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Create the page "Bot 2" on this wiki! See also the search results found.
- |<math>\bot</math>||\bot||<math>\vdash</math>||\vdash||<math>\dashv</math>||\dashv :<tt>(\frac{a}{x} )^2</tt>16 KB (2,315 words) - 19:35, 4 November 2024
- ...ath>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math> Then, <math>A_1=\frac{|AD| \times h_1}{2}</math>, thus <math>h_1=\frac{2}{3}A_1</math>3 KB (563 words) - 01:05, 25 November 2023
- Point <math>Q = OB \cap MQ</math>, <math>MQ \bot OB</math>. Solving: <math>x_{Q}=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right]</math>.8 KB (1,426 words) - 18:05, 10 November 2024
- M = (B + C)/2; ...(with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.6 KB (899 words) - 21:04, 10 November 2024
- ...l <math>AGLJ</math> is cyclic because <math>\angle JAL = \frac{\angle BAC}{2} = \angle YGM = \angle YGL = \angle JGL</math>. Again, <math>\triangle AJL ==Solution 2==7 KB (1,189 words) - 00:22, 19 November 2023
- ...e between the [[midpoint]]s of the two chords is 12. The quantity <math>OP^2</math> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> ..., <math>OE = \sqrt{25^2 - 15^2} = 20</math>, and <math>OF = \sqrt{25^2 - 7^2} = 24</math>.13 KB (1,989 words) - 15:35, 24 November 2024
- <math> \text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9 ==Problem 2==20 KB (3,108 words) - 13:14, 20 February 2020
- The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line <math>AKL</math> be <math>k</math> Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get5 KB (823 words) - 21:42, 15 June 2024
- ...ne{CB}</math>, so that they are mutually tangent. If <math>\overline{CD} \bot \overline{AB}</math>, then the ratio of the shaded area to the area of a ci <math>\textbf{(A)}\ 1:2\qquad2 KB (232 words) - 00:40, 16 August 2023
- real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),R=(c/2,-sqrt(25-(c/2)^2));8 KB (1,480 words) - 13:52, 5 August 2022
- <math>\angle POE = 2\alpha</math> and <math>\angle QOF = 2\beta</math> Now, <math>\angle POQ = 180 - 2(\alpha + \beta)</math>, and since <math>PAQO</math> is a cyclic quadrilater3 KB (481 words) - 09:18, 11 October 2024
- ...hat <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relativel O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);13 KB (2,252 words) - 10:32, 1 February 2024
- ...son DeleteSticky was disqualified, although it is likely DeleteSticky is a bot or a glitch in the system. 2 aaravdodhia11 KB (1,044 words) - 19:27, 21 September 2024
- ==Solution 2== ...math>y+x+\angle GBE=\frac{\pi}{2}</math>, or <math>\angle GBE = \frac{\pi}{2}-x-y</math>.6 KB (1,131 words) - 18:15, 6 October 2023
- ...math> is located at <math>(1+1, 5+4) = (2,9)</math>. The answer is <math>9-2 = 7 = \boxed{\textbf{(D)} ~7}</math>. ==Solution 2 (Complex Numbers)==7 KB (1,159 words) - 22:28, 9 November 2024
- ...angle BCA, </math> and <math>\angle XOY</math> are in the ratio <math>13 : 2 : 17, </math> the degree measure of <math>\angle BAC</math> can be written &= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theo10 KB (1,526 words) - 15:50, 25 December 2022
- ...,0.1)..(-0.8,-0.03)..(-0.4,-0.11)..(0,-0.15)..(0.4,-0.11)..(0.8,-0.03)..(1.2,0.1)); draw((0,2.4)--(0,-0.15));5 KB (890 words) - 19:31, 22 October 2024
- real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26);19 KB (3,240 words) - 16:09, 13 October 2024
