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  • * Intermediate is recommended for students who can expect to pass the AMC 10/12. ...es exposure to advanced topics like probability theory.Great for AMC 8 /10/12
    23 KB (3,038 words) - 18:33, 15 February 2025
  • *[[1995 AIME Problems/Problem 5|1995 AIME Problem 5]] *[[2002 AIME I Problems/Problem 12|2002 AIME I Problem 12]]
    5 KB (860 words) - 14:36, 10 December 2023
  • ...ate Tournament, achieve one of the top 10 scores in the state on the [[AMC 12]], or achieve one of the top 10 scores in the state on the [[AIME]], receiv ...owa ever since. The MN Gold team has finished in the top ten in Division A 12 times, winning the national championship in 1997. The MN Maroon team has f
    4 KB (680 words) - 15:45, 10 June 2015
  • :Subcase a: <math>T(n) = 1970 + u + 17 + u + 8 + u = 1995 + 3u = 2007 \Longrightarrow u = 4</math>. This exceeds our bounds, so no so S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \
    15 KB (2,558 words) - 22:08, 28 July 2024
  • '''1995 AHSME''' problems and solutions. The first link contains the full set of t * [[1995 AHSME Problems|Entire Exam]]
    2 KB (174 words) - 23:30, 1 October 2014
  • |year = 1995 [[1995 AHSME Problems/Problem 1|Solution]]
    17 KB (2,387 words) - 21:44, 26 May 2021
  • If <math>2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},</math> what is the value of <math>k</math>? == Problem 12 ==
    15 KB (2,222 words) - 09:40, 11 August 2020
  • * [[1996 AHSME Problems/Problem 12|Problem 12]] {{AHSME box|year=1996|before=[[1995 AHSME]]|after=[[1997 AHSME]]}}
    2 KB (174 words) - 23:30, 1 October 2014
  • ...\qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12 </math> dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot
    17 KB (2,590 words) - 12:38, 19 February 2020
  • <math> \text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)} ==Problem 12==
    15 KB (2,343 words) - 12:39, 19 February 2020
  • * [[1994 AHSME Problems/Problem 12|Problem 12]] {{AHSME box|year=1994|before=[[1993 AHSME]]|after=[[1995 AHSME]]}}
    2 KB (174 words) - 02:45, 29 September 2014
  • <math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20 </m draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle);
    14 KB (2,124 words) - 12:39, 19 February 2020
  • | [[AMC 12]]/[[AHSME]] * [//liquids.seas.harvard.edu/oleg/competition/prev.html ''Problems from 1995 to 2004 w/ Solutions'']
    15 KB (1,864 words) - 17:59, 15 February 2025
  • This problem is quite similar to [[1995_AIME_Problems/Problem_3|1995 AIME Problem 3]]. ...he can do so in <math>2^4 - 4 = 12</math> ways for each case . Thus, <math>12 \cdot 2</math> total paths for the subcase of staying in one direction. (Fo
    17 KB (2,801 words) - 14:35, 30 August 2024