Ahigh-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

Volume of a sphere = (4/3)πr³ 

At time 0, the radius of the sphere is 36 inches & each second after 0 the radius increases by 0.07 inches, so at any time t our radius is 36 + 0.07t 

Therefore at any time t our volume is (4/3)π(36 + 0.07t)³ so your equation is 

V = (4/3)π(36 + 0.07t)³ 

At t = 400, this becomes V = (4/3)π(36 + 0.07t)³ = (4/3)π(36 + 28)³ (4/3)π(64)³ = 1098066 cubic inches = 635.4 cubic feet

A’b’c’ are the coordinates to the new location of the transformed triangle abc. find the distance between the coordinates points to use for your calculations.