anonymous
  • 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?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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)
anonymous
  • anonymous
anonymous
  • 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

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anonymous
  • anonymous
Ok so V of cone=pi/3pi4^2x6
anonymous
  • anonymous
and V of sphere= 4/3pi4^3?
anonymous
  • anonymous
the cone is right. The sphere should be (4/3)π(0.4)^3
anonymous
  • anonymous
Oh alright, so then i subtract pi/3pi4^2x6-(4/3)pi(o.4)^3
anonymous
  • anonymous
yes
anonymous
  • anonymous
wait hold up. I put an extra π in the cone equation by mistake. There should only be one π there
anonymous
  • anonymous
so it would pi over 3? or pi r squared?
anonymous
  • anonymous
it would be pi*
anonymous
  • anonymous
anonymous
  • anonymous
It's (π/3)r²h so (π/3)(4^2)(6)
anonymous
  • anonymous
I got 66.87
anonymous
  • anonymous
No wait sorry
anonymous
  • anonymous
that's not what they're asking for lol
anonymous
  • anonymous
yeah, actually looking at your choices they just want you to set it up, not solve for a number
anonymous
  • anonymous
Yeah my bad, so the answer would be the 2nd option then
anonymous
  • anonymous
1 over 3(3.14)(4^2)(6) − 4 over 3(3.14)(0.4^3)
anonymous
  • anonymous
yes
anonymous
  • anonymous
alright thanks lol i have this super complex question if u can solve it i will be eternally grateful!!
anonymous
  • anonymous
i'll give it a try

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