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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?

Respuesta :

ANSWER

[tex]r = x - \frac{3}{2} [/tex]

EXPLANATION

The volume of the cone is

[tex]V=8\pi \: {x}^{2} + 24\pi \: x + 18\pi[/tex]

The height of the is 6 inches.

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

[tex]8\pi \: {x}^{2} + 24\pi \: x + 18\pi = \frac{1}{3} \times \pi \: {r}^{2} \times 6[/tex]

[tex]8\pi \: {x}^{2} + 24\pi \: x + 18\pi = 2 \pi \: {r}^{2} [/tex]

Divide through by 2π.

[tex]4 {x}^{2} + 12x + 9 = {r}^{2} [/tex]

The expression on the LHS is a perfect square trinomial.

[tex]( {x - \frac{3}{2} })^{2} = {r}^{2} [/tex]

[tex]r = x - \frac{3}{2} [/tex]