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• ...lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$. ...h of its endpoints, the number of edges $E$ is $\frac{3}{2}V = 72$.
5 KB (811 words) - 18:10, 25 January 2021
• Finally, we substitute $h$ into the volume equation to find $V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}$. ...ave the base area as $18\sqrt {133}$. Thus, the volume is $V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 5 7 KB (1,085 words) - 20:56, 28 December 2021 • ...th>(u,v)$ and $(p,q)$, then $u=2r-p$ and $v=2s-q$. So we start with the point they gave us and work backwards. We
4 KB (611 words) - 10:31, 23 August 2020
• ...$P$ pentagonal faces meet. What is the value of $100P+10T+V$? ...ge). Thus, $E=60$. Finally, using Euler's formula we have $V=E-30=30$.
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 $\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta$, we will be able to solve for $\cos \thet <cmath>\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc} 8 KB (1,172 words) - 13:34, 27 October 2021 • ...c{m}{n}$ be the [[probability]] that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$, where $m$ and $n$ are [[relatively p Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$, and le
4 KB (714 words) - 13:22, 14 October 2021
• ...coordinates of the vertex of the resulting pyramid. Call this point $V$. Clearly, the height of the pyramid is $z$. The desired v ...= QC[/itex]. We then use distance formula to find the distances from $V$ 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 $V = \frac{\pi}{3}r^2h$. In the second container, if we let $h',r'< From the formula [itex]V=\frac{\pi r^2h}{3}$, we can find that the volume of the container is
3 KB (544 words) - 21:20, 30 July 2017
• ...rea of [[pentagon]] $ABCDE$ is $451$. Find $u + v$. 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[/itex] perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$ ...r each pyramid(base times height divided by 3) we have $\dfrac{rF}{3}=V$. The surface area of the pyramid is $\dfrac{6\cdot{4}+6\cdot{2} 6 KB (937 words) - 16:34, 26 December 2021 • ...line{CA}$ and $\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN</mat 3 KB (585 words) - 10:12, 16 March 2016 • ...another identical wedge and sticking it to the existing one). Thus, [itex]V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$, so $n=\boxed{216}$.
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}[/itex] with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ 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
• $\int u\, dv=uv-\int v\,du$ ...math>u[/itex] will show up as $du$ and $dv$ as $v$ 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 $u, v : \mathbb{R \times R \to R}$ 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 $V$ to this path? MP("P",(-1,0),W);MP("V",(-.5,2.4),N);
3 KB (560 words) - 18:23, 10 March 2015
• | $\left(u(x)\times v(x)\right)'=u(x)v'(x)+u'(x)v(x)$ | $\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$
3 KB (504 words) - 18:23, 3 March 2010

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