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• Mock AIME 1 2006-2007 Problems/Problem 7
  ...[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}</cmath> ...[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.</cmath>
  3 KB (518 words) - 16:54, 25 November 2015
• Mock AIME 1 2006-2007 Problems/Problem 14
  *[[Mock AIME 1 2006-2007 Problems/Problem 15 | Next Problem]]
  3 KB (541 words) - 17:32, 22 November 2023
• Mock AIME 1 2006-2007/Problems
  ==Problem 15== [[Mock AIME 1 2006-2007 Problems/Problem 15|Solution]]
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems/Problem 6
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le ...irc\cos 35^\circ}=\frac{\sin 15^{\circ}(\cos 10^\circ-\cos 60^\circ)}{\cos 15^\circ(\cos 10^\circ+\cos 60^\circ)}</cmath>
  1 KB (157 words) - 10:51, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 14
  {{Mock AIME box|year=2006-2007|n=2|num-b=13|num-a=15}}
  2 KB (284 words) - 10:53, 4 April 2012
• Mock AIME 2 2006-2007 Problems
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le == Problem 15 ==
  5 KB (848 words) - 23:49, 25 February 2017
• Base numbers/Conversion
  ...t. The standard notation is to use the letters a=10,b=11,c=12,d=13,e=14,f=15.
  7 KB (1,177 words) - 15:56, 18 April 2020
• 1959 IMO Problems/Problem 1
  ...rac{46}{31} </math>. Again, both results are 1 off from being multiples of 15. ...ime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending
  5 KB (767 words) - 10:59, 23 July 2023
• 1971 Canadian MO Problems/Problem 1
  <math>(EP)(CE)=(BE)(ED)</math> and <math>2r-1=15.</math> Hence, <math>r=8.</math>
  680 bytes (114 words) - 21:38, 9 July 2019
• 1959 IMO Problems/Problem 4
  ...riangle), and <math>75^\circ</math> angle is constructed by constructing a 15 degree angle on top of the 60 degree point.
  6 KB (939 words) - 17:31, 15 July 2023
• 1995 AHSME Problems/Problem 8
  ...{16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 } </math> ...(0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6);
  1 KB (199 words) - 13:58, 5 July 2013
• Congruent (geometry)
  X=(10,0); Y=(13,2); Z=(15,0); X=(10,0); Y=(13,2); Z=(15,0);
  10 KB (1,655 words) - 21:43, 24 March 2022
• 2005 AMC 10A Problems/Problem 5
  ...would get <math>\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3</math> free windows.
  1 KB (182 words) - 03:42, 29 April 2023
• 2005 AMC 10A Problems/Problem 14
  {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}
  4 KB (694 words) - 19:25, 13 December 2021
• 2005 AMC 10A Problems/Problem 16
  {{AMC10 box|year=2005|ab=A|num-b=15|num-a=17}}
  2 KB (279 words) - 11:57, 17 July 2023
• Perfect cube
  * [[2005_AMC_10A_Problems/Problem_15 | 2005 AMC 10A Problem 15]]
  704 bytes (91 words) - 14:12, 24 August 2023
• System of equations
  ...Thus <center><math> 4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15. </math></center> ...uation from the second equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS. And thus we continue on in this way to find that <m
  5 KB (784 words) - 23:27, 30 July 2020
• MathPath
  | 12:15 - 1:15 || Lunch | 3:15 - 4:30 || Breakout Session
  2 KB (266 words) - 21:38, 13 December 2023
• 1970 Canadian MO Problems
  draw((16/5,12/5)--(16/5-.2,12/5-.15)--(16/5-.2+.15,12/5-.15-.2)--(16/5+.15,12/5-.2)--cycle,black);
  4 KB (604 words) - 04:32, 8 October 2014
• Reduced fraction
  ...15 and 27 are divisible by 3. So in order to reduce, we write <math>\frac{15}{27} = \frac{3 \cdot 5}{3\cdot 9} = \frac5 9</math>, and 5 and 9 are relati
  1 KB (225 words) - 13:22, 7 March 2021

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