Create the page "Power mean" on this wiki! See also the search results found.

• 1988 AHSME Problems/Problem 27
  By Power of a Point, <math>AB\cdot BE=BF^2\implies AB\cdot CD=BF^2</math>, so <math>
  1 KB (261 words) - 22:22, 13 August 2023
• 1998 CEMC Gauss (Grade 7) Problems
  ...in zero. The product of any pair, <math>p</math> and <math>q,</math> is a power of 10 (that is, 10, 100, 1000, 10 000 , ...). If <math>p > q</math> , the l
  10 KB (1,590 words) - 16:43, 29 January 2021
• 2015 UNCO Math Contest II Problems
  ...4, \ldots </math> A polynomial is quadratic if its highest power term has power two.
  7 KB (1,151 words) - 15:11, 20 August 2020
• 2015 UNCO Math Contest II Problems/Problem 3
  ...+c=5x^4+7x^2+19</math>. By matching coefficients (the coefficients of each power of x on both sides must be equal), we derive the system <math>a=5</math>,<m
  682 bytes (121 words) - 10:03, 6 March 2024
• 2015 AMC 12A Problems/Problem 21
  ...of the ellipse <math>(\sqrt{15},0), (-\sqrt{15},0),</math> So we can apply power of a point to find the diameter. Let <math>x</math> denote the length of th
  5 KB (836 words) - 16:09, 20 June 2020
• 2015 AMC 12A Problems/Problem 23
  First two integrals easily done by power rule
  12 KB (1,981 words) - 18:33, 3 September 2023
• 2015 AMC 12A Problems/Problem 25
  ...th>S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)</math>, which is a power of <math>3.</math> Hence, our desired sum is <math>(1 + 1 + 3 + 9 + 27 + 81
  8 KB (1,340 words) - 14:24, 23 November 2023
• 2015 AMC 12B Problems/Problem 24
  Then the power of point <math>R</math> with respect to <math>C_1</math> is and the power of point <math>R</math> with respect to <math>C_2</math> is
  7 KB (1,188 words) - 19:18, 19 October 2023
• 2015 AIME II Problems/Problem 11
  Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The an Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that
  9 KB (1,249 words) - 21:59, 26 November 2023
• 2015 AIME II Problems/Problem 15
  ...distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and By [[Power of a Point Theorem|Power of a Point]], <math>CH\cdot HG=AH\cdot HE</math>. <math>AH=16/5\cdot 5\sqrt
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2016 AMC 10A Problems/Problem 22
  ...d the number of factors of <math>81n^4</math>, we raise this to the fourth power and multiply it by <math>81</math>, and find the factors of that number. W
  3 KB (429 words) - 19:27, 30 December 2022
• 2016 AMC 10A Problems/Problem 24
  ...sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\ove
  24 KB (3,861 words) - 14:17, 26 April 2024
• 2016 AMC 10A Problems/Problem 25
  ...ath> can be written as <math>p_1^ap_2^bp_3^c\dots</math>, then the highest power of all prime numbers <math>p_1,p_2,p_3\dots</math> must divide into either ...5,50</math> is uncovered by <math>x</math> and <math>y</math> both at full power capacity.
  7 KB (1,135 words) - 09:28, 5 September 2022
• 2015 USAMO Problems/Problem 2
  ...respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies th ...e circumradius of <math>\triangle AST</math> be <math>R</math>. Taking the power of <math>A</math> with respect to <math>(N_9)</math>, we get <cmath>AN_9^2
  5 KB (902 words) - 09:58, 20 August 2021
• 2015 USAJMO Problems/Problem 3
  ...respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies th
  4 KB (760 words) - 16:45, 29 April 2020
• 2015 IMO Problems/Problem 2
  is a power of 2. (<i>A power of 2 is an integer of the form <math>2^n</math> where <math>n</math> is a n
  3 KB (610 words) - 22:51, 16 October 2015
• 2015 AMC 8 Problems/Problem 22
  ...or <math>60 = 2^2 \cdot 3^1 \cdot 5^1</math>. Now, we just add one to the power of each factor. So since we have powers of <math>2</math>, <math>1</math>,
  2 KB (394 words) - 23:39, 23 January 2024
• 2016 AMC 12A Problems/Problem 24
  ...math>f(x)</math> so that the acceleration <math>f''(x)=0</math>. Using the power rule,
  3 KB (568 words) - 17:14, 20 April 2024
• 2016 AMC 10B Problems/Problem 8
  ...\pmod{100}</math>, which equals <math>225^{1008}</math>. Next, since every power (greater than <math>0</math>) of every number ending in <math>25</math> wil
  2 KB (366 words) - 12:46, 2 July 2023
• 2016 AMC 12B Problems/Problem 25
  Since the product <math>a_1a_2\cdots a_k</math> is an integer, it must be a power of <math>2</math>, so the sum of the base-<math>2</math> logarithms must be
  5 KB (916 words) - 15:07, 6 July 2024

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)