Create the page "15!!!" on this wiki! See also the search results found.

• Mock MathCounts
  * <b>Add variety.</b> Nobody wants to see a sprint round with 15 counting problems or 20 algebra problems. Generally, algebra and geometry s
  26 KB (3,265 words) - 21:34, 20 March 2024
• 2007 iTest Problems
  ===Problem 15=== [[2007 iTest Problems/Problem 15|Solution]]
  30 KB (4,794 words) - 23:00, 8 May 2024
• 2005 Alabama ARML TST
  ...to an easy [[AIME]], Alabaman scores average about 5 right, while the top 15 people usually answer 8 or more correctly. The first link contains the full * [[2005 Alabama ARML TST Problems/Problem 15]]
  1 KB (170 words) - 01:01, 19 June 2018
• 2006 iTest
  *[[2006 iTest Problems/Problem 15|Problem 15]]
  3 KB (320 words) - 09:56, 23 April 2024
• Areteem Institute Summer Math Programs
  University of Chicago August 2 thru August 15.
  2 KB (373 words) - 19:58, 6 January 2015
• 2005 AMC 10A Problems
  == Problem 15 == [[2005 AMC 10A Problems/Problem 15|Solution]]
  14 KB (2,026 words) - 11:45, 12 July 2021
• Division
  For instance: <math>15 \div 2.5 = 150 \div 25 = 6.</math>
  2 KB (259 words) - 09:52, 23 January 2020
• 2004 AMC 12A Problems/Problem 7
  ...layers <math>A</math>, <math>B</math>, and <math>C</math> start with <math>15</math>, <math>14</math>, and <math>13</math> tokens, respectively. How man First round: <math>15,14,13</math> (given)
  4 KB (588 words) - 16:51, 24 March 2023
• 2003 AMC 10A
  * [[2003 AMC 10A Problems/Problem 15]]
  1 KB (165 words) - 18:48, 6 October 2014
• 2003 AMC 10A Problems
  A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24
  13 KB (1,900 words) - 22:27, 6 January 2021
• 1999 AMC 8 Problems/Problem 16
  {{AMC8 box|year=1999|num-b=15|num-a=17}}
  1 KB (167 words) - 20:30, 11 January 2024
• 2003 AMC 10A Problems/Problem 14
  <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 {{AMC10 box|year=2003|ab=A|num-b=13|num-a=15}}
  2 KB (336 words) - 15:49, 19 August 2023
• 2003 AMC 10A Problems/Problem 16
  {{AMC10 box|year=2003|ab=A|num-b=15|num-a=17}}
  1 KB (218 words) - 15:52, 19 August 2023
• 2003 AMC 10A Problems/Problem 19
  #REDIRECT[[2003 AMC 12A Problems/Problem 15]]
  45 bytes (5 words) - 18:31, 31 July 2011
• 1997 AJHSME Problems/Problem 24
  ...{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi</math> ...\pi - 15\pi = 10\pi</math>.Therefore the ratio of shaded:unshaded is <math>15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}</math>.
  2 KB (394 words) - 17:05, 20 October 2023
• 2004 AMC 10A
  * [[2004 AMC 10A Problems/Problem 15]]
  2 KB (182 words) - 01:29, 7 October 2014
• 2003 AMC 10A Problems/Problem 21
  <math>3*5=15</math> cases for case 2 10+15+3= <math>28</math> total
  7 KB (994 words) - 17:51, 11 April 2024
• 2003 AMC 10A Problems/Problem 22
  ...lugging this into our equation for line <math>GA</math> gives us <math>G=(-15,20)</math>, so <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
  9 KB (1,446 words) - 22:48, 8 May 2024
• 2004 AMC 10A Problems
  ...ns. Players <math>A</math>, <math>B</math>, and <math>C</math> start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the gam == Problem 15 ==
  15 KB (2,092 words) - 20:32, 15 April 2024
• 2004 AMC 12A Problems/Problem 15
  {{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

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)