I can't exactly SHOW you where to put the numbers, but I can teach you the process of how you'd do it.
First off, label your number line from 0-15, as it is the simplest. (You'd be counting by 1 per each line). Then, follow this process:
1) Look at the first digit of your value. Place your number according to your first digit. (So, you'd put 0.365 at the 0 line and 3.521 at the 3 line)
2) Look at the second digit of your value. Imagine that between the two main lines (0-1 and 3-4) that there is 10 smaller lines. Then, you can place your number according to your second digit. (So, you'd put 0.365 at the 0.3 line and 3.521 at the 3.5 line).
3) Look at the third digit of your value. Imagine that between the two smaller lines (0.3-0.4 and 3.5-3.6) that there is 10 smaller lines. Then, you can place your number according to your third digit. (So, you'd put 0.365 at the 0.36 line and 3.521 at the 3.52 line).
4) Look at the fourth digit of your value. Imagine that between the two even smaller lines (0.36-0.37 and 3.52-3.53) that there is 10 smaller lines. Then, you can place your number according to your fourth digit. (So you'd place 0.365 at the 0.365 line and 3.521 at the 3.521 line)
John would have 60 apples!
Hope this helps!!!!! And btw tell John to give me some of them apples lol XD
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
