The radius r, in inches, of a spherical balloon is related to the volume V by r(V)= cubed√3V/4pi

Air is pumped into the balloon so the volume after t seconds is given by V(t)=12+20t.

a. Find the expression for the composite function r(V(t)).

b. What is the exact time in seconds when the radius reaches 16 inches?

r = ∛(3V/(4π)) = ∛(3(12+20t)/(4π)) = ∛(3(3+5t)/π)

∛(3(3+5t)/π) = 16
3(3+5t)/π = 4096
3+5t = 4096π/3
t = (4096π - 9)/15

a. To find the composite function r(V(t)), we need to substitute V(t) into the function for r(V).

Substituting V(t) = 12 + 20t into the expression for r(V), we have:
r(V(t)) = (cubed root of 3V(t))/(4pi)

Simplifying further, we get:
r(V(t)) = (cubed root of 3(12 + 20t))/(4pi)
= (cubed root of (36 + 60t))/(4pi)

So, the expression for the composite function r(V(t)) is:
r(V(t)) = (cubed root of (36 + 60t))/(4pi)

b. To find the exact time when the radius reaches 16 inches, we need to solve the equation r(V(t)) = 16.

Substituting 16 for r(V(t)), we have:
16 = (cubed root of (36 + 60t))/(4pi)

Multiplying both sides by 4pi, we get:
64pi = cubed root of (36 + 60t)

Cube both sides of the equation to eliminate the cubed root:
(64pi)^3 = 36 + 60t

Simplifying further, we have:
262,144pi^3 = 36 + 60t

Rearranging the equation to solve for t, we get:
60t = 262,144pi^3 - 36

Dividing both sides by 60, we have:
t = (262,144pi^3 - 36)/60

Therefore, the exact time in seconds when the radius reaches 16 inches is:
t = (262,144pi^3 - 36)/60

a. To find the expression for the composite function r(V(t)), we need to substitute the expression for V(t) into the function r(V).

Given:
r(V) = (3V/4π)^(1/3)
V(t) = 12 + 20t

Substituting V(t) into r(V):
r(V(t)) = (3V(t)/4π)^(1/3)
= (3(12 + 20t)/4π)^(1/3)
= (36 + 60t)/(4π)^(1/3)
= (9 + 15t)/(π)^(1/3)

Therefore, the expression for the composite function r(V(t)) is (9 + 15t)/(π)^(1/3).

b. To find the exact time in seconds when the radius reaches 16 inches, we need to solve the equation r(V(t)) = 16.

Given:
r(V(t)) = (9 + 15t)/(π)^(1/3)

Setting r(V(t)) equal to 16:
(9 + 15t)/(π)^(1/3) = 16

Multiplying both sides by (π)^(1/3):
9 + 15t = 16(π)^(1/3)

Subtracting 9 from both sides:
15t = 16(π)^(1/3) - 9

Dividing both sides by 15:
t = (16(π)^(1/3) - 9)/15

Therefore, the exact time in seconds when the radius reaches 16 inches is (16(π)^(1/3) - 9)/15 seconds.

To find the expression for the composite function r(V(t)), we need to substitute the expression for V(t) into the equation for the radius function r(V).

a. Let's substitute V(t) = 12 + 20t into the equation for r(V):

r(V(t)) = (3V(t)/(4π))^(1/3)
= (3(12 + 20t)/(4π))^(1/3)
= (36 + 60t)/(4π)^(1/3)
= [(6(6 + 10t))/(4π)^(1/3)]
= [(3(2(3 + 5t)))/(4π)^(1/3)]
= [(3(2(3 + 5t)))/(2^(2/3)*π^(1/3))]

So the expression for the composite function r(V(t)) is [(3(2(3 + 5t)))/(2^(2/3)*π^(1/3))].

b. To find the exact time in seconds when the radius reaches 16 inches, we need to solve the equation r(V(t)) = 16.

[(3(2(3 + 5t)))/(2^(2/3)*π^(1/3))] = 16
(3(2(3 + 5t)))/(2^(2/3)*π^(1/3)) = 16

Now, we can solve for t by simplifying the equation:

(3(6 + 10t))/(2^(2/3)*π^(1/3)) = 16
(18 + 30t)/(2^(2/3)*π^(1/3)) = 16

We can now cross-multiply and solve for t:

(18 + 30t)*[2^(2/3)*π^(1/3)] = 16*[2^(2/3)*π^(1/3)]
18 + 30t = 16*(2^(2/3)*π^(1/3))
18 + 30t = 16*(2^(2/3))*(π^(1/3))
18 + 30t = 16*(2^(2/3))*(π^(1/3))

Now, divide both sides of the equation by 16*(2^(2/3))*(π^(1/3)) to isolate t:

t = [16*(2^(2/3))*(π^(1/3)) - 18] / 30

By substituting the numerical values for π and evaluating the expression on the right-hand side, you can find the exact time in seconds when the radius reaches 16 inches.