Algebra 1, Volume of Cone, Polynomials

Question

A cone with height h and radius r has volume V =1/3 π r^2h. If h is 9 in., and V is equal to 3π y^2 + 30πy + 75π in. ^3, what is the cone’s radius (r) in terms of y?

Answer 1

The first answer is definitely the route you should take. However, your question needs to be formatted more clearly to show that h is not a part of the exponent. V = 1/3 π r² h

3πy² + 30πy + 75π = 1/3 π r² (9)

Divide both sides by π:

3y² + 30y + 75 = 1/3 r² (9)

When consolidating, your 9 and 1/3 on the right can be combined to equal 9/3, or 3. Then divide by 3 and take the square root to get r by itself.

3y² + 30y + 75 = 3r²

y² + 10y + 25 = r²

^

r = √(y² + 10y + 25)

r = (y + 5) (y + 5)

Answer 2

That is a good question. Create an equation substituting the known volume for V and 9 for h:

3πy^2 + 30πy + 75π = 1/3*πr^18

Divide both sides by π:

3y^2 + 30y + 75 = 1/3*r^18

Multiply both sides by 3 to remove the fraction:

9y^2 + 90y + 225 = r^18

Then take the 18th root of both sides (that is, raise both sides to the 1/18 power) and you've solved for r. Hope this helps