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anonymous

  • one year ago

A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the funnel at any given time is 16.59375π cubic centimeters. Given this information, what is the volume of the partial cone that makes up the top part of the funnel? 15.75π cm3 17.25π cm3 16.33π cm3 12.5π cm3

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  1. anonymous
    • one year ago
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    The dimensions are 1.5 base, 1.5 beside the base 4 cm height it looks like and 6cm on the top of the circle

  2. Michele_Laino
    • one year ago
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    please can you make a drawing of your funnel?

  3. anonymous
    • one year ago
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    How ? It's like 6cm 4cm 1.5 1.5

  4. anonymous
    • one year ago
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    The 6cm is at the top of the circle

  5. Michele_Laino
    • one year ago
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    |dw:1434260848319:dw| like that?

  6. anonymous
    • one year ago
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    Yes there's a 1.5 inside the square and on the outside

  7. Michele_Laino
    • one year ago
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    please place your measures on my drawing

  8. anonymous
    • one year ago
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    6cm in the circle 4cm in the middle, 1.5 in the square and 1.5 on the left side of the square

  9. Michele_Laino
    • one year ago
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    |dw:1434261034237:dw|

  10. Michele_Laino
    • one year ago
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    am I right?

  11. anonymous
    • one year ago
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    Yes

  12. anonymous
    • one year ago
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    You need another 1.5 on the side of the 1.5

  13. Michele_Laino
    • one year ago
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    ok! the general formula for volume of a partial cone like this: |dw:1434261300118:dw| is: \[V = \left\{ {\left( {{R^2} + {r^2}} \right) + \sqrt {R \times r} } \right\}\frac{{\pi h}}{3}\]

  14. Michele_Laino
    • one year ago
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    like this: |dw:1434261500243:dw|

  15. anonymous
    • one year ago
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    I don't know the answer it's so hard and this my last question

  16. anonymous
    • one year ago
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    Yes

  17. Michele_Laino
    • one year ago
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    sorry I have made an eror the right formula for volume V is: \[V = \left\{ {\left( {{R^2} + {r^2}} \right) + \sqrt {{R^2} \times {r^2}} } \right\}\frac{{\pi h}}{3}\]

  18. Michele_Laino
    • one year ago
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    now we have 2 of such partial cones

  19. anonymous
    • one year ago
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    So what's the answer 15.75pi/cm^3 17.25 pi/cm^3 16.33pi/cm^3 12.5pi/cm^3

  20. Michele_Laino
    • one year ago
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    I'm sorry, I think that our funnel is like below: |dw:1434261952973:dw|

  21. anonymous
    • one year ago
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    This is due in 5 minutes please help!!

  22. Michele_Laino
    • one year ago
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    Now using may formula above, we can write the volume of our funnel as below: \[V = \left\{ {\left( {{6^2} + {{1.5}^2}} \right) + \sqrt {{6^2} \times {{1.5}^2}} } \right\}\frac{{\pi h}}{3} + \pi \times {1.5^2}\left( {4 - h} \right) = 16.59375\pi \] where h is like in my drawing: |dw:1434262276286:dw|

  23. Michele_Laino
    • one year ago
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    what is h?

  24. anonymous
    • one year ago
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    1.5

  25. anonymous
    • one year ago
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    No h is 4cm

  26. Michele_Laino
    • one year ago
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    please wait I'm checking my computations

  27. Michele_Laino
    • one year ago
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    h can not be equal to 4 cm

  28. anonymous
    • one year ago
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    Text me I can send you a picture of the problem on my laptop 859-550-7140

  29. Michele_Laino
    • one year ago
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    please make a screeshot of your picture and post it here as a file

  30. anonymous
    • one year ago
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    I do not know how

  31. Michele_Laino
    • one year ago
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    use the snipping tool of Microsoft Windows

  32. Michele_Laino
    • one year ago
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    with that snipping tool you are able to make a screenshot of your drawing, then post it here using the "Attach File" tab

  33. anonymous
    • one year ago
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    1 Attachment
  34. anonymous
    • one year ago
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    there you go

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