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  • ...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

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