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  • ...ypically contain a squared term such as <math>(x-3)^2</math>. However, the problem may be posed as to convert from an expanded form to a factored perfect squa <math>x^2+2x=28 </math> solve for x
    2 KB (422 words) - 15:20, 5 March 2023
  • * <math>28! = 304888344611713860501504000000</math> ([[2007 iTest Problems/Problem 6|Source]])
    10 KB (809 words) - 15:40, 17 March 2024
  • ...GREAT in binary). Here's one that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY! * [[2000 AMC 8 Problems/Problem 11]]
    10 KB (1,572 words) - 21:11, 22 September 2024
  • ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is
    8 KB (1,192 words) - 16:20, 16 June 2023
  • == Problem == ...it deleted. Now, we know that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math>
    4 KB (622 words) - 21:47, 13 October 2024
  • == Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]
    12 KB (1,792 words) - 12:06, 19 February 2020
  • == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]
    13 KB (1,948 words) - 09:35, 16 June 2024
  • == Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]
    10 KB (1,547 words) - 03:20, 9 October 2022
  • == Problem 1 == [[2003 AMC 12B Problems/Problem 1|Solution]]
    13 KB (1,987 words) - 17:53, 10 December 2022
  • == Problem == ...the Law of Cosines, <math>7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}</math>. In <math>\triangle
    2 KB (299 words) - 14:29, 5 July 2022
  • == Problem == ...>20</math> or greater, so there is a total probability of <math>\dfrac{14}{28}=\boxed{\textbf{(D) }\frac{1}{2}}</math>.
    4 KB (607 words) - 14:16, 23 June 2024
  • ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]
    13 KB (2,028 words) - 15:32, 22 March 2022
  • == Problem == MP("II", (8,-28), (0,0));
    3 KB (424 words) - 09:14, 17 December 2021
  • ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]
    17 KB (2,246 words) - 12:37, 19 February 2020
  • == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>
    4 KB (628 words) - 10:28, 14 April 2024
  • == Problem == pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
    4 KB (693 words) - 12:03, 28 December 2021
  • == Problem == ...h>n(n + 7)</math> for some positive integer <math>n</math>. When <math>n = 28</math>, this product is <math>980</math>, and since AIME answers are nonneg
    8 KB (1,249 words) - 20:25, 20 November 2024
  • == Problem == ...assign to <math>(0,1,0)</math>. (We will see how this correlates with the problem.) Then define for each lattice point <math>(i,j)</math> its triplet thus:
    5 KB (897 words) - 23:21, 28 July 2022
  • == Problem == ...and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>2
    5 KB (906 words) - 22:15, 6 January 2024
  • == Problem == ...\tan\angle{B}=\frac{56}{33}</math> so <math>BJ=\frac{33}{56}*6x=\frac{99x}{28}</math>.
    14 KB (2,340 words) - 15:38, 21 August 2024

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