Question

suppose a spherical snowball with a diameter of 6 centimeters is melted in a large bowl. the resulting water is then poured into a cone-shaped paper cup that is 10 centimeters deep and has a diameter of 6 centimeters. the water overflows the paper cup, as the volume of the snowball turns out to be more than that of the paper cup. how much greater is the snowball's volume than that of the cone-shaped cup?

Answer 1

Oooh fun!

Alright, 

formula for a sphere is (4$\pi$r^3)3
4*pi*6^3=2714.34/3
904.78

Now you can do this 2 ways. First is that a cone is 1/3 of a sphere with the same radius so just take 904.78 and multiply it by 1/3 and then by 2 to get 2/3 of it or do this.

cone~ (6^2*pi*10)/3
376.99

and subtract

904.78-376.99=527.79

527.79^3 greater. Check my math. I could be wrong.

Answer 2

Answer:

18.8 cm cubed

Step-by-step explanation:

So first thing is that we need to find the volume of both shapes. We have a cone and a sphere.

Volume of a sphere is: 4/3π r^3

Volume of a cone: 1/3π r^2 h

I found that the volume of the sphere is 113.04 and the volume of the cone is 94.2.

Now, we have to subtract:

113.04- 94.2=

18.8 cm cubed