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- By the [[Power of a Point Theorem]] on <math>E</math>, we get <math>EF = \frac{12^2}{27} = Using Power of a Point, we have6 KB (974 words) - 13:01, 29 September 2023
- ==Video Solution 1 (easy to digest) by Power Solve==2 KB (384 words) - 22:57, 17 February 2024
- ====4.1.4: Using the JCF to calculate the transition matrix to the power of any n, large or small==== ...then we must find the sum of the coefficients that share a variable with a power divisible by <math>3</math>.15 KB (2,406 words) - 23:56, 23 November 2023
- ...\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by [[power of a point]]. Using the fact that the ratio of corresponding sides in simil4 KB (658 words) - 19:15, 19 December 2021
- ...proach is to consider the graph of <math>f(x)</math>, which iterates every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math3 KB (545 words) - 23:41, 14 June 2023
- ...(or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divid ...can see that the number of zeros in a term more or less correlates to the power of <math>x^2</math>. Thus, we let <math>y=10x^2</math>. The equation then b6 KB (1,060 words) - 17:36, 26 April 2024
- ...her is (<math>6 = 2 \cdot 3</math>) a prime raised to the <math>5</math>th power, or two primes, one of which is squared. The smallest example of the former ...e now divide all of the odd factors from <math>n</math>; then we require a power of <math>2</math> with <math>\frac{18}{6} = 3</math> factors, namely <math>2 KB (397 words) - 15:55, 11 May 2022
- ...> or to <math>n+2^{m_n+1}</math> where <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge 2</math> i ...made from a number that is divisible by <math>2^M</math> (and by no higher power of 2). Thus we must have <math>M < i_0</math>, since otherwise a number div7 KB (1,280 words) - 17:23, 26 March 2016
- <math>z</math> if <math>f</math> has a convergent [[power series]] expansion on some its power series diverges when <math>\lvert x \rvert > 1</math>. But in the9 KB (1,537 words) - 21:04, 26 July 2017
- ...f Wayzata HS (1991-2006), founder and long-time problem writer of the ARML Power Contest, and Mike Reiners of Christ's Household of Faith School (2007-14),4 KB (680 words) - 16:45, 10 June 2015
- ...f [[ARML]], with 15-member teams competing in Individual, Team, Relay, and Power rounds, although the scoring is slightly different. Teams compete in A and1 KB (201 words) - 14:39, 25 June 2023
- ...est positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer.713 bytes (114 words) - 01:45, 19 August 2012
- ...h> is cyclic, since <math>\angle AEB=\angle AFB</math>. We then have, from Power of a Point, that <math>CE\cdot CA=CF\cdot CB</math>. In other words, <math> As above, use Power of a Point to compute <math>CF=2</math> and <math>FB=1</math>. Since triang3 KB (518 words) - 16:54, 25 November 2015
- Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}2 KB (311 words) - 10:53, 4 April 2012
- ** [[Arithmetic Mean-Geometric Mean | Arithmetic Mean-Geometric Mean Inequality]] ** [[Power Mean Inequality]]1 KB (100 words) - 16:22, 18 May 2021
- ...est positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer.3 KB (560 words) - 19:23, 10 March 2015
- ...] of the set (also known as the [[average]]). For example, the arithmetic mean of the members of the set {3, 5, 10} is ==Mean, Median, Mode==1 KB (148 words) - 15:28, 12 November 2023
- ...the circle at <math>P.</math> Let the radius be <math>r.</math> Applying [[power of a point]],680 bytes (114 words) - 21:38, 9 July 2019
- ...n</math> elements and let <math>\displaystyle \mathcal P (M)</math> be its power set. Find all functions <math>\displaystyle f : \mathcal P (M) \to \{ 0,1,211 KB (1,779 words) - 14:57, 7 May 2012
- * [[Power of a point]]1 KB (122 words) - 16:25, 18 May 2021