All categories

3 years ago

Solve for r , what is r equal to ?

Mathematics

1 answer:

Ivanshal [37]3 years ago

3 0

Answer:

$r= \sqrt[3]{\frac{3V}{4\pi}}$

Step-by-step explanation:

$V = \frac{4}{3} \pi r^3 \\\\3V = 4\pi r^3 \\\\r^3 = \frac{3V}{4\pi}\\\\\huge \purple {r= \sqrt[3]{\frac{3V}{4\pi}}}$

You might be interested in

Write the ratio as a fraction in lowest terms. Be sure to make necessary conversions:

agasfer [191]

Answer:12533

Step-by-step explanation:

4 0

2 years ago

Ashley went to HEB to buy the apples to make her famous apple pie. if she bought 22 pounds of apples for $1.43 per pound. How mu

PilotLPTM [1.2K]

You have to multiply the price by the quantity and you will find the total cost.

1.43 * 22 = 31.46

8 0

2 years ago

What is the domain and range of the set?<br> {(2,2.3),(5,2.3), (7,15)}

aleksandrvk [35]

Answer:

DOMAIN: 2, 5, 7

RANGE: 2.3, 2.3, 15

Step-by-step explanation:

Domain is always the first value of the set, and range is always the second value of the set. Hope it makes sense, and that it helped.

Have a good day! goodluck.

3 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

What is the measure of ABC and please show me how to do it

oksano4ka [1.4K]

Arc length from A to C (red line) is the given inside angle times 2, so AC = 50*2 = 100 degrees.

A full circle is 360 degrees, so arc ABC (blue line) would be 360 - 100 = 260 degrees.

The answer is D.

8 0

3 years ago