class of the associated vector bundle, i.e. the normal bundle to the reduced submanifold). This prompts new very attractive paths for research." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8986152,"math_prob":0.7939273,"size":1790,"snap":"2022-05-2022-21","text_gpt3_token_len":433,"char_repetition_ratio":0.12653975,"word_repetition_ratio":0.024911031,"special_character_ratio":0.21787709,"punctuation_ratio":0.08256881,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9862035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T21:08:03Z\",\"WARC-Record-ID\":\"<urn:uuid:208a5199-8248-40a8-8f49-0b2ef1ae23ee>\",\"Content-Length\":\"4358\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e123d6a4-6b23-42f8-92f9-855057855998>\",\"WARC-Concurrent-To\":\"<urn:uuid:f48b4f09-db02-449b-a0eb-09f67d2ea39d>\",\"WARC-IP-Address\":\"130.88.101.51\",\"WARC-Target-URI\":\"https://personalpages.manchester.ac.uk/staff/theodore.voronov/volumes.html\",\"WARC-Payload-Digest\":\"sha1:64MFJWMHI2DOVZ26JZMQQFVEBQDWXYN4\",\"WARC-Block-Digest\":\"sha1:IKXN2REU5PSDEZZXKWDRSOW6IBWZSM6R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304961.89_warc_CC-MAIN-20220126192506-20220126222506-00438.warc.gz\"}"}
https://socratic.org/questions/how-do-you-calculate-the-following-to-the-correct-number-of-significant-figures	[ "# How do you calculate the following to the correct number of significant figures?\n\n## a) $\\frac{1.27 g}{5.296 m L}$ b) $\\frac{12.235 g}{1.01 L}$ c) $\\frac{17.3 g + 2.785 g}{30.20 m L}$\n\nOct 2, 2016\n\na)$0.240 \\frac{g}{m L}$\n\nb)$12.1 \\frac{g}{L}$\n\nc)$0.666 \\frac{g}{m L}$\n\n#### Explanation:\n\nWhen dividing, the quotient should have the same number of significant figures as the smaller number of sig figs in either the dividend or divisor.\n\nWhen adding, the sum should have the same number of signficant figures as the \"least accurate\" place in the addends.\n\na) $\\frac{1.27 g}{5.296 m L} = 0.2398 \\frac{g}{m L}$\n\nwhich I will round to $0.240 \\frac{g}{m L}$ because $1.27$ has only 3 significant figures.\n\nb)$\\frac{12.235 g}{1.01 L} = \\frac{12.114 g}{L}$\n\nwhich I will round to $12.1 \\frac{g}{L}$ because $1.01$ has only 3 sig figs.\n\nc)$\\frac{17.3 g + 2.785 g}{30.20 m L}$\n\nFirst add the numbers in the numerator.\n\n$17.3 + 2.785 = 20.085$ which I will round to $20.1$ because the \"least accurate\" place in either of the addends is the tenths place in $17.3$\n\nNext, complete the division.\n\n$\\frac{20.1 g}{30.20 m L} = 0.6656 \\frac{g}{m L}$\n\nwhich I will round to $0.666 \\frac{g}{m L}$ because $20.1$ has only 3 sig figs." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6635625,"math_prob":0.9998404,"size":738,"snap":"2022-05-2022-21","text_gpt3_token_len":223,"char_repetition_ratio":0.13079019,"word_repetition_ratio":0.05982906,"special_character_ratio":0.2899729,"punctuation_ratio":0.13071896,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998033,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T23:40:49Z\",\"WARC-Record-ID\":\"<urn:uuid:58fb8209-c4aa-450d-b02d-aee187563ad6>\",\"Content-Length\":\"35857\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f60e4802-4f54-401e-ac2a-a418d01b7b5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:be1df3b9-cecc-4e66-a2ab-2cb5850e0dc6>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-calculate-the-following-to-the-correct-number-of-significant-figures\",\"WARC-Payload-Digest\":\"sha1:5UQ5FBDSJVCLI7ZYE6GCAA2GYG3AY66I\",\"WARC-Block-Digest\":\"sha1:AIFCCUQ3B5T42VEBO4UFLCVDU6VZLOOO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662550298.31_warc_CC-MAIN-20220522220714-20220523010714-00145.warc.gz\"}"}
https://fr.scribd.com/document/318542976/Algebra-Reteachings-Lessons-1-10	[ "Vous êtes sur la page 1sur 20\n\nName\n\nDate\n\nClass\n\nReteaching\nClassifying Real Numbers\n\nA set is a collection of objects. You use several different number sets as\n\nyou study algebra.\nNatural Numbers\n\n{1, 2, 3, 4, }\n\nWhole Numbers\n\n{0, 1, 2, 3, 4,}\n\nIntegers\n\n{, 4, 3, 2, 1, 0, 1, 2, 3, 4,}\na where a and b are integers\nNumbers that can be written as __\nb\nand b0. (As decimals they terminate or repeat.)\na . (As decimals, they neither\nNumbers that cannot be written as __\nb\nterminate nor repeat.)\n\nRational Numbers\nIrrational Numbers\nReal Numbers\n\nIdentify the subsets of real numbers to\n\nwhich 17 belongs.\n\nIdentify the subsets of real numbers to\n\n\u0002\nwhich 5 belongs.\na ,so it is\nThis number cannot be written as __\nb\nan irrational number. All irrational numbers are\n\u0002\nalso real numbers, so 5 belongs to these two\nsets.\n\nThis number can be used to count things,\n\nso it is a natural number. All natural numbers\nare also whole numbers, integers, rational\nnumbers, and real numbers, so 17 belongs\nto these five sets.\n\nPractice\nIdentify the subsets of real numbers to which each number belongs.\n4\n1. __\n9\n\n2. 21\n\na , so it\nThis number can be written as __\nb\nis a rational number and a\n\nThis number is the opposite of a natural\n\nreal\n\nnumber, so it is an integer,\n\nnumber.\n\nnumber, and a\n\nreal\n\nrational\n\nnumber.\n\n3. 1.5\n\n4. 63,298\n\nrational, real\n\nrational, real\n6. 5\n\n5. 0\n\nwhole, integers, rational, real\n\nirrational, real\n\u0002\n\n3\n7. ___\n11\n\n8. 33\n\nirrational, real\n\nrational, real\nSaxon. All rights reser ved.\n\nSaxon Algebra 1\n\nReteaching\n\ncontinued\nClosure is a property of a set. A set is closed under a given operation\nif the outcome of the operation on any two members of the set is also\na member of the set. A counterexample is an example that proves\na statement false. Intersection \u0002 \u0002 \u0003 and union \u0002 \u0003 \u0003 are two ways\nto combine sets.\nFind A \u0002 B and A \u0003 B.\nA \u0005 {1, 2, 3, 6, 7, 8}; B \u0005 {2, 4, 6}\n\nDetermine whether the statement is true\n\nor false. Give a counterexample if the\nstatement is false.\nThe set of integers is closed under\nmultiplication.\n\nA \u0002 B \u0002 {2, 6}\n\nStep 2: Find the elements that are in A or B.\n\nA \u0003 B \u0002 {1, 2, 3, 4, 6, 7, 8}\n\n\u00033 \u0004 5 \u0002 \u000315\n\u00037 \u0004 \u00032 \u0002 14\nThe product is always an integer, so the\nstatement is true.\n\nPractice\nDetermine whether the statement is true or false.\nGive a counterexample if the statement is false.\n9. The set of natural numbers is closed under multiplication.\n\n8\n1\u00049\u0002 9\n7 \u0004 5 \u0002 35\nnatural number , so the statement is true\n\nThe product is always a\n\n2\u00044\u0002\n\nFind A \u0002 B and A \u0003 B.\n10. A \u0002 {1, 2, 3, 5, 7, 11}; B \u0002 {1, 3, 5, 7, 9}\n\n11. A = {4, 8, 12}; B = {0, 2, 4, 6, 8, 10}\n\n5, 7}\nA \u0003 B \u0002 {1, 2, 3, 5 , 7 , 9 , 11}\n\nA \u0002 B \u0002 {4,\n\nA \u0002 B \u0002 {1, 3,\n\n8}\n\nA \u0003 B \u0002 {0, 2, 4,\n\n8 , 10 , 12 }\n\nDetermine whether each statement is true or false.\n\nGive a counterexample if the statement is false.\n12. The set of rational numbers is closed under subtraction.\n\ntrue\n\n\u000310 \u0004 (\u00032) \u0002 \u0005 5 \u0002 5\n\n13. The set of irrational numbers is closed under division. false;\n\n2 \u0006 2 \u0005 1, not irrational\n\nFind A \u0002 B and A \u0003 B.\n14. A \u0002 {2, 6, 10, 14, 18}; B \u0002 {0, 2, 4, 6, 8}\n\nSaxon. All rights reser ved.\n\nA \u0002 B \u0005 {2, 6};\nA \u0003 B \u0005 {0, 2, 4, 6, 8, 10, 14, 18}\n2\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\n\nUnderstanding Variables and Expressions\n\nConstants and variables are the building blocks of algebraic expressions.\n\nA constant is a quantity whose value does not change. A constant is\n\nusually a number. A variable can change, or vary, and is used to represent\nan unknown number. A letter is usually used to represent a variable.\n\nConstants\n7ab\n\n4c\n\nVariables\nIdentify the constants and variables in the\nexpression 9x 7.\n\nIdentify the constants and variables in the\n\nequation 8y xz 2.\n\nThe numbers 9 and 7 never change.\n\nThey are constants.\n\nThe numbers 8 and 2 never change.\n\nThey are constants.\n\nThe letter x represents an unknown\n\nnumber. It is a variable.\n\nThe letters x, y, and z represent\n\nunknown numbers. They are\nvariables.\n\nPractice\nIdentify the constants and variables.\n1. 11jk m\n\n2. 9z 2y 13\n\nThe number 11 is a constant because\n\nit never changes.\n\nThe numbers 9, 2, and 13\n\nare constants because they\nnever change.\n\nThe letters j, k, and m are\n\nvariables because they represent\nunknown numbers.\n\nThe letters z and y are variables\n\nbecause they represent unknown numbers.\n\n3. 3x 7y 20\n\n4. 9h 5fg 1\n\nconstants\n\nThe numbers 3, 7, and 20\n\nare constants because they never change.\n\nThe numbers 9, 5, and 1 are\n\nbecause they never change.\n\nThe letters x and y are variables\n\nbecause they represent unknow numbers.\n\nThe letters h, f, and g are variables\n\nbecause they represent unknown numbers.\n\n5. 24uv 3\nConstants:\n\n6. 100x 2wz\n\n24, 3\n\nSaxon. All rights reser ved.\n\nVariables:\n\nu, v\n\nConstants:\n3\n\n100, 2\n\nVariables:\n\nx, w, z\n\nSaxon Algebra 1\n\nReteaching\n\ncontinued\nConstants and variables can be multiplied to form a product. Each constant\nand variable in a product is a factor of the product. The product of a\nconstant and one or more variables is usually written without multiplication\nsymbols. For example, 6 \u0002 s \u0002 t is written as 6st.\nThe numeric factor in a product is called the coefficient. In the expression\n6st, the coefficient is 6 and the other factors are s and t.\nThe parts of an expression separated by or signs are called terms.\nIdentify the factors and coefficients\nin the expression 9ab.\n\n3xy \u0003 7uv \u0003 8.\n\nRecall: The terms are the parts of the\n\nexpression separated by or\nsigns.\n\nThe factors, or quantities, that are\n\nmultiplied are 9, a, and b.\nStep 2: Find the coefficients.\n\nThe coefficient is the numeric factor 9.\n\nPractice\nIdentify the factors and coefficients in each expression.\n7. 4gh\n\n8. yz\n\nThe factors, or quantities, that are\n\nmultiplied are \u00044, g, and h.\n\nThe factors, or quantities, that are\n\nmultiplied are y, and z .\n\n\u00044.\n\nThe coefficient is the numeric factor 1,\n\nwhich is implied in this case.\n\nIdentify the terms in each expression.\n\ny\n10. (9x 2) __\nxz 5xy\n\nb 8c\n9. 3a __\n4\nThe terms are the parts of the expression\nseparated by or signs. The terms are\n\nb , and 8c\n3a, __\n4\n\nThe terms are the parts of the expression\n\nseparated by or signs. The terms are .\n\ny\n(9x \u0003 2), __\nxz, and 5xy\n\nIdentify the factors and coefficients in each expression.\n\n3n\n11. ___\n5\nFactors:\n\n12. 7xyz\n\n3, n\n__\n5\n\n; Coefficient:\n\n3\n__\n5\n\nFactors:\n\n\u00047, x, y, z; Coefficient:\n\n\u00047\n\nIdentify the terms in each expression.\n\n(3r 1)\n13. 12p _______ prs\n7s\n\n(3r \u0004 1)\nTerms: 12p, _______\n7s\n\n7jk\n14. ____ 9mk 6kn\n8mn\n\n7jk 9mk, 6kn\n\n8mn\n\nTerms: _____,\n\n, prs\n4\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nSimplifying Expressions Using the Product Rule of Exponents\n\nYou have used multiplication to simplify expressions. Now you will use\nmultiplication to simplify expressions with exponents. An exponent can\nbe used to show repeated multiplication.\nSimplify the expression 3 4.\n\nThe base is 3 and the exponent is 4.\n\nThe exponent 4 indicates that the\nbase 3 is used as a factor\nfour times.\n\nThe base is 0.2 and the exponent is\n\n3. The exponent 3 indicates that the\nbase 0.2 is used as a factor\nthree times.\n\nStep 2: Write the power as repeated\n\nmultiplication.\n34\u0002 3 \u0002 3 \u0002 3 \u0002 3\n\nStep 2: Write the power as repeated\n\nmultiplication.\n3\n(0.2) \u0002 (0.2) (0.2) (0.2)\n\nStep 3: Simplify.\n3 4 \u0002 3 \u0002 3 \u0002 3 \u0002 3 \u0002 81\n\nStep 3: Simplify.\n(0.2) 3 \u0002 (0.2) (0.2) (0.2) \u0002 0.008\n\nPractice\nComplete the steps to simplify each expression.\n1. 4\n\n2. (0.5) 4\n\nThe base is\nis 3 .\n43\u0002 4 \u0002\n\n4\u00024\n\n0.5\n\n(0.5) 4 \u0002 (0.5)\n\n4\u00024\u00024\n\n43 \u0002\n\nThe base is\nis 4 .\n\n64\n\n(0.5) 4 \u0002\n\n\u0003 0.0625\n\n512\n\n3. 8 3 \u0002\n\n\u0002 \u0003\u0002\n\n2\n5. __\n5\n\n7. 7 4 \u0002\n\n4. (1.2) 2 \u0002\n\n16\n____\n\n6. 10 5 \u0002\n\n625\n\n\u0002 \u0003\n\n3\n8. __\n4\n\n2401\n\n9. (3.1) 2 \u0002\n\n9.61\n\n10. 10 9 \u0002\n\n\u0002 \u0003\n\n1\n___\n\n12. (1.5) 2 \u0002\n\n1\n11. __\n2\n\nSaxon. All rights reser ved.\n\n64\n5\n\n1.44\n100,000\n27\n___\n64\n1,000,000,000\n2.25\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nUse the product rule of exponents to find the product of powers whose\nbases are the same.\nProduct Rule of Exponents\nIf m and n are real numbers and x \u0002 0, then\nxm\u0002 xn\u0003 xm \u0004 n\nSimplify the expression x 2 \u0002 x 4 \u0002 x 3.\nStep 1: Identify the bases and the\nexponents.\nEach factor has x as the base.\nThe exponents are 2, 4, and 3.\nStep 2: Add the exponents to find the power\nof the product.\nx2 \u0002 x4 \u0002 x3 \u0003 x2\u0004 4 \u0004 3 \u0003 x9\n\nSimplify the expression x 2 \u0002 x 5\u0002 y 3 \u0002 y 4.\n\nStep 1: Identify the bases.\nThe first two factors have x as the base.\nThe last two factors have y as the base.\nStep 2: Add the exponents for the factors that\nhave x as the base. Then add the\nexponents for the factors that have\ny as the base.\n2\u00045\n\u0002 y3\u0004 4 \u0003 x7 y7\nx\n\nPractice\nSimplify each expression.\n3\n5\n13. y \u0002 y\n\n14. s 2 \u0002 s 7 \u0002 s 3 \u0002 t 4 \u0002 t 6\n\nare 3 and 5.\n\ny3\u0002 y5\u0003 y\n\n3\u00035\n\n\u0003y\n\nThe first three factors have S as the\n\nbase. The last two factors have\nt as the base.\n\n2\u00037 \u00033\n\n\u0002t\n\n4\u00036\n\n\u0003s\n\n12\n\n\u0002t\n\n10\n\nSimplify each expression.\n\n15. z 2 \u0002 z 3 \u0003\n17. u 15 \u0002 u 10 \u0003\n\nz5\n\n16. m 4 \u0002 m 2 \u0002 m 5 \u0003\n\nu 25\n\n18. z 20 \u0002 z 80 \u0003\n\n19. a 4 \u0002 a 5 \u0002 b 2 \u0002 b 3 \u0002 b 6 \u0003\n21. x 10 \u0002 x 4 \u0002 y 17 \u0002y 2 \u0003\n\na 9 \u0002b 11\n\nm 11\nz 100\n\n20. g 6 \u0002 g 2 \u0002 g 3 \u0002 h 4 \u0003\n22. b 4 \u0002 b 3 \u0002 f 2 \u0002 f 7 \u0002 g 5 \u0003\n\nx 14 \u0002y 19\n\ng 11 \u0002h 4\nb 7 \u0002f 9 \u0002g 5\n\n23. You are taking a true-false test that has two parts. There are 2 5\nways to answer the 5 questions in Part 1. There are 2 10 ways to\nanswer the 10 questions in Part 2. How many ways are there to\n\n32,768\nSaxon. All rights reser ved.\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nUsing Order of Operations\n\nYou have simplified expressions that contain one operation. Now you will\nuse the order of operations to simplify expressions that contain more\nthan one operation.\nOrder of Operations\n1. Work inside grouping symbols.\n2. Simplify powers and roots.\n3. Multiply and divide from left to right.\n4. Add and subtract from left to right.\n2\nWrite the expression 22 \u0002 (5 \u0003 3) \u0002 7 \u0003 2 . Then use the order of\noperations to simplify.\n\n22 \u0002 (5 \u0003 3) \u0002 \u00032\n\n\u0004 22 \u0002 2 \u0002 7 \u0003 2 2\n\nSimplify inside the parentheses.\n\n\u0004 22 \u0002 2 \u0002 7 \u00034\n\nSimplify exponents.\n\n\u0004 22 \u0002 14 \u0003 4\n\n\u0004 32\n\nAdd and subtract from left to right.\n\nPractice\nWrite the expression. Then use the order of operations to simplify.\n1. 3 \u0002 (5 \u0002 3 ) \u0003 8 \u0005 2\n\n2. 27 \u0005 3 2 \u0002 2 \u00035\n\n3 \u0002 (5 \u0002 3 ) \u0003 8 \u0005 2\n\u00043\u0002\n\n\u0004 24 \u0003\n\n27 \u0005\n\n\u00038\u00052\n\n\u0002 2 \u00035\n\n\u00022\u00035\n\n\u0004 20\nWrite the expression. Then use the order of operations to simplify.\n3. (6 \u0002 10) \u0005 2 \u0002 5 \u0003 1 \u0004\n5. 7 2 \u0002 4 2 \u0002 3 \u0004\n\n39\n\n97\n\n6. 3 \u0002 3 \u0005 3 \u0002 3 \u0004\n\n18\n\n7. 25 \u0003 8 \u0002 2 \u0002 3 2 \u0004\n9. 3 \u0002 6 \u0003 4 2 \u0005 2 \u0004\n\n8. 6 \u0003 10 \u0005 5 \u0004\n\n10\n\n10. 4 \u0002 7 \u0002 4 \u0005 2 2 \u0004\n\n29\n\n11. 40 \u0003 2 \u0002 3 2 \u0004\n\n22\n\n12. 8 \u0002 12 \u0005 6 \u0003 3 \u0004\n\n13. 5 \u0002 3 2 \u0003 13 \u0004\n\n32\n\n14. 21 \u0002 49 \u0005 7 \u0002 1 \u0004\n\nSaxon. All rights reser ved.\n\n22\n\n4. (6 \u0002 12) \u0005 3 \u0002 4 2 \u0004\n\n7\n29\nSaxon Algebra 1\n\nReteaching\n\ncontinued\nCompare the expressions. Use \u0002, \u0003, or \u0004.\n2 \u0005 4 2 \u0006 10\n\n\u0002 6 \u0006 2 4\u0006 2\n\nStep 1: Use the order of operations to simplify each expression.\n\n2 \u0005 4 2 \u000610\n6 \u0007 2 4 \u00062\n2 \u0005 16 \u000610\n3 4\u0006 2\n18 \u000610\n12 \u0006 2\n8\n10\nStep 2: Compare using \u0002, \u0003, or \u0004.\n8 \u0002 10\nPractice\nCompare the expressions. Use \u0002, \u0003, or \u0004.\n\n\u0002 20 \u00064 \u0005 3\n\n15. 6 5 \u0006 16 \u0006 8 \u0005 4\n6 5 \u000616 \u0007 8 \u0005 4\n30 \u0006\n\n16. 5 2 2 \u0005 28 \u0007 7\n\n20 \u00064 \u0005 3\n\n5 2 \u0005 28 \u0007 7\n\n20 \u0006 16 \u0005 3\n\n2 \u00054\n28 \u0005 4\n32\n\n\u0002 14 \u0007 7 \u0005 4 2\n\n14 \u0007 7 \u0005 4\n\n\u00066\n\n2 \u00054 8 \u00066\n2 \u0005 32 \u0006 6\n34 \u0006 6\n\n24\n\n\u00027\n\n\u00066\n\n14 \u0007 7 \u0005 4 2 \u0006 6\n\n4 \u0005 28 \u0007 7\n20 \u0005 28 \u0007 7\n20 \u0005 4\n\n\u00053\n\n28\n\n\u0002 28\n\n24 \u0002\n\n32 \u0003\n\nCompare the expressions. Use \u0002, \u0003, or \u0004.\n\n18. 3 5 \u0006 (4 \u0005 2) \u0006 3 \u0002\n\u0002 24\u00056\u00072\n\u0002 10 \u0007 2 \u0005 2 (7 \u0006 4)\n\u0002\n19. 6 (2 \u0005 4) \u0007 3 \u0006 5 \u0002\n\u0003 12 \u0007 (8 \u0007 2) (3 \u0007 3) 20. 3 \u0005 4 3 \u0005 4 \u0002\n\u0003 3 \u000543\u00064\n21. (20 \u0005 12) \u0007 (4 \u0005 4) \u0002\n\u0004 3 2 \u0007 (6 \u0006 3) 22. 42 \u0007 7 \u0005 2 \u0002\n\u0003 3 \u0007 (2 \u0005 1)\n23. 5 \u0006 3 5 \u0005 6 \u0002\n24. (5 \u0005 3) \u0007 2 \u0005 2 \u0002\n\u0002 6 \u0006 (2 \u0005 4) 3\n\u0003 (3 \u0005 6) \u0007 3\n25. 54 \u0005 36 \u0007 3 \u0006 30 \u0002\n\u0003 12 \u0005 3 3 \u0006 28 26. 20 \u0007 4 5 2 \u0007 10 \u0002\n\u0003 (2 \u0005 4) \u0007 12 \u0005 1\n27. (3 \u0005 6) \u0007 3 \u0002\n28. 6 \u0006 (2 \u0005 4) 3 \u0002\n\u0002 5 \u000635\u00056\n\u0003 (5 \u0005 3) \u0007 2 \u0005 2\n17. 2 (7 \u0005 3 ) \u0007 4\n\n29. A rectangular swimming pool is 5 feet deep, 17 feet long, and 12 feet\nwide. Its volume is (5 17 12) cubic feet. A circular swimming pool\nhas a radius of 8 feet and is 6 feet deep. Its volume is about\n2\n(3.14 8 6) cubic feet. What is the difference in volume\nbetween the two pools?\n\nSaxon. All rights reser ved.\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nFinding Absolute Value and Adding Real Numbers\n\nYou have classified real numbers and used them in expressions. Now\nyou will find the absolute values of real numbers.\nAbsolute Value\nThe absolute value of a number is the distance from the number to 0 on a number line.\nThe absolute value of 2 is written as \u0002 2 \u0002.\n\n\u0002 \u0002\n\n3 .\nSimplify __\n4\n\nSimplify \u0002 83 \u0002.\n\n3\n4\n\n-1 -3 -1 -1\n4\n2\n4\n\n1\n4\n\n1\n2\n\n3\n4\n\n8\u00023 \u0003 5\n\n5\n\n3 unit from 0 on the number line.\n\n3 is __\n\u0002__\n4 4\n3.\n3 is __\nThe absolute value of \u0002__\n4 4\n\n-1\n\n5 is 5 units from zero on the number line.\n\nThe absolute value of 8\u00023 is 5.\nPractice\nComplete the steps to simplify each expression.\n1. \u0002 \u00022.5 \u0002\n\n2. \u0002\u0002 10\u00027 \u0002\n10\u00027 \u0003\n\n2.5\n\n3\n3\n\n-3\n\n-2\n\n-1\n\n1\n-3 -2 1\n\n\u00022.5 .is\nline.\n\n2.5\n\n\u0002 \u00022.5 \u0002 \u0003\n\nnumber line.\n\u0002 10\u00027 \u0002 \u0003 3\n\n2.5\n\n\u0002\u0002 10\u00027 \u0002 \u0003\n\nSimplify.\n3. \u0002 8.2 \u0002 \u0003\n5.\n\n8.2\n\n1 \u0003\n2\u0002__\n3\n\n7. \u0002\u0002 12\u00024 \u0002 \u0003\n\n2\n1__\n3\n8\n\n4. \u0002 \u000221 \u0002 \u0003\n\n21\n\n6. \u0002\u0002 5.4 \u0002 \u0003\n\n5.4\n\n8.\n\n9 \u0003\n\u0002___\n11\n\n9\n___\n11\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nAdd the numbers absolute values and use\n\nthe same sign as the numbers.\n\nFind the difference of the numbers absolute\n\nvalues and use the sign of the number with the\ngreater absolute value.\n\nStep 1: Find the difference of the absolute\n\n17 \u0002 5 \u0003 22\nStep 2: Use the sign of the numbers. The\nsign is negative.\n\nvalues 23 \u0004 9 \u0003 14.\nStep 2: Find the sign of the number with the\nlargest absolute value. \u0002 23 \u0002 \u0005 \u0002 \u00049 \u0002, so\nthe sign is positive.\n\n(\u000417) \u0002 (\u00045) \u0003 \u000422\n\n23 \u0002 (\u00049) \u0003 14\n\nPractice\nComplete the steps to find each sum.\n\n\u0003 \u0004 \u0003 \u0004\n\n9. (\u000415) \u0002 11\n15 \u0004 11 \u0003\n\n1 \u0002\n1\n10. \u0004__\n\u0004__\n4\n2\n\n1 \u0002 __\n1\u0003\n__\n\n\u0002 \u000415 \u0002\u0005\u0002 11 \u0002 so the sign of the answer\n\nis\n\nnegative.\n\n3\n__\n4\n\nnegative,\nso the sign of the answer is negative.\n\nThe sign of the numbers is\n\n(\u000415) \u0002 11 \u0003\n\n\u0003\u0004__14 \u0004 \u0002 \u0004__12 \u0003\n\n3\n__\n4\n\nFind each sum.\n\n11. \u0003 \u000416 \u0004 \u0002 3 \u0003\n\n13\n\n3 \u0002 __\n4\u0003\n13. \u0004___\n10\n5\n\n1\n__\n\n14. 19 \u0002 (\u000436) \u0003\n\n15. 14.8 \u0002 \u0003 \u000411.2 \u0004 \u0003\n\n\u0003 \u0004 \u0003 \u0004\n\n5\n1 \u0002 \u0004__\n16. \u0004__\n3\n9\n\n3.6\n\n12\n\n17\n\u0003\n\n8\n__\n9\n\n17. Jenna saved \\$28 for a new CD. She bought it on sale for \\$19. Use\n\n\\$9\n18. In one hour, a hot-air balloon rose 600 feet. In the next hour, the\nballoon dropped 950 feet. Use addition to find the change in the\nelevation of the balloon in the two-hour period.\n\n350 ft\n\n10\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nSubtracting Real Numbers\n\nYou have added real numbers. Now you will subtract real numbers.\nTwo numbers with the same absolute value but different signs are called\nopposites. The opposite of a number is also called the additive inverse.\nThe sum of a number and its additive inverse is 0.\nTo Subtract Real Numbers\n\nFind the difference (6) 12.\n\n(\u00026) \u0003 (\u000212)\n\n(\u00028) \u0003 15\n\nStep 2: Find the difference of the absolute\n\n6 \u0003 12 \u0004 18\n\nvalues. 15 \u0002 8 \u0004 7\n\nStep 2: Find the sign of the number with the\n\nlargest absolute value. \u0002 15 \u0002 \u0005 \u0002 \u00028 \u0002, so\n\nsign is negative.\n\nthe sign is positive.\n\n(\u00026) \u0002 12 \u0004 \u000218\n\n(\u00028) \u0003 (\u000215) \u0004 7\n\nPractice\nComplete the steps to find each difference.\n2. (\u000219) \u0002 3\n\n1. (\u00026) \u0002 (\u000211)\n(\u00026) \u0002 (\u000211) \u0004 (\u00026) \u0003\n11 \u0002 (6) \u0004\n\n11\n\n(\u000219) \u0002 3 \u0004 (\u000219) \u0003\n\n19 \u0003 3 \u0004\n\n\u0002 \u000211 \u0002\u0005\u0002 \u00026 \u0002, so the sign of the answer\n\nis\n\n22\n\nnegative,\nso the sign of the answer is negative.\nThe sign of the numbers is\n\npositive .\n\n(\u00026) \u0002 (\u000211) \u0004\n\n(3)\n\n(\u000219) \u0002 3 \u0004\n\n\u000222\n\nFind each difference.\n\n3. \u0003 \u000217 \u0004 \u0002 (\u000212) \u0004\n\n\u0003 \u0004\n\n5 \u0002 __\n1\u0004\n5. \u0002__\n8\n4\n7. (\u00029.1) \u0002 2.6 \u0004\n\n4. 25 \u0002 (\u000232) \u0004\n\n7\n__\n8\n\n1 \u0004\n5 \u0002 \u0002___\n6. __\n12\n6\n\n11.7\n\n8. 7 \u0002 26 \u0004\n\n57\n11\n___\n12\n\n19\n\n9. For safety, scuba divers usually do not dive deeper than 40 meters\nbelow sea level. A diver in a helmet suit can safely dive about 21 meters\ndeeper than a scuba diver. What is the maximum safe depth for\na helmet suit diver in relation to sea level?\n\n\u000261 m\n\n11\n\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nClosure\nA set is closed under a given operation if the outcome of the operation on any two members\nof the set is also a member of the set.\nDetermine whether the statement is true or false. Give a counterexample if the statement\nis false.\nThe set of natural numbers is closed under subtraction.\nSubtract two natural numbers:\n3\u00022\u00031\n4\u00022\u00032\n2 \u0002 3 \u0003 \u00021\nThe last statement is a counterexample, so the statement is false.\nPractice\nComplete the steps to determine whether the statement is true or false.\nWrite a counterexample if the statement is false.\n10. The set A \u0003 {\u00021, 0, 1} is closed under subtraction.\nSubtract two numbers in the set:\n1\u00020\u0003\n\n0\u00021\u0003\n\n\u00021\n\n\u00021 \u0002 1 \u0003\n\n\u00022\n\nDetermine whether each statement is true or false. Give a\n\ncounterexample for false statements.\n11. The set of even integers, {..., \u00024, \u00022, 0, 2, 4, ...}, is closed under subtraction.\n\ntrue\n12. The set of irrational numbers is closed under subtraction.\n\nfalse; sample counterexample: 2 2 0\n\n13. The set of odd integers plus zero, {..., \u00025, \u00023, \u00021, 0, 1, 3, 5, ...} is\nclosed under subtraction.\n\nfalse; sample counterexample: 3 1 2\n\n14. The set of integers that are a multiple of 4, {..., \u000212, \u00028, \u00024, 0, 4, 8, 12, ...}\nis closed under subtraction.\n\ntrue\n\n12\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\n\nSimplifying and Comparing Expressions with Symbols of Inclusion\n\nYou have used the order of operations to simplify expressions.\nNow you will apply this concept to simplifying expressions within symbols\nof inclusion.\nSymbols of inclusion indicate which numbers, variables, and operations\nare parts of the same term. Some symbols of inclusion are fraction bars,\nabsolute value symbols, parentheses, braces, and brackets.\n\nSimplify 3[11 \u0002 (12 \u0003 9)2] \u0002 7. Justify each step.\n\n3[11 \u0002 (12 \u0003 9)2] \u0002 7\n\u0004 3[11 \u0002 32] \u0002 7\n\n\u0004 3[11 \u0002 9] \u0002 7\n\nEvaluate the exponent.\n\n\u0004 3 \u0004 20 \u0002 7\n\n\u0004 60 \u0002 7\n\nMultiply.\n\n\u0004 67\n\nPractice\nComplete the steps to simplify the expression.\n\n\u0004 4 \u0002 [2z \u0002 \u0002 5 \u0003 7 \u0002]\n_____\n1. 6z\n3\n6z\n\u0004 4 \u0002 [2z \u0002 \u0002 5 \u0003 7 \u0002]\n_____\n3\n6z\n\u0004 4 \u0002 [2z \u0002\u0002 \u00032 \u0002]\n_____\n3\n\nSubtract inside absolute\n\n6z\n\u0004 4 \u0002 [2z \u00022]\n_____\n3\n\nSimplify\n\n24z \u0002 2z \u0002 2\n____\n\nSimplify the\n\nnumerator.\n\nSimplify the\n\nfraction.\n\n8z\n\n\u0002 2z \u0002 2\n\n10z \u0002 2\n\nvalue symbols.\n\nabsolute value.\n\nSimplify.\n2. \u0002 11 \u0003 19 \u0002 \u0002 15 \u0004\n\n23\n\n4. [ (9 \u0003 2)2 \u0003 2(13 \u0003 4) ] \u0002 3 \u0004\n\n\u0004 2 \u0002 3(7 \u0003 2)2 \u0005 5 \u0004\n_____\n6. 14\n4\n\n3. 4(18 \u0003 3) \u0005 (14 \u0003 10 \u0002 \u0002 \u00036 \u0002) \u0004\n\n34\n\n5. 5[ 24 \u0003 (11 \u0003 9)3 ] \u0005 4 \u0004\n\n20\n\n\u00027\u00046 \u00053\u0004\n_____\n7. (16 \u0003 7)2 \u0003 3\n2\n\n22\n\n13\n\n17\n\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nCompare the expressions. Use \u0002, \u0003, or .\n\n[ 21 \u0004 3(9 \u0004 5)2 ] \u0005 50\n\n\u0002 (12 \u0004 7)\n\n\u0004 [ 19 \u0004 3(21 \u0004 15) ]\n\nSimplify each expression. Then compare.\n\n[ 21 \u0004 3(9 \u0004 5) 2 ] \u0005 50\n\u0006 [ 21 \u0004 3(4) 2 ] \u0005 50\n\u0006 [ 21 \u0004 3 \u0007 16 ] \u0005 50\n\u0006 [ 21 \u0004 48 ] \u0005 50\n\u0006 [ \u000427 ] \u0005 50\n\n(12 \u0004 7) 2\u0004[ 19 \u0004 3(21 \u0004 15) ]\n\n\u0006 (5) 2\u0004[ 19 \u0004 3(21 \u0004 15) ]\n\u0006 25 \u0004[ 19 \u0004 3(21 \u0004 15) ]\n\u0006 25 \u0004[ 19 \u0004 3(6) ]\n\u0006 25 \u0004[ 19 \u0004 18 ]\n\n\u0006 50\n\n\u0006 25 \u00041\n\u0006 24\n\n2\nSince 23 \u0002 24, [ 21 \u0004 3(9 \u0004 5) ] \u0005 50\n\n\u0002 (12 \u0004 7) \u0004[ 19 \u0004 3(21 \u0004 15) ].\n\n\u0002\n2\n\nPractice\nComplete the steps to simplify each expression. Compare the\nexpressions. Use \u0002, \u0003, or .\n2\n8. 2(17 \u0004 8) \u0004 [ 3 \u0004(8 \u0004 6) ]\n\n\u0002 [3 (\u00042)\n\n2(17 \u0004 8) \u0004 [ 3 2 \u0004 (8\u0004 6) ]\n\n2 ]\n\u0006 2(17 \u0004 8) \u0004 [ 9 \u0004 2 ]\n\u0006 2(17 \u0004 8) \u0004 7\n\u0006 2( 9 ) \u0004 7\n\u0006 18 \u0004 7\n\u0006 11\n\u0006 2(17 \u0004 8) \u0004 [ 3 2 \u0004\n\nSince11\n\n] \u0005 4(5 \u0005 2)\n\n[ 3 (\u00042) 3 ]\u0005 4(5 \u0005 2)\n\u0006 [ 3 ( 8 ) ]\u0005 4(5 \u0005 2)\n\u0006 (24) \u0005 4(5 \u0005 2)\n\u0006 24 \u0005 4( 7 )\n\u0006 24 \u0005 28\n\u0006 4\n\nCompare the expressions. Use \u0002, \u0003, or .\n\n2\n3\n9. 6 \u0004 [ 4(3 \u0005 2) \u0004 2 ]\n\n\u0003 5(9 \u0004 3) \u0005 4[ 6 \u0005 (\u00042) ]\n\u0002\n3\n\n\u0002 [ (14 \u0004 9) \u0004 6 ] \u0005 2\n\u0002\n11. 9 \u0005 [ 3 \u0002 19 \u0004 3 \u0003 \u0005 5 ] \u0004 4 \u0002\n\u0006 \u0002 23 3\u0004 8 \u0003 \u0004 [ 2(6 \u0004 3) + 3 ]\n4\n10. 3 [ 6 2 \u0004 5(22 \u0004 2 4) \u0005 1 ]\n______\n\n______\n\n14\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nUsing Unit Analysis to Convert Measures\n\nYou have used ratios to compare two quantities. Now you will use unit\nratios to convert measures into different units.\nUnit analysis is the process of converting measures into different units.\nThe peregrine falcon can reach speeds up to 200 miles per hour. How fast is this in\nyards per hour?\nStep 1: Identify the known and missing information.\n\n? yd\n200 mi \u0002 ______\n______\n1 hour\n\n1 hour\n\nStep 2: Equate units.\n\n1760 yd\n1mi .\nSo, the unit ratio is _______, or _______\n1 mi\n1760 yd\n\n1 mi \u0002 1,760 yd\n\nStep 3: Write the multiplication sentence. Then multiply.\n\n1760 yd\n200 mi \u0003 ________\n______\n1 hr\n\n1 mi\n1760 yd\n200 mi \u0003 _______\n______\n1 mi\n1 hr\n200 \u0003 1760 yd\n\u0002 ____________\n1 hr\n352,000 yd\n\u0002 __________\n1 hr\n\nCancel out common factors.\n\nMultiply.\nWrite the ratio of yards per hour.\n\nThe peregrine falcon can reach speeds up to 352,000 yards per hour.\n\nPractice\n1. Alberto Contador won the 2007 Tour de France with an average\nspeed of about 39 kilometers per hour. What was Albertos average\nspeed in meters per hour?\n\n1000 m 39\n39,000 m\n\u0003 1000 m \u0002 ___________\n39 km \u0003 _________\n______\n\u0002 ___________\n1hr\n\n1 hr\n\n1 km\n\n1 hr\n\n2. Some elephants can eat up to 660 pounds\n\nof food per day. How much food can an\nelephant eat in tons per day? One ton is\nequal to 2000 pounds.\n\n3. A sprinkler with a flow rate of 2 gallons per\n\nminute is watering a lawn. What is the flow\nrate of the sprinkler in gallons per hour?\n\n120 gal/h\n\n0.33 t/d\n\n39,000\n\n15\n\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nRemember that area is measured in square units and volume is measured\n\nin cubic units.\nA covered patio measures 6.25 yards by 5 yards. What is the area of the patio in\nsquare feet?\nStep 1: Find the area of the patio.\n\nSo, the conversion is 31.25 yd 2 ? ft 2.\n\n6.25 yd \u0002 5 yd \u0003 31.25 yd 2\nStep 2: Equate units.\n\n1 yd\n3 ft .\nSo, the unit ratio is ____, or ____\n1 yd\n3 ft\nStep 3: Write the multiplication sentence. Then multiply.\n3 ft \u0002 ____\n3 ft\n31.25 yd \u0002 yd \u0002 ____\n1 yd 1 yd\n1 yd \u0003 3 ft\n\n3 ft \u0002 ____\n3 ft\n31.25 yd \u0002 yd \u0002 ____\n1 yd 1 yd\n\n\u0002 3 ft \u0002 3 ft\n_____________\n\u0003 31.25\n1\n\nMultiply.\n\n\u0003 281.25 ft\n\nThe area of the outdoor patio is 281.25 square feet.\n\nPractice\n4. Mr. Greenes yard is 50 feet by 20 feet. He wants to buy sod to cover\nhis yard. Each piece of sod is 1-yard square. What is the area of\nMr. Greenes yard in square yards?\n50 ft \u0002 20 ft \u0003 1000\n\nft 2\n\n1 yd 1 yd 1000 \u00021 yd \u0002 1yd\n1000 ft \u0002 ft \u0002 ____ \u0002 ____ \u0002 _____________ \u0002 111\n\n3 ft 3 ft\n\n111\n\n5. An interior room is 12 feet by 17 feet.\n\nCarpet pieces are 1-yard square. How\nmany square yards of carpet must be\npurchased to cover the floor of the\nroom?\n\nsquare yards.\n\n6. A hose with a flow rate of 15 cubic feet\n\nper hour is filling a large aquarium.\nWhat is the flow rate of the hose in\ncubic inches per hour?\n\n25,920 in 3/h\n\n2 yd 2\n22__\n3\n\nyd 2\n\n16\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\nEvaluating and Comparing Algebraic Expressions\n\nYou have simplified expressions containing only numbers and operations.\n\nNow you will evaluate expressions that contain numbers and/or variables.\nThese expressions are called algebraic expressions.\nEvaluate the expression for n 2 and p 1. Evaluate the expression for a 1 and b 3.\n6p \u0002 3n \u0003 np\n\n2(b \u0003 a) 3 \u0002 5b 2\n\nexpression.\n\nStep 1: Substitute 1 for a and 3 for b in the\n\nexpression.\n\n61\u000232\u000321\n\n2(3 \u0003 1) 3 \u0002 5 3 2\n\noperations.\n\nStep 2: Simplify using the order of operations.\n\n2(3 \u0003 1) 3 \u0002 5 3 2\n\n61\u000232\u000321\n\n\u0004 2(2) 3 \u0002 5 3 2\n\n\u00046\u00026\u00032\n\n\u000428 \u000259\n\n\u00046\n\n\u0004 16 \u0002 45\n\u0004 61\n\nPractice\nComplete the steps to evaluate each expression for the given values.\n2\n2\n2. 2y \u0002 3x \u0003 4y; x \u0004 \u00033, y \u0004 5\n\n1. \u00038c \u0002 4a \u0002 ac; a \u0004 3, c \u0004 \u00032\n\u0003 8(\u00032) \u0002 4 3 \u0002\n\u0004 16 \u0002\n\u0004\n\n12\n\n(\u00032)\n\n2(\n\n\u00036\n\n) \u0002 3(\u00033) \u0003 4 5\n2\n\n\u0004 2(25)\u0002 3(9)\u0003\n\n22\n\n\u0004 50\u0002\n\n27\n\n20\n\n\u0003 20 \u0004\n\n57\n\nEvaluate each expression for the given values.\n\n3. 3b \u0002 ab \u0002 2; a \u0004 5, b \u0004 1\n\n4. 5(c \u0002 d) \u0002 6(c \u0002 2d); c \u0004 4, d \u0004 1\n\n10\n\n61\n\n5. 7x \u0003 2y \u0002 3xy; x \u0004 5, y \u0004 2\n\n6. 2st \u0002 t 2 \u0003 4s; s \u0004 \u00033, t \u0004 \u00032\n\n61\n\n28\n\n7. m \u0002 p 3 \u0003 7p; m \u0004 10, p \u0004 2\n\n8. q \u0003 r 2 \u0002 2(4 \u0002 q); q \u0004 6, r \u0004 \u00031\n\n25\n\n9. A cable company charges a \\$36 monthly fee and then \\$2.99 for each\nmovie ordered. They use the expression 36 \u0002 2.99m, where m is the\nnumber of movies ordered, to find the total amount to charge for each\nmonth. How much would the cable company charge for the month of\nJune if three movies were ordered?\n\n\\$44.97\n\n17\n\nSaxon Algebra 1\n\nReteaching\ncontinued\n\nb 2 \u0004 4ab \u0005 3b 3\n\n\u00052a b\n3\n\nStep 1: Substitute 2 for a and \u00051 for b in the expressions.\n\n(\u00051) 2 \u0004 4(2)(\u00051) \u0005 3(\u00051) 3\n\n\u00052(2) (\u00051)\n3\n\nStep 2: Simplify using the order of operations.\n\n(\u00051) 2 \u0004 4(2)(\u00051) \u0005 3(\u00051) 3\n\n\u0005 2(2) 3(\u00051)\n\n\u0006 1 \u0004 4(2)(\u00051) \u0005 3(\u00051)\n\n\u0006 \u00052(8) (\u00051)\n\n\u0006 1 \u0004 (8)(\u00051) \u0004 3\n\n\u0006 \u000516 (\u00051)\n\n\u0006 1 \u0004 (\u00058) \u0004 3\n\n\u0006 16\n\n\u0006 \u00054\nStep 3: Compare using \u0002, \u0003, or \u0006.\nSince \u00054 \u0002 16, b 2 \u0004 4ab \u0005 3b 3 \u0002 \u00052a 3b when a = 2 and b = \u00051.\nPractice\nComplete the steps to compare the expressions when x 7 and\ny 2. Use \u0002, \u0003, or .\n10. 2(x \u0004 y) \u0004 3x \u0005 0.5y\n2(x \u0004 y) \u0004 3x \u0005 0.5y\n\n12x \u0004 xy\n\n12x \u0004 xy\n\n\u0006 12(7) \u0004 (7)(2)\n\n\u0006 2(9) \u0004 3(7) \u0005 0.5(2)\n\n\u0006 84 \u0004 14\n\n\u0006 18 \u0004 21 \u0005\n\n\u0006 98\n\n\u0006 38\n2(x \u0004 y) \u0004 3x \u0005 0.5y \u0002 12x \u0004 xy when x \u0006 7 and y \u0006 2.\nCompare the expressions for the given values. Use <, >, or \u0006.\n\n\u0002 2a b \u0004 5b; a \u0006 \u00051, b \u0006 3\n\n12. \u00052h \u0004 (2h \u0004 k )\n\n\u0003 \u0005h \u0004 (2h \u0004 k ); h \u0006 \u00053, k \u0006 3\n13. 5(x \u0005 y) \u0004 2y\n\u0006 (x \u0004 y ) ; x \u0006 5, y \u0006 2\n14. k \u0004 2j\n\u0003 j \u0004 2k; j \u0006 3, k \u0006 7\n2 2\n11. 3a b \u0005 4b\n\n15. Cell phone company A charges a \\$30 monthly fee and 15 cents per minute.\nThey use the expression 30 \u0004 0.15m to find the total amount to charge\nfor each month. Cell phone company B charges a \\$25 monthly fee\nand 17 cents per minute. They use the expression 25 \u0004 0.17m\nto find the total amount to charge for each month. Which cell phone\ncompany charges less for 300 minutes during a month? A\n\n18\n\nSaxon Algebra 1\n\nName\n\nDate\n\nClass\n\nReteaching\n\n10\n\nYou have solved problems by adding and subtracting pairs of numbers.\nNow you will solve problems by adding and subtracting three or more\nrational numbers.\n\n\u0002 \u0003\n\n3 .\n2 __\n6 __\n1 __\nSimplify __\n7 7 7\n7\nStep 1: Write the problem as addition.\n\nSimplify 2.14 0.22 5.25 (3.81).\n\nStep 1: Write the problem as addition.\n\n\u0002 \u0003\n\n\u0002 2.14 \u0004 (\u00030.22) \u0004 5.25 \u0004 (\u00033.81)\n\n2 \u0004 __\n1 \u0004 __\n3\n6 \u0004 \u0003__\n\u0002 \u0003__\n7 7\n7\n7\nStep 2: Group the terms with like signs.\n\n\u0002 7\u0003\n\nStep 2: Group the terms with like signs.\n\n\u0002 2.14 \u0004 5.25 \u0004 (\u00030.22) \u0004 (\u00033.81)\n\n2 \u0004 \u0003__\n1 \u0004 __\n6 \u0004 __\n3\n\u0002 \u0003__\n\n7\n\n\u0002 7.39 \u0003 4.03 \u0002 3.36\n\n3 \u0004 __\n9 \u0002 __\n6\n\u0002 \u0003__\n7 7 7\nPractice\nComplete the steps to simplify each expression.\n\n\u0002 \u0003 \u0002 \u0003\n5\n\u0002 \u0002 \u00031 \u0003 \u0004 __ \u0004 2 \u0004 __\n9\n9\n9\n\u0002 49 \u0003\n5 \u00042\n4 \u0004 __\n\u0002 \u0002 \u0003 1 \u0003\u0004 __\n9\n\u0002 9\u0003 9 9\n\n5 \u0004 __\n2 \u0003 __\n1 \u0003 \u0003__\n4\n1. \u0003__\n9\n9\n9 9\n__\n\n\u0002 \u0003\n5\n__\n9\n\n\u0002 7.24 \u0004 (2.78) \u0004 (\n\n__\n\n__\n\n3.4 ) \u0004 5.12\n\n\u0002 7.24 \u0004 5.12 \u0004 (2.78) \u0004 (\n\n__\n\n2\n7 \u0002 __\n\u0004 __\n9\n9\n\n12.36\n\nSimplify.\n3 \u0002\n1 \u0003 __\n4 \u0003 \u0003__\n3. __\n5 5\n5\n\n\u0003 6.18 \u0002\n\n6.18\n\n\u0002 \u0003 0\n13\n6\n5. \u0003 2 \u0004 5 \u0003\u0002 \u00036 \u0003 \u0004 4 \u0002 ___ or 1__\n7\n7\n7 7\n7\n7\n\n3 ___\n3 \u0002\n4\n5 \u0003___\n__\n6. \u0003___\n\u0003\n13\n13 13 \u0003 \u000313\n\n__\n\n__\n\n__\n\n\u0002 \u0003\n\n10.49\n\n\u0002 \u0003\n\n5 \u0002\n3 \u0004__\n4 \u0003 __\n4. \u0003 \u0003__\n11\n11 11\n\n__\n\n6.02\n\n3.4 )\n\n\u0002 \u0003\n\n2\n___\n11\n5\n\u0003___\n13\n5.33\n\n10. 34.19 \u0003 (\u000321.7) \u0004 3.79 \u0003 15.2 \u0002\n\n44.48\n\n11. Mrs. Lewis has \\$156 in her checking account. She writes two checks\nfor \\$31.19 and \\$15.76 and makes one deposit for \\$119. What is her\nnew balance?\n\n\\$228.05\n\n19\n\nSaxon Algebra 1\n\nReteaching\n\n10\n\ncontinued\nOrder the numbers from least to greatest.\n7 , 0.6, __\n2, 1\n___\n5\n10\nStep 1: Place each number on a number line.\n2\n5\n\n-0.6\n-2\n\n-1\n\n7\n10\n\n1\n1\n\nStep 2: To order the numbers, read the numbers on the number line from left to right.\n2 , ___\n7,1\n0.6, __\n5 10\nPractice\nComplete the steps to order the numbers from least to greatest.\n4\n1, __\n12. 1.75, 0.3, __\n4 5\n-1.75\n-2\n\n-4\n5\n\n3 , 1.25, ___\n5 , 0.4\n13. __\n5\n10\n-1\n4\n\n-1\n\n0.3\n0\n\n-2\n\nare\n\n-5\n10\n\n-1.25\n\n4 , __\n1, 0.3.\n1.75, __\n5\n4\n\n-1\n\n0.4\n\n3\n5\n\ngreatest are\n\n3.\n5 , 0.4, __\n1.25, ___\n5\n10\n\nOrder from least to greatest.\n\n1 , __\n4, 1.4, 2\n14. __\n3 3\n\n9 , 1, 0.05\n15. 0.9, ___\n10\n\n5 , 3.025\n16. 5.8, 7, 3__\n8\n\n5, 1.2, 0.8\n3 , __\n17. __\n2 6\n\n3 , __\n7 , 1.45, 2.7\n18. 2__\n4 5\n\n8 , 1.5, 0.3\n19. 1.75, ___\n10\n\n4, __\n1\n2, 1.4, __\n3\n3\n\n9\n1, 0.9, 0.05, ___\n10\n\n5, 3.025\n7, 5.8, 3__\n8\n\n3 , 1.2, 0.8, __\n5\n__\n2\n6\n\n7 , 1.45\n3 , 2.7, __\n2__\n5\n4\n\n8\n1.75, 1.5, 0.3, ___\n10\n\n20. Terra is deep-sea diving. She descends 20 feet below sea level, ascends\n7 feet, descends 15 feet and ascends 4 feet. What is her current\nposition in relation to sea level?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69750786,"math_prob":0.9864321,"size":15689,"snap":"2019-26-2019-30","text_gpt3_token_len":6592,"char_repetition_ratio":0.12712783,"word_repetition_ratio":0.094000556,"special_character_ratio":0.4697559,"punctuation_ratio":0.15094776,"nsfw_num_words":4,"has_unicode_error":false,"math_prob_llama3":0.9623776,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T03:46:39Z\",\"WARC-Record-ID\":\"<urn:uuid:04b7b83d-f9b3-48a7-bc1f-020b88dae947>\",\"Content-Length\":\"335385\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f32c4c8-6e4f-43dc-83b1-bc508fac1143>\",\"WARC-Concurrent-To\":\"<urn:uuid:362a0fd8-3b5b-4553-a93e-df1c427a8ab2>\",\"WARC-IP-Address\":\"151.101.250.152\",\"WARC-Target-URI\":\"https://fr.scribd.com/document/318542976/Algebra-Reteachings-Lessons-1-10\",\"WARC-Payload-Digest\":\"sha1:YPKNKDVFRFFA36XJUEQ5FQEKOEFYTIVD\",\"WARC-Block-Digest\":\"sha1:TALXDPOIJUAGDJARHSULJZCBMCAE4NWV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998369.29_warc_CC-MAIN-20190617022938-20190617044938-00271.warc.gz\"}"}
https://math.stackexchange.com/questions/2737286/given-an-overdetermined-system-of-linear-equation-find-a-subset-that-can-be-sol	[ "# Given an overdetermined system of linear equation, find a subset that can be solved with exact solution\n\nGiven a set A such that A consists of an overdetermined system of linear equation.\n\nFind $$B \\subset A$$ such that B has x equations and x unknowns and has an exact solution.\n\nFor example:\n\nIn a system where you have 4 unknowns and 7 equations, you can solve this by trying all 4 distinct equations you can create from the 7 equations, and then see if it's solvable.\n\nBut the permutations become really big as your overdetermined system grows.\n\nIs there a correct way to do this? Is Linear Programming an option? & if so, how to change this into a linear programming problem?\n\n• By RREF we can select the independent equations. – user Apr 14 '18 at 20:37" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9546279,"math_prob":0.99414116,"size":572,"snap":"2020-45-2020-50","text_gpt3_token_len":129,"char_repetition_ratio":0.13028169,"word_repetition_ratio":0.0,"special_character_ratio":0.22377622,"punctuation_ratio":0.09565217,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99974996,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T07:59:57Z\",\"WARC-Record-ID\":\"<urn:uuid:ce9d626c-a647-49e6-b61c-af4bd8b4b159>\",\"Content-Length\":\"143553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea48f950-7ca8-4fc0-ab26-950c71dacd06>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b04595a-0f1b-4fbc-a1ae-a8dcbb7f611b>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2737286/given-an-overdetermined-system-of-linear-equation-find-a-subset-that-can-be-sol\",\"WARC-Payload-Digest\":\"sha1:PK7OAMZHX24YXN32H7K7XTNSADOEUXWR\",\"WARC-Block-Digest\":\"sha1:GD22QFVU7VPU2HI42VXEYQ7WPICARHOJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876136.24_warc_CC-MAIN-20201021064154-20201021094154-00692.warc.gz\"}"}
https://en.wikipedia.org/wiki/Matter_wave	[ "# Matter wave\n\nMatter waves are a central part of the theory of quantum mechanics, being an example of wave–particle duality. All matter exhibits wave-like behavior. For example, a beam of electrons can be diffracted just like a beam of light or a water wave. In most cases, however, the wavelength is too small to have a practical impact on day-to-day activities. Hence in our day-to-day lives with objects the size of tennis balls and people, matter waves are not relevant.\n\nThe concept that matter behaves like a wave was proposed by Louis de Broglie (/dəˈbrɔɪ/) in 1924. It is also referred to as the de Broglie hypothesis. Matter waves are referred to as de Broglie waves.\n\nThe de Broglie wavelength is the wavelength, λ, associated with a massive particle (i.e., a particle with mass, as opposed to a massless particle) and is related to its momentum, p, through the Planck constant, h:\n\n$\\lambda ={\\frac {h}{p}}={\\frac {h}{mv}}.$", null, "Wave-like behavior of matter was first experimentally demonstrated by George Paget Thomson's thin metal diffraction experiment, and independently in the Davisson–Germer experiment both using electrons, and it has also been confirmed for other elementary particles, neutral atoms and even molecules.\n\n## Historical context\n\nAt the end of the 19th century, light was thought to consist of waves of electromagnetic fields which propagated according to Maxwell's equations, while matter was thought to consist of localized particles (see history of wave and particle duality). In 1900, this division was exposed to doubt, when, investigating the theory of black-body radiation, Max Planck proposed that light is emitted in discrete quanta of energy. It was thoroughly challenged in 1905. Extending Planck's investigation in several ways, including its connection with the photoelectric effect, Albert Einstein proposed that light is also propagated and absorbed in quanta; now called photons. These quanta would have an energy given by the Planck–Einstein relation:\n\n$E=h\\nu$", null, "and a momentum\n\n$p={\\frac {E}{c}}={\\frac {h}{\\lambda }}$", null, "where ν (lowercase Greek letter nu) and λ (lowercase Greek letter lambda) denote the frequency and wavelength of the light, c the speed of light, and h the Planck constant. In the modern convention, frequency is symbolized by f as is done in the rest of this article. Einstein's postulate was confirmed experimentally by Robert Millikan and Arthur Compton over the next two decades.\n\n## de Broglie hypothesis", null, "Propagation of de Broglie waves in 1d – real part of the complex amplitude is blue, imaginary part is green. The probability (shown as the color opacity) of finding the particle at a given point x is spread out like a waveform; there is no definite position of the particle. As the amplitude increases above zero the curvature decreases, so the amplitude decreases again, and vice versa. The result is an alternating amplitude: a wave. Top: plane wave. Bottom: wave packet.\n\nDe Broglie, in his 1924 PhD thesis, proposed that just as light has both wave-like and particle-like properties, electrons also have wave-like properties. By rearranging the momentum equation stated in the above section, we find a relationship between the wavelength, λ associated with an electron and its momentum, p, through the Planck constant, h:\n\n$\\lambda ={\\frac {h}{p}}.$", null, "The relationship is now known to hold for all types of matter: all matter exhibits properties of both particles and waves.\n\nWhen I conceived the first basic ideas of wave mechanics in 1923–1924, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905.\n\n— de Broglie\n\nIn 1926, Erwin Schrödinger published an equation describing how a matter wave should evolve – the matter wave analogue of Maxwell's equations — and used it to derive the energy spectrum of hydrogen.\n\n## Experimental confirmation\n\nMatter waves were first experimentally confirmed to occur in George Paget Thomson's cathode ray diffraction experiment and the Davisson-Germer experiment for electrons, and the de Broglie hypothesis has been confirmed for other elementary particles. Furthermore, neutral atoms and even molecules have been shown to be wave-like.\n\n### Electrons\n\nIn 1927 at Bell Labs, Clinton Davisson and Lester Germer fired slow-moving electrons at a crystalline nickel target. The angular dependence of the diffracted electron intensity was measured, and was determined to have the same diffraction pattern as those predicted by Bragg for x-rays. At the same time George Paget Thomson at the University of Aberdeen was independently firing electrons at very thin metal foils to demonstrate the same effect. Before the acceptance of the de Broglie hypothesis, diffraction was a property that was thought to be exhibited only by waves. Therefore, the presence of any diffraction effects by matter demonstrated the wave-like nature of matter. When the de Broglie wavelength was inserted into the Bragg condition, the observed diffraction pattern was predicted, thereby experimentally confirming the de Broglie hypothesis for electrons.\n\nThis was a pivotal result in the development of quantum mechanics. Just as the photoelectric effect demonstrated the particle nature of light, the Davisson–Germer experiment showed the wave-nature of matter, and completed the theory of wave–particle duality. For physicists this idea was important because it meant that not only could any particle exhibit wave characteristics, but that one could use wave equations to describe phenomena in matter if one used the de Broglie wavelength.\n\n### Neutral atoms\n\nExperiments with Fresnel diffraction and an atomic mirror for specular reflection of neutral atoms confirm the application of the de Broglie hypothesis to atoms, i.e. the existence of atomic waves which undergo diffraction, interference and allow quantum reflection by the tails of the attractive potential. Advances in laser cooling have allowed cooling of neutral atoms down to nanokelvin temperatures. At these temperatures, the thermal de Broglie wavelengths come into the micrometre range. Using Bragg diffraction of atoms and a Ramsey interferometry technique, the de Broglie wavelength of cold sodium atoms was explicitly measured and found to be consistent with the temperature measured by a different method.\n\nThis effect has been used to demonstrate atomic holography, and it may allow the construction of an atom probe imaging system with nanometer resolution. The description of these phenomena is based on the wave properties of neutral atoms, confirming the de Broglie hypothesis.\n\nThe effect has also been used to explain the spatial version of the quantum Zeno effect, in which an otherwise unstable object may be stabilised by rapidly repeated observations.\n\n### Molecules\n\nRecent experiments even confirm the relations for molecules and even macromolecules that otherwise might be supposed too large to undergo quantum mechanical effects. In 1999, a research team in Vienna demonstrated diffraction for molecules as large as fullerenes. The researchers calculated a De Broglie wavelength of the most probable C60 velocity as 2.5 pm. More recent experiments prove the quantum nature of molecules made of 810 atoms and with a mass of 10,123 amu. As of 2019, this has been pushed to molecules of 25,000 amu.\n\nStill one step further than Louis de Broglie go theories which in quantum mechanics eliminate the concept of a pointlike classical particle and explain the observed facts by means of wavepackets of matter waves alone.\n\n## de Broglie relations\n\nThe de Broglie equations relate the wavelength λ to the momentum p, and frequency f to the total energy E of a free particle:\n\n{\\begin{aligned}&\\lambda =h/p\\\\&f=E/h\\end{aligned}}", null, "where h is the Planck constant. The equations can also be written as\n\n{\\begin{aligned}&\\mathbf {p} =\\hbar \\mathbf {k} \\\\&E=\\hbar \\omega \\\\\\end{aligned}}", null, "or \n\n{\\begin{aligned}&\\mathbf {p} =\\hbar \\mathbf {\\beta } \\\\&E=\\hbar \\omega \\\\\\end{aligned}}", null, "where ħ = h/2π is the reduced Planck constant, k is the wave vector, β is the phase constant, and ω is the angular frequency.\n\nIn each pair, the second equation is also referred to as the Planck–Einstein relation, since it was also proposed by Planck and Einstein.\n\n### Special relativity\n\nUsing two formulas from special relativity, one for the relativistic momentum and one for the relativistic mass energy\n\n$E=mc^{2}=\\gamma m_{0}c^{2}$", null, "${\\vec {p}}=m{\\vec {v}}=\\gamma m_{0}{\\vec {v}}$", null, "allows the equations to be written as\n\n{\\begin{aligned}&\\lambda =\\,\\,{\\frac {h}{\\gamma m_{0}v}}\\,=\\,{\\frac {h}{m_{0}v}}\\,\\,\\,\\,{\\sqrt {1-{\\frac {v^{2}}{c^{2}}}}}\\\\&f={\\frac {c}{\\lambda }}={\\frac {\\gamma \\,m_{0}vc}{h}}={\\frac {m_{0}c^{2}}{h}}{\\bigg /}{\\sqrt {{\\frac {c^{2}}{v^{2}}}-1}}\\end{aligned}}", null, "where $m_{0}$", null, "denotes the particle's rest mass, $v$", null, "its velocity, $\\gamma$", null, "the Lorentz factor, and $c$", null, "the speed of light in a vacuum. See below for details of the derivation of the de Broglie relations. Group velocity (equal to the particle's speed) should not be confused with phase velocity (equal to the product of the particle's frequency and its wavelength). In the case of a non-dispersive medium, they happen to be equal, but otherwise they are not.\n\n#### Group velocity\n\nAlbert Einstein first explained the wave–particle duality of light in 1905. Louis de Broglie hypothesized that any particle should also exhibit such a duality. The velocity of a particle, he concluded, should always equal the group velocity of the corresponding wave. The magnitude of the group velocity is equal to the particle's speed.\n\nBoth in relativistic and non-relativistic quantum physics, we can identify the group velocity of a particle's wave function with the particle velocity. Quantum mechanics has very accurately demonstrated this hypothesis, and the relation has been shown explicitly for particles as large as molecules.\n\nDe Broglie deduced that if the duality equations already known for light were the same for any particle, then his hypothesis would hold. This means that\n\n$v_{g}={\\frac {\\partial \\omega }{\\partial k}}={\\frac {\\partial (E/\\hbar )}{\\partial (p/\\hbar )}}={\\frac {\\partial E}{\\partial p}}$", null, "where E is the total energy of the particle, p is its momentum, ħ is the reduced Planck constant. For a free non-relativistic particle it follows that\n\n{\\begin{aligned}v_{g}&={\\frac {\\partial E}{\\partial p}}={\\frac {\\partial }{\\partial p}}\\left({\\frac {1}{2}}{\\frac {p^{2}}{m}}\\right)\\\\&={\\frac {p}{m}}\\\\&=v\\end{aligned}}", null, "where m is the mass of the particle and v its velocity.\n\nAlso in special relativity we find that\n\n{\\begin{aligned}v_{g}&={\\frac {\\partial E}{\\partial p}}={\\frac {\\partial }{\\partial p}}\\left({\\sqrt {p^{2}c^{2}+m_{0}^{2}c^{4}}}\\right)\\\\&={\\frac {pc^{2}}{\\sqrt {p^{2}c^{2}+m_{0}^{2}c^{4}}}}\\\\&={\\frac {pc^{2}}{E}}\\end{aligned}}", null, "where m0 is the rest mass of the particle and c is the speed of light in a vacuum. But (see below), using that the phase velocity is vp = E/p = c2/v, therefore\n\n{\\begin{aligned}v_{g}&={\\frac {pc^{2}}{E}}\\\\&={\\frac {c^{2}}{v_{p}}}\\\\&=v\\end{aligned}}", null, "where v is the velocity of the particle regardless of wave behavior.\n\n#### Phase velocity\n\nIn quantum mechanics, particles also behave as waves with complex phases. The phase velocity is equal to the product of the frequency multiplied by the wavelength.\n\nBy the de Broglie hypothesis, we see that\n\n$v_{\\mathrm {p} }={\\frac {\\omega }{k}}={\\frac {E/\\hbar }{p/\\hbar }}={\\frac {E}{p}}.$", null, "Using relativistic relations for energy and momentum, we have\n\n$v_{\\mathrm {p} }={\\frac {E}{p}}={\\frac {mc^{2}}{mv}}={\\frac {\\gamma m_{0}c^{2}}{\\gamma m_{0}v}}={\\frac {c^{2}}{v}}={\\frac {c}{\\beta }}$", null, "where E is the total energy of the particle (i.e. rest energy plus kinetic energy in the kinematic sense), p the momentum, $\\gamma$", null, "the Lorentz factor, c the speed of light, and β the speed as a fraction of c. The variable v can either be taken to be the speed of the particle or the group velocity of the corresponding matter wave. Since the particle speed $v", null, "for any particle that has mass (according to special relativity), the phase velocity of matter waves always exceeds c, i.e.\n\n$v_{\\mathrm {p} }>c,\\,$", null, "and as we can see, it approaches c when the particle speed is in the relativistic range. The superluminal phase velocity does not violate special relativity, because phase propagation carries no energy. See the article on Dispersion (optics) for details.\n\n### Four-vectors\n\nUsing four-vectors, the De Broglie relations form a single equation:\n\n$\\mathbf {P} =\\hbar \\mathbf {K}$", null, "which is frame-independent.\n\nLikewise, the relation between group/particle velocity and phase velocity is given in frame-independent form by:\n\n$\\mathbf {K} =\\left({\\frac {\\omega _{o}}{c^{2}}}\\right)\\mathbf {U}$", null, "where\n\nFour-momentum $\\mathbf {P} =\\left({\\frac {E}{c}},{\\vec {\\mathbf {p} }}\\right)$", null, "Four-wavevector $\\mathbf {K} =\\left({\\frac {\\omega }{c}},{\\vec {\\mathbf {k} }}\\right)=\\left({\\frac {\\omega }{c}},{\\frac {\\omega }{v_{p}}}\\mathbf {\\hat {n}} \\right)$", null, "Four-velocity $\\mathbf {U} =\\gamma (c,{\\vec {\\mathbf {u} }})=\\gamma (c,v_{g}{\\hat {\\mathbf {n} }})$", null, "## Interpretations\n\nThe physical reality underlying de Broglie waves is a subject of ongoing debate. Some theories treat either the particle or the wave aspect as its fundamental nature, seeking to explain the other as an emergent property. Some, such as the hidden variable theory, treat the wave and the particle as distinct entities. Yet others propose some intermediate entity that is neither quite wave nor quite particle but only appears as such when we measure one or the other property. The Copenhagen interpretation states that the nature of the underlying reality is unknowable and beyond the bounds of scientific inquiry.\n\nSchrödinger's quantum mechanical waves are conceptually different from ordinary physical waves such as water or sound. Ordinary physical waves are characterized by undulating real-number 'displacements' of dimensioned physical variables at each point of ordinary physical space at each instant of time. Schrödinger's \"waves\" are characterized by the undulating value of a dimensionless complex number at each point of an abstract multi-dimensional space, for example of configuration space.\n\nAt the Fifth Solvay Conference in 1927, Max Born and Werner Heisenberg reported as follows:\n\nIf one wishes to calculate the probabilities of excitation and ionization of atoms [M. Born, Zur Quantenmechanik der Stossvorgange, Z. f. Phys., 37 (1926), 863; [Quantenmechanik der Stossvorgange], ibid., 38 (1926), 803] then one must introduce the coordinates of the atomic electrons as variables on an equal footing with those of the colliding electron. The waves then propagate no longer in three-dimensional space but in multi-dimensional configuration space. From this one sees that the quantum mechanical waves are indeed something quite different from the light waves of the classical theory.\n\nAt the same conference, Erwin Schrödinger reported likewise.\n\nUnder [the name 'wave mechanics',] at present two theories are being carried on, which are indeed closely related but not identical. The first, which follows on directly from the famous doctoral thesis by L. de Broglie, concerns waves in three-dimensional space. Because of the strictly relativistic treatment that is adopted in this version from the outset, we shall refer to it as the four-dimensional wave mechanics. The other theory is more remote from Mr de Broglie's original ideas, insofar as it is based on a wave-like process in the space of position coordinates (q-space) of an arbitrary mechanical system.[Long footnote about manuscript not copied here.] We shall therefore call it the multi-dimensional wave mechanics. Of course this use of the q-space is to be seen only as a mathematical tool, as it is often applied also in the old mechanics; ultimately, in this version also, the process to be described is one in space and time. In truth, however, a complete unification of the two conceptions has not yet been achieved. Anything over and above the motion of a single electron could be treated so far only in the multi-dimensional version; also, this is the one that provides the mathematical solution to the problems posed by the Heisenberg-Born matrix mechanics.\n\nIn 1955, Heisenberg reiterated this:\n\nAn important step forward was made by the work of Born [Z. Phys., 37: 863, 1926 and 38: 803, 1926] in the summer of 1926. In this work, the wave in configuration space was interpreted as a probability wave, in order to explain collision processes on Schrödinger's theory. This hypothesis contained two important new features in comparison with that of Bohr, Kramers and Slater. The first of these was the assertion that, in considering \"probability waves\", we are concerned with processes not in ordinary three-dimensional space, but in an abstract configuration space (a fact which is, unfortunately, sometimes overlooked even today); the second was the recognition that the probability wave is related to an individual process.\n\nIt is mentioned above that the \"displaced quantity\" of the Schrödinger wave has values that are dimensionless complex numbers. One may ask what is the physical meaning of those numbers. According to Heisenberg, rather than being of some ordinary physical quantity such as, for example, Maxwell's electric field intensity, or mass density, the Schrödinger-wave packet's \"displaced quantity\" is probability amplitude. He wrote that instead of using the term 'wave packet', it is preferable to speak of a probability packet. The probability amplitude supports calculation of probability of location or momentum of discrete particles. Heisenberg recites Duane's account of particle diffraction by probabilistic quantal translation momentum transfer, which allows, for example in Young's two-slit experiment, each diffracted particle probabilistically to pass discretely through a particular slit. Thus one does not need necessarily think of the matter wave, as it were, as 'composed of smeared matter'.\n\nThese ideas may be expressed in ordinary language as follows. In the account of ordinary physical waves, a 'point' refers to a position in ordinary physical space at an instant of time, at which there is specified a 'displacement' of some physical quantity. But in the account of quantum mechanics, a 'point' refers to a configuration of the system at an instant of time, every particle of the system being in a sense present in every 'point' of configuration space, each particle at such a 'point' being located possibly at a different position in ordinary physical space. There is no explicit definite indication that, at an instant, this particle is 'here' and that particle is 'there' in some separate 'location' in configuration space. This conceptual difference entails that, in contrast to de Broglie's pre-quantum mechanical wave description, the quantum mechanical probability packet description does not directly and explicitly express the Aristotelian idea, referred to by Newton, that causal efficacy propagates through ordinary space by contact, nor the Einsteinian idea that such propagation is no faster than light. In contrast, these ideas are so expressed in the classical wave account, through the Green's function, though it is inadequate for the observed quantal phenomena. The physical reasoning for this was first recognized by Einstein.\n\n## De Broglie's phase wave and periodic phenomenon\n\nDe Broglie's thesis started from the hypothesis, \"that to each portion of energy with a proper mass m0 one may associate a periodic phenomenon of the frequency ν0, such that one finds: 0 = m0c2. The frequency ν0 is to be measured, of course, in the rest frame of the energy packet. This hypothesis is the basis of our theory.\"\n\nDe Broglie followed his initial hypothesis of a periodic phenomenon, with frequency ν0 , associated with the energy packet. He used the special theory of relativity to find, in the frame of the observer of the electron energy packet that is moving with velocity $v$", null, ", that its frequency was apparently reduced to\n\n$f=\\nu _{0}{\\sqrt {1-{\\frac {v^{2}}{c^{2}}}}}\\,.$", null, "Then\n\n$\\lambda f=E/p=v_{\\mathrm {p} }\\,.$", null, "using the same notation as above. The quantity $v_{\\mathrm {p} }$", null, "is the velocity of what de Broglie called the \"phase y wave\". Its wavelength is $\\lambda$", null, "and frequency $f$", null, ". De Broglie reasoned that his hypothetical intrinsic particle periodic phenomenon is in phase with that phase wave. This was his basic matter wave conception. He noted, as above, that $v_{\\mathrm {p} }>c$", null, ", and the phase wave does not transfer energy.\n\nWhile the concept of waves being associated with matter is correct, de Broglie did not leap directly to the final understanding of quantum mechanics with no missteps. There are conceptual problems with the approach that de Broglie took in his thesis that he was not able to resolve, despite trying a number of different fundamental hypotheses in different papers published while working on, and shortly after publishing, his thesis. These difficulties were resolved by Erwin Schrödinger, who developed the wave mechanics approach, starting from a somewhat different basic hypothesis." ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/2fe51c74da5131f00b9dc318ab67fb2e872d868f", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/c6c0386dc6d9530519404f95570fcc8548ed2326", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/8e47dfdf8f6b7a2e8b741aa57d1cc726d23a0e5d", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/2/21/Propagation_of_a_de_broglie_wave.svg/290px-Propagation_of_a_de_broglie_wave.svg.png", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/25ae59ad6fb050fbb5a803b838468a29e69c4615", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9c48c0c53eeae7a8471182b3c75e2a744ed8e462", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0db2065aaf54683b4a56b8901f973b6c4b05ac5d", null, 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http://bda.ath.cx/blog/2007/12/06/jordan-form/	[ "## Jordan Form\n\nThis is happening to thousands of Math graduate students all over the country:\n\nYou are taking a Linear Algebra class, and you are expected to be able to find the Jordan form of a matrix on the exams (and probably the qualifier). Your textbook (perhaps “Matrix Analysis” by Horn and Johnson?) describes the form, proves that every complex matrix is similar to a matrix in Jordan form, describes what can be extrapolated about a matrix given its Jordan form, all with detailed proofs. Then it goes on to hand wave about how to find the Jordan form of a matrix, and extrapolating the algorithm from the description takes a degree in re-constructive surgery. Your algebra textbook (Dummit & Foote?) has an even more confusing description.\n\nThe usual suspects like Wikipedia and Mathworld are completely useless. Google has never failed you like this before. Your textbook also explains why the Jordan form is completely useless in practical applications, which explains why nobody bothers to explain the algorithm and then requires you to know it.\n\nYou ask a senior student or your professor, and they explain about generalized eigenvectors. You grind through some examples, and finally it starts to make sense.\n\nOr you stumble across this everything2 article, which is the only published article on or off the Internet describing the algorithm in plain language. Actually it’s still pretty confusing, but as a grad student, you have a high complexity (pain) tolerance.\n\nIt helps to realize that the process is similar to diagonalization (and in fact is diagonalization if the matrix is diagonalizable) but instead of placing the eigenvectors as columns in the transformation matrix, you place the generalized eigenvector as columns (and order matters to some extent here).\n\nIf you don’t need the transformation matrix, you may be able to skip some of these steps: the geometric multiplicity of an eigenvalue gives the number of Jordan blocks associated with that eigenvalue, the algebraic multiplicity gives the sum of the sizes of all Jordan blocks associated with an eigenvalue, and the size of the blocks can be found by looking at… er… how many generalized eigenvalues are associated with each eigenvector. I don’t have my notes here so I’m getting a bit hazy on that part. Perhaps I will make this clear in a later edit…" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9278928,"math_prob":0.8587636,"size":2317,"snap":"2022-27-2022-33","text_gpt3_token_len":463,"char_repetition_ratio":0.106355384,"word_repetition_ratio":0.015789473,"special_character_ratio":0.18990074,"punctuation_ratio":0.0783848,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96161896,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T00:19:38Z\",\"WARC-Record-ID\":\"<urn:uuid:f3f78a65-89ae-48f4-a389-ec67a4f45b92>\",\"Content-Length\":\"25634\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ca5e1a8a-2745-4175-b826-7cbc3c171591>\",\"WARC-Concurrent-To\":\"<urn:uuid:38fcd1bd-f746-4a7a-9ff3-9f541ab19185>\",\"WARC-IP-Address\":\"207.192.72.39\",\"WARC-Target-URI\":\"http://bda.ath.cx/blog/2007/12/06/jordan-form/\",\"WARC-Payload-Digest\":\"sha1:EQXSXUTP22PC5SCRSJQBA3L3LA4AKWVS\",\"WARC-Block-Digest\":\"sha1:GFHQOMYMBKOX6JJJ5IPU67O2SZEP6VVK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104205534.63_warc_CC-MAIN-20220702222819-20220703012819-00443.warc.gz\"}"}
https://tex.stackexchange.com/questions/208043/ordinal-table-of-contents	[ "# Ordinal table of contents\n\nI want create table of contents As follows:\n\nChapter first (chapter name)\n\n1.1 (section name)\n\n1.2 (section name)\n\n1.2.1 (subsection name)\n\nChapter second (chapter name)\n\n2.1 (section name)\n\n2.1.1 (subsection name)\n\n2.2 (section name)\n\nPlease guide me.\n\nThanks in advance\n\n• Hi and welcome. Where are you stuck? Having the chapter unnumbered or having the sections numbered the way they are. Or producing a toc in the first place? – Johannes_B Oct 20 '14 at 7:49\n• @Johannes_B: I believe, its the ordinal numbers first, second etc, as the title of the question proposes. – user31729 Oct 20 '14 at 7:50\n\n## 1 Answer\n\nThis is a solution that depends on the book class, by removing patching the \\@chapter macro, changing the \\addcontentsline entry and setting it to \\Ordinalstring{chapter} from fmtcount package.\n\n\\documentclass{book}\n\n\\usepackage{fmtcount}\n\\usepackage{etoolbox}\n\n\\makeatletter\n\\patchcmd{\\@chapter}{\\addcontentsline{toc}{chapter}{\\protect\\numberline{\\thechapter}#1}}{\\addcontentsline{toc}{chapter}{\\protect{\\numberline{\\chaptername~\\Ordinalstring{chapter} #1}}}}{}{}\n\\makeatother\n\n\\begin{document}\n\\tableofcontents\n\n\\chapter{Some chapter name}\n\n\\section{Number one}\n\n\\subsection{Subsec Number one}\n\n\\chapter{Another chapter name}\n\n\\section{Number one from 2nd chapter}\n\n\\subsection{Subsec Number one}\n\n\\end{document}", null, "" ]	[ null, "https://i.stack.imgur.com/bu38n.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85258055,"math_prob":0.4725663,"size":405,"snap":"2019-35-2019-39","text_gpt3_token_len":124,"char_repetition_ratio":0.22693267,"word_repetition_ratio":0.0,"special_character_ratio":0.33827162,"punctuation_ratio":0.17391305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9781892,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T12:03:14Z\",\"WARC-Record-ID\":\"<urn:uuid:6536d3dd-920f-4c7a-9f61-b8f51eb91ea2>\",\"Content-Length\":\"133248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d99d9074-7877-40c2-a056-a387e55710e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:aedec196-811f-4fea-b1cc-a7a5286430cc>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/208043/ordinal-table-of-contents\",\"WARC-Payload-Digest\":\"sha1:QOKJYWWH32NGHODCIVUSNOHRDL57FNRR\",\"WARC-Block-Digest\":\"sha1:NIH3RRVJX37WSYUUQV5TDP2RS6UD6XAZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027323328.16_warc_CC-MAIN-20190825105643-20190825131643-00005.warc.gz\"}"}
https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=1971722&nr_roman=(M)(C)(M)(L)(X)(X)MDCCXXII&lang=en	[ "# Convert the Hindu-Arabic number 1,971,722 to a Roman number written with Roman numerals. Turn and write it using the Latin numeral system letters I, V, X, L, C, D, M. Learn by using the detailed explanations converter\n\n## The latest Hindu-Arabic numbers converted to Roman numerals\n\n How to convert: write the Hindu-Arabic number 1,971,722 using Roman numerals: (M)(C)(M)(L)(X)(X)MDCCXXII Jun 01 16:43 UTC (GMT) How to convert: write the Hindu-Arabic number 731,984 using Roman numerals: (D)(C)(C)(X)(X)(X)MCMLXXXIV Jun 01 16:43 UTC (GMT) How to convert: write the Hindu-Arabic number 1,705 using Roman numerals: MDCCV Jun 01 16:42 UTC (GMT) How to convert: write the Hindu-Arabic number 80 using Roman numerals: LXXX Jun 01 16:42 UTC (GMT) How to convert: write the Hindu-Arabic number 160 using Roman numerals: CLX Jun 01 16:42 UTC (GMT) How to convert: write the Hindu-Arabic number 12,799 using Roman numerals: (X)MMDCCXCIX Jun 01 16:42 UTC (GMT) How to convert: write the Hindu-Arabic number 220,918 using Roman numerals: (C)(C)(X)(X)CMXVIII Jun 01 16:41 UTC (GMT) How to convert: write the Hindu-Arabic number 95,384 using Roman numerals: (X)(C)(V)CCCLXXXIV Jun 01 16:40 UTC (GMT) How to convert: write the Hindu-Arabic number 138 using Roman numerals: CXXXVIII Jun 01 16:40 UTC (GMT) How to convert: write the Hindu-Arabic number 14 using Roman numerals: XIV Jun 01 16:40 UTC (GMT) How to convert: write the Hindu-Arabic number 1,794,974 using Roman numerals: (M)(D)(C)(C)(X)(C)M(V)CMLXXIV Jun 01 16:40 UTC (GMT) How to convert: write the Hindu-Arabic number 444,444 using Roman numerals: (C)(D)(X)(L)M(V)CDXLIV Jun 01 16:40 UTC (GMT) How to convert: write the Hindu-Arabic number 2,560,018 using Roman numerals: (M)(M)(D)(L)(X)XVIII Jun 01 16:40 UTC (GMT) All the Hindu-Arabic numbers converted to Roman numerals, online operations\n\n## The set of basic symbols of the Roman system of writing numerals\n\n• ### () M = 1,000,000 or \|M\| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.\n\n() These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: \"(\" and \")\", because:\n\n• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and\n• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line \"\|\" and the Roman numeral \"I\" (1).\n\n() An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate \"1,000 times\". See below...\n\nLogic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.\n\n() At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.\n\nThus, initially, the largest number that could be written using Roman numerals was:\n\n• MMMCMXCIX = 3,999." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8142505,"math_prob":0.9863552,"size":2440,"snap":"2023-14-2023-23","text_gpt3_token_len":794,"char_repetition_ratio":0.20566502,"word_repetition_ratio":0.25661376,"special_character_ratio":0.33237705,"punctuation_ratio":0.1574074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.955848,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T16:43:28Z\",\"WARC-Record-ID\":\"<urn:uuid:efeac498-7c9f-4745-b502-2a5c9662cc2c>\",\"Content-Length\":\"46779\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efd72836-6521-4688-ad5b-fc75194fcb14>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e8031c1-8cd9-4d2c-b3ab-9d22d177f7ef>\",\"WARC-IP-Address\":\"93.115.53.187\",\"WARC-Target-URI\":\"https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=1971722&nr_roman=(M)(C)(M)(L)(X)(X)MDCCXXII&lang=en\",\"WARC-Payload-Digest\":\"sha1:OZ2BWBXFO3B344DNQWMZYDHSCFZPO2HO\",\"WARC-Block-Digest\":\"sha1:VSPUM63YBYIF4LS6SQTOVYVHGIZ4FXII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647895.20_warc_CC-MAIN-20230601143134-20230601173134-00510.warc.gz\"}"}
https://shelah.logic.at/papers/211/	[ "# Sh:211\n\n• Shelah, S. (1992). The Hanf numbers of stationary logic. II. Comparison with other logics. Notre Dame J. Formal Logic, 33(1), 1–12.\n• Abstract:\nWe show that the ordering of the Hanf number of L_{\\omega,\\omega}(wo) (well ordering), L^c_{\\omega,\\omega} (quantification on countable sets), L_{\\omega, \\omega}(aa) (stationary logic) and second order logic, have no more restraints provable in ZFC than previously known (those independence proofs assume CON(ZFC) only). We also get results on corresponding logics for L_{\\lambda,\\mu}.\n• Current version: 1996-03-11_10 (20p) published version (12p)\nBib entry\n@article{Sh:211,\nauthor = {Shelah, Saharon},\ntitle = {{The Hanf numbers of stationary logic. II. Comparison with other logics}},\njournal = {Notre Dame J. Formal Logic},\nfjournal = {Notre Dame Journal of Formal Logic},\nvolume = {33},\nnumber = {1},\nyear = {1992},\npages = {1--12},\nissn = {0029-4527},\nmrnumber = {1149955},\nmrclass = {03C75 (03C55 03E35)},\ndoi = {10.1305/ndjfl/1093636007},\nnote = {\\href{https://arxiv.org/abs/math/9201243}{arXiv: math/9201243}},\narxiv_number = {math/9201243}\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73235613,"math_prob":0.95742744,"size":1166,"snap":"2020-45-2020-50","text_gpt3_token_len":382,"char_repetition_ratio":0.10499139,"word_repetition_ratio":0.06451613,"special_character_ratio":0.39536878,"punctuation_ratio":0.21681416,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9705289,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T12:34:45Z\",\"WARC-Record-ID\":\"<urn:uuid:8e5fe998-09c0-4976-8e4c-2a11ece5e4d9>\",\"Content-Length\":\"9089\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47e5ad0f-03b2-43c3-8a0c-79d7fa6fc183>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a644c89-b613-4ee7-8c65-25fc6dd59bb1>\",\"WARC-IP-Address\":\"78.47.24.141\",\"WARC-Target-URI\":\"https://shelah.logic.at/papers/211/\",\"WARC-Payload-Digest\":\"sha1:FDUXQQTAIGJIGMPTMCMAPIXWHAAPFOL5\",\"WARC-Block-Digest\":\"sha1:BK26UQDJSY2ZTXIVQJI4RK4QDN5B4BMZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141213431.41_warc_CC-MAIN-20201130100208-20201130130208-00692.warc.gz\"}"}
https://www.generationgenius.com/intro-to-finding-area/	[ "", null, "Read an Intro to Finding Area \| for kids 3rd, 4th, & 5th Grade\n1%\nIt was processed successfully!", null, "WHAT IS INTRO TO FINDING AREA?\n\nYou will learn what area is, how area is measured, how to find area by counting square units, and how to find the area of rectangles using addition or multiplication. You will also learn that area is additive and shapes can be composed and decomposed to find the area of shapes formed by rectangles.\n\nTo better understand finding area…\n\nWHAT IS INTRO TO FINDING AREA?. You will learn what area is, how area is measured, how to find area by counting square units, and how to find the area of rectangles using addition or multiplication. You will also learn that area is additive and shapes can be composed and decomposed to find the area of shapes formed by rectangles. To better understand finding area…\n\n## LET’S BREAK IT DOWN!\n\n### Somersaults", null, "Area is measured in square units. Let’s say you want to make a floor mat for a tumbling class. You put 5 rows that each have 2 square pads. You can count the number of squares to find the area you can tumble on and find that the area is 10 square units. Try this one yourself: You lay 24 square pads in a rectangular mat for a wrestling class. Then, you pick up 4 of the pads. What is the area of the new mat?\n\nSomersaults Area is measured in square units. Let’s say you want to make a floor mat for a tumbling class. You put 5 rows that each have 2 square pads. You can count the number of squares to find the area you can tumble on and find that the area is 10 square units. Try this one yourself: You lay 24 square pads in a rectangular mat for a wrestling class. Then, you pick up 4 of the pads. What is the area of the new mat?\n\n### Defenders of the Area", null, "Different shapes can have the same area, as long as the same number of square units are needed to fill the space. You can also add or remove square units for a larger or smaller area. Let’s say you have a cape that is made of 27 squares of fabric of the same size. It is 9 squares long and 3 squares wide. You remove 6 squares from the bottom of the cape and attach them to a side to use as a hood. The area of your cape is still 27 square units. Try this one yourself: You use 12 square units of fabric to make a cape that is 3 squares long and 4 squares wide. Your cape is not long enough, so you add 9 more squares at the bottom. What is the area of your cape now?\n\nDefenders of the Area Different shapes can have the same area, as long as the same number of square units are needed to fill the space. You can also add or remove square units for a larger or smaller area. Let’s say you have a cape that is made of 27 squares of fabric of the same size. It is 9 squares long and 3 squares wide. You remove 6 squares from the bottom of the cape and attach them to a side to use as a hood. The area of your cape is still 27 square units. Try this one yourself: You use 12 square units of fabric to make a cape that is 3 squares long and 4 squares wide. Your cape is not long enough, so you add 9 more squares at the bottom. What is the area of your cape now?\n\n### Poster Array", null, "The area of a rectangle or square can be thought of as an array, with rows and columns of square units. To find the number of square units in the shape, you can multiply the number of columns by the number of rows. This is the same as multiplying the length by the width. Let’s say you have square posters that you are using to cover a wall. The wall fits 5 posters high and 5 posters wide. You can multiply 5 times 5 to find that the total area of the wall is 25 square posters. Try this one yourself: One-foot square posters are used to cover a wall. The wall is 6 posters high and 4 posters wide. What is the area covered by the posters?\n\nPoster Array The area of a rectangle or square can be thought of as an array, with rows and columns of square units. To find the number of square units in the shape, you can multiply the number of columns by the number of rows. This is the same as multiplying the length by the width. Let’s say you have square posters that you are using to cover a wall. The wall fits 5 posters high and 5 posters wide. You can multiply 5 times 5 to find that the total area of the wall is 25 square posters. Try this one yourself: One-foot square posters are used to cover a wall. The wall is 6 posters high and 4 posters wide. What is the area covered by the posters?\n\n### Pixel Art", null, "Shapes are not always perfect squares or rectangles. Sometimes you can break shapes apart into squares and rectangles to find their area. Let’s say you have a T-shaped figure that can be broken into a 2 × 2 square and a 6 × 2 rectangle. You can multiply to find that the area of the square is 4 square units and the area of the rectangle is 12 square units. Then, you can add the two areas to find that the area of the full shape is 4 + 12 = 16 square units. Try this one yourself: You have a U-shaped figure that can be broken into 2 rectangles that are 6 units long and 3 units wide and one square with a side length of 3 units. What is the area of the figure?\n\nPixel Art Shapes are not always perfect squares or rectangles. Sometimes you can break shapes apart into squares and rectangles to find their area. Let’s say you have a T-shaped figure that can be broken into a 2 × 2 square and a 6 × 2 rectangle. You can multiply to find that the area of the square is 4 square units and the area of the rectangle is 12 square units. Then, you can add the two areas to find that the area of the full shape is 4 + 12 = 16 square units. Try this one yourself: You have a U-shaped figure that can be broken into 2 rectangles that are 6 units long and 3 units wide and one square with a side length of 3 units. What is the area of the figure?\n\n## INTRO TO FINDING AREA VOCABULARY\n\nArea\nArea is the amount of space inside a closed, two-dimensional shape.\nSquare\nA square is a polygon with four sides that are the same length and four right angles.\nRectangle\nA rectangle is a polygon with four sides and four right angles.\nUnit of measurement\nA way to communicate the size of a measurement.\nSquare unit\nA square unit is a square with side lengths of 1 unit that is used to measure the area of a shape.\nFormula\nA rule or fact written with mathematical symbols. The formula for the area of a rectangle is length times width.\nMade by arranging items or shapes into equal rows and columns." ]	[ null, "https://www.facebook.com/tr", null, "https://www.generationgenius.com/wp-content/uploads/2021/07/Intro-to-Finding-Area-THUMBNAIL-4-1024x576.jpg", null, "https://www.generationgenius.com/wp-content/uploads/2021/06/412reading1-600x338.jpeg", null, "https://www.generationgenius.com/wp-content/uploads/2021/06/412reading2-600x338.jpeg", null, "https://www.generationgenius.com/wp-content/uploads/2021/06/412reading3-600x338.jpeg", null, "https://www.generationgenius.com/wp-content/uploads/2021/06/412reading4-600x338.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9411131,"math_prob":0.9864858,"size":2794,"snap":"2022-40-2023-06","text_gpt3_token_len":664,"char_repetition_ratio":0.18996416,"word_repetition_ratio":0.060822897,"special_character_ratio":0.23836793,"punctuation_ratio":0.07804878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99228036,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,1,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T05:08:35Z\",\"WARC-Record-ID\":\"<urn:uuid:581c4766-db40-4234-b966-570f626588db>\",\"Content-Length\":\"1049687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38a0cd56-85e6-4911-900a-83031b649faa>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8b1505c-4ebf-410b-9a2c-b0e73aebf29e>\",\"WARC-IP-Address\":\"104.26.7.79\",\"WARC-Target-URI\":\"https://www.generationgenius.com/intro-to-finding-area/\",\"WARC-Payload-Digest\":\"sha1:YZL4KBD5VTNATBZUF5EOVFUM7POVZFDY\",\"WARC-Block-Digest\":\"sha1:XUUHQPKSH7VF75CKHTHBCPOMBWOMVFI6\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764501407.6_warc_CC-MAIN-20230209045525-20230209075525-00787.warc.gz\"}"}
https://www.wulnut.top/2020/01/15/python_ex34/	[ "python入门学习\n\n# 基本操作\n\n## type() and isinstance() 函数\n\ntype(object): 接收一个对象object来作为参数，返回这个参数的数据类型\nisinstance(object, class): 判断接收的对象object是否是给定的类型class的对象：如果是就返回True,如果不是返回False.\n\ntype(object):\n\nm = 120\nprint(\"m Type: \", type(m))\n\nm = \"大数据\"\nprint(\"m Type: \", type(m))", null, "isinstance(object, class):\n\na = 20\nprint(\"a是整型么？\", isinstance(a, int))", null, "1. type()不会认为子类对象时一种父类类型，不考虑继承关系，也就是说type()只检测当前该数据的数据类型\n2. isinstance()会认为子类队形时一种父类类型，会考虑继承关系，也就是说如果该数据时类的话isinstance()会\n检测父类的数据类型\n\n## eval()函数\n\neval()函数用来执行一个字符串表达式，并返回表达式的值，其一般格式为：\n\neval(expression[,globals[,locals]])\n\n\na = eval('2 + 3')\nprint(\"a: \", a)\n\na, b = eval(input(\"请输入两个数(用','隔开): \"))\nprint(\"a: \", a)\nprint(\"b: \", b)\n\n\n## 简单了解位运算符\n\nkey = input(\"请输入加密密匙：\")\nenc = input(\"请输入要加密的字符: \")\n\ndec = ord(key) ^ ord(enc)\nprint(\"加密结果:\",chr(dec))\n\nenc = ord(key) ^ dec\nprint(\"解密结果:\",chr(enc))\n\n\n1\n\n• ord()函数是对输入的字符转换成ASCII码\n• chr()函数是对输入的ASCII码(可以是十进制、十六进制)转换成对应的字符" ]	[ null, "https://www.wulnut.top/img/type.png", null, "https://www.wulnut.top/img/isinstance.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.58375424,"math_prob":0.9990244,"size":1100,"snap":"2022-27-2022-33","text_gpt3_token_len":602,"char_repetition_ratio":0.14051095,"word_repetition_ratio":0.0,"special_character_ratio":0.28454545,"punctuation_ratio":0.18079096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97168756,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T07:10:48Z\",\"WARC-Record-ID\":\"<urn:uuid:805b0636-a644-49df-aa4b-905f399164ff>\",\"Content-Length\":\"27655\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96457d97-4f14-4a83-9a1c-8fa1018d5c06>\",\"WARC-Concurrent-To\":\"<urn:uuid:927332de-c6a9-43b1-98c3-646653a8e3e3>\",\"WARC-IP-Address\":\"8.48.85.209\",\"WARC-Target-URI\":\"https://www.wulnut.top/2020/01/15/python_ex34/\",\"WARC-Payload-Digest\":\"sha1:76HOLE4HNXPIX5C6WSNBKI2TZNXGARZQ\",\"WARC-Block-Digest\":\"sha1:BZHSXDCDMJZE4YAEGX5QTUPOEGFLHO6D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103034877.9_warc_CC-MAIN-20220625065404-20220625095404-00614.warc.gz\"}"}
http://www.cut-the-knot.org/pythagoras/Munching/DiameterChord.shtml	[ "Why Diameter Is the Longest Chord?\n\nIndeed why? Why a diameter is the longest chord in a circle? I sometimes heard and several times read at popular math sites that the reason why a diameter is the longest chord is that a diameter passes through the center of the circle. While that is true that passing through the center has something to do with the length of a chord, the answer, as given, is vacuous. Since the definition of a diameter is a chord passing through the center of the circle, such an explanation actually reads: \"The diameter is the longest chord in a circle because it is a diameter of the circle.\" How much does that explain?\n\nLet's recollect the definitions:\n\nA segment of a straight line joining two points on a circle is called a chord; a chord that passes through the center of the circle is called a diameter. (Ambiguously, the word \"diameter\" also denotes the length of a diameter.)\n\nAs the statement is definitely not an axiom of geometry, it must be proved, i.e., logically derived from simpler statements and the definitions.\n\nIncidentally, Euclid proved that statement in the third book of his Elements as a more informative Proposition XV:\n\nOf straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote.\n\nOn a casual inspection, it seems obvious:", null, "Nonetheless, it has to be proved and Euclid proves the first part with the reference to the Triangle Inequality (I.20) and the second part to the Pythagorean theorem.", null, "Let $AB$ be a diameter of the circle with center $C$ and $DE$ a chord not through $C.$ Then, by the definition of the circle as the locus of points equidistant from the center, $CA = CB = CD = CE = R,$ the radius of the circle. Which makes $AB = 2R.$ (The diameter is twice as long as the radius.)\n\nOn the other hand, in $\\Delta CDE,$ by the triangle inequality,\n\n$DE\\lt CD + CE = R + R= 2R = AB.$\n\nThe same route is taken in my favorite Kiselev's Geometry. In another geometric classics, Lessons in Geometry by J. Hadamard the triangle inequality is applied differently, after establishing the fact that, for a point $P$ on the diameter $AB,$ one of the ends of the diameter is farthest from P among all points of the circle and the other is nearest.", null, "Pick point $P$ not on a circle, pass a line through $P$ and the center of the circle. The line will meet the circle in two points $A$ and $B.$ Let $A$ be the nearest of the two to $P.$ Then $AP$ is the (absolute) difference between $OP$ and $OA = R,$ $BP$ is the sum of $OP$ and $R.$ Let $M$ be any other point on the circle. In $\\Delta OMP,$\n\n$PM \\gt \|OP - OM\| = \|OP - R\| = \|OP - OA\| = AP.$\n\nAlso,\n\n$PM \\lt OP + OM = OP + R = BP.$\n\nThus, of all points on the circle, $A$ is the nearest while $B$ is the farthest from $P.$ Observe now that the two inequalities that express this fact are valid also for $P$ on the circle. Let $P$ coincide with $A.$ Then, for any $M \\ne B,$ $MA \\lt AB$, meaning that the diameter $AB$ is longer than the chord $MA.$\n\nA third proof makes use of the Pythagorean and Thales' theorems. If $AB$ is a diameter and $M$ a point on a circle different from $A$ and $B,$ then $\\Delta AMB$ is right at $M.$ The Pythagorean theorem tells us that $AM^{2} + BM^{2} = AB^{2},$ implying in particular that, say, $AM \\lt AB.$\n\nNote: Vincent Pantaloni observed that conversely,\n\nThe longest chord in a circle passes through the center of the circle and is, therefore, a diameter.\n\nIndeed, let $AB$ be the longest chord in circle $(O)$ centered at $O.$ Let $AO$ meet the circle the second time in $B'.$ Then\n\n$AB' = 2R = OA + OB \\ge AB,$\n\nimplying (since $AB$ was said to be the longest chord) $AB' = AB,$ so that $B = B'$ and $O$ lies on $AB.$\n\nThere is another proof exploiting the (all-around) symmetrical nature of the circle.\n\nLet $AB$ be a chord of circle $(O)$ that is not a diameter. Reflect the circle in $AB.$ Note that any axis of symmetry of a circle ought to pass through its center (because a perpendicular bisector of a chord always goes through the center of the circle.) Since $AB$ is not a diameter, it could not be an axis of symmetry either, thus the reflection gives us another circle $(O')$ in which $AB$ is also a chord. Moreover, $AB\\perp OO'.$", null, "Let $CC'$ and $DD'$ be the common tangents of the two circles. In particular, $OC\\perp CC'\\perp O'C'$ and similarly for the second tangent. $AB$ does not cross $CC'$ for, if it did, $OA$ would be longer than $OC,$ which could not be. So, there is a rectangle $CDD'C'$ and, within, line segment $AB$ parallel to, say, $CD,$ implying $AB\\lt CD.$\n\nReferences\n\n1. A. Givental, Kiselev's Geometry. Book I. PLANIMETRY, Sumizdat, 2006, p. 88.\n2. J. Hadamard, Lessons in Geometry , Education Development Center, Dec 2006, Corollary to Theorem 64.\n3. R. Simson, The Elements of Euclid, Eliborn Classics, 2005, p. 65", null, "" ]	[ null, "http://www.cut-the-knot.org/pythagoras/Munching/DiameterChord1.gif", null, "http://www.cut-the-knot.org/pythagoras/Munching/DiameterChord2.gif", null, "http://www.cut-the-knot.org/pythagoras/Munching/Hadamard.gif", null, "http://www.cut-the-knot.org/pythagoras/Munching/DiameterChord3.jpg", null, "http://www.cut-the-knot.org/gifs/tbow_sh.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9183774,"math_prob":0.9994717,"size":4927,"snap":"2019-26-2019-30","text_gpt3_token_len":1320,"char_repetition_ratio":0.16473696,"word_repetition_ratio":0.025442477,"special_character_ratio":0.27217373,"punctuation_ratio":0.12941177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99991655,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-18T19:24:40Z\",\"WARC-Record-ID\":\"<urn:uuid:e3dce6cd-fa93-4eb1-93e4-3cc3022af97e>\",\"Content-Length\":\"16739\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f684689-6d0b-47eb-a401-7bdc12c67cb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:1825c9f7-dd82-4fde-9a80-ea3926c7a781>\",\"WARC-IP-Address\":\"107.180.50.227\",\"WARC-Target-URI\":\"http://www.cut-the-knot.org/pythagoras/Munching/DiameterChord.shtml\",\"WARC-Payload-Digest\":\"sha1:2BSRO4IQEHCXJ2EQC5GUBHQFBGPGYWQ3\",\"WARC-Block-Digest\":\"sha1:Y5XBSG4WJAGZDAQ3KFLS2GDAXJ2JRWCP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998813.71_warc_CC-MAIN-20190618183446-20190618205446-00252.warc.gz\"}"}
http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20180416111102608709066	[ "# MathSciDoc: An Archive for Mathematician ∫\n\n#### Numerical Analysis and Scientific Computingmathscidoc:1804.25031\n\nJournal of Computational and Applied Mathematics, 2018\nTo avoid the order reduction when third order implicit-explicit Runge-Kutta time discretization is used together with the local discontinuous Galerkin (LDG) spatial discretization, for solving convection-diffusion problems with time-dependent Dirichlet boundary conditions, we propose a strategy of boundary treatment at each intermediate stage in this paper. The proposed strategy can achieve optimal order of accuracy by numerical verification. Also by suitably setting numerical flux on the boundary in the LDG methods, and by establishing an important relationship between the gradient and interface jump of the numerical solution with the independent numerical solution of the gradient and the given boundary conditions, we build up the unconditional stability of the corresponding scheme, in the sense that the time step is only required to be upper bounded by a suitable positive constant, which is independent of the mesh size.\nlocal discontinuous Galerkin method, implicit-explicit time discretization, convection-diffusion equation, Dirichlet boundary condition, order reduction\n```@inproceedings{haijin2018third," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63627875,"math_prob":0.85050595,"size":792,"snap":"2023-40-2023-50","text_gpt3_token_len":198,"char_repetition_ratio":0.087563455,"word_repetition_ratio":0.41791046,"special_character_ratio":0.23863636,"punctuation_ratio":0.20325203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9641451,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T20:59:21Z\",\"WARC-Record-ID\":\"<urn:uuid:e1cf82fa-0699-4c0b-9cb4-fb6b73945e31>\",\"Content-Length\":\"26614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6193049a-bf32-436c-8918-f7512a76d6e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:4487fc45-25ec-43f0-952e-b86a9c686a02>\",\"WARC-IP-Address\":\"101.6.6.219\",\"WARC-Target-URI\":\"http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20180416111102608709066\",\"WARC-Payload-Digest\":\"sha1:VJDWZEWSCZI6HH2SKGMFYOF6CAG2WEH3\",\"WARC-Block-Digest\":\"sha1:G7QWGLJU7C2GJO6VPFJQHRVWKKNIICP7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510326.82_warc_CC-MAIN-20230927203115-20230927233115-00258.warc.gz\"}"}
https://www.nature.com/articles/s41598-020-79507-4?error=cookies_not_supported&code=ca69ccae-cfb8-4e43-8d7f-7c0a8199f621	[ "## Introduction\n\nNumber theory has a reputation of “unreasonable effectiveness.” Perhaps the most famous example is the Fibonacci numbers, which are closely related to golden ratio, plant growth (Phyllotaxis), and DNA patterns. Over the past years, a few developments have begun to unfold the potentials of number theory in understanding complex networks1,2. This work employs the p-adic numbers for modeling complex networks based on the Erdős-Rényi (ER) random graph. The p-adic number system gives an extension of the ordinary arithmetic of rational numbers ($${\\mathbb {Q}}$$) in a way distinct from the common extension of $${\\mathbb {Q}}$$ to real numbers ($${\\mathbb {R}}$$) and complex numbers ($${\\mathbb {C}}$$). The p-adic number system can be applied in various scientific fields. One groundbreaking application in physics is the p-adic AdS/CFT3. Khrennikov et al.4 developed p-adic wavelet for modeling reaction-diffusion dynamics. Applications in biology include the models for hierarchical structures of protein5 and genetic code6.\n\n### Networks, behaviors, and health\n\nThere is growing interest on a unified theory of complex networks from various fields7,8,9. The formalism commenced with graph theory in mathematics. The emergence of giant component is essential for the evaluation of random graphs10,11,12. Network analysis revealed hidden structures in social and economic systems13,14. Exponential random graph15 and stochastic blockmodels16,17 were developed. Many statistical mechanics in physics, e.g., percolation18 and time series19, are modeled with complex networks. Network models are closely associated with epidemiology20,21 and public health22. Chains of affection23 is a classical work revealing the network structure as a critical factor in public health. The spatiotemporal dynamics20 is essential for understanding the epidemic processes.\n\nThe complex networks also play an essential role in the microscopic scale. Numerous researches in biology rely on the curation and archival storage of protein, genetic, and chemical interactions for all major model organism species and humans (e.g. BioGRID and STRING database). The disease network24,25, protein-protein interaction network, and gene network26 contributed to the network approaches to life science27. In the near future, the network advances in biology, such as drug targeting28,29 and network medicine30, might be critical for improving our health.\n\n### Erdős–Rényi model and extensions\n\nThe classical ER model has inspired the continuous development of a rich spectrum of sophisticated models. There are three major approaches:\n\n• Creating more complex structures such as multilayer networks31,32 and multiplex networks33,34. Complex networks often exhibit community or module structures7.\n\n• Manipulating the rules for constructing edges. The Achlioptas process picks two candidate edges each time18 for competitive graph-evolution. ER process, Bohman and Frieze process, and product rule process were compared with one another for analyzing explosive percolation35.\n\n• Projecting graphs onto social, geometric, or geographic structures. Major developments in that aspect include the hyperbolic networks36, spatial preferential attachment37, inhomogeneous random graph38,39, and spatial networks30.\n\n### Findings\n\nCommon attempts to modeling the hierarchical structures7 are reflected by various notions including subpopulation, subgraph, mixing pattern, community, and module. We postulate that hierarchical structures are naturally encrypted in a standard graph. Consequently, imposing additional structures onto a graph to enrich its behavior is not always necessary. The key is the p-adic absolute value40. According to Ostrowski’s theorem, every non-trivial absolute value on $${\\mathbb {Q}}$$ is either the usual real absolute value or the p-adic absolute value. Various hierarchical structures can be represented by p-adic integers as nodal indices. The network topology and relative strengths between connections are unified as p-adic distances between numbers.\n\nThe p-adic random graph (PARG), probably the simplest model of inhomogeneous networks, offers a flexible method to simulate various observations in complex networks, especially the phenomenon of multiple big components. Degree distribution is a key property for distinguishing random, free-scale, and small world networks. However, PARG indicates that two random graphs with identical degree distribution may produce significantly different component sizes.\n\nWe fit PARG to the component size distribution of the genetic interaction networks, and also to the joint distributions of big components in COVID-19 outbreaks. The experiments imply that the community structures are responsible for the multimodal distributions of the sizes of big components. The largest or the second largest component could be more stable at (multiple) specific sizes. Therefore, maintaining a local peak could be valuable for intervening the spreading processes.\n\nIn contrast to the celebrated ER network, another early prototype of random graph, the Rado graph, has been rarely revisited. The Rado graph employs the binary number system to encode the graph edges, using Ackermann coding of hereditarily finite sets. PARG explores the fundamental nature of integers to encode the probability of connecting a pair of nodes. The p-adic number system extends the ordinary arithmetic of rational numbers3,40. Our PARG model will focus on the p-adic metric on nonnegative integers. An integer’s r-adic (picking a prime number r) absolute value is the reciprocal of the largest power of r that divides it. For example, $$\|40\|_2=1/8$$ (let r=2, then the 2-adic absolute value of 40 equals 1/8), $$\|40\|_3=1$$, $$\|40\|_5=1/5$$. Such absolute value is the most significant example of ultrametrics40. Each node is naturally associated with an integer, i.e., its index in the ordered set of nodes, or an arbitrary (unique) integer can be assigned to each node. The probability of connecting any pair of nodes ij is proportional to the p-adic closeness between the two nodal indices $$v_i, v_j$$ :\n\n\\begin{aligned} p_{ij}= \\frac{p^}{ \|v_j -v_i\|_r} \\end{aligned}\n(1)\n\nfor $$i,j\\in [1,n],\\; i<j$$ (assuming $$i<j \\Leftrightarrow v_i<v_j$$). $$p^$$ is a constant as the probability in the ER sense. When comparing ER and PARG, we normalize the PARG probability so that $$\\sum _{ij} p_{ij} =n(n-1) p^/2$$. As a result, the number of edges in ER equals that in PARG. The p-adic distances encode a hierarchical structure, as shown in the circular tree-map in Fig. 1a. The distances between any pair of numbers from the same small circle and the same big circle are 1/9 and 1/3, respectively. If the two numbers are from different big circles, their distance is 1.\n\nThe digit format of a p-adic number is intuitive, for example, $$201_3=2\\cdot 3^0 +0\\cdot 3^1+1\\cdot 3^2=11$$. One can construct a set of integers (as the nodal indices) in their digit formats by\n\n\\begin{aligned} u_0 . u_1 . \\cdots .u_m \\;_r\\, {\\mathop {=}\\limits ^{ \\text {def}}}\\, \\{ a_0 a_1 \\cdots a_m \\;_r \\; \| \\; 0\\le a_i < u_i \\text { for } i=0,1,\\cdots ,m \\} \\end{aligned}\n(2)\n\nwhere r is the chosen prime number. We call the model a full PARG if $$u_0=u_1=\\cdots = u_m=r$$ (i.e., $$n= r^{m+1}$$), otherwise, we call it a regular PARG (i.e., $$n= \\prod _i u_i$$). The nodal indices can also be arbitrary digits under a given prime r, which leads to a general PARG. The expression (2) facilities the enumeration of hierarchical structures. For example, $$3.2.3_3$$ fully describes a hierarchical structure as shown in Fig. 1c. More examples are in illustrated in Fig. 1. A notation such as $$G(3.2.3_3, p^)$$ fully specifies a PARG.\n\nPARG implements a Bernoulli process on all pairs from a set of p-adic numbers. Let $$p_k$$ represent the probability of a randomly chosen node with degree k. In the ER model, $$p_k$$ follows a Binomial distribution. It becomes a Poisson distribution in the limit of large n. By contrast, n in PARG equals the number of individuals in observations. Because of the symmetric connectivity in regular PARG, $$p_k$$ can be obtained from the degree distribution of one node:\n\n\\begin{aligned} p_k= \\sum _{\\alpha _0,\\alpha _1,\\cdots ,\\alpha _m} \\prod _{i=0}^m (r^i p^)^{\\alpha _i} (1-r^i p^)^{d_i-\\alpha _i} \\left( {\\begin{array}{c}d_i\\\\ \\alpha _i\\end{array}}\\right) \\end{aligned}\n(3)\n\nwhere $$\\sum _{i=0}^m d_i =n-1$$ holds. $$d_i$$ denotes the number of links (from the chosen node) with probability $$r^ip^$$ according to (1-2). The numbers $$\\alpha _0,\\alpha _1,\\cdots ,\\alpha _m$$ denote all combinations satisfying $$k=\\sum _{i=0}^m \\alpha _i$$. Numeric computing of (3) indicates that $$p_k$$ in PARG and that in ER are almost identical. The two models can have the same degree distribution and the same number of edges. However, there could be significant differences in the size distributions of big components in PARG and ER (Fig. 2, Supplementary Note 1, Table S1 and S2). The largest components in PARG may exhibit multimodal distribution due to the hierarchical structure. This implies that certain sizes of (multiple) big components are more statistically stable than other sizes.\n\n## Sizes of big components\n\nThe relative sizes of multiple big components in complex networks were much less studied compared to the studies carried out on the giant component12,41. Analytical methods11,42 and generating functions10 have been widely employed for analyzing the component sizes. Rather than let $$n\\rightarrow \\infty$$, n in PARG is equal to the number of relevant individuals in observations. As a result, the sizes of simulated components are similar to that in ground truth. When n is finite, numeric random realizations are suited to evaluate the various probabilities about component sizes (see “Methods” section).\n\nWhen n is fixed, fitting ER model to empirical data only involves the single parameter p, while PARGs involve two kinds of parameters: the probability p and the hierarchical structures represented by (2). The configuration space of (2) is very vast, even when $$n<1000$$, so we opted for ad hoc heuristics to choose the hierarchical structures that fit relatively well to observations. The heuristics includes: (1) scaling the distances. For example, $$7.5.6_r$$ with $$r=7, 11, 13,\\cdots$$ represent the same hierarchical structure, though the distances between the levels are scaled. (2) Flat vs. deep hierarchy. For instance, both $$16.16_{17}$$ and $$2.2.2.2.2.2.2.2_2$$ refer to 256 nodes. The former is made of 16 groups (each contains 16 nodes), while the latter’s structure looks like a high tree.\n\nBased on the hierarchical structures in PARG, the following experiments analyze multiple big components, especially $$5\|C_2\|>\|C_1\|$$, as static structures observed in networks. The topics range from microscopic networks, such as biological networks26,28, to macroscopic networks, such as epidemics21,43.\n\n### Genetic interaction networks\n\nThe essential role of genetic interaction networks plays in biology has been lately revealed25. The essential genetic interaction network of yeast genes (theCellmap.org) contains 1,261 mutant strains. Their interactions have been characterized by Pearson correlation coefficient (PCC). Genes with highly correlated genetic interaction profiles (PCC>0.4) form clusters of specific pathways or protein complexes26. We set three PCC thresholds above 0.4 to obtain three graphs with distinct big components, as shown in Fig. 3g–i. We count the components sizes falling into the predefined intervals of Fibonacci numbers ($$b_{i+1}=b_i+b_{i-1}, b_0=3$$). Let $$\\theta _i$$ be the number of simulated components whose sizes fall between $$b_i$$ and $$b_{i+1}$$, and $$\\theta _i^$$ be that in ground truth. The error of a random realization is given by\n\n\\begin{aligned} \\sum _i \\left[ (b_{i+1}-b_i) (\\theta _i - \\theta _i^) \\right] ^2 \\end{aligned}\n(4)\n\nThe averaged error from many random realizations yields a relatively accurate evaluation. ER and PARG with distinct values of np lead to different errors (Fig. 3a–c). Equipped with (configurable) hierarchical structures, PARGs fit better to observations than ER. The component size distributions of the best fits are shown in Fig. 3d−f.\n\n### Protein-protein interaction\n\nExploring the protein interaction networks of proteins poses a major challenge in biomedicine. Protein-protein interaction (PPI) is crucial to understanding cellular pathways and human diseases44. The following experiment creates a graph from a set of 408 S. cerevisiae protein complexes as45. The graph nodes represent individual proteins from these complexes. An edge is constructed between two nodes (proteins) if they belong to the same complex. The graph includes 1,628 nodes and 11,249 edges.\n\nWe define the similar metric $$S_{duo}$$ and $$S_{tri}$$ (see “Methods” section) to compare the simulated component sizes with the ground truth. PARG fits better ($$S_{tri}$$=0.307) than the ER model ($$S_{tri}$$=0.112) to the PPI network, as shown in Fig. 4. The simulation data can be found in supplementary Table S3 and S4. It means that the chosen PARG has a higher probability that the sizes of its big components resemble those in the PPI network.\n\n### Instrumental resource street network\n\nThere has been growing attention to the impact of social networks on health46. For example, homeless youth is an active research field, including analysis and interventions. Social networks of homeless youth47,48 are vital for understanding and intervening the observed phenomena.\n\nThis experiment involves a social network about employment services utilization among homeless youth. The original research49 queried 136 homeless youth in Los Angeles in 2008. Four distinct networks were constructed from the same population, according to instrumental, emotional, employment services use, and sociometric relationship respectively. Only the instrumental network ($$\|C_1^\|$$=30, $$\|C_2^\|$$=13) satisfies $$5\|C_2^\|>\|C_1^\|$$, i.e., the second largest component is large enough. Regarding the similarity metric $$S_{duo}$$, the ER model attains the maximum similarity ($$S_{duo}$$=0.155) at $$p=1.08/n$$; while $$G(10.14_{353}, 1.75/n )$$ has a much higher similarity ($$S_{duo}$$ =0.316). The community structure in the PARG corresponds to the social or geographical networks of the homeless youth; although the map between the two is still elusive.\n\nWe also fit the hyperbolic networks36 and the Achlioptas process35 to the observed component sizes ($$\|C_1^\|$$=30, $$\|C_2^\|$$=13). Details can be found in “Methods” section. The random hyperbolic graph50,51 reaches the maximum similarity ($$S_{duo}$$=0.266) when $$C=-9.3$$, $$\\alpha =10$$, $$D=0.0684R$$. The Achlioptas process with product rule (PR) has the maximum similarity ($$S_{duo}$$=0.221) when the number of edges is equal to 100. So PARG outperforms the other two models in this case.\n\nCoronavirus52 has spread among many Chinese cities since the end of January 2020. The incubation period53 and possible mild symptoms54 made the prevention more complicated. Social networking sites (or local officials) reported traces of infected people. Relationships between the infected (and those who had close contacts with them) were also investigated. From a point of view of networks, the cities exhibited three distinct patterns: 1) No big components. Shenzen reported 416 confirmed cases by February 20, 2020. The largest cluster has only 9 people. 2) A giant component. Xinyu reported 110 confirmed cases by February 10, 2020. The giant component consists of 52 cases, nearly half of the infected population. The second largest component has only six cases. 3) Multiple big components. Tianjin reported 136 confirmed cases by February 3, 2020. The first, second, and third largest clusters contain 44, 17 and 11 cases, respectively, which are related to a huge department store, the railway, and a residential area, respectively.\n\nWe focus on Tianjin’s infection network (Supplementary Note 1, Table 5), which consists of multiple big components. A graph is created to visualize the relationship between the infected when the outbreak was around its peak, as shown in Fig. 5. The simulation data can be found in Supplementary Table S6 and S7. The similarity metric $$S_{tri}$$ could be biased when $$\|C_3^\|$$ is quite smaller than $$\|C_1^\|$$, so the similarity metric $$T_{tri}$$ (see “Methods” section) is employed in this case to fit the models to the observed clusters, as shown in Fig. 6. The PARG $$G(8.17_{59}, 1.49/n)$$ has the highest similarity $$T_{tri}$$=0.00916.\n\nWe also compared the results of the random hyperbolic graph and the Achlioptas process with that of PARG. The random hyperbolic graph reaches the maximum similarity ($$T_{tri}$$=0.0150) when $$C=-9.65$$, $$\\alpha =1$$, $$D=0.04294R$$. The Achlioptas process with Bohman Frieze (BF) rule has the maximum similarity ($$T_{tri}$$=0.00825) when the number of edges is equal to 93. So the random hyperbolic graph fits best to this case.\n\nA later investigation indicates that the 11 cases (in yellow, Fig. 5) are probably related to the department store as well. In this new perspective, the two big clusters form the largest component of 55 nodes. We employ the metric $$T_{duo}$$ to fit modes to this new observation ($$\|C_1^\|$$=55, $$\|C_2^\|$$=17), as shown in Fig 7. The simulation data can be found in Supplementary Table S8 and S9. The great variety of hierarchical structures enable PARG to fit relatively well to observations from distinct perspectives.\n\n## Discussion\n\nThe size distributions of big components in complex networks are attributed to the structure of physical world; the behaviors of agents (nodes) and the information transmitting between them; and the observer (how to look at the events). The ER model offered prominent findings of component sizes in networks, however, it rarely fits the joint distribution P($$\|C_1\|\\approx \|C_1^\|$$, $$\|C_2\|\\approx \|C_2^\|$$, $$\\cdots$$) in real observations. A successful strategy is introducing inhomogeneous structures or selective rules (for constructing edges) to the random graph to increase its versatility. PARG probably provides the simplest way to fully describe a hierarchical structure in an ER-like model.\n\nPARG interprets the n in ER model as the cardinality of a set of natural numbers (nodal indices). Consequently, the probability p can be weighted by distances between the nodal indices. In number theory (Ostrowski’s theorem), any non-trivial definition of absolute value on $${\\mathbb {Q}}$$ is either the conventional one or the p-adic absolute value. So, the p-adic ultrametric reveals the natural hierarchical structures hidden in any graph with indexed nodes. PARG blurs the boundaries between the topology approaches (e.g., multiplex networks) and the geometric approaches (e.g., hyperbolic networks).\n\nIn our PARG approach, n denotes the number of observed individuals, whereas in previous ER-like models $$n\\rightarrow \\infty$$. The limit of n facilities analytical approaches10,11, while a relatively small finite n is convenient for numerical random realizations. Random graph theories such as explosive percolation35 and synchronization55, deeply revealed the dynamics of the emerged components. By contrast, this work explores the sizes of resultant (static) components from observations or simulations. The results imply that the proportions between the big component sizes are closely associated with the hierarchical structures of complex networks. This implication is in contrast to previous emphasis10,27,56 on degree distribution as the fingerprint of network structures.\n\nWe fit PARGs and other random graphs to observations from various types of networks. The PARG outperforms the ER model and the Achlioptas process. The random hyperbolic model fits better than PARG to certain cases but worse than PARG to other cases. The simulations of PARG show that the size of big components, e.g., $$P(\|C_1\|=x)$$ and $$P(\|C_2\|=x)$$, can exhibit multimodal distribution (e.g., Figs. 2b and 7(b)) due to their modular structures. In the case of multimodal distribution, the first peak of $$P(\|C_2\|=x)$$ could be very close to the last peak of $$P(\|C_1\|=x)$$. Thus, one may distinguish the major mode from the minor modes when investigating the giant component. One present challenge in network epidemic modeling57 is designing network-based interventions. Current strategies include targeting high-degree nodes or central nodes. The multimodal distribution of $$P(\|C_i\|=x)$$ has implications in controlling the spreading processes in networks. Since $$P(\|C_i\|=x)$$ has more than one local peak, it might be possible to predict and maintain the big components’ growth around a local peak.\n\n## Methods\n\n### Random graph implementation\n\nAll random graphs are created through random experiments implemented in the Java programming language (Java 1.8 with Eclipse IDE). Given a connecting probability p (in the ER context), an edge is included in the graph if\n\n\\begin{aligned} p > rand \\end{aligned}\n\nwhere rand stands for a random number between 0 and 1, generated by the method nextDouble() from the java.util.Random class. The method generates a stream of pseudorandom numbers via linear congruential generator (LCG) with modulus $$2^{48}$$.\n\nA disjoint-set data structure (Union-Find algorithm) is employed to find all components in graph. N random realizations of ER or PARG yield an ensemble of binary numbers\n\n\\begin{aligned} \\delta _{ix}^j= {\\left\\{ \\begin{array}{ll} 1, \\text { if } \|C_i^j\|=x \\\\ 0, \\text { otherwise} \\\\ \\end{array}\\right. } \\end{aligned}\n\nfor $$x=1,2,\\cdots , n$$. $$\|C_i^j\|$$ denotes the size of the ith largest component in the jth realization. Then one can evaluate the size distribution of the big components by\n\n\\begin{aligned} P(\|C_i\|=x) =\\frac{1}{N} \\sum _{j=1}^N \\delta _{ix}^j \\end{aligned}\n\nOne can choose an a prior function to measure the similarity between the simulated components sizes and the observed sizes. For example, Poisson distribution can be used to define the similarity between $$\|C_i^j\|$$ and the observation :\n\n\\begin{aligned} \\Psi _i^j= \\frac{\|C_i^\|!\\; \|C_i^\|^{\\left( \|C_i^j\|- \|C_i^\|\\right) } }{\|C_i^j\|!} \\end{aligned}\n\nor one can use the normal distribution:\n\n\\begin{aligned} \\Psi _i^j= \\exp \\left( \\frac{ \\left( \|C_i^j\|-\|C_i^\|\\right) ^2}{s^2 \|C_i^\|^2} \\log \\frac{1}{2} \\right) \\end{aligned}\n\nwhere s is a constant (typically $$s=0.1$$, so that 10% deviation from the truth results in 1/2 similarity). Simulations in this work employed the later definition. The probability distribution of $$\|C_2\|$$ under the condition $$\|C_1\|\\approx \|C_1^\|$$ is given by\n\n\\begin{aligned} P(\|C_2\|=x \\; \\big \|\\; \|C_1\| \\approx \|C_1^\| ) =\\frac{1}{N} \\sum _{j=1}^N \\Psi _1^j \\delta _{2x}^j \\end{aligned}\n\nLikewise, $$P(\|C_3\|=x \\; \\big \|\\; \|C_1\| \\approx \|C_1^\| ) =\\frac{1}{N} \\sum _{j=1}^N \\Psi _1^j \\delta _{3x}^j$$.\n\nWe define the objective function\n\n\\begin{aligned} S_{duo}= \\frac{1}{2N} \\sum _{j=1}^N \\sum _{i=1}^2 \\Psi _i^j,\\;\\; S_{tri}= \\frac{1}{3N} \\sum _{j=1}^N \\sum _{i=1}^3 \\Psi _i^j \\end{aligned}\n\nfor a random graph to measure whether its first two/three largest component sizes are close to that in observation. When $$\|C_2^\|$$ or $$\|C_3^\|$$ is much smaller than $$\|C_1^\|$$, the following metric would be more appropriate:\n\n\\begin{aligned} T_{duo}= \\frac{1}{N} \\sum _{j=1}^N \\Psi _1^j \\Psi _2^j, \\;\\; T_{tri}= \\frac{1}{N} \\sum _{j=1}^N \\Psi _1^j \\Psi _2^j \\Psi _3^j \\end{aligned}\n\n### Genetic interaction network of yeast genes\n\nThe data of yeast genes is from https://thecellmap.org/ costanzo2016/. The networks in Fig. 3g–i are drawn from the data of the Essential $$\\times$$ Essential network, which involves 1,261 mutant strains. So, each network in Fig. 3g–i consists of 1261 nodes. The PPC values between the mutant strains are obtained from the genetic interaction profile similarity matrices. An edge is included in the graph if the corresponding PPC value is above the predefined threshold. The graphs are projected onto squares using our Java program, as shown in Fig. 3g–i. In a dynamical process, the agents (nodes) push away from each other, while each edge drags the two end nodes into a fixed range. The color (hue) of edges indicates the size of the relevant component.\n\n### Protein network of yeast genes\n\nThe data of S. cerevisiae protein complexes is obtained from the additional File 1 of45. We programmed a Java application to read the table, construct the 1,628 nodes (proteins) and 11,249 edges (a pair of nodes belong to a same complex), and find the components via Union-Find algorithm.\n\n### Random hyperbolic graph\n\nOur implementation follows the formulation in50,51. $$R=2\\ln n + C$$ denotes the radius of the disc. The probability density for the radial coordinate r of a point $$(r, \\phi )$$ is given by\n\n\\begin{aligned} \\alpha \\frac{ \\sinh (\\alpha r)}{ \\cosh (\\alpha R) -1} \\end{aligned}\n\nWe use inverse transform sampling to generate the radii of the points, i.e., $$r= \\frac{1}{a} arcosh (1+ \\cosh (\\alpha R) x -x)$$ where $$x\\in (0,1)$$ denotes a random number from the uniform distribution. Our experiments generate x by nextDouble() in Java. For each pair of nodes uv, a link is added to the graph if $$d(u,v) < D$$ where $$D \\in (0,R]$$ is a constant and d(uv) denotes the distance between the two points in the hyperbolic space.\n\n\\begin{aligned} \\cosh ( d(u,v)) = \\cosh r_u \\cosh r_v - cos(\\theta _u - \\theta _v) \\sinh r_u \\sinh r_v \\end{aligned}\n\n### Achlioptas process\n\nAt each time step of the Achlioptas process18,35, two edges $$e_1$$ and $$e_2$$ compete for addition. Suppose $$e_1$$ involves two components of size $$\|C_a\|,\|C_b\|$$; $$e_2$$ involves two components of size $$\|C_c\|,\|C_d\|$$, we consider three types of competing rules:\n\n• add $$e_1$$ if $$\|C_a\|+\|C_b\| < \|C_c\|+\|C_d\|$$, add $$e_2$$ otherwise.\n\n• product rule (PR): add $$e_1$$ if $$\|C_a\|\|C_b\| < \|C_c\|\|C_d\|$$, add $$e_2$$ otherwise.\n\n• Bohman Frieze (BF) rule: add $$e_1$$ if $$\|C_a\|=\|C_b\| =1$$, add $$e_2$$ otherwise.\n\nOur experiment treats the number of edges (added to the graph) as the parameter.\n\n## Conclusions\n\nA number theory approach to random graph is proposed. A set of n random numbers generates an n by n adjacency matrix whose binary elements follow the probability (1). Thus, a hierarchical structure is implemented through ultrametrics40,58. The simplicity of the digit form (2) for hierarchical structures of random graph facilitates the enumeration of different setting of clusters and hierarchies. In contrast to mapping complex structures or rules from real world to random graph, our PARG approach explores the complex structures in numbers2,59 which might be rich enough for modeling complicated observations. An alternative point of view suggests that a plain graph can unfold a hidden hierarchical structure, based on an indispensable definition of absolute value.\n\nThe proposed PARG model is more abstract than multilayer networks, multiplex networks, and social-geographical models, but more concrete than ER-like models without community structures. Therefore, a future framework of research would consist of two interconnected levels: (1) Searching for hyper parameters, such as hierarchical structures, in PARG, given empirical data and (2) Constructing the ad hoc realistic models, for example, the biomolecular environments in cell or the social-geographical structures in a city, which the PARG model is projected onto." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8468567,"math_prob":0.99581814,"size":36583,"snap":"2023-14-2023-23","text_gpt3_token_len":9258,"char_repetition_ratio":0.14027174,"word_repetition_ratio":0.014330332,"special_character_ratio":0.26351038,"punctuation_ratio":0.18718523,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99887156,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-26T01:02:02Z\",\"WARC-Record-ID\":\"<urn:uuid:27dc6248-0ed1-4cd7-8120-535eb5b6397b>\",\"Content-Length\":\"368248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:20cbaee3-d3a6-43c8-b91e-d4903f37e8c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d4763cb-164f-4dbe-b548-bbec71e3305a>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://www.nature.com/articles/s41598-020-79507-4?error=cookies_not_supported&code=ca69ccae-cfb8-4e43-8d7f-7c0a8199f621\",\"WARC-Payload-Digest\":\"sha1:VMQXC4ATFPQA7DO6VI4SZ34LOZDJE4MO\",\"WARC-Block-Digest\":\"sha1:V5QACJ7WWV2UGKJ6NSNAQNYKV6GYQSNM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945376.29_warc_CC-MAIN-20230325222822-20230326012822-00235.warc.gz\"}"}
https://docs.monai.io/en/stable/_modules/monai/apps/pathology/transforms/spatial/dictionary.html	[ "# Source code for monai.apps.pathology.transforms.spatial.dictionary\n\n```# Copyright (c) MONAI Consortium\n# you may not use this file except in compliance with the License.\n# You may obtain a copy of the License at\n# Unless required by applicable law or agreed to in writing, software\n# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.\n# See the License for the specific language governing permissions and\n\nimport copy\nfrom typing import Any, Dict, Hashable, List, Mapping, Optional, Tuple, Union\n\nfrom monai.config import KeysCollection\nfrom monai.config.type_definitions import NdarrayOrTensor\nfrom monai.transforms.transform import MapTransform, Randomizable\nfrom monai.utils import deprecated\n\nfrom .array import SplitOnGrid, TileOnGrid\n\n__all__ = [\"SplitOnGridd\", \"SplitOnGridD\", \"SplitOnGridDict\", \"TileOnGridd\", \"TileOnGridD\", \"TileOnGridDict\"]\n\nclass SplitOnGridd(MapTransform):\n\"\"\"\nSplit the image into patches based on the provided grid shape.\nThis transform works only with torch.Tensor inputs.\n\nArgs:\ngrid_size: a tuple or an integer define the shape of the grid upon which to extract patches.\nIf it's an integer, the value will be repeated for each dimension. Default is 2x2\npatch_size: a tuple or an integer that defines the output patch sizes.\nIf it's an integer, the value will be repeated for each dimension.\nThe default is (0, 0), where the patch size will be inferred from the grid shape.\n\nNote: the shape of the input image is inferred based on the first image used.\n\"\"\"\n\nbackend = SplitOnGrid.backend\n\ndef __init__(\nself,\nkeys: KeysCollection,\ngrid_size: Union[int, Tuple[int, int]] = (2, 2),\npatch_size: Optional[Union[int, Tuple[int, int]]] = None,\nallow_missing_keys: bool = False,\n):\nsuper().__init__(keys, allow_missing_keys)\nself.splitter = SplitOnGrid(grid_size=grid_size, patch_size=patch_size)\n\ndef __call__(self, data: Mapping[Hashable, NdarrayOrTensor]) -> Dict[Hashable, NdarrayOrTensor]:\nd = dict(data)\nfor key in self.key_iterator(d):\nd[key] = self.splitter(d[key])\nreturn d\n\n[docs]@deprecated(since=\"0.8\", msg_suffix=\"use `monai.transforms.GridPatchd` or `monai.transforms.RandGridPatchd` instead.\")\nclass TileOnGridd(Randomizable, MapTransform):\n\"\"\"\nTile the 2D image into patches on a grid and maintain a subset of it.\nThis transform works only with np.ndarray inputs for 2D images.\n\nArgs:\ntile_count: number of tiles to extract, if None extracts all non-background tiles\nDefaults to ``None``.\ntile_size: size of the square tile\nDefaults to ``256``.\nstep: step size\nDefaults to ``None`` (same as tile_size)\nrandom_offset: Randomize position of the grid, instead of starting from the top-left corner\nDefaults to ``False``.\nDefaults to ``False``.\nbackground_val: the background constant (e.g. 255 for white background)\nDefaults to ``255``.\nfilter_mode: mode must be in [\"min\", \"max\", \"random\"]. If total number of tiles is more than tile_size,\nthen sort by intensity sum, and take the smallest (for min), largest (for max) or random (for random) subset\nDefaults to ``min`` (which assumes background is high value)\n\n\"\"\"\n\nbackend = SplitOnGrid.backend\n\ndef __init__(\nself,\nkeys: KeysCollection,\ntile_count: Optional[int] = None,\ntile_size: int = 256,\nstep: Optional[int] = None,\nrandom_offset: bool = False,\nbackground_val: int = 255,\nfilter_mode: str = \"min\",\nallow_missing_keys: bool = False,\nreturn_list_of_dicts: bool = False,\n):\nsuper().__init__(keys, allow_missing_keys)\n\nself.return_list_of_dicts = return_list_of_dicts\nself.seed = None\n\nself.splitter = TileOnGrid(\ntile_count=tile_count,\ntile_size=tile_size,\nstep=step,\nrandom_offset=random_offset,\nbackground_val=background_val,\nfilter_mode=filter_mode,\n)\n\n[docs] def randomize(self, data: Any = None) -> None:\nself.seed = self.R.randint(10000) # type: ignore\n\ndef __call__(\nself, data: Mapping[Hashable, NdarrayOrTensor]\n) -> Union[Dict[Hashable, NdarrayOrTensor], List[Dict[Hashable, NdarrayOrTensor]]]:\n\nself.randomize()\n\nd = dict(data)\nfor key in self.key_iterator(d):\nself.splitter.set_random_state(seed=self.seed) # same random seed for all keys\nd[key] = self.splitter(d[key])\n\nif self.return_list_of_dicts:\nd_list = []\nfor i in range(len(d[self.keys])):\nd_list.append({k: d[k][i] if k in self.keys else copy.deepcopy(d[k]) for k in d.keys()})\nd = d_list # type: ignore\n\nreturn d\n\nSplitOnGridDict = SplitOnGridD = SplitOnGridd\nTileOnGridDict = TileOnGridD = TileOnGridd\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5926617,"math_prob":0.8150213,"size":4703,"snap":"2022-27-2022-33","text_gpt3_token_len":1205,"char_repetition_ratio":0.10683124,"word_repetition_ratio":0.06644518,"special_character_ratio":0.25770783,"punctuation_ratio":0.21933962,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95832705,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T05:17:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e84674d1-b00a-4aa4-a86b-7555576e20c9>\",\"Content-Length\":\"29905\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d5495e8-601b-4264-8e11-cf7290be1873>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4ea828c-cc8a-4ba9-8d34-82e0282d957b>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://docs.monai.io/en/stable/_modules/monai/apps/pathology/transforms/spatial/dictionary.html\",\"WARC-Payload-Digest\":\"sha1:2LJ53EQ4DXQ7UTTIGZ76YLXC7IJT22B5\",\"WARC-Block-Digest\":\"sha1:QNNXGGXMXLRVL34IWMVU3ODOXDN34XZI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103920118.49_warc_CC-MAIN-20220701034437-20220701064437-00405.warc.gz\"}"}
https://www.hackmath.net/en/math-problem/2096	[ "Tank 9\n\nThe tank with volume V is filled with one pump for three hours and by second pump for 5 hours. When both pumps will run simultaneously calculate:\n\na) how much of the total volume of the tank is filled in one hour\nb) for how long is the tank full\n\nResult\n\na = 0.53\nb = 1.875 h\n\nSolution:\n\n1*(1/3+1/5) = a\nb(1/3+1/5)=1\n\n15a = 8\n8b = 15\n\na = 815 ≈ 0.533333\nb = 158 = 1.875\n\nCalculated by our linear equations calculator.\n\nLeave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):", null, "Be the first to comment!", null, "To solve this verbal math problem are needed these knowledge from mathematics:\n\nDo you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?\n\nNext similar math problems:\n\n1. Viju", null, "viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju??\n2. Jane plants", null, "Jane plants flowers in the garden. If she planted 12 every hour instead of 9 flowers, she would finish with the job an hour earlier. How many flowers does she plant?\n3. Hectoliters of water", null, "The pool has a total of 126 hectoliters of water. The first pump draws 2.1 liters of water per second. A second pump pumps 3.5 liters of water per second. How long will it take both pumps to drain four-fifths of the water at the same time?\n4. Three figures - numbers", null, "The sum of three numbers, if each is 10% larger than the previous one, is 662. Determine the figures.\n5. Family parcels", null, "In father will he divided the land so that the older son had three bigger part than younger son. Later elder son gave 2.5 ha field to younger and they had both the same. Determine the area of family parcel.\n6. Three brothers", null, "The three brothers have a total of 42 years. Jan is five years younger than Peter and Peter is 2 years younger than Michael. How many years has each of them?\n7. Two numbers", null, "We have two numbers. Their sum is 140. One-fifth of the first number is equal to half the second number. Determine those unknown numbers.\n8. The larger", null, "The larger of two numbers is nine more than four times the smaller number. The sum of the two numbers is fifty-nine. Find the two numbers.\n9. Trees", null, "Along the road were planted 250 trees of two types. Cherry for 60 CZK apiece and apple 50 CZK apiece. The entire plantation cost 12,800 CZK. How many was cherries and apples?\n10. Six years", null, "In six years Jan will be twice as old as he was six years ago. How old is he?\n11. ATC camp", null, "The owner of the campsite offers 79 places in 22 cabins. How many of them are triple and quadruple?\n12. Theatro", null, "Theatrical performance was attended by 480 spectators. Women were in the audience 40 more than men and children 60 less than half of adult spectators. How many men, women and children attended a theater performance?\n13. Equations - simple", null, "Solve system of linear equations: x-2y=6 3x+2y=4\n14. Substitution", null, "solve equations by substitution: x+y= 11 y=5x-25\n15. Equations", null, "Solve following system of equations: 6(x+7)+4(y-5)=12 2(x+y)-3(-2x+4y)=-44\n16. The dormitory", null, "The dormitory accommodates 150 pupils in 42 rooms, some of which are triple and some are quadruple. Determine how many rooms are triple and how many quadruples.\n17. Blackberries", null, "Daniel, Jolana and Stano collected together 34 blackberries. Daniel collected 8 blackberries more than Jolana, Jolana 4 more than Stano. Determine the number blackberries each collected ." ]	[ null, "https://www.hackmath.net/hashover/images/first-comment.png", null, "https://www.hackmath.net/hashover/images/avatar.png", null, "https://www.hackmath.net/thumb/34/t_4134.jpg", null, "https://www.hackmath.net/thumb/59/t_7959.jpg", null, "https://www.hackmath.net/thumb/50/t_7850.jpg", null, "https://www.hackmath.net/thumb/68/t_2968.jpg", null, "https://www.hackmath.net/thumb/69/t_1869.jpg", null, "https://www.hackmath.net/thumb/90/t_6990.jpg", null, "https://www.hackmath.net/thumb/65/t_2665.jpg", null, "https://www.hackmath.net/thumb/79/t_6779.jpg", null, "https://www.hackmath.net/thumb/28/t_4728.jpg", null, "https://www.hackmath.net/thumb/95/t_4995.jpg", null, "https://www.hackmath.net/thumb/44/t_3644.jpg", null, "https://www.hackmath.net/thumb/67/t_1667.jpg", null, "https://www.hackmath.net/thumb/81/t_6881.jpg", null, "https://www.hackmath.net/thumb/18/t_5518.jpg", null, "https://www.hackmath.net/thumb/47/t_1447.jpg", null, "https://www.hackmath.net/thumb/78/t_7978.jpg", null, "https://www.hackmath.net/thumb/10/t_3710.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9478936,"math_prob":0.98651314,"size":3210,"snap":"2019-43-2019-47","text_gpt3_token_len":813,"char_repetition_ratio":0.11072988,"word_repetition_ratio":0.072164945,"special_character_ratio":0.24672897,"punctuation_ratio":0.102874435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9909783,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T09:37:45Z\",\"WARC-Record-ID\":\"<urn:uuid:66f49ec5-1b6c-4919-9dcb-3c652d9b3a80>\",\"Content-Length\":\"22028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d72fbeb4-7e96-4cc2-bf91-6348d61ed426>\",\"WARC-Concurrent-To\":\"<urn:uuid:922c8314-e7a8-48eb-bb33-c92eec2e4526>\",\"WARC-IP-Address\":\"104.24.104.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/math-problem/2096\",\"WARC-Payload-Digest\":\"sha1:EOBIHM2IFVU7VADPXH5BJDIZMQVCZOVA\",\"WARC-Block-Digest\":\"sha1:WGUV4W4NRR2JCB2B543OOABV6DUSYKP5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986666959.47_warc_CC-MAIN-20191016090425-20191016113925-00538.warc.gz\"}"}
https://geodorable.com/tag/trigonometric/	[ "## Trigonometric Functions Worksheet\n\nTrigonometric Functions Worksheet. Plus each one comes with an answer key. A really great activity for allowing students to understand the concepts of the solving trigonometric.", null, "14 Best Images of Basic Trigonometry Worksheet Trig Equations Worksheet from byveera.blogspot.com\n\nTrigonometric functions interactive worksheet on trigonometric functions for 11th standard id: Trigonometric functions the cosine and sine functions can be de ned geometrically by the co. Ambiguous case of the law of sines.\n\n## Trigonometric Ratios Worksheet\n\nTrigonometric Ratios Worksheet. In this guide, we will cover: For instance, you will be asked to provide the overall area of the triangle for yourself.\n\nFinding height using trigonometric ratios. Only three trigonometric ratios are sine cosine tangent. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you.\n\n## Trigonometric Ratios Worksheet 2 Answers\n\nTrigonometric Ratios Worksheet 2 Answers. Equations using reciprocal trig functions doc maths higher decimal places math worksheets. Pin on geometry resources what is sin 1.\n\nSo, if you wish to receive all. A = =(4)(2) sin 45 a= (2) (3) sin. 1) 2.7 m 2) 34.7° 3) 5.3 m 4) 8.7 m 5) 32.2°.\n\n## Trigonometric Functions Worksheet With Answers\n\nTrigonometric Functions Worksheet With Answers. 19 rows printable trigonometry worksheets with answers. Plus each one comes with an answer key.\n\nRight triangle trigonometry she loves math trigonometry right triangle love. At cazoom maths we provide a. After it finishes the last column of the present row, checking continues with the first column of the following row.\n\n## Trigonometric Functions On The Unit Circle Worksheet Answers\n\nTrigonometric Functions On The Unit Circle Worksheet Answers. The unit circle the two historical perspectives of trigonometry incorporate different methods for introducing the trigonometric functions. July corresponds to t = 7." ]	[ null, "https://i2.wp.com/www.worksheeto.com/postpic/2013/02/right-triangle-trig-ratios-worksheet_210476.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8249503,"math_prob":0.9739279,"size":2144,"snap":"2022-27-2022-33","text_gpt3_token_len":500,"char_repetition_ratio":0.19953272,"word_repetition_ratio":0.06128134,"special_character_ratio":0.21921642,"punctuation_ratio":0.13399504,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955313,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T07:50:44Z\",\"WARC-Record-ID\":\"<urn:uuid:0345cbd0-3bf6-4487-aef3-d08e76d806e3>\",\"Content-Length\":\"61267\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba92894e-19f6-4cce-8313-999800db2f29>\",\"WARC-Concurrent-To\":\"<urn:uuid:8170ed60-695a-4802-9200-6380f765a9e3>\",\"WARC-IP-Address\":\"172.67.147.40\",\"WARC-Target-URI\":\"https://geodorable.com/tag/trigonometric/\",\"WARC-Payload-Digest\":\"sha1:ZTUDD4JXD5EYTMP4PQJMNCEOKQ2UOAED\",\"WARC-Block-Digest\":\"sha1:E2Y2G3MGGUZUDASKM2YTPFSMXP6QHUUP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103034877.9_warc_CC-MAIN-20220625065404-20220625095404-00572.warc.gz\"}"}
https://www.codurance.com/publications/2019/01/14/active-pattern	[ "# Active Pattern\n\n14 Jan 2019", null, "## Jorge Gueorguiev Garcia\n\nSee author's bio and posts\n\nLast week I was pointed by someone to Active Patterns in F#. And it has been quite an interesting discovery.\n\nActive Patterns are used on F# to partition data. Those partitions then can be used with pattern matching. Microsoft's webpage compares them to discriminated unions.\n\nHow they can be used? Well, you could look at the above link, or just follow this post (or, in fact, do both).\n\nBelow is an initial version of a trivial function using pattern matching.\n\n``````let thisNumberTrait(number) =\nmatch number with\n\| x when x = 0 -> \"Is Zero!!\"\n\| x when x > 0 -> \"Is Positive!!\"\n\| x when x < 0 -> \"Is Negative!!\"\n\| _ -> \"I shouldn't be here\"\n``````\n\nA first improvement of the above code would be to put the logic on the guards behind functions, so we can arrive to this:\n\n``````let isZero(number) = number = 0\nlet isPositive(number) = number > 0\nlet isNegative(number) = number < 0\n\nlet thisNumberTrait(number) =\nmatch number with\n\| x when isZero(x) -> \"Is Zero!!\"\n\| x when isPositive(x) -> \"Is Positive!!\"\n\| x when isNegative(x) -> \"Is Negative!!\"\n\| _ -> \"I shouldn't be here\"\n``````\n\nBut what if we could remove the guard clauses completely? That is where Active Patterns come into play.\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(number) else None\nlet (\|Positive\|_\|) (number) = if number > 0 then Some(number) else None\nlet (\|Negative\|_\|) (number) = if number < 0 then Some(number) else None\n\nlet thisNumberTrait(number) =\nmatch number with\n\| Zero(x) -> \"Is Zero!!\"\n\| Positive(x) -> \"Is Positive!!\"\n\| Negative(x) -> \"Is Negative!!\"\n\| _ -> \"I shouldn't be here\"\n``````\n\nAs you can see there are a few differences on the code. If we concentrate on the logic functions, we can see that now the name of the function has been changed for the construct `(\| \| \|)`, and that now is it returning an Option type. Then, the pattern matching has replaced the `x when ...` code with the new Pattern `Zero(x)`. We complicate a bit the functions, we simplify the pattern matching.\n\nIf we look at what is the answer that we are providing, we realize that we are not using the number at all, so in fact we could change the line\n\n``````\| Zero(x) -> \"Is Zero!!\"\n``````\n\nfor\n\n``````\| Zero(_) -> \"Is Zero!!\"\n``````\n\nThe `x` is not the data that we pass to the Pattern, is the data that is returned! We pass the data implicilty. But wait, if I don't use the return data, do I need to return it at all? And the answer is no. So now we replace\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(number) else None\n``````\n\nwith\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(Zero) else None\n``````\n\nWe return the name of the pattern, instead of the data.\n\nWhich now means that on the match we can do the change from\n\n``````\| Zero(_) -> \"Is Zero!!\"\n``````\n\nto\n\n``````\| Zero -> \"Is Zero!!\"\n``````\n\nSo far we have returned the same data that we have passed, and not returned anything. But what if we want to return data of a different type? That is possible: Look below that `(\|Positive\|_\|)` returns a string now.\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(Zero) else None\nlet (\|Positive\|_\|) (number) = if number > 0 then Some(\"Positive\") else None\nlet (\|Negative\|_\|) (number) = if number < 0 then Some(number) else None\n\nlet thisNumberTrait(number) =\nmatch number with\n\| Zero -> \"Is Zero!!\"\n\| Positive(x) -> sprintf \"Is %s!!\" x\n\| Negative(_) -> \"Is Negative!!\"\n\| _ -> \"I shouldn't be here\"\n``````\n\nSo far I have been using Partial Patterns. That is, the data could not be defined as a partition. That is why there is an underscore (`(\|Zero\|_\|)`) and why we are returning an Option (Some\|None). But we could have a full Active Pattern, where the data must be inside one of the partitions. The changes are easy just as below)\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(Zero) else None\n``````\n\nto\n\n``````let (\|Zero\|NonZero\|) (number) = if number = 0 then Zero else NonZero\n``````\n\nPatterns can be combined using `&` for the and combination, and `\|` for the or combination.\n\n``````\| Positive(x) -> sprintf \"Is %s!!\" x\n``````\n\nwe could write\n\n``````\| NonZero & Positive(x) -> sprintg \"Is %s!!\" x\n``````\n\nNot that in this case makes any difference, but is a posibility.\n\nFinally, we could make single big pattern, replacing\n\n``````let (\|Zero\|_\|) (number) = if number = 0 then Some(Zero) else None let (\|Positive\|_\|) (number) = if number > 0 then Some(Positive) else None\nlet (\|Negative\|_\|) (number) = if number < 0 then Some(Negative) else None ``````\n\nwith\n\n``````let (\|Zero\|Positive\|Negative\|) (number) = if number = 0 then Zero elif number > 0 then Positive else Negative\n``````\n\nWhich in this case looks a bit pointless to me, as I'm nearly back to the original code. It will have it's uses, though.\n\nActive Patterns are an intersting construct, especially when using the Active Pattern multiple times." ]	[ null, "https://www.codurance.com/hubfs/Codurance_September2020/images/jorge_gueorguiev.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.849496,"math_prob":0.9925396,"size":4586,"snap":"2023-40-2023-50","text_gpt3_token_len":1247,"char_repetition_ratio":0.18245308,"word_repetition_ratio":0.2175981,"special_character_ratio":0.29241168,"punctuation_ratio":0.10789766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989717,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T10:29:48Z\",\"WARC-Record-ID\":\"<urn:uuid:7cc15d7d-2540-4539-bc3c-2b200e876783>\",\"Content-Length\":\"86313\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a0ffc51-d90b-4010-9790-ec5e8ce5d0ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:edd5fc15-a8b0-40df-9e55-4694b0b67435>\",\"WARC-IP-Address\":\"199.60.103.31\",\"WARC-Target-URI\":\"https://www.codurance.com/publications/2019/01/14/active-pattern\",\"WARC-Payload-Digest\":\"sha1:PTERLVWWORH4YYRGFT6XCDIM6YSNF7D3\",\"WARC-Block-Digest\":\"sha1:FAWW675JQA4RCBSCUVO7ZBLVW5ESN47C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511364.23_warc_CC-MAIN-20231004084230-20231004114230-00803.warc.gz\"}"}
https://madhavamathcompetition.com/2020/04/24/iii-tutorial-problems-symmetric-and-alternating-functions-rmo-and-iitjee-math/?shared=email&msg=fail	[ "III. Tutorial problems. Symmetric and alternating functions. RMO and IITJEE math\n\n1. Simplify:", null, "$(b^{-1}+c^{1})(b+c-a)+(c^{-1}+a^{-1})(c+a-b)+(a^{-1}+b^{-1})(a+b=c)$\n2. Simplify:", null, "$\\frac{(x-b)(x-c)}{(a-b)(a-c)} + \\frac{(x-c)(x-a)}{(b-c)(b-a)} + \\frac{(x-a)(x-b)}{(c-a)(c-b)}$\n3. Simplify:", null, "$\\frac{b^{2}+c^{2}-a^{2}}{(a-b)(a-c)} + \\frac{c^{2}+a^{2}-b^{2}}{(b-c)(b-a)} + \\frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}$\n4. Simplify:", null, "$\\frac{b-c}{1+bc} + \\frac{c-a}{1+ca} + \\frac{a-b}{1+ab}$\n5. Simplify:", null, "$\\frac{a(b-c)}{1+bc} + \\frac{b(c-a)}{1+ca} + \\frac{c(a-b)}{1+ab}$\n6. Factorize:", null, "$(b-c)^{2}(b+c-2a)+(c-a)^{2}(c+a-2b)+(a-b)^{2}(a+b-2c)$. Put", null, "$b-c=x, c-a=y, a-b=z$and", null, "$b+c-2a=y-z$\n7. Factorize:", null, "$8(a+b+c)^{2}-(b+c)^{2}-(c+a)^{2}-(a+b)^{2}$. Put", null, "$b+c=x, c+a=y, a+b=z$.\n8. Factorize:", null, "$(a+b+c)^{2}-(b+c-a)^{2}-(c+a-b)^{2}+(a+b-c)^{2}$\n9. Factorize:", null, "$(1-a^{2})(1-b^{2})(1-c^{2})+(a-bc)(b-ac)(c-ab)$\n10. Express the following substitutions as the product of transpositions: (i)", null, "$\\left(\\begin{array}{cccccc}123456\\\\654321\\end{array}\\right)$ (ii)", null, "$\\left(\\begin{array}{cccccc}123456\\\\246135\\end{array}\\right)$ (iii)", null, "$\\left(\\begin{array}{cccccc}123456\\\\641235\\end{array}\\right)$\n\nRegards,\n\nNalin Pithwa.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6995468,"math_prob":1.0000085,"size":493,"snap":"2022-05-2022-21","text_gpt3_token_len":121,"char_repetition_ratio":0.16768916,"word_repetition_ratio":0.0,"special_character_ratio":0.18661258,"punctuation_ratio":0.2804878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000005,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-29T01:42:21Z\",\"WARC-Record-ID\":\"<urn:uuid:64696431-bebd-4896-8e90-9c371c366a31>\",\"Content-Length\":\"86512\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80f2d10f-932b-449a-b7cd-40ff9ee2e836>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa7b6c4a-3f92-49bb-9bca-62f03f09c816>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://madhavamathcompetition.com/2020/04/24/iii-tutorial-problems-symmetric-and-alternating-functions-rmo-and-iitjee-math/?shared=email&msg=fail\",\"WARC-Payload-Digest\":\"sha1:WZ2OZIV2IK4XWHB3WPCFIOTH4OKIOJUH\",\"WARC-Block-Digest\":\"sha1:AEDHAZZ2XOW5HRZOQY2AD6HLN75JM5BT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320299894.32_warc_CC-MAIN-20220129002459-20220129032459-00538.warc.gz\"}"}
https://votetinadavis.com/valentine-math-worksheets.html	[ "# Valentine Math Worksheets", null, "### Free Valentine S Day Math Worksheet Made By Teachers Valentine Math Worksheet Math Valentines Valentine Worksheets", null, "### Pin By Super Teacher Worksheets On Holidays Super Teacher Worksheets Math Valentines Valentines School Math Coloring", null, "### Valentine S Day Math Activities Kindergarten Distance Learning Kindergarten Valentines Kindergarten Math Math Valentines", null, "### Math Printables For Valentine S Day Freebie Second Grade Common Core Math Valentines Math Printables Second Grade Math", null, "### Happy Valentine S Day This Is A Free Valentine S Day Addition Worksheet Just Download And Pr Valentine Worksheets Valentines School Valentines Day Activities", null, "### Valentine S Day Math And Literacy Centers With Printable Worksheets And A Math Worksheet Freebie Kindergarten Smarts Kindergarten Valentines Math Valentines Valentine Math Worksheet", null, "### These Valentines Day worksheets for addition in printable PDF format will sweeten up your math practice for February.\n\nValentine math worksheets. 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https://codereview.stackexchange.com/questions/196034/slow-flask-sqlalchemy-query-using-association-tables	[ "# Slow Flask-SQLAlchemy query using association tables\n\nI have two models in Flask-SQLAlchemy (Post and Comment) that have many-to-many relationship that is manifested in the third model (post_mentions):\n\npost_mentions = db.Table(\n'post_mentions',\ndb.Column('post_id', db.Integer, db.ForeignKey('posts.id'), primary_key=True),\n)\n\nclass Post(db.Model):\n__tablename__ = 'posts'\n\nid = db.Column(db.Integer, primary_key=True)\nname = db.Column(db.String, unique=True, nullable=False)\nmentions = db.relationship('Comment', secondary=post_mentions, lazy='dynamic')\n\ndef __eq__(self, other):\nreturn self.name.lower() == other.name.lower()\n\ndef __hash__(self):\nreturn hash(self.name.lower())\n\nclass Comment(db.Model):\n\nid = db.Column(db.Integer, primary_key=True)\ntext = db.Column(db.Text, nullable=False)\ncreated_at = db.Column(db.Integer, nullable=False)\n\n\nThere is also a /posts endpoint that triggers the following query:\n\n# flask and other imports\n\[email protected]('/posts')\ndef posts():\npage_num = request.args.get('page', 1)\nposts = models.Post.query.join(models.post_mentions)\\\n.group_by(models.post_mentions.columns.post_id)\\\n.order_by(func.count(models.post_mentions.columns.post_id).desc())\\\n.paginate(page=int(page_num), per_page=25)\nreturn render_template('posts.html', posts=posts)\n\n\nThere are more than 14k+ posts and 32k+ comments stored in SQLite database. As you can see from the snippet above, when someone hits /posts endpoint, SQLAlchemy loads all data at once to the memory and then subsequent queries (e.g. retrieving posts, comments to that posts, etc..) take sub-millisecond time, since data is being served from the memory without hitting the database. Initial load takes 10s+ on my laptop, which is, to put it mildly, suboptimal.\n\nSo the question is: Considering that users won't view 97+% of posts, how can I both order posts by number of mentions in comments and load them on demand instead of doing it in one swoop?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72887576,"math_prob":0.76346767,"size":1973,"snap":"2023-40-2023-50","text_gpt3_token_len":448,"char_repetition_ratio":0.13103098,"word_repetition_ratio":0.0,"special_character_ratio":0.25392804,"punctuation_ratio":0.2372449,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98399526,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T10:20:00Z\",\"WARC-Record-ID\":\"<urn:uuid:739ff544-ad6b-4530-8458-12c6dae51399>\",\"Content-Length\":\"153114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b59850a2-10ab-4c11-bec5-0decc440a214>\",\"WARC-Concurrent-To\":\"<urn:uuid:734c96c9-083b-4435-b8a3-925c44824510>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/196034/slow-flask-sqlalchemy-query-using-association-tables\",\"WARC-Payload-Digest\":\"sha1:Y4NLUSX4DLHKR4YBLHCY3YIRH2LN5BQA\",\"WARC-Block-Digest\":\"sha1:SN754L2RWXDUE4TSPCPAIVYPHBOZPIMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233508959.20_warc_CC-MAIN-20230925083430-20230925113430-00155.warc.gz\"}"}
https://edu-answer.com/mathematics/question515178130	[ "", null, ", 15.01.2022 14:20 amandasantiago2001\n\nWhat are the features of the function f(x) = -2 (1/2)^x+ 3 graphed below?", null, "", null, "", null, "Another question on Mathematics", null, "Mathematics, 21.06.2019 14:50\nSimplify 5 square root of 7 end root plus 12 square root of 6 end root minus 10 square root of 7 end root minus 5 square root of 6 . (1 point) 5 square root of 14 end root minus 7 square root of 12 5 square root of 7 end root minus 7 square root of 6 7 square root of 12 end root minus 5 square root of 14 7 square root of 6 end root minus 5 square root of 7", null, "Mathematics, 21.06.2019 15:30\nMary works for a company that ships packages and must measure the size of each box that needs to be shipped. mary measures a box and finds the length is 7 inches, the width is 14 inches, and the height is 15 inches. what is the volume of the box? [type your answer as a number.]", null, "Mathematics, 21.06.2019 19:10\nThe triangles in the diagram are congruent. if mzf = 40°, mza = 80°, and mzg = 60°, what is mzb?", null, "Mathematics, 21.06.2019 19:30\nOkay so i didn't get this problem petro bought 8 tickets to a basketball game he paid a total of \\$200 write an equation to determine whether each ticket cost \\$26 or \\$28 so i didn't get this question so yeahyou have a good day.\nWhat are the features of the function f(x) = -2 (1/2)^x+ 3 graphed below?...\nQuestions", null, "", null, "", null, "", null, "", null, "", null, "Spanish, 13.11.2020 23:20", null, "", null, "", null, "", null, "", null, "Mathematics, 13.11.2020 23:20", null, "", null, "", null, "Mathematics, 13.11.2020 23:20", null, "", null, "", null, "", null, "", null, "Chemistry, 13.11.2020 23:20", null, "Questions on the website: 14531560" ]	[ null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/User.png", null, "https://edu-answer.com/tpl/images/ask_question.png", null, "https://edu-answer.com/tpl/images/ask_question_mob.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/himiya.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/es.png", null, "https://edu-answer.com/tpl/images/cats/en.png", null, "https://edu-answer.com/tpl/images/cats/en.png", null, "https://edu-answer.com/tpl/images/cats/istoriya.png", null, "https://edu-answer.com/tpl/images/cats/istoriya.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/himiya.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/istoriya.png", null, "https://edu-answer.com/tpl/images/cats/obshestvoznanie.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null, "https://edu-answer.com/tpl/images/cats/en.png", null, "https://edu-answer.com/tpl/images/cats/himiya.png", null, "https://edu-answer.com/tpl/images/cats/mat.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82203346,"math_prob":0.9942285,"size":1498,"snap":"2022-05-2022-21","text_gpt3_token_len":511,"char_repetition_ratio":0.18674698,"word_repetition_ratio":0.2189781,"special_character_ratio":0.38317758,"punctuation_ratio":0.1750663,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947878,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T01:26:29Z\",\"WARC-Record-ID\":\"<urn:uuid:41571f07-0b53-40b1-af12-b5beaeb2990b>\",\"Content-Length\":\"71726\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a8badde-a54b-4a98-addc-fa7dd97226b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5e4379b-c96c-4cfa-9b02-d7a94ce3220b>\",\"WARC-IP-Address\":\"104.21.68.106\",\"WARC-Target-URI\":\"https://edu-answer.com/mathematics/question515178130\",\"WARC-Payload-Digest\":\"sha1:RZ2TNPKZP33V7J77IBQZFRLFBAE4EXPF\",\"WARC-Block-Digest\":\"sha1:NYZY7KIPTIFVIMNRBCB2O5GPH7KHTT6I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300658.84_warc_CC-MAIN-20220118002226-20220118032226-00431.warc.gz\"}"}
https://psychology.wikia.org/wiki/Structural_equation_modeling	[ "34,718 Pages\n\nThis banner appears on articles that are weak and whose contents should be approached with academic caution\n\n.\n\nStructural equation modeling (SEM) is a statistical technique for testing and estimating causal relationships using a combination of statistical data and qualitative causal assumptions. This view of SEM was articulated by the geneticist Sewall Wright (1921), the economists Trygve Haavelmo (1943) and Herbert Simon (1953), and formally defined by Judea Pearl (2000) using a calculus of counterfactuals.\n\nStructural Equation Models (SEM) encourages confirmatory rather than exploratory modeling; thus, it is suited to theory testing rather than theory development. It usually starts with a hypothesis, represents it as a model, operationalises the constructs of interest with a measurement instrument, and tests the model. The causal assumptions embedded in the model often have falsifiable implications which can be tested against the data. With an accepted theory or otherwise confirmed model, SEM can also be used inductively by specifying the model and using data to estimate the values of free parameters. Often the initial hypothesis requires adjustment in light of model evidence, but SEM is rarely used purely for exploration.\n\nAmong its strengths is the ability to model constructs as latent variables (variables which are not measured directly, but are estimated in the model from measured variables which are assumed to 'tap into' the latent variables). This allows the modeler to explicitly capture the unreliability of measurement in the model, which in theory allows the structural relations between latent variables to be accurately estimated. Factor analysis, path analysis and regression all represent special cases of SEM.\n\nIn SEM, the qualitative causal assumptions are represented by the missing variables in each equation, as well as vanishing covariances among some error terms. These assumptions are testable in experimental studies and must be confirmed judgmentally in observational studies.\n\nAn alternative technique for specifying Structural Equation Models using partial least squares path modeling has been implemented in software such as LVPLS (Latent Variable Partial Least Square), PLSGraph, SmartPLS (Ringle et al. 2005) and XLSTAT (Addinsoft, 2008). Some feel this is better suited to data exploration. More ambitiously, The TETRAD project aims to develop a way to automate the search for possible causal models from data.\n\n## Steps in performing SEM analysis\n\n### Model specification\n\nSince SEM is a confirmatory technique, the model must be specified correctly based on the type of analysis that the modeller is attempting to confirm. When building the correct model, the modeller uses two different kind of variables, namely exogenous and endogenous variables. The distinction between these two types of variables is whether the variable regresses on another variable or not. Like in a linear regression the dependent variable (DV) regresses on the independent variable (IV), meaning that the DV is being predicted by the IV. Within SEM modelling this means that the exogenous variable is the variable that another variable regresses on. Exogenous variables can be recognized in a graphical version of the model, as the variables sending out arrowheads, denoting which variable it is predicting. A variable that regresses on a variable is always an endogenous variable even if this same variable is used as an variable to be regressed on. Endogenous variables are recognized as the receivers of a arrowhead in the model. The fact that a variable can play a double role in a SEM model (independent as well dependent), makes that SEM is more useful than the linear regression, since instead of performing two regressions one SEM model will do. There are usually two main parts to SEM: the structural model showing potential causal dependencies between endogenous and exogenous variables, and the measurement model showing the relations between the latent variables and their indicators. Confirmatory factor analysis models, for example, contain only the measurement part; while path diagrams (to be distinct from linear regression) can be viewed as an SEM that only has the structural part. Specifying the model delineates causal (in fact counterfactual) relationships between variables that are thought to be possible (and therefore want to be 'free' to vary) and those relationships between variables that already have an estimated relationship, which can be gathered from previous studies (these relationships are 'fixed' in the model).\n\nA modeller will often specify a set of theoretically plausible models in order to assess whether the model proposed is the best of the set. Not only must the modeller account for the theoretical reasons for building the model as it is, but the modeller must also take into account the number of data points and the number of parameters that the model must estimate to identify the model. An identified model is a model where a specific parameter value uniquely identifies the model, and no other equivalent formulation can be given by a different parameter value. A data point is a variable with observed scores, like a variable containing the scores on a question or the number of times respondents buy a car. The parameter is the value of interest, which might be a regression coefficient between the exogenous and the endogenous variable or the factor loading (regression coefficient between a indicator and its factor). If the number of data points is smaller than the number of estimated parameters an unidentified model is the result, since there are too few reference points to account for all the variance in the model. The solution is to constrain one of the paths to zero, which means that it is no longer part of the model.\n\n### Estimation of free parameters\n\nParameter estimation is done comparing the actual covariance matrices representing the relationships between variables and the estimated covariance matrices of the best fitting model. This is obtained through numerical maximization of a fit criterion as provided by maximum likelihood, weighted least squares or asymptotically distribution-free methods.\n\nThis is often accomplished by using a specialized SEM analysis program, such as SPSS' AMOS, EQS, LISREL, Mplus, Mx, the sem package in R, or SAS PROC CALIS (more information on SAS PROC CALIS: see UCLA or UCR).\n\n### Assessment of fit\n\nUsing a SEM analysis program, one can compare the estimated matrices representing the relationships between variables in the model to the actual matrices. Formal statistical tests and fit indices have been developed for these purposes. Individual parameters of the model can also be examined within the estimated model in order to see how well the proposed model fits the driving theory. Most, though not all, estimation methods make such tests of the model possible.\n\nHowever, the model tests are only correct provided that the model is correct. Although this problem exists in all statistical hypothesis tests, its existence in SEM has led to a large body of discussion among SEM experts, leading to a large variety of different recommendations on the precise application of the various fit indices and hypothesis tests.\n\nFor each measure of fit, rules of thumb have evolved regarding what represents good fit between model and data. These rules of thumb often need to be updated based on contextual factors such as the sample size, the ratio of indicators to factors, and the overall size of the model. Measures of fit differ in several ways. Some of them reward more parsimonious models (i.e., those with more constrained parameters). Because different measures of fit capture different elements of the fit of the model, it is appropriate to report a selection of different fit measures.\n\nSome of the more commonly used measures of fit include:\n\n### Model modification\n\nThe model may need to be modified in order to improve the fit, thereby estimating the most likely relationships between variables. Many programs provide modification indices which report the improvement in fit that results from adding an additional path to the model. Modifications that improve model fit are then flagged as potential changes that can be made to the model. In addition to improvements in model fit, it is important that the modifications also make theoretical sense.\n\n### Interpretation and communication\n\nThe model is then interpreted and claims about the constructs are made based on the best fitting model.\n\nCaution should always be taken when making claims of causality even when experimentation or time-ordered studies have been done. The term causal model must be understood to mean: \"a model that conveys causal assumptions,\" not necessarily a model that produces validated causal conclusions. Collecting data at multiple time points and using an experimental or quasi-experimental design can help rule out certain rival hypotheses but even a randomized experiment cannot rule out all such threats to causal inference. Good fit by a model consistent with one causal hypothesis does not rule out equally good fit by another model consistent with a different causal hypothesis. However careful research design can help distinguish such rival hypotheses.\n\n### Replication and revalidation\n\nAll model modifications should be replicated and revalidated before interpreting and communicating the results.\n\n## Comparison to other methods\n\nIn machine learning, SEM may be viewed as a generalization of Linear-Gaussian Bayesian networks which drops the acyclicality constraint, i.e. which allows causal cycles." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85778433,"math_prob":0.9563857,"size":14205,"snap":"2021-31-2021-39","text_gpt3_token_len":2950,"char_repetition_ratio":0.1382297,"word_repetition_ratio":0.0036849377,"special_character_ratio":0.20478705,"punctuation_ratio":0.112769485,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98509204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T23:48:29Z\",\"WARC-Record-ID\":\"<urn:uuid:daae6fbb-54f3-4af2-8027-791964d2200c>\",\"Content-Length\":\"124519\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c8e278e-9038-48b9-9b79-51f6134914c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:c278fba7-16d6-4ebf-9306-6d07830e7c1d>\",\"WARC-IP-Address\":\"151.101.128.194\",\"WARC-Target-URI\":\"https://psychology.wikia.org/wiki/Structural_equation_modeling\",\"WARC-Payload-Digest\":\"sha1:6TTWPBIM3LOQ3QKCIY4OYVGB6ET5MSV7\",\"WARC-Block-Digest\":\"sha1:TATOK2G5ZNXPNSW4R5L3CTGTM3QKD6AL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154277.15_warc_CC-MAIN-20210801221329-20210802011329-00012.warc.gz\"}"}
https://www.studentsguide360.com/12th-chemistry-important-questions-unit-7-chemical-kinetics/	[ "", null, "# 12th Chemistry Important Questions Unit 7 Chemical Kinetics\n\nTN 12th Chemistry Important Questions Unit 7 Chemical Kinetics. 12th Chemistry Important Questions Chapter 7 Transition And Inner Transition Elements +2 Chemistry Important 2 Marks, 12th Chemistry Important 3 Marks, Chemistry Important 5 Marks Questions Based on reduced Syllabus 2021-2022. 12th Free Online Test (MCQs). HSC 12th Chemistry Revision Test Important Questions. 12th Tamil Full Guide.", null, "## Unit 7 Chemical Kinetics\n\n### 12th Chemistry All Units Important Questions\n\n• 1. Define average rate and instantaneous rate.\n• 2. Define rate law and rate constant.\n• 3. Derive integrated rate law for a zero-order reaction A –> product.\n• 4. Define the half-life of a reaction. Show that for a first-order reaction half-life is independent of initial concentration.\n• 5. What is an elementary reaction? Give the differences between the order and molecularity of a reaction.\n• 6. Write the rate law for the following reactions.\n(a) A reaction that is 3/2 order in x and zero-order in y.\n(b) A reaction that is second order in NO and first order in Br2.\n• 7. The rate law for a reaction of A, B, and C has been found to be\nrate = K[A]2[B][L]3/2. How would the rate of reaction change when\n(i) Concentration of [L] is quadrupled\n(ii) Concentration of both [A] and [B] are doubled (iii) Concentration of [A] is halved, (iv) Concentration of [A] is reduced to 1/3 and concentration of [L] is quadrupled.\n• 8. The rate of formation of a dimer in a second order reaction is 7.5×10-3molL-1s-1 at 0.05 mol L-1 monomer concentration. Calculate the rate constant.\n• 9. Write the Arrhenius equation and explain the terms involved.\n• 10. The decomposition of Cl2O7 at 500K in the gas phase to Cl2 and O2 is a first order reaction. After\n1 minute at 500K, the pressure of Cl2O2 falls from 0.08 to 0.04 atm. Calculate the rate constant in s-1.\n• 11. Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of an ester at different temperatures is given below.\n• 12. Explain pseudo first order reaction with an example.\n• 13. Identify the order for the following reactions\n(i) Rusting of Iron\n(iii) 2A + 3B –>AB products ; rate =k[A]1/2[B]2\n• 14. A gas phase reaction has energy of activation 200 kJ mol-1. If the frequency factor of the reaction is 1.6 x 1013s-1Calculate the rate constant at 600 K. (e-40.09 =3.8×10-18)\n• 15. The rate constant for a first order reaction is 1.54 x 10-3s-1. Calculate its half life time.\n• 16. The half life of the homogeneous gaseous reaction SO2Cl2–>SO2 + Cl2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO2Cl2 to be reduced to 1% of the initial value?\n• 17. The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180seconds?\n• 18. A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80%complete?\n• 19. The activation energy of a reaction is 225 k Cal mol-1 and the value of rate constant at 40°C is 1.8×10-5s-1 Calculate the frequency factor,A.\n• 20. A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?\n• 21. Define the rate of the reaction.\n• 22. Write the rate expression of the following reaction 2NH3 –> N2 + 3H2\n• 23. What are the differences between rate and rate constant of a reaction?\n• 24. What are the differences between order and molecularity?\n• 25. Give the two examples of first order reaction\n• 26. Define Zero order reaction?\n• 27. Derive the relationship between half life period and zero order rate constant\n• 28. Derive the rate constant for Zero order reaction\n• 29. Derive the rate constant for First order reaction\n• 30. Derive the relationship between half life period and First order rate constant\n• 31. Define half life period\n• 32. Define Activation energy\n• 33. Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.\n• 34. Why a negative sign is introduced in the rate expression?\n• 35. Give two examples for Zero order reaction.\n• 36. Derive Arhenius equation to calculate activation energy from the rate constant K1 and K2 at temperature T21 and T2 respectively." ]	[ null, "https://i0.wp.com/www.studentsguide360.com/wp-content/uploads/2021/12/12th-Chemistry-Important-Questions-Unit-1-METALLURGY.png", null, "https://i0.wp.com/www.studentsguide360.com/wp-content/uploads/2021/12/12th-Chemistry-Important-Questions-Unit-1-METALLURGY.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8477332,"math_prob":0.9806343,"size":4553,"snap":"2023-40-2023-50","text_gpt3_token_len":1219,"char_repetition_ratio":0.19344911,"word_repetition_ratio":0.05703422,"special_character_ratio":0.27564242,"punctuation_ratio":0.1168688,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9948554,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T16:07:39Z\",\"WARC-Record-ID\":\"<urn:uuid:9c2fa9f1-f90c-4ec1-88ec-bf64aa46a447>\",\"Content-Length\":\"190297\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b4c90b2-faaf-473a-84da-0e4a58f8132a>\",\"WARC-Concurrent-To\":\"<urn:uuid:1080930e-8fae-4438-9dd7-50405d8f97c8>\",\"WARC-IP-Address\":\"136.243.92.92\",\"WARC-Target-URI\":\"https://www.studentsguide360.com/12th-chemistry-important-questions-unit-7-chemical-kinetics/\",\"WARC-Payload-Digest\":\"sha1:WV5EFGEMSN667F5JX6SE3VPY4O5RVZDW\",\"WARC-Block-Digest\":\"sha1:G4RVZOADQTLSCCPGV5QCPTJHM4LQNAH2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100912.91_warc_CC-MAIN-20231209134916-20231209164916-00517.warc.gz\"}"}
https://www.brainbell.com/tutors/Perl/Listing_16_1_A_very_simple_calculator.htm	[ "# Listing 16.1. A very simple calculator\n\n```#!/usr/local/bin/perl\nuse CGI::Form;\n\\$q = new CGI::Form;\nprint \\$q->start_html(-title=>`A Very Simple Calculator');\nprint \"<H1>A Very Simple Calculator</H1>\\n\";\nif (\\$q->cgi->var(`REQUEST_METHOD') eq `GET') {\n\\$val=0;\n&printForm(\\$q,\\$val);\n} else {\n\\$val=\\$q->param(`hiddenValue');\n\\$modifier=\\$q->param(`Modifier');\nif (\\$modifier=~/^[\\d]+\\$/) {\n\\$op=\\$q->param(`Action');\n\\$val+=\\$modifier;\n} elsif (\\$op eq \"Subtract\") {\n\\$val-=\\$modifier;\n} elsif (\\$op eq \"Multiply\") {\n\\$val*=\\$modifier;\n} elsif (\\$op eq \"Divide\") {\n\\$val/=\\$modifier;\n}\n} else {\nprint \"<P><STRONG>Please enter a numeric value!</STRONG><BR><BR>\\n\";\n}\n\\$q->param(`hiddenValue',\\$val);\n&printForm(\\$q,\\$val);\n}\nprint \\$q->end_html();\nsub printForm {\nmy(\\$q,\\$val)=@_;\nprint \"<P>The current value is: \\$val\\n\";\nprint \"<P>Please enter a value and select an operation.\\n<BR>\";\nprint \\$q->start_multipart_form();\nprint \\$q->hidden(-name=>`hiddenValue',-value=>\\$val);\nprint \"<TABLE><TR><TD COLSPAN=4>\\n\";\nprint \\$q->textfield(-name=>`Modifier',-size=>12,-maxlength=>5);\nprint \"</TD></TR>\\n<TR><TD>\\n\";\nprint \"\\n</TD><TD>\\n\";\nprint \\$q->submit(-name=>`Action',-value=>`Subtract');\nprint \"\\n</TD><TD>\\n\";\nprint \\$q->submit(-name=>`Action',-value=>`Multiply');\nprint \"\\n</TD><TD>\\n\";\nprint \\$q->submit(-name=>`Action',-value=>`Divide');\nprint \"\\n</TD><TD>\\n\";\nprint \"</TR></TABLE>\\n\";\nprint \\$q->end_form;\n}\n```\n\nYou will note that in this example, we use the hidden field `Value` to retain the current value of the calculator. We do some very basic field validation to make sure the user actually gave us a number. Obviously, this calculator will accept only integer values. Remember that when the user leaves our CGI program and returns, all state is now lost. Figure 16.1 shows the calculator as it appears in the Web browser.\n\nLet's look at another example now in Listing 16.2 that will make use of a file to retain state about a certain client. We will use the `REMOTE_ADDR` CGI environment variable to distinguish between clients. In this example, we will be allowing users to enter our Web site and write whatever they like in a big text field and then store the contents of that field in a file for them to later come back to and modify. Additionally, we will keep track of how many times users submitted changes to their text and display that value to them. Our example will simply use the /tmp/visitors directory to store these files and not worry about cleaning up these files.\n\n`Figure 16.1.` The calculator example in the browser." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65465826,"math_prob":0.9540978,"size":2543,"snap":"2021-43-2021-49","text_gpt3_token_len":714,"char_repetition_ratio":0.14533281,"word_repetition_ratio":0.0,"special_character_ratio":0.31183642,"punctuation_ratio":0.1503006,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98389775,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T08:08:50Z\",\"WARC-Record-ID\":\"<urn:uuid:06d40ad2-e3a1-4dfe-b6c8-9736901f7e91>\",\"Content-Length\":\"33728\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a6c1a02-25cc-442c-bb28-57c5fda4a43c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2c94468-bd59-4978-bab5-6a558f18f059>\",\"WARC-IP-Address\":\"35.175.60.16\",\"WARC-Target-URI\":\"https://www.brainbell.com/tutors/Perl/Listing_16_1_A_very_simple_calculator.htm\",\"WARC-Payload-Digest\":\"sha1:RLIYX42N5Z6H3QW3H6VG3S7QTSFHP7B2\",\"WARC-Block-Digest\":\"sha1:HPMZM2J3HT6XC3LNVOFGDTU7MQ2FF5XO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585653.49_warc_CC-MAIN-20211023064718-20211023094718-00620.warc.gz\"}"}
https://iraqidinarchat.net/?p=45439	[ "# Rate of Arab and foreign currencies in Iraqi dinars on Wednesday 11-2-2016\n\n### Rate of Arab and foreign currencies in Iraqi dinars on Wednesday 11-2-2016\n\nPublished on: 11-2-2016, 09:30", null, "Rate of Arab and foreign currencies in Iraqi dinars on Wednesday 11-2-2016\n\nU.S. dollar\nUS \\$ 1 = 1,164.1200 Iraqi dinars\n1 Iraqi dinars = US \\$ 0.0009\nSale price of a hundred dollars = 129.900 Iraqi Dinars\nThe purchase price of a hundred dollars = 128.900 Iraqi Dinars\neuro\n1 euro = Iraqi dinars 1,288.7413\nIQD 1 = 0.0008 euros\nSterling pound\n£ 1 = 1,426.2681 Iraqi dinars\nIQD 1 = 0.0007 pounds\nIQD 1 = 0.0012 Canadian dollars\nAustralian Dollar\n1 AUD = 887.8279 dinars\nIQD 1 = 0.0011 Australian dollars\nJapanese Yen\n1 Japanese Yen = 11.2065 dinars\nIQD 1 = 0.0892 Japanese yen\nSwiss Franc\n1 Swiss Franc = 1,195.6861 Iraqi dinars\nIQD 1 = 0.0008 Swiss Franc\nTurkish lira\n1 Turkish lira = 373.2709 dinars\nIQD 1 = 0.0027 Turkish lira\nChinese yuan\n1 Chinese yuan = 172.0951 dinars\nIQD 1 = 0.0058 Chinese yuan\nThai Baht\n1 Thai Baht = 33.2568 dinars\nIQD 1 = 0.0301 Thai Baht\nMalaysian Ringgit\n1 Malaysian Ringgit = 277.5018 dinars\nIQD 1 = 0.0036 Malaysian Ringgit\nIndian Rupee\n1 Indian Rupee = 17.4200 dinars\nIQD 1 = 0.0574 Indian Rupee\nIranian Rial\n1 IRR = 0.0387 Iraqi dinars\nIQD 1 = 25.8393 Iranian Rial\nEgyptian Pound\n1 Egyptian Pound = 131.1330 dinars\nIQD 1 = 0.0076 Egyptian pounds\nSaudi riyal\n1 SAR = 310.7552 dinars\nIQD 1 = 0.0032 SAR\nUAE dirham\n1 AED = 317.0175 dinars\nIQD 1 = 0.0032 AED\nSudanese Pound\n1 SDG = 183.0064 dinars\nIQD 1 = 0.0055 Sudanese pounds\nAlgerian dinar\n1 DA = 10.6162 dinars\nIQD 1 = 0.0942 DA\nBahraini Dinar\n1 BD = 3,109.2949 Iraqi dinars\nIQD 1 = 0.0003 BD\nJordanian Dinar\nJD 1 = 1,647.7282 Iraqi dinars\nIQD 1 = 0.0006 JD\nKuwaiti dinar\n1 KD = 3,852.1509 Iraqi dinars\nIQD 1 = 0.0003 Kuwaiti dinars\nLebanese Pound\n1 LP = 0.7923 Iraqi dinars\nIQD 1 = 1.2621 LP\nIsraeli Shekel\n1 Israeli Shekel = 305.4150 dinars\nIQD 1 = 0.0033 Israeli Shekel\nLibyan dinar\n1 LD = 846.7559 dinars\nIQD 1 = 0.0012 LD\nMoroccan dirham\nIQD 1 = 0.0084 Moroccan dirhams\nMauritanian ounce\n1 UM = 3.2663 Iraqi dinars\nIQD 1 = 0.3062 UM\nSyrian Lira\n1 SYP = 5.4429 Iraqi dinars\nIQD 1 = 0.1837 LS\nSomali shilling\n1 Somali Shilling = 2.0260 Iraqi dinars\nIQD 1 = 0.4936 Somali Shilling\nOmani Rial\nRO 1 = 3,024.2539 Iraqi dinars\nIQD 1 = 0.0003 RO\nQatari Riyal\nQR 1 = 319.8923 dinars\nIQD 1 = 0.0031 QR\nTunisian Dinar\n1 TND = 520.6867 dinars\nIQD 1 = 0.0019 Tunisian dinars\nYemeni riyal\n1 YR = 4.6579 Iraqi dinars\nIQD 1 = 0.2147 YR\nDjiboutian franc\n1 Djibouti francs = 6.5316 Iraqi dinars\nIQD 1 = 0.1531 Djibouti francs\n\nskypressiq.net" ]	[ null, "https://iraqidinarchat.net/wp-content/uploads/2013/03/dinar-dollar.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5985002,"math_prob":0.9903253,"size":2676,"snap":"2020-45-2020-50","text_gpt3_token_len":1148,"char_repetition_ratio":0.27881736,"word_repetition_ratio":0.0952381,"special_character_ratio":0.43609866,"punctuation_ratio":0.1369637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95551556,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T18:05:48Z\",\"WARC-Record-ID\":\"<urn:uuid:ca01d8a1-4096-427e-8655-bca0ce1bad63>\",\"Content-Length\":\"42242\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25b06566-5655-4891-9b23-79db0824960d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f86dca0a-0575-4244-b236-1bbc12837a95>\",\"WARC-IP-Address\":\"104.192.220.131\",\"WARC-Target-URI\":\"https://iraqidinarchat.net/?p=45439\",\"WARC-Payload-Digest\":\"sha1:BOFNE56VQ2GRSBGNMAHEW3K24FS64Z3M\",\"WARC-Block-Digest\":\"sha1:EAWGV3K3VG7B3YB7HNK3PAZFPRLJMTH2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141681209.60_warc_CC-MAIN-20201201170219-20201201200219-00187.warc.gz\"}"}
https://c.happycodings.com/file-operations/c-file-io-for-binary-files.html	[ "", null, "# C Programming Code Examples\n\n## C > File Operations Code Examples\n\n### C FILE I/O for Binary files\n\nProgram Code Checks Alphabets by ASCII - C Language program Check Alphabets using ASCII code. Every printable & non-printable Symbol is treated as a Character and has an ASCII code. The ASCII code is unique integer\n\nNumber Decimal System to Binary System - C Language program code to using recursion finds a binary equivalent of a decimal number entered by the user. The user has to enter a decimal which has a base 10 and the program\n\nC Coding Print all the Paths from the Root - C language program Function to store all the paths from the root node to all leaf nodes in a array. Function which helps the print_path to recursively print all the nodes. Function to...\n\nC program code insert an element in array - Code insert an element in array at specified position. Insert element in an array at given position. The program should also print an error message if the insert position is invalid.\n\nC Code Implement a Queue using an Array - C program code ask the user for operation like insert, delete, display and exit. According to the option entered, access its respective function using switch statement and use the\n\nC++ Program Implements Fibonacci Heap - Link nodes in fibonnaci heap. Union nodes in fibonnaci heap. Extract min node in fibonnaci heap. Consolidate node in fibonnaci heap and Decrease key of nodes in fibonnaci heap. Find\n\nCode Sum of Prime numbers between 1- n - C program to find sum of all prime numbers between 1 to n using for loop. Prime numbers are positive integers greater than 1 that has only two divisors 1 and the number itself. For\n\nConvert Binary Number Octal Vice-Versa - You will learn to convert \"binary number\" to octal, and octal number to binary manually by creating a \"user-defined\" function. In this, first convert the binary number to \"decimal\"." ]	[ null, "https://happycodings.com/images/HappyCodings.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8109096,"math_prob":0.69934314,"size":1871,"snap":"2021-31-2021-39","text_gpt3_token_len":402,"char_repetition_ratio":0.1237279,"word_repetition_ratio":0.006153846,"special_character_ratio":0.20470336,"punctuation_ratio":0.07183908,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96844465,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-16T16:40:20Z\",\"WARC-Record-ID\":\"<urn:uuid:62bf9229-b354-4530-975f-4592e36eb383>\",\"Content-Length\":\"30363\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc63ce17-4e98-4007-963d-4388efc17b75>\",\"WARC-Concurrent-To\":\"<urn:uuid:e10772e0-7134-4186-beb0-ed14dbc7f5ad>\",\"WARC-IP-Address\":\"172.67.145.119\",\"WARC-Target-URI\":\"https://c.happycodings.com/file-operations/c-file-io-for-binary-files.html\",\"WARC-Payload-Digest\":\"sha1:PIVN2VXP3YHC42KNFGEKCVNVZ7XXWA2M\",\"WARC-Block-Digest\":\"sha1:ZMZFRGO5I3PN6TFWCYMMM6UV65I6C2IZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780053657.29_warc_CC-MAIN-20210916145123-20210916175123-00192.warc.gz\"}"}
https://es.mathworks.com/matlabcentral/cody/problems/2022-find-a-pythagorean-triple/solutions/1085653	[ "Cody\n\n# Problem 2022. Find a Pythagorean triple\n\nSolution 1085653\n\nSubmitted on 20 Dec 2016 by Pranav\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\na = 1; b = 2; c = 3; d = 4; flag_correct = false; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))\n\n2 Pass\na = 2; b = 3; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))\n\n3 Pass\na = 3; b = 4; c = 5; d = 6; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))\n\n4 Pass\na = 3; b = 4; c = 4.5; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))\n\n5 Pass\na = 3; b = 3.5; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5189801,"math_prob":0.9985649,"size":924,"snap":"2020-45-2020-50","text_gpt3_token_len":329,"char_repetition_ratio":0.1576087,"word_repetition_ratio":0.28387097,"special_character_ratio":0.36471862,"punctuation_ratio":0.25,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969409,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T03:32:15Z\",\"WARC-Record-ID\":\"<urn:uuid:8463989e-efa5-4c2a-a116-4308fa2ffdf7>\",\"Content-Length\":\"81278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6193f3d9-b587-49cd-a5f7-83d4be15a343>\",\"WARC-Concurrent-To\":\"<urn:uuid:b373bcde-3024-46ce-944b-368ead179370>\",\"WARC-IP-Address\":\"184.24.72.83\",\"WARC-Target-URI\":\"https://es.mathworks.com/matlabcentral/cody/problems/2022-find-a-pythagorean-triple/solutions/1085653\",\"WARC-Payload-Digest\":\"sha1:CL5ZUJX2AACZ3E5S3D7VAIKAYG5WJKZM\",\"WARC-Block-Digest\":\"sha1:HVMRK7NEOJEL2CQBJX2CCLAFJNQKG3ME\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896048.53_warc_CC-MAIN-20201028014458-20201028044458-00250.warc.gz\"}"}
https://planetcalc.com/7921/	[ "# Convert grams to liters and liters to grams\n\nThis online calculator converts grams to liters and liters to grams given a gas formula. It uses molar volume of a gas at STP (standard temperature and pressure)\n\n### Articles that describe this calculator", null, "#### Convert grams to liters and liters to grams\n\nDigits after the decimal point: 1\nVolume of the gas, liters\n\nMass of the gas, grams\n\nMolar mass of the gas\n\nMolar mass details\n\n### Calculators used by this calculator\n\nURL copied to clipboard\n\n#### Similar calculators\n\nPLANETCALC, Convert grams to liters and liters to grams" ]	[ null, "https://planetcalc.com/img/32x32i.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7381312,"math_prob":0.93435365,"size":513,"snap":"2022-40-2023-06","text_gpt3_token_len":114,"char_repetition_ratio":0.18664047,"word_repetition_ratio":0.13414635,"special_character_ratio":0.18128654,"punctuation_ratio":0.054945055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9673632,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T16:00:50Z\",\"WARC-Record-ID\":\"<urn:uuid:a171db1c-2849-4a1d-8201-5b4f8d8af9ef>\",\"Content-Length\":\"41574\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:443ac70d-ce48-4f33-aecc-3ba97d4fd4a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:334e3755-290a-4442-b1d5-a8670b5ae401>\",\"WARC-IP-Address\":\"104.217.251.114\",\"WARC-Target-URI\":\"https://planetcalc.com/7921/\",\"WARC-Payload-Digest\":\"sha1:LGWVQ7RTMO43IO634CVLW3QJ3E26QZJT\",\"WARC-Block-Digest\":\"sha1:AN3XXBIFQ2SCJDK4DISVD7JR7RY5YFRY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337516.13_warc_CC-MAIN-20221004152839-20221004182839-00655.warc.gz\"}"}