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anonymous

  • 5 years ago

Will someone please help me.. I don't know what i did wrong with a couple of problems please help

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  1. anonymous
    • 5 years ago
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    I'll help with one.

  2. anonymous
    • 5 years ago
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    okay it says to find the volume of the solid

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  3. anonymous
    • 5 years ago
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    Do you know the formula for the volume of a cone?

  4. anonymous
    • 5 years ago
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    not of the top of my head

  5. anonymous
    • 5 years ago
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    sorry, my computer just froze up it is (1/3)*Pi*r^2*h

  6. anonymous
    • 5 years ago
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    are 10 and 18 the heights, or are they the length of the lines they are next to?

  7. anonymous
    • 5 years ago
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    yes so \[1/3\times \pi \times 5^{2}\times10\]

  8. anonymous
    • 5 years ago
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    only if 10 is the height

  9. anonymous
    • 5 years ago
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    \[1/3\times \pi \times 5^{2}\times18\]

  10. anonymous
    • 5 years ago
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    10 is the height of the top cone and 18 is the height of the bottom cone

  11. anonymous
    • 5 years ago
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    Actually, I don't think so. If you do it that way you get 733.04, which got counted wrong on your test.

  12. anonymous
    • 5 years ago
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    I doubt you would get it wrong for being off by .04

  13. anonymous
    • 5 years ago
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    So, lets assume that 18 and 10 are the lengths of the lines they are next to, and not the heights.

  14. anonymous
    • 5 years ago
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    You following so far?

  15. anonymous
    • 5 years ago
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    yea

  16. anonymous
    • 5 years ago
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    Awesome. We need the heights to use our formula, so now we just need to figure out how to get them. Thinking in 3D is rather tricky, so I like to make this into something I can do in 2D.

  17. anonymous
    • 5 years ago
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    Notice how we have the radius and the hypotenuse of a right triangle?

  18. anonymous
    • 5 years ago
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    ya

  19. anonymous
    • 5 years ago
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    Well, we can use the good old Pythagorean formula to figure out the missing side.

  20. anonymous
    • 5 years ago
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    Sqrt[18^2-5^2] and Sqrt[10^2-5^2] are our two heights

  21. anonymous
    • 5 years ago
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    (1/3)Pi*( Sqrt[18^2-5^2] )^2 + (1/3)Pi*( Sqrt[10^2-5^2] )^2

  22. anonymous
    • 5 years ago
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    ^ That should give you the correct answer, which comes out to be...

  23. anonymous
    • 5 years ago
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    idk i dont have a calculator

  24. anonymous
    • 5 years ago
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    oops, I did that formula wrong :/

  25. anonymous
    • 5 years ago
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    (1/3)Pi*( Sqrt[18^2-5^2] )^2 + (1/3)Pi*( Sqrt[10^2-5^2] )^2 isn't right, but you get the idea

  26. anonymous
    • 5 years ago
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    I left out the radius

  27. anonymous
    • 5 years ago
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    You get it though, right?

  28. anonymous
    • 5 years ago
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    kinda

  29. anonymous
    • 5 years ago
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    (1/3)*Pi*r^2*h

  30. anonymous
    • 5 years ago
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    you did it right, if 10 and 18 had been the height. Unfortunatly, they were the side length (confusing problem)

  31. anonymous
    • 5 years ago
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    you just use the Pythagorean formula to get the heights, and do the same thing with them that you did with 10 and 18 the first time around.

  32. anonymous
    • 5 years ago
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    Well, I'm sleepy, so I'm off to bed. I have to wake up and do calculus D:

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