Convert 3.35 into a percent. Type in your number answer only. Ex

Mathematics

1 answer:

konstantin123 [22]3 years ago

7 0

3.35 *100 =335%
Hope it’s help

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315 - 4/15 equals what

yarga [219]

315 - 4/15 314 11/15

7 0

3 years ago

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Find the volume of the composite figure. Round to the nearest tenth. The base of each pyramid is a square with a side length of

nataly862011 [7]

Answer:

$75$ cubic inches

Step-by-step explanation:

Given:

The base of each pyramid is a square with a side length of 5 inches

Height of of pyramid is 9 inches.

To find: volume of the pyramid

Solution:

As the base of each pyramid is a square with a side length of 5 inches,

length of the pyramid (l) = base of the pyramid (b) = 5 inches

Height of of pyramid (h) = 9 inches

Volume of the pyramid $(V)=\frac{lbh}{3}$

On putting values of l, b and h, we get

$V=\frac{5\times 5\times 9}{3}=75$ cubic inches

5 0

3 years ago

Find the volume of the ellipsoid x^2+y^2+9z^2=64

yuradex [85]

Parameterize the ellipsoid using the augmented spherical coordinates:

$\begin{cases}x=\frac18\rho\cos\theta\sin\varphi\\\\y=\frac18\rho\sin\theta\sin\varphi\\\\z=\frac38\rho\cos\varphi\end{cases}$

Then the Jacobian for the change of coordinates is

$\mathbf J=\dfrac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}=\begin{bmatrix}\frac18\cos\theta\sin\varphi&-\frac18\rho\sin\theta\sin\varphi&\frac18\rho\cos\theta\cos\varphi\\\\\frac18\sin\theta\sin\varphi&\frac18\rho\cos\theta\sin\varphi&\frac18\rho\sin\theta\cos\varphi\\\frac38\cos\varphi&0&-\frac38\rho\sin\varphi\end{bmatrix}$

which has determinant

$\det\mathbf J=-\dfrac3{512}\rho^2\sin\varphi$

Then the volume of the ellipsoid is given by

$\displaystyle\iiint_E\mathrm dx\,\mathrm dy\,\mathrm dz=\iiint_E|\det\mathbf J|\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

where $E$ denotes the spaced contained by the ellipsoid. In particular, we have the definite integral and volume

$\displaystyle\frac3{512}\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\dfrac\pi{128}$