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

Answer:
Two Angles are Complementary when they add up to 90 degrees (a Right Angle)
Step-by-step explanation:
Please mark Brainliest :)
350 / 7 = 50
50 * 2 = 100
100 new members
100 / 10 = 10
10 * 3 = 30
30 are new female members
10 to the sixth power is 1,000,000. Any number to the tenth power is the number 1 in front of the number of zeros that is equal to the exponent.
Answer and Explanation:
Gina wanted to swim at the pool. She is allowed to spend at most $30. The cost to swim in the pool is 3 dollars per hour plus a flat fee of $3. How many hours can Gina swim without going over her spending limit?
This problem can be represented by
.
'x' would be the number of hours. "At most" means less than or equal to. So, the value of x would have to be less than or equal to 30.

So, Gina would have a maximum of 9 hours to swim in the pool.
Hope this helps.