Mathematics, 02.04.2020 06:00 jhitotw

If the volume of a cone is 168 pi inches and its altitude is 14 inches, what is the radius of the cone? A. 36 inches B. 12 inches C. 72 inches D. 6 inches

Answers: 3

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The radius of the cone is 6 inches, which i mentioned in option D, for the volume of a cone is 168 $\pi$ inches and its altitude is 14 inches.

Step-by-step explanation:

The given is.

Volume of the cone is 168 $\pi$

Altitude of the cone is 14 inches

Step:1

Formula for Volume of cone is,

$V = \pi r^{2} \frac{h}{3}$ ....................................(1)

Where,

V = 168 $\pi$ cubic inches

h = 14 inches

Step:2

From the Eqn (1),

168 $\pi$ = $\pi r^{2} \frac{14}{3}$ (∵ Cancel $\pi$ on both sides)

168 = $r^{2} \frac{14}{3}$

168 × 3 = $r^{2} 14$

504 = $r^{2} 14$

$r^{2}$ = 36

r = $\sqrt{36}$

r = 6 inches

Result:

Thus the radius of the cone is 6 inches, If the volume of a cone is 168 pi inches and its altitude is 14 inches.

Answer from: Quest

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2 polynomial long division

dividing : x4+5x3-4x2-8x+6

("dividend")

by : x-1 ("divisor")

dividend x4 + 5x3 - 4x2 - 8x + 6

- divisor * x3 x4 - x3

remainder 6x3 - 4x2 - 8x + 6

- divisor * 6x2 6x3 - 6x2

remainder 2x2 - 8x + 6

- divisor * 2x1 2x2 - 2x

remainder - 6x + 6

- divisor * -6x0 - 6x + 6

remainder 0

quotient : x3+6x2+2x-6 remainder: 0

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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