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OpenStudy (anonymous):
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?
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OpenStudy (anonymous):
it has something to do with rate of change just not sure how it goes
OpenStudy (anonymous):
rate of change = dT/dt = (3/4)t - 15 at t =10 rate = 30/4 - 15 = -7.5 degrees C / minute
OpenStudy (anonymous):
cooling at 7.5 degrees /min
OpenStudy (anonymous):
how did you get 3/4?
OpenStudy (anonymous):
oh i seee!
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