A high-altitude spherical weather balloon expands as it rises, due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.16 inches per second, and that r = 38 inches at time t = 0. Determine the equation that models the volume V of the balloon at time t, and find the volume when t = 280 seconds.

Question

anonymous · Answer

AV(t) = four pi times the quantity of thirty eight plus zero point one six t to the third power divided by three; 1,129,910.14 in3
 BV(t) = four pi times the quantity of thirty eight plus zero point one six t to the third power divided by three; 2,377,823.53 in3
C V(t) = 4π(0.16t)2; 25,221.21 in3
D V(t) = 4π(38 + 0.16t)2; 1,325,689.77 in3

anonymous · Answer

@jigglypuff314 do you know this,
i know you set it up as 4pi(38 + .16t)^3 / 3 and i get 756,884.74

jigglypuff314 · Answer

I believe the equation is correct, but you might want to look at the number you got again :)

try doing  38+(0.16*280) = r
first

then  4pi(r)^3 / 3 =

anonymous · Answer

i still get the same answer?