What's wrong with the problem: 5pints times 1cup divided by 2 pints = 2.5cups

Mathematics

1 answer:

lys-0071 [83]3 years ago

3 0

I don't really know how the table works, but there are two cups in a pint, so 5 pints * 0.5 pints would be 2.5 pints, and 2.5 pints divided by 2 pints would be 1.25 pints?

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Find the volume of a cone of radius 49cm and height 10cm

den301095 [7]

Answer:

Step-by-step explanation:

Given,

Radius ( r ) = 49 cm

Height (h) = 10 cm

Volume of cone=?

Now,

Applying the formula to find the volume of cone:

$\pi \: {r}^{2} \frac{h}{3}$

Plugging the values,

$\frac{22}{7} \times {49}^{2} \times \frac{10}{3}$

$\frac{22}{7} \times 2401 \times \frac{10}{3}$

Calculate

$25153.34 \: {cm}^{3}$

Hope this helps...

Good luck on your assignment..

3 0

2 years ago

Read 2 more answers

Use spherical coordinates. evaluate (9 − x2 − y2) dv, where h is the solid hemisphere x2 + y2 + z2 ≤ 4, z ≥ 0.

avanturin [10]

In spherical coordinates, we have

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

which gives volume element

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

and so the triple integral is given by

$\displaystyle\iiint_H(9-x^2-y^2)\,\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9-\rho^2\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$
$=\displaystyle2\pi\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9\rho^2\sin\varphi-\rho^4\sin^3\varphi)\,\mathrm d\rho\,\mathrm d\varphi$
$=\dfrac{592\pi}{15}$