Question

PLEASE ANSWER THESE TWO QUESTIONS!!!!!!!!

Answer 1

Problem 5

The diameter of the sphere must match with the side length of the cube, if we wanted the sphere to fit perfectly inside the cube.

So we'll need to find the side length of the cube based on the volume.

If the volume of the cube is 216 cubic inches, then the side length is
`\sqrt[3]{216} = 6` inches.

Or you could note that 6^3 = 6*6*6 = 216.

The cube root operation is the reverse of cubing.

Therefore, the diameter is 6 inches and the radius is 6/2 = 3 inches.

Let's calculate the volume of the sphere.

`V = (4/3)\pi*r^3\\\\V = (4/3)\pi*3^3\\\\V = 36\pi\\\\V \approx 113.097336\\\\`

I used my calculator's stored version of pi to get the most accuracy possible. If your teacher requires you to use something like the approximation pi = 3.14, then be sure to do that of course.

Subtract this from the volume of the cube to find the amount of empty space between the sphere and cube. This can be thought of as the amount of air inside the box not taken up by the ball.

216 - 113.097336 = 102.902664

This then rounds to 102.903 when rounding to the nearest thousandth.

Answer: 102.903 cubic inches

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Problem 6

The relevant formulas are:

`V_s = (4/3)\pi*r^3 = \text{volume of a sphere}\\\\V_c = \pi*r^2*h = \text{volume of a cylinder}\\\\`

Let's equate the two volumes and see what we can find

`V_s = V_c\\\\(4/3)\pi*r^3 = \pi*r^2*h\\\\(4/3)r^3 = r^2*h \ \ \text{ ... The pi terms cancel}\\\\4r^3 = 3r^2*h\\\\4r = 3h \ \ \text{ ... Divide both sides by } r^2\\\\r = 3h/4\\\\r = 0.75h`

Unfortunately we can't go any further because we have two variables, but only one equation. The rule is that we need the same number of equations as variables in order to solve for said variables.

If we used r = 3 from problem 5, then we could find h based on it. However, I'm not sure if this problem is connected in any way at all to the previous one.

I don't think there's enough info here to solve. You'll have to ask your teacher for further clarification. Sorry for the trouble.