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https://www.colorhexa.com/0206a3	[ "# #0206a3 Color Information\n\nIn a RGB color space, hex #0206a3 is composed of 0.8% red, 2.4% green and 63.9% blue. Whereas in a CMYK color space, it is composed of 98.8% cyan, 96.3% magenta, 0% yellow and 36.1% black. It has a hue angle of 238.5 degrees, a saturation of 97.6% and a lightness of 32.4%. #0206a3 color hex could be obtained by blending #040cff with #000047. Closest websafe color is: #000099.\n\n• R 1\n• G 2\n• B 64\nRGB color chart\n• C 99\n• M 96\n• Y 0\n• K 36\nCMYK color chart\n\n#0206a3 color description : Dark blue.\n\n# #0206a3 Color Conversion\n\nThe hexadecimal color #0206a3 has RGB values of R:2, G:6, B:163 and CMYK values of C:0.99, M:0.96, Y:0, K:0.36. Its decimal value is 132771.\n\nHex triplet RGB Decimal 0206a3 `#0206a3` 2, 6, 163 `rgb(2,6,163)` 0.8, 2.4, 63.9 `rgb(0.8%,2.4%,63.9%)` 99, 96, 0, 36 238.5°, 97.6, 32.4 `hsl(238.5,97.6%,32.4%)` 238.5°, 98.8, 63.9 000099 `#000099`\nCIE-LAB 19.17, 54.949, -76.148 6.7, 2.787, 34.833 0.151, 0.063, 2.787 19.17, 93.904, 305.814 19.17, -5.654, -75.859 16.694, 42.42, -112.025 00000010, 00000110, 10100011\n\n# Color Schemes with #0206a3\n\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #a39f02\n``#a39f02` `rgb(163,159,2)``\nComplementary Color\n• #0257a3\n``#0257a3` `rgb(2,87,163)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #4e02a3\n``#4e02a3` `rgb(78,2,163)``\nAnalogous Color\n• #57a302\n``#57a302` `rgb(87,163,2)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #a34e02\n``#a34e02` `rgb(163,78,2)``\nSplit Complementary Color\n• #06a302\n``#06a302` `rgb(6,163,2)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #a30206\n``#a30206` `rgb(163,2,6)``\n• #02a39f\n``#02a39f` `rgb(2,163,159)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #a30206\n``#a30206` `rgb(163,2,6)``\n• #a39f02\n``#a39f02` `rgb(163,159,2)``\n• #010357\n``#010357` `rgb(1,3,87)``\n• #010471\n``#010471` `rgb(1,4,113)``\n• #02058a\n``#02058a` `rgb(2,5,138)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #0207bc\n``#0207bc` `rgb(2,7,188)``\n• #0308d5\n``#0308d5` `rgb(3,8,213)``\n• #0309ef\n``#0309ef` `rgb(3,9,239)``\nMonochromatic Color\n\n# Alternatives to #0206a3\n\nBelow, you can see some colors close to #0206a3. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #022ea3\n``#022ea3` `rgb(2,46,163)``\n• #0221a3\n``#0221a3` `rgb(2,33,163)``\n• #0213a3\n``#0213a3` `rgb(2,19,163)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #0b02a3\n``#0b02a3` `rgb(11,2,163)``\n• #1902a3\n``#1902a3` `rgb(25,2,163)``\n• #2602a3\n``#2602a3` `rgb(38,2,163)``\nSimilar Colors\n\n# #0206a3 Preview\n\nThis text has a font color of #0206a3.\n\n``<span style=\"color:#0206a3;\">Text here</span>``\n#0206a3 background color\n\nThis paragraph has a background color of #0206a3.\n\n``<p style=\"background-color:#0206a3;\">Content here</p>``\n#0206a3 border color\n\nThis element has a border color of #0206a3.\n\n``<div style=\"border:1px solid #0206a3;\">Content here</div>``\nCSS codes\n``.text {color:#0206a3;}``\n``.background {background-color:#0206a3;}``\n``.border {border:1px solid #0206a3;}``\n\n# Shades and Tints of #0206a3\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000008 is the darkest color, while #f4f4ff is the lightest one.\n\n• #000008\n``#000008` `rgb(0,0,8)``\n• #00011b\n``#00011b` `rgb(0,1,27)``\n• #01022f\n``#01022f` `rgb(1,2,47)``\n• #010242\n``#010242` `rgb(1,2,66)``\n• #010355\n``#010355` `rgb(1,3,85)``\n• #010469\n``#010469` `rgb(1,4,105)``\n• #02057c\n``#02057c` `rgb(2,5,124)``\n• #020590\n``#020590` `rgb(2,5,144)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\n• #0207b6\n``#0207b6` `rgb(2,7,182)``\n• #0207ca\n``#0207ca` `rgb(2,7,202)``\n• #0308dd\n``#0308dd` `rgb(3,8,221)``\n• #0309f1\n``#0309f1` `rgb(3,9,241)``\n• #0b11fc\n``#0b11fc` `rgb(11,17,252)``\n• #1e24fc\n``#1e24fc` `rgb(30,36,252)``\n• #3237fc\n``#3237fc` `rgb(50,55,252)``\n• #454afd\n``#454afd` `rgb(69,74,253)``\n• #595dfd\n``#595dfd` `rgb(89,93,253)``\n• #6c70fd\n``#6c70fd` `rgb(108,112,253)``\n• #7f82fd\n``#7f82fd` `rgb(127,130,253)``\n• #9395fe\n``#9395fe` `rgb(147,149,254)``\n• #a6a8fe\n``#a6a8fe` `rgb(166,168,254)``\n• #b9bbfe\n``#b9bbfe` `rgb(185,187,254)``\n• #cdcefe\n``#cdcefe` `rgb(205,206,254)``\n• #e0e1ff\n``#e0e1ff` `rgb(224,225,255)``\n• #f4f4ff\n``#f4f4ff` `rgb(244,244,255)``\nTint Color Variation\n\n# Tones of #0206a3\n\nA tone is produced by adding gray to any pure hue. In this case, #4e4e57 is the less saturated color, while #0206a3 is the most saturated one.\n\n• #4e4e57\n``#4e4e57` `rgb(78,78,87)``\n• #48485d\n``#48485d` `rgb(72,72,93)``\n• #414264\n``#414264` `rgb(65,66,100)``\n• #3b3c6a\n``#3b3c6a` `rgb(59,60,106)``\n• #353670\n``#353670` `rgb(53,54,112)``\n• #2e3077\n``#2e3077` `rgb(46,48,119)``\n• #282a7d\n``#282a7d` `rgb(40,42,125)``\n• #222483\n``#222483` `rgb(34,36,131)``\n• #1b1e8a\n``#1b1e8a` `rgb(27,30,138)``\n• #151890\n``#151890` `rgb(21,24,144)``\n• #0f1296\n``#0f1296` `rgb(15,18,150)``\n• #080c9d\n``#080c9d` `rgb(8,12,157)``\n• #0206a3\n``#0206a3` `rgb(2,6,163)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0206a3 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5241456,"math_prob":0.7160255,"size":3635,"snap":"2020-24-2020-29","text_gpt3_token_len":1648,"char_repetition_ratio":0.13412283,"word_repetition_ratio":0.011111111,"special_character_ratio":0.56698763,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9933797,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T22:15:31Z\",\"WARC-Record-ID\":\"<urn:uuid:f3bcf975-5adf-42ea-8038-ca7844a84d1e>\",\"Content-Length\":\"36167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc926790-7119-4d95-b8c4-55cd7fe66418>\",\"WARC-Concurrent-To\":\"<urn:uuid:290272a1-7569-4d42-9855-ea21c32600c2>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0206a3\",\"WARC-Payload-Digest\":\"sha1:UC6F4Q4K62NRJ6A2FV7BAUAFFYUJFUX4\",\"WARC-Block-Digest\":\"sha1:ZMHJDGYGGT7F5VBA6R3UPOYMFLUY35IT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655883439.15_warc_CC-MAIN-20200703215640-20200704005640-00252.warc.gz\"}"}
https://quant.stackexchange.com/questions/2294/how-can-i-compare-distributions-using-only-mean-and-standard-deviation	[ "# How can I compare distributions using only mean and standard deviation?\n\nI only have means and standard deviations of samples of two random variables. What technique can I use to determine how similar the distributions these describe are? Assume that the values are built from very similar samples. I'm looking for a mechanism to detect when the distributions deviate from one another by some threshold. Access to historical observations is limited.\n\n• Are you asking how to determine which distributions the data follow based on the mean and standard deviation? – strimp099 Oct 31 '11 at 23:34\n\n## 2 Answers\n\nThere are a number of different tests that are generally used to compare samples to different distributions, such as Jarque-Bera, Anderson-Darling, and Kolmogorov–Smirnov (see this related question).\n\nIn your case, with just the standard deviation and mean, there isn't a whole lot to say. You need to assume a distribution (e.g. normal). You would be able to tell much more if you could at least get the skewness and kurtosis.\n\nBe careful, remember that the mean and the standard deviation don't tell you the whole story: http://en.wikipedia.org/wiki/Anscombe%27s_quartet" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9317061,"math_prob":0.7597182,"size":1117,"snap":"2019-13-2019-22","text_gpt3_token_len":245,"char_repetition_ratio":0.12578617,"word_repetition_ratio":0.0,"special_character_ratio":0.21038496,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97970206,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-24T07:04:01Z\",\"WARC-Record-ID\":\"<urn:uuid:85701402-c0e2-4cf9-b898-833fa261fef1>\",\"Content-Length\":\"137260\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c86585a1-c034-4a6e-b943-e4c2327954d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:99df15d0-2a2a-483a-a604-1ba37710c6b7>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/2294/how-can-i-compare-distributions-using-only-mean-and-standard-deviation\",\"WARC-Payload-Digest\":\"sha1:UF6S4L4D2NGJDQIXC5Y5QA6CUOCZQD4Z\",\"WARC-Block-Digest\":\"sha1:UP63CGXO4L53YRORSB5OINIJOUEFVBWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257553.63_warc_CC-MAIN-20190524064354-20190524090354-00353.warc.gz\"}"}
https://factorization.info/factors/0/factors-of-1058.html	[ "Factors of 1058", null, "We have all the information you will ever need about the Factors of 1058. We will provide you with the definition of Factors of 1058, show you how to find the Factors of 1058, give you all the Factors of 1058, tell you how many Factors 1058 has, and supply you with all the Factor Pairs of 1058 to prove that our answer is solved correctly.\n\nFactors of 1058 definition\nThe Factors of 1058 are all the integers (positive and negative whole numbers) that you can evenly divide into 1058. 1058 divided by a Factor of 1058 will equal another Factor of 1058.\n\nHow to find the Factors of 1058\nSince the Factors of 1058 are all the numbers that you can evenly divide into 1058, we simply need to divide 1058 by all numbers up to 1058 to see which ones result in an even quotient. When we did that, we found that these calculations resulted in an even quotient:\n\n1058 ÷ 1 = 1058\n1058 ÷ 2 = 529\n1058 ÷ 23 = 46\n1058 ÷ 46 = 23\n1058 ÷ 529 = 2\n1058 ÷ 1058 = 1\n\nThe Postive Factors of 1058 are therefore all the numbers we used to divide (divisors) above to get an even number. Here is the list of all Postive Factors of 1058 in numerical order:\n\n1, 2, 23, 46, 529, and 1058.\n\nFactors of 1058 include negative numbers. Therefore, all the Positive Factors of 1058 can be converted to negative numbers. The list of Negative Factors of 1058 are:\n\n-1, -2, -23, -46, -529, and -1058.\n\nHow many Factors of 1058?\nWhen we counted the Factors of 1058 that we listed above, we found that 1058 has 6 Positive Factors and 6 Negative Factors. Thus, the total number of Factors of 1058 is 12.\n\nFactor Pairs of 1058\nFactor Pairs of 1058 are combinations of two factors that when multiplied together equal 1058. Here are all the Positive Factor Pairs of 1058\n\n1 × 1058 = 1058\n2 × 529 = 1058\n23 × 46 = 1058\n46 × 23 = 1058\n529 × 2 = 1058\n1058 × 1 = 1058\n\nLike we said above, Factors of 1058 include negative numbers. Minus times minus equals plus, thus you can convert the Positive Factor Pair list above by simply putting a minus in front of every factor to get all the Negative Factor Pairs of 1058:\n\n-1 × -1058 = 1058\n-2 × -529 = 1058\n-23 × -46 = 1058\n-46 × -23 = 1058\n-529 × -2 = 1058\n-1058 × -1 = 1058\n\nFactor Calculator\nDo you need the factors for a particular number? You can submit a number below to find the factors for that number with detailed explanations like we did with Factors of 1058 above.\n\nFactors of 1059\nWe hope this step-by-step tutorial to teach you about Factors of 1058 was helpful. Do you want to see if you learned something? If so, give the next number on our list a try and then check your answer here.\n\nCopyright \| Privacy Policy \| Disclaimer \| Contact" ]	[ null, "https://factorization.info/images/factors-of.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8955977,"math_prob":0.9868463,"size":2651,"snap":"2021-21-2021-25","text_gpt3_token_len":771,"char_repetition_ratio":0.25047225,"word_repetition_ratio":0.026871402,"special_character_ratio":0.36099586,"punctuation_ratio":0.086142324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986565,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T14:44:46Z\",\"WARC-Record-ID\":\"<urn:uuid:8dcc8220-44ee-446b-98de-7d8ff6b238a4>\",\"Content-Length\":\"7742\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05dfe1db-5aaf-4208-b08d-70f25d9ca790>\",\"WARC-Concurrent-To\":\"<urn:uuid:604bd186-1a81-4240-b112-9d7cbb02e5ae>\",\"WARC-IP-Address\":\"13.32.204.77\",\"WARC-Target-URI\":\"https://factorization.info/factors/0/factors-of-1058.html\",\"WARC-Payload-Digest\":\"sha1:WJNH2TPQB44R7MNG7LECOJRDIAW4B4LI\",\"WARC-Block-Digest\":\"sha1:2XTBHAZSHDEGLMQQ76FYDKTAPEY4UNH2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487612537.23_warc_CC-MAIN-20210614135913-20210614165913-00428.warc.gz\"}"}
http://magictesseract.com/odd_tesseract	[ "# PRIME ORDER TESSERACTS\n\nCreation of prime order tesseracts is just an extension of the concepts developed for the creation of Prime Order Squares and Prime Order Cubes. The greater difficulty is the size of the figure and in visualization. Sums for the figures described below were checked for accuracy because I have not determined a general proof for these figures.\n\n## ORDER-3 TESSERACTS (ternary base lines)\n\nThe order-3 magic tesseract generator at right contains four sets of four base lines. Each set generates a base tesseract but the base tesseracts are not shown. Only their sum, after multiplication and addition of 1, is shown as the Order-3 Magic Tesseract. The sums section at the right side of the table determines the validity of the base lines entered. All six including the uniform integral distribution must be correct for a valid tesseract. Each base line must have a 0, 1, and 2 in any order. For a valid base tesseract, the sum of the middle numbers of the four base lines of each set must be 1, 4, or 7 (1 mod 3). Not all combinations of valid base tesseracts will yield a valid magic tesseract.\n\nThere are 58 order-3 magic tesseracts. Harvey Heinz has a listing of them and much more information about them on his Order-3 Magic Tesseracts page. Shown at right is an aspect of the tesseract designated as Index # 5 on that page. I am sure that all 58 can be generated by modifying the base lines in the figure although I have not done so. I am also sure that all 384 aspects of each of the 58 tesseracts can be generated with different arrangements of the base lines. There are 6^16 possible arrangements.\n\nAn aspect of each of the first four tesseracts on Heinz's Order-3 Magic Tesseracts page can be generated by changing all of the base lines in one of the base line sets to 012. Separate modification of each set will produce a different index numbered tesseract. The tesseract initially shown and these four are all the tesseracts that have a 1 as their lowest corner number and thus a zero for the first number of all base lines.\n\nTo obtain additional order-3 tesseracts some of the base lines must start with 1 or 2. If all of the base lines except one of the base lines in the last set start with zero then the order-3 tesseract will have as its lowest corner number 2 or 3 depending on what number the exception base line starts on. To make 4-9 the smallest corner number, the next to last base tesseract must have one of its base lines start with 1 or 2.\n\n## ORDER-5 TESSERACTS (quinary base lines)\n\nThere are only two quinary base lines that can be used to generate order-5 base figures. These are 01234 and 02413. They are generated from the equations x mod 5 and 2x mod 5 for 0 ≤ x ≤ 4. These lines can be shifted (12340, 23401, etc) and/or reversed (04321, 31420, etc.). But for purposes of building pan-magic figures, shifted and reversed lines are the same. An order-5 base pan tesseract must therefore be composed of all the same base line or some combination of the two different base lines.\n\nOnly the base tesseracts composed of the same base line produce an interesting result. These base tesseracts are pan-3-agonal. Combining base tesseracts with the same base line will yield an intermediate that does not have uniform integral distribution. Shifting base lines on one of the base tesseracts does not help, but reversing some lines will give tesseract combinations with uniform integral distribution. By systematically reversing lines an order-5 pan-3-agonal tesseract can be built.\n\nThe 5 tess worksheet of the TesseractLines Excel Spreadsheet available on the Downloads page gives the master base lines of two order-5 pan-3-agonal tesseracts. The first just has the pan feature while the second is also associated. The main quadragonals in the second tesseract also add to S. Sums for the tesseract are given in the spreadsheet.\n\n## ORDER-7 TESSERACTS\n\nThere are three base 7 base lines that can be used to generate order-7 base figures. These are 0123456, 0246135, and 0362514. They are generated from the equations x mod 7, 2x mod 7, and 3x mod 7 for 0 ≤ x ≤ 6. An order-7 base tesseract can therefore be composed of all the same base line, some combination of two different base lines, or a combination of all three base lines.\n\nWhen base tesseracts are made from four sets of four base seven base lines, both order-7 pan-3-agonal and pan-4-agonal magic tesseracts can be made. The pan-3-agonal base tesseracts as above can be made using four identical base lines, but they can also be made from three of one base line and one of another. Other combinations of three and one will yield a pan-4-agonal tesseract. Combinations of all three base lines or of two of one and two of another do not yield order-7 pan-x-agonal base tesseracts of any type. Combining four base tesseracts, 343(tesseract 1) + 49(tesseract 2) + 7(tesseract 3) + tesseract 4, can yield a pan-x-agonal tesseract if all four base tesseracts are pan-x-agonal and x is always the same. Uniform integral distribution must be ensured.\n\nThe 7 tess worksheet of the TesseractLines Excel Spreadsheet available on the Downloads page gives the master base lines of four order-7 pan-x-agonal tesseracts. Two are pan-3-agonal and two are pan-4-agonal. In each pair, one is just pan-x-agonal while the other is also associated. The main quadragonals of the associated pan-3-agonal tesseract also add to S. The sums for the figures are included in the spreadsheet.\n\n## ORDER-11 TESSERACTS\n\nThere are five base 11 base lines that can be used to generate order-11 base figures. They are generated from the equations x mod 11, 2x mod 11, 3x mod 11, 4x mod 11, and 5x mod 11 for 0 ≤ x ≤ 10. As more base lines become available, more base line combinations become possible and more pan-x-agonal tesseract types become possible.\n\nWith order-11 base lines it is possible to make pan-3-agonal, pan-4-agonal, and pan-3,4-agonal magic tesseracts. The base line combinations that will give the various types of pan-x-agonal base tesseracts become more varied as the order gets larger. I haven't determined a pattern except that base tesseracts made from four of the same base line always give pan-3-tesseracts.\n\nThe 11 tess worksheet of the TesseractLines Excel Spreadsheet available on the Downloads page gives the master base lines of six order-11 pan-x-agonal tesseracts. There are two pan-3-agonal, two pan-4-agonal, and two pan-3,4-agonal. In each pair, one is just pan-x-agonal while the other is also associated. The main quadragonals of the associated pan-3-agonal tesseract also add to S. A rudimentary sum checker for the figures is included in the spreadsheet. I can provide a better checker for anyone interested.\n\n## ORDER-13 TESSERACTS\n\nThere are six base 13 base lines for generating order-13 base figures. In addition to the three tesseract types that can be generated for order-11 tesseracts, order-13 pan-2-agonal and pan-2,4-agonal tesseracts can be built.\n\nThe 13 tess worksheet of the TesseractLines Excel Spreadsheet available on the Downloads page gives the master base lines of ten order-13 pan-x-agonal tesseracts. There are two pan-2-agonal, two pan-3-agonal, two pan-4-agonal, two pan-2,4-agonal, and two pan-3,4-agonal tesseracts. In each pair, one is just pan-x-agonal while the other is also associated. The main quadragonals of the associated pan-2-agonal and pan-3-agonal tesseracts also add to S. A rudimentary sum checker for the figures is included in the spreadsheet. I can provide a better checker for anyone interested.\n\n## ORDER-17 TESSERACTS\n\nNasik tesseracts for prime orders are possible when o ≥ 17. These are constructed from base tesseracts made from four different base lines. If the four different base lines are picked correctly they will yield a nasik base tesseract otherwise they will generate pan-2-agonal, pan-2,3-agonal, or pan-2,4-agonal tesseracts. Two order-17 nasik tesseracts are given in the 17 tess worksheet of the TesseractLines Excel Spreadsheet available on the Downloads page.\n\nFor order-17 and larger tesseracts, all seven pan-x-agonal types are possible. The 17 tess worksheet has two of each of the seven pan-x-agonal types. One of each pair is just the pan-x-agonal tesseract and one is also associated. For the pan-2-agonal, pan-3-agonal, and pan-2,3-agonal tesseracts, the associated tesseract also has all main quadragonals equal to S.\n\n## LARGER PRIME ORDER NASIK TESSERACTS\n\nIt should be possible to make a nasik tesseract for any prime order ≥ 17. I offer the following procedure for an order-o tesseract without proof. I believe it to be correct based primarily on a lack of failure to date.\n\nThe 4 base lines x mod o, 2x mod o, 4x mod o, and 8x mod o with 0 ≤ x < o should always give valid base tesseracts. By systematically moving base lines in this, set a valid nasik tesseract can be built. The process is as follows.\n\n1. Let the four base lines be: a = x mod o, b = 2x mod o, c = 4x mod o, and d = 8x mod o for 0 ≤ x < o.\n2. For the first base tesseract let the row base line be a, the column b, the pillar c, and the file d.\n3. For the second base tesseract let the row base line be b, the column c, the pillar d, and the file a.\n4. For the third base tesseract let the row base line be c, the column d, the pillar a, and the file b.\n5. For the last base tesseract let the row base line be d, the column a, the pillar b, and the file c.\n\nThe four base tesseracts can be combined as o3(base tesseract 1) + o2(base tesseract 2) + o(base tesseract 3) + base tesseract 4 to make the master base lines. The value at position nijkl in the magic tesseract is: nijkl = o3((row1i + column1j + pillar1k + file1l) mod o) + o2((row2i + column2j + pillar2k + file2l) mod o) + o((row3i + column3j + pillar3k + file3l) mod o) + ((row4i + column4j + pillar4k + file4l) mod o)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8964986,"math_prob":0.8496605,"size":9718,"snap":"2021-43-2021-49","text_gpt3_token_len":2408,"char_repetition_ratio":0.22493309,"word_repetition_ratio":0.13446397,"special_character_ratio":0.23224944,"punctuation_ratio":0.0840685,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9600697,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T12:29:32Z\",\"WARC-Record-ID\":\"<urn:uuid:2de795fe-f6e5-4bc5-b16d-6d99f64b8769>\",\"Content-Length\":\"32181\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96ccd367-61ee-4330-ab7a-0a3db1ca729e>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d06224e-98b7-4b07-b2f4-0b0036b07948>\",\"WARC-IP-Address\":\"45.56.73.13\",\"WARC-Target-URI\":\"http://magictesseract.com/odd_tesseract\",\"WARC-Payload-Digest\":\"sha1:ZOCT4EVS5GCNVBADLXP4ZRIJZETUWJJH\",\"WARC-Block-Digest\":\"sha1:IRSPUNTZYSCTJAAAVT5AHUNH5N32TCZX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585405.74_warc_CC-MAIN-20211021102435-20211021132435-00672.warc.gz\"}"}
https://www.hackmath.net/en/math-problem/8288?tag_id=154	[ "# Concentric circles\n\nThere is given a circle K with a radius r = 8 cm. How large must a radius have a smaller concentric circle that divides the circle K into two parts with the same area?\n\nCorrect result:\n\nr2 = 5.6569 cm\n\n#### Solution:", null, "We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!", null, "#### You need to know the following knowledge to solve this word math problem:\n\nWe encourage you to watch this tutorial video on this math problem:\n\n## Next similar math problems:\n\n• Maxwell’s inductance bridge", null, "The four arms of Maxwell’s inductance bridge are; Arm AB contains an Inductive coil of inductance L1 having resistance R1. Arms BC and CD contain non-inductive resistances of 200Ω and 100Ω respectively. Arm AD contains standard inductor of inductance L2 a\n• The projectile", null, "The projectile was fired horizontally from a height of h = 25 meters above the ground at a speed of v0 = 250 m/s. Find the range and flight time of the projectile.\n• As shown", null, "As shown, in △ ABC, ∠C = 90°, AD bisects ∠BAC, DE⊥AB to E, BE = 2, BC = 6, then the perimeter of △ BDE\n• What are 2", null, "What are the two more terms of the GP a, ax, ax2, ax3, __, __?\n• Find the 19", null, "Find the 1st term of the GP ___, -6, 18, -54.\n• If the 3", null, "If the 6th term of a GP is 4 and the 10th is 4/81, find common ratio r.\n• Up and down motion", null, "We throw the body from a height h = 5 m above the Earth vertically upwards v0 = 10 m/s. How long before we have to let the second body fall freely from the same height to hit the Earth at the same time?\n• Sum of GP members", null, "Determine the sum of the GP 30, 6, 1.2, to 5 terms. What is the sum of all terms (to infinity)?\n• Acute triangle", null, "In the acute triangle KLM, V is the intersection of its heights and X is the heel of height to the side KL. The axis of the angle XVL is parallel to the side LM and the angle MKL is 70°. What size are the KLM and KML angles?\n• Telco company", null, "The upstairs communications company offers customers a special long distance calling rate that includes a \\$0.10 per minute charge. Which of the following represents this fee scheduale where m represents the number of minutes and c is the overall cost of t\n• Trapezoid 25", null, "Trapezoid PART with AR\|\|PT has (angle P=x) and (angle A=2x) . In addition, PA = AR = RT = s. Find the length of the median of Trapezoid PART in terms of s.\n• Sphere cut", null, "A sphere segment is cut off from a sphere k with radius r = 1. The volume of the sphere inscribed in this segment is equal to 1/6 of the volume of the segment. What is the distance of the cutting plane from the center of the sphere?\n• What is 10", null, "What is the 5th term, if the 8th term is 80 and common ratio r =1/2?\n• Volume per time", null, "How long does fill take for a pump with a volume flow of 200 l per minute fill a cube-shaped tank up to 75% of its height if the length of the cube edge is 4 m?\n• Gardens", null, "The area of the square garden is 3/4 of the area of the triangular garden with sides of 80 m, 50 m, 50 m. How many meters of the fence do we need to fence a square garden?\n• Space diagonal angles", null, "Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD.", null, "How long is a ladder that touches on a wall 4 meters high and its lower part is 3 meters away from the wall?" ]	[ null, "https://www.hackmath.net/img/88/medzikruzie2.jpg", null, "https://www.hackmath.net/hashover/images/avatar.png", null, "https://www.hackmath.net/thumb/51/t_34151.jpg", null, "https://www.hackmath.net/thumb/91/t_33991.jpg", null, "https://www.hackmath.net/thumb/31/t_33831.jpg", null, "https://www.hackmath.net/thumb/31/t_33631.jpg", null, "https://www.hackmath.net/thumb/41/t_33641.jpg", null, "https://www.hackmath.net/thumb/1/t_33601.jpg", null, "https://www.hackmath.net/thumb/51/t_33951.jpg", null, "https://www.hackmath.net/thumb/71/t_33671.jpg", null, "https://www.hackmath.net/thumb/51/t_33851.jpg", null, "https://www.hackmath.net/thumb/81/t_33781.jpg", null, "https://www.hackmath.net/thumb/31/t_34031.jpg", null, "https://www.hackmath.net/thumb/31/t_33331.jpg", null, "https://www.hackmath.net/thumb/61/t_33661.jpg", null, "https://www.hackmath.net/thumb/81/t_34481.jpg", null, "https://www.hackmath.net/thumb/91/t_33891.jpg", null, "https://www.hackmath.net/thumb/81/t_32981.jpg", null, "https://www.hackmath.net/thumb/31/t_33231.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9038178,"math_prob":0.9868767,"size":3279,"snap":"2020-45-2020-50","text_gpt3_token_len":909,"char_repetition_ratio":0.123664126,"word_repetition_ratio":0.011976048,"special_character_ratio":0.27294907,"punctuation_ratio":0.091794156,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986958,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T06:24:49Z\",\"WARC-Record-ID\":\"<urn:uuid:d8d56784-0e19-41de-8a54-cd78e73cc28f>\",\"Content-Length\":\"63554\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2693305-930b-471e-a1df-5209e28c1997>\",\"WARC-Concurrent-To\":\"<urn:uuid:71af969f-b26a-422a-9613-73afc7a476e7>\",\"WARC-IP-Address\":\"104.24.104.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/math-problem/8288?tag_id=154\",\"WARC-Payload-Digest\":\"sha1:5AZJC3OIO6LYI4GYVWU5OIG6D42ZGEMD\",\"WARC-Block-Digest\":\"sha1:RV6BOKV2KGQO53TWNZQUVXLOG4OUMYVD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878921.41_warc_CC-MAIN-20201022053410-20201022083410-00393.warc.gz\"}"}
https://npirl.blogspot.com/2008/03/mathematical-sunday-with-seifert.html	[ "## Sunday, March 9, 2008\n\n### Mathematical Sunday with Seifert Surface\n\nAh, Seifert Surface! Every encounter with that postdoctoral mathematician is enlightening. Take today for example...\n\nBettina Tizzy: How are things at xyz? Teleport directly to Seifert's new sim, xyz, from here.\n\nSeifert Surface: Just made a sculpty thing which is the graph of the function xyz = 1.", null, "It is 24 prims all told. It would be 12 except that plane sculpties are \"one sided,\" so Seifert needed twice as many.\n\nOf course, I had to ask Seifert for a simplified explanation...\n\nSeifert Surface: You have coordinates for points in space, x,y,z. You can ask, what are the points x,y,z so that xyz = 1? That is, you multiply them together and get 1 so, eg: if x, y and z are all 1, it satisfies the equation. 1,1,1 is on my graph, so is 1,-1,-1\n\nBettina Tizzy: But we are at 500mts, so how do you get 1?\n\nSeifert Surface: Ah, yes, true. The center of this thing is assumed to be 0,0,0 for purposes of the thing, and it goes out to 8 and -8 in each direction... otherwise you wouldn't be able to see any of it. You know like the graph of y = x^2? If you see it on a computer screen, then the origin is not at the bottom left corner of the screen, and the scale is not 1 unit to 1 meter. You can scale and move the graph somewhere else, and it's still the graph of y=x^2. Same thing here. This is just a 3D version of a normal graph.\n\nBettina Tizzy: What applications can this be given?\n\nSeifert Surface: Well, learning about functions. Unfortunately, it's a real pain to make something like this at present... something that you could use to quickly graph things would be very useful for teaching. You could have an extension to prims, in the same vein as for sculpties. With sculpties, you send a texture to encode the data of the shape. Instead, you could send the mathematical formula for the shape. Then we'd be in business and, of course, animated stuff would work well, too.\n\nBettina Tizzy: What is this shield item?", null, "Seifert Surface: A prototype for the larger piece. I had a smaller range on the graph before, but it looks better with more. That one goes from 0 to 2. The big one is 0 to 8.", null, "Bettina Tizzy: Fits better than a glove! More like skin.\n\nSeifert Surface: Should be the exact same function. It fits!" ]	[ null, "https://3.bp.blogspot.com/_fEglkkU__oM/R9RWMMC0sfI/AAAAAAAABvM/uqZv5YtJh4U/s400/Seifert+Surface+in+Second+Life.jpg", null, "https://4.bp.blogspot.com/_fEglkkU__oM/R9RU2cC0sdI/AAAAAAAABu8/Q11uFx76Ras/s400/Seifert+1.jpg", null, "https://2.bp.blogspot.com/_fEglkkU__oM/R9RVl8C0seI/AAAAAAAABvE/HE8wMl6pg8c/s400/Seifert+fits.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9296832,"math_prob":0.8565953,"size":2545,"snap":"2021-04-2021-17","text_gpt3_token_len":681,"char_repetition_ratio":0.12908304,"word_repetition_ratio":0.047826085,"special_character_ratio":0.24950884,"punctuation_ratio":0.19354838,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9597099,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T07:04:05Z\",\"WARC-Record-ID\":\"<urn:uuid:508b6a93-63f7-45d6-953b-a7ce8913d156>\",\"Content-Length\":\"420820\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d9d49e1-319e-4615-8051-f8cc4b1b8d5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:78ac25a9-a8e4-4247-a192-f4be0a8d849e>\",\"WARC-IP-Address\":\"142.250.65.65\",\"WARC-Target-URI\":\"https://npirl.blogspot.com/2008/03/mathematical-sunday-with-seifert.html\",\"WARC-Payload-Digest\":\"sha1:VLV6NZQLCSZNHVYYZ2V6XYPQFICZHHJC\",\"WARC-Block-Digest\":\"sha1:CKZ5QX2COPON3KUUTHBUDVVYGF6GTBOM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088731.42_warc_CC-MAIN-20210416065116-20210416095116-00370.warc.gz\"}"}
https://zh.wikipedia.org/zh/%E5%8F%8D%E6%BC%94	[ "# 反演\n\n• $O$", null, "直線：直線\n• $O$", null, "：不過$O$", null, "的直線\n• 不過$O$", null, "的圓：圓\n• $O$", null, "的球：不過$O$", null, "的平面\n\n$x_{i}\\rightarrow {\\frac {k^{2}x_{i}}{\\sum _{j}x_{j}^{2}}}$", null, "• 立體投影法：可以取球面上任意一點為中心，球的直徑為$k$", null, "• 共軸圓：在平面取一系列共心圓，取一系列經過共心圓圓心的線，任意取一點為中心進行反演。" ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9d70e1d0d87e2ef1092ea1ffe2923d9933ff18fc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/3a059e3672617fe96cf0c87f0d25bf3e5c788865", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c9a2c7b599b37105512c5d570edc034056dd40", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.7082088,"math_prob":1.0000032,"size":446,"snap":"2023-40-2023-50","text_gpt3_token_len":338,"char_repetition_ratio":0.25339368,"word_repetition_ratio":0.0,"special_character_ratio":0.3632287,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999099,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T17:32:10Z\",\"WARC-Record-ID\":\"<urn:uuid:824bec63-46f2-44fb-b1d4-fcd518e61ba8>\",\"Content-Length\":\"72402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0f34331-57f6-4a50-b8f8-5fcd7f845c1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:06eccd67-40c5-4011-a9ae-fdccf97107f1>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://zh.wikipedia.org/zh/%E5%8F%8D%E6%BC%94\",\"WARC-Payload-Digest\":\"sha1:OVL5AMUNULM44K6CQUF2JSYSFZIWL7ER\",\"WARC-Block-Digest\":\"sha1:DRPESLMAWR7KBMPZVK3WBCL53AECYRCK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100762.64_warc_CC-MAIN-20231208144732-20231208174732-00880.warc.gz\"}"}
https://physics.stackexchange.com/questions/640905/should-quarks-be-grouped-together-in-feynman-diagrams	[ "# Should quarks be grouped together in Feynman diagrams?\n\nI have a questions regarding Feynman diagrams when it involves hadrons. I think the easiest way is considering this reactions:\n\n1) $$\\Lambda_{c}^{+} \\rightarrow p+\\bar{K}^{0}$$\n\nI need to draw the feynman diagram for this reaction. I know that the quark contents of the particles are the following: $$\\Lambda_{c}^{+}=u d c,\\; p=u u d, \\;\\bar{K}^{0}=s \\bar{d}$$.\nI know that it is a weak interaction and the c-quark decay directly to the s-quark or the d-quark + any other particles allowed by the appropriate laws. So after analyzing the possible decays (quark mixing and so on), I made the following feynman diagram:", null, "However I looked up the solution, and it was very similar but still different:", null, "The solution doesn't group together the quarks in the correct way, so that it looks like the quarks form the hadrons given in the interaction. Is that because it's not important that you group them together, as long the correct quark constituents are drawn on the right side? I mean if the reaction created the $$\\Lambda=uds$$, $$\\pi^{+}= u \\bar{d}$$, would they have the excact same feynman diagram?\n\nA Feynman diagram shouldn't show spectator particles. It shouldn't care about anything composite. The fact that the incoming and outgoing quarks remain trapped in a certain hadron isn't relevant to the interaction the diagram shows. You just want the classic $$2$$-in,-$$2$$-out$$^\\dagger$$ setup per diagram, so each vertex/edge is elementary, and you can explain the hadronic implications afterwards.\n$$^\\dagger$$ Where anti-in=out, anti-out=in." ]	[ null, "https://i.stack.imgur.com/zv5smm.png", null, "https://i.stack.imgur.com/UM5mqm.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9109427,"math_prob":0.98485327,"size":1087,"snap":"2022-27-2022-33","text_gpt3_token_len":277,"char_repetition_ratio":0.12188365,"word_repetition_ratio":0.0,"special_character_ratio":0.2474701,"punctuation_ratio":0.097222224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99832547,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T02:40:41Z\",\"WARC-Record-ID\":\"<urn:uuid:e7d0756a-6496-4e89-8829-b77c7547cb3f>\",\"Content-Length\":\"224236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e7b33d3-8d65-423e-a5f7-dfb29e4df92f>\",\"WARC-Concurrent-To\":\"<urn:uuid:91c51170-6db3-412f-a9c3-09babc67108e>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/640905/should-quarks-be-grouped-together-in-feynman-diagrams\",\"WARC-Payload-Digest\":\"sha1:WDAYPGVZZII767M2KRJDRESYL6VEPN2H\",\"WARC-Block-Digest\":\"sha1:Y7VU4HW7XSPUGHKUKBPTXWP5XKME4YYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573876.92_warc_CC-MAIN-20220820012448-20220820042448-00390.warc.gz\"}"}
https://whatisconvert.com/440-cubic-centimeters-in-teaspoons	[ "# What is 440 Cubic Centimeters in Teaspoons?\n\n## Convert 440 Cubic Centimeters to Teaspoons\n\nTo calculate 440 Cubic Centimeters to the corresponding value in Teaspoons, multiply the quantity in Cubic Centimeters by 0.20288413535365 (conversion factor). In this case we should multiply 440 Cubic Centimeters by 0.20288413535365 to get the equivalent result in Teaspoons:\n\n440 Cubic Centimeters x 0.20288413535365 = 89.269019555608 Teaspoons\n\n440 Cubic Centimeters is equivalent to 89.269019555608 Teaspoons.\n\n## How to convert from Cubic Centimeters to Teaspoons\n\nThe conversion factor from Cubic Centimeters to Teaspoons is 0.20288413535365. To find out how many Cubic Centimeters in Teaspoons, multiply by the conversion factor or use the Volume converter above. Four hundred forty Cubic Centimeters is equivalent to eighty-nine point two six nine Teaspoons.\n\n## Definition of Cubic Centimeter\n\nA cubic centimeter (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume which is derived from SI-unit cubic meter. One cubic centimeter is equal to 1⁄1,000,000 of a cubic meter, or 1⁄1,000 of a liter, or one milliliter; therefore, 1 cm3 ≡ 1 ml.\n\n## Definition of Teaspoon\n\nA teaspoon (occasionally \"teaspoonful\") is a unit of volume, especially widely used in cooking recipes and pharmaceutic prescriptions. It is abbreviated as tsp. or, less often, as t., ts., or tspn. In the United States one teaspoon as a unit of culinary measure is 1⁄3 tablespoon, that is, 4.92892159375 ml; it is exactly 1 1⁄3 US fluid drams, 1⁄6 US fl oz, 1⁄48 US cup, and 1⁄768 US liquid gallon and 77⁄256 or 0.30078125 cubic inches. For nutritional labeling on food packages in the US, the teaspoon is defined as precisely 5 ml.\n\n## Using the Cubic Centimeters to Teaspoons converter you can get answers to questions like the following:\n\n• How many Teaspoons are in 440 Cubic Centimeters?\n• 440 Cubic Centimeters is equal to how many Teaspoons?\n• How to convert 440 Cubic Centimeters to Teaspoons?\n• How many is 440 Cubic Centimeters in Teaspoons?\n• What is 440 Cubic Centimeters in Teaspoons?\n• How much is 440 Cubic Centimeters in Teaspoons?\n• How many tsp are in 440 cm3?\n• 440 cm3 is equal to how many tsp?\n• How to convert 440 cm3 to tsp?\n• How many is 440 cm3 in tsp?\n• What is 440 cm3 in tsp?\n• How much is 440 cm3 in tsp?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8368197,"math_prob":0.96591294,"size":2282,"snap":"2021-04-2021-17","text_gpt3_token_len":646,"char_repetition_ratio":0.22958736,"word_repetition_ratio":0.07631579,"special_character_ratio":0.2971078,"punctuation_ratio":0.13274336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98329514,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T22:05:43Z\",\"WARC-Record-ID\":\"<urn:uuid:33ed4aec-a842-4333-848e-eddcc1c2bdf9>\",\"Content-Length\":\"32358\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c70ba490-048f-44cb-aec4-67abd48f6755>\",\"WARC-Concurrent-To\":\"<urn:uuid:77ef9a29-8b10-4628-8259-09a9e55d05cf>\",\"WARC-IP-Address\":\"172.67.211.83\",\"WARC-Target-URI\":\"https://whatisconvert.com/440-cubic-centimeters-in-teaspoons\",\"WARC-Payload-Digest\":\"sha1:DJ32KAVMJNRN5K7Z52II5LI7XR3CDQMH\",\"WARC-Block-Digest\":\"sha1:O34CMOYGU25NIPP5DDCYSNL2WVX7MF6N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519784.35_warc_CC-MAIN-20210119201033-20210119231033-00051.warc.gz\"}"}
https://lists.boost.org/Archives/boost/1999/11/0882.php	[ "", null, "# Boost :\n\nFrom: scleary_at_[hidden]\nDate: 1999-11-09 09:47:10\n\n> Regarding what should be the default return value of the PRNG,\n> throughout I have assumed that it would be an integer, most likely\n> 32-bit. My PERSONAL preference, though, would be for a real,\n> either in [0,1] or [0,1). I find these ranges far more useful in far more\n> contexts than the straight 32-bit integer. Furthurmore, given the\n> unsuitability of the modulus technique to map the PRNG integer output\n> onto a certain range, this means that virtually all random number\n> requests, except those that request a 32-bit integer directly, will\n> first map the PRNG output to this range before remapping it\n> to whatever range is required. However, concerns about the cost of the\ndivision\n> involved, when such division may be uneccessary (imagine users who do\n> need the 32-bit integers -- if we go the [0,1] route, they will be\n> getting numbers that have been divided by 0xfffffff and then\n> multiplied by 0xfffffff: perverse!) led me to stick to the \"PRNG gives\n32-bit\n> integer\" assumption.\n>\n\nThere is another possibility. What if we design the random number classes\nto use fixed precision arithmetic internally? That would avoid the overhead\nof floating-point, and also have the default range be [0,1) (assuming all\nbits are fractional bits), and also have a direct mapping (no mult/div\nnecessary) to the range of the underlying unsigned integer type: [0,\nnumeric_limits<underlying_int_type>::max()].\n\n-Steve" ]	[ null, "https://lists.boost.org/boost/images/boost.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8495699,"math_prob":0.88979846,"size":1593,"snap":"2019-51-2020-05","text_gpt3_token_len":411,"char_repetition_ratio":0.10195091,"word_repetition_ratio":0.0,"special_character_ratio":0.25298178,"punctuation_ratio":0.14649682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96470875,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-25T16:49:57Z\",\"WARC-Record-ID\":\"<urn:uuid:c5f1212d-bc48-4aa3-a1b7-7220657db902>\",\"Content-Length\":\"12348\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72b2c527-5bd6-4c6a-bfa1-25cc03f6d78c>\",\"WARC-Concurrent-To\":\"<urn:uuid:73b14b81-e1ea-4f4d-8860-174cdf9d834b>\",\"WARC-IP-Address\":\"146.20.110.251\",\"WARC-Target-URI\":\"https://lists.boost.org/Archives/boost/1999/11/0882.php\",\"WARC-Payload-Digest\":\"sha1:DJPBJUSIR3QYKPE26A7S6SDUSTTIDNGC\",\"WARC-Block-Digest\":\"sha1:TTTAALFRPHX4PQGWB7MSOUS2A5KJVBAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251678287.60_warc_CC-MAIN-20200125161753-20200125190753-00175.warc.gz\"}"}
https://www.intmath.com/forum/differentiation-transcendental-28/derivative-of-log-function:143	[ "Search IntMath\nClose\n\n450+ Math Lessons written by Math Professors and Teachers\n\n5 Million+ Students Helped Each Year\n\n1200+ Articles Written by Math Educators and Enthusiasts\n\nSimplifying and Teaching Math for Over 23 Years\n\n# derivative of log function [Solved!]\n\n### My question\n\nIn the chapter Derivative of the Logarithmic Function, Example #6, is it necessary to apply the change of base to get to the right solution?\n\n### Relevant page\n\n5. Derivative of the Logarithmic Function\n\n### What I've done so far\n\nBased upon the fact that\n\ndy/dx = 1/[\"argument\" xx ln(base)] xx [d/dx(\"argument\")]\n\nmy solution was\n\ndy/dx = 1/[6x ln 2] * (6) = 1/[x ln 2]\n\nX\n\nIn the chapter Derivative of the Logarithmic Function, Example #6, is it necessary to apply the change of base to get to the right solution?\nRelevant page\n\n<a href=\"/differentiation-transcendental/5-derivative-logarithm.php\">5. Derivative of the Logarithmic Function</a>\n\nWhat I've done so far\n\nBased upon the fact that\n\ndy/dx = 1/[\"argument\" xx ln(base)] xx [d/dx(\"argument\")]\n\nmy solution was\n\ndy/dx = 1/[6x ln 2] * (6) = 1/[x ln 2]\n\n## Re: derivative of log function\n\n@Phinah\n\nPlease use the math entry system so I, and others, can read your question. I have edited it just now.\n\nI checked your answer by finding 1/(ln(2)) and it has the same value, 1.4427, so your approach appears to be fine!\n\nWhenever I see a log expression with a base other than e, I automatically change base. This means I only need to learn one formula and can apply it in many places. In this case, I like the look of your approach better! :-)\n\nX\n\n@Phinah\n\nPlease use the math entry system so I, and others, can read your question. I have edited it just now.\n\nI checked your answer by finding 1/(ln(2)) and it has the same value, 1.4427, so your approach appears to be fine!\n\nWhenever I see a log expression with a base other than e, I automatically change base. This means I only need to learn one formula and can apply it in many places. In this case, I like the look of your approach better! :-)\n\n## Re: derivative of log function\n\nOk got it! Thank you for the explanation.\n\nX\n\nOk got it! Thank you for the explanation." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8712275,"math_prob":0.49323344,"size":2026,"snap":"2022-27-2022-33","text_gpt3_token_len":595,"char_repetition_ratio":0.100890204,"word_repetition_ratio":0.7722222,"special_character_ratio":0.2941757,"punctuation_ratio":0.11751152,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.97989094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T12:58:45Z\",\"WARC-Record-ID\":\"<urn:uuid:3000d8eb-e807-403b-89cd-a1700db49d70>\",\"Content-Length\":\"97953\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a625ba87-0b41-4755-9c9c-2fd86bd29040>\",\"WARC-Concurrent-To\":\"<urn:uuid:13c6a290-b2d4-46d9-ba72-78567efb97ed>\",\"WARC-IP-Address\":\"172.67.217.164\",\"WARC-Target-URI\":\"https://www.intmath.com/forum/differentiation-transcendental-28/derivative-of-log-function:143\",\"WARC-Payload-Digest\":\"sha1:YSOO5MG2MRIQZFNVIKHPGRUVHIYPSWVL\",\"WARC-Block-Digest\":\"sha1:NUREUZPXI5MSPYX7PG7B5SHG3XRUJ5NU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571950.76_warc_CC-MAIN-20220813111851-20220813141851-00795.warc.gz\"}"}
https://www.enjoyalgorithms.com/blog/validate-binary-search-tree/	[ "# Validate Binary Search Tree: Check if a Binary Tree is BST or not\n\nKey takeaway: An excellent problem to understand the properties of binary search trees and learn step-by-step optimization using various approaches.\n\n### Let’s understand the problem\n\nGiven the root of a binary tree, write a program to check whether it is a valid binary search tree (BST) or not. A BST is valid if it has the following properties:\n\n• All nodes in the left subtree have values less than the node’s value.\n• All nodes in the right subtree have values greater than the node’s value\n• Both left and right subtrees are also binary search trees.\n\n### Examples", null, "### Discussed solution approaches\n\n• A wrong solution approach!\n• A brute force approach: Comparing each node value with the max of the left sub-tree and min of the right sub-tree.\n• An optimized version of the brute force approach: Tracking min and the max range of values allowed for the subtree rooted at each node.\n• Using inorder traversal and extra space\n• An efficient approach: In-place solution using inorder traversal\n\n### Wrong solution approach!\n\nAn unclear understanding of this problem can lead to a wrong approach. For example, Suppose we understood the BST property incorrectly and designed this solution: We visit each node recursively and compare the value stored in each node with the value stored on the left and right child. If the node value is greater than the left child value or less than the right child value, we return false. Otherwise, we recursively call the same function to verify the left sub-tree and right sub-tree.\n\n``````boolean isValidBST (TreeNode root)\n{\nif (root == NULL)\nreturn true\n\nif (root->left != NULL && root->left->value > root->value)\nreturn false\nif (root->right != NULL && root->right->value < root->value)\nreturn false\n\nboolean left = isValidBST(root->left)\nboolean right = isValidBST(root->right)\nif(left == true && right == true)\nreturn true\nelse\nreturn false\n}\n``````\n\n#### What is the mistake in the above solution?\n\nBased on the BST property, each node value must be greater than all the nodes in the left sub-tree and less than all the nodes in the right sub-tree. But the above solution just checks this property for the children of each node, not the entire left or right subtree. So this solution will give the wrong result for several input cases. Consider this example:", null, "### Solution 1: A brute force and correct approach\n\n#### Solution Idea\n\nNow the critical question is: how can we fix the above solution? Here is an idea: A tree can not be BST if the left subtree contains any value greater than the node’s value and the right subtree contains any value smaller than the node’s value. In other words, the node value must be less than the maximum in the left sub-tree and greater than the minimum in the right sub-tree.\n\nNote: In a BST, the rightmost node would be the node with the maximum value and the leftmost node would be the node with the minimum value.\n\n#### Solution Steps\n\n• Base case: If (root == NULL ), we return true.\n• We find max in the left subtree and min in the right subtree using two helper functions i.e. leftMax = bstMax(root->left) and rightMax = bstMin(root->right).\n• If leftMax > root->value or rightMax < root->value, then root node violates the BST property and we return false.\n• Otherwise, the root node satisfies the BST property. So we recursively check that left and right subtree are correct BST or not. For this, we call the same function with root->left and root->right as the input parameters.\n``````boolean left = isValidBST(root->left)\nboolean right = isValidBST(root->right)\n``````\n• If both left and right are true, then we return true. Otherwise, we return false. Think!\n\n#### Solution Pseudocode\n\n``````boolean isValidBST(TreeNode root)\n{\nif (root == NULL)\nreturn true\nint leftMax = bstMax(root->left)\nint rightMax = bstMin(root->right)\nif (leftMax > root->value \|\| rightMax < root->value)\nreturn false\n\nboolean left = isValidBST(root->left)\nboolean right = isValidBST(root->right)\n\nif(left == true && right == true)\nreturn true\nelse\nreturn false\n}\nint bstMax(TreeNode root)\n{\nwhile(root->right != NULL)\nroot = root->right\nreturn root->value\n}\nint bstMin(TreeNode root)\n{\nwhile(root->left != NULL)\nroot = root->left\nreturn root->value\n}\n``````\n\n#### Time complexity analysis\n\nWe are traversing the tree recursively and calling bstMax(root->left) and bstMin(root->right) for each node. The worst-case time complexity of finding max in a BST = O(n), the worst-case time complexity of finding min in a BST = O(n). Think!\n\nIf we look closely, the time complexity will depend on the tree's structure. The worst-case situation will occur when the tree is high unbalanced i.e., either left-skewed (All nodes have a left child except the single leaf node) or a right-skewed tree (All nodes have a right child except the single leaf node). So worst-case time complexity to validate BST = nO(n) = O(n^2).\n\nThe above code is recursive. So here are some critical questions for the learning purpose! We will add answers to these questions later in this blog.\n\n• The best-case scenario will occur when the tree is balanced.\n• The recurrence relation for the worst case is T(n) = T(n - 1) + O(n)\n• The recurrence relation for the best case is T(n) = 2T(n/2) + O(logn)\n• Proof that the best case time complexity is O(n)! Hint: Use the first case of the master theorem.\n• What would be the space complexity?\n\n### Solution 2: An optimized version of the brute force approach\n\nWe need to traverse some nodes several times in the above method. Can we think of solving this problem efficiently? Can we think of solving the problem without finding the max and min for each node?\n\n#### Solution Idea\n\nIn a BST, the value stored in a subtree rooted at any node lies in a particular range. In other words, there is an upper (rangeMax) and a lower (rangeMin) limit of values that lie in a specific subtree. For example:\n\n• The tree with a root node lies in the range (INTMIN, INTMAX).\n• The left subtree lies in the range (INT_MIN, root-> value -1].\n• Right subtree range is [root->value + 1, INT_MAX), and so on.\n\nOne solution would be to traverse the tree recursively and keep track of the min and max range of values allowed for the subtree rooted at each node. We pass the acceptable range as a function argument while recursing for the left and right subtree.\n\nHere we need to go through each node only once, and the initial values for the min and max range should be INTMIN and INTMAX. If node->value < rangeMin or node->value > rangeMax, then node’s value falls outside the valid range and violates the min-max constraints. So we return false. Otherwise, we recursively check left and right subtrees with an updated range.", null, "#### Solution Pseudocode\n\n``````boolean isBST(TreeNode root)\n{\nreturn isValidBST(root, INT_MIN, INT_MAX))\n}\nboolean isValidBST(TreeNode root, int rangeMin, int rangeMax)\n{\nif(root == NULL)\nreturn true\nif (root->value < rangeMin \|\| root->value > rangeMax)\nreturn false\n\nboolean left = isValidBST(root->left, rangeMin, root->value -1)\nboolean right = isValidBST(root->right, root->value + 1, rangeMax)\n\nif(left == true && right == true)\nreturn true\nelse\nreturn false\n}\n``````\n\n#### Time and space complexity analysis\n\nWe visit each node only once and perform an O(1) operation with each node. So time complexity = n O(1) = O(n). Space complexity depends on the size of the recursion call stack, which is equal to the height of the tree. So space complexity = O(h). What would be the worst and best case of space complexity? Think!\n\n### Solution 3: Using inorder traversal and extra space\n\n#### Solution Idea\n\nThe inorder traversal of a binary search tree explores node values in sorted order. So another idea would be to traverse the tree using inorder traversal and store node values in an ArrayList or vector. Now we traverse the ArrayList or vector using a loop to check whether it is sorted or not. If sorted, then the tree is a BST; otherwise, not.\n\n#### Solution Pseudocode\n\n``````boolean isValidBST (TreeNode root)\n{\nvector<int> sorted\ninorderTraversal(root, &sorted)\nfor(int i = 1; i < sorted.size(); i = i + 1)\n{\nif (sorted[i] < sorted[i - 1])\nreturn false\n}\nreturn true\n}\nvoid inorderTraversal(TreeNode root, int & sorted)\n{\nif (root == NULL)\nreturn\ninorderTraversal(root->left, sorted)\nsorted.push_back(root->value)\ninorderTraversal(root->right, sorted)\n}\n``````\n\n#### Time and space complexity analysis\n\nTime complexity = Time complexity of inorder traversal for storing elements in array list + Time complexity of the single loop to check sorted arraylist = O(n) + O(n) = O(n). Space complexity = O(n) for storing elements in an arraylist + O(h) for recursion stack space = O(n + h)\n\n### Solution 4: In-place solution using inorder traversal\n\n#### Solution Idea\n\nThe critical question is: can we think to optimize the above approach and avoid using extra space? As we know that an inorder traversal of a BST returns the nodes in sorted order. So one idea would be to keep track of previously visited nodes while traversing the tree. If the current node value is less than the previous node value, then the tree is not BST.\n\n#### Solution pseudocode: Recursive inorder traversal\n\n``````boolean isBST(TreeNode root)\n{\nTreeNode prev = NULL\nreturn isValidBST(root, &prev)\n}\nboolean isValidBST(TreeNode root, TreeNode &prev)\n{\nif (root == NULL)\nreturn true\nboolean left = isValidBST(root->left, prev)\nif (prev != NULL && root->value < prev->value)\nreturn false\nprev = root\nboolean right = isValidBST(root->right, prev)\nif(left == true && right == true)\nreturn true\nelse\nreturn false\n}\n``````\n\n#### Solution pseudocode: Iterative inorder traversal using stack\n\n``````boolean isValidBST(TreeNode root)\n{\nif (root == NULL)\nreturn true\nStack<TreeNode> stack\nTreeNode curr = root\nTreeNode prev = NULL\nwhile (curr != NULL \|\| stack.isEmpty()== false)\n{\nif (curr != NULL)\n{\nstack.push(curr)\ncurr = curr->left\n}\nelse\n{\ncurr = stack.pop()\nif(prev != NULL && curr->value < prev->data)\nreturn false\nprev = curr\ncurr = curr->right\n}\n}\nreturn true\n}\n``````\n\n#### Time and space complexity analysis\n\nWe visit each node only once and perform an O(1) operation with each node. So time Complexity = n* O(1) = O(n). Space complexity depends on the depth of the size of the recursion call stack, which is equal to the height of the tree. So space complexity = O(h).\n\n### Critical Ideas to think about!\n\n• How can we modify the above solution to work for duplicate values?\n• In the 4th approach using recursive inorder traversal, can we think of implementing without using the pointer by reference?\n• Can we think of implementing 2nd approach, using pointers by reference instead of passing the value?\n• Is there a way to solve this problem using some other approach?\n\n### Suggested problems to practice\n\n• Check if the given binary tree is a strict binary tree or not\n• Check whether the given binary is perfect or not\n• Check if a binary tree is a complete tree or not\n• Check if a binary tree is a subtree of another binary subtree\n• Check if the given binary tree is Heap or not\n• Check if the given binary tree is SumTree or not\n\nPlease write in the message below if you have alternative approaches or find an error/bug in the above approaches. Enjoy learning, Enjoy algorithms!" ]	[ null, "https://cdn-images-1.medium.com/max/720/1VDjeWgeN5ajFNHH80_Xbeg.png", null, "https://cdn-images-1.medium.com/max/720/1iwA_hbrSne4h3OG04drO3A.png", null, "https://cdn-images-1.medium.com/max/1080/1*9coFI61ERPnK64idFcBtNQ.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7411096,"math_prob":0.9820952,"size":9324,"snap":"2022-27-2022-33","text_gpt3_token_len":2146,"char_repetition_ratio":0.14774679,"word_repetition_ratio":0.12666245,"special_character_ratio":0.23734449,"punctuation_ratio":0.09002849,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9975709,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T08:13:44Z\",\"WARC-Record-ID\":\"<urn:uuid:2e33d502-9011-4645-93be-1be83d36d2ff>\",\"Content-Length\":\"115813\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bb3e68f-23a6-4cc4-9e3e-84cec9718ffd>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d4bbf96-d89b-40ca-994b-b0bb3f637752>\",\"WARC-IP-Address\":\"34.196.254.27\",\"WARC-Target-URI\":\"https://www.enjoyalgorithms.com/blog/validate-binary-search-tree/\",\"WARC-Payload-Digest\":\"sha1:225B7T6OUA34TXDPIMBBF65J2J7N3EWT\",\"WARC-Block-Digest\":\"sha1:RAFXSH235YJS6WA6HEOTM2PTYGBRY22U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215805.66_warc_CC-MAIN-20220703073750-20220703103750-00250.warc.gz\"}"}
https://adotb.xyz/hannah/	[ "# Lesson Plan: 25th July\n\n## Integration by substitution\n\nWe’d cover integration by substitution first. At the A Levels, any integration that needs to be done by substitution will have the substitution given to us. We can then follow these next general steps (we will assume that $x$ is our original variable and $u$ is our new variable.\n\n1. Differentiate the given substitution (to change the “$\\mathrm{d}x$” into “$\\mathrm{d}u$”.\n2. Substitute both $x$ and “$\\mathrm{d}x$” into the “$u$”-counterparts.\n3. After substitution and simplification, we should now have an integral that can be done relatively simply.\n4. Change all $u$s back to the original variables $x$’s.\n\n### Example 1\n\n20d) Use the substitution $u=e^{2x}$ to find $\\displaystyle \\int \\frac{e^{2x}}{\\sqrt{1-e^{4x}}} , \\mathrm{d}x$.\n\nStep 1: $\\displaystyle \\frac{\\mathrm{d}u}{\\mathrm{d}x} = 2 e^{2x}$\n\n“$\\displaystyle \\mathrm{d}x = \\frac{1}{2e^{2x}} , \\mathrm{d}u$”\n\nStep 2:\n\n$\\displaystyle \\int \\frac{e^{2x}}{\\sqrt{1-e^{4x}}} , \\mathrm{d}x =\\int \\frac{u}{\\sqrt{1-u^2}} \\left( \\frac{1}{2u} , \\mathrm{d}u\\right)$\n\nStep 3:\n$\\displaystyle = \\frac{1}{2} \\int \\frac{1}{\\sqrt{1-u^2}} , \\mathrm{d}u = \\frac{1}{2} \\sin^{-1} u + c$\n\nStep 4:\n$\\displaystyle = \\frac{1}{2} \\sin^{-1} (e^{2x}) +c$\n\n### Example 2:\n\nFor trigo substitutions there may be a few extra tricks, such as the use of our Pythagorean/square formulas, and using the right angled triangle to simplify certain steps (in step 4 in this example: I’d draw it out for you during class).\n\nModified 17e) Use the substitution $x = \\sec \\theta$ to find $\\displaystyle \\int \\frac{1}{\\sqrt{x^2-1}} , \\mathrm{d}x$.\n\nStep 1: $\\displaystyle \\frac{\\mathrm{d}x}{\\mathrm{d}\\theta} = \\sec \\theta \\tan \\theta$\n\n“$\\displaystyle \\mathrm{d}x = \\sec \\theta \\tan \\theta , \\mathrm{d}\\theta$”\n\nStep 2:\n\n$\\displaystyle \\int \\frac{1}{\\sqrt{x^2-1}} , \\mathrm{d}x =\\int \\frac{1}{\\sqrt{\\sec^2 \\theta – 1}} \\left( \\sec \\theta \\tan \\theta , \\mathrm{d}\\theta \\right)$\n\nStep 3:\n$\\displaystyle =\\int \\frac{1}{\\sqrt{\\tan^2 \\theta}} \\sec \\theta \\tan \\theta , \\mathrm{d}\\theta = \\int \\sec \\theta , \\mathrm{d}\\theta = \\ln \| \\sec \\theta + \\tan \\theta \|+ c$\n\nStep 4:\n$\\displaystyle = \\ln \| x + \\sqrt{x^2-1} \| +c$\n\n## Integration by parts\n\n$\\displaystyle \\int u \\frac{\\mathrm{d}v}{\\mathrm{d}x} , \\mathrm{d}x = uv – \\int \\frac{\\mathrm{d}u}{\\mathrm{d}x}v,\\mathrm{d}x$\n\nIntegration by parts is analogous to “product” rule in differentiation, except that the formula is a lot more complicated, and there is still an integration sign left after applying the formula.\n\nSuccessfully applying integration by parts involve making the second term simpler such that we can now integrate it.\n\n### Example 1\n\n14d) $\\displaystyle \\int x \\cos x , \\mathrm{d}x$\n\n$\\displaystyle \\int x \\cos x , \\mathrm{d}x = x \\sin x – \\int 1 \\sin x , \\mathrm{d}x = x \\sin x + \\cos x + c$\n\n### A discussion on order\n\nWe will discuss more about order in class, and how it is important in this technique.\nIf no hints are given, a useful heuristic is LIATE\n\n• L: Logarithm: $\\ln x$\n• I: Inverse Trigo: $\\sin^{-1}x, \\tan^{-1}x$\n• A: Algebra: $1, x, x^2$\n• T: Trigo: $\\sin x, \\cos x$\n• E: Exponential\n\nRoughly we group them into 3:\n\n• LI: Almost always ‘differentiated’ (i.e. the ‘$u$’)\n• TE: Almost always ‘integrated’ (i.e. the ‘$\\frac{\\mathrm{d}v}{\\mathrm{d}x}$’)\n• A: KIV: depending on who they’re paird with" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6580546,"math_prob":0.9999298,"size":3812,"snap":"2021-31-2021-39","text_gpt3_token_len":1379,"char_repetition_ratio":0.19353992,"word_repetition_ratio":0.08361775,"special_character_ratio":0.3549318,"punctuation_ratio":0.14321926,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999752,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T07:24:49Z\",\"WARC-Record-ID\":\"<urn:uuid:80b264ab-aed8-4411-ac04-b1f009d60f6e>\",\"Content-Length\":\"42554\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02bb04e1-4116-4a1c-b42b-923cb43a7233>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d711174-a429-4ce5-95f8-33fd4555c2bd>\",\"WARC-IP-Address\":\"173.236.228.211\",\"WARC-Target-URI\":\"https://adotb.xyz/hannah/\",\"WARC-Payload-Digest\":\"sha1:SSHYJCS6SQU2LYXUIFTUIW6K4AEK4AJQ\",\"WARC-Block-Digest\":\"sha1:SPAWLC3EMJXWEZJ77JACAKPBJLJA6YIM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057830.70_warc_CC-MAIN-20210926053229-20210926083229-00485.warc.gz\"}"}
https://ltwiki.org/index.php?title=B_sources_(common_examples)&oldid=1990	[ "# B sources (common examples)\n\n(diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff)\n\nFor a listing of all b-source functions see B sources (complete reference).\n\n### When, why and how to use tripdt and tripdv with explanation\n\nHelp states, \"If the voltage across a source changes by more than tripdv volts in tripdt seconds, that simulation time step is rejected.\"\n\nThis is rather terse, but it seems to mean, when attempting a solution for the next transient simulation step, in addition to comparing the non linear error of the next step to reltol and other step size acceptance limits, the simulation engine will perform the following test:\n\nIF (the voltage change across the b-source > tripdv AND the time change across the simulation step > tripdt) THEN reject the simulation step size, i.e. make it smaller and try again. (This is quite different than just testing if the magnitude of slope of the voltage across the source gets too large.)\n\nBy experimentation, it seems Mike's description is not exactly correct, at least for the last time step while leaving the state change constraint where it seems the allowable tripdt limit is increased by two. Here is a netlist that illustrates LTspice invoking the time step state change constraint.\n\n```V1 1 0 SINE(0 1 1k) ; input test source\nB1 2 0 V=buf(V(1)) tripdv=0.1 tripdt=10n ; b-source equivalent to a BUF a-device (with state transition time accuracy control)\n.opt plotwinsize=0\n.tran 5m\n```\n\nThe u() function is digital in nature so changing tripdv over a wide range has little effect just so long as it is a bit smaller than the state change of the unit step function. However, tripdt directly controls the maximum allowable time change through the state change except the last of these time changes may be twice the limit (no doubt due to the standard way LTspice increases time steps by doubling).\n\nThe problem with the tripdv/tripdt constraint arises when the b-source voltage is analog rather than digital in nature because it is very possible to choose these limits such that normal voltage changes exceed them. That causes an effect much like a very small maximum time step and the simulation may run very slowly.\n\nIn order to differentiate the effects on time step size and accuracy that stem from b-source tripdv and tripdt from the normal step size control algorithms, it is probably a good idea to disable or slacken the normal step controls. If maximum step size is not specified, it seems LTspice may select a max step size as a fraction of the end time and perhaps reltol and some other parameters.\n\nFor b-source step size control experimentation, I would recommend loosening up reltol and setting the step size to the same as the stop time. Also waveform compression should be disabled.\n\n```.opt reltol=10m plotwinsize=0\n```\n\nIf you put yourself in the \"shoes\" of the simulator solver, you can see that the shape of a signal can effect how easy it is to locate activity (pin down state changes or high dv/dt) in a b-source signal. For example, first consider a square wave for which you want the simulator to locate and reproduce the transitions very precisely, yet run as fast as possible between transitions. After an edge event the solver keeps increasing the step size because nothing is changing. At some point the voltage changes and the solver has to reject steps to back up in time in order to locate the latest edge event (otherwise the points across the edge would remain widely spaced and when the plot engine connects the dots, the square wave may end up looking more like a triangle wave).\n\nWhen you make a square wave with the standard voltage source, the solver knows ahead of time that it is a square wave and where the edges should be in time, but if you make a square wave with a b-source, I don't think it has any way to interpret whatever arbitrary function you have assigned it, so it must rely on tripdv and tripdt to recognize an event. However, a square wave change is easy to see because even if the edge is missed, the level has changed and remains changed for at least several time steps, so the solvers knows an event occurred that it should back up and locate.\n\nNow consider how much more difficult it is for the solver to \"see\" a narrow pulse type waveform. If the solver doesn't know ahead of time where the edges should be, it could easily blow right by the pulse, never knowing it should have occurred at all. The points on either side of the pulse will be correct so the plot engine will just connect the dots as a flat line. Unlike the case for the square wave, tightening up tripdv and triptdt will not help unless a time point chances to fall within the narrow active region of the pulse (then the edges of that pulse may be sharpened by tightening tripdv and tripdt, but most pulses will still be overlooked).\n\nSo what to do if you need to make narrow b-source pulses? First make a rectangular or saw tooth wave and key its edge from whatever will also drive the pulse.\n\nNote that a transient simulation is not continuous with time, but is actually a collection of consecutive solutions to the circuit at a series of discrete time steps. In order to make the simulation run as quickly as possible, the simulation engine constantly monitors the nonlinear error between adjacent steps and will adjust step size dynamically for a step size that doesn't lead to an unacceptable error. If the error is too small, the step size is still used, but the next time step is made larger. If the error is too large, the step size is rejected, i.e., the solution results are discarded and another solution is attempted with a smaller step size.\n\nIt seems that even if tripdt and tripdv are not specified, LTspice still applies an internal default test to accept or reject a step based on whether or not the behavioral source changes too much in one time step. Perhaps this internal default function is separate from the tripdv/tripdt test, or perhaps a default value for tripdt is estimated based on something like the the simulation end time.\n\n### Replicating a-devices with b-sources\n\nIt is a trivial exercise to replicate most digital functions by directly applying the logical operators available to b-sources in LTspice. One need only take the initiative to read their descriptions in the Help file and write the obvious expressions. This applies to buffers, inverters, and-gates, or-gates and xor-gates. With a little thought, one can easily build most more complex digital functions with b-sources as well. However, digital functions that are based on temporal states are more problematic. This class of functions includes flip-flops, edge triggered flip-flops, sample-and-hold devices and the like.\n\nFor example, b-sources do not directly provide a sample-and-hold function, yet a very efficient implementation is possible with a roundabout application of the Verilog integration function, sdt(x,ic,r), which provides integration of a variable with optional initial condition and optional reset to that initial condition (assert).\n\nWith sdt(0,ic,r), when the integrand is set to zero, the initial condition (which may vary during the simulation) will be continuously sampled as long as the reset is high and then held as long as reset is low, thus producing an ideal sample-and-hold with zero acquisition time.\n\nBuilding an edge triggered digital device also requires a similar indirect application of the Verilog differentiation function, ddt(x). To detect the positive going edge of an input clock pulse stream, it must first be squared up (ideally buffered to a logical one or zero), then differentiated and buffered again to produce a pulse only on the positive edge. Here is the SPICE code:\n\n#### * Edge triggered b-source logic and integrated averaging in LTspice\n\nThese following functions are intended to be used in b-sources. The first two replicate edge triggered logic such as is used in the a-device DFLOP clock input. The last function is more complicated as it includes three behavioral integrators to provide a running integrated average of x (starting on the falling edge of sampling pulse s) that is then held until the next rising edge of sampling pulse s. In normal use, x would be an analog input and the sampling pulse s would be a series of very narrow 1 volt positive digital pulse.\n\n```.func up(s) buf( ddt(s)) ; generates pulse on rising edge of s\n.func dn(s) buf(-ddt(s)) ; generates pulse on falling edge of s\n\n* The following averages x between sample pulses s\n.func ave(x,s) sdt(0,sdt(x,0,dn(s))/sdt(1,0,dn(s)),up(s))\n```\n```B1 0 1 I= ave(V(x),V(s)) Rpar=1 Cpar=1n ; source for x and s not shown\n```\n\n#### * D Flip-Flop (with Q-not feedback to create divide by two counter)\n\n```* Divide by two counter (D flip-flop with Q-not data feedback)\nVclk clk 0 PULSE(0 1 1u 1u 1u 1u 10u) ; input pulse source\nBdflop 0 Q I=sdt(0, inv(V(Q)), buf(ddt(buf(V(clk))))) Rpar=1 Cpar=1nF ; D Flip-Flop\n.tran 50u\n```\n\nNote that the b-source has been Nortonized into a one amp current source driving a 1nF capacitor in parallel with a one ohm resistor. This provides a picosecond time constant delay to output changes. This small delay is a convergence aid that is necessary so that the solver is able to resolve state changes of the b-source that depend directly on its own output.\n\nAlthough this b-source divide-by-two counter is robust and efficient, it is no match in speed to the equivalent LTspice native a-device (either the DFLOP or the even more versatile COUNTER device). LTspice's native a-device will always outperform the best b-source equivalent implementation, so this entire exercise is somewhat of an academic study.\n\nBecause they are general purpose devices, b-sources are computationally burdened with carrying a full Jacobian (although this can be turned off - often with disastrous results - with the NoJacob parameter). Also, since b-sources are largely analog devices, they do not have a built-in provision to recognize state changes like the digital a-devices do (yes, this can be less effectively approximated with b-sources by specifying the tripdv and tripdt parameters). The bottom line is that in LTspice it is always better to use a-devices whenever they fit the requirement.\n\n#### * Replicating the MODULATOR a-device\n\nIn the netlist below a MODULATOR a-device is set up with 1V=100Hz and 0V=0Hz. It is then driven by V1 to sweep from 50Hz to 150Hz over 1 second. The frequency control input is V(f) and its swept outputs are V(sinA) and V(cosA).\n\n```.param fo=100\nV1 f 0 PWL(0 0.5 1 1.5)\nA1 f 0 0 0 0 cosA sinA 0 MODULATOR Mark={fo} Space=0\n```\n\nThe b-sources below produce the same swept waveforms on V(sinB) and V(cosB). R1 and R2 duplicate the output impedance of the a-device and C1 and C2 are important for waveform fidelity and must be selected based on the frequency range of interest. They are needed to \"trick\" LTspice into producing enough waveform data points as the frequency increases. (Because b-sources are general purpose devices, it is often a problem to get them to correctly control step size.)\n\n```.param wo=2Pifo\nB1 0 sinB I=sin(wosdt(V(f))) Rpar=1 Cpar=1µ\nB2 0 cosB I=cos(wosdt(V(f))) Rpar=1 Cpar=1µ\n```\n\n### Building other functions with b-sources\n\n#### * State Machine Helper Logic\n\nThe internal parameter “time” is available to state machine logic. It would very useful to create a new parameter called “StateTime” that would reset to zero at each state change. This would make it much more straightforward to create transitional “timer” states. Note, assuming the state value is made available as an output, it is possible using a b-source to create a \"state-time\" node that is based on the state value as follows:\n\n```B1 st 0 V=sdt(1,0,buf(abs(ddt(V(state))))) ; node \"st\" is \"state time\"\n```\n\nAnother useful function might be to add a parameter called something like “StateChange” that would go high only during state transitions and be low otherwise. Note that this incorporated in the b-source function above as buf(abs(ddt(V(state)))).\n\n#### * Up/Down Counter in a single b-source\n\n```B1 0 cnt I=sdt(0, round(V(cnt)+buf(ddt(V(up)))-buf(ddt(V(dn)))), buf(ddt(V(up)+V(dn))) ) Rpar=1 Cpar=10n\n```\n\nThe b-source is a current source Nortonized into voltage source with a 1 ohm output impedance. The b-source parallel capacitor must be 2x the slowest rise time of the up/down inputs. In between up and down count pulses the sdt function holds the last count.\n\nThe up and down inputs, V(up) V(dn), use LTspice's standard 1 volt logic levels and trigger on the positive edges.\n\nThe cnt (count) output is an integer count that starts from zero. I haven't shown a reset, but it would be very easy to add.\n\n#### * Other, Notes, etc.\n\nI've got an equation in an arbitrary behavioral source that uses the variable TIME to plot. What I can't figure out is how to cause this to repeat at some interval.\n\nTime is a ramp that never resets. What you need is a replacement for time in your equation that resets to zero at your repeat interval, i.e., a sawtooth function.\n\nYou could set up a standalone voltage source to make the sawtooth and replace \"time\" in your equation with the sawtooth node, \"v(x)\" or you could do this directly in the equation by replacing \"time\" with either of the following expressions:\n\n• time-int(time/m)m ; where m is the modulus (repeat interval)\n• idtmod(1,0,m) ; integrates 1dt with 0 offset at modulus m\n\nIf you want, you could define such expressions with a .function statement in order to keep your b-source equation from getting too messy.\n\n```.function a(m)= time-int(time/m)*m\n.func b(m)= idtmod(1,0,m) ; .func is the short name\n.param r=4 ; r is the actual reset modulus you wish to use\n```\n\nNow, within your b-source equation, you could replace each instance of \"time\" with either \"a(r)\" or \"b(r)\" (or change the names to suit what makes sense to you).\n\nNOTES:\n\n• Explain how to make a discretized (pseudo-digital) source.\n• Demo various kinds of output limiters (tanh, limit, etc.) and explain why and where these are well worth using.\n• Show how to convert a BV source into an equivalent BI source (with parallel capacitor) and explain why this is generally a much better form to use.\n\nother??" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9078416,"math_prob":0.9524726,"size":14171,"snap":"2022-05-2022-21","text_gpt3_token_len":3279,"char_repetition_ratio":0.12938519,"word_repetition_ratio":0.009193481,"special_character_ratio":0.22059135,"punctuation_ratio":0.08065668,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97451067,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T00:59:06Z\",\"WARC-Record-ID\":\"<urn:uuid:06ba570e-d643-4f7b-bd9f-11fc90549458>\",\"Content-Length\":\"33353\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:452ae60b-2b42-4f39-9281-821f34ab8c8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ad95356-09da-4ce0-9f08-1bbae94bf4af>\",\"WARC-IP-Address\":\"74.208.236.221\",\"WARC-Target-URI\":\"https://ltwiki.org/index.php?title=B_sources_(common_examples)&oldid=1990\",\"WARC-Payload-Digest\":\"sha1:ZY4ABJZ5O3B5ZTKXHYCCQ5DUTI6RNJAZ\",\"WARC-Block-Digest\":\"sha1:ZF7ZTKIIFOQ6MWNYWQ3VJXQKLEQVK3QH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663011588.83_warc_CC-MAIN-20220528000300-20220528030300-00393.warc.gz\"}"}
https://copyassignment.com/simple-gui-calculator-using-tkinter-in-python/	[ "# GUI Calculator Using Tkinter In Python", null, "We will make a Simple GUI Calculator Using Tkinter In Python which is a python module to make GUI applications simply so that’s why we will use it to make a GUI Calculator Using tkinter\n\nHere we are going to provide you with the source code with a complete explanation so that you can easily make your GUI Calculator Using tkinter\n\n## Source Code\n\n```# importing the tkinter module\nfrom tkinter import *\n\n# initializing the tkinter\nroot = Tk()\n\n# setting the width and height of the gui\nroot.geometry(\"430x500\") # x is small case here\n\n# declaring an empty string variable\nexpression = \"\"\n\n# defining function which will set expressions and answers to the user\ndef setexpression(num):\nglobal expression\nexpression = expression + str(num)\nvalue.set(expression)\n\n# defining a function to calculate the expression entered by the user\ndef calculator():\ntry:\nglobal expression\nexcept:\nvalue.set(\"Enter correct expression\")\nexpression = \"\"\n\n# function to clear everything in expression\ndef clear():\nglobal expression\nexpression = \"\"\nvalue.set(expression)\n\n# declaring font variables as (\"Language\", size)\nlarge_font = ('Verdana', 15)\nsmall_font = ('Verdana', 10)\n\n# declaring variable to take value of expression entered by the user\nvalue = StringVar(value=\"Enter expression\")\n\n# entry widget to take expression from user and to show\n# calculations\nEntry(root, textvariable=value, font=large_font).grid(row=0,\n\n# Now, there are some most basic buttons which should be present\n# in a calculator\n# here, each button is calling the setexpression function which\n# is used to set values in the entry widget entered by the user\n# on pressing the buttons 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, .\nButton(root, text=\"+\", fg=\"red\", command=lambda:\nButton(root, text=\"-\", fg=\"red\", command=lambda:\nButton(root, text=\"X\", fg=\"red\", command=lambda:\nButton(root, text=\"/\", fg=\"red\", command=lambda:\nButton(root, text=\"1\", fg=\"red\", command=lambda:\nButton(root, text=\"2\", fg=\"red\", command=lambda:\nButton(root, text=\"3\", fg=\"red\", command=lambda:\nButton(root, text=\"4\", fg=\"red\", command=lambda:\nButton(root, text=\"5\", fg=\"red\", command=lambda:\nsetexpression(\"5\"), height=4, width=8).grid(row=2, column=2)\nButton(root, text=\"6\", fg=\"red\", command=lambda:\nButton(root, text=\"7\", fg=\"red\", command=lambda:\nButton(root, text=\"8\", fg=\"red\", command=lambda:\nButton(root, text=\"9\", fg=\"red\", command=lambda:\nButton(root, text=\"0\", fg=\"red\", command=lambda:\nButton(root, text=\".\", fg=\"red\", command=lambda:\n\n# \"=\" button to call the calculator button which will return and\n# show the calculated value in the entry widget\nButton(root, text=\"=\", fg=\"red\", command=calculator, height=4,\n\n# \"Clear\" button to call the clear function which will clear the\n# entry widget so that the user can start clculating again\nButton(root, text=\"Clear\", fg=\"red\", command=clear, height=4,\n\n# .mainloop() is used when the code is ready to run\nroot.mainloop()```\n\n## Output:\n\nKeep updated for more amazing content like this", null, "" ]	[ null, "https://copyassignment.com/wp-content/uploads/2022/05/Copyassignment.com_-675x380.png", null, "https://secure.gravatar.com/avatar/2ab349f6058efe42f43c38dfa93a62c7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.548464,"math_prob":0.9065661,"size":5175,"snap":"2023-14-2023-23","text_gpt3_token_len":1355,"char_repetition_ratio":0.19686714,"word_repetition_ratio":0.020895522,"special_character_ratio":0.2879227,"punctuation_ratio":0.18341464,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930706,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-31T22:38:02Z\",\"WARC-Record-ID\":\"<urn:uuid:64e7c4a4-c156-4802-a11c-e6f707c1bfdc>\",\"Content-Length\":\"140045\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5012459b-1aba-46d0-a6ad-da110e163769>\",\"WARC-Concurrent-To\":\"<urn:uuid:e460fc29-3a8b-42e4-a6b9-fba8aa00d539>\",\"WARC-IP-Address\":\"194.163.38.220\",\"WARC-Target-URI\":\"https://copyassignment.com/simple-gui-calculator-using-tkinter-in-python/\",\"WARC-Payload-Digest\":\"sha1:GNJ6HG3LNJF4F6YREUDVTPFGOO2NOIZV\",\"WARC-Block-Digest\":\"sha1:OAW63QY4E5X3A243A5NZUETSSMDTAAH4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647459.8_warc_CC-MAIN-20230531214247-20230601004247-00292.warc.gz\"}"}
https://www.accountingformanagement.org/internal-rate-of-return-method/	[ "# Internal rate of return method\n\nLike net present value method, internal rate of return (IRR) method also takes into account the time value of money. It analyzes an investment project by comparing the internal rate of return to the minimum required rate of return of the company.\n\nThe internal rate of return sometime known as yield on project is the rate at which an investment project promises to generate a return during its useful life. It is the discount rate at which the present value of a project’s net cash inflows becomes equal to the present value of its net cash outflows. In other words, internal rate of return is the discount rate at which a project’s net present value becomes equal to zero.\n\nThe minimum required rate of return is set by management. Most of the time, it is the cost of capital of the company.\n\nUnder this method, If the internal rate of return promised by the investment project is greater than or equal to the minimum required rate of return, the project is considered acceptable otherwise the project is rejected. Internal rate of return method is also known as time-adjusted rate of return method.\n\nTo understand how computations are made and how a proposed investment is accepted or rejected under this method, consider the following example:\n\n## Example:\n\nThe management of VGA Textile Company is considering to replace an old machine with a new one. The new machine will be capable of performing some tasks much faster than the old one. The installation of machine will cost \\$8,475 and will reduce the annual labor cost by \\$1,500. The useful life of the machine will be 10 years with no salvage value. The minimum required rate of return is 15%.\n\nRequired: Should VGA Textile Company purchase the machine? Use internal rate of return (IRR) method for your conclusion.\n\n### Solution:\n\nTo conclude whether the proposal should be accepted or not, the internal rate of return promised by machine would be found out first and then compared to the company’s minimum required rate of return.\n\nThe first step in finding out the internal rate of return is to compute a discount factor called internal rate of return factor. It is computed by dividing the investment required for the project by net annual cash inflow to be generated by the project. The formula is given below:\n\nFormula of internal rate of return factor:", null, "In our example, the required investment is \\$8,475 and the net annual cost saving is \\$1,500. The cost saving is equivalent to revenue and would, therefore, be treated as net cash inflow. Using this information, the internal rate of return factor can be computed as follows:\n\nInternal rate of return factor = \\$8,475 /\\$1,500\n\n= 5.650\n\nAfter computing the internal rate of return factor, the next step is to locate this discount factor in “present value of an annuity of \\$1 in arrears table“. Since the useful life of the machine is 10 years, the factor would be found in 10-period line or row. After finding this factor, see the rate of return written at the top of the column in which factor 5.650 is written. It is 12%. It means the internal rate of return promised by the project is 12%. The final step is to compare it with the minimum required rate of return of the VGA Textile Company. That is 15%.\n\nConclusion:\n\nAccording to internal rate of return method, the proposal is not acceptable because the internal rate of return promised by the proposal (12%) is less than the minimum required rate of return (15%).\n\nNotice that the internal rate of return promised by the proposal is a discount rate that equates the present value of cash inflows with the present value of cash out flows as proved by the following computation:", null, "*Value from “present value of an annuity of \\$1 in arrears table“.\n\nShow your love for us by sharing our contents.\n\n### 23 Comments on Internal rate of return method\n\n1.", null, "tejaswini\n\nThanking u but I want some more examples.\n\n2.", null, "Accounting for Management\n\nYou can see exercises and problems sections of the website.\n\n3.", null, "Ameli\n\nWhat about how to compare to exclusive projects?\n\n4.", null, "patricia\n\nhow can i find the PVF@ more than 21 for IRR?\n\n5.", null, "Mercy\n\nhow do i calculate the IRR when given the salvage value.\n\n6.", null, "abel afore markson\n\nwhat is the best formula for calculating internal rate of return?\n\n7.", null, "Buster\n\nIt seems as though all the examples are on investment in fixed asset. What about investment in services e.g. investing in the stock market. Does the same approach holds?\n\n8.", null, "Rose Anne\n\nwhat if i’m not going to use the table in locating the discount factor?\n\n9.", null, "Wasem solomon\n\nDe formula u gave n eg. Is rather compounding issues\n\n10.", null, "kate jerry\n\nHow do I calculate for 19% discount rate using four figure table, so as to ascertain irr?\n\n11.", null, "Elida\n\nIt’s a plsreuae to find someone who can think so clearly\n\n12.", null, "sumit\n\nYou can solve like our class does?\n\n13.", null, "James\n\nhow do we compute for IRR and NPV if no rate is given, only the current amount of a planned purchase of a machine, its life in years, savings it can give the company and its depreciation amount?\n\n14.", null, "nehecristy\n\na project cost N556,000 to initiate and it will generate annual cashflows(receipt less payment) of N200,000 for a period of 5 yrs. the expected scrap value is 56000. depreciation is provided on straight line method.cost of capital is 22%\nrequired: arr, npv, irr, pbp, discounted pvb.\npls help me. dis is an assignment for 15 marks in my college tanks.\n\n15.", null, "TWIZEYE JULIAS\n\nOK\n\n16.", null, "Siva Kumar\n\nHow can i get Rate of Return for indefinite period?\n\n17.", null, "George\n\nIt’s strange to see internal rate of return computed analytically, are you sure this is the correct approach? Seems to be different from what is claimed in a lot of other sources: that it can’t be computed analytically and you need approximations to make it work, e.g. https://www.gigacalculator.com/calculators/irr-calculator.php . Is your “internal rate of return factor” the same or different than the classic “internal rate of return” concept???\n\n18.", null, "Kulwa\n\nwhere can I get examples that show estimates flow of the net benefits for for two projects with the the same discount rate? example 5% and 10%\n\n19.", null, "NAGGITA VICTORIA\n\nMore examples are needed\n\n20.", null, "banky lawrence\n\nsuppose you have a 4yr project that costs \\$500. the cash flow over the 4 yr life will be \\$100,\\$200,\\$300 and \\$400 from yr 1-4 respectively. whats the IRR of the project?\n\n21.", null, "julius moses\n\nthanks alot for that precise example\n\n22.", null, "Etimu Simon.S.E\n\nWhat is the cost of capital of the company? How does it relate to the minimum rate of return?\nWhat is the realistic way for the company to determine the minimum rate of return?\n\n23.", null, "Manoharan\n\nIRR will work out for our loan availment – example if I get a loan of Rs.1 crore repayment in 10 equal instalment on 18% PA flat interest, what will be by nett IRR" ]	[ null, "https://www.accountingformanagement.org/wp-content/uploads/2012/12/internal-rate-of-return-method-img1-1.png", null, "https://www.accountingformanagement.org/wp-content/uploads/2012/12/internal-rate-of-return-method-img2.png", null, "https://secure.gravatar.com/avatar/1062f0b57130562e3337f82db137de2a", null, "https://secure.gravatar.com/avatar/d7ef5cc2cdb2a97b68047754a9d909c4", null, "https://secure.gravatar.com/avatar/7b523443d1eb604c554528db92fae334", null, "https://secure.gravatar.com/avatar/8480a11f9b2a5ed04dbfa08381426a86", null, "https://secure.gravatar.com/avatar/e1a19076bf3179583aa5c04496761251", null, "https://secure.gravatar.com/avatar/50641466b5266e17cc7e11f03f93b69b", null, "https://secure.gravatar.com/avatar/9c928749cf622bbd470f2704dc2caf92", null, "https://secure.gravatar.com/avatar/1ed19cb1d477f2bd4612c1fbe362af86", null, "https://secure.gravatar.com/avatar/73e556d8a3553d8f3decaa4f404d66e3", null, "https://secure.gravatar.com/avatar/b758ea3d5976f988dd3c2a1b84d8c112", null, "https://secure.gravatar.com/avatar/aceb00cfb7b555dacdc849fc8a16d1b1", null, "https://secure.gravatar.com/avatar/b6c4e3f9f751d78e0c12c172a1b486f7", null, "https://secure.gravatar.com/avatar/5d2a98676c89e6ba72042f3aee87b29f", null, "https://secure.gravatar.com/avatar/ac2cd3bb1c30025c2ad4030ca4a605d6", null, "https://secure.gravatar.com/avatar/33cb471716b428b62b70977e64900945", null, "https://secure.gravatar.com/avatar/06fce4a6856f5cbd5e17c0eafe648e65", null, "https://secure.gravatar.com/avatar/8bd62768ed307735c1636902bdbc67f4", null, "https://secure.gravatar.com/avatar/42b7cc7c9627235bfe2d1ca5de31a3a2", null, "https://secure.gravatar.com/avatar/b28527998f40d8a71b1f6bd8e00bbbc2", null, "https://secure.gravatar.com/avatar/fbb464e194ef6b63431f754abfdb21ea", null, "https://secure.gravatar.com/avatar/888fa08e9139aa6e3a6db95eaa96b240", null, "https://secure.gravatar.com/avatar/784bae96d4b8c7ffeeb97475f7200164", null, "https://secure.gravatar.com/avatar/5f787248307d0e81a37f2383c4c25ec5", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92589235,"math_prob":0.92121834,"size":6382,"snap":"2019-51-2020-05","text_gpt3_token_len":1396,"char_repetition_ratio":0.18548134,"word_repetition_ratio":0.072256915,"special_character_ratio":0.22187401,"punctuation_ratio":0.09553641,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.985197,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,2,null,2,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T09:13:37Z\",\"WARC-Record-ID\":\"<urn:uuid:7b3faa44-1f87-4ec0-b70f-f83d15d16889>\",\"Content-Length\":\"77831\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6899c25-f283-4d1a-b9a5-ecea40d776b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3c18f18-36f1-4a63-8ebc-751295178e77>\",\"WARC-IP-Address\":\"198.96.95.58\",\"WARC-Target-URI\":\"https://www.accountingformanagement.org/internal-rate-of-return-method/\",\"WARC-Payload-Digest\":\"sha1:MS7Y3MISLHHT75KBVFIKDKC6TO4YBKJC\",\"WARC-Block-Digest\":\"sha1:RCAGKWNATDNKDR3PTBNRRFY3XR4RBLUH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601628.36_warc_CC-MAIN-20200121074002-20200121103002-00353.warc.gz\"}"}
https://inquiryintoinquiry.com/2019/07/26/animated-logical-graphs-25/	[ "## Animated Logical Graphs • 25\n\nLet’s examine the formal operation table for the third in our series of reflective forms to see if we can elicit the general pattern:", null, "$\\begin{array}{\|{3}{c}\|\|c\|} \\multicolumn{4}{c}{\\text{Formal Operation Table} ~ \\texttt{(} a \\texttt{,} b \\texttt{,} c \\texttt{)}} \\\\[4pt] \\hline a & b & c & \\texttt{(}a\\texttt{,}b\\texttt{,}c\\texttt{)} \\\\ \\hline\\hline \\texttt{Space}&\\texttt{Space}&\\texttt{Space}&\\texttt{Cross} \\\\ \\texttt{Space}&\\texttt{Space}&\\texttt{Cross}&\\texttt{Space} \\\\ \\texttt{Space}&\\texttt{Cross}&\\texttt{Space}&\\texttt{Space} \\\\ \\texttt{Space}&\\texttt{Cross}&\\texttt{Cross}&\\texttt{Cross} \\\\ \\hline \\texttt{Cross}&\\texttt{Space}&\\texttt{Space}&\\texttt{Space} \\\\ \\texttt{Cross}&\\texttt{Space}&\\texttt{Cross}&\\texttt{Cross} \\\\ \\texttt{Cross}&\\texttt{Cross}&\\texttt{Space}&\\texttt{Cross} \\\\ \\texttt{Cross}&\\texttt{Cross}&\\texttt{Cross}&\\texttt{Cross} \\\\ \\hline \\end{array}$\n\nOr, thinking in terms of the corresponding cactus graphs, writing", null, "${}^{\\backprime\\backprime} \\texttt{o} {}^{\\prime\\prime}$ for a blank node and", null, "${}^{\\backprime\\backprime} \\texttt{\|} {}^{\\prime\\prime}$ for a terminal edge, we get the following Table:", null, "$\\begin{array}{\|{3}{c}\|\|c\|} \\multicolumn{4}{c}{\\text{Formal Operation Table} ~ \\texttt{(} a \\texttt{,} b \\texttt{,} c \\texttt{)}} \\\\[4pt] \\hline \\quad a \\quad & \\quad b \\quad & \\quad c \\quad & \\texttt{(}a\\texttt{,}b\\texttt{,}c\\texttt{)} \\\\ \\hline\\hline \\texttt{o} & \\texttt{o} & \\texttt{o} & \\texttt{\|} \\\\ \\texttt{o} & \\texttt{o} & \\texttt{\|} & \\texttt{o} \\\\ \\texttt{o} & \\texttt{\|} & \\texttt{o} & \\texttt{o} \\\\ \\texttt{o} & \\texttt{\|} & \\texttt{\|} & \\texttt{\|} \\\\ \\hline \\texttt{\|} & \\texttt{o} & \\texttt{o} & \\texttt{o} \\\\ \\texttt{\|} & \\texttt{o} & \\texttt{\|} & \\texttt{\|} \\\\ \\texttt{\|} & \\texttt{\|} & \\texttt{o} & \\texttt{\|} \\\\ \\texttt{\|} & \\texttt{\|} & \\texttt{\|} & \\texttt{\|} \\\\ \\hline \\end{array}$\n\nEvidently, the rule is that", null, "${}^{\\backprime\\backprime} \\texttt{(} a \\texttt{,} b \\texttt{,} c \\texttt{)} {}^{\\prime\\prime}$ denotes the value denoted by", null, "${}^{\\backprime\\backprime} \\texttt{o} {}^{\\prime\\prime}$ if and only if exactly one of the variables", null, "$a, b, c$ has the value denoted by", null, "${}^{\\backprime\\backprime} \\texttt{\|} {}^{\\prime\\prime},$ otherwise", null, "${}^{\\backprime\\backprime} \\texttt{(} a \\texttt{,} b \\texttt{,} c \\texttt{)} {}^{\\prime\\prime}$ denotes the value denoted by", null, "${}^{\\backprime\\backprime} \\texttt{\|} {}^{\\prime\\prime}.$ Examining the whole series of reflective forms shows this is the general rule.\n\n• In the Entitative Interpretation", null, "$(\\mathrm{En}),$ where", null, "$\\texttt{o}$ = false and", null, "$\\texttt{\|}$ = true,", null, "${}^{\\backprime\\backprime} \\texttt{(} x_1 \\texttt{,} \\ldots \\texttt{,} x_k \\texttt{)} {}^{\\prime\\prime}$ translates as “not just one of the", null, "$x_j$ is true”.\n• In the Existential Interpretation", null, "$(\\mathrm{Ex}),$ where", null, "$\\texttt{o}$ = true and", null, "$\\texttt{\|}$ = false,", null, "${}^{\\backprime\\backprime} \\texttt{(} x_1 \\texttt{,} \\ldots \\texttt{,} x_k \\texttt{)} {}^{\\prime\\prime}$ translates as “just one of the", null, "$x_j$ is not true”.\n\n### 3 Responses to Animated Logical Graphs • 25\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86466396,"math_prob":0.99803764,"size":1026,"snap":"2020-45-2020-50","text_gpt3_token_len":229,"char_repetition_ratio":0.118395306,"word_repetition_ratio":0.16292135,"special_character_ratio":0.21442495,"punctuation_ratio":0.082417585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998883,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T19:34:53Z\",\"WARC-Record-ID\":\"<urn:uuid:dd7e3d4b-82d3-46ce-ab23-d625ebb25060>\",\"Content-Length\":\"208013\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c031065-22cb-4297-b1ce-c8ae4baff3c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:12835d26-3e84-409f-b6b6-3a68f1aa8f0d>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://inquiryintoinquiry.com/2019/07/26/animated-logical-graphs-25/\",\"WARC-Payload-Digest\":\"sha1:X2KYG2YBQC645FOYEMJTG3FALRHI35XL\",\"WARC-Block-Digest\":\"sha1:CC5LHWZTJMIPDJQ6YACJOX2YT6PZSWHI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141715252.96_warc_CC-MAIN-20201202175113-20201202205113-00434.warc.gz\"}"}
http://docs.wradlib.org/en/latest/generated/wradlib.georef.rect.get_radolan_coords.html	[ "wradlib.georef.rect.get_radolan_coords(lon, lat, trig=False)\nParameters: lon (float, numpy.ndarray of floats) – longitude lat (float, numpy.ndarray of floats) – latitude trig (boolean) – if True, uses trigonometric formulas for calculation, otherwise osr transformations if False, uses osr spatial reference system to transform between projections trig is recommended to be False, however, the two ways of computation are expected to be equivalent." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.597477,"math_prob":0.9753325,"size":570,"snap":"2019-26-2019-30","text_gpt3_token_len":138,"char_repetition_ratio":0.09717315,"word_repetition_ratio":0.027027028,"special_character_ratio":0.2122807,"punctuation_ratio":0.21782178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9798745,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T04:55:53Z\",\"WARC-Record-ID\":\"<urn:uuid:427c0bb3-5f0d-4977-aaa5-2427cb8f7e7e>\",\"Content-Length\":\"16496\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e420cfa-32fd-4153-847b-a97c0095ffa7>\",\"WARC-Concurrent-To\":\"<urn:uuid:193f267f-f0bd-490f-99cb-647d4f75e234>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"http://docs.wradlib.org/en/latest/generated/wradlib.georef.rect.get_radolan_coords.html\",\"WARC-Payload-Digest\":\"sha1:62WPLSQ7NYIJBP2QRNNJZPF5X55GPAOS\",\"WARC-Block-Digest\":\"sha1:VN6QS6MP5BKD73RSCKIK3LYRYASB6I22\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999141.54_warc_CC-MAIN-20190620044948-20190620070948-00439.warc.gz\"}"}
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.lapacklin.ztbcon.html	[ "# naginterfaces.library.lapacklin.ztbcon¶\n\nnaginterfaces.library.lapacklin.ztbcon(norm, uplo, diag, kd, ab)[source]\n\nztbcon estimates the condition number of a complex triangular band matrix.\n\nFor full information please refer to the NAG Library document for f07vu\n\nhttps://www.nag.com/numeric/nl/nagdoc_27.3/flhtml/f07/f07vuf.html\n\nParameters\nnormstr, length 1\n\nIndicates whether or is estimated.\n\nor\n\nis estimated.\n\nis estimated.\n\nuplostr, length 1\n\nSpecifies whether is upper or lower triangular.\n\nis upper triangular.\n\nis lower triangular.\n\ndiagstr, length 1\n\nIndicates whether is a nonunit or unit triangular matrix.\n\nis a nonunit triangular matrix.\n\nis a unit triangular matrix; the diagonal elements are not referenced and are assumed to be .\n\nkdint\n\n, the number of superdiagonals of the matrix if , or the number of subdiagonals if .\n\nabcomplex, array-like, shape\n\nThe triangular band matrix .\n\nReturns\nrcondfloat\n\nAn estimate of the reciprocal of the condition number of . is set to zero if exact singularity is detected or the estimate underflows. If is less than machine precision, is singular to working precision.\n\nRaises\nNagValueError\n(errno )\n\nOn entry, error in parameter .\n\nConstraint: , or .\n\n(errno )\n\nOn entry, error in parameter .\n\nConstraint: or .\n\n(errno )\n\nOn entry, error in parameter .\n\nConstraint: or .\n\n(errno )\n\nOn entry, error in parameter .\n\nConstraint: .\n\n(errno )\n\nOn entry, error in parameter .\n\nConstraint: .\n\nNotes\n\nztbcon estimates the condition number of a complex triangular band matrix , in either the -norm or the -norm:\n\nNote that .\n\nBecause the condition number is infinite if is singular, the function actually returns an estimate of the reciprocal of the condition number.\n\nThe function computes or exactly, and uses Higham’s implementation of Hager’s method (see Higham (1988)) to estimate or .\n\nReferences\n\nHigham, N J, 1988, FORTRAN codes for estimating the one-norm of a real or complex matrix, with applications to condition estimation, ACM Trans. Math. Software (14), 381–396" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5780781,"math_prob":0.9188004,"size":2015,"snap":"2021-43-2021-49","text_gpt3_token_len":494,"char_repetition_ratio":0.13674788,"word_repetition_ratio":0.21766561,"special_character_ratio":0.23672457,"punctuation_ratio":0.19788918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.977854,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T09:55:18Z\",\"WARC-Record-ID\":\"<urn:uuid:cc6af155-1173-48ca-9b5a-02b0dcc5e6be>\",\"Content-Length\":\"91976\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e75bb7a2-884c-4d68-9613-4d00cc129cfe>\",\"WARC-Concurrent-To\":\"<urn:uuid:b772d38b-f5a2-43a8-b2b3-7c2eebafb055>\",\"WARC-IP-Address\":\"78.129.168.4\",\"WARC-Target-URI\":\"https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.lapacklin.ztbcon.html\",\"WARC-Payload-Digest\":\"sha1:MRGPHJ262Q74HW5BUCEXQ3243QJ27ZEA\",\"WARC-Block-Digest\":\"sha1:GSCXCQS2OUIO2SOF6JSKRT52S67IUJQ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585246.50_warc_CC-MAIN-20211019074128-20211019104128-00159.warc.gz\"}"}
https://www.exceldemy.com/macro-to-print-specific-sheets-in-excel/	[ "# How to Print Specific Sheets Using VBA Macro in Excel (4 Ways)\n\nIn Microsoft Excel, you will have to print sheets for various purposes. It is one of the familiar things while working with Excel. Now, we macro to print specific sheets in excel use normally our built-in print button or the print option of Excel to print sheets. However, you can also use VBA macro to print sheets in Excel. In this tutorial, you will learn to print specific sheets using VBA macro in Excel.\n\nThis tutorial will be on point with other resources and proper illustrations. So, read the whole article.\n\n## 4 Ways to Print Specific Sheets Using VBA Macro\n\nIn the following sections, we are going to provide you with two simple but effective VBA macros that you can imply in your worksheets. We recommend you learn and apply all these methods to your workbook. It will surely develop your Excel knowledge.\n\n### 1. Print Specific Sheets by Sheet Number Using VBA Macro\n\nNow, you can print multiple sheets using the sheet number. But this method will print specific sheets that are contiguous. We will use the basic PrintOut syntax.\n\nFirst, open your Visual Basic Editor. After that, insert a module. Then type the following code:\n\n``````Sub print_sheets_by_number()\n\nFor i = 3 To 6\nWorksheets(i).PrintOut\nNext i\n\nEnd Sub``````\n\nWe are using a For-Next loop here. This code will print the sheets from 3 to 6. But, remember, it will print them one by one. After running the macro, it will ask you to save your file.\n\nNow, after running the code, Excel will save these specific sheets as pdf. And, it will ask you to save the print output file:", null, "Also, it will show the following dialog box:", null, "After completing all of these, you will get your print preview.\n\nRead More: How to Print Page Number in Excel (5 Easy Ways)\n\n### 2. Macro to Print Specific Sheets by Sheet Name\n\nAnother way to print specific sheets using a macro is to print them by the sheet name. This method is pretty obvious but effective. If you have limited sheets in your Excel workbook, you can use the following code:\n\n``````Sub print_sheets_by_name()\n\nWorksheets(\"Dataset\").PrintOut\nWorksheets(\"AutoSum\").PrintOut\nWorksheets(\"MIN\").PrintOut\nWorksheets(\"SMALL\").PrintOut\n\nEnd Sub``````\n\nAs you can see, we have used the sheet name to print those multiple and specific sheets from the Excel workbook. But, this process is pretty hectic. You have to remember the sheet names.\n\n### 3. Print Specific Sheets Using Array\n\nNow, you can use an array to print specific sheets. You have to enter the sheet names in the array like the following:\n\n``````Sub print_multiple_sheets()\n\nWorksheets(Array(\"Dataset\", \"AutoSum\", \"MIN\", \"SMALL\")).PrintOut\n\nEnd Sub``````\n\n### 4. Print Specific Sheets by a Button in Excel\n\nNow, another way to print specific sheets is to create a print button in Excel. First, we will create a button. After that, we will insert a code for that button. If you click that button, it will print specific sheets. Why is this helpful? We will use this because it is hazard-free. Just one click will do all of these.\n\n📌 Steps\n\n• First, go to the Developer If you don’t have it, enable the developer tab in the ribbon.\n• Now, from the Controls group, click on Insert.", null, "• After that, from the ActiveX Controls, click on the Button", null, "• Now, place the button on your worksheet.", null, "• To change the name of the button, right-click on the button. After that, click on Properties.", null, "• From the Properties dialog box, change the caption of the button. It will change the name of the button.", null, "• As you can see, our print button is ready.", null, "• Now, double click on the button while Design Mode is on. You will see the following VBA code.", null, "• Now, type the following code:\n``````Private Sub CommandButton1_Click()\n\nActiveSheet.PrintOut\n\nEnd Sub``````\n\nThis code will print the active sheet that you are working on. Now, from here, you can follow either of the two ways:\n\n• Copy the same button and paste it into all the worksheets. In this way, you will be able to print any sheet you want.\n• Or you can use the macro of the previous two methods and use them in this button. In this way, you will print specific sheets with one button.\n\n## Other Useful Macros to Print Sheets in Excel\n\nIn the following sections, we are going to provide you with some very essential VBA macros that you can use in a lot of scenarios. You can use all of these codes in your workbook. It will work nicely. Let’s get into it.\n\n### 1. Print Specific Sheets into Single Page\n\nThe following code will print the sheet “Dataset” exactly one page wide and tall.\n\n``````Sub print_single_page()\n\nWith Worksheets(\"Dataset\").PageSetup\n.Zoom = False\n.FitToPagesTall = 1\n.FitToPagesWide = 1\nEnd With\n\nEnd Sub``````\n\n### 2. Print Sheets with Comments\n\nThe following code will print sheets with comments:\n\n``````Sub print_sheets_with_comments()\n\nApplication.DisplayCommentIndicator = xlCommentAndIndicator\n\nWith ActiveSheet\n.PrintOut\nEnd With\n\nEnd Sub``````\n\n### 3. Macro to Print Hidden Sheets in Excel\n\nIf you have some specific hidden sheets in your Excel workbook, use the following code to print them:\n\n``````Sub print_hidden_sheet()\n\nDim current_visible As Long\nDim working_sheet As Worksheet\n\nFor Each working_sheet In ActiveWorkbook.Worksheets\nWith working_sheet\ncurrent_visible = .Visible\nIf current_visible >= 0 Then\n.Visible = xlSheetVisible\n.PrintOut\n.Visible = current_visible\nEnd If\nEnd With\nNext working_sheet\n\nEnd Sub``````\n\n### 4. Macro to Print both Hidden and Visible Sheets\n\nBy the following code, you can print all the hidden and visible sheets:\n\n``````Sub print_hidden_visible_sheet()\n\nDim current_visible As Long\nDim working_sheet As Worksheet\n\nFor Each working_sheet In ActiveWorkbook.Worksheets\nWith working_sheet\ncurrent_visible = .Visible\n.Visible = xlSheetVisible\n.PrintOut\n.Visible = current_visible\nEnd With\nNext working_sheet\n\nEnd Sub``````\n\n### 5. Print Multiple Excel Worksheets with Macro\n\nWe have already shown the code above. Use the sheet names in the array to print multiple sheets in Excel:\n\n``````Sub print_multiple_sheets()\n\nWorksheets(Array(\"Dataset\", \"AutoSum\", \"MIN\", \"SMALL\")).PrintOut\n\nEnd Sub``````\n\n### 6. Print All Worksheets\n\nTo print all the worksheets of your Excel workbook, use the following code:\n\n``````Sub print_all_sheets()\n\nWorksheets.PrintOut\n\nEnd Sub``````\n\n### 7. Print Entire Workbook\n\nTo print the whole workbook, follow any the of the following VBA codes:\n\nUsing ActiveWorkbook:\n\n``````Sub print_active_workbook()\n\nActiveWorkbook.PrintOut\n\nEnd Sub``````\n\nUsing ThisWorkbook:\n\n``````Sub print_this_workbook()\n\nThisWorkbook.PrintOut\n\nEnd Sub``````\n\n### 8. VBA Code to Print a Specific Sheet\n\nMention the sheet name to print a specific sheet from your workbook.\n\n``````Sub print_specific_sheet()\n\nSheets(\"Dataset\").PrintOut\n\nEnd Sub``````\n\n### 9. Print Active Sheet in Excel\n\nPrint the active sheet using the following code:\n\n``````Sub print_active_sheet()\n\nActiveSheet.PrintOut\n\nEnd Sub``````\n\n### 10. Print Selected Sheets\n\nBy the following code, you can print the specifically selected worksheets:\n\n``````Sub print_selected_sheet()\n\nActiveWindow.SelectedSheets.PrintOut\n\nEnd Sub``````\n\n### 11. Print a Selection from a Sheet\n\nIf you want to print a specific selection, use the following code:\n\n``````Sub print_selection()\n\nSelection.PrintOut\n\nEnd Sub``````\n\n### 12. Excel VBA to Print a Range\n\nThe following code will help you to print a specific range from a worksheet.\n\n``````Sub print_range()\n\nRange(\"B4:C11\").PrintOut\n\nEnd Sub``````\n\n### 13. Print Preview\n\nTo print a preview, use the following code:\n\n``````Sub print_preview()\n\nActiveSheet.PrintOut preview:=True\n\nEnd Sub``````\n\n### 14. Print Specific Sheets by Taking User Input\n\nNow, you can take the sheet number or sheet name as input from the user. The following code will do that with ease.\n\n#### 14.1 Take Sheet Number as User Input\n\nThe following code will take the sheet number as input from the user:\n\n``````Sub print_user_input_number()\n\nDim sheet_number As Integer\n\nsheet_number = Application.InputBox(\"Enter Sheet Number to Print:\")\nWorksheets(sheet_number).PrintOut\n\nEnd Sub``````\n\n#### 14.2 Take Sheet Name as User Input\n\nYou can print a specific sheet by taking the sheet name as user input:\n\n``````Sub print_user_input_name()\n\nDim sheet_name As String\n\nsheet_name = Application.InputBox(\"Enter Sheet Name to Print:\")\nWorksheets(sheet_name).PrintOut\n\nEnd Sub``````\n\n### 15. Excel VBA to Print Preview a Selected Range in Excel\n\nThe following code will give you the preview, along with the print option:\n\n``````Sub print_preview_selected_range()\n\nSheets(\"Dataset\").Range(\"B4:C11\").PrintPreview\n\nEnd Sub``````\n\n## 💬 Things to Remember\n\n✎ Here, the macros are based on our practice workbook. If you are using a different workbook, change the sheet name, number, and range.\n\n## Conclusion\n\nTo conclude, I hope this tutorial has provided you with a piece of useful knowledge about the Date in VBA codes. We recommend you learn and apply all these instructions to your dataset. Download the practice workbook and try these yourself. Also, feel free to give feedback in the comment section. Your valuable feedback keeps us motivated to create tutorials like this.\n\nDon’t forget to check our website Exceldemy.com for various Excel-related problems and solutions.\n\nKeep learning new methods and keep growing!\n\n## Related Articles", null, "#### Shanto\n\nHello! I am Shanto. An Excel & VBA Content Developer. My goal is to provide our readers with great tutorials on various Excel-related problems. I hope our easy but effective tutorials will enrich your knowledge. I have completed my BSc in Computer Science & Engineering from Daffodil International University. Working with data was always my passion. Love to work with data, analyze those, and find patterns. Also, love to research. Always look for challenges to keep me growing.\n\nWe will be happy to hear your thoughts", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20700%20530'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20343%20211'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20700%20616'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20700%20544'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20469%20583'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20469%20583'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20700%20516'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20681%20520'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20511%20226'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2069%2069'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20160%2050'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7848603,"math_prob":0.4614554,"size":9260,"snap":"2022-27-2022-33","text_gpt3_token_len":2102,"char_repetition_ratio":0.19328004,"word_repetition_ratio":0.079622135,"special_character_ratio":0.21695465,"punctuation_ratio":0.13145809,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.96173096,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T06:54:52Z\",\"WARC-Record-ID\":\"<urn:uuid:2f2ec9aa-39d3-4838-82a2-d58a453ef212>\",\"Content-Length\":\"211995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dea39993-98e3-4cfd-938a-cdefcdf95bd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:863afb7a-e27e-4039-9b4b-53f0e8301616>\",\"WARC-IP-Address\":\"172.67.190.223\",\"WARC-Target-URI\":\"https://www.exceldemy.com/macro-to-print-specific-sheets-in-excel/\",\"WARC-Payload-Digest\":\"sha1:RJ3QKX5OMDSSVEU5XFWDXWPHEN6LVSQM\",\"WARC-Block-Digest\":\"sha1:PAX4NVCB7F7ZX37RSCSHCKE7AJHS6M5A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571909.51_warc_CC-MAIN-20220813051311-20220813081311-00048.warc.gz\"}"}
https://bitcoinj.org/javadoc/0.16.1/org/bitcoinj/crypto/LazyECPoint.html	[ "## Class LazyECPoint\n\n• java.lang.Object\n• org.bitcoinj.crypto.LazyECPoint\n\n• ```public class LazyECPoint\nextends java.lang.Object```\nA wrapper around ECPoint that delays decoding of the point for as long as possible. This is useful because point encode/decode in Bouncy Castle is quite slow especially on Dalvik, as it often involves decompression/recompression.\n• ### Constructor Summary\n\nConstructors\nConstructor Description\n```LazyECPoint(org.bouncycastle.math.ec.ECCurve curve, byte[] bits)```\nConstruct a LazyECPoint from a public key.\n```LazyECPoint(org.bouncycastle.math.ec.ECPoint point, boolean compressed)```\nConstruct a LazyECPoint from an already decoded point.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method Description\n`org.bouncycastle.math.ec.ECPoint` `add(org.bouncycastle.math.ec.ECPoint b)`\n`LazyECPoint` `compress()`\nReturns a compressed version of this elliptic curve point.\n`LazyECPoint` `decompress()`\nReturns a decompressed version of this elliptic curve point.\n`boolean` `equals(java.lang.Object o)`\n`boolean` `equals(org.bouncycastle.math.ec.ECPoint other)`\n`org.bouncycastle.math.ec.ECPoint` `get()`\n`org.bouncycastle.math.ec.ECFieldElement` `getAffineXCoord()`\n`org.bouncycastle.math.ec.ECFieldElement` `getAffineYCoord()`\n`org.bouncycastle.math.ec.ECCurve` `getCurve()`\n`org.bouncycastle.math.ec.ECPoint` `getDetachedPoint()`\n`byte[]` `getEncoded()`\n`byte[]` `getEncoded(boolean compressed)`\n`org.bouncycastle.math.ec.ECFieldElement` `getX()`\n`org.bouncycastle.math.ec.ECFieldElement` `getXCoord()`\n`org.bouncycastle.math.ec.ECFieldElement` `getY()`\n`org.bouncycastle.math.ec.ECFieldElement` `getYCoord()`\n`org.bouncycastle.math.ec.ECFieldElement` `getZCoord(int index)`\n`org.bouncycastle.math.ec.ECFieldElement[]` `getZCoords()`\n`int` `hashCode()`\n`boolean` `isCompressed()`\n`boolean` `isInfinity()`\n`boolean` `isNormalized()`\n`boolean` `isValid()`\n`org.bouncycastle.math.ec.ECPoint` `multiply(java.math.BigInteger k)`\n`org.bouncycastle.math.ec.ECPoint` `negate()`\n`org.bouncycastle.math.ec.ECPoint` `normalize()`\n`org.bouncycastle.math.ec.ECPoint` `scaleX(org.bouncycastle.math.ec.ECFieldElement scale)`\n`org.bouncycastle.math.ec.ECPoint` `scaleY(org.bouncycastle.math.ec.ECFieldElement scale)`\n`org.bouncycastle.math.ec.ECPoint` `subtract(org.bouncycastle.math.ec.ECPoint b)`\n`org.bouncycastle.math.ec.ECPoint` `threeTimes()`\n`org.bouncycastle.math.ec.ECPoint` `timesPow2(int e)`\n`org.bouncycastle.math.ec.ECPoint` `twice()`\n`org.bouncycastle.math.ec.ECPoint` `twicePlus(org.bouncycastle.math.ec.ECPoint b)`\n• ### Methods inherited from class java.lang.Object\n\n`clone, finalize, getClass, notify, notifyAll, toString, wait, wait, wait`\n• ### Constructor Detail\n\n• #### LazyECPoint\n\n```public LazyECPoint(org.bouncycastle.math.ec.ECCurve curve,\nbyte[] bits)```\nConstruct a LazyECPoint from a public key. Due to the delayed decoding of the point the validation of the public key is delayed too, e.g. until a getter is called.\nParameters:\n`curve` - a curve the point is on\n`bits` - public key bytes\n• #### LazyECPoint\n\n```public LazyECPoint(org.bouncycastle.math.ec.ECPoint point,\nboolean compressed)```\nConstruct a LazyECPoint from an already decoded point.\nParameters:\n`point` - the wrapped point\n`compressed` - true if the represented public key is compressed\n• ### Method Detail\n\n• #### compress\n\n`public LazyECPoint compress()`\nReturns a compressed version of this elliptic curve point. Returns the same point if it's already compressed. See the `ECKey` class docs for a discussion of point compression.\n• #### decompress\n\n`public LazyECPoint decompress()`\nReturns a decompressed version of this elliptic curve point. Returns the same point if it's already compressed. See the `ECKey` class docs for a discussion of point compression.\n• #### get\n\n`public org.bouncycastle.math.ec.ECPoint get()`\n• #### getEncoded\n\n`public byte[] getEncoded()`\n• #### getDetachedPoint\n\n`public org.bouncycastle.math.ec.ECPoint getDetachedPoint()`\n• #### isInfinity\n\n`public boolean isInfinity()`\n• #### timesPow2\n\n`public org.bouncycastle.math.ec.ECPoint timesPow2(int e)`\n• #### getYCoord\n\n`public org.bouncycastle.math.ec.ECFieldElement getYCoord()`\n• #### getZCoords\n\n`public org.bouncycastle.math.ec.ECFieldElement[] getZCoords()`\n• #### isNormalized\n\n`public boolean isNormalized()`\n• #### isCompressed\n\n`public boolean isCompressed()`\n• #### multiply\n\n`public org.bouncycastle.math.ec.ECPoint multiply(java.math.BigInteger k)`\n• #### subtract\n\n`public org.bouncycastle.math.ec.ECPoint subtract(org.bouncycastle.math.ec.ECPoint b)`\n• #### isValid\n\n`public boolean isValid()`\n• #### scaleY\n\n`public org.bouncycastle.math.ec.ECPoint scaleY(org.bouncycastle.math.ec.ECFieldElement scale)`\n• #### getXCoord\n\n`public org.bouncycastle.math.ec.ECFieldElement getXCoord()`\n• #### scaleX\n\n`public org.bouncycastle.math.ec.ECPoint scaleX(org.bouncycastle.math.ec.ECFieldElement scale)`\n• #### equals\n\n`public boolean equals(org.bouncycastle.math.ec.ECPoint other)`\n• #### negate\n\n`public org.bouncycastle.math.ec.ECPoint negate()`\n• #### threeTimes\n\n`public org.bouncycastle.math.ec.ECPoint threeTimes()`\n• #### getZCoord\n\n`public org.bouncycastle.math.ec.ECFieldElement getZCoord(int index)`\n• #### getEncoded\n\n`public byte[] getEncoded(boolean compressed)`\n\n`public org.bouncycastle.math.ec.ECPoint add(org.bouncycastle.math.ec.ECPoint b)`\n• #### twicePlus\n\n`public org.bouncycastle.math.ec.ECPoint twicePlus(org.bouncycastle.math.ec.ECPoint b)`\n• #### getCurve\n\n`public org.bouncycastle.math.ec.ECCurve getCurve()`\n• #### normalize\n\n`public org.bouncycastle.math.ec.ECPoint normalize()`\n• #### getY\n\n`public org.bouncycastle.math.ec.ECFieldElement getY()`\n• #### twice\n\n`public org.bouncycastle.math.ec.ECPoint twice()`\n• #### getAffineYCoord\n\n`public org.bouncycastle.math.ec.ECFieldElement getAffineYCoord()`\n• #### getAffineXCoord\n\n`public org.bouncycastle.math.ec.ECFieldElement getAffineXCoord()`\n• #### getX\n\n`public org.bouncycastle.math.ec.ECFieldElement getX()`\n• #### equals\n\n`public boolean equals(java.lang.Object o)`\nOverrides:\n`equals` in class `java.lang.Object`\n• #### hashCode\n\n`public int hashCode()`\nOverrides:\n`hashCode` in class `java.lang.Object`" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5137608,"math_prob":0.48143488,"size":5662,"snap":"2022-40-2023-06","text_gpt3_token_len":1420,"char_repetition_ratio":0.3435843,"word_repetition_ratio":0.17751479,"special_character_ratio":0.17979513,"punctuation_ratio":0.27586207,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956111,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-01T06:04:15Z\",\"WARC-Record-ID\":\"<urn:uuid:31e5e2fd-70de-4450-b53e-dd131de6cd2a>\",\"Content-Length\":\"29582\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:139c4745-f28d-4326-8e03-6216afa1dfe7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f225c062-bfe9-46f3-85c4-3bce46eeaa3f>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://bitcoinj.org/javadoc/0.16.1/org/bitcoinj/crypto/LazyECPoint.html\",\"WARC-Payload-Digest\":\"sha1:V2VWP5BF6MDA7GL3CR7EGL54H5WFIVDB\",\"WARC-Block-Digest\":\"sha1:5DXZGYQJZU67KCKAQD6BEOZGZNJXJZAK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335530.56_warc_CC-MAIN-20221001035148-20221001065148-00612.warc.gz\"}"}
https://donaanacosta.com/qa/what-number-multiplied-by-itself-gives-2809.html	[ "", null, "# What Number Multiplied By Itself Gives 2809?\n\n## Is 77 a triangular number?\n\n0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666….\n\n## Which number multiplied by itself gives 64?\n\n88 is the number multiplied by itself is 64.\n\n## What is 4 multiplied by itself?\n\nWhat is the “Exponent 4” of a number? The “Exponent 4” of a number is the number multiplied by itself 4 times. It is written as number4. Saying “3 to the exponent 4” or 34 is the same as saying 3 times 3 times 3 times 3 (equals 81).\n\n## What number multiplied by itself gives 121?\n\n11√121 is, as we all know, 11. I think together with the number pyramid that’s quite a clue.\n\n## IS 500 a perfect square?\n\nIs 500 a perfect square number? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. … Thus, the square root of 500 is not an integer, and therefore 500 is not a square number.\n\n## What can equal 48?\n\nAnswer and Explanation: The factor pairs of the number 48 are: 1 x 48, 2 x 24, 3 x 16, 4 x 12, and 6 x 8. When you multiply each of these factor pairs together, you get 48….\n\n## What number when multiplied by itself is 11 greater than the preceding number when it is multiplied by itself?\n\nSo, using that formula, you can just put 2x – 1 = 11, giving x = 6. Let the number be x. Thus, the number is 6.\n\n## What are the triangular numbers from 1 to 100?\n\nThe triangular numbers up to 100 are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 — so what’s next? Perfect Numbers These are the numbers which equal the sum of all of their smaller factors. They are few and far between — in fact, nobody knows how many there are. Only 47 perfect numbers are currently known.\n\n## Which number multiplied by itself gives 12345678987654321?\n\n3 1’s gives 12321. so to get 12345678987654321 we need to get square of 9 1’s(111,111,111).\n\n## What number multiplied by itself gives 100?\n\nWhat number, when multiplied by itself, gives 100? The answer is both 10 and −10. The radical symbol for square roots is √0. It is used to represent the unique non-negative square root of a number, also called the principal square root.\n\n## What is the 8th triangular number?\n\nWhen we add these all together, our eighth triangular number is 36. This means that the eighth equilateral triangle in the sequence has 36 dots.\n\n## Which number multiplied by itself gives 625?\n\n25 x 25 = 625 What number multiplied by itself equals 626?\n\n## What multiplied by itself equals 196?\n\n14 x 14 = 196 What number multiplied by itself equals 197?\n\n## What times what equals 12345678987654321?\n\nBecause 111111111 times 111111111 equals 12345678987654321 according to the laws of multiplication for real numbers.\n\n## What is 4 by the power of 5?\n\nAnswer and Explanation: What is 5 to the 4th power? That would be the same as 5 multiplied by itself 4 times. In other words, it would be: 5 x 5 x 5 x 5.\n\n## What numbers are powers of 3?\n\nIn the powers of 3 table, the ones digits form the repeating pattern 3,9,7,1,3,9,7,1,… .\n\n## When a number is multiplied by itself?\n\nIn mathematics, we call multiplying a number by itself “squaring” the number. We call the result of squaring a whole number a square or a perfect square. A perfect square is any number that can be written as a whole number raised to the power of 2. For example, 9 is a perfect square.\n\n## What is a square of 12?\n\n“Is it possible to recreate this up to 12×12?”0 Squared=09 Squared=8110 Squared=10011 Squared=12112 Squared=1448 more rows\n\n## What number multiplied by itself gives 144 as product?\n\nA square roots calculator finds the number that, when multiplied by itself, would give you the number you are starting out with. For example, the square root of 144 is 12, because 12 times 12 equals 144. Of course, -12 times -12 is also 144. Therefore, every number actually has two square roots.\n\n## Why is 1 a triangular number?\n\nTriangular numbers have that name because, if drawn as dots they can form a triangle. But 1 is just a single dot, so it can’t be a triangular number, can it???" ]	[ null, "https://mc.yandex.ru/watch/66677209", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9039485,"math_prob":0.99918187,"size":5036,"snap":"2020-45-2020-50","text_gpt3_token_len":1415,"char_repetition_ratio":0.24403815,"word_repetition_ratio":0.1875,"special_character_ratio":0.33598095,"punctuation_ratio":0.16710876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995894,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T14:41:34Z\",\"WARC-Record-ID\":\"<urn:uuid:137cdd94-62e9-4ca1-8da3-8ad6bc1107cc>\",\"Content-Length\":\"41467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f3bea776-453e-4c35-8f2c-fd282b0d0c83>\",\"WARC-Concurrent-To\":\"<urn:uuid:82bcc3d4-be2b-4eb2-be2f-17bb0540b991>\",\"WARC-IP-Address\":\"87.236.16.235\",\"WARC-Target-URI\":\"https://donaanacosta.com/qa/what-number-multiplied-by-itself-gives-2809.html\",\"WARC-Payload-Digest\":\"sha1:3XUXFE7DHPVFL5I7OPROC7ZSYBOKNVEN\",\"WARC-Block-Digest\":\"sha1:P6O5K3QYQ33BAV3KHCKSYKHRXQJGJ5X2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107898577.79_warc_CC-MAIN-20201028132718-20201028162718-00337.warc.gz\"}"}
https://www.greenlighttestprep.com/module/gre-quantitative-comparison/video/1099	[ "# Lesson: QC Strategy - Looking for Equality\n\n## Comment on QC Strategy - Looking for Equality\n\n### For the last example where\n\nFor the last example where Quantity A had (x-16)(x-7)(x-4) and where Quantity B had (x-9)(x-11)(x-7); why didn't you divide both Quantities by (x-7) since it was present in both cases, and then performed \"FOIL\" to better cancel out x2 and -20x from both sides leaving you with 64 for Quantity A and 99 for Quantity B?\n\nI understand this video is about looking for Equality, but in regards to Equality; (in pausing the video before watching the explanation), I saw what both sides had equally in common (which was (x-7) being multiplied by both sides).\n\nWhy then wasn't B the right answer if we used your Matching Operations Strategy verses plugging in possible answers for x (especially when in the Matching Operations' Video the last example where QA was 2x and QB was 3x, that the answer wasn't B because we couldn't be sure of what X was; just like how we're unsure of what x could be in this particular example); wouldn't it be better to try to cancel out all the variables if possible (using the Matching Operations Strategy) to derive to the correct answer?\n\nThank you in advance; I am truly appreciative.", null, "### You need to be very careful\n\nYou need to be very careful when dividing both quantities by a variable or an expression involving variables. First off, you might be dividing both quantities by zero, which can be problematic.\nTo illustrate what I mean, consider what would happen if Quantity A was (0)(3) and Quantity B was (0)(2). Clearly, the two quantities are equal. However, if we divide both sides by 0, we \"seemingly\" get Quantity A: 3 and Quantity B: 2.\nThat's the risk you run by dividing both quantities by (x-7) since it's possible that you could be dividing both quantities by 0.\nAlso, if you watch the matching operations video (https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...), you'll see that we cannot divide both quantities by a negative value. How can we be certain that (x-7) is not a negative value.\nFor these reasons, we can't simply divide both quantities by (x-7)\nI hope that helps.\n\n### Yes it helps alot, thank you\n\nYes it helps alot, thank you so much :)\n\n### In the previous video, it\n\nIn the previous video, it said that if you use different \"nice\" numbers there is a chance that even though in the first few answers both quantities are equal, there is still a chance that they are not, in this video, if you find two conflicting results in the beginning is there no need to keep testing?", null, "### Yes, that's the downside of\n\nYes, that's the downside of testing numbers. If you plug in different numbers but keep getting the SAME result (e.g., on the first test you find that quantity A and quantity B are equal, and on the second test you find that quantity A and quantity B are equal), it's still possible that the two quantities are not always equal. So, you CAN'T be absolutely certain the answer is C.\n\nHowever, if you plug in two different sets of values and get CONFLICTING results (e.g., on the first test you find that quantity A and quantity B are equal, and on the second test you find that quantity A is greater than quantity B), then you can be CERTAIN that the answer is D.\n\n### These strategies are so\n\nThese strategies are so supremely awesome! I'm so totally in love with your videos right now!", null, "### That's great to hear. Thanks\n\nThat's great to hear. Thanks for taking the time to say that!\n\n### In the last example, I simply\n\nIn the last example, I simply eliminated the common binomial (x-7) and then plugged in a value for X which made the calculations faster as I was left with two binomials on each side as opposed to three. Not sure why that wasn't done.", null, "### Great question!\n\nGreat question!\n\nThe strategy you describe will get you in trouble. To understand why, let's examine what you mean by \"eliminate.\" Presumably, you eliminated (x-7) by dividing each side by (x-7). The problem with this strategy is that we don't know the value of x. So, (x-7) can be positive, negative, or zero, and each of these cases alters the outcome when we divide by (x-7).\n\nFor example, what if the two quantities were as follows:\nQuantity A: 3(x-7)\nQuantity B: 2(x-7)\n\nIf we divide both quantities by (x-7), we get:\nQuantity A: 3\nQuantity B: 2\nSo, is the correct answer B?\n\nNo, the correct answer is D. If we plug x = 1 into 3(x-7) and 2(x-7), we see that Quantity A is greater. If we plug plug x = 8 into 3(x-7) and 2(x-7), we see that Quantity B is greater. If we plug in x = 7, the quantities are equal.\n\nFor more on acceptable operations you can perform on each quantity, see https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...\n\n### Last example we can solve by\n\nLast example we can solve by following way,\na=(x-16)(x-7)(x-4),b=(x-9)(x-11)(x-7)\na=(x-16)(x-4),b=(x-9)(x-11)\na=x2-20x+64,b=x2-20x+99\na=x2+64,b=x2+99\nThen B must be greater,So by this process how the answer will be D,Please Sir...", null, "### There's a problem with your\n\nThere's a problem with your second step where you eliminated the (x-7) from both quantities. Presumably, you divided both quantities by (x-7)\n\nSince we don't know the value of x, it's possible that you have inadvertently divided both quantities by zero, which will negatively impact your solution.\n\nAlso, since we don't know the value of x, it may be the case that (x - 7) is negative, and we have a rule about dividing both quantities by a negative value (for more on this, see: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...\n\nConsider this example:\nQUANTITY A: 3x\nQUANTITY B: 2x\nIf x = 0, then the two quantities are EQUAL\nIf x = 1, then quantity A is greater\nIf x = -1, then quantity B is greater\nSo, the correct answer here is D.\n\nNow, let's see what happens if we break our rule and divide both sides by x.\nWe get:\nQUANTITY A: 3\nQUANTITY B: 2\nThis would SUGGEST that the correct answer is A, but this is not the case.\n\nDoes that help?\n\n### Best ever explanation\n\nBest ever explanation", null, "Thanks!\n\n### U r an awesome teacher.I like\n\nU r an awesome teacher.I like the way u teach.", null, "### Thanks for the kind words,\n\nThanks for the kind words, Seema!\n\n### Quantity A : \|x + y\|\n\nQuantity A : \|x + y\|\nQuantity B : \|x\| + \|y\|\nIn this test you did try different values for x and y ? why ? why we did not choose same value like 1 and then plug in number ? why x= 1 & y =2 but at first step we did plug in x= 0 and y=0 ?\nI am a little confuse!!!! because I thought we should plugin these type of numbers :\n0, 1, -1, 1/2, -1/2, 10,-10 but same for both variable!!! not different for each!!", null, "There's no need (or requirement) to choose the same values for each variable. If anything, choosing the same values would make the \"plugging in values\" strategy much less effective.\n\nFor example, if we kept choosing identical values for x and y for the above questions, we'd incorrectly conclude that the two quantities are always EQUAL.\n\nDoes that help?\n\nCheers,\nBrent\n\nyes" ]	[ null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null, "https://www.greenlighttestprep.com/sites/default/files/styles/thumbnail/public/pictures/picture-1-1600118168.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9155769,"math_prob":0.97797966,"size":6599,"snap":"2020-45-2020-50","text_gpt3_token_len":1711,"char_repetition_ratio":0.13813496,"word_repetition_ratio":0.077586204,"special_character_ratio":0.26155478,"punctuation_ratio":0.13112392,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99690026,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T13:32:19Z\",\"WARC-Record-ID\":\"<urn:uuid:60f19f8d-eba4-49f2-8a09-27152a83304d>\",\"Content-Length\":\"43386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19aa4a13-4250-4787-bb61-20e01a113a05>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e374a81-3650-4bdb-8345-97d5aab524df>\",\"WARC-IP-Address\":\"108.160.144.133\",\"WARC-Target-URI\":\"https://www.greenlighttestprep.com/module/gre-quantitative-comparison/video/1099\",\"WARC-Payload-Digest\":\"sha1:PDL7OENJF72O6WSMVS4UFWC47XB3TZ5K\",\"WARC-Block-Digest\":\"sha1:SUQE52X6JZO4XUIVLNRUXESVOX4SA7AI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876500.43_warc_CC-MAIN-20201021122208-20201021152208-00390.warc.gz\"}"}
https://zbmath.org/?q=an:1196.65031	[ "# zbMATH — the first resource for mathematics\n\nAn approach for determining the matrix of limiting state probabilities in discrete Markov processes. (English) Zbl 1196.65031\nSummary: A new approach for determining the matrix of limiting state probabilities in Markov processes is proposed and a polynomial time algorithm for calculating this matrix is presented. The computational complexity of the algorithm is $$O(n^{4})$$ where $$n$$ is the number of the states of the discrete system.\n\n##### MSC:\n 65C40 Numerical analysis or methods applied to Markov chains 60J22 Computational methods in Markov chains 90C39 Dynamic programming 90C40 Markov and semi-Markov decision processes 65Y20 Complexity and performance of numerical algorithms\nFull Text:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72567797,"math_prob":0.966013,"size":1151,"snap":"2021-31-2021-39","text_gpt3_token_len":301,"char_repetition_ratio":0.10636443,"word_repetition_ratio":0.14193548,"special_character_ratio":0.26238054,"punctuation_ratio":0.19339623,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994291,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T12:44:28Z\",\"WARC-Record-ID\":\"<urn:uuid:84fb829c-f953-4f8d-8eaf-8bed1a229794>\",\"Content-Length\":\"46917\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1fe0cc41-b46e-4595-9b60-666cedfcef52>\",\"WARC-Concurrent-To\":\"<urn:uuid:110b37dc-db8e-4ba3-8e43-bc2f2104fe3f>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:1196.65031\",\"WARC-Payload-Digest\":\"sha1:Z6IZU5ZG3QJTCWO3FZPRGJWHSAYSI54K\",\"WARC-Block-Digest\":\"sha1:WGET5NQOP6PSBAW4XKG4IVQUOBGB3ELM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150264.90_warc_CC-MAIN-20210724094631-20210724124631-00537.warc.gz\"}"}
https://ounces-to-grams.appspot.com/pl/430-uncja-na-gram.html	[ "Ounces To Grams\n\n# 430 oz to g430 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 430 ounce to grams?\n\n 430 oz * 28.349523125 g = 12190.2949437 g 1 oz\nA common question is How many ounce in 430 gram? And the answer is 15.1678036383 oz in 430 g. Likewise the question how many gram in 430 ounce has the answer of 12190.2949437 g in 430 oz.\n\n## How much are 430 ounces in grams?\n\n430 ounces equal 12190.2949437 grams (430oz = 12190.2949437g). Converting 430 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 430 oz to g.\n\n## Convert 430 oz to common mass\n\nUnitMass\nMicrogram12190294943.8 µg\nMilligram12190294.9437 mg\nGram12190.2949437 g\nOunce430.0 oz\nPound26.875 lbs\nKilogram12.1902949438 kg\nStone1.9196428571 st\nUS ton0.0134375 ton\nTonne0.0121902949 t\nImperial ton0.0119977679 Long tons\n\n## What is 430 ounces in g?\n\nTo convert 430 oz to g multiply the mass in ounces by 28.349523125. The 430 oz in g formula is [g] = 430 * 28.349523125. Thus, for 430 ounces in gram we get 12190.2949437 g.\n\n## 430 Ounce Conversion Table", null, "## Alternative spelling\n\n430 Ounces to g, 430 Ounces in g, 430 oz in g, 430 Ounces to Gram, 430 Ounces in Gram, 430 Ounce to Gram, 430 Ounce to g, 430 Ounce in g," ]	[ null, "https://ounces-to-grams.appspot.com/image/430.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77981126,"math_prob":0.9010965,"size":649,"snap":"2023-14-2023-23","text_gpt3_token_len":215,"char_repetition_ratio":0.19534884,"word_repetition_ratio":0.0,"special_character_ratio":0.4422188,"punctuation_ratio":0.1503268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97694814,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T11:37:12Z\",\"WARC-Record-ID\":\"<urn:uuid:6882a9d7-2349-4c5c-b12a-610029dc7d3f>\",\"Content-Length\":\"28193\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36410840-e4ff-4070-bc47-9ea0bc0a9e9f>\",\"WARC-Concurrent-To\":\"<urn:uuid:2516428b-a4c4-40f7-be81-918d52a7b369>\",\"WARC-IP-Address\":\"172.253.63.153\",\"WARC-Target-URI\":\"https://ounces-to-grams.appspot.com/pl/430-uncja-na-gram.html\",\"WARC-Payload-Digest\":\"sha1:BYOAD3H2Q7WFNR5YSJ6K6DY6MTA266XA\",\"WARC-Block-Digest\":\"sha1:7QVPLCNLNUBJDUN7BLJYEM3R4QLJFUIZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945144.17_warc_CC-MAIN-20230323100829-20230323130829-00602.warc.gz\"}"}
https://assignmentshark.com/blog/sample-of-a-python-user-defined-function/	[ "# Sample of a Python User Defined Function\n\nIn the following sample, we are going to talk about Python user defined function. A function in Python is an object that takes arguments and returns a value. A function is a block of organized, reusable code that is used to perform a specific task. Functions provide better modularity of the application and significantly increase the level of code reuse. There are some rules for creating functions in Python. You will learn about these rules if you read through the following sample.\n\nWe advise all students who are starting to study programming to read it through. Also, you can find examples of completed assignments in other disciplines. After you read our sample, you will be able to work with user defined functions in Python. You will see that such functions can be easily created by you. So, don’t wait anymore and read the following example – it will surely be helpful for you.\n\nPython Functions\n\nIn this particular paper we are going to discuss general structure of user-defined functions in Python, along with the overview of the most interesting built-in ones.\n\nUser-defined function\n\nIn Python, in order to define some function, you just need to write def and the name of the function. It is very simple. The only thing that should be followed strictly is indentation, because in Python there are no braces that specify the beginning and ending of the method. You just need to keep your code in the appropriate place to avoid errors.\n\n```def simpleOne(val):\nfor i in range(val):\nprint(\"It's a new day\")\n\nsimpleOne(4)\n```\n\nOutput:\n\n```\"It's a new day\"\n\"It's a new day\"\n\"It's a new day\"\n\"It's a new day\"\n```\n\nThis program just outputs string for as many times as you specify in its parameter.\n\nAs you see, the function is called by simply writing its name with the parameter if there is one.\n\nBuilt-in functions\n\nThe first built-in function that appears when learning Python is print(). It simply outputs the contents of what is inside this method. Even the example above uses it.\n\nIn the next example, two functions are used: list() and zip(). The first returns the list of separate items depending on what is in the arguments list in the parentheses. Zip() method combines items from several tuples by their index.\n\n```a = ['Sam', 'Dean']\nb = [1992, 2013]\nc = [1, 2, 3, 4, 5]\nprint(list(zip(a, b, c)))\n```\n\nOutput:\n\n```[('Sam', 1992, 1), ('Dean', 2013, 2)]\n```\n\nIn this example we are integrating all the existing tuples, and since there can only be two mutual pairs, you can observe them in output.\n\n```a = ['Sam', 'Dean']\nb = [1992, 2013]\nc = [1, 2, 3, 4, 5]\nfor l,m in zip(a, b):\nd[l]=m\nform = l + \":{ \" + l + \"}\"\nprint(form.format(**d))\n```\n\nOutput:\n\n```\"Sam: 1992\"\n\"Dean: 2013\"\n```\n\nIn the next example several functions are used at the same time: input(), split(), range(), len().\n\nInput method is for accepting user types in console. Split function is for separating some string with the specified in the arguments symbol or character, thus creating more independent units. Range(), in this example, is used for setting the upper limit of loops. Len() returns the size of some variable that is in the parentheses.\n\nAll in all, this small program receives input from the user, separates it by blank space, and then outputs each individual value on the new line.\n\n```print(\"Enter some words and numbers\")\ns = input()\nsome = s.split(\" \")\nfor i in range((len(some))):\nprint(some[i])\n```\n\nYou might wonder what is the data type of that variable ‘some’. You can guess or use the function type() to find out for sure. So, if we add one line of code to the above line, what we receive is:", null, "```print(\"The data type of this is \", type(some))" ]	[ null, "https://assignmentshark.com/blog/wp-content/uploads/2019/04/python-output1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88945824,"math_prob":0.9355237,"size":3744,"snap":"2020-34-2020-40","text_gpt3_token_len":871,"char_repetition_ratio":0.11764706,"word_repetition_ratio":0.055214725,"special_character_ratio":0.25320512,"punctuation_ratio":0.12941177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97340894,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-26T20:23:01Z\",\"WARC-Record-ID\":\"<urn:uuid:e834451c-17fc-4d4e-a671-1d6c2039e821>\",\"Content-Length\":\"41549\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69e7bb14-bf59-4163-8f04-cd06f36dd46e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9f5d4f7-7644-401f-b494-05efa1eda1b0>\",\"WARC-IP-Address\":\"69.65.28.148\",\"WARC-Target-URI\":\"https://assignmentshark.com/blog/sample-of-a-python-user-defined-function/\",\"WARC-Payload-Digest\":\"sha1:W5PBSRTI7B5VZQ6B3XE72IRXGOVOUXWZ\",\"WARC-Block-Digest\":\"sha1:RG4LPJWBXHMTY4ZPJZJORABHXUFIY273\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400245109.69_warc_CC-MAIN-20200926200523-20200926230523-00010.warc.gz\"}"}
https://wikimili.com/en/Geometric_terms_of_location	[ "# Geometric terms of location\n\nLast updated\n\nGeometric terms of location describe directions or positions relative to the shape of an object. These terms are used in descriptions of engineering, physics, and other sciences, as well as ordinary day to day discourse.\n\nThough these terms by themselves may be somewhat ambiguous, they are usually used in a context in which their meaning is clear. For example, when referring to a drive shaft it is clear what is meant by axial or radial directions. Or, in a free body diagram, one may similarly infer a sense of orientation by the forces or other vectors represented.\n\n## Examples\n\nCommon geometric terms of location are:\n\n• Axial – along the center of a round body, or the axis of rotation of a body\n• Azimuthal or circumferential – following around a curve or circumference of an object. For instance, the pattern of cells in Taylor–Couette flow varies along the azimuth of the experiment.\n• Collinear - in the same line\n• Degree of freedom - axis direction, see six degrees of freedom\n• Elevation - along a curve from a point on the horizon to the zenith, directly overhead.\n• Depression – along a curve from a point on the horizon to the nadir, directly below.\n• Lateral – spanning the width of a body. The distinction between width and length may be unclear out of context.\n• Lineal – following along a given path. The shape of the path is not necessarily straight (compare to linear). For instance, a length of rope might be measured in lineal meters or feet. See arc length.\n• Longitudinal – spanning the length of a body.\n• Orthogonal – at right angles to a line, or more generally, on a different axis.\n• Parallel - in the same direction\n• Perpendicular - at right angles to, synonym to orthogonal\n• Radial – along a direction pointing along a radius from the center of an object, or perpendicular to a curved path.\n• Tangential – intersecting a curve at a point and parallel to the curve at that point.\n• Transverse – orthogonal to a specified direction, such as a particle trajectory or an axis of rotation.\n• Vertical – spanning the length of a body.\n\n## Related Research Articles", null, "In plane geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angles formed by two rays lie in a plane, but this plane does not have to be a Euclidean plane. Angles are also formed by the intersection of two planes in Euclidean and other spaces. These are called dihedral angles. Angles formed by the intersection of two curves in a plane are defined as the angle determined by the tangent rays at the point of intersection. Similar statements hold in space, for example, the spherical angle formed by two great circles on a sphere is the dihedral angle between the planes determined by the great circles.", null, "In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuthal angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith, measured from a fixed reference direction on that plane. It can be seen as the three-dimensional version of the polar coordinate system.", null, "A rotation is a circular movement of an object around a center of rotation. A three-dimensional object can always be rotated around an infinite number of imaginary lines called rotation axes. If the axis passes through the body's center of mass, the body is said to rotate upon itself, or spin. A rotation about an external point, e.g. the Earth about the Sun, is called a revolution or orbital revolution, typically when it is produced by gravity. The axis is called a pole.", null, "Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations. In a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave. A simple example of a polarized transverse wave is vibrations traveling along a taut string (see image); for example, in a musical instrument like a guitar string. Depending on how the string is plucked, the vibrations can be in a vertical direction, horizontal direction, or at any angle perpendicular to the string. In contrast, in longitudinal waves, such as sound waves in a liquid or gas, the displacement of the particles in the oscillation is always in the direction of propagation, so these waves do not exhibit polarization. Transverse waves that exhibit polarization include electromagnetic waves such as light and radio waves, gravitational waves, and transverse sound waves in solids.", null, "Orbital inclination measures the tilt of an object's orbit around a celestial body. It is expressed as the angle between a reference plane and the orbital plane or axis of direction of the orbiting object.", null, "In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle. The property extends to other related geometric objects.", null, "In astronomy, an analemma is a diagram showing the position of the Sun in the sky, as seen from a fixed location on Earth at the same mean solar time, as that position varies over the course of a year. The diagram will resemble the figure 8. Globes of Earth often display an analemma.", null, "A gear or cogwheel is a rotating machine part having cut teeth or, in the case of a cogwheel, inserted teeth, which mesh with another toothed part to transmit torque. Geared devices can change the speed, torque, and direction of a power source. Gears almost always produce a change in torque, creating a mechanical advantage, through their gear ratio, and thus may be considered a simple machine. The teeth on the two meshing gears all have the same shape. Two or more meshing gears, working in a sequence, are called a gear train or a transmission. A gear can mesh with a linear toothed part, called a rack, producing translation instead of rotation.", null, "A circle of latitude on Earth is an abstract east–west circle connecting all locations around Earth at a given latitude.", null, "In mathematics and physics, the right-hand rule is a common mnemonic for understanding orientation of axes in three-dimensional space.\n\nIn physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body.", null, "Rotational symmetry, also known as radial symmetry in biology, is the property a shape has when it looks the same after some rotation by a partial turn. An object's degree of rotational symmetry is the number of distinct orientations in which it looks exactly the same for each rotation.", null, "In classical geometry, a radius of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length. The name comes from the Latin radius, meaning ray but also the spoke of a chariot wheel. The plural of radius can be either radii or the conventional English plural radiuses. The typical abbreviation and mathematical variable name for radius is r. By extension, the diameter d is defined as twice the radius:\n\nIn optics a ray is an idealized model of light, obtained by choosing a line that is perpendicular to the wavefronts of the actual light, and that points in the direction of energy flow. Rays are used to model the propagation of light through an optical system, by dividing the real light field up into discrete rays that can be computationally propagated through the system by the techniques of ray tracing. This allows even very complex optical systems to be analyzed mathematically or simulated by computer. Ray tracing uses approximate solutions to Maxwell's equations that are valid as long as the light waves propagate through and around objects whose dimensions are much greater than the light's wavelength. Ray theory does not describe phenomena such as diffraction, which require wave theory. Some wave phenomena such as interference can be modeled in limited circumstances by adding phase to the ray model.\n\nIn Gaussian optics, the cardinal points consist of three pairs of points located on the optical axis of a rotationally symmetric, focal, optical system. These are the focal points, the principal points, and the nodal points. For ideal systems, the basic imaging properties such as image size, location, and orientation are completely determined by the locations of the cardinal points; in fact only four points are necessary: the focal points and either the principal or nodal points. The only ideal system that has been achieved in practice is the plane mirror, however the cardinal points are widely used to approximate the behavior of real optical systems. Cardinal points provide a way to analytically simplify a system with many components, allowing the imaging characteristics of the system to be approximately determined with simple calculations.\n\nIn geometry, a plane of rotation is an abstract object used to describe or visualize rotations in space. In three dimensions it is an alternative to the axis of rotation, but unlike the axis of rotation it can be used in other dimensions, such as two, four or more dimensions.", null, "The Rayleigh sky model describes the observed polarization pattern of the daytime sky. Within the atmosphere Rayleigh scattering of light from air molecules, water, dust, and aerosols causes the sky's light to have a defined polarization pattern. The same elastic scattering processes cause the sky to be blue. The polarization is characterized at each wavelength by its degree of polarization, and orientation.\n\nIn astronomy, geography, and related sciences and contexts, a direction or plane passing by a given point is said to be vertical if it contains the local gravity direction at that point. Conversely, a direction or plane is said to be horizontal if it is perpendicular to the vertical direction. In general, something that is vertical can be drawn from up to down, such as the y-axis in the Cartesian coordinate system.", null, "In geometry, an object has symmetry if there is an operation or transformation that maps the figure/object onto itself. Thus, a symmetry can be thought of as an immunity to change. For instance, a circle rotated about its center will have the same shape and size as the original circle, as all points before and after the transform would be indistinguishable. A circle is thus said to be symmetric under rotation or to have rotational symmetry. If the isometry is the reflection of a plane figure about a line, then the figure is said to have reflectional symmetry or line symmetry; it is also possible for a figure/object to have more than one line of symmetry." ]	[ null, "https://upload.wikimedia.org/wikipedia/commons/thumb/2/21/Two_rays_and_one_vertex.png/320px-Two_rays_and_one_vertex.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/4/4f/3D_Spherical.svg/320px-3D_Spherical.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/0/02/Rotating_Sphere.gif", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Polarizacio.jpg/320px-Polarizacio.jpg", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Orbit1.svg/320px-Orbit1.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/8/84/Perpendicular-coloured.svg/320px-Perpendicular-coloured.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Analemma_fishburn.tif/lossy-page1-254px-Analemma_fishburn.tif.jpg", null, "https://upload.wikimedia.org/wikipedia/commons/1/14/Gears_animation.gif", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/5/56/World_map_longlat.svg/320px-World_map_longlat.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/d/d2/Right_hand_rule_cross_product.svg/320px-Right_hand_rule_cross_product.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/3/36/The_armoured_triskelion_on_the_flag_of_the_Isle_of_Man.svg/320px-The_armoured_triskelion_on_the_flag_of_the_Isle_of_Man.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/0/03/Circle-withsegments.svg/317px-Circle-withsegments.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/b/b1/Degpolred.jpg/320px-Degpolred.jpg", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/6/68/Simetria-bilateria.svg/320px-Simetria-bilateria.svg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9369503,"math_prob":0.9672462,"size":9739,"snap":"2020-24-2020-29","text_gpt3_token_len":1902,"char_repetition_ratio":0.12922445,"word_repetition_ratio":0.0024630541,"special_character_ratio":0.18852039,"punctuation_ratio":0.09845985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833327,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,6,null,null,null,null,null,8,null,null,null,null,null,2,null,6,null,1,null,3,null,5,null,null,null,1,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-05T06:14:10Z\",\"WARC-Record-ID\":\"<urn:uuid:c38557ab-d978-4302-9fe3-bcd647362c09>\",\"Content-Length\":\"28230\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c066fd8-a414-42a7-820f-192d91dad289>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2cf679d-5da8-4eca-b97d-01243327d23e>\",\"WARC-IP-Address\":\"172.67.148.56\",\"WARC-Target-URI\":\"https://wikimili.com/en/Geometric_terms_of_location\",\"WARC-Payload-Digest\":\"sha1:FAZXEENX3I2O3GXYG46Z7RL7N5P5L5VH\",\"WARC-Block-Digest\":\"sha1:H3YR53KEKMD4ITRRFW36CKWNPF3BGGE3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348493151.92_warc_CC-MAIN-20200605045722-20200605075722-00388.warc.gz\"}"}
http://mike-the-strike.net/Soaring_Forecast/December2009.htm	[ "Daily report for December 2009\n```Averages\\Extremes for day :01\n------------------------------------------------------------\n\nAverage temperature = 52.5°F\nAverage humidity = 37%\nAverage dewpoint = 26.2°F\nAverage barometer = 29.9 in.\nAverage windspeed = 2.4 mph\nAverage gustspeed = 2.7 mph\nAverage direction = 333° (NNW)\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 01 at time 00:00\nMaximum temperature = 60.8°F on day 01 at time 15:33\nMinimum temperature = 44.4°F on day 01 at time 07:51\nMaximum humidity = 50% on day 01 at time 08:35\nMinimum humidity = 30% on day 01 at time 18:34\nMaximum pressure = 29.929 in. on day 01 at time 09:29\nMinimum pressure = 29.808 in. on day 01 at time 19:25\nMaximum windspeed = 5.8 mph on day 01 at time 15:43\nMaximum gust speed = 9 mph from 225 °( SW) on day 01 at time 15:33\nMaximum heat index = 60.8°F on day 01 at time 15:33\n\nClick here for today´s 24 hour graph :1 :12 :2009\n\nAverages\\Extremes for day :02\n------------------------------------------------------------\n\nAverage temperature = 53.7°F\nAverage humidity = 33%\nAverage dewpoint = 25.3°F\nAverage barometer = 29.8 in.\nAverage windspeed = 2.0 mph\nAverage gustspeed = 2.4 mph\nAverage direction = 58° (ENE)\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 02 at time 23:57\nMaximum temperature = 60.8°F on day 02 at time 15:38\nMinimum temperature = 46.0°F on day 02 at time 23:38\nMaximum humidity = 46% on day 02 at time 08:03\nMinimum humidity = 27% on day 02 at time 15:38\nMaximum pressure = 29.899 in. on day 02 at time 11:08\nMinimum pressure = 29.691 in. on day 02 at time 17:38\nMaximum windspeed = 4.6 mph on day 02 at time 10:31\nMaximum gust speed = 8 mph from 090 °( E ) on day 02 at time 09:27\nMaximum heat index = 60.8°F on day 02 at time 15:38\n\nClick here for today´s 24 hour graph :2 :12 :2009\n\nAverages\\Extremes for day :03\n------------------------------------------------------------\n\nAverage temperature = 52.2°F\nAverage humidity = 29%\nAverage dewpoint = 17.7°F\nAverage barometer = 29.8 in.\nAverage windspeed = 3.1 mph\nAverage gustspeed = 4.0 mph\nAverage direction = 29° (NNE)\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 03 at time 23:57\nMaximum temperature = 60.6°F on day 03 at time 15:38\nMinimum temperature = 45.1°F on day 03 at time 23:57\nMaximum humidity = 44% on day 03 at time 07:38\nMinimum humidity = 5% on day 03 at time 23:57\nMaximum pressure = 29.901 in. on day 03 at time 23:57\nMinimum pressure = 29.705 in. on day 03 at time 00:38\nMaximum windspeed = 6.9 mph on day 03 at time 23:57\nMaximum gust speed = 8 mph from 090 °( E ) on day 03 at time 23:57\nMaximum heat index = 60.6°F on day 03 at time 15:38\n\nClick here for today´s 24 hour graph :3 :12 :2009\n\nAverages\\Extremes for day :04\n------------------------------------------------------------\n\nAverage temperature = 45.9°F\nAverage humidity = 6%\nAverage dewpoint = -17.3°F\nAverage barometer = 30.0 in.\nAverage windspeed = 3.8 mph\nAverage gustspeed = 5.0 mph\nAverage direction = 70° (ENE)\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 04 at time 23:57\nMaximum temperature = 54.5°F on day 04 at time 15:38\nMinimum temperature = 38.7°F on day 04 at time 07:38\nMaximum humidity = 12% on day 04 at time 19:38\nMinimum humidity = 5% on day 04 at time 15:38\nMaximum pressure = 30.076 in. on day 04 at time 10:38\nMinimum pressure = 29.901 in. on day 04 at time 00:38\nMaximum windspeed = 8.1 mph on day 04 at time 10:38\nMaximum gust speed = 13 mph from 068 °(ENE) on day 04 at time 10:38\nMaximum heat index = 54.5°F on day 04 at time 15:38\n\nClick here for today´s 24 hour graph :4 :12 :2009\n\nAverages\\Extremes for day :05\n------------------------------------------------------------\n\nAverage temperature = 46.8°F\nAverage humidity = 11%\nAverage dewpoint = -7.6°F\nAverage barometer = 29.8 in.\nAverage windspeed = 3.4 mph\nAverage gustspeed = 4.4 mph\nAverage direction = 93° ( E )\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 05 at time 23:57\nMaximum temperature = 54.7°F on day 05 at time 14:55\nMinimum temperature = 41.2°F on day 05 at time 06:38\nMaximum humidity = 21% on day 05 at time 22:19\nMinimum humidity = 6% on day 05 at time 08:38\nMaximum pressure = 29.919 in. on day 05 at time 00:38\nMinimum pressure = 29.743 in. on day 05 at time 23:09\nMaximum windspeed = 6.9 mph on day 05 at time 11:21\nMaximum gust speed = 12 mph from 045 °( NE) on day 05 at time 11:01\nMaximum heat index = 54.7°F on day 05 at time 14:55\n\nClick here for today´s 24 hour graph :5 :12 :2009\n\nAverages\\Extremes for day :06\n------------------------------------------------------------\n\nAverage temperature = 47.3°F\nAverage humidity = 34%\nAverage dewpoint = 19.7°F\nAverage barometer = 29.8 in.\nAverage windspeed = 1.4 mph\nAverage gustspeed = 1.7 mph\nAverage direction = 187° ( S )\nRainfall for month = 0.00 in.\nRainfall for year = 2.85 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 06 at time 23:57\nMaximum temperature = 54.1°F on day 06 at time 14:43\nMinimum temperature = 37.4°F on day 06 at time 03:39\nMaximum humidity = 47% on day 06 at time 09:35\nMinimum humidity = 16% on day 06 at time 00:51\nMaximum pressure = 29.902 in. on day 06 at time 11:10\nMinimum pressure = 29.754 in. on day 06 at time 00:29\nMaximum windspeed = 4.6 mph on day 06 at time 16:16\nMaximum gust speed = 9 mph from 203 °(SSW) on day 06 at time 14:08\nMaximum heat index = 54.1°F on day 06 at time 14:43\n\nClick here for today´s 24 hour graph :6 :12 :2009\n\nAverages\\Extremes for day :07\n------------------------------------------------------------\n\nAverage temperature = 49.2°F\nAverage humidity = 68%\nAverage dewpoint = 38.6°F\nAverage barometer = 29.7 in.\nAverage windspeed = 6.3 mph\nAverage gustspeed = 8.0 mph\nAverage direction = 188° ( S )\nRainfall for month = 0.83 in.\nRainfall for year = 3.68 in.\nRainfall for day = 0.83 in.\nMaximum rain per minute = 0.04 in. on day 07 at time 14:36\nMaximum temperature = 50.5°F on day 07 at time 09:25\nMinimum temperature = 46.2°F on day 07 at time 23:51\nMaximum humidity = 84% on day 07 at time 23:29\nMinimum humidity = 44% on day 07 at time 00:53\nMaximum pressure = 29.867 in. on day 07 at time 00:30\nMinimum pressure = 29.379 in. on day 07 at time 23:29\nMaximum windspeed = 20.7 mph on day 07 at time 23:27\nMaximum gust speed = 38 mph from 180 °( S ) on day 07 at time 22:47\nMaximum heat index = 50.5°F on day 07 at time 09:25\n\nClick here for today´s 24 hour graph :7 :12 :2009\n\nAverages\\Extremes for day :08\n------------------------------------------------------------\n\nAverage temperature = 43.5°F\nAverage humidity = 50%\nAverage dewpoint = 25.4°F\nAverage barometer = 29.8 in.\nAverage windspeed = 4.3 mph\nAverage gustspeed = 5.2 mph\nAverage direction = 212° (SSW)\nRainfall for month = 0.86 in.\nRainfall for year = 3.71 in.\nRainfall for day = 0.03 in.\nMaximum rain per minute = 0.03 in. on day 08 at time 03:16\nMaximum temperature = 48.0°F on day 08 at time 14:47\nMinimum temperature = 36.7°F on day 08 at time 23:53\nMaximum humidity = 84% on day 08 at time 00:19\nMinimum humidity = 33% on day 08 at time 16:43\nMaximum pressure = 29.973 in. on day 08 at time 22:49\nMinimum pressure = 29.459 in. on day 08 at time 00:09\nMaximum windspeed = 12.7 mph on day 08 at time 02:59\nMaximum gust speed = 25 mph from 270 °( W ) on day 08 at time 03:51\nMaximum heat index = 48.0°F on day 08 at time 14:47\n\nClick here for today´s 24 hour graph :8 :12 :2009\n\nAverages\\Extremes for day :09\n------------------------------------------------------------\n\nAverage temperature = 43.2°F\nAverage humidity = 41%\nAverage dewpoint = 20.5°F\nAverage barometer = 29.9 in.\nAverage windspeed = 2.0 mph\nAverage gustspeed = 2.3 mph\nAverage direction = 51° ( NE)\nRainfall for month = 0.86 in.\nRainfall for year = 3.71 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 09 at time 23:57\nMaximum temperature = 51.3°F on day 09 at time 14:41\nMinimum temperature = 36.1°F on day 09 at time 00:57\nMaximum humidity = 50% on day 09 at time 23:57\nMinimum humidity = 31% on day 09 at time 14:15\nMaximum pressure = 30.008 in. on day 09 at time 11:02\nMinimum pressure = 29.838 in. on day 09 at time 04:38\nMaximum windspeed = 4.6 mph on day 09 at time 15:15\nMaximum gust speed = 8 mph from 225 °( SW) on day 09 at time 15:37\nMaximum heat index = 51.3°F on day 09 at time 14:41\n\nClick here for today´s 24 hour graph :9 :12 :2009\n\nAverages\\Extremes for day :10\n------------------------------------------------------------\n\nAverage temperature = 44.3°F\nAverage humidity = 40%\nAverage dewpoint = 21.0°F\nAverage barometer = 29.9 in.\nAverage windspeed = 2.1 mph\nAverage gustspeed = 2.7 mph\nAverage direction = 29° (NNE)\nRainfall for month = 0.86 in.\nRainfall for year = 3.71 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 10 at time 23:57\nMaximum temperature = 52.7°F on day 10 at time 16:24\nMinimum temperature = 37.8°F on day 10 at time 03:09\nMaximum humidity = 51% on day 10 at time 01:29\nMinimum humidity = 30% on day 10 at time 16:38\nMaximum pressure = 30.014 in. on day 10 at time 10:30\nMinimum pressure = 29.846 in. on day 10 at time 04:38\nMaximum windspeed = 4.6 mph on day 10 at time 11:25\nMaximum gust speed = 7 mph from 045 °( NE) on day 10 at time 11:30\nMaximum heat index = 52.7°F on day 10 at time 16:24\n\nClick here for today´s 24 hour graph :10 :12 :2009\n\nAverages\\Extremes for day :11\n------------------------------------------------------------\n\nAverage temperature = 47.7°F\nAverage humidity = 39%\nAverage dewpoint = 23.7°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.0 mph\nAverage gustspeed = 2.4 mph\nAverage direction = 18° (NNE)\nRainfall for month = 0.86 in.\nRainfall for year = 3.71 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 11 at time 23:57\nMaximum temperature = 56.8°F on day 11 at time 15:39\nMinimum temperature = 40.6°F on day 11 at time 04:38\nMaximum humidity = 46% on day 11 at time 23:57\nMinimum humidity = 30% on day 11 at time 17:11\nMaximum pressure = 30.050 in. on day 11 at time 10:10\nMinimum pressure = 29.883 in. on day 11 at time 20:38\nMaximum windspeed = 4.6 mph on day 11 at time 12:06\nMaximum gust speed = 9 mph from 045 °( NE) on day 11 at time 10:58\nMaximum heat index = 56.8°F on day 11 at time 15:39\n\nClick here for today´s 24 hour graph :11 :12 :2009\n\nAverages\\Extremes for day :12\n------------------------------------------------------------\n\nAverage temperature = 51.2°F\nAverage humidity = 42%\nAverage dewpoint = 28.5°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.2 mph\nAverage gustspeed = 2.6 mph\nAverage direction = 35° ( NE)\nRainfall for month = 0.86 in.\nRainfall for year = 3.71 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 12 at time 23:57\nMaximum temperature = 57.4°F on day 12 at time 15:33\nMinimum temperature = 45.0°F on day 12 at time 01:38\nMaximum humidity = 49% on day 12 at time 21:53\nMinimum humidity = 31% on day 12 at time 15:07\nMaximum pressure = 30.103 in. on day 12 at time 11:09\nMinimum pressure = 29.879 in. on day 12 at time 04:38\nMaximum windspeed = 4.6 mph on day 12 at time 12:05\nMaximum gust speed = 8 mph from 045 °( NE) on day 12 at time 11:45\nMaximum heat index = 57.4°F on day 12 at time 15:33\n\nClick here for today´s 24 hour graph :12 :12 :2009\n\nAverages\\Extremes for day :13\n------------------------------------------------------------\n\nAverage temperature = 52.0°F\nAverage humidity = 58%\nAverage dewpoint = 37.4°F\nAverage barometer = 29.9 in.\nAverage windspeed = 2.4 mph\nAverage gustspeed = 2.9 mph\nAverage direction = 166° (SSE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.06 in.\nMaximum rain per minute = 0.03 in. on day 13 at time 13:22\nMaximum temperature = 57.4°F on day 13 at time 10:47\nMinimum temperature = 46.4°F on day 13 at time 23:49\nMaximum humidity = 78% on day 13 at time 22:09\nMinimum humidity = 43% on day 13 at time 11:11\nMaximum pressure = 29.985 in. on day 13 at time 00:09\nMinimum pressure = 29.890 in. on day 13 at time 05:49\nMaximum windspeed = 6.9 mph on day 13 at time 10:39\nMaximum gust speed = 12 mph from 225 °( SW) on day 13 at time 10:33\nMaximum heat index = 57.4°F on day 13 at time 10:47\n\nClick here for today´s 24 hour graph :13 :12 :2009\n\nAverages\\Extremes for day :14\n------------------------------------------------------------\n\nAverage temperature = 51.3°F\nAverage humidity = 55%\nAverage dewpoint = 34.8°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.0 mph\nAverage gustspeed = 2.5 mph\nAverage direction = 13° (NNE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 14 at time 23:57\nMaximum temperature = 59.0°F on day 14 at time 15:09\nMinimum temperature = 45.5°F on day 14 at time 07:57\nMaximum humidity = 76% on day 14 at time 02:05\nMinimum humidity = 34% on day 14 at time 16:39\nMaximum pressure = 30.100 in. on day 14 at time 23:49\nMinimum pressure = 29.911 in. on day 14 at time 04:29\nMaximum windspeed = 4.6 mph on day 14 at time 13:19\nMaximum gust speed = 7 mph from 293 °(WNW) on day 14 at time 13:41\nMaximum heat index = 59.0°F on day 14 at time 15:09\n\nClick here for today´s 24 hour graph :14 :12 :2009\n\nAverages\\Extremes for day :15\n------------------------------------------------------------\n\nAverage temperature = 52.0°F\nAverage humidity = 29%\nAverage dewpoint = 20.1°F\nAverage barometer = 30.2 in.\nAverage windspeed = 3.5 mph\nAverage gustspeed = 4.2 mph\nAverage direction = 35° ( NE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 15 at time 23:57\nMaximum temperature = 62.2°F on day 15 at time 15:17\nMinimum temperature = 44.1°F on day 15 at time 05:29\nMaximum humidity = 40% on day 15 at time 00:51\nMinimum humidity = 16% on day 15 at time 16:37\nMaximum pressure = 30.230 in. on day 15 at time 11:09\nMinimum pressure = 30.097 in. on day 15 at time 00:10\nMaximum windspeed = 8.1 mph on day 15 at time 09:35\nMaximum gust speed = 13 mph from 068 °(ENE) on day 15 at time 09:37\nMaximum heat index = 62.2°F on day 15 at time 15:17\n\nClick here for today´s 24 hour graph :15 :12 :2009\n\nAverages\\Extremes for day :16\n------------------------------------------------------------\n\nAverage temperature = 56.1°F\nAverage humidity = 27%\nAverage dewpoint = 21.9°F\nAverage barometer = 30.1 in.\nAverage windspeed = 4.4 mph\nAverage gustspeed = 5.4 mph\nAverage direction = 39° ( NE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 16 at time 23:57\nMaximum temperature = 65.5°F on day 16 at time 15:21\nMinimum temperature = 48.2°F on day 16 at time 05:17\nMaximum humidity = 31% on day 16 at time 23:51\nMinimum humidity = 21% on day 16 at time 16:43\nMaximum pressure = 30.191 in. on day 16 at time 11:09\nMinimum pressure = 30.070 in. on day 16 at time 23:57\nMaximum windspeed = 8.1 mph on day 16 at time 12:17\nMaximum gust speed = 14 mph from 023 °(NNE) on day 16 at time 11:07\nMaximum heat index = 74.1°F on day 16 at time 14:39\n\nClick here for today´s 24 hour graph :16 :12 :2009\n\nAverages\\Extremes for day :17\n------------------------------------------------------------\n\nAverage temperature = 57.2°F\nAverage humidity = 28%\nAverage dewpoint = 23.7°F\nAverage barometer = 30.0 in.\nAverage windspeed = 3.2 mph\nAverage gustspeed = 3.9 mph\nAverage direction = 28° (NNE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 17 at time 23:57\nMaximum temperature = 66.4°F on day 17 at time 15:41\nMinimum temperature = 50.2°F on day 17 at time 07:01\nMaximum humidity = 33% on day 17 at time 07:55\nMinimum humidity = 20% on day 17 at time 14:55\nMaximum pressure = 30.082 in. on day 17 at time 10:49\nMinimum pressure = 29.991 in. on day 17 at time 17:09\nMaximum windspeed = 6.9 mph on day 17 at time 09:01\nMaximum gust speed = 10 mph from 045 °( NE) on day 17 at time 10:21\nMaximum heat index = 74.1°F on day 17 at time 16:47\n\nClick here for today´s 24 hour graph :17 :12 :2009\n\nAverages\\Extremes for day :18\n------------------------------------------------------------\n\nAverage temperature = 57.7°F\nAverage humidity = 22%\nAverage dewpoint = 19.0°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.6 mph\nAverage gustspeed = 3.2 mph\nAverage direction = 17° (NNE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 18 at time 23:57\nMaximum temperature = 66.2°F on day 18 at time 15:25\nMinimum temperature = 50.0°F on day 18 at time 07:47\nMaximum humidity = 30% on day 18 at time 20:51\nMinimum humidity = 16% on day 18 at time 11:19\nMaximum pressure = 30.038 in. on day 18 at time 10:09\nMinimum pressure = 29.926 in. on day 18 at time 18:09\nMaximum windspeed = 5.8 mph on day 18 at time 05:27\nMaximum gust speed = 10 mph from 360 °( N ) on day 18 at time 04:41\nMaximum heat index = 73.4°F on day 18 at time 14:35\n\nClick here for today´s 24 hour graph :18 :12 :2009\n\nAverages\\Extremes for day :19\n------------------------------------------------------------\n\nAverage temperature = 56.7°F\nAverage humidity = 19%\nAverage dewpoint = 14.1°F\nAverage barometer = 30.0 in.\nAverage windspeed = 4.8 mph\nAverage gustspeed = 5.9 mph\nAverage direction = 34° ( NE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 19 at time 23:57\nMaximum temperature = 65.3°F on day 19 at time 15:39\nMinimum temperature = 50.0°F on day 19 at time 06:17\nMaximum humidity = 28% on day 19 at time 06:25\nMinimum humidity = 12% on day 19 at time 17:07\nMaximum pressure = 30.082 in. on day 19 at time 23:49\nMinimum pressure = 29.952 in. on day 19 at time 00:49\nMaximum windspeed = 9.2 mph on day 19 at time 09:54\nMaximum gust speed = 16 mph from 360 °( N ) on day 19 at time 14:07\nMaximum heat index = 69.6°F on day 19 at time 15:39\n\nClick here for today´s 24 hour graph :19 :12 :2009\n\nAverages\\Extremes for day :20\n------------------------------------------------------------\n\nAverage temperature = 57.4°F\nAverage humidity = 20%\nAverage dewpoint = 16.3°F\nAverage barometer = 30.1 in.\nAverage windspeed = 5.1 mph\nAverage gustspeed = 6.1 mph\nAverage direction = 42° ( NE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 20 at time 23:57\nMaximum temperature = 63.9°F on day 20 at time 13:55\nMinimum temperature = 49.3°F on day 20 at time 07:45\nMaximum humidity = 22% on day 20 at time 23:57\nMinimum humidity = 17% on day 20 at time 14:07\nMaximum pressure = 30.153 in. on day 20 at time 10:49\nMinimum pressure = 30.047 in. on day 20 at time 23:09\nMaximum windspeed = 9.2 mph on day 20 at time 04:59\nMaximum gust speed = 14 mph from 045 °( NE) on day 20 at time 05:01\nMaximum heat index = 63.9°F on day 20 at time 13:55\n\nClick here for today´s 24 hour graph :20 :12 :2009\n\nAverages\\Extremes for day :21\n------------------------------------------------------------\n\nAverage temperature = 57.4°F\nAverage humidity = 25%\nAverage dewpoint = 21.2°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.4 mph\nAverage gustspeed = 3.0 mph\nAverage direction = 31° (NNE)\nRainfall for month = 0.92 in.\nRainfall for year = 3.77 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 21 at time 23:57\nMaximum temperature = 62.6°F on day 21 at time 13:15\nMinimum temperature = 52.2°F on day 21 at time 23:57\nMaximum humidity = 35% on day 21 at time 23:57\nMinimum humidity = 21% on day 21 at time 10:35\nMaximum pressure = 30.059 in. on day 21 at time 01:29\nMinimum pressure = 29.896 in. on day 21 at time 23:57\nMaximum windspeed = 5.8 mph on day 21 at time 10:05\nMaximum gust speed = 9 mph from 023 °(NNE) on day 21 at time 09:15\nMaximum heat index = 62.6°F on day 21 at time 13:15\n\nClick here for today´s 24 hour graph :21 :12 :2009\n\nAverages\\Extremes for day :22\n------------------------------------------------------------\n\nAverage temperature = 50.3°F\nAverage humidity = 44%\nAverage dewpoint = 28.1°F\nAverage barometer = 29.7 in.\nAverage windspeed = 3.1 mph\nAverage gustspeed = 3.8 mph\nAverage direction = 204° (SSW)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.16 in.\nMaximum rain per minute = 0.04 in. on day 22 at time 14:44\nMaximum temperature = 56.8°F on day 22 at time 13:21\nMinimum temperature = 44.6°F on day 22 at time 23:55\nMaximum humidity = 67% on day 22 at time 16:09\nMinimum humidity = 30% on day 22 at time 13:49\nMaximum pressure = 29.896 in. on day 22 at time 00:09\nMinimum pressure = 29.554 in. on day 22 at time 14:49\nMaximum windspeed = 11.5 mph on day 22 at time 15:09\nMaximum gust speed = 17 mph from 248 °(WSW) on day 22 at time 15:01\nMaximum heat index = 56.8°F on day 22 at time 13:21\n\nClick here for today´s 24 hour graph :22 :12 :2009\n\nAverages\\Extremes for day :23\n------------------------------------------------------------\n\nAverage temperature = 43.9°F\nAverage humidity = 55%\nAverage dewpoint = 28.7°F\nAverage barometer = 29.8 in.\nAverage windspeed = 1.5 mph\nAverage gustspeed = 1.9 mph\nAverage direction = 196° (SSW)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 23 at time 23:57\nMaximum temperature = 49.1°F on day 23 at time 16:23\nMinimum temperature = 38.8°F on day 23 at time 08:31\nMaximum humidity = 63% on day 23 at time 04:53\nMinimum humidity = 48% on day 23 at time 17:39\nMaximum pressure = 29.991 in. on day 23 at time 23:57\nMinimum pressure = 29.678 in. on day 23 at time 00:09\nMaximum windspeed = 4.6 mph on day 23 at time 16:03\nMaximum gust speed = 8 mph from 203 °(SSW) on day 23 at time 15:11\nMaximum heat index = 49.1°F on day 23 at time 16:23\n\nClick here for today´s 24 hour graph :23 :12 :2009\n\nAverages\\Extremes for day :24\n------------------------------------------------------------\n\nAverage temperature = 44.6°F\nAverage humidity = 47%\nAverage dewpoint = 24.6°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.3 mph\nAverage gustspeed = 2.9 mph\nAverage direction = 13° (NNE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 24 at time 23:57\nMaximum temperature = 53.2°F on day 24 at time 15:39\nMinimum temperature = 36.9°F on day 24 at time 07:21\nMaximum humidity = 68% on day 24 at time 07:01\nMinimum humidity = 30% on day 24 at time 23:57\nMaximum pressure = 30.085 in. on day 24 at time 10:09\nMinimum pressure = 29.991 in. on day 24 at time 00:09\nMaximum windspeed = 4.6 mph on day 24 at time 23:43\nMaximum gust speed = 10 mph from 360 °( N ) on day 24 at time 13:13\nMaximum heat index = 53.2°F on day 24 at time 15:39\n\nClick here for today´s 24 hour graph :24 :12 :2009\n\nAverages\\Extremes for day :25\n------------------------------------------------------------\n\nAverage temperature = 43.9°F\nAverage humidity = 29%\nAverage dewpoint = 13.4°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.5 mph\nAverage gustspeed = 3.0 mph\nAverage direction = 19° (NNE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 25 at time 23:57\nMaximum temperature = 51.4°F on day 25 at time 15:03\nMinimum temperature = 37.2°F on day 25 at time 07:33\nMaximum humidity = 39% on day 25 at time 23:57\nMinimum humidity = 21% on day 25 at time 12:57\nMaximum pressure = 30.023 in. on day 25 at time 00:09\nMinimum pressure = 29.893 in. on day 25 at time 17:29\nMaximum windspeed = 4.6 mph on day 25 at time 14:23\nMaximum gust speed = 8 mph from 225 °( SW) on day 25 at time 15:41\nMaximum heat index = 51.4°F on day 25 at time 15:03\n\nClick here for today´s 24 hour graph :25 :12 :2009\n\nAverages\\Extremes for day :26\n------------------------------------------------------------\n\nAverage temperature = 44.8°F\nAverage humidity = 27%\nAverage dewpoint = 11.9°F\nAverage barometer = 30.0 in.\nAverage windspeed = 2.8 mph\nAverage gustspeed = 3.4 mph\nAverage direction = 38° ( NE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 26 at time 23:57\nMaximum temperature = 52.9°F on day 26 at time 15:31\nMinimum temperature = 36.9°F on day 26 at time 07:31\nMaximum humidity = 40% on day 26 at time 01:23\nMinimum humidity = 14% on day 26 at time 15:59\nMaximum pressure = 30.103 in. on day 26 at time 23:29\nMinimum pressure = 29.955 in. on day 26 at time 01:09\nMaximum windspeed = 6.9 mph on day 26 at time 11:51\nMaximum gust speed = 12 mph from 090 °( E ) on day 26 at time 11:59\nMaximum heat index = 52.9°F on day 26 at time 15:31\n\nClick here for today´s 24 hour graph :26 :12 :2009\n\nAverages\\Extremes for day :27\n------------------------------------------------------------\n\nAverage temperature = 47.3°F\nAverage humidity = 22%\nAverage dewpoint = 10.1°F\nAverage barometer = 30.1 in.\nAverage windspeed = 3.6 mph\nAverage gustspeed = 4.5 mph\nAverage direction = 43° ( NE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 27 at time 23:57\nMaximum temperature = 55.6°F on day 27 at time 15:39\nMinimum temperature = 40.6°F on day 27 at time 07:19\nMaximum humidity = 27% on day 27 at time 00:09\nMinimum humidity = 15% on day 27 at time 15:17\nMaximum pressure = 30.159 in. on day 27 at time 11:09\nMinimum pressure = 30.067 in. on day 27 at time 02:09\nMaximum windspeed = 8.1 mph on day 27 at time 08:47\nMaximum gust speed = 12 mph from 090 °( E ) on day 27 at time 08:37\nMaximum heat index = 55.6°F on day 27 at time 15:39\n\nClick here for today´s 24 hour graph :27 :12 :2009\n\nAverages\\Extremes for day :28\n------------------------------------------------------------\n\nAverage temperature = 50.7°F\nAverage humidity = 19%\nAverage dewpoint = 9.6°F\nAverage barometer = 30.0 in.\nAverage windspeed = 5.7 mph\nAverage gustspeed = 7.0 mph\nAverage direction = 39° ( NE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 28 at time 23:50\nMaximum temperature = 55.8°F on day 28 at time 16:01\nMinimum temperature = 45.1°F on day 28 at time 02:57\nMaximum humidity = 23% on day 28 at time 04:11\nMinimum humidity = 13% on day 28 at time 15:17\nMaximum pressure = 30.100 in. on day 28 at time 00:09\nMinimum pressure = 29.935 in. on day 28 at time 18:09\nMaximum windspeed = 12.7 mph on day 28 at time 11:07\nMaximum gust speed = 20 mph from 023 °(NNE) on day 28 at time 10:45\nMaximum heat index = 55.8°F on day 28 at time 16:01\n\nClick here for today´s 24 hour graph :28 :12 :2009\n\nAverages\\Extremes for day :29\n------------------------------------------------------------\n\nAverage temperature = 49.2°F\nAverage humidity = 22%\nAverage dewpoint = 12.1°F\nAverage barometer = 30.0 in.\nAverage windspeed = 4.7 mph\nAverage gustspeed = 5.4 mph\nAverage direction = 35° ( NE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 29 at time 00:00\nMaximum temperature = 50.0°F on day 29 at time 02:35\nMinimum temperature = 48.4°F on day 29 at time 00:00\nMaximum humidity = 23% on day 29 at time 00:00\nMinimum humidity = 21% on day 29 at time 00:25\nMaximum pressure = 29.976 in. on day 29 at time 00:00\nMinimum pressure = 29.938 in. on day 29 at time 02:49\nMaximum windspeed = 6.9 mph on day 29 at time 00:00\nMaximum gust speed = 9 mph from 068 °(ENE) on day 29 at time 06:07\nMaximum heat index = 50.0°F on day 29 at time 02:35\n\nClick here for today´s 24 hour graph :29 :12 :2009\n\nAverages\\Extremes for day :30\n------------------------------------------------------------\n\nAverage temperature = 48.4°F\nAverage humidity = 23%\nAverage dewpoint = 12.2°F\nAverage barometer = 30.0 in.\nAverage windspeed = 6.9 mph\nAverage gustspeed = 5.8 mph\nAverage direction = 60° (ENE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 30 at time 00:00\nMaximum temperature = 48.4°F on day 30 at time 00:00\nMinimum temperature = 48.4°F on day 30 at time 00:00\nMaximum humidity = 23% on day 30 at time 00:00\nMinimum humidity = 23% on day 30 at time 00:00\nMaximum pressure = 29.976 in. on day 30 at time 00:00\nMinimum pressure = 29.976 in. on day 30 at time 00:00\nMaximum windspeed = 6.9 mph on day 30 at time 00:00\nMaximum gust speed = 6 mph from 068 °(ENE) on day 30 at time 00:00\nMaximum heat index = 48.4°F on day 30 at time 00:00\n\nClick here for today´s 24 hour graph :30 :12 :2009\n\nAverages\\Extremes for day :31\n------------------------------------------------------------\n\nAverage temperature = 51.3°F\nAverage humidity = 20%\nAverage dewpoint = 11.5°F\nAverage barometer = 30.3 in.\nAverage windspeed = 3.5 mph\nAverage gustspeed = 4.3 mph\nAverage direction = 34° ( NE)\nRainfall for month = 1.08 in.\nRainfall for year = 3.93 in.\nRainfall for day = 0.00 in.\nMaximum rain per minute = 0.00 in. on day 31 at time 23:50\nMaximum temperature = 56.7°F on day 31 at time 16:19\nMinimum temperature = 43.5°F on day 31 at time 12:00\nMaximum humidity = 46% on day 31 at time 12:00\nMinimum humidity = 15% on day 31 at time 17:03\nMaximum pressure = 30.292 in. on day 31 at time 23:49\nMinimum pressure = 29.976 in. on day 31 at time 04:00\nMaximum windspeed = 8.1 mph on day 31 at time 13:27\nMaximum gust speed = 13 mph from 045 °( NE) on day 31 at time 13:17\nMaximum heat index = 56.7°F on day 31 at time 16:19\n\nClick here for today´s 24 hour graph :31 :12 :2009\n\n---------------------------------------------------------------------------------------------\nAverages\\Extremes for the month of December 2009\n\n---------------------------------------------------------------------------------------------\nAverage temperature = 50.0°F\nAverage humidity = 34%\nAverage dewpoint = 19.6°F\nAverage barometer = 29.951 in.\nAverage windspeed = 3.2 mph\nAverage gustspeed = 3.9 mph\nAverage direction = 46° ( NE)\nRainfall for month = 1.083 in.\nRainfall for year = 3.933 in.\nMaximum rain per minute = 0.035 in on day 22 at time 14:44\nMaximum temperature = 66.4°F on day 17 at time 15:41\nMinimum temperature = 36.1°F on day 09 at time 00:57\nMaximum humidity = 84% on day 08 at time 00:19\nMinimum humidity = 5% on day 04 at time 15:38\nMaximum pressure = 30.29 in. on day 31 at time 23:49\nMinimum pressure = 29.38 in. on day 07 at time 23:29\nMaximum windspeed = 20.7 mph from 180°( S ) on day 07 at time 23:27\nMaximum gust speed = 38.0 mph from 203°(SSW) on day 07 at time 22:47\nMaximum heat index = 74.1°F on day 16 at time 14:39\nAvg daily max temp :56.8°F\nAvg daily min temp :43.6°F\nTotal windrun = 2149.7miles\n-----------------------------------\nDaily rain totals\n-----------------------------------\n00.83 in. on day 7\n00.03 in. on day 8\n00.06 in. on day 13\n00.16 in. on day 22\n```" ]	[ null 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https://ithems.riken.jp/ja/members/ken-furukawa	[ "2020-10-01 - 特別研究員 (兼務)\n\n## 自己紹介\n\nI am Ken Furukawa. My research area is the mathematical theory of partial differential equations related to the fluid dynamics and diffusion phenomena. More specifically, we studied mathematically rigorous justification of the derivation of the primitive equations by the Navier-Stokes equations. The primitive equations have nice properties on well-posedness and have a strong connection with the Navier-Stokes equations. We obtained some results on the well-posedness of the Navier-Stokes equations in this research.\n\nRecently, I have been interested in the mathematical aspect of data assimilation. Data assimilation is very useful to obtain a plausible forecast and is also closely related to our lives. However, mathematically rigorous studies of data assimilation from the mathematical theory of PDE are under development. I will study data assimilation from the view PDE point of view.\n\n## 関連イベント", null, "### Maximal Regularity and Partial Differential Equations\n\n2020年9月8日16:00 - 18:10 セミナー iTHEMS数学セミナー\n\n## 関連ニュース", null, "### Self-introduction: Ken Furukawa\n\n2020-10-01 今週の注目人物" ]	[ null, "https://ithems.riken.jp/storage/app/uploads/public/606/fe0/8ee/thumb_1813_50_50_0_0_crop.jpg", null, "https://ithems.riken.jp/storage/app/uploads/public/606/fe0/8ee/thumb_1813_50_50_0_0_crop.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6867935,"math_prob":0.88304794,"size":1196,"snap":"2021-31-2021-39","text_gpt3_token_len":414,"char_repetition_ratio":0.13087249,"word_repetition_ratio":0.0,"special_character_ratio":0.20735785,"punctuation_ratio":0.08121827,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9648402,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,8,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T04:30:05Z\",\"WARC-Record-ID\":\"<urn:uuid:73eed02f-956d-4159-8ba8-bc86ec91d7d5>\",\"Content-Length\":\"45328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d9b6d1d-dc6a-4d57-9091-8877f59199b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:61f9f817-54fd-4bd5-8495-f50b75a27a6d>\",\"WARC-IP-Address\":\"160.16.232.176\",\"WARC-Target-URI\":\"https://ithems.riken.jp/ja/members/ken-furukawa\",\"WARC-Payload-Digest\":\"sha1:3XWQIN2ZYIKJ7MPN6G766KGDCPI3KVE6\",\"WARC-Block-Digest\":\"sha1:HC735225VVCT4YSTE44MUJXRT6KIBSXW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780054023.35_warc_CC-MAIN-20210917024943-20210917054943-00641.warc.gz\"}"}
https://cbp.tnw.utwente.nl/PolymeerDictaat/node24.html	[ "", null, "", null, "", null, "", null, "", null, "Next: Appendix A Up: The Gaussian chain Previous: Green's function method\n\n# One Gaussian chain in a box\n\nAs an example of the use of the Green's function method, we calculate the pressure exerted by a Gaussian chain on the walls of a confining box .\n\nThe potential is zero everywhere inside the box, and infinite everywhere outside the box. Then", null, "at the walls of the box. Solving Eq. (3.18) yields", null, "(3.21)", null, "(3.22)\n\nwhere", null, "(3.23)", null, "(3.24)\n\n Z =", null, "(3.25) Zi =", null, "(3.26)\n\nIntroducing the eigenfunctions and eigenvalues from Eqs. (3.23) and (3.24) we get", null, "(3.27)\n\nwhere the prime at the summation sign indicates that only odd n should occur in the sum.\n\nNow look at two limits\n\ni.", null, "The polymer is much smaller than the box\n Zi =", null, "(3.28) Z =", null, "(3.29)\n\nThe pressure on wall one is", null, "(3.30)\n\ni.e. independent of the wall number, and equal to the ideal gas result.\n\nii.", null, "The polymer is very much constrained by the box", null, "(3.31)\n\n P1 =", null, "(3.32) =", null, "(3.33)\n\nIn this case the pressure on the different walls depends on the size of the box orthogonal to the wall.", null, "", null, "", null, "", null, "", null, "Next: Appendix A Up: The Gaussian chain Previous: Green's function method\nW.J. Briels" ]	[ null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/next_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/up_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/previous_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/contents_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/index_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img308.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img309.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img310.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img311.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img312.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img313.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img314.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img315.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img316.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img317.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img318.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img319.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img320.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img321.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img322.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/img323.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/next_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/up_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/previous_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/contents_motif.gif", null, "https://cbp.tnw.utwente.nl/PolymeerDictaat/index_motif.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7613385,"math_prob":0.9921858,"size":1160,"snap":"2021-04-2021-17","text_gpt3_token_len":363,"char_repetition_ratio":0.15484428,"word_repetition_ratio":0.065727696,"special_character_ratio":0.3336207,"punctuation_ratio":0.15662651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964356,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-15T23:41:25Z\",\"WARC-Record-ID\":\"<urn:uuid:fbdb959a-92b1-43d0-8333-18bb2b8f6637>\",\"Content-Length\":\"12168\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdca970c-a83d-4e62-bfa7-0da4385598a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:e87db482-7efe-4b5f-bacc-c7820b1084ee>\",\"WARC-IP-Address\":\"130.89.1.50\",\"WARC-Target-URI\":\"https://cbp.tnw.utwente.nl/PolymeerDictaat/node24.html\",\"WARC-Payload-Digest\":\"sha1:7NJ3F6F4QRPV2B5UAE2MQMDVSKSUMWI6\",\"WARC-Block-Digest\":\"sha1:BSJZRZN7FX6OG64JIEE34VIYUOBYYSAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703497681.4_warc_CC-MAIN-20210115224908-20210116014908-00348.warc.gz\"}"}
https://www.indiabix.com/electronics-and-communication-engineering/exam-question-papers/213005	[ "# Electronics and Communication Engineering - Exam Questions Papers\n\n21.\n\nTwo discrete time systems with impulse responses h1[n] = δ[n - 1] and h2[n] = δ[n - 2] are connected in cascade. The overall impulse response of the cascaded system is\n\n A. δ[n - 1] + δ[n - 2] B. δ[n - 4] C. δ[n - 3] D. δ[n - 1] + δ[n - 2]\n\nExplanation:\n\nh1(n) → delay by 1, h2(n) → delay by 2,\n\nh1(n) * h2(n) → delay by 3.\n\n22.\n\nThe magnitude plot of a composite signal x(t) = e2jt + e3jt is\n\n A. Half wave rectified sinusoid B. Full wave rectified sinusoid C. Exponentially increasing sinusoid D. Exponentially decreasing sinusoid\n\nExplanation:\n\nx(t) = ej2 . 5t (e-j0.5t + e-j0 . 5t)\n\nx(t) = 2ej2.5t cos (0.5)t", null, "The magnitude of x(t) is\n\n\|x(t)\| = 2\|cos (0.5)t\|.\n\n23.\n\nA message signal given by", null, "is amplitude modulated with a carrier of frequency ωc to generate s(t) = [1 + m(t)] cos ωct\n\n A. 8.33% B. 11.11% C. 20% D. 25%\n\nExplanation:", null, ".\n\n24.\n\nA 16 bit counter type ADC uses 1 MHz clock then its maximum conversion time is __________ .\n\n A. 16 μs B. 15 ms C. 6.6 ms D. 66 ms\n\nExplanation:\n\nMaximum conversion time of counter type ADC = (216 - 1) x 1 μs = 65.5 ms » 66 ms.\n\n25.\n\nA first order system will never be able to give a __________ response.\n\n1. Band pass\n2. Band reject\n3. All pass\nChoose the correct option\n\n A. 1, 2 and 3 are true B. 1 and 3 are true, 2 is false C. 1 and 2 are false, 3 is true D. 1 and 2 are true, 3 is false\n\nExplanation:\n\nThe ideal response of a Band pass filter BRF and APF is as shown", null, "So for BPF we require 0 at ω = 0, p i.e. at least 2 zeroes are required in the Transform.\n\nHence order has to be atleast 2.\n\nSimilarly for band reject filter order should be at least 2. Whereas APF has a constant amplitude.\n\nHence Zeroes in the function need not matter.\n\nAPF function is given as", null, "Hence first order can provide APF.\n\n#### Current Affairs 2022\n\nInterview Questions and Answers" ]	[ null, "https://www.indiabix.com/_files/images/electronics-and-communication-engineering/exam-question-papers/10-08-26.png", null, "https://www.indiabix.com/_files/images/electronics-and-communication-engineering/exam-question-papers/6-05-47.png", null, "https://www.indiabix.com/_files/images/electronics-and-communication-engineering/exam-question-papers/6-08-47.png", null, "https://www.indiabix.com/_files/images/electronics-and-communication-engineering/exam-question-papers/10-07-19.png", null, "https://www.indiabix.com/_files/images/electronics-and-communication-engineering/exam-question-papers/10-07-19-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80009264,"math_prob":0.99113077,"size":2285,"snap":"2022-40-2023-06","text_gpt3_token_len":662,"char_repetition_ratio":0.35773784,"word_repetition_ratio":0.18061674,"special_character_ratio":0.32341358,"punctuation_ratio":0.094382025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980972,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T15:10:54Z\",\"WARC-Record-ID\":\"<urn:uuid:773a024c-0bd6-46c4-9467-4dcff64c44b1>\",\"Content-Length\":\"39718\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e6d9b47-aa8b-4b76-b91d-1ce1d5bd47bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:9fefc9f0-9234-4c77-830b-19c50195d56a>\",\"WARC-IP-Address\":\"35.244.11.196\",\"WARC-Target-URI\":\"https://www.indiabix.com/electronics-and-communication-engineering/exam-question-papers/213005\",\"WARC-Payload-Digest\":\"sha1:JZE4BZWYNTQCX4H76RGJO557YP2ZECAR\",\"WARC-Block-Digest\":\"sha1:5XVEHVE2BGXPU7FOYPUJ2QTI3OV7CNWS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337631.84_warc_CC-MAIN-20221005140739-20221005170739-00790.warc.gz\"}"}
https://rdrr.io/cran/OneArmPhaseTwoStudy/man/get_p_exact_subset.html	[ "# get_p_exact_subset: Calculates the exact p value. In OneArmPhaseTwoStudy: Planning, Conducting, and Analysing Single-Arm Phase II Studies\n\n## Description\n\nCalculates the exact p value for a given subset design.\n\n## Usage\n\n `1` ```get_p_exact_subset(t, u, r1, n1, n, pc0, pt0, sub1 = setupSub1Design()) ```\n\n## Arguments\n\n `t` observed responses in the subset endpoint. `u` observed responses in the superset endpoint. `r1` critical value for the first stage. `n1` sample size for the first stage. `n` overall sample size. `pc0` the response probability for the subset endpoint under the null hypothesis. `pt0` the response probability for the superset endpoint under the null hypothesis. `sub1` \"sub1\"-object used to calculate the p value in c++ .\n\n`setupSub1Design`\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13``` ```#Setup \"sub1\"-object sub1 <- setupSub1Design(pc0 = 0.5, pt0 = 0.6) #Calculate a subset design design <- getSolutionsSub1(sub1, skipN1 = FALSE)\\$Solutions[4,] #Assuming 9 responses in the subset endpoint and 13 responses #in the superset endpoint were observed. t = 9 u = 13 p_val <- get_p_exact_subset(t, u, design\\$r1, design\\$n1, design\\$n, design\\$pc0, design\\$pt0, sub1) p_val ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7074942,"math_prob":0.96634877,"size":1139,"snap":"2023-14-2023-23","text_gpt3_token_len":343,"char_repetition_ratio":0.15330397,"word_repetition_ratio":0.06315789,"special_character_ratio":0.30289727,"punctuation_ratio":0.14883721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T08:18:21Z\",\"WARC-Record-ID\":\"<urn:uuid:c9904b63-b33c-4456-b1f3-3e12573b25ee>\",\"Content-Length\":\"33164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25770859-7ee6-4291-ae69-14989767096e>\",\"WARC-Concurrent-To\":\"<urn:uuid:33554e81-e5f3-4d40-9392-a3ae406b41ff>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/OneArmPhaseTwoStudy/man/get_p_exact_subset.html\",\"WARC-Payload-Digest\":\"sha1:45CD2RLWVNBGDV3PYJON4VRY4CHNFOL7\",\"WARC-Block-Digest\":\"sha1:UGMJYNWI4XU6C55N3H5454VOASQ7SGYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649177.24_warc_CC-MAIN-20230603064842-20230603094842-00263.warc.gz\"}"}
https://percentage-calculator.net/what-percent-of-x-is-y/what-percent-of-220-is-66.php	[ "# What percent of 220 is 66?\n\nAnswer: 30 percent of 220 is 66\n\n## Fastest method for calculating what percent of 220 is 66\n\nAssume the unknown value is 'Y', and 66 of 220 can be written as:\n\nY = 66 / 220\n\nBy multiplying both numerator and denominator by 100 we will get:\n\nY = 66 / 220 x 100 / 100 = 30 / 100\n\nY = 30%\n\nAnswer: 30 percent of 220 is 66\n\nIf you want to use a calculator, simply enter 66÷220x100 and you will get your answer which is 30\n\nYou may also be interested in:\n\nHere is a calculator to solve percentage calculations such as what percent of 220 is 66. You can solve this type of calculation with your own values by entering them into the calculator's fields, and click 'Calculate' to get the result and explanation.\n\nWhat percent of\nis\n?\n%\n\n## Have time and want to learn the details?\n\nLet's solve the equation for Y by first rewriting it as: 100% / 220 = Y% / 66\n\nDrop the percentage marks to simplify your calculations: 100 / 220 = Y / 66\n\nMultiply both sides by 66 to isolate Y on the right side of the equation: 66 ( 100 / 220 ) = Y\n\nComputing the left side, we get: 30 = Y\n\nThis leaves us with our final answer: 30 percent of 220 is 66" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91749704,"math_prob":0.99719936,"size":977,"snap":"2020-34-2020-40","text_gpt3_token_len":278,"char_repetition_ratio":0.14491265,"word_repetition_ratio":0.04040404,"special_character_ratio":0.34493348,"punctuation_ratio":0.078431375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99942267,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T22:28:20Z\",\"WARC-Record-ID\":\"<urn:uuid:f2a84990-94ba-49a2-985a-6e2ebb59a222>\",\"Content-Length\":\"59199\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49c71799-2ebb-4135-96f9-b06570001ef2>\",\"WARC-Concurrent-To\":\"<urn:uuid:955f4671-e476-4411-a4d3-1936a95717a8>\",\"WARC-IP-Address\":\"68.66.224.6\",\"WARC-Target-URI\":\"https://percentage-calculator.net/what-percent-of-x-is-y/what-percent-of-220-is-66.php\",\"WARC-Payload-Digest\":\"sha1:7WZXB33QGZBM76WHPLOBAGECNV7XU6FL\",\"WARC-Block-Digest\":\"sha1:DZZBOAVLSLOREEK5TI2CH66YOIFP7HLB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402132335.99_warc_CC-MAIN-20201001210429-20201002000429-00103.warc.gz\"}"}
https://apps.dtic.mil/sti/citations/ADA289424	[ "# Abstract:\n\nThis study uses linearized equations of motion for a rigid body with an attached spring-mass-damper to maximize the decay rate of a satellites coning motion. An analysis of the numerical eigenvalues is presented which leads to an optimal relationship between relevant parameters damper placement, spring constant, damping coefficient, system moments of inertia, and damper mass fraction. The coupled systems eigenvalues do not provide truly critical damping, thus the real eigenvalue parts are minimized in order to achieve damping which requires the minimum amount of time. A comparison between this optimal design method and a classical method concludes a noticeable improvement in damping performance for optimized systems.\n\n# Subject Categories:\n\n• Unmanned Spacecraft\n• Operations Research" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8090806,"math_prob":0.855893,"size":1121,"snap":"2021-21-2021-25","text_gpt3_token_len":224,"char_repetition_ratio":0.09489705,"word_repetition_ratio":0.0,"special_character_ratio":0.1837645,"punctuation_ratio":0.11891892,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9667872,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T22:48:57Z\",\"WARC-Record-ID\":\"<urn:uuid:2715e87a-6bb1-48d2-ba81-1451ff965841>\",\"Content-Length\":\"87675\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d8bb55e-99da-4275-b32b-9278bc055b78>\",\"WARC-Concurrent-To\":\"<urn:uuid:55c045c9-642a-4a7d-92e8-e781e2e006ef>\",\"WARC-IP-Address\":\"131.84.180.30\",\"WARC-Target-URI\":\"https://apps.dtic.mil/sti/citations/ADA289424\",\"WARC-Payload-Digest\":\"sha1:55VDYZTZB2TX64XFHK5YDHGPZDL5ZRA5\",\"WARC-Block-Digest\":\"sha1:YZV7EUCQJH7B4RXIUXUS42IWJZGTL6QH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488504838.98_warc_CC-MAIN-20210621212241-20210622002241-00320.warc.gz\"}"}
https://proofwiki.org/wiki/Definition:Plane_Angle	[ "# Definition:Angle\n\n(Redirected from Definition:Plane Angle)\n\n## Definition\n\nGiven two intersecting lines or line segments, the amount of rotation about the intersection required to bring one into correspondence with the other is called the angle between them.\n\nIn the words of Euclid:\n\nA plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.\n\n### Rectilineal\n\nIn the words of Euclid:\n\nAnd when the lines containing the angle are straight, the angle is called rectilineal.\n\nThus the distinction is made between straight-line angles and curved-line angles.\n\nMost of the time the fact that angles are rectilineal is taken for granted.\n\n### Subtend\n\nLet $AB$ be a line segment and $C$ be a point:", null, "The line segment $AB$ is said to subtend the angle $\\angle ACB$.\n\nTwo angles are adjacent if they have an intersecting line in common:", null, "### Containment\n\nThe two lines whose intersection forms an angle are said to contain that angle.\n\n### Vertex\n\nThe point at which the lines containing an angle meet is known as the vertex of that angle.\n\n## Measurement\n\nThe usual units of measurement for angle are as follows:\n\n### Degree\n\nThe degree (of arc) is a measurement of plane angles, symbolized by $\\degrees$.\n\n $\\displaystyle$ $\\displaystyle 1$ degree $\\displaystyle$ $=$ $\\displaystyle 60$ minutes $\\displaystyle$ $=$ $\\displaystyle 60 \\times 60 = 3600$ seconds $\\displaystyle$ $=$ $\\displaystyle \\dfrac 1 {360}$ full angle (by definition)\n\n### Minute\n\nThe minute (of arc) is a measurement of plane angles, symbolized by $'$.\n\n$1$ minute $= 60$ seconds (of arc) $= \\dfrac 1 {60}$ of a degree of arc.\n\n### Second\n\nThe second (of arc) is a measurement of plane angles, symbolized by $''$.\n\n$1''$ is defined as $\\dfrac 1 {60}$ of a minute of arc, or $\\dfrac 1 {3600}$ of a degree of arc.\n\nThe radian is a measure of plane angles symbolized either by the word $\\radians$ or without any unit.\n\nRadians are pure numbers, as they are ratios of lengths. The addition of $\\radians$ is merely for clarification.\n\n$1 \\radians$ is the angle subtended at the center of a circle by an arc whose length is equal to the radius:", null, "## Types of Angle\n\nAngles can be divided into categories:\n\n### Zero Angle\n\nThe zero angle is an angle the measure of which is $0$ regardless of the unit of measurement.\n\n### Acute Angle\n\nAn acute angle is an angle which has a measure between that of a right angle and that of a zero angle.\n\n### Right Angle\n\nA right angle is an angle that is equal to half of a straight angle.\n\n### Obtuse Angle\n\nAn obtuse angle is an angle which has a measurement between those of a right angle and a straight angle.\n\n### Straight Angle\n\nA straight angle is defined to be the angle formed by the two parts of a straight line from a point on that line.\n\n### Reflex Angle\n\nA reflex angle is an angle which has a measure between that of a straight angle and that of a full angle.\n\n### Full Angle\n\nA full angle is an angle equivalent to one full rotation.\n\nIt is possible to consider angles outside the $\\left[{0^\\circ \\,.\\,.\\, 360^\\circ}\\right]$ or $\\left[{0 \\,.\\,.\\, 2 \\pi}\\right]$ range.\n\nHowever, in geometric contexts it is usually preferable to convert these to angles inside this range by adding or subtracting multiples of a full angle.\n\n## Directed versus Undirected Angles\n\nThe most basic definition of angle is an undirected angle on the interval $\\left[{0^\\circ \\,.\\,.\\, 180^\\circ}\\right]$ or $\\left[{0 \\,.\\,.\\, \\pi}\\right]$.\n\nThis definition is often insufficient, in cases such as the external angles of a polygon.\n\nTherefore, angles are most commonly defined in one of two ways:\n\n$(1): \\quad$ Undirected angles on the interval $\\left[{0^\\circ \\,.\\,.\\, 360^\\circ}\\right]$ or $\\left[{0 \\,.\\,.\\, 2 \\pi}\\right]$.\n$(2): \\quad$ Directed angles, with the positive direction being counterclockwise from a given line (or, if no line is specified, from the $x$-axis).\nThis definition is more commonly found in applied mathematics, such as in surveying, navigation, or, more colloquially, in a $720^\\circ$ degree spin in skateboarding, skiing, etc.\n\n## Also see\n\n• Results about angles can be found here." ]	[ null, "https://proofwiki.org/w/images/thumb/3/30/Subtend.png/250px-Subtend.png", null, "https://proofwiki.org/w/images/thumb/7/70/AdjacentAngles.png/250px-AdjacentAngles.png", null, "https://proofwiki.org/w/images/thumb/8/87/Radian.png/300px-Radian.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8871571,"math_prob":0.9972763,"size":4278,"snap":"2019-26-2019-30","text_gpt3_token_len":1104,"char_repetition_ratio":0.14599906,"word_repetition_ratio":0.082058415,"special_character_ratio":0.27045348,"punctuation_ratio":0.11793612,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99982446,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,7,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T21:50:36Z\",\"WARC-Record-ID\":\"<urn:uuid:d85ea2e8-3607-43b6-a07f-30f4d77fc9f8>\",\"Content-Length\":\"50284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f65c07af-35fc-44a6-8283-1014d9e0e02e>\",\"WARC-Concurrent-To\":\"<urn:uuid:400ab829-d006-4c21-a015-11745d8fd9e9>\",\"WARC-IP-Address\":\"104.27.168.113\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Definition:Plane_Angle\",\"WARC-Payload-Digest\":\"sha1:B6T4Z5CFHEEFKUX2ECL2LERHSGYHYMTF\",\"WARC-Block-Digest\":\"sha1:4O6NY2KKIXIB45LCAXVUV2M4IRVLO4JF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525829.33_warc_CC-MAIN-20190718211312-20190718233312-00243.warc.gz\"}"}
https://www.circuitbread.com/textbooks/introduction-to-electricity-magnetism-and-circuits/magnetic-forces-and-fields/chapter-8-review	[ "", null, "# Chapter 8 Review\n\n## Key Terms\n\ncosmic rays\ncomprised of particles that originate mainly from outside the solar system and reach Earth\n\ncyclotron\ndevice used to accelerate charged particles to large kinetic energies\n\ndees\nlarge metal containers used in cyclotrons that serve contain a stream of charged particles as their speed is increased\n\ngauss\n\nunit of the magnetic field strength;\n\nHall effect\ncreation of voltage across a current-carrying conductor by a magnetic field\n\nhelical motion\nsuperposition of circular motion with a straight-line motion that is followed by a charged particle moving in a region of magnetic field at an angle to the field\n\nmagnetic dipole\nclosed-current loop\n\nmagnetic dipole moment\nterm\n\nof the magnetic dipole, also called\n\nmagnetic field lines\ncontinuous curves that show the direction of a magnetic field; these lines point in the same direction as a compass points, toward the magnetic south pole of a bar magnet\n\nmagnetic force\nforce applied to a charged particle moving through a magnetic field\n\nmass spectrometer\ndevice that separates ions according to their charge-to-mass ratios\n\nmotor (dc)\nloop of wire in a magnetic field; when current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft; electrical energy is converted into mechanical work in the process\n\nnorth magnetic pole\ncurrently where a compass points to north, near the geographic North Pole; this is the effective south pole of a bar magnet but has flipped between the effective north and south poles of a bar magnet multiple times over the age of Earth\n\nright-hand rule-1\nusing your right hand to determine the direction of either the magnetic force, velocity of a charged particle, or magnetic field\n\nsouth magnetic pole\ncurrently where a compass points to the south, near the geographic South Pole; this is the effective north pole of a bar magnet but has flipped just like the north magnetic pole\n\ntesla\nSI unit for magnetic field:\n\nvelocity selector\napparatus where the crossed electric and magnetic fields produce equal and opposite forces on a charged particle moving with a specific velocity; this particle moves through the velocity selector not affected by either field while particles moving with different velocities are deflected by the apparatus\n\n## Key Equations\n\n Force on a charge in a magnetic field", null, "Magnitude of magnetic force", null, "Radius of a particle’s path in a magnetic field", null, "Period of a particle’s motion in a magnetic field", null, "Force on a current-carrying wire in a uniform magnetic field", null, "Magnetic dipole moment", null, "Torque on a current loop", null, "Energy of a magnetic dipole", null, "Drift velocity in crossed electric and magnetic fields", null, "Hall potential", null, "Hall potential in terms of drift velocity", null, "Charge-to-mass ratio in a mass spectrometer", null, "Maximum speed of a particle in a cyclotron", null, "## Summary\n\n#### 8.1 Magnetism and Its Historical Discoveries\n\n• Magnets have two types of magnetic poles, called the north magnetic pole and the south magnetic pole. North magnetic poles are those that are attracted toward Earth’s geographic North Pole.\n• Like poles repel and unlike poles attract.\n• Discoveries of how magnets respond to currents by Oersted and others created a framework that led to the invention of modern electronic devices, electric motors, and magnetic imaging technology.\n\n#### 8.2 Magnetic Fields and Lines\n\n• Charges moving across a magnetic field experience a force determined by", null, "The force is perpendicular to the plane formed by", null, "and", null, "• The direction of the force on a moving charge is given by the right hand rule 1 (RHR-1): Sweep your fingers in a velocity, magnetic field plane. Start by pointing them in the direction of velocity and sweep towards the magnetic field. Your thumb points in the direction of the magnetic force for positive charges.\n• Magnetic fields can be pictorially represented by magnetic field lines, which have the following properties:\n1. The field is tangent to the magnetic field line.\n2. Field strength is proportional to the line density.\n3. Field lines cannot cross.\n4. Field lines form continuous, closed loops.\n• Magnetic poles always occur in pairs of north and south—it is not possible to isolate north and south poles.\n\n#### 8.3 Motion of a Charged Particle in a Magnetic Field\n\n• A magnetic force can supply centripetal force and cause a charged particle to move in a circular path of radius", null, "• The period of circular motion for a charged particle moving in a magnetic field perpendicular to the plane of motion is", null, "• Helical motion results if the velocity of the charged particle has a component parallel to the magnetic field as well as a component perpendicular to the magnetic field.\n\n#### 8.4 Magnetic Force on a Current-Carrying Conductor\n\n• An electrical current produces a magnetic field around the wire.\n• The directionality of the magnetic field produced is determined by the right hand rule-2, where your thumb points in the direction of the current and your fingers wrap around the wire in the direction of the magnetic field.\n• The magnetic force on current-carrying conductors is given by", null, "where", null, "is the current and", null, "is the length of a wire in a uniform magnetic field", null, "#### 8.5 Force and Torque on a Current Loop\n\n• The net force on a current-carrying loop of any plane shape in a uniform magnetic field is zero.\n• The net torque", null, "on a current-carrying loop of any shape in a uniform magnetic field is calculated using", null, "where", null, "is the magnetic dipole moment and", null, "is the magnetic field strength.\n• The magnetic dipole moment", null, "is the product of the number of turns of wire", null, "the current in the loop", null, "and the area of the loop", null, "or", null, "#### 8.6 The Hall Effect\n\n• Perpendicular electric and magnetic fields exert equal and opposite forces for a specific velocity of entering particles, thereby acting as a velocity selector. The velocity that passes through undeflected is calculated by\n• The Hall effect can be used to measure the sign of the majority of charge carriers for metals. It can also be used to measure a magnetic field.\n\n#### 8.7 Applications of Magnetic Forces and Fields\n\n• A mass spectrometer is a device that separates ions according to their charge-to-mass ratios by first sending them through a velocity selector, then a uniform magnetic field.\n• Cyclotrons are used to accelerate charged particles to large kinetic energies through applied electric and magnetic fields.\n\n8.1 a.\n\nb.\n\nc.\n\nd.\n\n8.2 a.\n\ntoward the south; b.\n\n8.3 a. bends upward; b. bends downward\n\n8.4 a. perpendicular ; b. anti-aligned\n\n8.5 a.\n\nb.\n\n8.6\n\n## Conceptual Questions\n\n#### 8.2 Magnetic Fields and Lines\n\n1. Discuss the similarities and differences between the electrical force on a charge and the magnetic force on a charge.\n\n2. (a) Is it possible for the magnetic force on a charge moving in a magnetic field to be zero? (b) Is it possible for the electric force on a charge moving in an electric field to be zero? (c) Is it possible for the resultant of the electric and magnetic forces on a charge moving simultaneously through both fields to be zero?\n\n#### 8.3 Motion of a Charged Particle in a Magnetic Field\n\n3. At a given instant, an electron and a proton are moving with the same velocity in a constant magnetic field. Compare the magnetic forces on these particles. Compare their accelerations.\n\n4. Does increasing the magnitude of a uniform magnetic field through which a charge is traveling necessarily mean increasing the magnetic force on the charge? Does changing the direction of the field necessarily mean a change in the force on the charge?\n\n5. An electron passes through a magnetic field without being deflected. What do you conclude about the magnetic field?\n\n6. If a charged particle moves in a straight line, can you conclude that there is no magnetic field present?\n\n7. How could you determine which pole of an electromagnet is north and which pole is south?\n\n#### 8.4 Magnetic Force on a Current-Carrying Conductor\n\n8. Describe the error that results from accidently using your left rather than your right hand when determining the direction of a magnetic force.\n\n9. Considering the magnetic force law, are the velocity and magnetic field always perpendicular? Are the force and velocity always perpendicular? What about the force and magnetic field?\n\n10. Why can a nearby magnet distort a cathode ray tube television picture?\n\n11. A magnetic field exerts a force on the moving electrons in a current carrying wire. What exerts the force on a wire?\n\n12. There are regions where the magnetic field of earth is almost perpendicular to the surface of Earth. What difficulty does this cause in the use of a compass?\n\n#### 8.6 The Hall Effect\n\n13. Hall potentials are much larger for poor conductors than for good conductors. Why?\n\n#### 8.7 Applications of Magnetic Forces and Fields\n\n14. Describe the primary function of the electric field and the magnetic field in a cyclotron.\n\n## Problems\n\n#### 8.2 Magnetic Fields and Lines\n\n15. What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases?\n\n16. Repeat previous exercise for a negative charge.\n\n17. What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases, assuming it moves perpendicular to\n\n18. Repeat previous exercise for a positive charge.\n\n19. What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases, assuming\n\nis perpendicular to\n\n?\n\n20. Repeat previous exercise for a negative charge.\n\n21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a\n\ncharge and flies due west at a speed of\n\nover Earth’s south magnetic pole, where the\n\nmagnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.\n\n22. (a) A cosmic ray proton moving toward Earth at\n\nexperiences a magnetic force of\n\nWhat is the strength of the magnetic field if there is a\n\nangle between it and the proton’s velocity? (b) Is the value obtained in part a. consistent with the known strength of Earth’s magnetic field on its surface? Discuss.\n\n23. An electron moving at\n\nin a\n\nmagnetic field experiences a magnetic force of\n\nWhat angle does the velocity of the electron make with the magnetic field? There are two answers.\n\n24. (a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than\n\nWhat is the greatest the charge can be if it moves at a maximum speed of\n\nin Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.\n\n#### 8.3 Motion of a Charged Particle in a Magnetic Field\n\n25. A cosmic-ray electron moves at\n\nperpendicular to Earth’s magnetic field at an altitude where the field strength is\n\nWhat is the radius of the circular path the electron follows?\n\n26. (a) Viewers of Star Trek have heard of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at\n\nin a circular path\n\nin radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility?\n\n27. (a) An oxygen-\n\nion with a mass of\n\ntravels at\n\nperpendicular to a\n\nmagnetic field, which makes it move in a circular arc with a\n\nradius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.\n\n28. An electron in a TV CRT moves with a speed of\n\nin a direction perpendicular to Earth’s field, which has a strength of\n\n(a) What strength electric field must be applied perpendicular to the Earth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by\n\nwhat is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)\n\n29. (a) At what speed will a proton move in a circular path of the same radius as the electron in the previous exercise? (b) What would the radius of the path be if the proton had the same speed as the electron? (c) What would the radius be if the proton had the same kinetic energy as the electron? (d) The same momentum?\n\n30. (a) What voltage will accelerate electrons to a speed of\n\n? (b) Find the radius of curvature of the path of a proton accelerated through this potential in a\n\nfield and compare this with the radius of curvature of an electron accelerated through the same potential.\n\n31. An alpha-particle\n\ntravels in a circular path of radius\n\nin a uniform magnetic field of magnitude\n\n(a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?\n\n32. A particle of charge\n\nand mass\n\nis accelerated from rest through a potential difference\n\nafter which it encounters a uniform magnetic field\n\nIf the particle moves in a plane perpendicular to\n\nwhat is the radius of its circular orbit?\n\n#### 8.4 Magnetic Force on a Current-Carrying Conductor\n\n33. What is the direction of the magnetic force on the current in each of the six cases?\n\n34. What is the direction of a current that experiences the magnetic force shown in each of the three cases, assuming the current runs perpendicular to\n\n?\n\n35. What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases, assuming\n\nis perpendicular to\n\n?\n\n36. (a) What is the force per meter on a lightning bolt at the equator that carries\n\nperpendicular to Earth’s\n\n? (b) What is the direction of the force if the current is straight up and Earth’s field direction is due north, parallel to the ground?\n\n37. (a) A dc power line for a light-rail system carries\n\nat an angle of\n\nto Earth’s\n\nfield. What is the force on a\n\nsection of this line? (b) Discuss practical concerns this presents, if any.\n\n38. A wire carrying a\n\ncurrent passes between the poles of a strong magnet that is perpendicular to its field and experiences a\n\nforce on the\n\nof wire in the field. What is the average field strength?\n\n#### 8.5 Force and Torque on a Current Loop\n\n39. (a) By how many percent is the torque of a motor decreased if its permanent magnets lose\n\nof their strength? (b) How many percent would the current need to be increased to return the torque to original values?\n\n40. (a) What is the maximum torque on a\n\n-turn square loop of wire\n\non a side that carries a\n\ncurrent in a\n\nfield? (b) What is the torque when\n\nis\n\n?\n\n41. Find the current through a loop needed to create a maximum torque of\n\nThe loop has\n\nsquare turns that are\n\non a side and is in a uniform\n\nmagnetic field.\n\n42. Calculate the magnetic field strength needed on a\n\n-turn square loop\n\non a side to create a maximum torque of\n\nif the loop is carrying\n\n43. Since the equation for torque on a current-carrying loop is τ = NIAB sin\n\nthe units of\n\nmust equal units of\n\nVerify this.\n\n44. (a) At what angle\n\nis the torque on a current loop\n\nof maximum? (b)\n\nof maximum? (c)\n\nof maximum?\n\n45. A proton has a magnetic field due to its spin. The field is similar to that created by a circular current loop\n\nin radius with a current of\n\nFind the maximum torque on a proton in a\n\nfield. (This is a significant torque on a small particle.)\n\n46. (a) A\n\nis vertical, with its axis on an east-west line. A current of\n\ncirculates clockwise in the loop when viewed from the east. Earth’s field here is due north, parallel to the ground, with a strength of\n\nWhat are the direction and magnitude of the torque on the loop? (b) Does this device have any practical applications as a motor?\n\n47. Repeat the previous problem, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where Earth’s field is north, but at an angle\n\nbelow the horizontal and with a strength of\n\n#### 8.6 The Hall Effect\n\n48. A strip of copper is placed in a uniform magnetic field of magnitude\n\nThe Hall electric field is measured to be\n\n(a) What is the drift speed of the conduction electrons? (b) Assuming that\n\nelectrons per cubic meter and that the cross-sectional area of the strip is\n\ncalculate the current in the strip. (c) What is the Hall coefficient\n\n?\n\n49. The cross-sectional dimensions of the copper strip shown are\n\nby\n\nThe strip carries a current of\n\nand it is placed in a magnetic field of magnitude\n\nWhat are the value and polarity of the Hall potential in the copper strip?\n\n50. The magnitudes of the electric and magnetic fields in a velocity selector are\n\nand\n\nrespectively. (a) What speed must a proton have to pass through the selector? (b) Also calculate the speeds required for an alpha-particle and a singly ionized\n\natom to pass through the selector.\n\n51. A charged particle moves through a velocity selector at constant velocity. In the selector,\n\nand\n\nWhen the electric field is turned off, the charged particle travels in a circular path of radius\n\nDetermine the charge-to-mass ratio of the particle.\n\n52. A Hall probe gives a reading of\n\nfor a current of\n\nwhen it is placed in a magnetic field of\n\nWhat is the magnetic field in a region where the reading is\n\nfor\n\nof current?\n\n#### 8.7 Applications of Magnetic Forces and Fields\n\n53. A physicist is designing a cyclotron to accelerate protons to one-tenth the speed of light. The magnetic field will have a strength of\n\nDetermine (a) the rotational period of the circulating protons and (b) the maximum radius of the protons’ orbit.\n\n54. The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are\n\nand\n\nand the strength of the magnetic field that separates the ions is\n\nA stream of singly charged\n\nions is found to bend in a circular arc of radius\n\nWhat is the mass of the\n\nions?\n\n55. The magnetic field in a cyclotron is\n\nand the maximum orbital radius of the circulating protons is\n\n(a) What is the kinetic energy of the protons when they are ejected from the cyclotron? (b) What is this energy in\n\n? (c) Through what potential difference would a proton have to be accelerated to acquire this kinetic energy? (d) What is the period of the voltage source used to accelerate the protons? (e) Repeat the calculations for alpha-particles.\n\n56. A mass spectrometer is being used to separate common oxygen-\n\nfrom the much rarer oxygen-\n\ntaken from a sample of old glacial ice. (The relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) The ratio of the masses of these two ions is\n\nto\n\nthe mass of oxygen-\n\nis\n\nand they are singly charged and travel at\n\nin a\n\nmagnetic field. What is the separation between their paths when they hit a target after traversing a semicircle?\n\n57. (a) Triply charged uranium-\n\nand uranium-\n\nions are being separated in a mass spectrometer. (The much rarer uranium-\n\nis used as reactor fuel.) The masses of the ions are\n\nand\n\nrespectively, and they travel at\n\nin a\n\nfield. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-\n\nfrom uranium-\n\n58. Calculate the magnetic force on a hypothetical particle of charge\n\nmoving with a velocity of\n\nin a magnetic field of\n\n59. Repeat the previous problem with a new magnetic field of\n\n60. An electron is projected into a uniform magnetic field\n\nwith a velocity of\n\nWhat is the magnetic force on the electron?\n\n61. The mass and charge of a water droplet are\n\nand\n\nrespectively. If the droplet is given an initial horizontal velocity of\n\nwhat magnetic field will keep it moving in this direction? Why must gravity be considered here?\n\n62. Four different proton velocities are given. For each case, determine the magnetic force on the proton in terms of\n\nand\n\n63. An electron of kinetic energy\n\npasses between parallel plates that are\n\napart and kept at a potential difference of\n\nWhat is the strength of the uniform magnetic field\n\nthat will allow the electron to travel undeflected through the plates? Assume\n\nand\n\nare perpendicular.\n\n64. An alpha-particle\n\nmoving with a velocity\n\nenters a region where\n\nand\n\nWhat is the initial force on it?\n\n65. An electron moving with a velocity\n\nenters a region where there is a uniform electric field and a uniform magnetic field. The magnetic field is given by\n\nIf the electron travels through a region without being deflected, what is the electric field?\n\n66. At a particular instant, an electron is traveling west to east with a kinetic energy of\n\nEarth’s magnetic field has a horizontal component of\n\nnorth and a vertical component of\n\ndown. (a) What is the path of the electron? (b) What is the radius of curvature of the path?\n\n67. Repeat the calculations of the previous problem for a proton with the same kinetic energy.\n\n68. What magnetic field is required in order to confine a proton moving with a speed of\n\nto a circular orbit of radius\n\n?\n\n69. An electron and a proton move with the same speed in a plane perpendicular to a uniform magnetic field. Compare the radii and periods of their orbits.\n\n70. A proton and an alpha-particle have the same kinetic energy and both move in a plane perpendicular to a uniform magnetic field. Compare the periods of their orbits.\n\n71. A singly charged ion takes\n\nto complete eight revolutions in a uniform magnetic field of magnitude\n\nWhat is the mass of the ion?\n\n72. A particle moving downward at a speed of\n\nenters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If\n\nand the charge-to-mass ratio\n\nof the particle is\n\nwhat is the radius of the path? (c) What is the speed of the particle after it has moved in the field for\n\n? for\n\n?\n\n73. A proton, deuteron, and an alpha-particle are all accelerated through the same potential difference. They then enter the same magnetic field, moving perpendicular to it. Compute the ratios of the radii of their circular paths. Assume that\n\nand\n\n74. A singly charged ion is moving in a uniform magnetic field of\n\ncompletes\n\nrevolutions in\n\nIdentify the ion.\n\n75. Two particles have the same linear momentum, but particle\n\nhas four times the charge of particle\n\nIf both particles move in a plane perpendicular to a uniform magnetic field, what is the ratio\n\nof the radii of their circular orbits?\n\n76. A uniform magnetic field of magnitude\n\nis directed parallel to the\n\n-axis. A proton enters the field with a velocity\n\nand travels in a helical path with a radius of\n\n(a) What is the value of\n\n? (b) What is the time required for one trip around the helix? (c) Where is the proton\n\nafter entering the field?\n\n77. An electron moving at\n\nenters a magnetic field that makes a\n\nangle with the\n\n-axis of magnitude\n\nCalculate the (a) pitch and (b) radius of the trajectory.\n\n78. (a) A\n\n-long section of cable carrying current to a car starter motor makes an angle of\n\nwith Earth’s\n\nfield. What is the current when the wire experiences a force of\n\n? (b) If you run the wire between the poles of a strong horseshoe magnet, subjecting\n\nof it to a\n\nwhat force is exerted on this segment of wire?\n\n79. (a) What is the angle between a wire carrying an\n\ncurrent and the\n\nfield it is in if\n\nof the wire experiences a magnetic force of\n\n? (b) What is the force on the wire if it is rotated to make an angle of\n\nwith the field?\n\n80. A\n\n-long segment of wire lies along the\n\n-axis and carries a current of\n\nin the positive\n\n-direction. Around the wire is the magnetic field of\n\nFind the magnetic force on this segment.\n\n81. A\n\nsection of a long, straight wire carries a current of\n\nwhile in a uniform magnetic field of magnitude\n\nCalculate the magnitude of the force on the section if the angle between the field and the direction of the current is (a)\n\n(b)\n\n(c)\n\nor (d)\n\n82. An electromagnet produces a magnetic field of magnitude\n\nthroughout a cylindrical region of radius\n\nA straight wire carrying a current of\n\npasses through the field as shown in the accompanying figure. What is the magnetic force on the wire?\n\n83. The current loop shown in the accompanying figure lies in the plane of the page, as does the magnetic field. Determine the net force and the net torque on the loop if\n\nand\n\n84. A circular coil of radius\n\nis wound with five turns and carries a current of\n\nIf the coil is placed in a uniform magnetic field of strength\n\nwhat is the maximum torque on it?\n\n85. A circular coil of wire of radius\n\nhas\n\nturns and carries a current of\n\nThe coil lies in a magnetic field of magnitude\n\nthat is directed parallel to the plane of the coil. (a) What is the magnetic dipole moment of the coil? (b) What is the torque on the coil?\n\n86. A current-carrying coil in a magnetic field experiences a torque that is\n\nof the maximum possible torque. What is the angle between the magnetic field and the normal to the plane of the coil?\n\n87. A\n\nby\n\nrectangular current loop carries a current of\n\nWhat is the magnetic dipole moment of the loop?\n\n88. A circular coil with\n\n(a) What current through the coil results in a magnetic dipole moment of\n\n? (b) What is the maximum torque that the coil will experience in a uniform field of strength\n\n? (c) If the angle between\n\nand\n\nis\n\nwhat is the magnitude of the torque on the coil? (d) What is the magnetic potential energy of coil for this orientation?\n\n89. The current through a circular wire loop of radius\n\nis\n\n(a) Calculate the magnetic dipole moment of the loop. (b) What is the torque on the loop if it is in a uniform\n\nmagnetic field such that\n\nand\n\nare directed at\n\nto each other? (c) For this position, what is the potential energy of the dipole?\n\n90. A wire of length\n\nis wound into a single-turn planar loop. The loop carries a current of\n\nand it is placed in a uniform magnetic field of strength\n\n(a) What is the maximum torque that the loop will experience if it is square? (b) If it is circular? (c) At what angle relative to\n\nwould the normal to the circular coil have to be oriented so that the torque on it would be the same as the maximum torque on the square coil?\n\n91. Consider an electron rotating in a circular orbit of radius\n\nShow that the magnitudes of the magnetic dipole moment\n\nand the angular momentum\n\nof the electron are related by:\n\n92. The Hall effect is to be used to find the sign of charge carriers in a semiconductor sample. The probe is placed between the poles of a magnet so that magnetic field is pointed up. A current is passed through a rectangular sample placed horizontally. As current is passed through the sample in the east direction, the north side of the sample is found to be at a higher potential than the south side. Decide if the number density of charge carriers is positively or negatively charged.\n\n93. The density of charge carriers for copper is\n\nelectrons per cubic meter. What will be the Hall voltage reading from a probe made up of\n\ncopper plate when a current of\n\nis passed through it in a magnetic field of\n\nperpendicular to the\n\n94. The Hall effect is to be used to find the density of charge carriers in an unknown material. A Hall voltage\n\nfor\n\ncurrent is observed in a\n\nmagnetic field for a rectangular sample with length\n\nwidth\n\nand height\n\nDetermine the density of the charge carriers.\n\n95. Show that the Hall voltage across wires made of the same material, carrying identical currents, and subjected to the same magnetic field is inversely proportional to their diameters. (Hint: Consider how drift velocity depends on wire diameter.)\n\n96. A velocity selector in a mass spectrometer uses a\n\nmagnetic field. (a) What electric field strength is needed to select a speed of\n\n? (b) What is the voltage between the plates if they are separated by\n\n?\n\n97. Find the radius of curvature of the path of a\n\nproton moving perpendicularly to the\n\nfield of a cyclotron.\n\n98. Unreasonable results To construct a non-mechanical water meter, a\n\nmagnetic field is placed across the supply water pipe to a home and the Hall voltage is recorded. (a) Find the flow rate through a\n\n-diameter pipe if the Hall voltage is\n\n(b) What would the Hall voltage be for the same flow rate through a\n\n-diameter pipe with the same field applied?\n\n99. Unreasonable results A charged particle having mass\n\n(that of a helium atom) moving at\n\nperpendicular to a\n\nmagnetic field travels in a circular path of radius\n\n(a) What is the charge of the particle? (b) What is unreasonable about this result? (c) Which assumptions are responsible?\n\n100. Unreasonable results An inventor wants to generate\n\npower by moving a\n\n-long wire perpendicular to Earth’s\n\nfield. (a) Find the speed with which the wire must move. (b) What is unreasonable about this result? (c) Which assumption is responsible?\n\n101. Unreasonable results Frustrated by the small Hall voltage obtained in blood flow measurements, a medical physicist decides to increase the applied magnetic field strength to get a\n\noutput for blood moving at\n\nin a\n\n-diameter vessel. (a) What magnetic field strength is needed? (b) What is unreasonable about this result? (c) Which premise is responsible?\n\n## Challenge Problems\n\n102. A particle of charge\n\nand mass\n\nmoves with velocity\n\npointed in the\n\n-direction as it crosses the\n\n-axis at\n\nat a particular time. There is a negative charge\n\nfixed at the origin, and there exists a uniform magnetic field\n\npointed in the\n\n-direction. It is found that the particle describes a circle of radius\n\nFind\n\nin terms of the given quantities.\n\n103. A proton of speed\n\nenters a region of uniform magnetic field of\n\nat an angle of\n\nto the magnetic field. In the region of magnetic field proton describes a helical path with radius\n\nand pitch\n\n(distance between loops). Find\n\nand\n\n104. A particle’s path is bent when it passes through a region of non-zero magnetic field although its speed remains unchanged. This is very useful for “beam steering” in particle accelerators. Consider a proton of speed\n\nentering a region of uniform magnetic field\n\nover a\n\n-wide region. Magnetic field is perpendicular to the velocity of the particle. By how much angle will the path of the proton be bent? (Hint: The particle comes out tangent to a circle.)\n\n105. In a region a non-uniform magnetic field exists such that\n\nand\n\nwhere\n\nis a constant. At some time\n\na wire of length\n\nis carrying a current\n\nis located along the\n\n-axis from origin to\n\nFind the magnetic force on the wire at this instant in time.\n\n106. A copper rod of mass\n\nand length\n\nis hung from the ceiling using two springs of spring constant\n\nA uniform magnetic field of magnitude\n\npointing perpendicular to the rod and spring (coming out of the page in the figure) exists in a region of space covering a length\n\nof the copper rod. The ends of the rod are then connected by flexible copper wire across the terminals of a battery of voltage\n\nDetermine the change in the length of the springs when\n\ncurrent\n\nruns through the copper rod in the direction shown in figure. (Ignore any force by the flexible wire.)\n\n107. The accompanied figure shows an arrangement for measuring mass of ions by an instrument called the mass spectrometer. An ion of mass\n\nand charge\n\nis produced essentially at rest in source\n\na chamber in which a gas discharge is taking place. The ion is accelerated by a potential difference\n\nand allowed to enter a region of constant magnetic field\n\nIn the uniform magnetic field region, the ion moves in a semicircular path striking a photographic plate at a distance\n\nfrom the entry point. Derive a formula for mass\n\nin terms of\n\nand\n\nand pivoted along a central support. The two ends of the wire are touching a brush that is connected to a dc power source. The structure is between the poles of a magnet such that we can assume there is a uniform magnetic field on the wire. In terms of a coordinate system with origin at the center of the ring, magnetic field is\n\nand the ring rotates about the\n\n-axis. Find the torque on the ring when it is not in the\n\n-plane.\n\n109. A long-rigid wire lies along the\n\n-axis and carries a current of\n\nin the positive\n\n-direction. Around the wire is the magnetic field\n\nwith\n\nin meters and\n\nin millitesla. Calculate the magnetic force on the segment of wire between\n\nand\n\n110. A circular loop of wire of area\n\ncarries a current of\n\nAt a particular instant, the loop lies in the\n\n-plane and is subjected to a magnetic field\n\nAs viewed from above the\n\n-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? 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https://www.dezyre.com/recipes/calculate-time-taken-by-each-step-for-loop	[ "How to calculate the time taken by each step in a for loop?\nMACHINE LEARNING RECIPES DATA CLEANING PYTHON DATA MUNGING PANDAS CHEATSHEET ALL TAGS\n\n# How to calculate the time taken by each step in a for loop?\n\nThis recipe helps you calculate the time taken by each step in a for loop\n\n0\n\n## Recipe Objective\n\nTime elapsed to execute a function is a crucial element in a data scientist's job specifically when you have to deal with large datasets. Since there are different approaches to do the same task, time required to these tasks plays an important role when we deal with large dataset. This recipe demonstrates us to calculate the time taken by each step in a for loop.\n\nWe will use Sys.time() function to carry out this task\n\n## Example:\n\nWe will calculate the squares of numbers from 1 to 4 using for loop and calculate the time required by each step\n\n``` for (i in 1:4){ # start time stamp start = Sys.time() print(i^2) # end time stamp end =Sys.time() #Time required for each step elapsed_time = end - start print(elapsed_time) } ```\n``` 1\nTime difference of 0 secs\n 4\nTime difference of 0 secs\n 9\nTime difference of 0 secs\n 16\nTime difference of 0 secs\n```\n\n#### Relevant Projects\n\n##### Time Series Forecasting with LSTM Neural Network Python\nDeep Learning Project- Learn to apply deep learning paradigm to forecast univariate time series data.\n\n##### Credit Card Fraud Detection as a Classification Problem\nIn this data science project, we will predict the credit card fraud in the transactional dataset using some of the predictive models.\n\n##### Machine Learning project for Retail Price Optimization\nIn this machine learning pricing project, we implement a retail price optimization algorithm using regression trees. This is one of the first steps to building a dynamic pricing model.\n\n##### Resume parsing with Machine learning - NLP with Python OCR and Spacy\nIn this machine learning resume parser example we use the popular Spacy NLP python library for OCR and text classification.\n\n##### Build a Similar Images Finder with Python, Keras, and Tensorflow\nBuild your own image similarity application using Python to search and find images of products that are similar to any given product. You will implement the K-Nearest Neighbor algorithm to find products with maximum similarity.\n\n##### Data Science Project - Instacart Market Basket Analysis\nData Science Project - Build a recommendation engine which will predict the products to be purchased by an Instacart consumer again.\n\n##### Ecommerce product reviews - Pairwise ranking and sentiment analysis\nThis project analyzes a dataset containing ecommerce product reviews. The goal is to use machine learning models to perform sentiment analysis on product reviews and rank them based on relevance. Reviews play a key role in product recommendation systems.\n\n##### Customer Market Basket Analysis using Apriori and Fpgrowth algorithms\nIn this data science project, you will learn how to perform market basket analysis with the application of Apriori and FP growth algorithms based on the concept of association rule learning.\n\n##### Mercari Price Suggestion Challenge Data Science Project\nData Science Project in Python- Build a machine learning algorithm that automatically suggests the right product prices.\n\n##### Customer Churn Prediction Analysis using Ensemble Techniques\nIn this machine learning churn project, we implement a churn prediction model in python using ensemble techniques." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84341156,"math_prob":0.8084257,"size":838,"snap":"2021-21-2021-25","text_gpt3_token_len":206,"char_repetition_ratio":0.12829736,"word_repetition_ratio":0.046357617,"special_character_ratio":0.25417662,"punctuation_ratio":0.048780486,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9554511,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T19:22:42Z\",\"WARC-Record-ID\":\"<urn:uuid:66fb0acf-e58b-4f61-87c2-7e0d08cb5a30>\",\"Content-Length\":\"67627\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3479781e-57d3-40a3-89be-6a1cd1789f1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:0b1983fb-f787-4b9d-a1c8-7e997f332485>\",\"WARC-IP-Address\":\"184.173.29.42\",\"WARC-Target-URI\":\"https://www.dezyre.com/recipes/calculate-time-taken-by-each-step-for-loop\",\"WARC-Payload-Digest\":\"sha1:TNEM6WL3J4KQ4NRCWAKMEBN7M6NE2S4N\",\"WARC-Block-Digest\":\"sha1:KMZM4HBMRO7WSJGPLVDZEXMHHLAFMCT3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487621519.32_warc_CC-MAIN-20210615180356-20210615210356-00544.warc.gz\"}"}
http://arxiv-export-lb.library.cornell.edu/abs/2301.13781v2	[ "math.PR\n\n# Title: Scaling limits for fractional polyharmonic Gaussian fields\n\nAbstract: This work is concerned with fractional Gaussian fields, i.e. Gaussian fields whose covariance operator is given by the inverse fractional Laplacian $(-\\Delta)^{-s}$ (where, in particular, we include the case $s >1$). We define a lattice discretization of these fields and show that their scaling limits -- with respect to the optimal Besov space topology (up to an endpoint case) -- are the original continuous fields. As a byproduct, in dimension $d<2s$, we prove the convergence in distribution of the maximum of the fields. A key tool in the proof is a sharp error estimate for the natural finite difference scheme for $(-\\Delta)^s$ under minimal regularity assumptions, which is also of independent interest.\n Comments: v2: minor corrections, additional references Subjects: Probability (math.PR); Analysis of PDEs (math.AP); Numerical Analysis (math.NA) Cite as: arXiv:2301.13781 [math.PR] (or arXiv:2301.13781v2 [math.PR] for this version)\n\n## Submission history\n\nFrom: Florian Schweiger [view email]\n[v1] Tue, 31 Jan 2023 17:31:18 GMT (3911kb,D)\n[v2] Mon, 20 Feb 2023 10:16:02 GMT (3912kb,D)\n\nLink back to: arXiv, form interface, contact." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8262166,"math_prob":0.9158847,"size":1211,"snap":"2023-14-2023-23","text_gpt3_token_len":301,"char_repetition_ratio":0.09113505,"word_repetition_ratio":0.0,"special_character_ratio":0.26424444,"punctuation_ratio":0.15350877,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95803756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T20:35:47Z\",\"WARC-Record-ID\":\"<urn:uuid:698ef187-3b3c-4c40-935c-ffd5a16da09c>\",\"Content-Length\":\"16297\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43035f0a-5e7e-43d0-86b2-0c8759edcc49>\",\"WARC-Concurrent-To\":\"<urn:uuid:0565269f-82b0-462c-b913-10346ec028b9>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"http://arxiv-export-lb.library.cornell.edu/abs/2301.13781v2\",\"WARC-Payload-Digest\":\"sha1:TWFJRTZ5RXJNU2ZOZCAOCYFWGWHPHBR3\",\"WARC-Block-Digest\":\"sha1:OSGO6BKDUDJFQTXIZ4XMR435VRNAJ6U2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650264.9_warc_CC-MAIN-20230604193207-20230604223207-00609.warc.gz\"}"}
https://help.geogebra.org/topic/how-to-change-the-origin-of-vector	[ "# How to change the origin of vector.\n\nDrako shared this question 3 years ago\n\nI have a vector (4;50°), start in (0,0), but I want start in other point, for example in (4,8).", null, "1\n\nYou can define\n\nu = (4;50°)\n\nA=(4,8)\n\nB=A+u\n\nv=Vector(A,B)\n\nNow you can drag A or the arrow of v wherever you want, the defined translation stays the same as u\n\nchris", null, "1\n\nOh, the process is too long if I have many vectors.\n\nI already knew this method. But, I wanted some more efficient.", null, "", null, "1\n\nHi\n\nYou can move the vector\n\nJust catch him !", null, "1\n\nHello,\n\nwhy don't you construct your own tools?\n\nYour first tool \"Vec_Cartesian\" needs the input objects\n\npoint, number_one, number_two.\n\n\"Point\" is the origin of the vector, and the numbers are the coordinates.\n\nIf you use the name Vec_Cartesian\n\nfor the tool, the vector from (3,2) to (8,4) has the syntax\n\nVec_Cartesian[(3,2),5,2].\n\nYour second tool \"Vec_Polar\" needs the input objects\n\nIf you use Vec_Polar as the name, the vector (4;30°) with origin (6,7) would have the syntax\n\nVec_Polar[(6,7),4,30°]" ]	[ null, "https://accounts.geogebra.org/api/avatar.php", null, "https://accounts.geogebra.org/api/avatar.php", null, "https://help.geogebra.org/public/avatars/default-avatar.svg", null, "https://accounts.geogebra.org/api/avatar.php", null, "https://accounts.geogebra.org/api/avatar.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8276283,"math_prob":0.8989396,"size":1242,"snap":"2021-21-2021-25","text_gpt3_token_len":374,"char_repetition_ratio":0.104200326,"word_repetition_ratio":0.009569378,"special_character_ratio":0.29307568,"punctuation_ratio":0.14716981,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9832206,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-10T07:39:29Z\",\"WARC-Record-ID\":\"<urn:uuid:c3bdc533-ff9d-4960-8ae5-0d873e3e3485>\",\"Content-Length\":\"69107\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5dce5d4e-dc0a-4840-b991-64ad9c46f703>\",\"WARC-Concurrent-To\":\"<urn:uuid:8100f890-2c19-43d6-a0e3-e44d41c5fd62>\",\"WARC-IP-Address\":\"140.78.116.223\",\"WARC-Target-URI\":\"https://help.geogebra.org/topic/how-to-change-the-origin-of-vector\",\"WARC-Payload-Digest\":\"sha1:356S7S5WYTA32F6EHTGZUTYEQ4ERKTED\",\"WARC-Block-Digest\":\"sha1:V4WL45DKNUQRAU55CEEMQDTOTUTL2CUV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989115.2_warc_CC-MAIN-20210510064318-20210510094318-00624.warc.gz\"}"}
https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-61/issue-1/Large-time-behavior-of-solutions-to-Schr%C3%B6dinger-equations-with-a/10.2969/jmsj/06110039.full	[ "Translator Disclaimer\nJanuary, 2009 Large time behavior of solutions to Schrödinger equations with a dissipative nonlinearity for arbitrarily large initial data\nNaoyasu KITA, Akihiro SHIMOMURA\nJ. Math. Soc. Japan 61(1): 39-64 (January, 2009). DOI: 10.2969/jmsj/06110039\n\n## Abstract\n\nWe study the asymptotic behavior in time of solutions to the Cauchy problem of nonlinear Schrödinger equations with a long-range dissipative nonlinearity given by $\\lambda \|u\|^{p-1} u$ in one space dimension, where $1 (namely, $p$ is a critical or subcritical exponent) and $\\lambda$ is a complex constant satisfying Im $\\lambda <0$ and $\\big( (p-1)/2\\sqrt{p} \\big) \| Re \\lambda \| \\le \| Im \\lambda \|$ . We present the time decay estimates and the large-time asymptotics of the solution for arbitrarily large initial data, when “$p=3$ ” or “$p<3$ and $p$ is suitably close to $3$ ”.\n\n## Citation\n\nNaoyasu KITA. Akihiro SHIMOMURA. \"Large time behavior of solutions to Schrödinger equations with a dissipative nonlinearity for arbitrarily large initial data.\" J. Math. Soc. Japan 61 (1) 39 - 64, January, 2009. https://doi.org/10.2969/jmsj/06110039\n\n## Information\n\nPublished: January, 2009\nFirst available in Project Euclid: 9 February 2009\n\nzbMATH: 1167.35043\nMathSciNet: MR2272871\nDigital Object Identifier: 10.2969/jmsj/06110039\n\nSubjects:\nPrimary: 35Q55\nSecondary: 35B40\n\nKeywords: asymptotic behavior for large data , critical or subcritical nonlinearity , dissipative nonlinearity , nonlinear Schrödinger equation", null, "", null, "" ]	[ null, "https://projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png", null, "https://projecteuclid.org/images/journals/cover_jmsj.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82077515,"math_prob":0.91300726,"size":727,"snap":"2023-14-2023-23","text_gpt3_token_len":183,"char_repetition_ratio":0.102351315,"word_repetition_ratio":0.0,"special_character_ratio":0.23246217,"punctuation_ratio":0.13178295,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96311516,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T19:37:33Z\",\"WARC-Record-ID\":\"<urn:uuid:26635c19-7fdb-4847-8954-ea4c026064e7>\",\"Content-Length\":\"148337\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76a600e3-6e18-49bc-b44a-994938050033>\",\"WARC-Concurrent-To\":\"<urn:uuid:9baf57de-e05a-4a21-beb6-8f3b5fa77030>\",\"WARC-IP-Address\":\"45.60.100.145\",\"WARC-Target-URI\":\"https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-61/issue-1/Large-time-behavior-of-solutions-to-Schr%C3%B6dinger-equations-with-a/10.2969/jmsj/06110039.full\",\"WARC-Payload-Digest\":\"sha1:RWEBQGJT3CVQA75DRQS275BBNBIHSISA\",\"WARC-Block-Digest\":\"sha1:ZMKHMZAXESQEI4NL4HFIO43VVKMTII7J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653071.58_warc_CC-MAIN-20230606182640-20230606212640-00544.warc.gz\"}"}