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anonymous

  • 5 years ago

a tank shape cone with diameter 40ft and depth 20ft filled rate of 10cu ft/min how fast depth is changing when filled to depth of 10 ft

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  1. bahrom7893
    • 5 years ago
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    when is this due?

  2. anonymous
    • 5 years ago
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    tuesday i know dont have any clue now looking at the book is not helping me notes and online :-(

  3. bahrom7893
    • 5 years ago
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    hey i will solve this, but im helpin someone out right now, so if u want send me an email at bahrom.cfo@gmail.com and i will email u the solution as soon as im done with the other girl/guy

  4. anonymous
    • 5 years ago
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    thank you so much ;=)

  5. bahrom7893
    • 5 years ago
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    np.. i might even answer here as soon as im done with that other person

  6. anonymous
    • 5 years ago
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    ok if not i will send you and email i still have time for tutoring thank you ;-)

  7. bahrom7893
    • 5 years ago
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    So: You are given: r = 20 ft (diameter is 40ft, radius is half of 40 = 20ft) h = 20 ft dV/dt = + 10ft^3/min (change in volume is positive as it is being filled up)

  8. bahrom7893
    • 5 years ago
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    now when depth or h is 10, what is dh/dt (change in depth)

  9. bahrom7893
    • 5 years ago
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    So let me rewrite this neatly: r = 20ft h = 20 ft dV/dt = 10 ft^3/min dh/dt at h = 10 is ?

  10. bahrom7893
    • 5 years ago
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    hey i have to graph some stuff for this, can u go here: http://www.twiddla.com/496286

  11. bahrom7893
    • 5 years ago
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    click on that link, i still don't see u there

  12. bahrom7893
    • 5 years ago
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    u here?

  13. anonymous
    • 5 years ago
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    dont see anything on the white board

  14. anonymous
    • 5 years ago
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    thankkkkkkkk youuuuuuuuuuuuuuuuuuuuuuuu for the help

  15. bahrom7893
    • 5 years ago
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    You are veryyyy welcome!!!

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