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  • pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); Let <math>\theta = \angle DBA</math>. Then <math>\angle CAB = \angle DBC = 2\theta</math>, <math>\angle AOB = 180 - 3\theta</math>, and <math>\angle ACB
    5 KB (710 words) - 20:04, 14 September 2020
  • On Monday, Taye has <math>$2</math>. Every day, he either gains <math>$3</math> or doubles the amount o ...have <math>8</math> dollars or <math>7</math> dollars. Now, you have <math>2</math> values for each of these. For <math>10</math> dollars, you have <mat
    3 KB (476 words) - 16:29, 19 January 2025
  • The [[equation]] <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n} 2000x^6+100x^5+10x^3+x-2&=0\
    7 KB (1,098 words) - 23:33, 20 January 2025
  • ...margins using the geometry package, use the line <code>\usepackage[margin=2.5cm]{geometry}</code>. Check out the [http://tug.ctan.org/tex-archive/macro ==Document Formatting==
    29 KB (4,929 words) - 12:26, 12 September 2024
  • ...packages for use with LaTeX. These packages allow much of the mathematical formatting we have introduced on these pages, as well as much, much more. The packages ...n you installed MiKTeX, this folder is probably '''C:\Program Files\MiKTeX 2.9\tex\latex'''. (If you have an older MiKTeX installation, this folder is p
    10 KB (1,656 words) - 08:33, 21 September 2024
  • ...a^{-k})^2,</math> from which <cmath>a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.</cmath> We apply this result twice to get the answer: a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \
    4 KB (663 words) - 06:32, 4 November 2022
  • ...k and copypaste the solution, save. There. If anyone can do the wiki table formatting more elegantly, be my guest; after all, this is wikiiii. ...es to separate rows (Enter key) is encouraged, they won't affect the final formatting -->
    22 KB (3,358 words) - 14:17, 18 July 2017
  • <cmath>\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)</cmath> <b>Case </b>2) <math>d>0</math> (We can ignore <math>d<0</math> because of lemma)
    4 KB (768 words) - 22:15, 31 March 2010
  • ...<math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \ ...</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math>
    8 KB (1,347 words) - 20:14, 9 November 2024
  • ...e sheet language used to describe the presentation semantics (the look and formatting) of a document written in a markup language. 2. Black-on-white is the most preferred style, although white-on-black (rever
    5 KB (846 words) - 21:23, 13 July 2022
  • For a given integer <math>n\ge 2,</math> let <math>\{a_1,a_2,…,a_m\}</math> be the set of positive integer ==Solution (NO FORMATTING)==
    5 KB (939 words) - 21:00, 11 February 2024
  • In the second iteration, i = 2, so Python prints 2. '''Find <math>\boldsymbol{\sum_{n=1}^{50} 2^{n}.}</math>'''
    38 KB (6,076 words) - 18:09, 20 February 2025
  • ...0 \gamma</math>, where <math>\gcd(\alpha,3)=1</math> and <math>\gcd(\gamma,2)=1</math>. Subtracting the two equations, <math>38=7\alpha-20\gamma</math>. ...{3}</math> for <math>k \equiv{1} \pmod{3}</math> only, so <math>k \equiv{0,2} \pmod{3}</math>, giving us our answer. Since the problem asks for the sum
    17 KB (2,515 words) - 09:51, 4 February 2025
  • <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? <cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.</cmath>
    2 KB (341 words) - 14:02, 8 June 2023
  • ==Problem 2 == ...rac{110}{44}</math> simplifies to <math>\frac{5}{2}</math>, which is <math>2.5</math>;
    2 KB (239 words) - 18:33, 20 January 2025
  • Find the sum of all positive integers <math>n</math> such that when <math>1^3+2^3+3^3+\cdots +n^3</math> is divided by <math>n+5</math>, the remainder is < <cmath>1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2</cmath>
    6 KB (918 words) - 09:27, 21 December 2024
  • ...duct<cmath>\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24} ...\dots \cdot (63+1)(63-1)}{(4+2)(4-2)(5+2)(5-2)\cdot \dots \cdot (63+2)(63-2)} \
    5 KB (599 words) - 20:03, 15 February 2025
  • ...necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins. In general, t ...= 65</math> to <math>69</math>, a similar issue arises, as <math>25 \times 2 + 10 + 5 \times 1</math> is not as optimal as <math>25 + 4 \times 10</math>
    3 KB (517 words) - 21:09, 15 February 2025
  • <math>\lfloor{x}\rfloor^2-3x+2=0</math>, where <math>\lfloor{x}\rfloor</math> denotes the largest integer ...A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0</math>
    11 KB (1,927 words) - 13:40, 1 December 2024
  • ==Solution 2 (Elimination)== ~michaelwang13675 (Formatting)
    3 KB (427 words) - 00:37, 9 October 2024

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