The volume of a sphere is increasing at a constant rate of 141 cubic feet per minute. at the instant when the radius of the sphere is 11 feet, what is the rate of change of the radius? the volume of a sphere can be found with the equation v=4/3pi r^3 . round your answer to three decimal places.

0.093 ft/min

Step-by-step explanation:

V = 4/3 π r³

Take derivative with respect to time:

dV/dt = 4π r² dr/dt

Plug in values:

141 = 4π (11)² dr/dt

dr/dt = 141 / (484π)

dr/dt ≈ 0.093

The radius is increasing at 0.093 ft/min.

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !