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http://tasks.illustrativemathematics.org/content-standards/6/RP/A/tasks/2051	[ "# Hunger Games versus Divergent\n\nAlignments to Content Standards: 6.RP.A 6.RP.A.3\n\nThe 150 students at Skokie School were asked if they prefer seeing the movie Hunger Games or Divergent. The data showed that 100 preferred Hunger Games and 50 preferred Divergent.\n\n1. Look at the following statements and decide if each accurately reports the results of the survey and explain how you know.\n1. At Skokie School, 1/3 of the students prefer Hunger Games.\n2. Students prefer Hunger Games to Divergent in a ratio of 2 to 1.\n3. The ratio of students who prefer Divergent to students who prefer Hunger Games is 1 to 2.\n4. The number of students who prefer Hunger Games is 50 more than the number of students who prefer Divergent.\n5. The number of students who prefer Hunger Games is two times the number of students who prefer Divergent\n2. Compare statements (iv) and (v) above. In what ways is the information given by these statements similar? In what ways is it different? Explain.\n\n## IM Commentary\n\nThe goal of this task is to introduce ratio language and have students apply reasoning about ratios in a context. In order to answer the questions, students need to understand the difference between a fraction and a ratio and need to analyze the part to whole and the part to part relationships in a ratio.\n\nThis is an engaging introductory lesson for a unit on ratio and proportional relationships. Students can relate to the context and the numbers are \"friendly\" enough that students can focus on the concepts without getting bogged down in calculations.\n\nThis task was written as part of a collaborative project between Illustrative Mathematics, the Smarter Balanced Digital Library, the Teaching Channel, and Desmos.\n\n## Solution\n\n1. There are 150 students who took the survey so 1/3 of the students would be 50 students. But 100 of the students surveyed preferred Hunger Games to Divergent so this is not true. It is an easy mistake to make, however, because it is true that 1/3 more of the 150 students prefer Hunger Games compared to those who prefer Divergent.\n2. The ratio of students who prefer Hunger Games to students who prefer Divergent is 100:50. This is equivalent to the ratio 2:1 as we can see by multiplying 2 and 1 by 50. This can also be shown in steps with a ratio table:\n\nStudents who prefer Hunger Games Students who prefer Divergent\n100 50\n10 5\n2 1\n\nThe second line comes from the first by multiplying both entries by $\\frac{1}{10}$ and then the third row is the second row multiplied by $\\frac{1}{5}$.\n\n3. The ratio of students who prefer Divergent to students who prefer Hunger Games is 50:100. This is equivalent to 1:2 as we can see by multiplying 1 and 2 by 50. This can be shown with the ratio table of part (b) or with a double number line as below:", null, "The given information is furthest to the right and then the equivalent ratios, calculated by multiplying by $\\frac{1}{10}$ and then $\\frac{1}{5}$, are to the left closer to 0.\n\n4. Since 100 students prefer Hunger Games and 50 prefer Divergent and 100 = 50 + 50 it is true that 50 more students prefer Hunger Games.\n5. Since 100 students prefer Hunger Games and 50 prefer Divergent and 100 = 2 $\\times$ 50 it is true that the number of students who Hunger Games is twice the number who prefer Divergent.\n1. The two statements give the same information if we know how many students took the survey. If we know that 150 students took the survey then both (d) and (e) tell us that 100 students preferred Hunger Games and 50 preferred Divergent. On the other hand, without this information only (e) tells us the ratio of students who preferred Hunger Games to Divergent" ]	[ null, "http://s3.amazonaws.com/illustrativemathematics/images/000/003/600/large/hunger_d3f54aba52826b9ebf9253bded5b6d43.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9283837,"math_prob":0.934792,"size":4180,"snap":"2022-27-2022-33","text_gpt3_token_len":978,"char_repetition_ratio":0.25981802,"word_repetition_ratio":0.4871795,"special_character_ratio":0.23923445,"punctuation_ratio":0.07061791,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946167,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T04:13:47Z\",\"WARC-Record-ID\":\"<urn:uuid:0dade9e8-afad-4693-8710-2ee8bdd2b948>\",\"Content-Length\":\"29370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9fcd6e3-5ced-4b9b-8bd1-6d11b2bbe736>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f73ed10-3c46-43d7-9529-36d147336daf>\",\"WARC-IP-Address\":\"54.204.143.11\",\"WARC-Target-URI\":\"http://tasks.illustrativemathematics.org/content-standards/6/RP/A/tasks/2051\",\"WARC-Payload-Digest\":\"sha1:XAYPW3QJJHDN7EUVMY7GDWGFTZ6DK4FD\",\"WARC-Block-Digest\":\"sha1:F5ZP5PNBWYUYN65OWWVVAA3Z3G4RQKZP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104683683.99_warc_CC-MAIN-20220707033101-20220707063101-00384.warc.gz\"}"}
http://math.eretrandre.org/tetrationforum/printthread.php?tid=936	[ "", null, "Representations by 2sinh^[0.5] - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Representations by 2sinh^[0.5] (/showthread.php?tid=936) Representations by 2sinh^[0.5] - tommy1729 - 11/15/2014 As mentioned before Im considering number theory connections to tetration. One of those ideas is representations. Every integer M is the sum of 3 triangular number ... or 4 squares. Also every integer M is the sum of at most O(ln(M)) powers of 2. This is all classical and pretty well known. But there are functions that have growth between polynomials and powers of 2. So since 2sinh is close to exp and 2sinh^[0.5](0) = 0 , it is natural to ask 2S(n) := floor 2sinh^[0.5](n) 2S-1(n) := floor 2sinh^[-0.5](n) 2S numbers := numbers of type 2S(n) 2S-1 numbers := numbers of type 2S-1(n) 1) Every positive integer M is the sum of at most A(M) 2S numbers. A(M) = ?? 2) Every positive integer M is the sum of at most 2S(M) B numbers. B numbers := B(n) = ?? Of course we want sharp bounds on A(M) and B(n). ( A(M) = 4 + 2^M works fine but is not intresting for instance ) regards tommy1729 RE: Representations by 2sinh^[0.5] - tommy1729 - 11/16/2014 To estimate A(M) I use the following Tommy's density estimate * Let f(n) be a strictly increasing integer function such that f(n)-f(n-1) is also a strictly increasing integer function. Then to represent a positive density of primes between 2 and M we need to take T_f(M) elements of f(n). T_f(M) is about ln(M)/ln(f^[-1](M)). This is an upper estimate. * In this case to represent a positive density of primes between 2 and M we then need about ln(M)/ln^[3/2](M) 2S numbers. This is a brute upper estimate. A(M) is estimated as sqrt( ln(M)/ln^[3/2](M) ). Improvement should be possible. regards tommy1729" ]	[ null, "https://math.eretrandre.org/tetrationforum/images/TetrationForum_Kristof.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77457833,"math_prob":0.9942781,"size":2049,"snap":"2020-24-2020-29","text_gpt3_token_len":636,"char_repetition_ratio":0.11540342,"word_repetition_ratio":0.1373802,"special_character_ratio":0.32308444,"punctuation_ratio":0.13023256,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99916005,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-30T22:35:05Z\",\"WARC-Record-ID\":\"<urn:uuid:4f39b234-2ad1-4456-b104-9bce29002afa>\",\"Content-Length\":\"4400\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e85a5ed-3f82-4d60-9d7c-7af5f1d24f50>\",\"WARC-Concurrent-To\":\"<urn:uuid:f82db9d9-56ff-419b-ace5-5b3a15472bd6>\",\"WARC-IP-Address\":\"109.237.132.18\",\"WARC-Target-URI\":\"http://math.eretrandre.org/tetrationforum/printthread.php?tid=936\",\"WARC-Payload-Digest\":\"sha1:OMIXB2ETQSTTGNJO6Y5CQO5RQGH3I2ZP\",\"WARC-Block-Digest\":\"sha1:7TK3NN25E7K2OAW2PJ5V3MMRS4LUJXAU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347410352.47_warc_CC-MAIN-20200530200643-20200530230643-00292.warc.gz\"}"}
http://gre.kmf.com/question/all/641?keyword=&page=7	[ "4\n\n#### Quantity B\n\nk", null, "x+y+z\n\n#### Quantity B\n\n360", null, "S is an integer that is greater than 2.\n\n#### Quantity A\n\nThe average (arithmetic mean) of $\\frac{1}{S^{2}}$, $\\frac{1}{S}$, $S$, and $S^{2}$\n\n#### Quantity B\n\n2S\n\nWhich of the following are positive numbers?\n\nIndicate all such numbers.\nWhich of the following double inequalities involving $\\sqrt{0.8}$ is true?\nOne of the solutions of the equation $6x^{2}-x-35=0$ is $\\frac{n}{2}$, where n is a positive integer. What is the value of n?\n\nn=_____\nFor how many of the five response categories in the table was the number of responses less than 750?\nBased on the information given, which of the following statements are true?\n\nIndicate all such statements.\nIf 24 percent of the homeowners who were asked Question 2 responded both \"house is comfortable\" and \"house is in good condition\", what percent of the homeowners who were asked Question 2 gave neither of these responses?\nA college professor took attendance for the first 10 days of a class last semester. The professor noticed that Sarah attended class on 8 of those days, Andrew attended class on 7 of those days, Jeff attended class on 6 of those days, and on only 1 of those days did all three students attend class. On how many of the 10 days did at least two of the three students attend class?\n\n_____days\nIn the xy-plane, what is the x-intercept of the line that passes through the point (2, -4) and is perpendicular to the line 6x-3y=4?\n\nOn the number line, Z is the midpoint of line segment XY.\n\n#### Quantity A\n\nThe coordinate of Z\n\n#### Quantity B\n\n2.3", null, "x ≠ y and y ≠ 0\n\n#### Quantity A\n\n$\\frac{x+y}{y}$\n\n#### Quantity B\n\n$\\frac{x^{2}-y^{2}}{xy-y^{2}}$\n\n#### Quantity A\n\nThe remainder when 754,975,376 is divided by 4\n\n#### Quantity B\n\nThe remainder when 701,864,294 is divided by 4\n\nThe operation ▽ is defined by a▽b=$b^{a}$ for all nonzero integers a and b.\n\nk is a nonzero integer.\n\n(2▽1)▽k\n\n#### Quantity B\n\nk▽(4▽1)\n\nList M: r+3, r+2, r+1, r, r-1, r-2, r-3\n\n#### Quantity A\n\nThe standard deviation of the numbers in list M\n\nr\n\nx\n\n#### Quantity B\n\ny", null, "Suppose a, b, c are different integers, and the repeating decimal 0.abc=$\\frac{m}{n}$, where 0 < m < n < 100, m and n are both integers, then\n\nn\n\n#### Quantity B\n\n39\n\nJeremy bicycled a distance of 16 miles in x minutes. He averaged 15 miles per hour for the first 8 miles he bicycled.\n\n#### Quantity A\n\nThe number of minutes he took to bicycle the last 8 miles\n\n#### Quantity B\n\nx-32\n\nWhat is the least value of x such that (2x-1)(x-6) - (x-3)(x-6) = 0 ?\n1 2 ... 4 5 6 7 8 9 10 ... 12 13\n\n25000 +道题目\n\n136本备考书籍" ]	[ null, "http://img.kmf.com/kaomanfen/img/gre/bdef2394993a72a0b9b026e97e326f25.png", null, "http://img.kmf.com/kaomanfen/img/gre/0a7f5010b24c0e365474f34a3605d32b.png", null, "http://img.kmf.com/kaomanfen/img/gre/cc48812753f892cd3fa8259280246cab.png", null, "http://img.kmf.com/kaomanfen/img/gre/25e59f98db572e9e6b92bde44d9e6a5e.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9092123,"math_prob":0.9938803,"size":2541,"snap":"2021-43-2021-49","text_gpt3_token_len":790,"char_repetition_ratio":0.16081987,"word_repetition_ratio":0.026258206,"special_character_ratio":0.31483668,"punctuation_ratio":0.09386282,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998419,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,5,null,5,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T04:51:28Z\",\"WARC-Record-ID\":\"<urn:uuid:fe10809d-fa12-44e5-afce-fc5824c1cbdd>\",\"Content-Length\":\"62323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:282ae0e9-5157-4c1c-ae06-a9fe2d5af5b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:92d7e083-568c-464d-800a-eee5cc87b6f2>\",\"WARC-IP-Address\":\"47.89.156.56\",\"WARC-Target-URI\":\"http://gre.kmf.com/question/all/641?keyword=&page=7\",\"WARC-Payload-Digest\":\"sha1:RJWIYTORSHM2Z6UHWJK5XNX7K5FZEGU7\",\"WARC-Block-Digest\":\"sha1:3FBWTAVLX6ZD425YK7ADQN62FPPPIZWQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585302.56_warc_CC-MAIN-20211020024111-20211020054111-00405.warc.gz\"}"}
https://metanumbers.com/639628	[ "# 639628 (number)\n\n639,628 (six hundred thirty-nine thousand six hundred twenty-eight) is an even six-digits composite number following 639627 and preceding 639629. In scientific notation, it is written as 6.39628 × 105. The sum of its digits is 34. It has a total of 4 prime factors and 12 positive divisors. There are 290,720 positive integers (up to 639628) that are relatively prime to 639628.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 34\n• Digital Root 7\n\n## Name\n\nShort name 639 thousand 628 six hundred thirty-nine thousand six hundred twenty-eight\n\n## Notation\n\nScientific notation 6.39628 × 105 639.628 × 103\n\n## Prime Factorization of 639628\n\nPrime Factorization 22 × 11 × 14537\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 319814 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 639,628 is 22 × 11 × 14537. Since it has a total of 4 prime factors, 639,628 is a composite number.\n\n## Divisors of 639628\n\n12 divisors\n\n Even divisors 8 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.22119e+06 Sum of all the positive divisors of n s(n) 581564 Sum of the proper positive divisors of n A(n) 101766 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 799.767 Returns the nth root of the product of n divisors H(n) 6.28528 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 639,628 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 639,628) is 1,221,192, the average is 101,766.\n\n## Other Arithmetic Functions (n = 639628)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 290720 Total number of positive integers not greater than n that are coprime to n λ(n) 72680 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 51934 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 290,720 positive integers (less than 639,628) that are coprime with 639,628. And there are approximately 51,934 prime numbers less than or equal to 639,628.\n\n## Divisibility of 639628\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 3 4 3 4 7\n\nThe number 639,628 is divisible by 2 and 4.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (639628)\n\nBase System Value\n2 Binary 10011100001010001100\n3 Ternary 1012111101221\n4 Quaternary 2130022030\n5 Quinary 130432003\n6 Senary 21413124\n8 Octal 2341214\n10 Decimal 639628\n12 Duodecimal 26a1a4\n20 Vigesimal 3jj18\n36 Base36 dpjg\n\n## Basic calculations (n = 639628)\n\n### Multiplication\n\nn×y\n n×2 1279256 1918884 2558512 3198140\n\n### Division\n\nn÷y\n n÷2 319814 213209 159907 127926\n\n### Exponentiation\n\nny\n n2 409123978384 261687152045801152 167382429688751699251456 107062488736956871888810298368\n\n### Nth Root\n\ny√n\n 2√n 799.767 86.1607 28.2802 14.4939\n\n## 639628 as geometric shapes\n\n### Circle\n\n Diameter 1.27926e+06 4.0189e+06 1.2853e+12\n\n### Sphere\n\n Volume 1.09615e+18 5.1412e+12 4.0189e+06\n\n### Square\n\nLength = n\n Perimeter 2.55851e+06 4.09124e+11 904571\n\n### Cube\n\nLength = n\n Surface area 2.45474e+12 2.61687e+17 1.10787e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 1.91888e+06 1.77156e+11 553934\n\n### Triangular Pyramid\n\nLength = n\n Surface area 7.08624e+11 3.08401e+16 522254\n\n## Cryptographic Hash Functions\n\nmd5 1981baaee6f2be15f26b1af0e88bf46c 4c66718effd60de7deca4781ba802efd47c743e9 8be712525ac2b1bd4de6cfc6c7faa13346f2180ff02faa09c87554bb4595d860 a768258442c037abe7ad3a7c21d7b90d033948152de972ba759561c6db8349a3f9745e77550becb16ff4e33329b587b9a787c65f0644250366f58ae6b6500201 d04e96554d27ca14b2362f6b2772d746a23808dd" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6190904,"math_prob":0.9805859,"size":4640,"snap":"2021-43-2021-49","text_gpt3_token_len":1630,"char_repetition_ratio":0.11863676,"word_repetition_ratio":0.02827381,"special_character_ratio":0.46530172,"punctuation_ratio":0.07682458,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962232,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T02:35:27Z\",\"WARC-Record-ID\":\"<urn:uuid:527cdf84-64ca-4a6a-8ef2-698547e70c6c>\",\"Content-Length\":\"39442\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11b2aa66-ca25-4834-9a4c-380d7dad04d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9f5dd18-eb25-4152-872e-1a030cb51930>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/639628\",\"WARC-Payload-Digest\":\"sha1:SJULOKRFE4DYCQ2F5B3HMANYE4OURMR7\",\"WARC-Block-Digest\":\"sha1:ZWWVF27ZSUYYZBJH6TFJKCY2K3DBMU6X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358443.87_warc_CC-MAIN-20211128013650-20211128043650-00359.warc.gz\"}"}
https://www.mapleprimes.com/questions/141579-Percolation-Theory-And-Cylinder-Random-Packing	[ "# Question:Percolation theory and cylinder random packing\n\n## Question:Percolation theory and cylinder random packing\n\nMaple\n\nHi,\n\nI'm working at the moment on percolation theory and, in order to get interesting images I use Maple to produce random spheres packing using the code below:\n\nwith(RandomTools):\nwith(plots):\n\nx:=Generate(list(integer(range=0..20),100)): # random coordinate for x parameter\ny:=Generate(list(integer(range=0..20),100)): # random coordinate for y parameter\nz:=Generate(list(integer(range=0..20),100)): # random coordinate for z parameter\n\nP:=(x,y,z,R)->[x+Rcos(phi)cos(theta),y+Rcos(phi)sin(theta),z+Rsin(phi)]: # equation for one sphere\n\nS:=(i,j,k,R)-> plot3d(P(i,j,k,R),phi=-Pi/2..Pi/2,theta=0..2Pi): # 3D plot in space\n\ndisplay3d({S(x,y,z,R),S(x,y,z,R), and so and so....}); ## I finally plot everything in one space.\n\nThe result is not bad actually even if the method would be faster with a procedure... Here I only have to export the result in COLLADA format and I can play with it on a 3D render like Blender... and also do the calculations forthe percolation, on Maple.\n\nNow, just to compare I would like to obtain the same kind of random packing using capped cylinders randomly distributed in the space with various radius and length. The problem is not coming from the spacial distribution but all the cylinders are on the same direction. Does somebody has an idea to also work with random directions ?\n\nI tried this method but it doesn't work:\n\n1. Random distribution of discs\n\n2. Calculation of normal vector for all of them\n\n3. Ptotting of cylinders using the description of this normal vector\n\nI was quiet happy when I begin but It's seems really much more difficult than expected... The rotation can also be an interesting thing but I don't know how to introduce it...\n\nCan you help me ?\n\nThanks,\n\nLisa", null, "" ]	[ null, "https://www.mapleprimes.com/images/ajax-loader.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.839668,"math_prob":0.93780017,"size":1797,"snap":"2023-40-2023-50","text_gpt3_token_len":463,"char_repetition_ratio":0.098717235,"word_repetition_ratio":0.0,"special_character_ratio":0.27601558,"punctuation_ratio":0.19952494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962922,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T22:45:37Z\",\"WARC-Record-ID\":\"<urn:uuid:81822b8b-ecee-429e-a451-075363386fd2>\",\"Content-Length\":\"109028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73d6ed30-5236-47f5-a36d-9d5f0d79050b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cbf1616-e620-4012-892e-55bc966b6cc2>\",\"WARC-IP-Address\":\"199.71.183.11\",\"WARC-Target-URI\":\"https://www.mapleprimes.com/questions/141579-Percolation-Theory-And-Cylinder-Random-Packing\",\"WARC-Payload-Digest\":\"sha1:OEIUGAAI4OUUVYJPTUU2CO6BDTMTVZ3K\",\"WARC-Block-Digest\":\"sha1:75MVY6X4Z7ST7SSW7UZ5COHMB7DZX7XB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506423.70_warc_CC-MAIN-20230922202444-20230922232444-00595.warc.gz\"}"}
https://homework.cpm.org/category/CON_FOUND/textbook/caac/chapter/11/lesson/11.1.5/problem/11-64	[ "", null, "", null, "### Home > CAAC > Chapter 11 > Lesson 11.1.5 > Problem11-64\n\n11-64.\n\nSolve each of the following equations or systems.\n\n1. $x^2−1=15$\n\nHow many solutions should you be looking for?\n\n$x=\\pm4$\n\n1. $y=3x−2$\n$y=4x+3$\n\nUse the Equal Values method.\nMake sure to solve for both $x$ and $y$.\n\n1. $x^2−2x−8=0$\n\nFactor and use the Zero Product Property.\n\n1. $2x^2=−x+7$\n\nStart by setting the equation equal to $0$." ]	[ null, "https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png", null, 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https://discusstest.codechef.com/t/help-me-with-chef-and-sub-array-problem/15310	[ "", null, "# Help me with Chef and Sub Array problem?\n\nMy solution: https://www.codechef.com/viewsolution/14291195\n\nfor(j=n-1;j>-1;j–){\nif(check(myvector[j].second,i,y,n)) break;\n}\n\nYou can try using binary search here. Also using segment trees is much easier.\n\nI will help you with AC solution. Assume you have n elements and k be the length of the frame.\n\nFOR SUBTASK 1 : n,k<1000 -> here you can use simple Brute Force technique and do what the question tells you to do,i.e, when you encounter a ‘!’ character,rotate your array in o(n) and when you encounter a ‘?’ calculate the maximum of all the k sized subarrays and print it, it will alse take o(n). Thus the overall complexity will be o(np) which would pass for this subtask but not the other one.\n\nFOR SUBTASK 2: for given n-sized array of zeroes and ones, what i did was create an array of size 2n in which after the last element from the input, we insert again from the first element at the preceeding places,i.e\nif the elements are given “0 1 0 0 1 1” then we insert these from position 0 to position 5 of the array and again from the 6th position to 11th position we insert given elements in the given order.\nThen we create a new array(say csum[]) whose ith index stores the sum of first i elements of the above created array.Now, we know that after operating ‘!’ l times the start index would be ((n-l)%n)th element .\nlet this start index be st.\n\nSince we have stored the cumulative sum in our csum array we now have to answer the maximum sum that can be obtained in a k frame from index st to (st+n). By precalculating this for all values of st from 1 to n,we can answer each query in o(1) time and we dont even need to rotate the array,just increase the variable l when ‘!’ is encountered. For precalculation, rather than explaining that process myself I would recommend you to read this on Sliding window maximum.\n\nI hope it helped a little, if not, forgive me, its only my first answer to someone’s question in here.\n\nAny doubts, ask me freely in the comment section.\n\nMy solution\n\n2 Likes\n//" ]	[ null, "https://s3.amazonaws.com/discoursestaging/original/2X/4/45bf0c3f75fc1a2cdf5d9042041a80fa6dd3106b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.879537,"math_prob":0.9402745,"size":2366,"snap":"2021-31-2021-39","text_gpt3_token_len":592,"char_repetition_ratio":0.096104994,"word_repetition_ratio":0.11764706,"special_character_ratio":0.2476754,"punctuation_ratio":0.12878788,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924114,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T15:24:09Z\",\"WARC-Record-ID\":\"<urn:uuid:0954bf7d-348a-42ed-a410-ee24240cc238>\",\"Content-Length\":\"17224\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a7dbbc8-6075-4065-b93a-47120d79e933>\",\"WARC-Concurrent-To\":\"<urn:uuid:2163bb3b-cf11-4c93-9361-e3283051e071>\",\"WARC-IP-Address\":\"54.165.214.21\",\"WARC-Target-URI\":\"https://discusstest.codechef.com/t/help-me-with-chef-and-sub-array-problem/15310\",\"WARC-Payload-Digest\":\"sha1:KRVSSIYLFNJT6Y555GYR4XS5URMUA6CF\",\"WARC-Block-Digest\":\"sha1:66MHXESPJUBUOKIIJH7O5VMQ6Y3BYU5L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153392.43_warc_CC-MAIN-20210727135323-20210727165323-00422.warc.gz\"}"}
https://socratic.org/questions/554ebc5d581e2a6b7340ff54	[ "# Question #0ff54\n\nMay 10, 2015\n\nThe standard enthalpy of formation for a compound expresses the change in enthalpy (it can be positive or negative) when 1 mole of that compound is formed from the most stable form of its elements (at 1 atm and 273.15 K).\n\nStandard enthalpies of formation are thus calculated for reactions that produce 1 mole of a specific compound. For example, you could write the balanced chemical equation for your synthesis reaction like this\n\n${N}_{2 \\left(g\\right)} + 2 {O}_{2 \\left(g\\right)} \\to 2 N {O}_{2 \\left(g\\right)}$\n\nThis reaction will also have a change in enthalpy, $\\Delta {H}_{\\text{rxn}}$, but it not be equal to the standard enthalpy of formation, $\\Delta {H}_{\\text{f}}^{\\circ}$, because it produces 2 moles of nitrogen dioxide, instead of only 1 mole.\n\nTo get the reaction for which $\\Delta {H}_{\\text{rxn\" = DeltaH_\"f}}^{\\circ}$ you need to divide the above equation by 2, in order to get only 1 mole of the product\n\n$\\text{1/2} {N}_{2 \\left(g\\right)} + {O}_{2 \\left(g\\right)} \\to N {O}_{2 \\left(g\\right)}$\n\nThat's how you can determine the reaction for which the change in enthalpy is equal to the standard enthalpy of formation for any compound.\n\nHere's another example. Assume you want the reaction that produces $\\Delta {H}_{\\text{f}}^{\\circ}$ for iron (II) sulfide, $F e S$. You start from\n\n$8 F e + {S}_{8} \\to 8 F e S$\n\nNotice that you produce 8 moles of iron (II) sulfide, but you only need to produce 1 mole $\\to$ divide all the coefficients by 8\n\n$F e + \\text{1/8} {S}_{8} \\to F e S$\n\nThis reaction has $\\Delta {H}_{\\text{rxn\" = DeltaH_\"f}}^{\\circ}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9053486,"math_prob":0.99893945,"size":900,"snap":"2019-51-2020-05","text_gpt3_token_len":201,"char_repetition_ratio":0.15848215,"word_repetition_ratio":0.013333334,"special_character_ratio":0.20444444,"punctuation_ratio":0.049079753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996567,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T13:09:14Z\",\"WARC-Record-ID\":\"<urn:uuid:a3a04504-6978-4a71-8e88-ef476ca21a37>\",\"Content-Length\":\"36533\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3cf9cdc-c1f0-49df-943b-2eb4beecb88f>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b60840c-c58f-4be2-91a5-7e1c9ce6eb00>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/554ebc5d581e2a6b7340ff54\",\"WARC-Payload-Digest\":\"sha1:OXUSMKI67FGVSDUCLYR5ICKPA44HVZDY\",\"WARC-Block-Digest\":\"sha1:QTIQKWEESMFVGPH4QUHWZJEHYSSPOAQM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540555616.2_warc_CC-MAIN-20191213122716-20191213150716-00125.warc.gz\"}"}
https://encyclopediaofmath.org/wiki/Quasi-regular_ring	[ "# Quasi-regular ring\n\nA ring in which every element is quasi-regular. An element $a$ of an alternative (in particular, associative) ring $R$ is called quasi-regular if there is an element $a'\\in R$ such that\n$$a+a'+aa'=a+a'+a'a=0.$$\nThe element $a'$ is called the quasi-inverse of $a$. If $R$ is a ring with identity 1, then an element $a\\in R$ is quasi-regular with quasi-inverse $a'$ if and only if the element $1+a$ is invertible in $R$ with inverse $1+a'$. Every nilpotent element is quasi-regular. In an associative ring the set of all quasi-regular elements forms a group with respect to the operation of cyclic composition: $x\\cdot y=x+y+xy$. An important example of a quasi-regular ring is the ring of (non-commutative) formal power series without constant terms. There exist simple associative quasi-regular rings ." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7866167,"math_prob":0.9993579,"size":1140,"snap":"2023-14-2023-23","text_gpt3_token_len":328,"char_repetition_ratio":0.18309858,"word_repetition_ratio":0.0,"special_character_ratio":0.28245613,"punctuation_ratio":0.13983051,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99908555,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T04:32:50Z\",\"WARC-Record-ID\":\"<urn:uuid:2d95df00-25ca-4df7-aed5-39a943af5fc1>\",\"Content-Length\":\"15122\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5d7f7c7-8bac-4310-8776-8ba947aafe0b>\",\"WARC-Concurrent-To\":\"<urn:uuid:ccf494f5-40ef-4f84-818b-05bb938f51db>\",\"WARC-IP-Address\":\"34.96.94.55\",\"WARC-Target-URI\":\"https://encyclopediaofmath.org/wiki/Quasi-regular_ring\",\"WARC-Payload-Digest\":\"sha1:3H32YDUVRWABBBXSIQAMB4XQEMFRTV4Q\",\"WARC-Block-Digest\":\"sha1:KDO3ZXDY77F6CYS4RUQESCX2ACSH7PAV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649105.40_warc_CC-MAIN-20230603032950-20230603062950-00322.warc.gz\"}"}
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https://www.physicsforums.com/tags/projectile-motion/	[ "# Projectile motion Definition and 182 Discussions\n\nProjectile motion is a form of motion experienced by a launched object. Ballistics (Greek: βάλλειν, romanized: ba'llein, lit. 'to throw') is the science of dynamics that deals with the flight, behavior and effects of projectiles, especially bullets, unguided bombs, rockets, or the like; the science or art of designing and accelerating projectiles so as to achieve a desired performance.\n\nView More On Wikipedia.org\n1. ### Roll and Drop!\n\nVertical position (x component): y(t) = y0 +v0 * t -1/2gt^2 Position yx(t) = 0cm = 40cm + 0 t -1/2 981cm/s/s * t^2 edit: I replaced v0 = 0 and got t = 0.286s, which is incorrect according to the submission. This didn't work so I thought maybe by finding the angle of the throw would help...\n2. ### Circular motion to projectile motion\n\nSo first I found the velocity of the ball at the bottom of the swing from the force equations, which I got to be 4.9 m/s and this is only in the x-direction. Then using the projectile motion for delta y I found time, which is 0.2s. Then using that time I found the delta x to be 0.98m. I just...\n3. ### What is the minimum height h that is required to generate the greatest projectile range?\n\nI am calculating it like this: 𝑦=ℎ0−0.5𝑔𝑡^2=0→ℎ0=0.5𝑔𝑡^2→𝑡=sqrt(2ℎ0/g) 𝑥=𝑣0𝑡→ substituting t →𝑥=𝑣0sqrt(2ℎ0/g) 𝑑𝑥/𝑑ℎ0=𝑣0/(𝑔sqrt(2h0/g))=0 for maximum ℎ0=0. confused. can someone tell me how I am calculating this wrong?\n4. ### I If we throw a ball with initial velocity v0, is there a force?\n\nIf we throw a ball with initial velocity v0, and the ball progresses in a projectile motion, ignoring air resistance, will there be a force in the x direction? If so, what is that force in the x direction's value mathematically? I know there is a force in the y direction due to gravity.\n5. ### Kinematic Equations in Projectile Motion (this approach is not working)\n\nGivens: Vyi=12.5 m/s Vyf=-12.5 m/s (at the same horizontal level) ay=-9.81 m/s^2 Δy= zero m (as the displacement on the y-axis, when the projectile reaches the same horizontal level, is zero m) Δt=? When I use Δy=[(vyi+vyf)/2]Δt I get the time as undefined. Δt= 2Δy/(vyi+vyf) = 20 m/(12.5...\n6. ### Relationship between horizontal range and angle of launch\n\nI would like to conduct an experiment asking the question \"how does the angle at which a projectile is launched vary with respect to the horizontal range it covers?\" Ultimately, I'd like to prove that the horizontal range is directly proportional to sin(2(theta)). This will be done with the aid...\n7. ### Stopping a Bullet\n\n(a) ##u_{min}=\\big(1+\\frac{m_2}{m_1}\\big)\\sqrt{2\\mu_k g d}## (b) ##x_f=\\sqrt{\\frac{2h}{g}\\Big(\\big(\\frac{m_1}{m_1+m_2}u\\big)^2-2\\mu_k g d\\Big)}## Can someone check please?\n8. ### B Projectile motion — Thinking about forces on a curve ball\n\nWhen a ball is thrown such that it moves in a curved trajectory in the horizontal plane, it amuses me to think of its dynamics. In motion of a ball thrown upwards the force of gravity gives it a parabolic trajectory However when the ball is thrown to curve and hit a target, (in the horizontal...\n9. ### Tricky conceptual Projectile motion question\n\nSo far all I have determined is the equations of motion for the two and that is as follows. It is trivial that y(t)=v1sin(Q)t -gt^2/2 and that x(t)=v2cos(Q)t. Now the angle that is anticlockwise from the negative horizontal of the robber is 90 - Q using basic trigonometry, using this we can...\n10. ### Show that a projectile lands at a distance ##R = \\frac{2v_0^2 sin \\theta cos(\\theta + \\phi)}{g cos^2 \\phi}##\n\n##V_x = V_0 cos \\theta ## ##x = V_0 cos \\theta t## ##V_y = V_0 cos \\theta ## ##y = V_0 cos \\theta t## ##F_x = m\\ddot{x}## ##-mgsin \\phi = m\\ddot{x}## ##\\dot{x} = -gtsin\\phi + V_x## ##x = -\\frac{1}{2} gt^2 sin \\phi + V_x t## ##x = -\\frac{1}{2} gt^2 sin \\phi + v_0 cos\\theta t## ##F_y =...\n\n12. ### Projectile Intercept Math & Trigonometry\n\nLooking for some guidance on how to set up the equations for a projectile intercept given that you have perfect information about the target velocity, size and weather conditions in a 3D scenario, it's for an amateur videogame that I'm developing in my spare time For simplicity sake let's...\n13. ### Projectile motion with (constant) wind velocity\n\nLet me start be making a small sketch of the problem, shown to the right. If the range of the projectile on a still day ##R = v_{0x}T##, then on the windy day the range becomes ##R+2H = v'_{0x}T = (v_{0x}+v_w) T##. Since the maximum height attained by the projectile ##H =...\n14. ### Maximum range on ground for an elevated cannon\n\nI sketch a diagram for the problem and show it to the right. My approach is going to be along the following lines : (1) Use the equation of motion in the ##x## direction to express the total time of flight ##T## as a function of the range and initial angle ##T = T(R, \\theta_0)##. (2) Plug this...\n15. ### Runge-Kutta Projectile Approximation From Initial Conditions\n\nHi everyone. I'm a new member, great to be here:) I have a few questions that I wanted to ask you guys regarding the method by which we implement the Runge-Kutta approximation of Projectile Motion if we should do it using a numerical iterative method with a Spreadsheet like Excel. I have...\n16. ### Projectile motion in 2D\n\nI know the conventional method for solving this question using the formula for maximum range of a projectile in an inclined plane, but since it is an objective problem, if we consider a non general case where α=0, then clearly we can see that (see attachment) only one option matches which...\n17. ### Find the range of an object coming off an inclined plane\n\nHey Everyone, my physics teacher has assigned us a task which involves predicting the range of a ball falling down an inclined plane into a free-fall, the equation for the final velocity of the ball down the ramp, accounting for rotational velocity has been provided, this is the initial velocity...\n\n19. ### Will the pebble meet with the block according to the given condition?\n\nQuestion 1: I have used v= Aωcos(ωt+δ) where A= 0.2 m, ω= π/3, t=1 and δ=0. Are the values right in this case? I am confused. Question 2: From question 1 I have got the value of V which is 9 m/s. By using v= ω√(A^2-x^2), I have got the value of x. Now, do I need to add it with 2.5(distance...\n20. ### Maximum Distance a Projectile Moves Up an Inclined Plane\n\nthe red line is the initial velocity, the grey parabola is the path of the projectile. hi there...I'm kinda stuck at the part b of this problem. I can do part a with no problem. can anybody explain to me how to do the differentiation needed to solve part b?? by explain I mean explain the...\n21. ### Projectile motion hw help please\n\nI intended to finish the question with the equation of linear motion with constant acceleration, but it didn't work out. And I have no idea about the t^3 and t^4 of the position. How can I find the x component of the acceleration at time 3.4 s ? Where is the acceleration rate?\n22. ### How do you find the initial velocity of a projectile given angle/distance?\n\nI tried resolving the information given into vertical and horizontal components. I then tried to find time, as this is how I would find the initial velocity. However, I am unsure of how to use the angle in this problem to help solve it. I am also unsure of how to find the initial velocity only...\n23. ### The Determination of the Launch Angle of a Projectile\n\nHi, I am new here to the forum and I am having trouble with a project that I am undertaking with some friends. We are trying to build a firefighting robot. I am trying to derive an expression to solve for the launch angle theta of the water so that at x (meters), the projectile will be at 0.33...\n24. ### At what angle do we have to kick a ball up the hill for n-rebounds?\n\nI decided to try and find a solution in a green (tilted) coordinate system. I started solving this problem with thinking about 1-rebound: ##⟹y=0, α=\\text{angle under which we kick a ball}##; ##y=sin\\alpha v_0t-\\frac{1}{2}gt^2##; because I'm trying to solve this in a tilted system, I have to...\n25. ### Projectile motion of a two-point rigid body\n\nI would like to patch some gaps in my physics background. For example, I've been trying to come up with the sollution to the following: I have a model rigid body made up of two mass points and a massless rod connecting them. I throw the body with initial velocity under some angle of elevation...\n26. ### Pendulum projectile problem\n\nTried to find time in seconds in order to use the formula d=vt and find the distance in the x plane in which the target must be placed, to no avail.\n27. ### Comp Sci Ball rebounding off of a wall\n\nMy attempts involved using suvat equations to determine the rebound distance : S = 0.5 (u + v)t With u being 50 and v being 0 t being time taken to fall down (Height of impact / gravitational acceleration) t = 48.41 / 9.81 Plugging the numbers in gives S = 123.365m This is where i get...\n28. ### Projectile Motion at an angle\n\nVx = 10cos(23) = 9.21m/s Vy = 10sin(23) = 3.91m/s -30 = 3.91t + 1/2(-9.8)t^2 0 = t(3.91 - 4.9t) + 30 4.9t = 33.91 t = 6.92s Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2 Delta X = 298.38m Vf^2 = 9.21 + 2(9.8)(298.38) Vf^2 = 5857.458 Vf = 76.53m/s\n29. ### Need to find the spring constant to achieve Max Velocity\n\nHello All. I am mentoring a high school student in my area with his class project for school. He has chosen he wants to launch an object (in our case, a softball) into a 5' diameter area. The idea is to build basically an oversized slingshot using an extension spring as the source of energy. We...\n30. ### Projectile Motion with Unknown Initial Height\n\nFirst, I tried solving for the total time of flight, which I got as 100 = 5cos25t --> t=22 s Since we know the height at which the object lands, but not at which it is launched, I tried setting up the equation as: yf = 40 - y0 = y0 + 5sin25(22) - 1/2(9.8)(22)^2 However, I got y0 = 1183 m...\n31. ### A golf ball being hit on the moon vs Earth\n\nI am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this? I did 1/69.8=1.63 for g on the moon...\n32. ### Project Motion/Trigonometry Question\n\nMy reasoning and answer is wrong, but I cannot figure out why. Perhaps it is strange, perhaps not, but I want to figure out why my initial method of solving this problem did yield an incorrect answer. I began by creating an equation and drawing a right triangle. x is the horizontal part of...\n33. ### How does the release angle of a catapult affect the object's velocity?\n\n[Moderator: repeated text removed.]\n34. ### How to find the max height a projectile can reach on a hillside\n\nHere's a fully typed version of the problem with a diagram My attempt: Given the angle of the hill, I know that the horizontal displacement of the arrow and my vertical height on the hill are related by ##Δx=d+\\frac h {Tan(60)}## ...(1) where d is the distance of the enemy from the base...\n35. ### Impact of several variables on resulting projectile motion trajectory\n\nI was told to generate these variables (m, C, alpha, wind velocity) normally distributed and compare the random data with the result and then tell, which of the variables has the most impact. Here I am stuck, tried to compare variances, kurtosis and skewness of the data (the original variables...\n36. ### Projectile Motion: Path's length?\n\nSo I just learned about projectile motion. I understand why you can study it as two independent straight line motions . But this can give you a way to calculate total velocities or accelerations, just by adding its individual component of each vector. If the initial position of the projectile...\n37. ### Projectile Motion Experimental Error\n\nHomework Statement I got an experimental vertical acceleration of -12 m/s^2 of projectile motion for an experiment I did at home where I just had to throw a ball at around 45 degree to the horizontal up in the air and record it's motion, then analyze the motion via computer software. Obviously...\n38. ### Boomerang Problem, need help solving it\n\nHomework Statement A boomerang is thrown with an initial linear velocity of 5 m/s at an angle of 30 degrees vertically. The initial angular velocity is ##2\\frac{revolutions}{s}## At its peak, it has a displacement about the z axis of 2 meters and about the x axis of 10 meters. The force applied...\n39. ### Need to find final Height, Equation not Working\n\nHomework Statement Homework Equations The kinematic equations--namely, Sf = S0 + V0Δt The Attempt at a Solution [/B] I am a bit confused as it seems this problem is very straight-forward. My known variables: X0 = 0m Y0 = 1.7m Δt = 3.92s V0 = 29m/s Θ = 60ο Yf = ? So, I just use the above...\n40. ### Projectile Motion: Pitcher throwing a ball to a catcher\n\nHomework Statement determine the acute angle (in radians) of appropriate elevation in the throwing of a ball, if the initial velocity is 20 m / s, g=9.81 m/s^2 and the distance in x is 40m. The ball leaves the hand of the pitcher with an elevation equal to 1.8m and the catcher receives it at...\n41. ### AP Physics C Momentum Problem: A dart launched horizonally by a spring gun\n\nHomework Statement A boy launches a 20 g dart horizontally by a spring gun from a balcony 45 m above the ground. The dart lands 15 m away from the balcony. If the length of the gun’s barrel is 10 cm, what is the average horizontal force applied by the spring? (A) 1.0N (B) 2.0 N (C) 2.5 N...\n42.", null, "### Why Vertical Velocity changes & Horizontal does not?\n\nProjectile motion involves velocity in vertical and horizontal direction. It is important to understand that the two are independent of each other. Solving problems becomes much easier with this simple understanding.\n43. ### Varying Gravity and Air Resistance in projectile motion\n\nSalutations, I have been trying to approach a case about projectile motion considering variation of gravity acceleration and air resistance: A spherical baseball with mass \"m\" is hit with inclination angle $\\theta$ and launching velocity $v_0$, then, the wind has a drag force equals to ##F=kv##...\n44. ### Meaning of a phrase in a kinematics problem\n\n1. Problem statement: \"An object is thrown horizontally with an initial speed of 10 m/s. How far will it drop in 4 seconds?\" Homework Equations Δx = v0x.t Δy = v0y.t + 1/2(-g)t2 The Attempt at a Solution [/B] When I first read it, I thought that 'how far' means Δx, since when someone says...\n45. ### Calculating spring constant of a spring loaded cannon\n\nHomework Statement if you wanted to build a spring launched cannon that will shoot you over a building that is 35 m high and 30 m wide, and the cannon is being shot at 60 degrees. If the cannon can be no more than 2 m long, what spring constant do you need in the spring to make this work? here...\n46. ### Where on the hill does the projectile land?\n\nHomework Statement A projectile is shot at a hill, the base of which is 320 m away. The projectile is shot at 60° above the horizontal with an initial speed of 79 m/s. The hill can be approximated by a plane sloped at 19° to the horizontal. The equation of the straight line forming the hill is...\n47. ### What is conceptual physics?\n\nI have been wondering, what is conceptual physics? I remember taking a class in high school that was physics oriented, for example two trains leave a station at different speeds, and arrive at a central point, where do they overlap. Also there were trig functions on how to find the height of a...\n48. ### Using the Vy versus t graph to determine the y-component of the acceleration of the puck\n\nHomework Statement We had a lab on projectile motion and one of the questions was this: Use your Vy versus t graph to determine the y-component of the acceleration of the puck. Should this be equal to the acceleration due to gravity (9.8 m/s)? Explain why it is or why it is not. The...\n49. ### Projectile motion: Two objects are kicked upward at different angles...\n\nHomework Statement Two objects are kicked upward at different angles. Object A travels a greater horizontal distance than object B. Both reach the same maximum height. Which of the following statements about the objects are true? You may select more than one. 1.Object A is in the air for a...\n50. ### How to get R= 2 ( square root ) h1h2\n\nProblem : A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2..." ]	[ null, "https://www.physicsforums.com/data/xfmg/thumbnail/6/6221-c5f5512e84cc27261469f1ebbed5b29c.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9346973,"math_prob":0.93927306,"size":12894,"snap":"2022-27-2022-33","text_gpt3_token_len":3328,"char_repetition_ratio":0.13204034,"word_repetition_ratio":0.0042698546,"special_character_ratio":0.25825965,"punctuation_ratio":0.12623763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980126,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T13:41:50Z\",\"WARC-Record-ID\":\"<urn:uuid:93aca8f1-eb03-49c6-bb96-64946c2fb140>\",\"Content-Length\":\"122669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f6d9a02-d66e-46dc-b15a-60c836bcd44c>\",\"WARC-Concurrent-To\":\"<urn:uuid:cba8a8ad-21cf-487d-a70c-59badab6390c>\",\"WARC-IP-Address\":\"172.67.68.135\",\"WARC-Target-URI\":\"https://www.physicsforums.com/tags/projectile-motion/\",\"WARC-Payload-Digest\":\"sha1:YHRSKWUI77ORNASB4WRNHWOP3WTOSJFA\",\"WARC-Block-Digest\":\"sha1:XPA4KQS4BKD4V34QWXJX2NHZNCTC57GR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104141372.60_warc_CC-MAIN-20220702131941-20220702161941-00129.warc.gz\"}"}
https://old.lucentinnovation.com/blogs/technology-posts/working-with-arrays-in-javascript	[ "", null, "# Working with arrays in JavaScript\n\nArrays are a fundamental data structure in JavaScript, and they are used to store and manipulate a collection of values. An array is an ordered list of values, and each value in the array is called an element.\n\nIn JavaScript, you can create an array using the \"array literal\" syntax, which uses square brackets to enclose a list of values separated by commas. For example:\n\n```var numbers = [1, 2, 3, 4, 5]; ```\n\nIn this example, the array \"numbers\" contains five elements: 1, 2, 3, 4, and 5.\n\nYou can access the elements of an array using their index, which is the position of the element in the array. In JavaScript, arrays are zero-indexed, which means that the first element has an index of 0. For example:\n\n```console.log(numbers); // outputs 1 console.log(numbers); // outputs 5 ```\n\nYou can also use loops to iterate over the elements of an array. For example:\n\n```for (var i = 0; i < numbers.length; i++) { console.log(numbers[i]); ```\n`}`\n\nThis for loop will output each element of the \"numbers\" array to the console.\n\nArrays also have a number of built-in methods that can be used to manipulate the elements of the array. For example, you can use the \"push\" method to add an element to the end of the array:\n\n```numbers.push(6); ```\n\nYou can also use the \"pop\" method to remove the last element of the array:\n\n```numbers.pop(); ```\n\nThese are just a few examples of the many ways you can work with arrays in JavaScript. Arrays are a powerful data structure that are used to store and manipulate collections of data in a variety of applications.\n\nCheck out the rest of our series on Javascript by reading our other articles:" ]	[ null, "https://old.lucentinnovation.com/cdn/shop/articles/Introduction_to_JavScript_and_its_history_6_30x.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7895705,"math_prob":0.91692346,"size":2347,"snap":"2023-40-2023-50","text_gpt3_token_len":517,"char_repetition_ratio":0.17456253,"word_repetition_ratio":0.010362694,"special_character_ratio":0.2258202,"punctuation_ratio":0.13140312,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9631555,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T18:27:54Z\",\"WARC-Record-ID\":\"<urn:uuid:cc894223-54a8-4dca-b0ca-c9ffd12e8491>\",\"Content-Length\":\"68412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab557ec5-f469-409e-a937-6aa2a20be503>\",\"WARC-Concurrent-To\":\"<urn:uuid:708146c4-0ab7-4d5d-80fe-929afe0c5ec4>\",\"WARC-IP-Address\":\"23.227.38.74\",\"WARC-Target-URI\":\"https://old.lucentinnovation.com/blogs/technology-posts/working-with-arrays-in-javascript\",\"WARC-Payload-Digest\":\"sha1:JTJSE7H2YR3244UKYT7N3FYF775FQFAZ\",\"WARC-Block-Digest\":\"sha1:ZA22V2IQZISRVWVXRS3QIT556ZASTOJP\",\"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-20231001173415-20231001203415-00365.warc.gz\"}"}
http://ijfs.usb.ac.ir/article_4790.html	[ "", null, "# Fuzzy universal algebras on \\$L\\$-sets\n\nDocument Type: Research Paper\n\nAuthors\n\nDepartment of Mathematics, Ocean University of China, 238 Songling Road, 266100, Qingdao, P.R.China\n\nAbstract\n\nThis paper attempts to generalize universal algebras on classical sets to \\$L\\$-sets when \\$L\\$ is a GL-quantale. Some basic notions of fuzzy universal algebra on an \\$L\\$-set are introduced, such as subalgebra, quotient algebra, homomorphism, congruence, and direct product etc. The properties of them are studied. \\$L\\$-valued power algebra is also introduced and it is shown there is an onto homomorphism from \\$P(A)/R^{+}\\$ to \\$P(A/R)\\$ for any congruence \\$R\\$ on \\$L\\$-set \\$A\\$.\n\nKeywords\n\n### References\n\n C. Brink, Power structures and logic, Queastions Mathematicae, 9 (1986), 69–94.\n C. Brink, Power structures, Algebra Universalis, 30 (1993), 177–216.\n R. B˘elohl´avek, V. Vychodil, Fuzzy equational logic, Archive for Mathematical Logic, 41 (2002), 83–90.\n R. B˘elohl´avek, V. Vychodil, Algebras with fuzzy equalities, Fuzzy Sets and Systems, 157 (2006), 161–201.\n I. Bo˘snjak, R. Madar´az, S. Roz´alia, Good quotient relations and power algebras, Novi Sad Journal of Mathematics,\n29 (1999), 71–84.\n I. Bo˘snjak, R. Madar´asz, Power algebras and generalized quotient algebras, Algebra Universalis, 45 (2001), 179–189.\n I. Boˇsnjak, R. Madar´asz, G. Vojvodi´c, Algebras of fuzzy sets, Fuzzy Sets and Systems, 160 (2009), 2979–2988.\n I. Boˇsnjak, R. Madar´asz, On the composition of fuzzy power relations, Fuzzy Sets and Systems, 271 (2015), 81–87.\n A. B. Chakraborty, S. S. Khare, Fuzzy homomorphism and algebraic structures, Fuzzy Sets and Systems, 59 (1993),\n211–221.\n C. L. Chang, Fuzzy topological spaces, Journal of Mathematical Analysis and Applications, 24 (1968), 182–190.\n M. Demirci, Foundations of fuzzy functions and vague algebra based on many-valued equivalence relations, part I:\nfuzzy functions and their applications, International Journal of General Systems, 32(2) (2003), 123–155.\n M. Demirci, Foundations of fuzzy functions and vague algebra based on many-valued equivalence relations, part II:\nvague algebraic notions, International Journal of General Systems, 32(2) (2003), 157–175.\n M. Demirci, Foundations of fuzzy functions and vague algebra based on many-valued equivalence relations, part III:\nconstructions of vague algebraic notions and vague arithmetic operations, International Journal of General Systems,\n32(2) (2003), 177–201.\n G. Georgescu, Fuzzy power structures, Archive for Mathematical Logic, 47 (2008), 233–261.\n G. Gr¨atzer, Universal Algebra, 2nd ed. Springer-Verlag, 1979.\n U. H¨ohle, Commutative, residuated l-monoids, in: U. H¨ohle, E.P. Klement(Eds.), Non-Classical Logics and Their\nApplications to Fuzzy Subsets: A Handbook on the Mathematical Foundations of Fuzzy Set Theory, Kluwer Academic\nPublishers, Dordrecht, 1995, 53–105.\n J. Ignjatovi´ca, M. ´ Ciri´ca, S. Bogdanovi´cb, Fuzzy homomorphisms of algebras, Fuzzy Sets and Systems, 160 (2009), 2345–2365.\n H. Lai, D. Zhang, Good fuzzy preorders on fuzzy power structures, Archive for Mathematical Logic, 49 (2010),\n469–489.\n\n F. Li, Y. Yue, L-valued fuzzy rough sets, Iranian Journal of Fuzzy Systems, 16 (2019), 111-127.\n R. Madar´az, Remarks on power structures, Algebra Universalis, 34 (1995), 179–184.\n V. Murali, Fuzzy congruence relations, Fuzzy Sets and Systems, 41 (1991), 359–369.\n B. Pang, F. G,Shi, Fuzzy counterparts of hull operators and interval operators in the framework of L-convex spaces,\nFuzzy Sets and Systems, 2018, in press, DOI: http://doi.org/10.1016/j.fss.2018. 05. 012.\n B. Pang, Y. Zhao, Characterizations of L-convex spaces, Iranian Journal of Fuzzy Systems, 13(4) (2016), 51–61.\n Q. Pu, D. Zhang, Preordered sets valued in a GL-monoid, Fuzzy Sets and Systems, 187 (2012), 1–32.\n M. A. Samhan, Fuzzy quotient algebras and fuzzy factor congruences, Fuzzy Sets and Systems, 73 (1995), 269–277.\n L. Shen, Adjunctions in Quantaloid-enriched Categories, Ph.D Thesis, Sichuan University 2014.\n Z. Xiu, B. Pang, Base axioms and subbase axioms in M-fuzzifying convex spaces, Iranian Journal of Fuzzy Systems,\n15(2) (2018), 75–87.\n W. Yao, Y. She, L. Lu, Metric-based L-fuzzy rough sets: Approximation operators and de nable sets, Knowledge-\nBased Systems, 163 (2019), 91-102.\n L. A. Zadeh, Fuzzy sets, Information and Control, 8 (1965), 338–353." ]	[ null, "http://ijfs.usb.ac.ir/data/ijfs/coversheet/head_en.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7135285,"math_prob":0.8156631,"size":3769,"snap":"2019-43-2019-47","text_gpt3_token_len":1261,"char_repetition_ratio":0.15856573,"word_repetition_ratio":0.10594796,"special_character_ratio":0.3592465,"punctuation_ratio":0.28455284,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97560084,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T10:53:55Z\",\"WARC-Record-ID\":\"<urn:uuid:5ccf6e80-b939-4b1e-a4ad-f35db3784e17>\",\"Content-Length\":\"43720\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4da030c0-b477-4788-8869-512ceed20d32>\",\"WARC-Concurrent-To\":\"<urn:uuid:b07b78d6-bdd5-4ea0-980b-ba609fd4e257>\",\"WARC-IP-Address\":\"80.191.172.29\",\"WARC-Target-URI\":\"http://ijfs.usb.ac.ir/article_4790.html\",\"WARC-Payload-Digest\":\"sha1:4LF3YC54D7VG7O54JN2TCM4T7IESXDJG\",\"WARC-Block-Digest\":\"sha1:M3F5IFAB4XPFTMTQIOWWZYFF7UM6TSUC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668618.8_warc_CC-MAIN-20191115093159-20191115121159-00099.warc.gz\"}"}
https://gmatclub.com/forum/during-a-certain-game-after-each-turn-a-player-s-points-are-doubled-273984.html	[ "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 18 Oct 2019, 06:43", null, "GMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nNot interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.", null, "", null, "During a certain game, after each turn, a player’s points are doubled\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\nHide Tags\n\nMath Expert", null, "V\nJoined: 02 Sep 2009\nPosts: 58453\nDuring a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\n13", null, "00:00\n\nDifficulty:", null, "", null, "", null, "95% (hard)\n\nQuestion Stats:", null, "24% (01:57) correct", null, "76% (02:08) wrong", null, "based on 63 sessions\n\nHideShow timer Statistics\n\nDuring a certain game, after each turn, a player’s points are doubled and then reduced by 80. Players can only have a whole number of points and take turns until their score reaches exactly zero. Assuming no other points are gained or earned, how many points did player A start with?\n\n(1) Player A takes exactly 4 turns.\n\n(2) Player A’s starting score was not a multiple of 2.\n\n_________________\nMath Expert", null, "V\nJoined: 02 Aug 2009\nPosts: 7978\nDuring a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\n2\nDuring a certain game, after each turn, a player’s points are doubled and then reduced by 80. Players can only have a whole number of points and take turns until their score reaches exactly zero. Assuming no other points are gained or earned, how many points did player A start with?\n\nLet the initial point be x, so the points after one turn $$2x-80=2^1x-801=2^1x-80(2^1-1)$$ and the next $$2(2x-80)-80=4x-240=2^2x-803=2^2x-80(2^2-1)$$and so, the nth turn $$= 2^nx-80(2^n-1)$$\n\n(1) Player A takes exactly 4 turns.\nso after 4th turn points = 0..\nso two ways\na) 4th turn - $$2^4x-80(2^4-1)=0.........16x-8015=0......16x=8015......x=515=75$$\nb) work backwards - after 4th turn - 0, so before 4th turn - 0+80/2=40, before 3rd turn - (40+80)/2=60, before 2nd turn - (60+80)/2=70, before first turn - (70+80)/2=75\nsufficient\n\n(2) Player A’s starting score was not a multiple of 2.\nA straight logical answer is YES..\nsince we have a term being multiplied by 2, there will be only one term which will be a non-multiple of 2, because a term prior to it will be a decimal.\n\nother way..\nnth term = $$2^nx-80(2^n-1)=0........2^n(80-x)=80$$\nx is an ODD integer, and 80-x has to be a multiple of 5, so 80-x can be 5, 15 , 25 ....\nOnly $$2^n5 = 80$$ when n is 4, so $$80-x=5...x=80-5=75$$\n\nwork backwards -\nafter last turn - 0,\nso before that turn - $$\\frac{0+80}{2}=40$$,\nbefore that turn - $$\\frac{(40+80)}{2}=60$$,\nbefore that turn - $$\\frac{(60+80)}{2}=70$$,\nbefore that turn - $$\\frac{(70+80)}{2}=75$$\nbefore that - $$\\frac{(75+80)}{2}=77.5$$... but this is not a whole number\nso he started with 75\nsufficient\n\nD\n_________________\nManager", null, "", null, "B\nJoined: 09 Oct 2015\nPosts: 226\nRe: During a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\nis there any other way to solve statement b ?\nMath Expert", null, "V\nJoined: 02 Aug 2009\nPosts: 7978\nRe: During a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\n1\nrahulkashyap wrote:\nis there any other way to solve statement b ?\n\nA straight logical answer is YES..\nsince we have a term being multiplied by 2, there will be only one term which will be a non-multiple of 2, because a term prior to it will be a decimal.\n\nother way..\nnth term = $$2^nx-80(2^n-1)=0........2^n(80-x)=80$$\nx is an ODD integer, and 80-x has to be a multiple of 5, so 80-x can be 5, 15 , 25 ....\nOnly $$2^n5 = 80$$ when n is 4, so $$80-x=5...x=80-5=75$$\n_________________", null, "S\nJoined: 06 Mar 2017\nPosts: 189\nConcentration: Operations, General Management\nRe: During a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\nBunuel wrote:\nDuring a certain game, after each turn, a player’s points are doubled and then reduced by 80. Players can only have a whole number of points and take turns until their score reaches exactly zero. Assuming no other points are gained or earned, how many points did player A start with?\n\n(1) Player A takes exactly 4 turns.\n\n(2) Player A’s starting score was not a multiple of 2.\n\nBunuel VeritasKarishma Pls help to solve statement B.\nVeritas Prep GMAT Instructor", null, "V\nJoined: 16 Oct 2010\nPosts: 9706\nLocation: Pune, India\nRe: During a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\n1\nsiddreal wrote:\nBunuel wrote:\nDuring a certain game, after each turn, a player’s points are doubled and then reduced by 80. Players can only have a whole number of points and take turns until their score reaches exactly zero. Assuming no other points are gained or earned, how many points did player A start with?\n\n(1) Player A takes exactly 4 turns.\n\n(2) Player A’s starting score was not a multiple of 2.\n\nBunuel VeritasKarishma Pls help to solve statement B.\n\nWhen the player completes the game (reaches a score of 0), we know what if his number of points before that were N,\n\n2N - 80 = 0\nN = 40\n\nNow, can we say what the score was in the previous turn? Sure. We got 40 by doubling the previous score and subtracting 80.\n\n2M - 80 = 40\nM = 60\n\nNow, can we say what the score was in the previous turn? Sure. We got 60 by doubling the previous score and subtracting 80.\n\n2L - 80 = 60\nL = 70\n\nNow, can we say what the score was in the previous turn? Sure. We got 70 by doubling the previous score and subtracting 80.\n\n2K - 80 = 70\nK = 75\n(not an even score)\n\nNow, can we say what the score was in the previous turn? Sure. We got 75 by doubling the previous score and subtracting 80.\n\n2J - 80 = 75\nJ = 77.5\n\nThe reason we did not need to do all this calculation backwards was this - Every time, we are adding 80 to the current score and dividing by 2. As long as our current score is even,\n(Even + 80) = Even, so on dividing by 2 we get an integer.\nThe moment out current score becomes odd (75 in our case), the previous score will not be an integer because Odd + 80 = Odd. So on dividing by 2, we get a decimal. But every score must be an integer. This tells us that there is only one odd score that anyone can have and that is in the beginning of the game and it is 75 only (as calculated above). But since this is a DS question, we didn't really need to do the calculation as long as we know that we will get a single unique value of the starting score.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\nNon-Human User", null, "Joined: 09 Sep 2013\nPosts: 13257\nRe: During a certain game, after each turn, a player’s points are doubled [#permalink]\n\nShow Tags\n\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________", null, "Re: During a certain game, after each turn, a player’s points are doubled [#permalink] 11 Sep 2019, 11:11\nDisplay posts from previous: Sort by\n\nDuring a certain game, after each turn, a player’s points are doubled\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne", null, "", null, "" ]	[ null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/profile/close.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/profile/close.png", null, "https://gmatclub.com/forum/styles/gmatclub_light/theme/images/search/close.png", null, "https://cdn.gmatclub.com/cdn/files/forum/images/avatars/upload/avatar_73391.jpg", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_play.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_difficult_blue.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_difficult_blue.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_difficult_blue.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_separator.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_separator.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/timer_separator.png", null, "https://cdn.gmatclub.com/cdn/files/forum/images/no_avatar.svg", null, "https://cdn.gmatclub.com/cdn/files/forum/images/ranks/rank_phpbb_3.svg", null, "https://cdn.gmatclub.com/cdn/files/forum/images/no_avatar.svg", null, "https://cdn.gmatclub.com/cdn/files/forum/images/no_avatar.svg", null, "https://cdn.gmatclub.com/cdn/files/forum/images/avatars/gallery/Galery1/pirat4.gif", null, "https://cdn.gmatclub.com/cdn/files/forum/images/avatars/upload/avatar_116290.jpg", null, "https://cdn.gmatclub.com/cdn/files/forum/images/avatars/upload/avatar_362269.gif", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/viewtopic/posts_bot.png", null, "https://www.facebook.com/tr", null, "https://www.googleadservices.com/pagead/conversion/1071875456/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96582055,"math_prob":0.9646415,"size":2112,"snap":"2019-43-2019-47","text_gpt3_token_len":559,"char_repetition_ratio":0.14326376,"word_repetition_ratio":0.14184397,"special_character_ratio":0.2817235,"punctuation_ratio":0.095132746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99309254,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T13:43:11Z\",\"WARC-Record-ID\":\"<urn:uuid:57896afd-5b9b-46ba-972b-197de6824b5a>\",\"Content-Length\":\"848700\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9c615c9-58e2-4988-bbda-31841e4cb255>\",\"WARC-Concurrent-To\":\"<urn:uuid:31348a2c-d695-494f-9a1d-5f70c89eaab9>\",\"WARC-IP-Address\":\"198.11.238.98\",\"WARC-Target-URI\":\"https://gmatclub.com/forum/during-a-certain-game-after-each-turn-a-player-s-points-are-doubled-273984.html\",\"WARC-Payload-Digest\":\"sha1:KDFGVY2NVYCFB2J2LYRIXFDYNM33RJTE\",\"WARC-Block-Digest\":\"sha1:IMVK4LL3LSDPYK7UGA3M3BE4ESPKKXUK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986682998.59_warc_CC-MAIN-20191018131050-20191018154550-00135.warc.gz\"}"}
https://www.sanfoundry.com/hash-tables-quadratic-probing-multiple-choice-questions-answers-mcqs/	[ "# Hash Tables with Quadratic Probing Multiple Choice Questions and Answers (MCQs)\n\n«\n»\n\nThis set of Data Structures & Algorithms Multiple Choice Questions & Answers (MCQs) focuses on “Hash Tables with Quadratic Probing”.\n\n1. Which of the following schemes does quadratic probing come under?\na) rehashing\nb) extended hashing\nc) separate chaining\n\nExplanation: Quadratic probing comes under open addressing scheme to resolve collisions in hash tables.\n\n2. Quadratic probing overcomes primary collision.\na) True\nb) False\n\nExplanation: Quadratic probing can overcome primary collision that occurs in linear probing but a secondary collision occurs in quadratic probing.\n\n3. What kind of deletion is implemented by hashing using open addressing?\na) active deletion\nb) standard deletion\nc) lazy deletion\nd) no deletion\n\nExplanation: Standard deletion cannot be performed in an open addressing hash table, because the cells might have caused collision. Hence, the hash tables implement lazy deletion.\n\n4. In quadratic probing, if the table size is prime, a new element cannot be inserted if the table is half full.\na) True\nb) False\n\nExplanation: In quadratic probing, if the table size is prime, we can insert a new element even though table is exactly half filled. We can’t insert element if table size is more than half filled.\n\n5. Which of the following is the correct function definition for quadratic probing?\na) F(i)=i2\nb) F(i)=i\nc) F(i)=i+1\nd) F(i)=i2+1\n\nExplanation: The function of quadratic probing is defined as F(i)=i2. The function of linear probing is defined as F(i)=i.\n\n6. How many constraints are to be met to successfully implement quadratic probing?\na) 1\nb) 2\nc) 3\nd) 4\n\nExplanation: 2 requirements are to be met with respect to table size. The table size should be a prime number and the table size should not be more than half full.\n\n7. Which among the following is the best technique to handle collision?\nb) Linear probing\nc) Double hashing\nd) Separate chaining\n\nExplanation: Quadratic probing handles primary collision occurring in the linear probing method. Although secondary collision occurs in quadratic probing, it can be removed by extra multiplications and divisions.\n\n8. Which of the following techniques offer better cache performance?\nb) Linear probing\nc) Double hashing\nd) Rehashing\n\nExplanation: Linear probing offers better cache performance than quadratic probing and also it preserves locality of reference.\n\n9. What is the formula used in quadratic probing?\na) Hash key = key mod table size\nb) Hash key=(hash(x)+F(i)) mod table size\nc) Hash key=(hash(x)+F(i2)) mod table size\nd) H(x) = x mod 17\n\nExplanation: Hash key=(hash(x)+F(i2)) mod table size is the formula for quadratic probing. Hash key = (hash(x)+F(i)) mod table size is the formula for linear probing.\n\n10. For the given hash table, in what location will the element 58 be hashed using quadratic probing?\n\n 0 49 1 2 3 4 5 6 7 8 18 9 89\n\na) 1\nb) 2\nc) 7\nd) 6\n\nExplanation: Initially, 58 collides at position 8. Another collision occurs one cell away. Hence, F(i2)=4. Using quadratic probing formula, the location is obtained as 2.\n\nSanfoundry Global Education & Learning Series – Data Structure.\n\nTo practice all areas of Data Structure, here is complete set of 1000+ Multiple Choice Questions and Answers.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20150%20150%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7625901,"math_prob":0.9703844,"size":3859,"snap":"2022-05-2022-21","text_gpt3_token_len":973,"char_repetition_ratio":0.16420233,"word_repetition_ratio":0.0989011,"special_character_ratio":0.24177247,"punctuation_ratio":0.113667116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T03:07:08Z\",\"WARC-Record-ID\":\"<urn:uuid:75237a8d-76cb-48ee-b287-ddbe48af1cad>\",\"Content-Length\":\"152227\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:13f41589-b50e-4402-9f52-1634954cdaca>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ef33406-aa56-4d38-9dee-3557c9fe0ec1>\",\"WARC-IP-Address\":\"104.25.132.119\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/hash-tables-quadratic-probing-multiple-choice-questions-answers-mcqs/\",\"WARC-Payload-Digest\":\"sha1:EDCB3ZKHNJRU5436F6P52ZM7RUQNLTVM\",\"WARC-Block-Digest\":\"sha1:ULOC3ZH5LS7CFHYLCELVNP3GELG54MOI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522741.25_warc_CC-MAIN-20220519010618-20220519040618-00166.warc.gz\"}"}
https://www.cee-m.fr/?publication=a-mixed-control-problem-of-the-management-of-natural-resources	[ "# A mixed control problem of the management of natural resources\n\n• Home\n• A mixed control problem of the management of natural resources", null, "7 December 2015", null, "By CEE-M\n\nIn this paper, we study a mixed optimal control problem in which both continuous controls and impulse controls are admissible. The optimal solution of this problem can be characterized via the Hamilton-Jacobi-Bellman equation. However, the resolution of this equation can be cumbersome and in most cases only numerical solutions are possible. This is why we consider an example with particular functional forms. Building on the solutions of the pure continuous control and the pure impulse control problem, we propose a candidate for the optimal solution of the mixed control problem. We prove that this candidate verifies the Hamilton-Jacobi-Bellman equation. As conjectured by Clark, jumping to the steady state and then staying there is one possible optimal solution, but we show that the profit functions of the continuous control and impulse control sub-models need to be related in a particular manner to reach such solution. Although the above results were obtained with particular functional forms, we are working to prove the optimality of Clark’s policy for general functional forms, maintaining the relationship between the profit functions" ]	[ null, "https://www.cee-m.fr/wp-content/themes/ceem-theme/resources/img/Icone_calendar.png", null, "https://www.cee-m.fr/wp-content/themes/ceem-theme/resources/img/Icone_user.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9223335,"math_prob":0.9345003,"size":1239,"snap":"2022-40-2023-06","text_gpt3_token_len":229,"char_repetition_ratio":0.14898786,"word_repetition_ratio":0.0,"special_character_ratio":0.17352703,"punctuation_ratio":0.06635071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97061926,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T19:41:47Z\",\"WARC-Record-ID\":\"<urn:uuid:fd854293-e95d-44e9-b234-f21037a291a3>\",\"Content-Length\":\"27004\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6da9a46b-3d87-4b55-ad36-a67b592af99b>\",\"WARC-Concurrent-To\":\"<urn:uuid:06c3b89e-c7c8-41a7-b9f4-22930fa4443f>\",\"WARC-IP-Address\":\"188.165.4.35\",\"WARC-Target-URI\":\"https://www.cee-m.fr/?publication=a-mixed-control-problem-of-the-management-of-natural-resources\",\"WARC-Payload-Digest\":\"sha1:7XGV4APY3S7VDROYAU5TNGLFP675FA2R\",\"WARC-Block-Digest\":\"sha1:OP6SQC3LH5JYBSRGZTOFV2LDDVMGWIOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500074.73_warc_CC-MAIN-20230203185547-20230203215547-00174.warc.gz\"}"}
https://webglfundamentals.org/webgl/lessons/zh_cn/webgl-3d-lighting-directional.html	[ "# WebGLFundamentals.org\n\nFix, Fork, Contribute\n\n# WebGL 三维方向光源\n\ndrag the points\n\nrotate the direction\n\n## 法向量\n\n``````function setNormals(gl) {\nvar normals = new Float32Array([\n// 正面左竖\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n\n// 正面上横\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n\n// 正面中横\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n0, 0, 1,\n\n// 背面左竖\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n\n// 背面上横\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n\n// 背面中横\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n0, 0, -1,\n\n// 顶部\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n\n// 上横右面\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n\n// 上横下面\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n\n// 上横和中横之间\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n\n// 中横上面\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n0, 1, 0,\n\n// 中横右面\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n\n// 中横底面\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n\n// 底部右侧\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n1, 0, 0,\n\n// 底面\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n0, -1, 0,\n\n// 左面\n-1, 0, 0,\n-1, 0, 0,\n-1, 0, 0,\n-1, 0, 0,\n-1, 0, 0,\n-1, 0, 0]);\ngl.bufferData(gl.ARRAY_BUFFER, normals, gl.STATIC_DRAW);\n}\n``````\n\n``````// 找顶点着色器中的属性\nvar positionLocation = gl.getAttribLocation(program, \"a_position\");\n-var colorLocation = gl.getAttribLocation(program, \"a_color\");\n+var normalLocation = gl.getAttribLocation(program, \"a_normal\");\n\n...\n\n-// 创建一个缓冲存储颜色\n-var colorBuffer = gl.createBuffer();\n-// 绑定到 ARRAY_BUFFER\n-gl.bindBuffer(gl.ARRAY_BUFFER, colorBuffer);\n-// 将几何数据放入缓冲\n-setColors(gl);\n\n+// 创建缓冲存储法向量\n+var normalBuffer = gl.createBuffer();\n+// 绑定到 ARRAY_BUFFER (可以看作 ARRAY_BUFFER = normalBuffer)\n+gl.bindBuffer(gl.ARRAY_BUFFER, normalBuffer);\n+// 将法向量存入缓冲\n+setNormals(gl);\n``````\n\n``````-// 启用颜色属性\n-gl.enableVertexAttribArray(colorLocation);\n-\n-// 绑定颜色缓冲\n-gl.bindBuffer(gl.ARRAY_BUFFER, colorBuffer);\n-\n-// 告诉颜色属性怎么从 colorBuffer (ARRAY_BUFFER) 中读取颜色值\n-var size = 3; // 每次迭代使用3个单位的数据\n-var type = gl.UNSIGNED_BYTE; // 单位数据类型是无符号 8 位整数\n-var normalize = true; // 标准化数据 (从 0-255 转换到 0.0-1.0)\n-var stride = 0; // 0 = 移动距离 * 单位距离长度sizeof(type) 每次迭代跳多少距离到下一个数据\n-var offset = 0; // 从绑定缓冲的起始处开始\n-gl.vertexAttribPointer(\n- colorLocation, size, type, normalize, stride, offset)\n\n+// 启用法向量属性\n+gl.enableVertexAttribArray(normalLocation);\n+\n+// 绑定法向量缓冲\n+gl.bindBuffer(gl.ARRAY_BUFFER, normalBuffer);\n+\n+// 告诉法向量属性怎么从 normalBuffer (ARRAY_BUFFER) 中读取值\n+var size = 3; // 每次迭代使用3个单位的数据\n+var type = gl.FLOAT; // 单位数据类型是 32 位浮点型\n+var normalize = false; // 单位化 (从 0-255 转换到 0-1)\n+var stride = 0; // 0 = 移动距离 * 单位距离长度sizeof(type) 每次迭代跳多少距离到下一个数据\n+var offset = 0; // 从绑定缓冲的起始处开始\n+gl.vertexAttribPointer(\n+ normalLocation, size, type, normalize, stride, offset)\n``````\n\n``````attribute vec4 a_position;\n-attribute vec4 a_color;\n+attribute vec3 a_normal;\n\nuniform mat4 u_matrix;\n\n-varying vec4 v_color;\n+varying vec3 v_normal;\n\nvoid main() {\n// 将位置和矩阵相乘\ngl_Position = u_matrix * a_position;\n\n- // 将颜色传到片断着色器\n- v_color = a_color;\n\n+ // 将法向量传到片断着色器\n+ v_normal = a_normal;\n}\n``````\n\n``````precision mediump float;\n\n// 从顶点着色器中传入的值\n-varying vec4 v_color;\n+varying vec3 v_normal;\n\n+uniform vec3 u_reverseLightDirection;\n+uniform vec4 u_color;\n\nvoid main() {\n+ // 由于 v_normal 是插值出来的，和有可能不是单位向量，\n+ // 可以用 normalize 将其单位化。\n+ vec3 normal = normalize(v_normal);\n+\n+ float light = dot(normal, u_reverseLightDirection);\n\n* gl_FragColor = u_color;\n\n+ // 将颜色部分（不包括 alpha）和光照相乘\n+ gl_FragColor.rgb = light;\n}\n``````\n\n`````` // 寻找全局变量\nvar matrixLocation = gl.getUniformLocation(program, \"u_matrix\");\n+ var colorLocation = gl.getUniformLocation(program, \"u_color\");\n+ var reverseLightDirectionLocation =\n+ gl.getUniformLocation(program, \"u_reverseLightDirection\");\n``````\n\n`````` // 设置矩阵\ngl.uniformMatrix4fv(matrixLocation, false, worldViewProjectionMatrix);\n\n+ // 设置使用的颜色\n+ gl.uniform4fv(colorLocation, [0.2, 1, 0.2, 1]); // green\n+\n+ // 设置光线方向\n+ gl.uniform3fv(reverseLightDirectionLocation, m4.normalize([0.5, 0.7, 1]));\n``````\n\n``````attribute vec4 a_position;\nattribute vec3 a_normal;\n\nuniform mat4 u_worldViewProjection;\n+uniform mat4 u_world;\n\nvarying vec3 v_normal;\n\nvoid main() {\n// 将位置和矩阵相乘\n* gl_Position = u_worldViewProjection * a_position;\n\n* // 重定向法向量并传递给片断着色器\n* v_normal = mat3(u_world) * a_normal;\n}\n``````\n\n`````` // 寻找全局变量\n* var worldViewProjectionLocation =\n* gl.getUniformLocation(program, \"u_worldViewProjection\");\n+ var worldLocation = gl.getUniformLocation(program, \"u_world\");\n``````\n\n``````// 设置矩阵\ngl.uniformMatrix4fv(\n* worldViewProjectionLocation, false,\n* worldViewProjectionMatrix);\ngl.uniformMatrix4fv(worldLocation, false, worldMatrix);\n``````\n\nclick to toggle normals\n\n``````attribute vec4 a_position;\nattribute vec3 a_normal;\n\nuniform mat4 u_worldViewProjection;\nuniform mat4 u_worldInverseTranspose;\n\nvarying vec3 v_normal;\n\nvoid main() {\n// 将位置和矩阵相乘\ngl_Position = u_worldViewProjection * a_position;\n\n// 重定向法向量并传递给片断着色器\n* v_normal = mat3(u_worldInverseTranspose) * a_normal;\n}\n``````\n\n``````- var worldLocation = gl.getUniformLocation(program, \"u_world\");\n+ var worldInverseTransposeLocation =\n+ gl.getUniformLocation(program, \"u_worldInverseTranspose\");\n``````\n\n``````var worldViewProjectionMatrix = m4.multiply(viewProjectionMatrix, worldMatrix);\nvar worldInverseMatrix = m4.inverse(worldMatrix);\nvar worldInverseTransposeMatrix = m4.transpose(worldInverseMatrix);\n\n// 设置矩阵\ngl.uniformMatrix4fv(\nworldViewProjectionLocation, false,\nworldViewProjectionMatrix);\n-gl.uniformMatrix4fv(worldLocation, false, worldMatrix);\n+gl.uniformMatrix4fv(\n+ worldInverseTransposeLocation, false,\n+ worldInverseTransposeMatrix);\n``````\n\n``````var m4 = {\ntranspose: function(m) {\nreturn [\nm, m, m, m,\nm, m, m, m,\nm, m, m, m,\nm, m, m, m,\n];\n},\n\n...\n``````\n\n### mat3(u_worldInverseTranspose) * a_normal 的可选方案\n\n```v_normal = mat3(u_worldInverseTranspose) * a_normal;\n```\n\n```v_normal = (u_worldInverseTranspose * vec4(a_normal, 0)).xyz;\n```" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.5477472,"math_prob":0.9988437,"size":7211,"snap":"2020-24-2020-29","text_gpt3_token_len":4247,"char_repetition_ratio":0.20743722,"word_repetition_ratio":0.29320988,"special_character_ratio":0.35931218,"punctuation_ratio":0.3429752,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95098484,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-01T11:59:53Z\",\"WARC-Record-ID\":\"<urn:uuid:8f6ab950-d8a8-4949-9f24-109607b012e5>\",\"Content-Length\":\"37999\",\"Content-Type\":\"application/http; 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https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_1	[ "# 2008 IMO Problems/Problem 1\n\n## Problem\n\nAn acute-angled triangle", null, "$ABC$ has orthocentre", null, "$H$. The circle passing through", null, "$H$ with centre the midpoint of", null, "$BC$ intersects the line", null, "$BC$ at", null, "$A_1$ and", null, "$A_2$. Similarly, the circle passing through", null, "$H$ with centre the midpoint of", null, "$CA$ intersects the line", null, "$CA$ at", null, "$B_1$ and", null, "$B_2$, and the circle passing through", null, "$H$ with centre the midpoint of", null, "$AB$ intersects the line", null, "$AB$ at", null, "$C_1$ and", null, "$C_2$. Show that", null, "$A_1$,", null, "$A_2$,", null, "$B_1$,", null, "$B_2$,", null, "$C_1$,", null, "$C_2$ lie on a circle.\n\n## Solution\n\nLet", null, "$M_A$,", null, "$M_B$, and", null, "$M_C$ be the midpoints of sides", null, "$BC$,", null, "$CA$, and", null, "$AB$, respectively. It's not hard to see that", null, "$M_BM_C\\parallel BC$. We also have that", null, "$AH\\perp BC$, so", null, "$AH \\perp M_BM_C$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that", null, "$H$ is on the radical axis of the circles centered at", null, "$M_B$ and", null, "$M_C$, so", null, "$A$ is too. We then have", null, "$AC_1\\cdot AC_2=AB_2\\cdot AB_1\\Rightarrow \\frac{AB_2}{AC_1}=\\frac{AC_2}{AB_1}$. This implies that", null, "$\\triangle AB_2C_1\\sim \\triangle AC_2B_1$, so", null, "$\\angle AB_2C_1=\\angle AC_2B_1$. Therefore", null, "$\\angle C_1B_2B_1=180^{\\circ}-\\angle AB_2C_1=180^{\\circ}-\\angle AC_2B_1$. This shows that quadrilateral", null, "$C_1C_2B_1B_2$ is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments", null, "$C_1C_2$ and", null, "$B_1B_2$. However, these are just the perpendicular bisectors of", null, "$AB$ and", null, "$CA$, which meet at the circumcenter of", null, "$ABC$, so the circumcenter of", null, "$C_1C_2B_1B_2$ is the circumcenter of triangle", null, "$ABC$. Similarly, the circumcenters of", null, "$A_1A_2B_1B_2$ and", null, "$C_1C_2A_1A_2$ are coincident with the circumcenter of", null, "$ABC$. The desired result follows." ]	[ null, "https://latex.artofproblemsolving.com/e/2/a/e2a559986ed5a0ffc5654bd367c29dfc92913c36.png ", null, "https://latex.artofproblemsolving.com/b/1/9/b1902d279ba37d60bdce4e0e987b7cd19d48974e.png ", null, "https://latex.artofproblemsolving.com/b/1/9/b1902d279ba37d60bdce4e0e987b7cd19d48974e.png ", null, "https://latex.artofproblemsolving.com/6/c/5/6c52a41dcbd739f1d026c5d4f181438b75b76976.png ", null, "https://latex.artofproblemsolving.com/6/c/5/6c52a41dcbd739f1d026c5d4f181438b75b76976.png ", null, "https://latex.artofproblemsolving.com/3/6/c/36c7f3f828cf12ab9cafeffad282468a95219600.png ", null, "https://latex.artofproblemsolving.com/e/3/a/e3a8ae66ded7f33f9385a37c9cadfb82720013d8.png ", null, "https://latex.artofproblemsolving.com/b/1/9/b1902d279ba37d60bdce4e0e987b7cd19d48974e.png ", null, "https://latex.artofproblemsolving.com/e/6/5/e657259c0995d2c36d46744bb4aa81471e6575ec.png ", null, 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https://www.codevscolor.com/kotlin-check-string-numeric/	[ "# Kotlin program to check if a string is numeric", null, "## Kotlin check if a string is numeric :\n\nIn this post, I will show you how to check if a string is numeric or not in Kotlin. For example, “1.2” is a numeric string but 1.2x is not. We will learn different ways to solve this problem.\n\n### Method 1: Using string.toDouble() :\n\nKotlin provides a couple of functions for the string class. We can parse a string as an integer or as a double. toInt() method parse one string as an integer and returns the result. Similarly, toDouble parse one string as Double.\n\nFor our problem, we will try to parse the string as double i.e. we will use toDouble method. This method throws one exception, NumberFormatException if the parsing fails. So, for a string, if this method can parse its value, we can say that this string is numeric. Take a look at the below example program :\n\n``````fun isNumber(str: String) = try {\nstr.toDouble()\ntrue\n} catch (e: NumberFormatException) {\nfalse\n}\n\nfun main() {\nval strArray = arrayOf(\"1\", \"-2\", \"0\", \"1.2\", \"-1.2\", \"xy\", \"+-1.2\")\n\nstrArray.forEach {\nprintln(\"\\$it => \\${isNumber(it)}\")\n}\n}``````\n\nHere, isNumber method takes one String as argument and returns one boolean. In the try block, it tries to convert the string to a double using toDouble method. If the conversion is successful, i.e. if it is a number, it returns true. Else, it catches one exception and returns false.\n\nThe above program will print the below output :\n\n``````1 => true\n-2 => true\n0 => true\n1.2 => true\n-1.2 => true\nxy => false\n+-1.2 => false``````\n\n### Method 2: Using toDoubleOrNull :\n\ntoDoubleOrNull is another version of toDouble. It converts one string to double, if the conversion is successful, it returns that converted value and if it fails, it returns null. So, we don’t have to use any try-catch block here :\n\n``````fun isNumber(str: String) = str.toDoubleOrNull()?.let { true } ?: false\n\nfun main() {\nval strArray = arrayOf(\"1\", \"-2\", \"0\", \"1.2\", \"-1.2\", \"xy\", \"+-1.2\")\n\nstrArray.forEach {\nprintln(\"\\$it => \\${isNumber(it)}\")\n}\n}``````\n\nIt will print the same output as the above program. Simpler and cleaner." ]	[ null, "data:image/png;base64,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", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6879524,"math_prob":0.9354898,"size":2375,"snap":"2020-24-2020-29","text_gpt3_token_len":610,"char_repetition_ratio":0.12779418,"word_repetition_ratio":0.121140145,"special_character_ratio":0.2825263,"punctuation_ratio":0.16901408,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97867715,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-15T07:35:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a858495e-599a-49b5-bbe9-195d82b46701>\",\"Content-Length\":\"103128\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37442af3-7dbd-4237-8859-8185dc22d262>\",\"WARC-Concurrent-To\":\"<urn:uuid:32baa5bd-5393-4f75-b261-cc2f057338e4>\",\"WARC-IP-Address\":\"35.175.60.16\",\"WARC-Target-URI\":\"https://www.codevscolor.com/kotlin-check-string-numeric/\",\"WARC-Payload-Digest\":\"sha1:LCMSIUU6NCOCCT5R5S2TJZ5435G3AK7F\",\"WARC-Block-Digest\":\"sha1:LHXP3MNTRJ2V6TA4N7CN4D65VJJEHNPA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657163613.94_warc_CC-MAIN-20200715070409-20200715100409-00567.warc.gz\"}"}
https://www.scienceforums.net/profile/30917-ahmeeeeeeeeeed/	[ "", null, "", null, "# ahmeeeeeeeeeed\n\nSenior Members\n\n39\n\n## Community Reputation\n\n0 Neutral\n\n• Rank\nQuark\n1. I may think that the limit (r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta)) has a problem of the sin^2 theta downwards , as it may become zero and the limit won't be zero / positive value any more but then it is 0/0 , so what to do next to know whether this (x/x) will turn eventually to be zero in the limit or not ?! And question two , is actually the question in the above comment : why does wolfram alpha say the value of the limit is zero despite the result of m/(1+m^2)^2\n2. Hello I have the limit lim (x^9 * y) / (x^6 + y^2)^2 (x,y)---> (0,0) when I use polar the final result is limit = lim (r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta)) r--->0 and substituting r = 0 , it will give zero * I tried it on wolfram alpha and it gave zero http://www.wolframalpha.com/input/?i=limit+%28x^9+y%29%2F%28x^6+%2By^2%29^2++as++%28x%2Cy%29+--%3E+%280%2C0%29 but when I use cartezean and try the path y = mx^3 the result turns to be m/(1+m^2)^2 which depends on m so what is it ?!\n3. hello About Einestien's train example I can't get what it really does with the constancy of the speed of light , what would be the difference if , instead of two bolts , we have two balls thrown at the same angle ?! the observer in the train would still see the front ball first Or we are using light because it is what determines simultaneity in our eyes , which means when I see two balls passing by me at different times , I don't have to conclude they weren't thrown simultaneously but when I see Light doing that I will ? I need basically to know what difference does the constancy of the speed of light make here thanks in advance ;\n4. thanks for response , I put photos of my text book just to let you see the question just as I see it , and for my notes I'll explain it as ( sorry , I was answering just as I would answer in the exam. the question is asking for the velocity of the pin (B) at the instant (phi) equals 45 degrees the pin is B , and is attached to the arm AB , and allowed to move just in the circular groove , the circular groove itself moves in the (i cap) direction with a constant velocity = 60 m/s as the X velocity is constant = 60 m/s , the circular groove and also the pin have it so what I'm seeking is the velocity of the pin tangential to the circle ... I treated this problem as a joint kinematices problem , so I devided it into two parts the first part , I'm treating the pin only as a part of arm( AB) , and so I used polar co-ordinates R cap and theta cap where V = dR/dt in the direction of R cap , and R omega in the direction of theta cap = (.3)* omega since the length of arm AB is constant then dR/dt = zero and the (theta cap) velocity is R * omega We also have the X velocity of the pin so V = (.3 * omega) theta cap + 60 (i cap) ----------------> equation one ----- secondly I treated the pin as part of the circular part we see that both the groove and the pin move in the positive x direction with the same speed , so we can neglect this speed and consider the motion of the pin just circular motion I chose intrinsic co-ordinates with ( tcap) tangential and (n cap) pointing towards the center , I could also use polar again with this . We see that V in the direction tangential in the only velocity compnent here , and it is what I seeking then I resuluted the unit vectors (t cap) and (i cap) in the directions of R cap and theta cap using the geomitry of the figure , inorder to equate like quantities , that resulted in the tangential speed equals (- 61.5) t cap which means opposite in sense to the unit vector (t cap) , which also means that the pin is going up ! .. I can't find this logical as i think under this situation the pin should go down the groove not up the groove and here came my question here . sorry I can't understand this \" putting off some of the heavy hitters in maths\" Yes , A is stationary , B is the pin , the question as asking for the tangential velcoity of the pin , but of cource getting the \\omega of the pin in the half -circular groove would give us the tangential speed .. I'm using polar co-ordinates for arm AB as i said above.. yes the x velocity of the pin is the same value (60 m/s) but I'm seeking the tangential velocity also. that's what I expect , but my way of answering showed that at the instant phi = 45 the tangential velocity well be up not down , showing that phi ( at that instance) is seeking to increase not decrease , which I think is wrong and that's why I asked the question from the beginning ... --------------------------- so what I am seeking now is would you please tell me how I could find the tangential velocity of the pin B at the instant phi = 45 degrees from the givings in the textbook ? I want both its magnitude and direction , and I really want to help me find where is the mistake in my calculations ? thanks in advance well this is not mine if this is what you mean http://www.scienceforums.net/topic/65674-crank-slider-mechanism-velocity-of-piston-angular-velocity-of-link-hp-needed/\n5. Hello , here is a problem http://www.flickr.com/photos/76599498@N07/7070964675/ when I solve it , the pin turns to be going upwards , which I can't find logical , what do you see ,please ? here's my answer http://www.flickr.com/photos/76599498@N07/7070966251/ http://www.flickr.com/photos/76599498@N07/7070966875/\n6. no here the threading of nuts at 3:45 , but I was asking about very tiny nuts , but I think no problem in manufacturing a small machine of the same type\n7. But it gave me the Y in terms of X , not the position in terms of time\n8. thanks, I found a vedio on youtube that die forging is the used method\n9. Hello at this problem the car is moving around the circle with constant speed , and it is asking for d^2R/dt^2 for the radar And d^2(theta)/dt^2 for the radar at the instance the speed of car is vertical downwards as shown the speed equals 120 km/hour And the radius of the circle is 450 so I said a® = (d^2R/dt^2) - R (omega^2) but the only acceleration is the central which equals (v^2/ Radius) and in the same direction so we can get d^2R/dt^2 . and when I try to get the (alpha)(d^2 theta / d t^2) a (theta) = R(alpha) + 2 (omega) (dR/dt) and a(theta) equals zero and here is my question , what is the value of dR/dt ??? I think it has to be zero , since the velocity is vertical and the unit vector R cap is horizontal so 1- Is Alpha zero ? 2- can dR/dT equal zero at that situation , while d^R/dt^2 has a value ? I think no problem but I wanted to make sure I'm right in this solution\n10. hello, a problem says http://www.flickr.com/photos/76599498@N07/6895404760/ I couldn't know how to start but that's what I did http://www.flickr.com/photos/76599498@N07/6895405486/ http://www.flickr.com/photos/76599498@N07/6895406078/ http://www.flickr.com/photos/76599498@N07/6895403992/ but I don't think this is the way , since I got the acceleration at a special point . I want to ask some questions 1-the acceleration magnitude is variable , due to the variation of the radius od curv. , isn't it ? If so , what is meant by determining the acceleration without specifing a certain moment ? 2- this problem is under the title ( intrinstic co-ordinates) , If I were to use these co-ordinates what is the key to start with ? I can think of the tangential unit vector to be the same as the direction of the derivative of Y with respect to X , but how to employ this ? 3- the radius of curv. at some points turned to be negative , is that ordinary or did I just did some calculation wrong ?\n11. Hello, A Question says Assign a proper manufacturing process for nuts and bolts Can I say die forging then threading or die casting then threading How are they mass produced ? and is there tiny lathes to thread nuts ?\n×" ]	[ null, "https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_pattern.png", null, "https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9309308,"math_prob":0.9045206,"size":8387,"snap":"2020-34-2020-40","text_gpt3_token_len":2192,"char_repetition_ratio":0.12883216,"word_repetition_ratio":0.030012771,"special_character_ratio":0.28305712,"punctuation_ratio":0.09116972,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9957568,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T04:40:01Z\",\"WARC-Record-ID\":\"<urn:uuid:9461694d-e4c5-46d1-a34d-44e45e3a5b80>\",\"Content-Length\":\"91573\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbcf8eb7-fa53-475b-abc9-ec4f6ba020ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:40528807-0c2c-4770-b7d2-7aa94ba62c70>\",\"WARC-IP-Address\":\"94.229.79.58\",\"WARC-Target-URI\":\"https://www.scienceforums.net/profile/30917-ahmeeeeeeeeeed/\",\"WARC-Payload-Digest\":\"sha1:CDS57SMEGO5DCHSILCVTTOY62VRJJ3I4\",\"WARC-Block-Digest\":\"sha1:5V2N2Q2GKNULHGCNI2FJACS7WUP3JNXR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737152.0_warc_CC-MAIN-20200807025719-20200807055719-00359.warc.gz\"}"}
https://www.sanfoundry.com/java-program-print-odd-elements-odd-index-number/	[ "# Java Program to Print Odd Elements at Odd Index\n\n«\n»\n\nThis is the Java Program to Print Odd Elements at Odd Index Number.\n\nProblem Description\n\nGiven an array of integers, print the odd numbers present at odd index numbers.\nExample:\n\nArray = [2,1,4,4,5,7]\n\nOutput = 1,7\n\nProblem Solution\n\nIterate through the array, and check whether the current index is even or odd. If it is odd then check the element at that index to be even or odd, print the element if it is odd.\n\nProgram/Source Code\nSanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!\n\nHere is the source code of the Java Program to Print Odd Elements at Odd Index Number. The program is successfully compiled and tested using IDE IntelliJ Idea in Windows 7. The program output is also shown below.\n\n1. ` `\n2. `//Java Program to Print Odd Elements at Odd Index Number`\n3. ` `\n4. `import java.io.BufferedReader;`\n5. `import java.io.InputStreamReader;`\n6. ` `\n7. `public class OddIndexedOddElements {`\n8. ` // Function to print Odd Elements present at Odd Index Number`\n9. ` static void printOddIndexedElements(int[] array){`\n10. ` int i=0;`\n11. ` for(i=0; i<array.length; i++){`\n12. ` if(i%2 != 0 && array[i]%2 !=0){`\n13. ` System.out.print(array[i] + \" \");`\n14. ` }`\n15. ` }`\n16. ` }`\n17. ` `\n18. ` // Function to read user input`\n19. ` public static void main(String[] args) {`\n20. ` BufferedReader br = new BufferedReader(new InputStreamReader(System.in));`\n21. ` int size;`\n22. ` System.out.println(\"Enter the size of the array\");`\n23. ` try {`\n24. ` size = Integer.parseInt(br.readLine());`\n25. ` } catch (Exception e) {`\n26. ` System.out.println(\"Invalid Input\");`\n27. ` return;`\n28. ` }`\n29. ` int[] array = new int[size];`\n30. ` System.out.println(\"Enter array elements\");`\n31. ` int i;`\n32. ` for (i = 0; i < array.length; i++) {`\n33. ` try {`\n34. ` array[i] = Integer.parseInt(br.readLine());`\n35. ` } catch (Exception e) {`\n36. ` System.out.println(\"An error occurred\");`\n37. ` return;`\n38. ` }`\n39. ` }`\n40. ` System.out.println(\"Output\");`\n41. ` printOddIndexedElements(array);`\n42. ` }`\n43. `}`\nProgram Explanation\n\n1. In function printOddIndexedElements(), the loop for(i=0; i<array.length; i++) iterates through the array.\n2. Then using the condition if(i%2 != 0 && array[i]%2 != 0), we first check whether the current index is odd.\n3. If the index is odd, then we check the element at that index (the condition after the && operator), if the element is odd, then display the element.\n\nTime Complexity: O(n) where n is the number of elements in the array.\n\nRuntime Test Cases\n```\nCase 1 (Positive test Case - having odd elements at odd index):\n\nEnter the size of the array\n6\nEnter array elements\n2\n1\n4\n4\n5\n7\nOutput\n1 7\n\nCase 2 (Negative test Case - no odd element present at odd index):\n\nEnter the size of the array\n5\nEnter array elements\n1\n2\n3\n4\n5\nOutput```\n\nSanfoundry Global Education & Learning Series – Java Programs.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20150%20150%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5726395,"math_prob":0.9534853,"size":2818,"snap":"2023-14-2023-23","text_gpt3_token_len":727,"char_repetition_ratio":0.13859275,"word_repetition_ratio":0.13449024,"special_character_ratio":0.28069553,"punctuation_ratio":0.18072289,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98662764,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T04:49:17Z\",\"WARC-Record-ID\":\"<urn:uuid:debf9009-dc91-494c-82e4-56c4ddd548a1>\",\"Content-Length\":\"145540\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f609460e-335f-4f11-993a-7751e95161c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:b99c2027-3899-471a-8d01-1b8e4551a3c7>\",\"WARC-IP-Address\":\"172.67.82.182\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/java-program-print-odd-elements-odd-index-number/\",\"WARC-Payload-Digest\":\"sha1:R4VNODEGTL2PUAKQVTMUOAYZDGBZF5IX\",\"WARC-Block-Digest\":\"sha1:LMPFGG2GNWUQBVXRV6HEQN2QZ35Q32MF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649439.65_warc_CC-MAIN-20230604025306-20230604055306-00701.warc.gz\"}"}
http://theblogreaders.com/tag/decimal/	[ "Aptitude\n\n## Decimal Shortcut Methods\n\n1% = 1/100 = 0.01 2% = 2/100 = 0.02 = 1/50 3% = 3/100 = 0.03 4% = 4/100 = 0.04 = 1/25" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8239559,"math_prob":1.0000097,"size":176,"snap":"2021-31-2021-39","text_gpt3_token_len":87,"char_repetition_ratio":0.20348836,"word_repetition_ratio":0.0,"special_character_ratio":0.64204544,"punctuation_ratio":0.13953489,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T02:00:43Z\",\"WARC-Record-ID\":\"<urn:uuid:f94014ef-5c15-418a-8bf6-c123be33eeba>\",\"Content-Length\":\"39550\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d66efb6e-1f60-4f7a-a86f-a1a8c2b982a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:73488b67-75f4-40f8-b1b4-504a57353ccb>\",\"WARC-IP-Address\":\"166.62.10.227\",\"WARC-Target-URI\":\"http://theblogreaders.com/tag/decimal/\",\"WARC-Payload-Digest\":\"sha1:CFWE5W7AOUPTLGK3FNF6A5U64MVGLSVV\",\"WARC-Block-Digest\":\"sha1:ASWO5CCNFLTSNLMNSBRRILP4FB4LDIRO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154042.23_warc_CC-MAIN-20210731011529-20210731041529-00689.warc.gz\"}"}
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/subtopic/decimals	[ "# Decimals Articles: Excel and Google Sheets\n\nAll resources related to Decimals for Excel and Google Sheets.\n\n### Decimals\n\nHow to Convert Decimals to Fraction – Excelchat\nWe can convert decimal to fraction and fraction to decimal by using the Number ribbon. This tutorial will guide all levels of Excel users on how to insert fraction in Excel. Figure 1: How to Convert Decimal to Fraction How to Put a Fraction in Excel To convert a..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82679415,"math_prob":0.70779854,"size":367,"snap":"2021-31-2021-39","text_gpt3_token_len":76,"char_repetition_ratio":0.15977961,"word_repetition_ratio":0.0,"special_character_ratio":0.21253406,"punctuation_ratio":0.13157895,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9922825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T11:43:04Z\",\"WARC-Record-ID\":\"<urn:uuid:07d5e0c8-e82c-491c-80af-5505a51dce09>\",\"Content-Length\":\"29332\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a597a6b7-558b-4cbc-9650-eabad36b2d94>\",\"WARC-Concurrent-To\":\"<urn:uuid:607addc0-4d11-4bb6-8962-05ed233794da>\",\"WARC-IP-Address\":\"99.84.176.104\",\"WARC-Target-URI\":\"https://www.got-it.ai/solutions/excel-chat/excel-tutorial/subtopic/decimals\",\"WARC-Payload-Digest\":\"sha1:AZTZSLUG3E67KYZCK55NMADONNCE5FJ5\",\"WARC-Block-Digest\":\"sha1:TRRK74NB3T4LIBHABNQJVI2UZGZ7YWSR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057421.82_warc_CC-MAIN-20210923104706-20210923134706-00423.warc.gz\"}"}
http://nullege.com/codes/search/DateTime.DateTime.Day	[ "Did I find the right examples for you? yes no Crawl my project Python Jobs\n\n# DateTime.DateTime.Day\n\nAll Samples(46) \| Call(46) \| Derive(0) \| Import(0)\n\n``` ui_class = 'greyed'\ndaylist.append(dict(total=formatTime(day_total),\nday_of_week=date.Day(),\nstyle=ui_class))\nelse:\ndaylist.append(dict(total=None, day_of_week=date.Day(),\n```\n\n``` while pos >= 0:\n# decrease pos when the day name is found\nif ref_date.Day() == day_name:\npos = pos - 1\n\n```\n```\n#cool, we find the day\nif ref_date.Day() == day_name and pos == 0:\nreturn ref_date\n\n```\n\n``` if day_total > 0:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5999713,"math_prob":0.9477768,"size":924,"snap":"2020-34-2020-40","text_gpt3_token_len":256,"char_repetition_ratio":0.16847827,"word_repetition_ratio":0.0,"special_character_ratio":0.24567099,"punctuation_ratio":0.21764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98229986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T00:26:13Z\",\"WARC-Record-ID\":\"<urn:uuid:360241b5-f1b0-4aa1-ba73-a820e06d75e4>\",\"Content-Length\":\"15135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57ff5b59-83db-4fdf-8f28-5530e0f09170>\",\"WARC-Concurrent-To\":\"<urn:uuid:1318b3f9-00c8-4f2d-ab31-0cb59dc11881>\",\"WARC-IP-Address\":\"159.89.157.124\",\"WARC-Target-URI\":\"http://nullege.com/codes/search/DateTime.DateTime.Day\",\"WARC-Payload-Digest\":\"sha1:QRMZQLGIE5CCR6DJ4Y7FSPNH5NIBQJTP\",\"WARC-Block-Digest\":\"sha1:QEHWTXNNDYMNEXGSVHQWYNJVVEKVZDEC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400208095.31_warc_CC-MAIN-20200922224013-20200923014013-00111.warc.gz\"}"}
https://www.vexorian.com/2012/05/google-codejam-round-2-yay.html	[ "## Saturday, May 26, 2012\n\n### Google Codejam round 2 - yay!\n\nI am just glad I managed to advance to round 3. Round 3 is going to be quite brutal for a guy that barely got the 467-th position, but still.\n\n## Problem A - Swinging Wild\n\nProblem statement\n\nThe first degree of difficulty was in understanding the statement. Let us simplify imagining there is a vine of length 0 at position D. We want to know if we can reach this last vine.\n\nLet us say we just reached vine i. There is a maximum length we can use in that vine. For Vine 0, it is d. For the later vines, this length depends on the position of the previous vine. So, if we used vine j to reach vine i, the maximum length we can use for vine i is min(length i, d[i] - d[j]).\n\nThe first idea is a dynamic programming one. f(i, j) returns whether it is possible to reach the last vine if you have just reached vine i and the previous one was j. Then you can get the maximum length you can use for vine i. This is the maximum distance between i and the next vine you use. Thus you can pick any of those vines within reach, and continue the recursion until i = the last one. This is enough to solve A-small.\n\nBut we want to solve A-large, which would not allow such a O(n^3) complexity. Let us instead define maxlength[i], the maximum length possible at which we can reach the i-th vine. You can see that this maximum depends only on the vines before i. Once we know this maximum length, we can use i to find possible lengths for later vines.\n\nFor example, for vine 0, the maximum length is d. Pick each vine j within d distance of vine 0, then a possible length for vine j is min( d[j] - d, length[j] ). After this step, the maximum length for vine 1 and vine 0 is already know, and we can use vine 1's maximum length to update more vines. This approach is O(n^2) in time complexity and needs only O(n) memory.\n\nint N; int d; int l; int D; int maxlen; bool solve() { d[N] = D; l[N] = 0; for (int i=0; i<=N; i++) { maxlen[i] = -1; } maxlen = d; for (int i=0; i<N; i++) { if (maxlen[i] > -1) { for (int j=i+1; j<=N; j++) { if (d[j] - d[i] <= maxlen[i]) { maxlen[j] = std::max( maxlen[j], std::min(d[j] - d[i], l[j]) ); } else { break; } } } } return (maxlen[N] > -1); }\n\nThis problem made me feel confident, because I solved it rather quickly. I was not expecting the next problems to each be very complicated :).\n\n## Problem C - Mountain view\n\nProblem statement\n\nIt seemed like C-small was less tricky than B-small, because of the accuracy rate from other competitors. So I first tried to solve this one. I actually barely got a integer programming solution, and I am still not sure I actually know how to code an integer programming solver.\n\nAfter the match, I found solutions that merely pick random numbers for the (at most 10) heights and verify that the answer is correct. Then you can output the correct answer. If after enough random attempts, no answer is found , it is impossible. The small number of peaks makes the probability to find a correct answer (if it exists) quite large.\n\nI feel so lame for not thinking of this.\n\n## Problem B - Aerobics\n\nProblem statement\n\nSo, my initial idea was to consider the circles as squares. And then the problem is just to place those squares. Somewhere in the rectangle. My theory was that if you sort the square lengths in non-ascending order, and then always placed each square in the closest valid position to the bottom-left, you would find a solution (The area is at least 5 times larger than the area of the circles, so it is quite unlikely this solution won't work). Indeed, even in cases where your circles have quite large radius, but the rectangle has only 1 as width, this is possible.\n\nYou can also verify that this approach needs only integral coordinates. Somehow, during the match I thought that it was possible to have 0.5 coordinates, and thus I multiply everything by 2 and other unnecessary things.\n\nI was pretty sure that would work, the issue is how to get the best location. With W,L <= 109, we cannot just try each coordinate for the center until we find one that does not intersect. So much that I even tried thinking of a different strategy. For a second, I thought of random (ironic as random would have helped in the other problem, but not this one).\n\nI wish I noticed the obvious solution to this predicament in less time than a whole hour: mundane binary search. For each square you want to place, binary search for the minimum x at which there is enough space to place it. How do you know if there is enough space at a given x? Simply use another binary search, but this time for y. Since you place the squares in order, and the previous squares are all together in the bottom-left position, both situations are binary search friendly.\n\nint N; int W, L; int x; int y; int r; pair<int,int> ri; // If the center of \"square\" i was (px,py), would it intersect with the // previously placed squares? bool intersect(int px, int py, int i) { for (int j=0; j<i; j++) { if ( (x[j] + r[ri[j].second] > px - r[ri[i].second]) && (y[j] + r[ri[j].second] > py - r[ri[i].second]) ) { return true; } } return false; } int gety(int px, int i) { int loy = -1; int hiy = L; while (loy+1 < hiy) { int hay = loy + (hiy - loy)/2; if (intersect(px,hay,i)) { loy = hay; } else { hiy = hay; } } return hiy; } void solve() { for (int i=0; i<N; i++) { ri[i] = make_pair(-r[i],i); } sort(ri, ri+N); x = 0; y = 0; for (int i=1; i<N; i++) { // Binary search for x int lox = -1; int hix = W; int picky; while (lox+1 < hix) { int hax = lox + (hix - lox)/2; int py = gety(hax, i); if (intersect(hax, py, i)) { lox = hax; } else { hix = hax; } } int py = gety(hix, i); // I used this assert, because maybe the strategy did not work, this // assert would alert me when running the large input... assert(! intersect(hix, py, i) ); x[i] = hix; y[i] = py; } for (int j=0; j<N; j++) { for (int i=0; i<N; i++) { if (ri[i].second == j) { cout << x[i] << \" \"<<y[i]; if (j < N - 1) { cout << \" \"; } } } } cout << endl; }\n\nThen I noticed that this approach works for B-large as well. So I just submitted it.\n\n## The last minutes\n\nBy the time I finished B, I had around 50 minutes left. But I had no idea what to do. Kept trying to think of something for C. Read D and then got baffled and tried to think of something too. There was a time at which I thought to solve D-small with a random solution. I just wish I was so creative with C...\n\nAs time progressed, my ranking got closer and closer to 500. The very last minute it eached 508 and then 520... Some people failed the large inputs in some problems (not lucky for them, but lucky for me). And I advanced to round 3.\n\nTo be honest I was not even expecting to be in the top 1000 today.\n\nI really liked this match. B , C and D were all really heavy weight problems in \"interesting-ness\" and difficulty and A was a good distraction.", null, "vexorian said...\n\nJust read the analysis and noticed that A allows you to go backwards. Yet my solutions don't go backwards. It is not a important fact for A-small, because the f(i,j) function can be turned into a graph easily. But for A-large it matters.\n\nWell, just need to proof that going backwards is never going to improve your chances to reach the best vine:\n\nImagine you are in vine i, and that you can go backwards and then return to increase the length of the i-th vine. If that is possible, then there is another vine k, that is reachable from vine i (with its current length) and that allows you to reach i at a larger length. But if k can be reached backwards from i, then there should be a path that allows you to directly go from 0 to k with a length as large or larger than needed. Ok... the proof goes like that.\n\nI was really lucky for not noticing that the statement allows you to go backwards :)", null, "carlos arias said...\n\nque buen analisis el que le haces a estos problemas del codeJam que nunca la pone facil...saludos" ]	[ null, "https://resources.blogblog.com/img/blank.gif", null, "https://resources.blogblog.com/img/blank.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92296636,"math_prob":0.9565718,"size":7810,"snap":"2021-43-2021-49","text_gpt3_token_len":2081,"char_repetition_ratio":0.10581604,"word_repetition_ratio":0.007920792,"special_character_ratio":0.28898847,"punctuation_ratio":0.13431986,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9907111,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T17:59:39Z\",\"WARC-Record-ID\":\"<urn:uuid:bcb382b5-c623-414a-8e89-3c3392d66cb2>\",\"Content-Length\":\"129493\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06933af9-6984-48b7-b74f-12b82c9c6c61>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c1e9a97-a7b0-4795-9791-44fb338947a6>\",\"WARC-IP-Address\":\"142.250.188.51\",\"WARC-Target-URI\":\"https://www.vexorian.com/2012/05/google-codejam-round-2-yay.html\",\"WARC-Payload-Digest\":\"sha1:X7JJETWQYFPYDK5VPRY3N2Z3J2JQO7B7\",\"WARC-Block-Digest\":\"sha1:3LAHFLY3DSLPU4RSBBFOUTJB3K7LBWYJ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358570.48_warc_CC-MAIN-20211128164634-20211128194634-00053.warc.gz\"}"}
https://forums.minecraftforge.net/profile/118546-diamondminer88/content/	[ "Search In\n• More options...\nFind results that contain...\nFind results in...", null, "", null, "DiamondMiner88\n\nMembers\n\n186\n\nEverything posted by DiamondMiner88\n\n1. Thank you very much, that got me off the ground really fast! The last things I would like to ask is, I managed to make an outline that shows though blocks, however the highlighted rectangle does not. EDIT: If i disable drawing the highlight aka #drawCube then the outline stops being able to be seen though blocks however is still seen if you have a direct path to it. no idea why this is happening And the drawing is locked to a block, it can't be not centered on a block. Is there a way around this? Heres an example: outline.mp4 Heres the code: @SubscribeEvent public static void renderWorldLastEvent(RenderWorldLastEvent event) { MatrixStack ms = event.getMatrixStack(); IRenderTypeBuffer.Impl buffers = IRenderTypeBuffer.getImpl(Tessellator.getInstance().getBuffer()); RenderSystem.disableCull(); ms.push(); for (Entity e : Minecraft.getInstance().world.getAllEntities()) { if (!Minecraft.getInstance().player.equals(e) && !(e instanceof ItemEntity) && !(e instanceof ExperienceOrbEntity)) { drawCube(ms, buffers, new AxisAlignedBB(e.getPosition()).expand(0, 1, 0), new Color(0, 255, 0, 127)); draw3dOutline(ms, buffers, new AxisAlignedBB(e.getPosition()).expand(0, 1, 0), new Color(255, 255, 255, 255)); } } draw3dOutline(ms, buffers, new AxisAlignedBB(new BlockPos(0, 10, 0)), new Color(255, 255, 255, 255)); ms.pop(); buffers.finish(); } public static void drawCube(MatrixStack ms, IRenderTypeBuffer buffers, AxisAlignedBB aabb, Color color) { draw3dRectangle(ms, buffers, aabb, color, \"TOP\"); draw3dRectangle(ms, buffers, aabb, color, \"BOTTOM\"); draw3dRectangle(ms, buffers, aabb, color, \"NORTH\"); draw3dRectangle(ms, buffers, aabb, color, \"EAST\"); draw3dRectangle(ms, buffers, aabb, color, \"SOUTH\"); draw3dRectangle(ms, buffers, aabb, color, \"WEST\"); } public static void draw3dRectangle(MatrixStack ms, IRenderTypeBuffer buffers, AxisAlignedBB aabb, Color color, String side) { int r = color.getRed(); int g = color.getGreen(); int b = color.getBlue(); int a = color.getAlpha(); double renderPosX = Minecraft.getInstance().getRenderManager().info.getProjectedView().getX(); double renderPosY = Minecraft.getInstance().getRenderManager().info.getProjectedView().getY(); double renderPosZ = Minecraft.getInstance().getRenderManager().info.getProjectedView().getZ(); ms.push(); ms.translate(aabb.minX - renderPosX, aabb.minY - renderPosY, aabb.minZ - renderPosZ); IVertexBuilder buffer = buffers.getBuffer(RenderType.makeType(HypixelClient.MODID + \":rectangle_highlight\", DefaultVertexFormats.POSITION_COLOR, GL11.GL_QUADS, 256, false, true, RenderType.State.getBuilder().transparency(ObfuscationReflectionHelper.getPrivateValue(RenderState.class, null, \"field_228515_g_\")).cull(new RenderState.CullState(false)).build(false))); Matrix4f mat = ms.getLast().getMatrix(); float x = (float) (aabb.maxX - aabb.minX); float y = (float) (aabb.maxY - aabb.minY); float z = (float) (aabb.maxZ - aabb.minZ); switch (side) { case \"TOP\": buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); break; case \"BOTTOM\": buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); break; case \"NORTH\": buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); break; case \"EAST\": buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); break; case \"SOUTH\": buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); break; case \"WEST\": buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); break; } ms.pop(); } public static void draw3dOutline(MatrixStack ms, IRenderTypeBuffer buffers, AxisAlignedBB aabb, Color color) { int r = color.getRed(); int g = color.getGreen(); int b = color.getBlue(); int a = color.getAlpha(); double renderPosX = Minecraft.getInstance().getRenderManager().info.getProjectedView().getX(); double renderPosY = Minecraft.getInstance().getRenderManager().info.getProjectedView().getY(); double renderPosZ = Minecraft.getInstance().getRenderManager().info.getProjectedView().getZ(); ms.push(); ms.translate(aabb.minX - renderPosX, aabb.minY - renderPosY, aabb.minZ - renderPosZ); RenderType.State glState = RenderType.State.getBuilder().line(new RenderState.LineState(OptionalDouble.of(1))).layer(ObfuscationReflectionHelper.getPrivateValue(RenderState.class, null, \"field_228500_J_\")).transparency(ObfuscationReflectionHelper.getPrivateValue(RenderState.class, null, \"field_228515_g_\")).writeMask(new RenderState.WriteMaskState(true, false)).depthTest(new RenderState.DepthTestState(GL11.GL_ALWAYS)).build(false); IVertexBuilder buffer = buffers.getBuffer(RenderType.makeType(HypixelClient.MODID + \":line_1_no_depth\", DefaultVertexFormats.POSITION_COLOR, GL11.GL_LINES, 128, glState)); Matrix4f mat = ms.getLast().getMatrix(); float x = (float) (aabb.maxX - aabb.minX); float y = (float) (aabb.maxY - aabb.minY); float z = (float) (aabb.maxZ - aabb.minZ); // Top edges buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); // Bottom edges buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); // Side edges buffer.pos(mat, x, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, 0).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, y, z).color(r, g, b, a).endVertex(); buffer.pos(mat, 0, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, 0, z).color(r, g, b, a).endVertex(); buffer.pos(mat, x, y, z).color(r, g, b, a).endVertex(); ms.pop(); }\n2. I already tested nearly all of them that returned a Vec3d, and got the same exact results as in the screenshot; just a line that won't stay in one position Even with the current calculation i get the exact same thing With some of the methods in the screenshot below the hand appears detached\n3. 1.12 is no longer supported on this forum. Please update to a modern version of Minecraft to receive support.\n4. bump\n5. I don't know why this doesn't work, it looks like the line is relative to the camera position, but it only shows if you're near 0,0,0 in the world @SubscribeEvent public static void renderWorldLastEvent(RenderWorldLastEvent event) { ClientPlayerEntity player = Minecraft.getInstance().player; GL11.glPushMatrix(); GL11.glPushAttrib(GL11.GL_ENABLE_BIT); double d0 = player.prevPosX + (player.getPosX() - player.prevPosX) * (double)event.getPartialTicks(); double d1 = player.prevPosY + (player.getPosY() - player.prevPosY) * (double)event.getPartialTicks(); double d2 = player.prevPosZ + (player.getPosZ() - player.prevPosZ) * (double)event.getPartialTicks(); Vec3d player_pos = new Vec3d(d0, d1, d2); GL11.glTranslated(-player_pos.x, -player_pos.y, -player_pos.z); GL11.glDisable(GL11.GL_LIGHTING); GL11.glDisable(GL11.GL_TEXTURE_2D); GL11.glDisable(GL11.GL_DEPTH_TEST); Vec3d blockA = new Vec3d (0,0,0); Vec3d blockB = new Vec3d (0,10,0); GL11.glColor4f(1,1,1,1); GL11.glBegin(GL11.GL_LINE_STRIP); GL11.glVertex3d(blockA.x, blockA.y, blockA.z); GL11.glVertex3d(blockB.x, blockB.y, blockB.z); GL11.glEnd(); GL11.glPopAttrib(); GL11.glPopMatrix(); } Its supposed to draw a line from 0,0,0 (that bedrock) to 0,10,0 but i get something like this with the line moving randomly as i move around\n6. I would like to make an block outline that's visible though other blocks, however i could not find a way to do it after searching for over 2 hours The closest I've gotten is this thread however that's for 1.11.2 and the renderBox method won't work in 1.15.2 I'm new to rendering stuff so i have no idea how it works either.\n7. Unless they re-wrote Minecraft to make it optimized they don't split workload across cores If they did they're definitely not gonna distribute it\n8. No i did use a type hierarchy. i use intellij\n9. Thank you! It took me a while to figure out but i found out that its not TextComponentString, its StringTextComponent\n10. Sorry im still at a loss. I can't find TextComponentString, which for all the examples i see is required\n11. I feel stupider with every message, how do i use ITextComponent? I cant find a way\n12. Sorry how do make a ServerPlayerEntity? And this is a client only mod so this won't go though to the server correct?\n13. Thanks it works now, but now im trying to send a chat message to the player that only the player can see. Im trying to use this post's 2nd comment. Now in 1.14 i can't find the EntityPlayer class.\n14. Sorry i haven't modded for a long time, so i forgot how to use Events. This doesn't work public static void clientChatEvent(ClientChatEvent event) { System.out.println(event.getMessage()); }\n15. Did you install the 1.12.2 installer?\n16. How do i intercept what the player sends though chat to a server? I'm trying to make a bot using baritone and if the player sends like %start i need to stop it from going though to the server.\n17. Thank you!\n18. Look in the same directory as where the installer is located for a log file that is generated after each time the installer is run. Please post its contents though Pastebin or a Github gist.\n19. Below DaemonUmbra's post is this text: For players asking for assistance with Forge please expand the spoiler below and read the appropriate section(s) in its/their entirety. Below that says \"Reveal hidden contents\" that expands and has a section titled Logs. Read its contents\n20. Wont even load the world for me to run it. Ill try that.\n21. So i accidentally spawned in 20000 or more items. I have no idea how to remove them. Is there a quick fix like using nbt explorer and removing all entity in world?\n22. How do i create JSON shaped recipes that include universal filled buckets? My fluid name is molten_glass_black with the MODID of character_mod\n23. IGNORE POST bump\n24. BTW there is already a mod for changing the menu through a json config file. Its called CustomMainMenu and there's a version for 1.12.2\n25. How do i create JSON shaped recipes that include universal filled buckets. My fluid name is molten_glass_black with the MODID of character_mod\n×\n\n• Activity\n\n×\n• Create New..." ]	[ null, "https://forums.minecraftforge.net/uploads/set_resources_2/84c1e40ea0e759e3f1505eb1788ddf3c_pattern.png", null, "data:image/svg+xml,%3Csvg%20xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg%22%20viewBox%3D%220%200%201024%201024%22%20style%3D%22background%3A%2389c462%22%3E%3Cg%3E%3Ctext%20text-anchor%3D%22middle%22%20dy%3D%22.35em%22%20x%3D%22512%22%20y%3D%22512%22%20fill%3D%22%23ffffff%22%20font-size%3D%22700%22%20font-family%3D%22-apple-system%2C%20BlinkMacSystemFont%2C%20Roboto%2C%20Helvetica%2C%20Arial%2C%20sans-serif%22%3ED%3C%2Ftext%3E%3C%2Fg%3E%3C%2Fsvg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5595311,"math_prob":0.8049663,"size":13572,"snap":"2022-05-2022-21","text_gpt3_token_len":3944,"char_repetition_ratio":0.2189711,"word_repetition_ratio":0.344207,"special_character_ratio":0.289051,"punctuation_ratio":0.29655838,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973099,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T08:04:00Z\",\"WARC-Record-ID\":\"<urn:uuid:9c824236-6ee0-470f-b085-5b8644a9690a>\",\"Content-Length\":\"138559\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3951564-ee7d-413c-a19a-b0bacb15459a>\",\"WARC-Concurrent-To\":\"<urn:uuid:7751d8fe-b6a4-460b-a893-cce8bff574ed>\",\"WARC-IP-Address\":\"51.79.83.165\",\"WARC-Target-URI\":\"https://forums.minecraftforge.net/profile/118546-diamondminer88/content/\",\"WARC-Payload-Digest\":\"sha1:GBCT4DPGGJXLPEXJE332E2LAZMC4KJ5S\",\"WARC-Block-Digest\":\"sha1:KJPCY4WDZKYDXKPY27CMP5VFUTXXVECA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305242.48_warc_CC-MAIN-20220127072916-20220127102916-00492.warc.gz\"}"}
https://www.esaral.com/q/an-inverted-cone-has-a-depth-of-43873	[ "", null, "# An inverted cone has a depth of\n\nQuestion:\n\nAn inverted cone has a depth of $40 \\mathrm{~cm}$ and a base of radius $5 \\mathrm{~cm}$. Water is poured into it at a rate of $1.5$ cubic centimetres per minute. Find the rate at which the level of water in the cone is rising when the depth is $4 \\mathrm{~cm}$.\n\nSolution:", null, "" ]	[ null, "https://www.facebook.com/tr", null, "https://www.esaral.com/media/uploads/2022/03/26/image91414.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8606836,"math_prob":0.99912935,"size":281,"snap":"2023-14-2023-23","text_gpt3_token_len":81,"char_repetition_ratio":0.15162455,"word_repetition_ratio":0.0,"special_character_ratio":0.29537368,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9882238,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T21:44:13Z\",\"WARC-Record-ID\":\"<urn:uuid:762cdd05-0102-478c-bb0f-346d0f1cddad>\",\"Content-Length\":\"22873\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e13b76a9-3c8a-40b2-8b54-c9b16d51aaf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:c6f2fc50-e012-4ecc-be7f-c01ca4013fa4>\",\"WARC-IP-Address\":\"104.21.61.187\",\"WARC-Target-URI\":\"https://www.esaral.com/q/an-inverted-cone-has-a-depth-of-43873\",\"WARC-Payload-Digest\":\"sha1:I753PEGVUI2CWY6HRHEOSZXK436664OP\",\"WARC-Block-Digest\":\"sha1:EZ2UGO6ORAKKUHMF53YJW36FW7Y4QK7C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648209.30_warc_CC-MAIN-20230601211701-20230602001701-00689.warc.gz\"}"}
https://whatisconvert.com/199-cubic-meters-in-imperial-teaspoons	[ "# What is 199 Cubic Meters in Imperial Teaspoons?\n\n## Convert 199 Cubic Meters to Imperial Teaspoons\n\nTo calculate 199 Cubic Meters to the corresponding value in Imperial Teaspoons, multiply the quantity in Cubic Meters by 281560.63782283 (conversion factor). In this case we should multiply 199 Cubic Meters by 281560.63782283 to get the equivalent result in Imperial Teaspoons:\n\n199 Cubic Meters x 281560.63782283 = 56030566.926744 Imperial Teaspoons\n\n199 Cubic Meters is equivalent to 56030566.926744 Imperial Teaspoons.\n\n## How to convert from Cubic Meters to Imperial Teaspoons\n\nThe conversion factor from Cubic Meters to Imperial Teaspoons is 281560.63782283. To find out how many Cubic Meters in Imperial Teaspoons, multiply by the conversion factor or use the Volume converter above. One hundred ninety-nine Cubic Meters is equivalent to fifty-six million thirty thousand five hundred sixty-six point nine two seven Imperial Teaspoons.\n\n## Definition of Cubic Meter\n\nThe cubic meter (also written \"cubic metre\", symbol: m3) is the SI derived unit of volume. It is defined as the volume of a cube with edges one meter in length. Another name, not widely used any more, is the kilolitre. It is sometimes abbreviated to cu m, m3, M3, m^3, m**3, CBM, cbm.\n\n## Definition of Imperial Teaspoon\n\nAn Imperial teaspoon (usually abbreviated tsp.) is a unit of volume in the Imperial System. It widely used in cooking recipes and pharmaceutic prescriptions. An imperial teaspoon equals 5.9 ml.\n\n## Using the Cubic Meters to Imperial Teaspoons converter you can get answers to questions like the following:\n\n• How many Imperial Teaspoons are in 199 Cubic Meters?\n• 199 Cubic Meters is equal to how many Imperial Teaspoons?\n• How to convert 199 Cubic Meters to Imperial Teaspoons?\n• How many is 199 Cubic Meters in Imperial Teaspoons?\n• What is 199 Cubic Meters in Imperial Teaspoons?\n• How much is 199 Cubic Meters in Imperial Teaspoons?\n• How many uk tsp are in 199 m3?\n• 199 m3 is equal to how many uk tsp?\n• How to convert 199 m3 to uk tsp?\n• How many is 199 m3 in uk tsp?\n• What is 199 m3 in uk tsp?\n• How much is 199 m3 in uk tsp?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8110468,"math_prob":0.9229977,"size":2069,"snap":"2021-21-2021-25","text_gpt3_token_len":547,"char_repetition_ratio":0.25714287,"word_repetition_ratio":0.12320917,"special_character_ratio":0.27694538,"punctuation_ratio":0.11557789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98394704,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-12T23:09:36Z\",\"WARC-Record-ID\":\"<urn:uuid:0050fa52-bd1b-4a8f-85d2-5fdeafd30e52>\",\"Content-Length\":\"32679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5a24387-694c-4f26-8803-c4cdfc8fe0ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbe9d5e3-323f-4c34-a863-20158ec43c41>\",\"WARC-IP-Address\":\"104.21.23.139\",\"WARC-Target-URI\":\"https://whatisconvert.com/199-cubic-meters-in-imperial-teaspoons\",\"WARC-Payload-Digest\":\"sha1:JCDD52XLYBANZJ2EWQC2H7PVNHVLJY66\",\"WARC-Block-Digest\":\"sha1:R2SUXLSCILCEE5QXTPOMKK567CF6C27M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991413.30_warc_CC-MAIN-20210512224016-20210513014016-00171.warc.gz\"}"}
https://scipost.org/submissions/2108.04781v1/	[ "# Single-Logarithmic Corrections to Small-$x$ Helicity Evolution\n\n### Submission summary\n\n As Contributors: Yossathorn Tawabutr Arxiv Link: https://arxiv.org/abs/2108.04781v1 (pdf) Date submitted: 2021-08-11 03:10 Submitted by: Tawabutr, Yossathorn Submitted to: SciPost Physics Proceedings Proceedings issue: DIS2021 Academic field: Physics Specialties: High-Energy Physics - Experiment High-Energy Physics - Phenomenology Nuclear Physics - Experiment Nuclear Physics - Theory Approach: Theoretical\n\n### Abstract\n\nThe small-$x$ quark helicity evolution equations at double-logarithmic order, with the kernel $\\sim\\alpha_s\\ln^2(1/x)$, have been derived previously. In this work, we derive the single-logarithmic corrections to the equations, to order $\\alpha_s\\ln(1/x)$ of the evolution kernel. The new equations include the effects of the running coupling and the unpolarized small-$x$ evolution, both of which are parametrically significant at single-logarithmic order. The large-$N_c$ and large-$N_c\\& N_f$ approximations to the equation are computed. (Here, $N_c$ and $N_f$ are the numbers of quark colors and flavors, respectively.) Their solutions will provide more precise estimates of the quark helicity distribution at small $x$, contributing to the resolution of the proton spin puzzle.\n\n###### Current status:\nEditor-in-charge assigned" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7514205,"math_prob":0.91903174,"size":1303,"snap":"2021-43-2021-49","text_gpt3_token_len":339,"char_repetition_ratio":0.11085451,"word_repetition_ratio":0.0,"special_character_ratio":0.23100537,"punctuation_ratio":0.13364056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928321,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T18:38:44Z\",\"WARC-Record-ID\":\"<urn:uuid:b02ea56c-d528-46c7-b5f1-d89016c9aec8>\",\"Content-Length\":\"22218\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5df382f-8d08-4f4b-8f57-eaa9be06b6f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:62ae0407-fffa-44af-a5cd-cf38dd1ce9a8>\",\"WARC-IP-Address\":\"142.93.224.252\",\"WARC-Target-URI\":\"https://scipost.org/submissions/2108.04781v1/\",\"WARC-Payload-Digest\":\"sha1:OXW557FFU3AKULDZR6S7WAXJ65GHRKD5\",\"WARC-Block-Digest\":\"sha1:KTREV3KEPXEXXOWCIQA3LDAAGT62KDTH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584913.24_warc_CC-MAIN-20211016170013-20211016200013-00023.warc.gz\"}"}
https://exploringphysics.com/about-the-app/eunit-2-electrical-circuits/	[ "### eUnit 2 Electrical Circuits", null, "Building on knowledge from previous units, students will build and analyze simple circuits, and compare and contrast series and parallel circuits. Students will calculate current, voltage and resistance in a circuit. The mathematical representation of the connection between voltage, current, and resistance will be determined. By the end of the unit, students will apply their knowledge to electrical widgets that they have in their household. By the end of the unit students will be able to answer the following questions:\n\n• What are the factors that affect the resistance of a resistor?\n• How does one use current and voltage to qualitatively predict the brightness of a bulb in a circuit?\n• How are voltage, current and resistance related?\n• How does one describe the voltage across a resistor and the current through a resistor in a series circuit?\n• How does one describe the voltage across a resistor and the current through a resistor in a parallel circuit?\n• How does one calculate the power expended by a resistor?\n• How does one calculate the energy expended by a resistor?\n\nBuy materials required for teaching eUnit 2 hands-on labs.\n\nSuggested timeline for teaching Unit 2\n\nBig Ideas\n1. Voltage is proportional to the electric current and the resistance of the resistor.\n2. The voltage between two points in a circuit is the same for every circuit element(s) connected between the same two points (parallel circuit).\n3. The current in a circuit is the same at any two points in a circuit if there are no junctions between those two points (series circuit).\n\nLearning Goals and Objectives\nBy the end of this unit, the students should be able to:\n\n1. Investigate working electrical circuits. (DOK3) a) Describe how electrical devices work. b) Use a multimeter to measure current, voltage and resistance.\n2. Describe an electrical circuit, using multiple representations (i.e., pictorial diagrams, verbal descriptions, schematics, and mathematical models, etc.). (DOK3) a) Create a verbal, schematic, pictorial or mathematical description of a physical electrical circuit. b) Construct an electrical circuit from the verbal, schematic, mathematical, or pictorial representation of a circuit. c) Describe verbally and mathematically the relationship between current, voltage and resistance.\n3. Design and conduct an experiment to determine how different electrodes and electrolytes affect the voltage produced by a wet-cell battery. (DOK4) a) Define voltage and measure a battery’s voltage. b) Identify the essential components of a wet-cell battery. c) Construct batteries using different combinations of electrodes and electrolytes. d) Measure the voltage of the batteries constructed.\n4. Design and conduct an experiment to determine the parameters that affect the resistance of a wire. (DOK4) a) Define resistance and measure it in an electrical circuit. b) Conduct experiments to determine how the length, the diameter, and the resistivity of a wire affect its resistance and develop a mathematical relationship. c) Apply the relationship verbally and mathematically.\n5. Design and conduct an experiment to determine the relationship between current, resistance and voltage within an electrical circuit. (DOK4) a) Define current and measure it in an electrical circuit. b) Design and conduct a quantitative experiment to determine a mathematical relationship among current, voltage, and resistance (Ohm’s Law, V=IR). c) Apply Ohm’s Law verbally and mathematically.\n6. Mathematically, graphically and physically determine voltage, current and resistance in a series circuit. (DOK3) a) Compare and contrast the characteristics of a one-bulb circuit and a two-bulb series circuit. b) Measure the current throughout a series circuit. c) Measure the voltage of the battery and across each resistor in a series circuit and obtain a relationship among them. d) Graphically represent changes in voltage and current in a series circuit. e) Measure the equivalent resistance of a series circuit, and obtain a mathematical relationship (RS = R1 + R2 + R3 + …..) . f) Calculate the voltage, current and resistance for various series circuits. g) Calculate the voltage, current and/or resistance for various parallel circuits. h) Predict the behaviors of various mixed series/parallel circuits and short circuits.\n7. Calculate the power and electrical energy used in a given electrical circuit. (DOK2) a) Relate voltage and current to power. b) Calculate the electrical power in watts, using the equation P=VI. c) Relate power to energy and time. d) Calculate the electrical energy in joules or kilowatt-hours for various circuits.\n8. Design and conduct experiments to study parameters that affect electrical circuits. (DOK4) a) Make observations. b) Operationally define the variables. c) Generate a hypothesis. d) List the steps of the procedure used to collect the data. e) Collect the data and organize it into a table. f) Convert metric units as needed. g) Graph and interpret the results. h) Develop conclusions based on results." ]	[ null, "http://exploringphysics.com/wp-content/uploads/2018/11/UNIT2_COVER_300px-186x300.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86493796,"math_prob":0.9920392,"size":5047,"snap":"2019-51-2020-05","text_gpt3_token_len":1018,"char_repetition_ratio":0.18342257,"word_repetition_ratio":0.1,"special_character_ratio":0.19694868,"punctuation_ratio":0.11225659,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99899334,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T18:26:40Z\",\"WARC-Record-ID\":\"<urn:uuid:3e48fc0f-73ce-41bc-9b26-87e3aea0015b>\",\"Content-Length\":\"48309\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e292299-69d1-406e-b099-c3ce1d7197f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:67ebfe92-cbac-48fb-8523-0b2c04aba1bc>\",\"WARC-IP-Address\":\"107.180.57.159\",\"WARC-Target-URI\":\"https://exploringphysics.com/about-the-app/eunit-2-electrical-circuits/\",\"WARC-Payload-Digest\":\"sha1:55QWXPVHP2DP5WMBZ2774JO726VWR6AM\",\"WARC-Block-Digest\":\"sha1:2WQQOMG5374KKUNAQY5BGNQIBLFI3QMN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250604849.31_warc_CC-MAIN-20200121162615-20200121191615-00030.warc.gz\"}"}
https://golden-ratio.eye-of-revelation.org/GreatTriangle-6.html	[ "First of all, must underline that the original word \"section\" – since from the Latin: sectio aurea – represents a Portion of a Unit; therefore the main symbol Phi should be referred to the decimal irrational number .618… and not to its counterpart 1.618…, which is phi: a unit + a portion added won't be a section ( the later \"ratio\" definition comes more flexible, though). This being said, it's worth to be introduced an alternative quick way to figure out their ratio. It is such as they can be applied to any number or value N, almost like a unique parameter, simply alternating the modes, always getting: N × f = N / j and evenly N / f = N × j That brings to: F × j = 1. Conceptually, the Percentage of [a Unit + the same Percentage of itself] is eq. One. This also means that for the unique property of these two values around the Unit, any number either multiplied by F or divided by j gives the same result; as well as any number divided by F or multiplied by j !", null, "Then each subsequent reduction or sequential increase retain a constant Proportion between the parts (or elements), in a perennial reciprocal balance." ]	[ null, "https://golden-ratio.eye-of-revelation.org/image/Goldencurves.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9422318,"math_prob":0.97755116,"size":1057,"snap":"2023-40-2023-50","text_gpt3_token_len":227,"char_repetition_ratio":0.09306743,"word_repetition_ratio":0.0,"special_character_ratio":0.22138126,"punctuation_ratio":0.1127451,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751275,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T06:35:07Z\",\"WARC-Record-ID\":\"<urn:uuid:614c2b95-9f6d-46a7-abb4-efb5413708e2>\",\"Content-Length\":\"3516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b71b3a4-5067-441c-b283-9a47567a570d>\",\"WARC-Concurrent-To\":\"<urn:uuid:21f0c9f7-761a-4e8f-b47e-5fa4772e2691>\",\"WARC-IP-Address\":\"107.190.132.18\",\"WARC-Target-URI\":\"https://golden-ratio.eye-of-revelation.org/GreatTriangle-6.html\",\"WARC-Payload-Digest\":\"sha1:YAQZQLBH75BWK4ZBHBDFD3ZLXYT74TWR\",\"WARC-Block-Digest\":\"sha1:ROFHM7WWCFTTDY6LSYMUVNKQLK6XVYDJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506686.80_warc_CC-MAIN-20230925051501-20230925081501-00178.warc.gz\"}"}
http://abcstudyguide.com/python-numbers/	[ "# Python Numbers\n\nIn Python there are three types of numbers:\n\n• Integer: Integer (or int) number type indicates a whole number.\n• Float: If the number has a decimal portion, it is a floating-point number (or float).\n• Complex: Complex numbers have real and imaginary parts.\n```x = 5 # Integer\ny = 3.14 # Float\nz = 2 + 3j # Complex Number\n```\n\n## Number Type Conversion\n\nTo convert from one number type to another, you can use `int()`, `float()`, or `complex()` functions.\n\n```x = 5\ny = 3.14\nz = 2 + 3j\n\nt = float(x) # converting from int(x) to float (t = 5.0)\nw = int(y) # converting from float(y) to int (w = 3)\nu = complex(x) # converting from int(x) to complex\n```\n\n## Arithmetic Operators\n\nYou can use the following arithmetic operators in numbers:\n\n## str() Function\n\nTo concatenate a number with a string in Python, first you need to convert number to string using `str()` function, otherwise Python will give `TypeError`.\n\nTypeError: can only concatenate str (not “int”) to str\n\n```age = 32\nprint(\"Hello, I'm \" + age + \" years old.\")\n```\n\n`str()` function converts number to string\n\n```print(\"Hello, I'm \" + str(age) + \" years old.\")\n# Hello, I'm 32 years old.\n```\nSHARE" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63270116,"math_prob":0.9914918,"size":1427,"snap":"2020-34-2020-40","text_gpt3_token_len":436,"char_repetition_ratio":0.11946592,"word_repetition_ratio":0.013793103,"special_character_ratio":0.3651016,"punctuation_ratio":0.09737828,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994382,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T18:37:18Z\",\"WARC-Record-ID\":\"<urn:uuid:c7a03239-ede7-4473-b690-f780cbad7e96>\",\"Content-Length\":\"54464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f07e6d8-7a90-4bee-a463-a40599e5fb61>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8e5d552-976d-4ed9-aae6-a4a23ee49d55>\",\"WARC-IP-Address\":\"107.180.12.188\",\"WARC-Target-URI\":\"http://abcstudyguide.com/python-numbers/\",\"WARC-Payload-Digest\":\"sha1:KCBOC37HHBY4LNFKKWAT7MBQAC3BS3TE\",\"WARC-Block-Digest\":\"sha1:A6FEXGGUHDJ7AWIDBDBAIRWDM4APLSWD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401604940.65_warc_CC-MAIN-20200928171446-20200928201446-00766.warc.gz\"}"}
https://nightlies.apache.org/flink/flink-docs-release-1.7/dev/libs/ml/cross_validation.html	[ "This documentation is for an out-of-date version of Apache Flink. We recommend you use the latest stable version.\n$$\\newcommand{\\R}{\\mathbb{R}} \\newcommand{\\E}{\\mathbb{E}} \\newcommand{\\x}{\\mathbf{x}} \\newcommand{\\y}{\\mathbf{y}} \\newcommand{\\wv}{\\mathbf{w}} \\newcommand{\\av}{\\mathbf{\\alpha}} \\newcommand{\\bv}{\\mathbf{b}} \\newcommand{\\N}{\\mathbb{N}} \\newcommand{\\id}{\\mathbf{I}} \\newcommand{\\ind}{\\mathbf{1}} \\newcommand{\\0}{\\mathbf{0}} \\newcommand{\\unit}{\\mathbf{e}} \\newcommand{\\one}{\\mathbf{1}} \\newcommand{\\zero}{\\mathbf{0}} \\newcommand\\rfrac{^{#1}\\!/_{#2}} \\newcommand{\\norm}{\\left\\lVert#1\\right\\rVert}$$\n\n# Cross Validation\n\n## Description\n\nA prevalent problem when utilizing machine learning algorithms is overfitting, or when an algorithm “memorizes” the training data but does a poor job extrapolating to out of sample cases. A common method for dealing with the overfitting problem is to hold back some subset of data from the original training algorithm and then measure the fit algorithm’s performance on this hold-out set. This is commonly known as cross validation. A model is trained on one subset of data and then validated on another set of data.\n\n## Cross Validation Strategies\n\nThere are several strategies for holding out data. FlinkML has convenience methods for\n\n• Train-Test Splits\n• Train-Test-Holdout Splits\n• K-Fold Splits\n• Multi-Random Splits\n\n### Train-Test Splits\n\nThe simplest method of splitting is the trainTestSplit. This split takes a DataSet and a parameter fraction. The fraction indicates the portion of the DataSet that should be allocated to the training set. This split also takes two additional optional parameters, precise and seed.\n\nBy default, the Split is done by randomly deciding whether or not an observation is assigned to the training DataSet with probability = fraction. When precise is true however, additional steps are taken to ensure the training set is as close as possible to the length of the DataSet $\\cdot$ fraction.\n\nThe method returns a new TrainTestDataSet object which has a .training attribute containing the training DataSet and a .testing attribute containing the testing DataSet.\n\n### Train-Test-Holdout Splits\n\nIn some cases, algorithms have been known to ‘learn’ the testing set. To combat this issue, a train-test-hold out strategy introduces a secondary holdout set, aptly called the holdout set.\n\nTraditionally, training and testing would be done to train an algorithms as normal and then a final test of the algorithm on the holdout set would be done. Ideally, prediction errors/model scores in the holdout set would not be significantly different than those observed in the testing set.\n\nIn a train-test-holdout strategy we sacrifice the sample size of the initial fitting algorithm for increased confidence that our model is not over-fit.\n\nWhen using trainTestHoldout splitter, the fraction Double is replaced by a fraction array of length three. The first element corresponds to the portion to be used for training, second for testing, and third for holdout. The weights of this array are relative, e.g. an array Array(3.0, 2.0, 1.0) would results in approximately 50% of the observations being in the training set, 33% of the observations in the testing set, and 17% of the observations in holdout set.\n\n### K-Fold Splits\n\nIn a k-fold strategy, the DataSet is split into k equal subsets. Then for each of the k subsets, a TrainTestDataSet is created where the subset is the .training DataSet, and the remaining subsets are the .testing set.\n\nFor each training set, an algorithm is trained and then is evaluated based on the predictions based on the associated testing set. When an algorithm that has consistent grades (e.g. prediction errors) across held out datasets we can have some confidence that our approach (e.g. choice of algorithm / algorithm parameters / number of iterations) is robust against overfitting.\n\nK-Fold Cross Validation\n\n### Multi-Random Splits\n\nThe multi-random strategy can be thought of as a more general form of the train-test-holdout strategy. In fact, .trainTestHoldoutSplit is a simple wrapper for multiRandomSplit which also packages the datasets into a trainTestHoldoutDataSet object.\n\nThe first major difference, is that multiRandomSplit takes an array of fractions of any length. E.g. one can create multiple holdout sets. Alternatively, one could think of kFoldSplit as a wrapper for multiRandomSplit (which it is), the difference being kFoldSplit creates subsets of approximately equal size, where multiRandomSplit will create subsets of any size.\n\nThe second major difference is that multiRandomSplit returns an array of DataSets, equal in size and proportion to the fraction array that it was passed as an argument.\n\n## Parameters\n\nThe various Splitter methods share many parameters.\n\nParameter Type Description Used by Method\ninput DataSet[Any] DataSet to be split. randomSplit\nmultiRandomSplit\nkFoldSplit\ntrainTestSplit\ntrainTestHoldoutSplit\nseed Long\n\nUsed for seeding the random number generator which sorts DataSets into other DataSets.\n\nrandomSplit\nmultiRandomSplit\nkFoldSplit\ntrainTestSplit\ntrainTestHoldoutSplit\nprecise Boolean When true, make additional effort to make DataSets as close to the prescribed proportions as possible. randomSplit\ntrainTestSplit\nfraction Double The portion of the input to assign to the first or .training DataSet. Must be in the range (0,1) randomSplit\ntrainTestSplit\nfracArray Array[Double] An array that prescribes the proportions of the output datasets (proportions need not sum to 1 or be within the range (0,1)) multiRandomSplit\ntrainTestHoldoutSplit\nkFolds Int The number of subsets to break the input DataSet into. kFoldSplit" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87005657,"math_prob":0.9498542,"size":4815,"snap":"2023-40-2023-50","text_gpt3_token_len":1048,"char_repetition_ratio":0.14674704,"word_repetition_ratio":0.008,"special_character_ratio":0.19231568,"punctuation_ratio":0.0989273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97722864,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T13:48:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b2416049-b4e3-47d0-937f-dcaafc6f5e9a>\",\"Content-Length\":\"57151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d59fbca5-fd8c-4415-854d-aecb4ad0d23b>\",\"WARC-Concurrent-To\":\"<urn:uuid:97eb7b1e-7e65-4a51-87bf-8c00aeb913f7>\",\"WARC-IP-Address\":\"95.217.87.228\",\"WARC-Target-URI\":\"https://nightlies.apache.org/flink/flink-docs-release-1.7/dev/libs/ml/cross_validation.html\",\"WARC-Payload-Digest\":\"sha1:LVCGIPH7F42NFCRWSIQMYUF6YKI2RV4X\",\"WARC-Block-Digest\":\"sha1:X5OYZQ4CNWIIW5GBOEWZHO6VYDEMANYR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511106.1_warc_CC-MAIN-20231003124522-20231003154522-00759.warc.gz\"}"}
https://de.maplesoft.com/support/help/addons/view.aspx?path=LinearFunctionalSystems/SeriesSolution	[ "", null, "SeriesSolution - Maple Help\n\nLinearFunctionalSystems\n\n SeriesSolution\n return the formal series solution of a linear functional system of equations", null, "Calling Sequence SeriesSolution(sys, vars, method) SeriesSolution(A, b, x, case, method) SeriesSolution(A, x, case, method)", null, "Parameters\n\n sys - list of equations; linear functional system vars - list of function variables such as [y1(x), y2(x), ...]; variables to solve for A - Matrix with rational elements b - Vector with rational elements x - independent variable case - name indicating the case of the system; one of 'differential', 'difference', or 'qdifference' method - (optional) name indicating the version of EG-eliminations to use; one of 'quasimodular' or 'ordinary', the latter being the default", null, "Description\n\n • The SeriesSolution function returns the initial terms of the formal series solutions from the specified linear functional system of equations with polynomial coefficients. If such a solution does not exist, then the empty list $\\left[\\right]$ is returned.\n The system parameter is entered either in list form (a list of equations sys and a list of function variables vars to solve for), or in matrix form (matrix A, vector b, and the independent variable x, where the vector b is optional).\n The matrix form specifies the system $\\mathrm{Ly}\\left(x\\right)=\\mathrm{Ay}\\left(x\\right)+b$, where L is the operator (either differential, difference, or q-difference), $y\\left(x\\right)$ is the vector of the functions to solve for, A is a rational matrix, and b is a rational vector (right-hand side).\n For the matrix from of the calling sequence, the case of the system must be specified as one of 'differential', 'difference', or 'qdifference'. If b is not specified, the system is assumed to be homogeneous.\n • The function computes the matrix recurrence system corresponding to the given system. This matrix recurrence system is represented by its explicit matrix (the matrix n by n*m, where n is the order of the system, with the leading and trailing matrix being of size n by n). Then, the function triangularizes the leading matrix using LinearFunctionalSystems[MatrixTriangularization] in order to bound the number of the initial terms of the solution in such a way that the recurrences for the rest terms' coefficients have an invertible leading matrix and then builds these initial terms.\n • The solution is the list of series expansions in x, corresponding to vars. The order term (for example $\\mathrm{O}\\left({x}^{6}\\right)$) is the last term in the series.\n The solution involves arbitrary constants of the form _c1, _c2, etc.\n • The solution has an attribute which is a table with the following indices:\n\n 'initial' - the list of initial terms 'degree' - the formal degree of the initial terms 'recurrence' - the corresponding recurrence 'coefficients' - the coefficients of the initial terms in a proper basis (depending on the case) 'lead' - the leading shift of the recurrence 'trail' - the trailing shift of the recurrence 'variable' - the independent variable of the given system 'the_case' - 'differential', 'difference' or 'qdifference' 'homogeneous' - true if the given system is homogeneous, false otherwise 'index' - the index of the last arbitrary constant 'q_par' - the q parameter used\n\n Note: This data is used by LinearFunctionalSystems[ExtendSeries] in order to extend the number of computed initial terms.\n • The error conditions associated with SeriesSolution are the same as those which are generated by LinearFunctionalSystems[Properties].\n • This function is part of the LinearFunctionalSystems package, and so it can be used in the form SeriesSolution(..) only after executing the command with(LinearFunctionalSystems). However, it can always be accessed through the long form of the command by using the form LinearFunctionalSystems[SeriesSolution](..).", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{LinearFunctionalSystems}\\right):$\n > $\\mathrm{sys}≔\\left[\\frac{ⅆ}{ⅆx}\\mathrm{y1}\\left(x\\right)-\\mathrm{y2}\\left(x\\right),\\frac{ⅆ}{ⅆx}\\mathrm{y2}\\left(x\\right)-\\mathrm{y3}\\left(x\\right)-\\mathrm{y4}\\left(x\\right),\\frac{ⅆ}{ⅆx}\\mathrm{y3}\\left(x\\right)-\\mathrm{y5}\\left(x\\right),\\frac{ⅆ}{ⅆx}\\mathrm{y4}\\left(x\\right)-2\\mathrm{y1}\\left(x\\right)-2x\\mathrm{y2}\\left(x\\right)-\\mathrm{y5}\\left(x\\right),\\frac{ⅆ}{ⅆx}\\mathrm{y5}\\left(x\\right)-{x}^{2}\\mathrm{y1}\\left(x\\right)-2x\\mathrm{y3}\\left(x\\right)-\\mathrm{y6}\\left(x\\right),\\frac{ⅆ}{ⅆx}\\mathrm{y6}\\left(x\\right)-{x}^{2}\\mathrm{y2}\\left(x\\right)+2\\mathrm{y3}\\left(x\\right)\\right]:$\n > $\\mathrm{vars}≔\\left[\\mathrm{y1}\\left(x\\right),\\mathrm{y2}\\left(x\\right),\\mathrm{y3}\\left(x\\right),\\mathrm{y4}\\left(x\\right),\\mathrm{y5}\\left(x\\right),\\mathrm{y6}\\left(x\\right)\\right]:$\n > $\\mathrm{SeriesSolution}\\left(\\mathrm{sys},\\mathrm{vars}\\right)$\n $\\left[{3}{}{{\\mathrm{_c}}}_{{2}}{+}{{\\mathrm{_c}}}_{{5}}{+}{\\mathrm{O}}{}\\left({x}\\right){,}{3}{}{{\\mathrm{_c}}}_{{6}}{+}{{\\mathrm{_c}}}_{{3}}{+}{\\mathrm{O}}{}\\left({x}\\right){,}{-}\\frac{{{\\mathrm{_c}}}_{{4}}}{{2}}{+}{\\mathrm{O}}{}\\left({x}\\right){,}{{\\mathrm{_c}}}_{{1}}{+}\\frac{{{\\mathrm{_c}}}_{{4}}}{{2}}{+}{\\mathrm{O}}{}\\left({x}\\right){,}{-}{{\\mathrm{_c}}}_{{5}}{+}{\\mathrm{O}}{}\\left({x}\\right){,}{{\\mathrm{_c}}}_{{3}}{+}{\\mathrm{O}}{}\\left({x}\\right)\\right]$ (1)\n > $\\mathrm{sys}≔\\left[\\mathrm{y2}\\left(x\\right){x}^{2}+3\\mathrm{y2}\\left(x\\right)x+2\\mathrm{y2}\\left(x\\right)-2\\mathrm{y1}\\left(x\\right){x}^{2}-4\\mathrm{y1}\\left(x\\right)x+\\mathrm{y1}\\left(x+1\\right){x}^{2}+\\mathrm{y1}\\left(x+1\\right)x,\\mathrm{y2}\\left(x+1\\right)-\\mathrm{y1}\\left(x\\right)\\right]:$\n > $\\mathrm{vars}≔\\left[\\mathrm{y1}\\left(x\\right),\\mathrm{y2}\\left(x\\right)\\right]:$\n > $\\mathrm{SeriesSolution}\\left(\\mathrm{sys},\\mathrm{vars}\\right)$\n $\\left[{{\\mathrm{_c}}}_{{1}}{+}{x}{}\\left({2}{}{{\\mathrm{_c}}}_{{2}}{+}{{\\mathrm{_c}}}_{{1}}\\right){+}{\\mathrm{O}}{}\\left({{x}}^{{2}}\\right){,}{x}{}{{\\mathrm{_c}}}_{{1}}{+}{\\mathrm{O}}{}\\left({{x}}^{{2}}\\right)\\right]$ (2)", null, "References\n\n Abramov, S. A. \"EG-Eliminations.\" Journal of Difference Equations and Applications, (1999): 393-433." ]	[ null, "https://bat.bing.com/action/0", null, "https://de.maplesoft.com/support/help/addons/arrow_down.gif", null, "https://de.maplesoft.com/support/help/addons/arrow_down.gif", null, "https://de.maplesoft.com/support/help/addons/arrow_down.gif", null, "https://de.maplesoft.com/support/help/addons/arrow_down.gif", null, "https://de.maplesoft.com/support/help/addons/arrow_down.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.764654,"math_prob":0.99978584,"size":4507,"snap":"2023-14-2023-23","text_gpt3_token_len":1318,"char_repetition_ratio":0.16899845,"word_repetition_ratio":0.010016695,"special_character_ratio":0.21167074,"punctuation_ratio":0.11737089,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993673,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T12:06:06Z\",\"WARC-Record-ID\":\"<urn:uuid:8018731e-085e-4e3e-b0ef-831f1110a662>\",\"Content-Length\":\"332294\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:202f409b-150b-4174-b078-e6d92495891f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d398c90c-0abc-4e73-894e-6f4f39eb627a>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://de.maplesoft.com/support/help/addons/view.aspx?path=LinearFunctionalSystems/SeriesSolution\",\"WARC-Payload-Digest\":\"sha1:GEKQCKUSTE3OHUTDK2JTBZMBE2COFPCC\",\"WARC-Block-Digest\":\"sha1:NMOJ2JJPFFBPPKQJZKC4BNKWUMAOK4LJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644855.6_warc_CC-MAIN-20230529105815-20230529135815-00535.warc.gz\"}"}
https://www.math-only-math.com/Conversion-of-Mixed-Fractions-into-Improper-Fractions.html	[ "# Conversion of Mixed Fractions into Improper Fractions\n\nIn conversion of mixed fractions into improper fractions, we may follow the following steps:\n\nStep I:\n\nObtain the mixed fraction. Let the mixed fraction be 22/5\n\nStep II:\n\nIdentify the whole number and the numerator (top) and denominator (bottom) of the proper fraction.\n\nStep III:\n\nMultiply the whole number by the denominator of the proper fraction and add the result to the numerator of the proper fraction.\n\nStep IV:\n\nWrite the fraction having numerator equal to the number obtained in step III and denominator same as the denominator of the fraction in step II. Thus,\n\nFor Example:\n\nExpress each of the following mixed fractions as improper fractions:\n\n(i) Convert 8$$\\frac{4}{7}$$ into an improper fraction.\n\n8$$\\frac{4}{7}$$ means 8 whole and $$\\frac{4}{7}$$.\n\n$$\\frac{4}{7}$$\n\nSolution:\n\n8$$\\frac{4}{7}$$ = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + $$\\frac{4}{7}$$\n\nor, 8$$\\frac{4}{7}$$ = $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{7}{7}$$ + $$\\frac{4}{7}$$ = $$\\frac{60}{7}$$, [$$\\frac{7}{7}$$ means 1)\n\nWe can also convert a mixed number into an improper fraction as follows.\n\nFirst multiply the whole number by denominator. Here (8 × 7) + 4 = 60. Now, put the sum as the numerator of the required improper fraction and the denominator remains the same.\n\n8$$\\frac{4}{7}$$ = $$\\frac{(8 × 7) + 4}{7}$$ = $$\\frac{56 + 4}{7}$$ = $$\\frac{60}{7}$$\n\nThus, 8$$\\frac{4}{7}$$ = $$\\frac{60}{7}$$\n\n(ii) 3$$\\frac{2}{7}$$\n\n= $$\\frac{(3 × 7) + 2}{7}$$\n\n= $$\\frac{21 + 2}{7}$$\n\n= $$\\frac{23}{7}$$\n\n(iii) 4$$\\frac{5}{9}$$\n\n= $$\\frac{(4 × 9) + 5}{9}$$\n\n= $$\\frac{36 + 5}{9}$$\n\n= $$\\frac{41}{9}$$\n\n(iv) 3$$\\frac{2}{5}$$\n\n= $$\\frac{(3 × 5) + 2}{5}$$\n\n= $$\\frac{15 + 2}{5}$$\n\n= $$\\frac{17}{5}$$\n\n(v) 7$$\\frac{1}{4}$$\n\n= $$\\frac{(7 × 4) + 1}{4}$$\n\n= $$\\frac{28 + 1}{4}$$\n\n= $$\\frac{29}{4}$$\n\nRepresentations of Fractions on a Number Line\n\nFraction as Division\n\nTypes of Fractions\n\nConversion of Mixed Fractions into Improper Fractions\n\nConversion of Improper Fractions into Mixed Fractions\n\nEquivalent Fractions\n\nFractions in Lowest Terms\n\nLike and Unlike Fractions\n\nComparing Like Fractions\n\nComparing Unlike Fractions\n\nAddition and Subtraction of Like Fractions\n\nAddition and Subtraction of Unlike Fractions\n\nInserting a Fraction between Two Given Fractions" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7937118,"math_prob":1.0000094,"size":2533,"snap":"2019-51-2020-05","text_gpt3_token_len":850,"char_repetition_ratio":0.24752867,"word_repetition_ratio":0.0867347,"special_character_ratio":0.39715752,"punctuation_ratio":0.066523604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T07:00:30Z\",\"WARC-Record-ID\":\"<urn:uuid:6a406342-b49e-4e5c-929e-c038b5c02a44>\",\"Content-Length\":\"33792\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:279cc86a-e664-49f6-b5d1-630b4d1ac6b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:851d729b-bbc7-4e51-9606-ae92d9d6ed5c>\",\"WARC-IP-Address\":\"173.247.219.53\",\"WARC-Target-URI\":\"https://www.math-only-math.com/Conversion-of-Mixed-Fractions-into-Improper-Fractions.html\",\"WARC-Payload-Digest\":\"sha1:SLQTX62AAYDW3B6ACL7MSNB66S67RIKU\",\"WARC-Block-Digest\":\"sha1:36S2VLXI6FEVYPF6ZMIFVZ6FMYLEXNUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251687725.76_warc_CC-MAIN-20200126043644-20200126073644-00498.warc.gz\"}"}
http://musictheoryfundamentals.com/Quiz/modesQuiz.php	[ "# Modes Quiz\n\n1. Between which scale degrees do the half steps naturally occur in Ionian mode?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n2. A major scale is in what mode?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n3. Which mode begins on the 2nd scale degree?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n4. What mode begins on the 6th scale degree?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n5. On what scale degree does the locrian mode begin?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n6. A minor scale starts on what scale degree?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n7. A natural minor scale is in what mode?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n8. On what scale degree does the Mixolydian mode begin?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n9. What mode has naturally occurring half steps between the 3rd & 4th and 7th & 8th scale degrees?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n10. What mode has naturally occurring half steps between the 3rd & 4th and 6th & 7th scale degrees?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n11. What mode begins on the 4th scale degree?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n12. What mode on the leading tone of the major scale?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n13. What mode begins on the 3rd scale degree?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n14. What mode has naturally occurring half steps between the 1st & 2nd and 4th & 5th scale degrees?\n\ncheck answer\n\nCorrect!\n\nIncorrect.\n\n15. What mode has naturally occurring half steps between the 4th & 5th and 7th & 8th scale degrees?\n\ncheck answer\n\nCorrect!\n\nIncorrect." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91738707,"math_prob":0.8093207,"size":1237,"snap":"2021-21-2021-25","text_gpt3_token_len":306,"char_repetition_ratio":0.24493106,"word_repetition_ratio":0.4549763,"special_character_ratio":0.24413905,"punctuation_ratio":0.18987341,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9731355,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T23:43:47Z\",\"WARC-Record-ID\":\"<urn:uuid:b3a232cf-2a1f-47b0-bd18-f3c8458bdeff>\",\"Content-Length\":\"30206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd474041-cd2a-4ec4-8526-844cc71420af>\",\"WARC-Concurrent-To\":\"<urn:uuid:85a79fcc-73e3-49b8-8e16-df4921166953>\",\"WARC-IP-Address\":\"173.201.188.238\",\"WARC-Target-URI\":\"http://musictheoryfundamentals.com/Quiz/modesQuiz.php\",\"WARC-Payload-Digest\":\"sha1:XY4EEUYESWRW7WZ72BZE7YO5RVPPJJ6N\",\"WARC-Block-Digest\":\"sha1:BKKDD66PT5GLDHLSM3Y2O3RCET5VTTFG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487653461.74_warc_CC-MAIN-20210619233720-20210620023720-00426.warc.gz\"}"}
http://www.numbersaplenty.com/70179099	[ "Search a number\nBaseRepresentation\nbin1000010111011…\n…01100100011011\n311220001110121220\n410023231210123\n5120431212344\n610544103123\n71511340534\noct413554433\n9156043556\n1070179099\n1136683661\n121b604aa3\n13117021c3\n14946b68b\n156263c19\nhex42ed91b\n\n70179099 has 4 divisors (see below), whose sum is σ = 93572136. Its totient is φ = 46786064.\n\nThe previous prime is 70179089. The next prime is 70179107. The reversal of 70179099 is 99097107.\n\nIt can be divided in two parts, 701 and 79099, that added together give a triangular number (79800 = T399).\n\nIt is a semiprime because it is the product of two primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 70179099 - 27 = 70178971 is a prime.\n\nIt is a self number, because there is not a number n which added to its sum of digits gives 70179099.\n\nIt is not an unprimeable number, because it can be changed into a prime (70179029) by changing a digit.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 11696514 + ... + 11696519.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (23393034).\n\nAlmost surely, 270179099 is an apocalyptic number.\n\n70179099 is a deficient number, since it is larger than the sum of its proper divisors (23393037).\n\n70179099 is a wasteful number, since it uses less digits than its factorization.\n\n70179099 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 23393036.\n\nThe product of its (nonzero) digits is 35721, while the sum is 42.\n\nThe square root of 70179099 is about 8377.2966403250. The cubic root of 70179099 is about 412.4797154396.\n\nThe spelling of 70179099 in words is \"seventy million, one hundred seventy-nine thousand, ninety-nine\".\n\nDivisors: 1 3 23393033 70179099" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8565869,"math_prob":0.94905335,"size":1885,"snap":"2020-10-2020-16","text_gpt3_token_len":573,"char_repetition_ratio":0.15842637,"word_repetition_ratio":0.006289308,"special_character_ratio":0.4509284,"punctuation_ratio":0.1322314,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99628884,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T21:10:09Z\",\"WARC-Record-ID\":\"<urn:uuid:8f0ce1c9-fcc0-4ad0-b297-78b20eb99888>\",\"Content-Length\":\"8699\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad2b64c3-cea3-4297-ab34-de93c63714d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:ceaa8629-2add-417d-ad38-eb0ed6c69723>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"http://www.numbersaplenty.com/70179099\",\"WARC-Payload-Digest\":\"sha1:U3O6CQC5FNCVNVZ7EEIVHIDAVCLOVYVO\",\"WARC-Block-Digest\":\"sha1:YYUIVQ7MMIZ5QNVVMYSWBGGB3MB6CKXK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145981.35_warc_CC-MAIN-20200224193815-20200224223815-00521.warc.gz\"}"}
http://electricalacademia.com/transformer/single-phase-transformer-construction-theory/	[ "Home / Transformer / Single-Phase Transformer \| Construction \| Theory\n\n# Single-Phase Transformer \| Construction \| Theory\n\nWant create site? Find Free WordPress Themes and plugins.\n\nIn the same way that single-phase AC and three-phase AC are different, although both are alternating current electricity, single-phase and three-phase transformers cannot be interchanged. Nevertheless, most of what can be said for single-phase transformers is later on extended to three-phase transformers because they work based on the same principle.\n\nTransformers work based on the mutual induction between two windings. When electricity is applied to a coil a magnetic field is developed. Because in AC the voltage (and thus the current) continuously alters, the resulting magnetic field is variable. However, if a coil is placed in a varying magnetic field, it is similar to a wire that moves in a magnetic field; then a voltage is induced in it.\n\n## Construction of Single-Phase Transformer\n\nIn the simplest form, a single-phase transformer is made up of two windings which share a common core. The core must be closed and its material is a ferromagnetic metal; it can be of various shapes, but the shape does not play a significant role.\n\nSome shapes are preferable to the others, either because of manufacturing or because they can result in a better performance. The most common shapes for a core are rectangular and circular.\n\nFrom the manufacturing viewpoint, a rectangular shape is easier to work with. Figure 1shows a transformer with a rectangular core. In this configuration, called a core-type transformer, the two windings are on the opposite sides (legs) of the core. This is not necessary, and both can be on the same leg.\n\nIn fact, most of the small transformers have a core as shown in Figure 2, and both windings are placed—usually, side by side—on the middle leg. This is a called a shell-type transformer.\n\nCore-type transformer: Category for the construction of a transformer in which the core has the shape of a rectangular frame, and the windings are normally on the opposite sides of the rectangle.\n\nShell-type transformer: Category for the construction of a transformer in which the core has the shape of figure 8 (i.e., with a bridge in the middle of a rectangular frame).", null, "Figure 1 Basic structure of a single-phase transformer (core type). (a) Schematics. (b) Pictorial representation.\n\nThe two windings are called “primary” and “secondary,” depending on their role. The primary winding is connected to the voltage that we want to change. The secondary winding provides the required voltage.\n\nIn other words, the primary winding is connected to the input voltage and the secondary winding is the output and is connected to the output circuitry. To show a transformer in a circuit, the transformer symbol is used, as depicted in Figure 3.", null, "Figure 2 Most of transformers have a shell-type structure. (a) Schematics of a shell-type transformer. (b) Pictorial view.\n\nIf the primary winding has more turns than the secondary winding, the transformer decreases the voltage and it is a step-down transformer. If the primary winding has fewer turns than the secondary winding, the output voltage is higher than the input voltage and the transformer is called a step-up transformer.", null, "Figure 3 Symbol for a transformer.\n\nThe numbers of wire turn in the primary and secondary windings are represented by N1 and N2, respectively.\n\nStep-down transformer: Transformer with secondary winding having fewer turns than the primary winding, thus decreasing voltage.\n\nStep-up transformer: Transformer with secondary winding having more turns than the primary winding, thus increasing voltage.\n\nFor a step-down transformer, N1 > N2 and V1 > V2.\n\nFor a step-up transformer, N1 < N2 and V1 < V2.\n\nNote that a wire can be wound around a core in two directions, as shown in Figure 4. As a result, depending on the directions of both windings in a transformer, at a given instant and compared to the primary voltage, the current in the secondary winding can be in one or the opposite direction.", null, "Figure 4 Two possible ways of winding a wire around a core.\n\nIn other words, the secondary voltage can be in phase with the primary voltage, or it can be 180° out of phase with it. It is, therefore, important to pay attention to the direction of turns in the winding of a transformer.\n\nIn Figure 2, for instance, both windings have the same direction. To show the polarity in the drawings of transformer circuits, a dot is put at one side of each winding indication the phase relationship, illustrating also the relative directions of the primary and secondary currents. This is indicated in Figure 5.", null, "Figure 5 Number of turns N, voltage V and current I in the primary and secondary circuits of a transformer.\n\nThus, for example, considering a sinusoidal waveform, when the side with a dot in the primary winding is at its maximum value the side with a dot in the secondary winding is at its maximum value.\n\nIn practice, the terminals are marked by letters such as H and X on the primary and secondary referring to the dot side to indicate this polarity. Figure 5 also indicates the number of turns N, the voltage V, and the current I and its direction for a given instant, identified by subscripts 1 and 2 for the primary and the secondary circuits, respectively.\n\n## Multi-Output Transformers\n\nIt is possible to have multiple secondary windings on the same transformer. This provides various voltages from one source voltage, thus reducing the cost and space. This is very common in electronic devices such as a television that need various voltages for their operation. This is done in very small transformers.\n\nIn larger transformers and three-phase transformers, it is not normally necessary to have multiple voltages. Figure 6 shows the schematics of a multi-output transformer.", null, "Figure 6 Multi-output transformer.\n\nFigure 7 shows a very small, 20 W, transformer. It converts 120 V into 6 and 12 V. In this sense, it is a two-output transformer. A 4 kVA industrial transformer is shown in Figure 8.", null, "Figure 7 A 20 W single-phase transformer.", null, "Figure 8 A 4 kVA single-phase transformer.\n\nDid you find apk for android? You can find new Free Android Games and apps.\n\n### About Ahmad Faizan", null, "Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA." ]	[ null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-1-666x300.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-2-486x300.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-3-119x300.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-4.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-5.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-6.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-7-399x300.jpg", null, "http://electricalacademia.com/wp-content/uploads/2018/11/10-8-200x300.jpg", null, "http://1.gravatar.com/avatar/12ca3fae784a44fa8f8e9aca2b9fa673", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93374515,"math_prob":0.9427251,"size":6107,"snap":"2019-35-2019-39","text_gpt3_token_len":1271,"char_repetition_ratio":0.19859086,"word_repetition_ratio":0.053254437,"special_character_ratio":0.20435566,"punctuation_ratio":0.108639866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95538217,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T17:42:20Z\",\"WARC-Record-ID\":\"<urn:uuid:6d9c25cc-32b4-445c-bb1e-ac7bb0f18d4b>\",\"Content-Length\":\"86084\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a339a148-6e43-47d1-b9ab-968fdb539e0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:dce4a053-260f-4e25-a4a1-87208e726689>\",\"WARC-IP-Address\":\"107.180.46.155\",\"WARC-Target-URI\":\"http://electricalacademia.com/transformer/single-phase-transformer-construction-theory/\",\"WARC-Payload-Digest\":\"sha1:RIWU4XUAJUHSFZG3KPF7G7KPWXS7HI3F\",\"WARC-Block-Digest\":\"sha1:U4VXB5FG5C2W7Q23CE32FQXFOKD72RVC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027321351.87_warc_CC-MAIN-20190824172818-20190824194818-00421.warc.gz\"}"}
https://whatisconvert.com/462-square-kilometers-in-square-miles	[ "# What is 462 Square Kilometers in Square Miles?\n\n## Convert 462 Square Kilometers to Square Miles\n\nTo calculate 462 Square Kilometers to the corresponding value in Square Miles, multiply the quantity in Square Kilometers by 0.38610215854185 (conversion factor). In this case we should multiply 462 Square Kilometers by 0.38610215854185 to get the equivalent result in Square Miles:\n\n462 Square Kilometers x 0.38610215854185 = 178.37919724633 Square Miles\n\n462 Square Kilometers is equivalent to 178.37919724633 Square Miles.\n\n## How to convert from Square Kilometers to Square Miles\n\nThe conversion factor from Square Kilometers to Square Miles is 0.38610215854185. To find out how many Square Kilometers in Square Miles, multiply by the conversion factor or use the Area converter above. Four hundred sixty-two Square Kilometers is equivalent to one hundred seventy-eight point three seven nine Square Miles.\n\n## Definition of Square Kilometer\n\nSquare kilometre (International spelling as used by the International Bureau of Weights and Measures) or square kilometer (American spelling), symbol km2, is a multiple of the square metre, the SI unit of area or surface area. 1 km2 is equal to 1,000,000 square metres (m2) or 100 hectares (ha). It is also approximately equal to 0.3861 square miles or 247.1 acres.\n\n## Definition of Square Mile\n\nThe square mile (abbreviated as sq mi and sometimes as mi²) is an imperial and US unit of measure for an area equal to the area of a square with a side length of one statute mile. It should not be confused with miles square, which refers to a square region with each side having the specified length. For instance, 20 miles square (20 × 20 miles) has an area equal to 400 square miles; a rectangle of 10 × 40 miles likewise has an area of 400 square miles, but it is not 20 miles square. One square mile is equal to 4,014,489,600 square inches, 27,878,400 square feet or 3,097,600 square yards.\n\n## Using the Square Kilometers to Square Miles converter you can get answers to questions like the following:\n\n• How many Square Miles are in 462 Square Kilometers?\n• 462 Square Kilometers is equal to how many Square Miles?\n• How to convert 462 Square Kilometers to Square Miles?\n• How many is 462 Square Kilometers in Square Miles?\n• What is 462 Square Kilometers in Square Miles?\n• How much is 462 Square Kilometers in Square Miles?\n• How many mi2 are in 462 km2?\n• 462 km2 is equal to how many mi2?\n• How to convert 462 km2 to mi2?\n• How many is 462 km2 in mi2?\n• What is 462 km2 in mi2?\n• How much is 462 km2 in mi2?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8788691,"math_prob":0.99729145,"size":2499,"snap":"2022-05-2022-21","text_gpt3_token_len":619,"char_repetition_ratio":0.2480962,"word_repetition_ratio":0.09090909,"special_character_ratio":0.29371747,"punctuation_ratio":0.10816327,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99796665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T01:38:58Z\",\"WARC-Record-ID\":\"<urn:uuid:8aa1f489-6f8b-41c8-80ef-c6165079481e>\",\"Content-Length\":\"31470\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed3ca107-5c76-4fb4-8f71-a25a1cac0ed5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4861ef51-708e-43ac-b29b-6675d9e2d0f5>\",\"WARC-IP-Address\":\"172.67.133.29\",\"WARC-Target-URI\":\"https://whatisconvert.com/462-square-kilometers-in-square-miles\",\"WARC-Payload-Digest\":\"sha1:OOQPAACAIHAEMQ3O6Z36RMN3YFOKLJLC\",\"WARC-Block-Digest\":\"sha1:VRX4ZJ6QVQJCNQR554R3E2Z4UFKFZVTY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662543264.49_warc_CC-MAIN-20220522001016-20220522031016-00000.warc.gz\"}"}
https://physics.stackexchange.com/questions/250977/sweet-spot-of-rod-pendulum-problem-clarification	[ "# “Sweet Spot” of Rod-Pendulum - Problem Clarification\n\nI came across this problem in a book (shortened for brevity):\n\nConsider a rod of mass $m$ pivoted about one end, with the other end to rotate. Let the center of mass be a distance $a$ from the pivot point $I$ be the moment of inertia of the rod about an axis which we will consider rotations in. A particle comes in and hits the rod at a distance $b$ below the pivot point, imparting an impulse $F\\Delta t=\\xi$ on the rod. (a) Find the linear and angular momentum of the rod right after the time $\\Delta t$, and (b) Calculate the impulse imparted on the pivot point.\n\nMy problem is with (b). What the does \"impulse imparted on the pivot point\" even mean? I would think the pivot point is fixed, so it should have experienced no net impulse, but that's incorrect.\n\n• if the rod were floating free, there would be a rotational moment depending on where you hit it. The rod would not rotate if you hit it at the center of mass and would have a forward motion without rotation. If it is being held in place by a pivot, the pivot point would \"feel' an impulse in either direction, depending on where the rod is hit. – Peter R Apr 21 '16 at 0:46\n• @PeterR Ah okay, so here it would the time-integral of the normal/restoring force that the pivot exerts on the rod, right? – Arturo don Juan Apr 21 '16 at 0:53\n• @PeterR why don't you make that an answer.... – Floris Apr 21 '16 at 0:59\n• I am not sure what you mean by a normal/restoring force, but depending on where it's hit, the rod would exert an impulsive force on the pivot and the pivot would exert an equal but opposite force. – Peter R Apr 21 '16 at 1:00\n\nIn order to maintain the constraint of the pivot during the impact, a reaction impulse is needed. See the figure below for what I mean.", null, "At the center of mass the velocity is $v = a\\,\\omega$. This is a result of the two impulses $$(F-R) \\Delta t = m\\, a\\, \\omega$$\n\nIf the angular velocity is $\\omega$ then the net impulsive moments at the center of mass are\n\n$$(b F + a R) \\Delta t = I \\omega$$\n\nThese two equations are solved for the unknown reaction $R$ and motion of the rod $\\omega$.\n\n\\begin{aligned} R \\Delta t & = \\frac{I-m\\,ab}{I+m a^2} F \\Delta t \\\\ \\omega & = \\frac{a+b}{I+m a^2} F \\Delta t \\end{aligned}\n\nOnly when $b=\\frac{I}{m\\,a}$ the pivot reaction is zero. That is considered the instant axis of percussion of the rod about the pivot (sweet spot).\n\nAs you have hypothesized in the comments, this will be a time integral of a force at the pivot point. As I read the problem, it would actually be the force exerted by the rod on the pivot that you're integrating, not the other way around.\n\nMore formally, the \"impulse imparted on the pivot point\" is a vector representing the total linear momentum transferred to the anchor from the rod during the collision.\n\n(The \"collision\" here refers to an infinitessimal and not entirely realistic period of time in which energy and momentum redistribute themselves amongst the particle, the rod, and the anchor according to certain rules which are used for most basic thought experiments about rigid body collisions.)" ]	[ null, "https://i.stack.imgur.com/lPHUd.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90518063,"math_prob":0.9946911,"size":760,"snap":"2019-51-2020-05","text_gpt3_token_len":188,"char_repetition_ratio":0.15343915,"word_repetition_ratio":0.0,"special_character_ratio":0.23947369,"punctuation_ratio":0.08227848,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991025,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T17:25:34Z\",\"WARC-Record-ID\":\"<urn:uuid:51ce5934-9065-448c-b329-1cdd2debf1fc>\",\"Content-Length\":\"140603\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6bbb7ad-ce3d-4004-a209-321ff0fef2a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec287821-0d5e-4c80-9121-ff87d0febff1>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/250977/sweet-spot-of-rod-pendulum-problem-clarification\",\"WARC-Payload-Digest\":\"sha1:FOF2GT4WYFPVVYJWPJ7L2YQ7TMZUDQ5D\",\"WARC-Block-Digest\":\"sha1:ZUXKNY6EAY375JWB4X2HQKEOAZ2ZBG2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540528457.66_warc_CC-MAIN-20191210152154-20191210180154-00250.warc.gz\"}"}