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- ...be the number associated with <math>P_i</math> after the renumbering. It is found that <math>x_1 = y_2,</math> <math>x_2 = y_1,</math> <math>x_3 = y_4, <cmath>\begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\3 KB (493 words) - 13:51, 22 July 2020
- ...[[midpoint]] of the segment they determine also belongs to <math>S</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...h>, <math>(0,2)</math>, <math>(2,0)</math>, <math>(1,3)</math>, and <math>(3,1)</math>, <math>8</math> possibilities.8 KB (1,187 words) - 02:40, 28 November 2020
- ...<math>q</math>, and <math>r</math> are positive and satisfy <math>p+q+r=2/3</math> and <math>p^2+q^2+r^2=2/5</math>. The ratio of the area of triangle We let <math>[\ldots]</math> denote area; then the desired value is4 KB (673 words) - 20:15, 21 February 2024
- ...e because its base-<math>7</math> representation is <math>102</math>. What is the largest 7-10 double? <cmath>2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}</cmath> (This is because the digits in <math>N</math> ' s base 7 representation make a numbe3 KB (502 words) - 11:28, 9 December 2023
- ...h>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is [[parallel]] to <math>\overline{BC}</math> and contains the center of the [ .../math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \fr9 KB (1,540 words) - 08:31, 1 December 2022
- ...rolled four times. The [[probability]] that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form <ma The red path corresponds to the sequence of rolls <math>2, 3, 5, 5</math>. This establishes a [[bijection]] between valid dice roll seq11 KB (1,729 words) - 20:50, 28 November 2023
- ...tude is contained in the y-axis, and the square of the length of each side is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relat path e = xscale(2)*unitcircle; real x = -8/13*3^.5;6 KB (1,043 words) - 10:09, 15 January 2024
- ...ath>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. ...and <math>\angle TCA=75^{\circ}</math>, meaning <math>\triangle TAC</math> is an isosceles triangle and <math>AC=24</math>.3 KB (534 words) - 03:22, 23 January 2023
- ...nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. ...but <math>x^{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as2 KB (335 words) - 18:38, 9 February 2023
- ...h>\mathcal{S}</math>, and the mean of <math>\mathcal{S}\cup\{2001\}</math> is <math>27</math> more than the mean of <math>\mathcal{S}</math>. Find the me ...1</math> as it is from <math>2001</math>. Thus, the mean of <math>S</math> is1 KB (225 words) - 07:57, 4 November 2022
- ...ore the case <math>b = 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>. == Solution 3 ==4 KB (687 words) - 18:37, 27 November 2022
- ...h>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> ...h>. Since <math>ABFG</math> is an isosceles trapezoid and <math>CDE</math> is an isosceles triangle, we have symmetry about the <math>xz</math>-plane.7 KB (1,181 words) - 20:32, 8 January 2024
- ...th> and that 2002 is the largest element of <math>\mathcal{S},</math> what is the greatest number of elements that <math>\mathcal{S}</math> can have? ...mod <math>n</math>. Since they are positive integers, the largest element is at least <math>n^2+1</math>, the <math>(n+1)</math>th positive integer cong2 KB (267 words) - 19:18, 21 June 2021
- ...<math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. ...= black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));6 KB (974 words) - 13:01, 29 September 2023
- ...{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>AE=3</math> and <math>AF=10</math>. Given that <math>EB=9</math> and <math>FC=27 ...ey share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>.4 KB (643 words) - 22:44, 8 August 2023
- Note that it is impossible for any of <math>h,t,u</math> to be <math>1</math>, since then e ...t be <math>3</math> as well. This configuration works, so <math>333</math> is paintable.4 KB (749 words) - 19:44, 25 April 2024
- (1) <math>a_1,a_2,a_3\cdots</math> is a nondecreasing sequence of positive integers (3) <math>a_9=k</math>1 KB (205 words) - 19:54, 4 July 2013
- ...[[Binomial Expansion]] is valid for exponents that are not integers. That is, for all real numbers <math>x,y</math> and <math>r</math> with <math>|x|>|y ...=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots</cmath>2 KB (316 words) - 19:54, 4 July 2013
- ...12 combinations of two distinct vertices that form a square side only form 3 squares, and all 12 combinations of two vertices that form a square diagona ...onals, meaning we counted them 6 times. Therefore, our answer is <math>198-3(6-1)=198-15=\boxed{183}.</math>1 KB (220 words) - 20:50, 12 November 2022
- Since <math>m</math> is an integer, <math>t+1 = 29</math>, so <math>t=28</math>. It quickly follows ...=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>.2 KB (320 words) - 07:55, 4 November 2022
