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- ...ution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>. When <math>x=5</math>, <math>B=6, 9, ..., 93\Rightarrow 15</math> triples..4 KB (661 words) - 23:14, 26 May 2023
- ==Problem 15== \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\8 KB (1,424 words) - 17:13, 28 June 2024
- 46 bytes (5 words) - 01:01, 5 January 2014
- <math>x_6 = 15</math> <math>x_{14} = 105</math> <math>x_7 = 30</math> <math>x_{15} = 210</math>4 KB (770 words) - 17:44, 11 October 2023
- 1 KB (231 words) - 01:39, 16 August 2023
- 1 KB (245 words) - 08:45, 30 January 2018
- 2 KB (295 words) - 03:58, 29 December 2022
- 1 KB (169 words) - 23:17, 2 January 2014
- ...12A Problems|2014 AMC 12A #11]] and [[2014 AMC 10A Problems|2014 AMC 10A #15]]}} ...hour late if he continues at this speed. He increases his speed by <math>15</math> miles per hour for the rest of the way to the airport and arrives <m3 KB (409 words) - 04:18, 20 June 2022
- 1 KB (214 words) - 02:53, 28 May 2021
- 4 KB (612 words) - 22:42, 2 August 2021
- 3 KB (536 words) - 21:02, 31 October 2022
- 1 KB (203 words) - 14:09, 4 July 2024
- == Problem 15 == ...theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</math>.10 KB (1,643 words) - 22:30, 28 January 2024
- 897 bytes (140 words) - 13:39, 6 November 2020
- 821 bytes (134 words) - 02:46, 20 February 2018
- <math>\text{(A) } 15\quad1 KB (195 words) - 09:24, 1 January 2024
- 643 bytes (92 words) - 05:37, 4 February 2016
- 2 KB (250 words) - 18:29, 21 June 2018
- 1 KB (226 words) - 22:26, 13 July 2019
Page text matches
- ...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...ree HT, four TH, and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT sub5 KB (847 words) - 15:48, 21 August 2023
- ...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? == Problem 15 ==6 KB (869 words) - 15:34, 22 August 2023
- == Problem 15 == [[1988 AIME Problems/Problem 15|Solution]]6 KB (902 words) - 08:57, 19 June 2021
- == Problem 15 == [[1989 AIME Problems/Problem 15|Solution]]7 KB (1,045 words) - 20:47, 14 December 2023
- A triangle has vertices <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The equation of the bisector == Problem 15 ==6 KB (870 words) - 10:14, 19 June 2021
- ...<math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{} == Problem 15 ==7 KB (1,106 words) - 22:05, 7 June 2021
- \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 == Problem 15 ==8 KB (1,117 words) - 05:32, 11 November 2023
- ...n{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(a) the winner caught <math>15</math> fish;8 KB (1,275 words) - 06:55, 2 September 2021
- The increasing sequence <math>3, 15, 24, 48, \ldots\,</math> consists of those positive multiples of 3 that are == Problem 15 ==7 KB (1,141 words) - 07:37, 7 September 2018
- == Problem 15 == [[1995 AIME Problems/Problem 15|Solution]]6 KB (1,000 words) - 00:25, 27 March 2024
- ...>, <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 == Problem 15 ==6 KB (931 words) - 17:49, 21 December 2018
- == Problem 15 == [[1997 AIME Problems/Problem 15|Solution]]7 KB (1,098 words) - 17:08, 25 June 2020
- ...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]] == Problem 15 ==7 KB (1,084 words) - 02:01, 28 November 2023
- ...sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the tangent of angle <math>PAB</math> is <math>m/n,</math> wher == Problem 15 ==7 KB (1,094 words) - 13:39, 16 August 2020
- == Problem 15 == [[2000 AIME I Problems/Problem 15|Solution]]7 KB (1,204 words) - 03:40, 4 January 2023
- In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{A == Problem 15 ==7 KB (1,212 words) - 22:16, 17 December 2023
- == Problem 15 == [[2002 AIME I Problems/Problem 15|Solution]]8 KB (1,374 words) - 21:09, 27 July 2023
- == Problem 15 == [[2003 AIME I Problems/Problem 15|Solution]]6 KB (965 words) - 16:36, 8 September 2019
- Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10! ...</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to triangle <math>ABC</ma6 KB (947 words) - 21:11, 19 February 2019
- == Problem 15 == [[2001 AIME II Problems/Problem 15|Solution]]8 KB (1,282 words) - 21:12, 19 February 2019
