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:

Question

anonymous · Answer

The relation between radius and the volume is a cubic relation. If the radius is reducing at 0.5 cm/hr, the volume must decline at a constant rate of (0.5cm/hr)^3

lopus · Answer

$$volume=\frac{ 2 }{ 3 }\pi R^{3}   $$
$$\frac{ 2 }{ 3 }\pi (0.5*t)^{3}$$
final t= 12

lopus · Answer

$$144\pi \frac{cm ^{3} }{hr }$$
is possible?

anonymous · Answer

V = ( 4π/3 ) r³

lopus · Answer

the answer is :
A. 50pi cm3/hr
B. 60 pi ..
C. 72 pi ..
D. 120pi .. 
E. 144pi ..

anonymous · Answer

V= 4pi/3 * r^3
dV/dt = 4pi *r^2 *dr/dt
dV/dt = 4pi*(6)^2 *.5
=72pi

lopus · Answer

6?