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lopus

  • 3 years ago

to melt a snowball, with initial radius of 12 cm, its radius decreases at a constant rate of 0.5 cm / hr. if it starts to melt when t = 0, the rate of decline in the volume of the snowball after 12 hours is: the answer is : A. 50pi cm3/hr B. 60 pi .. C. 72 pi .. D. 120pi .. E. 144pi ..

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  1. lopus
    • 3 years ago
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    \[v=\frac{ 4 }{ 3 }*R ^{3}\]

  2. anonymous
    • 3 years ago
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    \[V=\frac{4}{3}\pi r^3\] if i remember correctly taking derivatives you get \[V'=4\pi r^2r'\]

  3. anonymous
    • 3 years ago
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    you are told \(r'=-0.5\) and you need \(V'\)

  4. anonymous
    • 3 years ago
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    oh and i guess after 12 hours \(r=6\) so make the replacements and you are done

  5. lopus
    • 3 years ago
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    why 6? can you explain me

  6. lopus
    • 3 years ago
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    is 4pi(36)*0.5 = 72

  7. lopus
    • 3 years ago
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    why not 4pi(0.5*12)^2=144

  8. anonymous
    • 3 years ago
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    \(r=6\) cm because it starts at 12 cm and decreases at a rate of 0.5 cm per hour

  9. Skaematik
    • 3 years ago
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    do you get this, OP?

  10. lopus
    • 3 years ago
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    is 4pi(36)*0.5 = 72 why not 4pi(0.5*12)^2=144

  11. Skaematik
    • 3 years ago
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    |dw:1353259541168:dw|

  12. lopus
    • 3 years ago
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    why not 4pi(0.5*12)^2=144

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