The radius r of a sphere is increasing at a rate of 7 inches per minute. (a) find the rate of change of the volume when r = 10 inches.

We need to find the rate of change of the volume of a sphere whose radius increases at 7 inches per minute.

The rate of change of the volume when r = 10 inches is 8792 in^3/min

The first thing we can define is:

r = 10 in + (7 in/min)*x

Where r is the radius of the sphere, and x is the number of minutes.

The volume of the sphere will be:

V = (4/3)*3.14*r^3

Let's write this as a function of x:

V = (4/3)*3.14*( 10 in + (7 in/min)*x)^3

Now we need to differentiate it with respect to x, we will get:

V' = [3*(4/3)*3.14*( 10 in + (7 in/min)*x)^2 ]*(7 in/min)

V' = (7 in/min)*4*3.14*(10 in + (7 in/min)*x)^2

Now we need to evaluate this in r = 10 inches, remember that this happens for x = 0, then we have:

V'(0) = (7 in/min)*4*3.14*(10 in + (7 in/min)*0)^2

        =  (7 in/min)*4*3.14*(10 in)^2 =  8792 in^3/min

The rate of change of the volume when r = 10 inches is 8792 in^3/min

If you want to learn more, you can read:

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To find rate of change, differentiate with respect to time. Then fill in the given values and evaluate.

The rate of change of volume is 2800π in³/min ≈ 8796 in³/min.

answer: the 4 kilogram bag

step-by-step explanation: this is because if you divide the number of kilograms into the total it will tell you how much each individual kilogram would be worth. 2 kilos is 7 and change 3 would be around 6 ish and the 4 kilo would be 4.6 and change and the 5 kilos would be like 4.88

center= (-2,0)

radius= sqrt2/2

step-by-step explanation: