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• 1963 TMTA High School Algebra I Contest Problem 16
  The factors of <math>\frac{4}{9} + \frac{1}{3}x + \frac{3}{16}x^2</math> are ...x) \quad \text{(B)} \quad (2/9-x/4)(2-3x/4) \quad \text{(C)} \quad (4/9-3x/16)(1-x)</math>
  1 KB (190 words) - 10:12, 2 February 2021

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• ICTM Math Contest
  *2016 Regional - Saturday 2/27/16 *2016 State Finals - Saturday 5/7/16
  8 KB (1,182 words) - 14:26, 3 April 2024
• Chemistry competitions
  Nearly 16,000 U.S. students participate in Local Chemistry Olympiad competitions. Ove *''General Chemistry; Volumes I & II'' by Charles E.Mortimer
  2 KB (258 words) - 19:31, 8 March 2023
• AMC historical results
  ===AIME I=== *AIME Floor: 85.5 (top ~16%)
  17 KB (1,913 words) - 21:59, 26 May 2024
• Simon's Favorite Factoring Trick
  .... Take out the <math>x</math>: <math>x(y-5)-4(y-5)=0+20</math> (notice how I artificially grouped up the <math>y</math> terms by adding <math>4*5=20</ma Now, <math>(x-4)(y-5)=20</math> (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs t
  7 KB (1,130 words) - 18:48, 3 June 2024
• Factorial
  ...tive integer]]s as <math>n!=n \cdot (n-1) \cdots 2 \cdot 1 = \prod_{i=1}^n i</math>. Alternatively, a [[recursion|recursive definition]] for the factor * <math>16! = 20922789888000</math>
  10 KB (809 words) - 16:40, 17 March 2024
• Discriminant
  <math>D(p)=a_n^{2n-2}\prod_{i<j}^{n}(r_i-r_j)^2</math> ...)=a^2+16a-80=0</math>. The sum of these values of <math>a</math> is <math>-16</math>.
  4 KB (734 words) - 19:19, 10 October 2023
• Overcounting
  ...divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</math> numbers that are divisible by either 2 or 3. One person will handshake with all others except himself,i.e, he will handshake with <math>n-1</math> people.Now since the total of pe
  4 KB (635 words) - 12:19, 2 January 2022
• Constructive counting
  ''[[2001 AMC 12 Problems/Problem 16 | 2001 AMC 12 Problem 16]]: A spider has one sock and one shoe for each of its eight legs. In how ma ...st leg, the location of its sock and shoe's two boxes are given by <math>_{16} C_2</math>; but on permuting them, we know that the sock appears first in
  12 KB (1,896 words) - 23:55, 27 December 2023
• Complementary counting
  ''[[2002 AIME I Problems/Problem 1 | 2002 AIME I Problem 1]] Many states use a sequence of three letters followed by a seque ...AMC 12B Problems/Problem 22 | 2008 AMC 12B Problem 22]]: A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking spa
  8 KB (1,192 words) - 17:20, 16 June 2023
• Ellipse
  *The ellipse with axis lengths <math>14</math> and <math>16</math> has the general equation of <math>\frac{x^2}{a^2}+\frac{y^2}{b^2}=1< ...re relatively prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])
  5 KB (892 words) - 21:52, 1 May 2021
• Fibonacci sequence
  ...e [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <div style="text-align:right">([[1990 AIME Problems/Problem ...r]]s such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>. <div style="text-align:right">([[1988 AIME Problems/Problem 13
  6 KB (957 words) - 23:49, 7 March 2024
• Power of a Point Theorem
  ''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions. ...nd intersect at a right angle at point <math>E</math>. Given that <math>BE=16</math>, <math>DE=4</math>, and <math>AD=5</math>, find <math>CE</math>.
  5 KB (827 words) - 17:52, 12 June 2024
• Mock AMC
  [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9575 16-20] ! scope="row" | '''Mock AMC I'''
  51 KB (6,175 words) - 20:58, 6 December 2023
• Modular arithmetic/Introduction
  for (int i = 0; i < 12; ++i) label (pic, (string) i, b);
  15 KB (2,396 words) - 20:24, 21 February 2024
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} [[2006 AIME I Problems/Problem 1|Solution]]
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  Suppose <math>b_{i} = \frac {x_{i}}3</math>. ...= \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 13
  ...s]] <math> x. </math> For example, <math> g(20)=4 </math> and <math> g(16)=16. </math> For each positive integer <math> n, </math> let <math> S_n=\sum_{k ...rac{n+1}{16} \le 62</math>, so <math>n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2</math>.
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 5
  <cmath>2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5</cmath> {{AIME box|year=2006|n=I|num-b=4|num-a=6}}
  3 KB (439 words) - 18:24, 10 March 2015
• 2022 AIME I Problems/Problem 1
  ...ctively. The graphs of both polynomials pass through the two points <math>(16,54)</math> and <math>(20,53).</math> Find <math>P(0) + Q(0).</math> R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\
  4 KB (670 words) - 13:03, 13 November 2023
• 2021 JMC 10
  ...ts, P_8</math> stand in a line, and person <math>P_i</math> calls <math>P_{i+1}</math> a liar where <math>P_1 = P_9.</math> Out of these eight people, h ...{(B) } 9 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 16</math>
  12 KB (1,784 words) - 16:49, 1 April 2021

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