Question

The temperature (in °C) of an object at time t (in minutes) is T(t) = 3/8t^2 − 15t + 240 for 0 ≤ t ≤ 20. At what rate is the object cooling at t = 10?

Answer 1

it has something to do with rate of change just not sure how it goes

Answer 2

rate of change = dT/dt = (3/4)t - 15
at t =10 rate = 30/4 - 15 = -7.5 degrees C / minute

Answer 3

cooling at 7.5 degrees /min

Answer 4

how did you get 3/4?

Answer 5

oh i seee!