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Create the page "15!!!" on this wiki! See also the search results found.
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- 2 KB (177 words) - 22:24, 15 December 2023
- 1 KB (191 words) - 19:16, 10 March 2024
- ==Problem 15==4 KB (575 words) - 04:48, 15 June 2024
- Case 2: <math>k = \tan 15^\circ</math>, <math>\tan 45^\circ</math>, <math>\tan 75^\circ</math>. So we only count for <math>k = \tan 15^\circ</math>.8 KB (1,395 words) - 17:26, 9 February 2024
- 3 bytes (2 words) - 23:06, 14 April 2024
- 0 bytes (0 words) - 23:09, 28 January 2024
- 280 bytes (34 words) - 00:11, 15 February 2024
Page text matches
- label("$y$",(15,-4),N); label("$y$",(15,-4),N);3 KB (528 words) - 18:29, 7 May 2024
- ...[[set]]s of two or more consecutive positive integers have a sum of <math>15</math>? ...then their average (which could be a fraction) must be a divisor of <math>15</math>. If the number of integers in the list is odd, then the average must3 KB (450 words) - 02:00, 13 January 2024
- <math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math>3 KB (429 words) - 18:14, 26 September 2020
- *[[2007 AIME II Problems/Problem 15]]2 KB (321 words) - 22:54, 20 June 2024
- **[[2007 iTest Problems/Problem 15|Problem 15]]3 KB (305 words) - 15:10, 5 November 2023
- ...p high school mathematics students in Minnesota. Each year, four teams of 15 students are selected out of a group of approximately 75-80 trainees. The4 KB (680 words) - 16:45, 10 June 2015
- ...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? == Problem 15 ==14 KB (2,059 words) - 01:17, 30 January 2024
- == Problem 15 == [[2005 AMC 10B Problems/Problem 15|Solution]]12 KB (1,874 words) - 21:20, 23 December 2020
- * [[2005 AMC 10A Problems/Problem 15]]2 KB (182 words) - 18:09, 6 October 2014
- University of Chicago August 2 thru August 15.2 KB (370 words) - 19:48, 6 January 2015
- The format of NYSML is identical to that of [[ARML]], with 15-member teams competing in Individual, Team, Relay, and Power rounds, althou1 KB (201 words) - 14:39, 25 June 2023
- .... This sum is written in summation notation as <math>\sum_{k=1}^5 5k=5+10+15+20+25</math>. In this case, 1 is the lower limit of summation, the number2 KB (335 words) - 17:17, 8 February 2024
- \qquad \mathrm{(B) \ } 8/\sqrt{15} ...frac{3}{4}\sqrt{15})} = \frac{6\cdot\frac{4}{3}}{\sqrt{15}} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>.2 KB (219 words) - 09:57, 31 August 2012
- == Problem 15 == [[University of South Carolina High School Math Contest/1993 Exam/Problem 15|Solution]]14 KB (2,102 words) - 22:03, 26 October 2018
- * [[Mock AIME 2 Pre 2005 Problems/Problem 15|Problem 15]]2 KB (181 words) - 10:58, 18 March 2015
- * [[Mock AIME 7 Pre 2005 Problems/Problem 15|Problem 15]]1 KB (146 words) - 16:33, 14 October 2022
- * [[Mock AIME 1 2005-2006/Problem 15|Problem 15]]1 KB (135 words) - 17:41, 21 January 2017
- ** [[Mock AIME 1 2006-2007 Problems/Problem 15|Problem 15]]1 KB (155 words) - 16:06, 3 April 2012
- * [[Mock AIME 2 2006-2007 Problems/Problem 15|Problem 15]]1 KB (145 words) - 10:55, 4 April 2012
- ...or do a bit of casework to find that the only solution is <math>x = 8, y = 15</math>. In the second case, we have <math>17^2 = y^2 - x^2 = (y - x)(y + x ...se, this value may be either <math>\frac {17}{8}</math> or <math>\frac{17}{15}</math>. In the second, it may be either <math>\frac{145}{144}</math> or <2 KB (329 words) - 15:53, 3 April 2012
