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lopus
 4 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 ..
lopus
 4 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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lopus
 4 years ago
Best ResponseYou've already chosen the best response.0\[v=\frac{ 4 }{ 3 }*R ^{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[V=\frac{4}{3}\pi r^3\] if i remember correctly taking derivatives you get \[V'=4\pi r^2r'\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are told \(r'=0.5\) and you need \(V'\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh and i guess after 12 hours \(r=6\) so make the replacements and you are done

lopus
 4 years ago
Best ResponseYou've already chosen the best response.0why 6? can you explain me

lopus
 4 years ago
Best ResponseYou've already chosen the best response.0why not 4pi(0.5*12)^2=144

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(r=6\) cm because it starts at 12 cm and decreases at a rate of 0.5 cm per hour

lopus
 4 years ago
Best ResponseYou've already chosen the best response.0is 4pi(36)*0.5 = 72 why not 4pi(0.5*12)^2=144

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353259541168:dw

lopus
 4 years ago
Best ResponseYou've already chosen the best response.0why not 4pi(0.5*12)^2=144
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