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:
1st angle = X
2nd angle = X + 7
3rd angle = X + 18
Then (sum. them): 3X + 25 = 180 ==> X = 51.67 degree
Therefore,
1st angle = 51.67 degree
2nd angle = 58.67 degree
3rd angle = 69.67 degree
Use a variable, x, to stand in for the number.
5+x is less than 17.
x is less than 12
Hope this helps!
Answer:
z=-3
Step-by-step explanation:
-5z+9=24 so, subtract 9 from both sides becuase of liked terms to get -5z=15, then divide both sides by -5 to get z = -3 hope it helps:)