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- <cmath>x+\frac{b}{2a}=\pm\sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{\pm\sqrt{b^{2}-4ac}}{2a}</cmath> <cmath>x=-\frac{b}{2a}+\frac{\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}</cmath>2 KB (269 words) - 18:39, 10 December 2024
- <cmath>\sqrt[3]{4^2+\frac{5}{3}+\pi} \approx 2.75</cmath>.1 KB (181 words) - 15:53, 14 January 2025
- ...umber | composite]], then it has a prime [[divisor]] not exceeding <math>\sqrt n</math>.1 KB (212 words) - 20:16, 7 December 2007
- ...ex number itself. It has a [[magnitude]] of 1, and can be written as <math>1 \text{cis } \left(\frac{\pi}{2}\right)</math>. Any [[complex number]] can b #<math>i^1=\sqrt{-1}</math>2 KB (321 words) - 14:57, 5 September 2008
- * (Alternative definition) <math>|x| = \sqrt{x^2}</math> ...ex number]]s <math>z</math>, the absolute value is defined as <math>|z| = \sqrt{x^2+y^2}</math>, where <math>x</math> and <math>y</math> are the real and i2 KB (368 words) - 09:37, 5 January 2009
- ...<+\infty</math>, the [[inequality]] <math>\left|x-\frac pq\right|\ge \frac 1{q^M}</math> holds for all sufficiently large denominators <math>q</math>. Suppose that there exist <math>0<\beta<\gamma<1</math>, <math>1<Q<+\infty</math>8 KB (1,431 words) - 12:48, 26 January 2008
- ...setminus</math>||\setminus||<math>\wr</math>||\wr||<math>\sqrt{x}</math>||\sqrt{x} ...rc}||<math>\triangledown</math>||\triangledown||<math>\sqrt[n]{x}</math>||\sqrt[n]{x}16 KB (2,315 words) - 19:35, 4 November 2024
- \mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \mathrm{(C)}\ 8\sqrt {1003}2 KB (339 words) - 12:15, 12 July 2015
- ...he 30-60-90 triangle, or the Pythagorean Theorem, to find that <math> x = \sqrt{3} </math> units.2 KB (270 words) - 19:05, 16 January 2025
- <math>\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}</mat496 bytes (62 words) - 22:16, 24 February 2007
- ...xact opposite side of the cone whose distance from the vertex is <math>375\sqrt{2}.</math> Find the least distance that the fly could have crawled. ...<math>R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800</math>. Setting <math>\theta R=C\implies 800\theta=1200\pi\impl2 KB (268 words) - 21:20, 23 March 2023
- ...he sum of the solutions to the equation <math>\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}</math>? Let <math>y = \sqrt[4]{x}</math>. Then we have688 bytes (104 words) - 12:34, 22 July 2020
- C = (21*sqrt(3),0);1 KB (212 words) - 12:37, 11 December 2024
- ...ontsize(10)); label("$7$",A--C,2*dir(210),fontsize(10)); label("$18$",A--D,1.5*dir(30),fontsize(10)); label("$36$",(3,0),up,fontsize(10)); ...h>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}</math>, where <math>a</math>, <math>b</math>, and2 KB (376 words) - 12:49, 1 August 2022
- <cmath>d=\frac{-3a + \sqrt{9a^2+120a}}{2}.</cmath> <cmath>a=\frac{-120 + \sqrt{14400+36x^2}}{18}</cmath>5 KB (921 words) - 22:21, 22 January 2023
- ...qrt{b}+\sqrt{b+60}=\sqrt{c}</math>. Square both sides to get <math>2b+60+2\sqrt{b^2+60b}=c</math>. Thus, <math>b^2+60b</math> must be a square, so we have1 KB (218 words) - 13:14, 25 June 2021
- ...al numbers <math>a</math> and <math>b</math>, define <math>a \diamond b = \sqrt{a^2 + b^2}</math>. What is the value of ...\textbf{(B) } \frac{17}{2} \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13\sqrt{2} \qquad \textbf{(E) } 26</math>833 bytes (110 words) - 12:58, 24 July 2022
- ...formula above can be simplified with Heron's Formula, yielding <math>r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math> *The [[area]] of the [[triangle]] by [[Heron's Formula]] is <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>.2 KB (384 words) - 17:38, 9 March 2023
- ...a triangle with legs of length <math>a</math> and <math>b</math> is <math>\sqrt{a^2 + b^2}</math>.810 bytes (133 words) - 18:02, 15 October 2018
- <math>\sqrt{\frac{(y_2-y_1)^2}{m^2+1}}</math>2 KB (326 words) - 11:11, 21 May 2009