Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bubble gum nested at the bottom of the cone. The radius of the snow cone is 4 inches, and the height of the cone is 6 inches. If the diameter of the bubble gum is 0.8 inches, which of the following can be used to calculate the volume of the cone that can be filled with flavored ice?

OpenStudy (anonymous):

1 over 3(3.14)(62)(4) − 4 over 3(3.14)(0.43) 1 over 3(3.14)(42)(6) − 4 over 3(3.14)(0.43) 1 over 3(3.14)(62)(4) − 4 over 3(3.14)(0.83) 1 over 3(3.14)(42)(6) − 4 over 3(3.14)(0.83)

OpenStudy (anonymous):

@peachpi

OpenStudy (anonymous):

you want to subtract the volume of the sphere from the volume of the cone. cone: \[V=\frac{ \pi }{ 3 }\pi r^2h\] r = radius, h = height sphere: \[V=\frac{ 4 }{ 3 }\pi r^3\] r = radius

OpenStudy (anonymous):

Ok so V of cone=pi/3pi4^2x6

OpenStudy (anonymous):

and V of sphere= 4/3pi4^3?

OpenStudy (anonymous):

the cone is right. The sphere should be (4/3)π(0.4)^3

OpenStudy (anonymous):

Oh alright, so then i subtract pi/3pi4^2x6-(4/3)pi(o.4)^3

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

wait hold up. I put an extra π in the cone equation by mistake. There should only be one π there

OpenStudy (anonymous):

so it would pi over 3? or pi r squared?

OpenStudy (anonymous):

it would be pi*

OpenStudy (anonymous):

@peachpi

OpenStudy (anonymous):

It's (π/3)r²h so (π/3)(4^2)(6)

OpenStudy (anonymous):

I got 66.87

OpenStudy (anonymous):

No wait sorry

OpenStudy (anonymous):

that's not what they're asking for lol

OpenStudy (anonymous):

yeah, actually looking at your choices they just want you to set it up, not solve for a number

OpenStudy (anonymous):

Yeah my bad, so the answer would be the 2nd option then

OpenStudy (anonymous):

1 over 3(3.14)(4^2)(6) − 4 over 3(3.14)(0.4^3)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

alright thanks lol i have this super complex question if u can solve it i will be eternally grateful!!

OpenStudy (anonymous):

i'll give it a try

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!