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- ==Problem 15==3 KB (496 words) - 00:28, 6 May 2024
- 2 KB (261 words) - 20:04, 3 December 2023
- 4 KB (696 words) - 12:38, 13 September 2021
- ...torname{lcm}(3,5)=15</math> before erasing. So, we first group <math>\frac{15}{5}=3</math> copies of the current cycle into one, then erase: As a quick confirmation, one cycle should have length <math>15\cdot\left(1-\frac{1}{3}\right)=10</math> at this point.</li><p>10 KB (1,471 words) - 13:57, 30 October 2023
- 46 bytes (5 words) - 21:15, 12 February 2020
- ...satisfy (7) and <math>a^2 + b^2 < 1000</math>, we have <math>1 \leq b \leq 15</math>.8 KB (1,299 words) - 17:37, 3 June 2023
- 2 KB (304 words) - 01:19, 12 July 2021
- 4 KB (577 words) - 19:18, 28 October 2022
- ...the legs. One of the acute angles of the triangle is: <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qqua ...rect answer, one of the acute angles of the triangle will measure to <math>15</math> degrees. This implies that the other acute angle of the triangle wou2 KB (263 words) - 11:33, 12 August 2020
- ...aw segments <math>XM</math>, and <math>YM</math>. We have <math>MT=3\sqrt{15}</math>. ...larly, Ptolemy's theorem in <math>YTMC</math> gives<cmath>16TX=11TY+3\sqrt{15}CY</cmath> while Pythagoras' theorem in <math>\triangle CYT</math> gives <m7 KB (1,221 words) - 16:46, 29 January 2023
- 10 KB (1,742 words) - 02:31, 13 November 2023
- 625 bytes (105 words) - 10:53, 6 August 2020
- == Problem 15==1 KB (177 words) - 10:44, 15 February 2021
- so <math>(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})</math>. Hence, <cmath> abc = (1-2\sin^2(\alpha))(1-2\sin^2(\beta))(1-2\sin^2(\gamma))=\cos(15^{\circ})\cos(45^{\circ})\cos(75^{\circ})=\frac{\sqrt{2}}{8}, </cmath>15 KB (2,208 words) - 01:25, 1 February 2024
- ==Problem 15==523 bytes (80 words) - 11:58, 1 September 2020
- 7 KB (1,026 words) - 13:43, 5 May 2024
- 2 KB (311 words) - 19:01, 24 May 2023
- 4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & 15 \\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & 15 \\ [1ex]8 KB (1,183 words) - 00:36, 27 May 2024
- ...20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}}4 KB (552 words) - 22:16, 5 June 2024
- 2 KB (264 words) - 17:57, 4 October 2020
Page text matches
- {{AIME box|year=1994|num-b=13|num-a=15}}2 KB (303 words) - 00:03, 28 December 2017
- ..., and <math>15</math>'s. Because <math>0</math>, <math>6</math>, and <math>15</math> are all multiples of <math>3</math>, the change will always be a mul4 KB (645 words) - 15:12, 15 July 2019
- draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10));4 KB (721 words) - 16:14, 8 March 2021
- The increasing [[sequence]] <math>3, 15, 24, 48, \ldots\,</math> consists of those [[positive]] multiples of 3 that946 bytes (139 words) - 21:05, 1 September 2023
- ...ve. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. ...e roll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,6 KB (979 words) - 13:20, 11 April 2022
- {{AIME box|year=1995|num-b=13|num-a=15}}3 KB (484 words) - 13:11, 14 January 2023
- For <math>y=3</math>, we have <math>15,27,\cdots ,99</math>, or <math>8</math> cases.4 KB (646 words) - 17:37, 1 January 2024
- ...\Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 10 ...cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context).5 KB (710 words) - 21:04, 14 September 2020
- {{AIME box|year=1996|num-b=13|num-a=15}}5 KB (923 words) - 21:21, 22 September 2023
- ...>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</mat pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);3 KB (521 words) - 01:18, 25 February 2016
- <math>\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165</math>.5 KB (879 words) - 11:23, 5 September 2021
- Hence, the answer is <math>\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}</math>6 KB (1,022 words) - 20:23, 17 April 2021
- <math>\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}</math>3 KB (461 words) - 00:33, 16 May 2024
- {{AIME box|year=1997|num-b=13|num-a=15}}5 KB (875 words) - 01:25, 19 June 2024
- ...+ 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>=4 KB (586 words) - 21:53, 30 December 2023
- {{AIME box|year=1998|num-b=13|num-a=15}}2 KB (390 words) - 21:05, 29 May 2023
- ...th>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the area of the polygon that is triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);7 KB (1,084 words) - 11:48, 13 August 2023
- real m=60-12*sqrt(15); <math>60 - m = 12\sqrt{15}</math><br />4 KB (624 words) - 19:00, 19 June 2024
- <math>\sum_{i=1}^{15} i=\frac{(15)(16)}{2}</math> ordered pairs. For <math>x > 15</math>, <math>y</math> must follow <math>x < y\le 30</math>. Hence, there a6 KB (913 words) - 16:34, 6 August 2020
- ...has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) <math>q^2+r^2=15^2</math>7 KB (1,169 words) - 15:28, 13 May 2024
