Question

A cone with height h and radius r has volume V = 13πr^2h. If the cone has a height of 6 in. and volume V = 8πx^2 + 24πx + 18π, what is its radius r in terms of x?

Answer 1

ANSWER

$r = x - \frac{3}{2}$

EXPLANATION

The volume of the cone is

$V=8\pi \: {x}^{2} + 24\pi \: x + 18\pi$

The height of the is 6 inches.

We put the values into the volume of cone to get.

$8\pi \: {x}^{2} + 24\pi \: x + 18\pi = \frac{1}{3} \times \pi \: {r}^{2} \times 6$

$8\pi \: {x}^{2} + 24\pi \: x + 18\pi = 2 \pi \: {r}^{2}$

Divide through by 2π.

$4 {x}^{2} + 12x + 9 = {r}^{2}$

The expression on the LHS is a perfect square trinomial.

$( {x - \frac{3}{2} })^{2} = {r}^{2}$

$r = x - \frac{3}{2}$

Answer 2

Answer:

Its NOT r=3x+2

Step-by-step explanation:

srry, it doesnt tell me the correct answer. hope this helps