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3 years ago

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The volume, V, of a cone is calculated using the formula V 1 Tr?h, where r is the radius of the base and h 3 is the height of th

e cone. Rearrange the formula to solve for the radius (r).

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer

r = 20.25

this is the awnser i know im only in 6th grade but im good at math

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The spinner below is divided into 5 equal sections. What is the probability of landing on a red or green section?

Artyom0805 [142]

Fraction: 2/5

Percentage: 40%

Decimal: 0.4

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3 years ago

Please look at the image below

Ierofanga [76]

Answer:

third and last choices

DPR and RPD

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3 years ago

- 5.8c + 4.2 - 3.1 + 1.4c − 5.8c + 4.2 − 3.1 + 1.4c

Cerrena [4.2K]

Answer:

-8.8c + 2.2 OR SOLVED c = 0.25

Step-by-step explanation:

<u>Step 1: Combine like terms</u>

- 5.8c + 4.2 - 3.1 + 1.4c − 5.8c + 4.2 − 3.1 + 1.4c

<em>-8.8c + 2.2</em>

<em />

<u>Step 2: If needed solve for c</u>

-8.8c + 2.2 - 2.2 = 0 - 2.2

-8.8c / -8.8 = -2.2 / -8.8

<em>c = 0.25</em>

<em />

Answer: -8.8c + 2.2 OR SOLVED c = 0.25

8 0

3 years ago

Read 2 more answers

If a roasting is 600mm x 500mm how many 25cm cake time will fit on it?

nydimaria [60]

So assuming that 25cm is 25 square cm because you cannot have a 2 dimentional cake so

600mm by 500mm
to find the area, multiply legnth by width

mm=milimeters
cm=centimeters
this is in mm so
1cm=10mm
1mm=1/10cm
to convert from mm to cm, you divide by 10
we must divide by 10 first
legnth =600mm
600/10=60cm
legnth=60cm

width=500mm
500/10=50cm
width=50cm

area=60 times 50=3000 square cm or cm^2
so 3000 is the area
25 is the area of each cake
divide 3000 into 25 and find how many cakes will fit

3000/25=120

120 cakes will fit

3 0

3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

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3 years ago