Answer:
- 7
Step-by-step explanation:
The average rate of change of f(x) in the closed interval [ a, b ] is

Here [ a, b ] = [ - 5, - 2 ]
f(b) = f(- 2) = (- 2)² - 4 = 4 - 4 = 0
f(a) = f(- 5) = (- 5)² - 4 = 25 - 4 = 21, thus
average rate of change =
=
= - 7
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

x>7
Step-by-step explanation:
when dividing by a (-)
the inequality sign changes
46/100
The numeric value is in the hundredths to covert it to a fraction you'll need to keep the same value so the fraction will be 46/100
If you're looking to simplify the fraction will be 23/50