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- ...ation of Adobe Flash player which was required on the old FTW. For the win beta will not include countdown rounds and a chat box but there are some new fea For The Win beta was released on January 4, 2021.823 bytes (146 words) - 09:49, 17 August 2021
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- * [[Phi Beta Kappa High School Award]] - ask your high school counselor or career center1 KB (197 words) - 14:00, 16 July 2010
- Let <math>\alpha</math>, <math>\beta</math> be cuts We say that <math>\alpha<\beta</math> iff <math>\alpha\subset\beta</math>3 KB (496 words) - 22:22, 5 January 2022
- Substituting <math>\alpha = x+y</math> and <math>\beta = x-y</math> into the product-to-sum identities yields the sum-to-product i8 KB (1,397 words) - 20:55, 20 January 2024
- Let <math>\alpha,\beta,\gamma,...,\omega</math> be the roots of a given polynomial <math>P(x)=a_nx <math>P(\alpha)=P(\beta)=P(\gamma)=...=P(\omega)=0</math>4 KB (690 words) - 12:11, 20 February 2024
- ...and <math>|qx-(\tilde\beta P-\tilde\alpha v)|\le\tilde\alpha|ux+v|+\tilde\beta|Qx-P|\le ...eta}Q\le \frac {3a}Q\le \frac {6a^2}q</math>. Thus, setting <math>p=\tilde\beta P-\tilde\alpha v</math>, we get <math>\left|x-\frac pq\right|<\frac {6a^2}{7 KB (1,290 words) - 11:18, 30 May 2019
- Suppose that there exist <math>0<\beta<\gamma<1</math>, <math>1<Q<+\infty</math> ...ft|P_n-Q_n x\right|<\gamma^n</math>. Then, for every <math>M>\frac{\log(Q/\beta)}{\log(1/\gamma)}</math>, the inequality <math>\left|x-\frac pq\right|<\fra8 KB (1,431 words) - 12:48, 26 January 2008
- ...ath>a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q</math>. [[Young's Inequality]] gives us4 KB (774 words) - 11:12, 29 October 2016
- | beta51 KB (6,135 words) - 05:44, 2 November 2024
- ** [https://artofproblemsolving.com/community/c5h63721p380055 beta Mock AIME]7 KB (889 words) - 06:52, 24 October 2024
- ...ation on <math>\pi_1(X)</math> given by <math>(\alpha\beta)(t) = \alpha(t)\beta(t)</math>, which is "compatible" with the concatenation in the following re3 KB (479 words) - 14:35, 1 December 2015
- |<math>\alpha</math>||\alpha||<math>\beta</math>||\beta||<math>\gamma</math>||\gamma||<math>\delta</math>||\delta16 KB (2,315 words) - 19:35, 4 November 2024
- Start with <math>e^{i(\alpha + \beta)} = (e^{i\alpha})(e^{i\beta})</math>, and apply Euler's forumla both sides: ...a) + i \sin(\alpha + \beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta).</math>3 KB (452 words) - 22:17, 4 January 2021
- | <math>\beta</math> | Beta1 KB (216 words) - 09:58, 12 July 2006
- ...s ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is <math> m/n, </math> where <math> m </math> and <math> n </ Let <math>q</math> be the number of questions Beta takes on day 1 and <math>a</math> be the number he gets right. Let <math>b<3 KB (436 words) - 17:31, 9 January 2024
- ...500} = \frac{3}{5}</math>. The largest possible two-day success ratio that Beta could achieve is <math> m/n, </math> where <math> m </math> and <math> n </9 KB (1,434 words) - 12:34, 29 December 2021
- ...ath> be the three sides of a triangle, and let <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math>, be the angles opposite them. If <math>a^2+b^2= <center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center>7 KB (1,045 words) - 19:47, 14 December 2023
- ...he rule: angle of incidence equals angle of reflection. Given that <math>\beta=\alpha/10=1.994^\circ\,</math> and <math>AB=BC,\,</math> determine the numb7 KB (1,141 words) - 06:37, 7 September 2018
- ...ath>\alpha + \beta</math> [[radian]]s, respectively, where <math>\alpha + \beta < \pi</math>. If <math>\cos \alpha</math>, which is a [[positive]] [[ration label("\(\alpha+\beta\)",(0.08,0.08),NE,fontsize(8));5 KB (763 words) - 15:20, 28 September 2019
- label("$\beta$", NN, 2*dir(.5*degrees(B-NN)));</asy></center> ...<math>AC = b</math>; let <math>\alpha = \angle MAC</math>, and let <math>\beta = \angle BNC</math>.11 KB (1,722 words) - 08:49, 13 September 2023
- ...ta}{1 - \tan \alpha \cdot \tan \beta}</math>, we get <math>\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}</math>.1 KB (178 words) - 22:25, 20 November 2023