The base of a box, built from a model, has an area that can be represented by the function B(x) = 4x^2 + 8, where x is the width of the model's base. The height of the box can be represented by the function H(x) = x + 2. Which function best describes C(x), the capacity of the box?

A. C(x) = 4x^3 + 8x^2 + 8x + 16
B. C(x) = 4x^3 + 16
C. C(x) = 4x^3 + 16x^2 + 16
D. C(x) = 4x^3 + 32x^2 + 2x + 16

A) C(x)=4x³+8x²+8x+16

Step-by-step explanation:

(4x²+8)(x+2)=

A) 4x³+8x²+8x+16

The capacity of the box would be the volume.

Multiply the area by the height:

4x^2 + 8 times x+2

Multiply each term by each term:

4x^2 * x = 4x^3

4x^2 * 2 = 8x^2

8 * x = 8x

8*2 = 16

Combine the terms to get:

4x^3 + 8x^2+ 8x +16

The answer is A. C(x) = 4x^3 + 8x^2 + 8x + 16

answer: the answer is fre sha voca do

step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.