The two parabolas intersect for

and so the base of each solid is the set

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas,
. But since -2 ≤ x ≤ 2, this reduces to
.
a. Square cross sections will contribute a volume of

where ∆x is the thickness of the section. Then the volume would be

where we take advantage of symmetry in the first line.
b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

We end up with the same integral as before except for the leading constant:

Using the result of part (a), the volume is

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

and using the result of part (a) again, the volume is

Step-by-step explanation:
This is in the form of

in this case, x is 1 and y is 7. So ourbpoint would be (1,7.)
Answer:
7
Step-by-step explanation:
Substitute x = -4 :
f(-4) = 3 - (-4)
= 3 + 4
= 7
Answer:
Iced Tea
Step-by-step explanation:
water is not normally 7.99$ so i'm guessing Iced tea depending on the fact that it could be a large or extra large which could get to the total of 7.99$