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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 ..
 one year ago
 one year 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 ..
 one year ago
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
\[V=\frac{4}{3}\pi r^3\] if i remember correctly taking derivatives you get \[V'=4\pi r^2r'\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
you are told \(r'=0.5\) and you need \(V'\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
oh and i guess after 12 hours \(r=6\) so make the replacements and you are done
 one year ago

lopusBest ResponseYou've already chosen the best response.0
why 6? can you explain me
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
\(r=6\) cm because it starts at 12 cm and decreases at a rate of 0.5 cm per hour
 one year ago

SkaematikBest ResponseYou've already chosen the best response.0
do you get this, OP?
 one year ago

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

SkaematikBest ResponseYou've already chosen the best response.0
dw:1353259541168:dw
 one year ago

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