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• 2010 AMC 12A Problems/Problem 15
  <math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{
  2 KB (255 words) - 21:47, 3 July 2013
• 2010 AIME I Problems/Problem 15
  ...BC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such th draw((0,0)--(15,0));
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2007 AMC 8 Problems/Problem 15
  1 KB (198 words) - 15:53, 2 July 2024
• 2010 AMC 10A Problems/Problem 15
  46 bytes (5 words) - 13:25, 26 May 2020
• Northeastern WOOTers Mock AIME I Problems/Problem 15
  == Problem 15 ==
  1 KB (226 words) - 21:35, 7 August 2021
• 2010 AIME II Problems/Problem 15
  == Problem 15 == ...math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> wit
  9 KB (1,537 words) - 14:45, 5 July 2024
• 2010 AMC 12B Problems/Problem 15
  ...ath>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples.
  2 KB (360 words) - 17:29, 26 May 2023
• 1999 AMC 8 Problems/Problem 15
  3 KB (516 words) - 14:50, 21 December 2022
• 2010 AMC 10B Problems/Problem 15
  1 KB (210 words) - 02:44, 26 September 2020
• 2011 AMC 12A Problems/Problem 15
  2 KB (376 words) - 23:14, 5 January 2024
• 2011 AMC 10A Problems/Problem 15
  == Problem 15 ==
  2 KB (264 words) - 15:31, 3 September 2022
• 2001 AMC 10 Problems/Problem 15
  ...e street at an angle. The length of the curb between the stripes is <math> 15 </math> feet and each stripe is <math> 50 </math> feet long. Find the dista ...)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math>
  2 KB (253 words) - 11:57, 20 October 2020
• 2011 AIME I Problems/Problem 15
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 AIME II Problems/Problem 15
  ...ber <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfl P(15) &= 171 \\
  8 KB (1,273 words) - 14:03, 7 January 2023
• 2001 AMC 8 Problems/Problem 15
  922 bytes (135 words) - 00:38, 5 July 2013
• 2011 AMC 10B Problems/Problem 15
  2 KB (287 words) - 01:48, 26 June 2022
• 2008 AMC 12B Problems/Problem 15
  3 KB (505 words) - 22:58, 7 October 2021
• 1999 AHSME Problems/Problem 15
  931 bytes (144 words) - 19:36, 1 May 2023
• 2011 AMC 12B Problems/Problem 15
  ==Problem 15==
  2 KB (284 words) - 20:54, 29 May 2023
• 2007 AMC 10B Problems/Problem 15
  ...12B Problems|2007 AMC 12B #11]] and [[2007 AMC 10B Problems|2007 AMC 10B #15]]}}
  997 bytes (137 words) - 12:22, 4 July 2013

Page text matches

• 2000 AMC 12 Problems
  == Problem 15 == [[2000 AMC 12 Problems/Problem 15|Solution]]
  13 KB (1,948 words) - 10:35, 16 June 2024
• 2001 AMC 12 Problems
  ...ee numbers is <math>10</math> more than the least of the numbers and <math>15</math> == Problem 15 ==
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  == Problem 15 == [[2002 AMC 12B Problems/Problem 15|Solution]]
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2003 AMC 12B Problems
  <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> ...{(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  ...{B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</math> <math>(\mathrm {A}) 13\qquad (\mathrm {B}) 14 \qquad (\mathrm {C}) 15 \qquad (\mathrm {D}) 16 \qquad (\mathrm {E}) 17</math>
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  ...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point == Problem 15 ==
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 4
  \text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25
  1 KB (152 words) - 16:11, 8 December 2013
• 2006 AMC 12B Problems/Problem 10
  ...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? ...third side has length <math>15</math>, and so the perimeter is <math>21+7+15=43 \Rightarrow \boxed{\text {(A)}}</math>.
  977 bytes (156 words) - 13:57, 19 January 2021
• 2006 AMC 12B Problems/Problem 14
  ...)=23</math>. The problem asks for the total cost of jam, or <math>N(5J)=11(15)=165</math> cents, or <math>1.65\implies\mathrm{(D)}</math> {{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 16
  {{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
  1 KB (210 words) - 12:36, 2 July 2024
• 2006 AMC 12B Problems/Problem 20
  \mathrm{(D)}\ \frac 15 .../(1-r)</math> for the sum of an infinite geometric sequence, we get <math>(15/100)/(1-(1/10)) = \boxed{\frac 16}</math>.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12A Problems/Problem 14
  {{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
  3 KB (442 words) - 03:13, 8 August 2022
• 2006 AMC 12A Problems/Problem 16
  {{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
  2 KB (286 words) - 10:16, 19 December 2021
• 2006 AMC 12A Problems/Problem 22
  ...e OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}</math>. Since <math>OA</math> is a radius and <math>OP</math> can b ...>, then we get a [[right triangle]]. Using simple trigonometry, <math>\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}</math>.
  2 KB (343 words) - 15:39, 14 June 2023
• 2006 AMC 12A Problems/Problem 25
  ...-[[empty set | empty]] [[subset]]s <math>S</math> of <math>\{1,2,3,\ldots ,15\}</math> have the following two properties? <math>{15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \ch
  8 KB (1,405 words) - 11:52, 27 September 2022
• 2005 AMC 12B Problems/Problem 6
  ...(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, whi \implies k^2+2k-15=0
  2 KB (299 words) - 15:29, 5 July 2022
• 2005 AMC 12B Problems/Problem 9
  ...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86</math>.
  2 KB (280 words) - 15:35, 16 December 2021
• 2005 AMC 12B Problems/Problem 11
  ...12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}} ...th>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you have <math>2</math> ten
  4 KB (607 words) - 15:16, 23 June 2024
• 2005 AMC 12B Problems/Problem 12
  {{AMC10 box|year=2005|ab=B|num-b=15|num-a=17}}
  2 KB (317 words) - 12:27, 16 December 2021
• 2005 AMC 12B Problems/Problem 14
  {{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}}
  2 KB (278 words) - 21:12, 24 December 2020

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