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  • <math>\delta x'(t)=v(t)</math> <math>v'(t)=a(t)</math>
    9 KB (1,355 words) - 06:29, 29 September 2021
  • Case V) <math>a+b=5c\Rightarrow (5a-1)(5b-1)=126</math> for which there are 2 solu
    2 KB (332 words) - 08:37, 30 December 2021
  • ...{R} </math> be an [[inner product]]. Then for any <math> \mathbf{a,b} \in V </math>,
    11 KB (1,952 words) - 15:38, 29 December 2021
  • ...system at all, used certain letters to represent certain values (e.g. I=1, V=5, X=10, L=50, C=100, D=500, M=1000). Imagine how difficult it would be to
    4 KB (547 words) - 16:23, 30 December 2020
  • ...\,\,y\,\,z\,\,...)</math>. The magnitude of a vector, denoted <math>\|\vec{v}\|</math>, is found simply by ...d by them, <math>\|\vec{v}+\vec{w}\|^2=\|\vec{v}\|^2+\|\vec{w}\|^2+2\|\vec{v}\|\|\vec{w}\|\cos\theta</math>.
    7 KB (1,265 words) - 12:22, 14 July 2021
  • ...aQ</math> and <math>|qx-(\tilde\beta P-\tilde\alpha v)|\le\tilde\alpha|ux+v|+\tilde\beta|Qx-P|\le ...\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}{q^2}</math>.
    7 KB (1,290 words) - 11:18, 30 May 2019
  • ...and let <math>I</math> be a [[prime ideal]] of <math>R</math>. Then <math>V(I)=\{p\in\mathbb{A}^n\mid f(p)=0\mathrm{\ for\ all\ } f\in I\}</math> is ca
    2 KB (361 words) - 00:59, 24 January 2020
  • ...of [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>.
    970 bytes (132 words) - 21:36, 1 February 2021
  • ! scope="row" | '''Mock AMC V'''
    58 KB (7,011 words) - 21:23, 25 January 2022
  • Let <math>U=2\cdot 2004^{2005}</math>, <math>V=2004^{2005}</math>, <math>W=2003\cdot 2004^{2004}</math>, <math>X=2\cdot 20 <math>\text{(A) } U-V \qquad \text{(B) } V-W \qquad \text{(C) } W-X \qquad \text{(D) } X-Y \qquad \text{(E) } Y-Z \qqu
    13 KB (1,953 words) - 21:24, 22 November 2021
  • ...ngles of a pentagon. Suppose that <math>v < w < x < y < z</math> and <math>v, w, x, y, </math> and <math>z</math> form an arithmetic sequence. Find the
    10 KB (1,548 words) - 12:06, 19 February 2020
  • Our original solid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surf Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where
    5 KB (839 words) - 21:12, 16 December 2015
  • ...>P^{}_{}</math> pentagonal faces meet. What is the value of <math>100P+10T+V\,</math>?
    8 KB (1,275 words) - 05:55, 2 September 2021
  • .... Let <math>m/n</math> be the probability that <math>\sqrt{2+\sqrt{3}}\le |v+w|</math>, where <math>m</math> and <math>n</math> are relatively prime pos
    7 KB (1,098 words) - 16:08, 25 June 2020
  • ...he area of pentagon <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
    7 KB (1,208 words) - 18:16, 2 January 2022
  • ...ine{UV}</math> with <math>U</math> on <math>\overline{PQ}</math> and <math>V</math> on <math>\overline{QR}</math> such that <math>\overline{UV}</math> i
    8 KB (1,282 words) - 20:12, 19 February 2019
  • ...Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahed <math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}<
    5 KB (865 words) - 09:17, 20 January 2021
  • ...rom vertex <math>V</math> and ending at vertex <math>A,</math> where <math>V\in\{A,B,C,D\}</math> and <math>k</math> is a positive integer. We wish to f ...math>V</math> to <math>A</math> and the paths from <math>A</math> to <math>V</math> have one-to-one correspondence. So, we must get <cmath>A_k+B_k+C_k+D
    12 KB (2,017 words) - 17:17, 26 January 2022
  • ...th>h = 15</math>, <math>l = 5</math>, <math>w = 10</math>. Therefore <math>V = 5 \cdot 10 \cdot 15 = \boxed{750}</math>
    2 KB (346 words) - 12:13, 22 July 2020
  • ...(x)</math> are also roots of <math>f(x)</math>. Let these roots be <math>u,v</math>. We get the system If we multiply the first equation by <math>v^{16}</math> and the second by <math>u^{16}</math> we get <cmath>\begin{alig
    8 KB (1,350 words) - 13:13, 17 September 2021
  • ...lies on exactly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>. ...h of its endpoints, the number of edges <math>E</math> is <math>\frac{3}{2}V = 72</math>.
    5 KB (811 words) - 18:10, 25 January 2021
  • Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>. ...ave the base area as <math>18\sqrt {133}</math>. Thus, the volume is <math>V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 5
    7 KB (1,085 words) - 20:56, 28 December 2021
  • ...th>(u,v)</math> and <math>(p,q)</math>, then <math>u=2r-p</math> and <math>v=2s-q</math>. So we start with the point they gave us and work backwards. We
    4 KB (611 words) - 10:31, 23 August 2020
  • ...<math>P</math> pentagonal faces meet. What is the value of <math>100P+10T+V</math>? ...ge). Thus, <math>E=60</math>. Finally, using Euler's formula we have <math>V=E-30=30</math>.
    4 KB (623 words) - 19:32, 15 February 2021
  • ...(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} ...(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);}
    4 KB (721 words) - 15:14, 8 March 2021
  • ...as <math>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</math>, we will be able to solve for <math>\cos \thet <cmath>\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}
    8 KB (1,172 words) - 13:34, 27 October 2021
  • ...c{m}{n}</math> be the [[probability]] that <math>\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>m</math> and <math>n</math> are [[relatively p Now, let <math>v</math> be the root corresponding to <math>m\theta=2m\pi/1997</math>, and le
    4 KB (714 words) - 13:22, 14 October 2021
  • ...coordinates of the vertex of the resulting pyramid. Call this point <math>V</math>. Clearly, the height of the pyramid is <math>z</math>. The desired v ...= QC</math>. We then use distance formula to find the distances from <math>V</math> to each of the vertices of the medial triangle. We thus arrive at a
    5 KB (805 words) - 21:34, 28 May 2021
  • (Computational) The volume of a cone can be found by <math>V = \frac{\pi}{3}r^2h</math>. In the second container, if we let <math>h',r'< From the formula <math>V=\frac{\pi r^2h}{3}</math>, we can find that the volume of the container is
    3 KB (544 words) - 21:20, 30 July 2017
  • ...rea of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>. D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cy
    3 KB (434 words) - 21:43, 16 May 2021
  • ...ath>P</math> perpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math> ...r each pyramid(base times height divided by 3) we have <math>\dfrac{rF}{3}=V</math>. The surface area of the pyramid is <math>\dfrac{6\cdot{4}+6\cdot{2}
    6 KB (937 words) - 16:34, 26 December 2021
  • ...line{CA}</math> and <math>\overline{AB}</math>, respectively. Let <math>U,V</math> be the intersections of line <math>EF</math> with line <math>MN</mat
    3 KB (585 words) - 10:12, 16 March 2016
  • ...another identical wedge and sticking it to the existing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=216\pi</math>, so <math>n=\boxed{216}</math>.
    941 bytes (159 words) - 02:39, 6 December 2019
  • triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); ...-U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);
    4 KB (518 words) - 14:01, 31 December 2021
  • ...ine{UV}</math> with <math>U</math> on <math>\overline{PQ}</math> and <math>V</math> on <math>\overline{QR}</math> such that <math>\overline{UV}</math> i pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10);
    6 KB (896 words) - 09:13, 22 May 2020
  • <math>\int u\, dv=uv-\int v\,du</math> ...math>u</math> will show up as <math>du</math> and <math>dv</math> as <math>v</math> in the integral on the RHS, u should be chosen such that it has an "
    1 KB (231 words) - 15:19, 18 May 2021
  • Specifically, let <math>u, v : \mathbb{R \times R \to R}</math> be definted <cmath> u(x,y) = \text{Re}\,f(x+iy), \qquad v(x,y) = \text{Im}\,f(x+iy) . </cmath>
    9 KB (1,537 words) - 20:04, 26 July 2017
  • https://www.youtube.com/watch?v=BBD66Q3KXuI ...enter connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full
    4 KB (719 words) - 18:41, 25 November 2020
  • the vertex <math>V</math> to this path? MP("P",(-1,0),W);MP("V",(-.5,2.4),N);
    3 KB (560 words) - 18:23, 10 March 2015
  • | <math>\left(u(x)\times v(x)\right)'=u(x)v'(x)+u'(x)v(x)</math> | <math>\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}</math>
    3 KB (504 words) - 18:23, 3 March 2010
  • *Given a weighted, undirected graph <math>G = (V,E)</math> and two vertices <math>s, t \in E</math>, does there exist a path
    6 KB (1,104 words) - 14:11, 25 October 2017
  • ...for which <cmath>\left\vert\sum_{j=m+1}^n(a_j-(v+1))\right\vert\le (T-v)\,v \le \left(\frac T 2\right)^2</cmath>for all integers <math>m</math> and <ma ...is:<cmath>\sum_{i=1}^v (T-v+i) - \sum_{i=1}^v i=\sum_{i=1}^v (T-v)=(T-v)\,v\;. \quad \blacksquare</cmath>
    4 KB (833 words) - 00:33, 31 December 2019
  • ...s had to make change on a purchase of LXIV dollars with bills marked L, X, V and I when handed XC dollars.
    2 KB (365 words) - 19:42, 20 February 2019
  • ...group over a set <math>I</math> and <math>uv = vu</math>, then <math>u^m = v^n</math>, for some [[integer]]s <math>m</math> and <math>n</math>.
    2 KB (454 words) - 16:54, 16 March 2012
  • \text{(V) } 2007 \quad </math> <math>\text{(V) Ying} \quad
    33 KB (5,143 words) - 19:49, 28 December 2021
  • ...ng subset. Hence <cmath>F(n,r)=\frac{1}{\binom{n}{r}} \sum_{v \in B} \deg (v)= \frac{n+1}{r+1}.</cmath>
    5 KB (879 words) - 10:18, 27 June 2020
  • 5 - '''V''' ''(quinque)''
    865 bytes (140 words) - 12:58, 24 March 2019
  • ..., then the [[greatest common factor]] of <math>2^u + 1 </math> and <math>2^v + 1 </math> is 3. ...ath>t </math>, contradicting our assumption that <math>u </math> and <math>v </math> are relatively prime.
    10 KB (1,739 words) - 05:38, 12 November 2019
  • label("V", (2, 6), NE);
    13 KB (1,968 words) - 14:16, 21 October 2021
  • ...ale the triangle with the circumradius by a [[line]]ar scale factor, <math>v</math>. :<math>\frac{65}{8}v=8u</math>
    8 KB (1,321 words) - 11:38, 15 January 2022

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