The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. (Note: The volume of a sphere with radius r is V=4/3 pi r³.) At the time when the radius of the sphere is 10 centimeters, what is the rate of increase of its volume?

The rate of the volume increase will be

Step-by-step explanation:

Let's take the derivative with respect to time on each side of the volume equation.

Now, we just need to put all the values on the rate equation.

We know that:

dR/dt is 0.04 cm/s  

And we need to know what is dV/dt when R = 10 cm.

Therefore using the equation of the volume rate:

I hope it helps you!

your answer is 23/2 - 13.5

step-by-step explanation:

7/2 evaluates to

multiply x and 4.5

multiply x and 1

the x just gets copied along.

the answer is x

x

4.5*x evaluates to 4.5x

multiply 3 and 4.5x

multiply 1 and x

the x just gets copied along.

x

4.5*x*3 evaluates to 13.5x

7/2-4.5*x*3 evaluates to

  + 8 =

the answer is

7/2-4.5*x*3+8 evaluates to

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.