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- {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}}2 KB (309 words) - 22:27, 15 August 2023
- ** [[2001 AMC 8 Problems/Problem 15]]1 KB (138 words) - 10:26, 22 August 2013
- <math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math> <math>\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72</math>13 KB (1,990 words) - 09:51, 19 June 2024
- ==Problem 15== [[Image:AIME 1985 Problem 15.png]]7 KB (1,071 words) - 19:24, 23 February 2024
- <math>(8, 15, 17)</math> <nowiki>*</nowiki> <math>(9, 12, 15)</math>4 KB (684 words) - 16:45, 1 August 2020
- ...g tag, for PoTW on the Wiki front page--><onlyinclude>A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed The volume of the original box is <math>15\cdot10\cdot8=1200.</math>1 KB (183 words) - 15:36, 19 August 2023
- This is not too bad with casework. Notice that <math>1*60=2*30=3*20=4*15=5*12=6*10=60</math>. Hence, <math>60</math> has <math>12</math> factors, of2 KB (326 words) - 15:40, 19 August 2023
- {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #15]] and [[2003 AMC 10A Problems|2003 AMC 10A #19]]}}3 KB (393 words) - 19:06, 3 June 2024
- ...|<font color="#FF 69 B4">Maha</font><font color="#FF00FF">maya</font>]] 21:15, 21 May 2012 (EDT)5 KB (923 words) - 19:51, 21 January 2024
- ...ofproblemsolving.com/Forum/viewtopic.php?p=432791#432791 AIME 1991 Problem 15]1 KB (240 words) - 16:49, 29 December 2021
- ...n as Georgia Tech). The 2014 competition will occur on Saturday, February 15, 2014. ...ation is online at [http://hsmc.gatech.edu]. The cost is <nowiki>$</nowiki>15 per contestant, which includes lunch and a T-shirt for the contestant. One3 KB (475 words) - 21:51, 31 December 2013
- ...e, most integers have many factorizations into 2 parts: <math>30 = 2 \cdot 15 = 3 \cdot 10 = 5 \cdot 6</math>. Thus, the Fundamental Theorem of Arithmet2 KB (376 words) - 23:28, 4 August 2022
- ...7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? <math>\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60</math>13 KB (1,968 words) - 18:32, 29 February 2024
- path p=(0,0)..(20,15)..(40,-5)..(50,0);6 KB (871 words) - 21:14, 12 June 2023
- ...al [[Modular arithmetic/Introduction |modulo]] 7 residues. To avoid having 15 with the same residue, 14 numbers with different modulo 7 residues can be p ...2}\approx 1.414</math>, we get <math>\frac{\sqrt{29-14.14}}{2}=\frac{\sqrt{15.86}}{2}<\frac{4}{2}=2</math>.10 KB (1,617 words) - 01:34, 26 October 2021
- **[[2021 Fall AMC 10B Problems/Problem 15|Problem 15]]2 KB (205 words) - 10:53, 1 December 2021
- ...h>, then if <math>BY = b</math> we have a system of equations. <math>a+b = 15, b+41-a = 52</math>. We can then solve for <math>a</math>, and since <math>5 KB (818 words) - 11:05, 7 June 2022
- **Probability: <math>{6\choose4}/64 = \frac{15}{64}</math> <math>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</math>8 KB (1,368 words) - 19:02, 20 July 2024
- O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.14213562373095 draw(Circle((0,0),15));7 KB (1,147 words) - 20:57, 9 July 2024
- Let <math>N</math> denote the number of permutations of the <math>15</math>-character string <math>AAAABBBBBCCCCCC</math> such that == Problem 15 ==6 KB (1,100 words) - 22:35, 9 January 2016
