anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
Michele_Laino
  • Michele_Laino
please can you make a drawing of your funnel?
anonymous
  • anonymous
How ? It's like 6cm 4cm 1.5 1.5

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anonymous
  • anonymous
The 6cm is at the top of the circle
Michele_Laino
  • Michele_Laino
|dw:1434260848319:dw| like that?
anonymous
  • anonymous
Yes there's a 1.5 inside the square and on the outside
Michele_Laino
  • Michele_Laino
please place your measures on my drawing
anonymous
  • anonymous
6cm in the circle 4cm in the middle, 1.5 in the square and 1.5 on the left side of the square
Michele_Laino
  • Michele_Laino
|dw:1434261034237:dw|
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
You need another 1.5 on the side of the 1.5
Michele_Laino
  • Michele_Laino
ok! the general formula for volume of a partial cone like this: |dw:1434261300118:dw| is: \[V = \left\{ {\left( {{R^2} + {r^2}} \right) + \sqrt {R \times r} } \right\}\frac{{\pi h}}{3}\]
Michele_Laino
  • Michele_Laino
like this: |dw:1434261500243:dw|
anonymous
  • anonymous
I don't know the answer it's so hard and this my last question
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
sorry I have made an eror the right formula for volume V is: \[V = \left\{ {\left( {{R^2} + {r^2}} \right) + \sqrt {{R^2} \times {r^2}} } \right\}\frac{{\pi h}}{3}\]
Michele_Laino
  • Michele_Laino
now we have 2 of such partial cones
anonymous
  • anonymous
So what's the answer 15.75pi/cm^3 17.25 pi/cm^3 16.33pi/cm^3 12.5pi/cm^3
Michele_Laino
  • Michele_Laino
I'm sorry, I think that our funnel is like below: |dw:1434261952973:dw|
anonymous
  • anonymous
This is due in 5 minutes please help!!
Michele_Laino
  • Michele_Laino
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|
Michele_Laino
  • Michele_Laino
what is h?
anonymous
  • anonymous
1.5
anonymous
  • anonymous
No h is 4cm
Michele_Laino
  • Michele_Laino
please wait I'm checking my computations
Michele_Laino
  • Michele_Laino
h can not be equal to 4 cm
anonymous
  • anonymous
Text me I can send you a picture of the problem on my laptop 859-550-7140
Michele_Laino
  • Michele_Laino
please make a screeshot of your picture and post it here as a file
anonymous
  • anonymous
I do not know how
Michele_Laino
  • Michele_Laino
use the snipping tool of Microsoft Windows
Michele_Laino
  • Michele_Laino
with that snipping tool you are able to make a screenshot of your drawing, then post it here using the "Attach File" tab
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
there you go

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