Search results
Create the page "(2b)v~(2b)" on this wiki! See also the search results found.
- .../math> bisects the angle between <math>\mathbf{u}</math> and <math>\mathbf{v}</math>. ...=x.</math> Therefore, we have <math>\frac{3b-a}{b-a} = -1,</math> so <math>2b=a.</math>16 KB (2,526 words) - 00:53, 6 May 2023
- <center><cmath>p(\hat{x})=p(a_0+a_1p)=p(a_0)+p'(a_0)a_1p+(a_1p)^2b</cmath></center> ...>, (2) <math>\hat{x}\equiv x\pmod{p^{n-k}}</math> and (3) <math>k=v(p'(x))=v(p'(\hat{x}))</math>.13 KB (2,298 words) - 23:34, 28 May 2023
- x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-47 KB (1,280 words) - 01:22, 6 February 2024
- x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-44 KB (684 words) - 01:55, 6 February 2024
- https://youtube.com/watch?v=LAuyU2OuVzE ...and <math>w=\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}</math>, so <cmath>\frac{v}{w}=\frac{b^2z_0}{c^2y_0},</cmath> which is the required condition for <mat14 KB (2,600 words) - 23:37, 10 March 2024
- https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel &=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{a^2+b^2}\right) \\12 KB (1,957 words) - 09:04, 18 May 2024
- ...frac{a+c}{a+2b+c} \implies \frac {B'D}{BD} = \frac {BB' - BD}{BD} = \frac{2b}{a+c} = 2\frac {B'I}{BI}.</cmath> The points <math>U \in AA''</math> and <math>V \in CC''</math> be such points that <math>L \in UV, UV \perp BB''.</math>14 KB (2,490 words) - 06:51, 6 June 2024
- DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. <cmath>\left\{\begin{array}{l} u = \sqrt {2} (x + y),\\v = 3y.\end{array}\right.</cmath>32 KB (5,375 words) - 11:43, 5 May 2024
