Me ! 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.

Find r as a function of time t.

r starts at 38 inches and steadily increases at 0.16 inch per second. So in t seconds r would have increased by 0.16t when added to the initial 38 inches will give:

r(t) = 38 + 0.16t

Volume of a sphere = 4/3(pi)r^3. Since radius changes with time, volume will change with time too.

V(t) = 4/3(pi)r^3 = 4/3(pi)(38 + 0.16t)^3

Put t=280 and solve to get V

V=2377823.53

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: