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• 1950 AHSME Problems/Problem 45
  ...lso counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>. ...the number of diagonals of a polygon with <math>n</math> sides is <math>n(n-3)/2</math>. Taking <math>n=100</math>, we see that the number of diagonals
  1 KB (217 words) - 21:42, 3 May 2024
• 1953 AHSME Problems/Problem 45
  Since both lengths are positive, the [[AM-GM Inequality]] is satisfied. The correct relationship between <math>a</math {{AHSME 50p box|year=1953|num-b=44|num-a=46}}
  689 bytes (111 words) - 23:02, 14 February 2020
• 1951 AHSME Problems/Problem 45
  == Problem 45 == {{AHSME 50p box|year=1951|num-b=44|num-a=46}}
  1 KB (194 words) - 12:27, 5 July 2013
• 1952 AHSME Problems/Problem 45
  Using the RMS-AM-GM-HM inequality, we can see that the answer is <math>\fbox{E}</math>. {{AHSME 50p box|year=1952|num-b=44|num-a=46}}
  624 bytes (104 words) - 17:36, 9 July 2015
• 1958 AHSME Problems/Problem 45
  ...ars and <math> y</math> cents, <math> x</math> and <math> y</math> both two-digit numbers. In error it is cashed for <math> y</math> dollars and <math> {{AHSME 50p box|year=1958|num-b=44|num-a=46}}
  795 bytes (124 words) - 06:29, 3 October 2014
• 2007 iTest Problems/Problem 45
  {{iTest box|year=2007|num-b=44|num-a=46}}
  3 KB (446 words) - 05:07, 16 June 2018
• 1955 AHSME Problems/Problem 45
  == Problem 45== ...h>, we get <math>r^2+2-2r = 2</math>, we can simplify this to get <math>r(r-2)=0</math>, so <math>r=0</math> or <math>2</math>. But since <math>r\neq 0<
  1 KB (232 words) - 17:23, 2 July 2020
• 2008 iTest Problems/Problem 45
  {{2008 iTest box|num-b=44|num-a=46}}
  2 KB (287 words) - 18:54, 12 July 2018
• 1959 AHSME Problems/Problem 45
  34 bytes (6 words) - 10:15, 14 June 2019
• 1954 AHSME Problems/Problem 45
  {{AHSME 50p box|year=1954|num-b=44|num-a=46}}
  1 KB (193 words) - 10:42, 6 July 2020
• 1956 AHSME Problems/Problem 45
  542 bytes (86 words) - 21:00, 27 March 2021
• 45-45-90 triangle
  a 45-45-90 triangle has 2 angles with the value of 45 and one that is 90 degrees. it is special because it has a ratio for it's s
  232 bytes (45 words) - 14:29, 7 January 2022

Page text matches

• 2005 AMC 12A Problems/Problem 16
  pair P0=O0+9*dir(-45), P3=O3+dir(70); [[Image:2005_12A_AMC-16b.png]]
  2 KB (307 words) - 15:30, 30 March 2024
• 2016 AMC 10A Problems/Problem 8
  ...bf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math> .../math> as the amount of money Foolish Fox started with we have <math>2(2(2x-40)-40)-40=0.</math> Solving this we get <math>\boxed{\textbf{(C) }35}</math
  1 KB (169 words) - 14:59, 8 August 2021
• MATHCOUNTS
  ...ter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}} ...dents have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The first 20 problems are usu
  10 KB (1,497 words) - 11:42, 10 March 2024
• Chemistry competitions
  ...National Chemistry Olympiad national exam (USNCO) is a 3-part, 4 hour and 45 minute exam administered in mid or late April by ACS Local Sections. Approx ...www.amazon.com/gp/product/0618221565/ref=pd_lpo_k2_dp_k2a_1_img/102-5655201-2084940?%5Fencoding=UTF8 ''Chemistry''] by Steven S. Zumdahl, Susan A. Zumda
  2 KB (258 words) - 19:31, 8 March 2023
• United States of America Mathematical Talent Search
  ...f|difficulty=3-6|breakdown=<u>Problem 1-2</u>: 3-4<br><u>Problem 3-5</u>: 5-6}} ...ed to high scorers at the end of the year. These typically include a free t-shirt, along with other prizes like books or software of the participant's c
  4 KB (613 words) - 13:08, 18 July 2023
• 1955 AHSME Problems/Problem 38
  ...ogether, we get: <math>2(a+b+c+d)=90</math>. This means that <math>a+b+c+d=45</math>.
  1 KB (200 words) - 23:35, 28 August 2020
• Joining an ARML team
  ...as well as practices of previous years' team rounds. Please email Xinke Guo-Xue at xinkeguoxue@gmail.com, or message Xinke's AoPS account "hurdler", if ...room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit
  21 KB (3,500 words) - 18:41, 23 April 2024
• Factorial
  ...[[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>45! = 119622220865480194561963161495657715064383733760000000000</math>
  10 KB (809 words) - 16:40, 17 March 2024
• North Carolina MathCounts
  ...Eli Park (20), Brian Zhang, Sukrith Velmineti, Eric Wu, Coach: Ann Chapoton-Genna ...rry Zhao (13), Benjamin Wu (51), Jeffrey Liu, David Li, Coach: Ann Chapoton-Genna
  4 KB (580 words) - 15:33, 2 April 2024
• Constructive counting
  ''How many four-digit numbers are there?'' '''Solution''': We can construct a four-digit by picking the first digit, then the second, and so on until the fourt
  12 KB (1,896 words) - 23:55, 27 December 2023
• Complementary counting
  ...see that there are <math>19</math> of them. Thus, our answer is is <math>99-19 = 80</math>. <math>\square</math> ...2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is a 2 or a 3?''
  8 KB (1,192 words) - 17:20, 16 June 2023
• Composite number
  ...10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 35 36 38 39 40 42 44 45 46 48 49 50 51 52 54 55 56 57 58 60 62 63 64 65 66 68 69 70 72 74 75 76 77
  6 KB (350 words) - 12:58, 26 September 2023
• Multiple
  ...umber of multiples. As an example, some of the multiples of 15 are 15, 30, 45, 60, and 75.
  860 bytes (142 words) - 22:51, 26 January 2021
• Math Day at the Beach
  This round lasts 45 minutes and consists of 15 multiple-choice questions. Scoring consists of: This round lasts for 25 minutes and consists of 5 short-answer questions. Your score is 5 times the number of correct answers, for a
  4 KB (644 words) - 12:56, 29 March 2017
• 2011 AMC 10B Problems/Problem 22
  ...f the triangle is the diagonal of the pyramid's base. This is a <math>45-45-90</math> triangle. Also, we can let the dimensions of the rectangle be <mat ...because the rectangle is perpendicular to the base, and they share a <math>45^\circ</math> angle with the larger triangle. Therefore, the legs of the rig
  4 KB (691 words) - 18:38, 19 September 2021
• 2011 AMC 10B Problems/Problem 4
  LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, eac ...nardo half of the difference, which is <math>\boxed{\textbf{(C) } \;\frac{B-A}{2}}</math>
  1 KB (249 words) - 13:05, 24 January 2024
• Mathematical Olympiad Summer Program
  The '''Mathematical Olympiad Program''' (abbreviated '''MOP''') is a 3-week intensive problem solving camp held at the Carnegie Mellon University t The other important purpose of MOP is to train younger students in Olympiad-level problem solving and broaden their mathematical horizons.
  6 KB (936 words) - 10:37, 27 November 2023
• Right triangle
  ...ies. One of these is the [[isosceles triangle|isosceles]] <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle, where the hypotenuse is equal to <math> label("$45^{\circ}$", A, 6*dir(290));
  3 KB (499 words) - 23:41, 11 June 2022
• 2006 AIME I Problems/Problem 15
  Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possi ...2006\implies (k + 1)^2 = 2007</math> and <math>44^2 = 1936 < 2007 < 2025 = 45^2</math>. Plugging in <math>k = 44</math> yields <math>(3/2)(2025 - 2007) =
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 8
  ...B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); ...ath>44.5^2</math> is also less than <math>2006</math>, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5
  5 KB (730 words) - 15:05, 15 January 2024

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