Math
posted by piper .
With time, t, in minutes, the temperature, H, in degrees Celsius, of a bottle of water put in the refrigerator at time t=o is given by
H = 4 + 16e.02t
How fast is the water cooling initially? After 10 minutes? Include units in your answer

Do you have a typo?
Should it not be H = 4 + 16e^(.02t) ?
if so, then ....
dH/dt =  .32e^(.02t)
Initially means t = 0
dH/dt = .32e^0 =  .32° / minute
when t=10
dH/dt = .32e^(.2) = .26° / min
check my arithmetic
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