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White raven [17]
3 years ago
6

How do I solve 1/8x=5

Mathematics
2 answers:
Aneli [31]3 years ago
8 0
1/8x = 5

1/8x = 5/1

Take reciprocal of both sides

8x/1 = 1/5

8x = 1/5              Divide both sides by 8

x = 1/5 ÷ 8

x = 1/5 * 1/8

x = 1/40
gizmo_the_mogwai [7]3 years ago
7 0
You multiply both sides by the reciprocal (flip fraction) 
1/8--->8/1=8
So now multiply both sides by 8
1/8*8=0 (cancels out) 
x=5*8
x=40
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Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ
MrRissso [65]

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So,  the volume is:

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R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq  4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace

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V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4  \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}  \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\

we get:

V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

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