Question

can someone explain this..

A can of soda at 82°F is placed in a refrigerator that maintains a constant temperature of 39°F. The temperature T of the soda t minutes after it is placed in the refrigerator is given by

T(t) = 39 + 43e ^ −0.058t

Find the temperature, to the nearest degree, of the soda 14 minutes after it is placed in the refrigerator.

and

When, to the nearest minute, will the temperature of the soda be 50°F?

Answer 1

Well they give you the equation of
\[T(t)=39+43e ^{-0.058t}\]
So for the first part it gives you t = 14 minutes since t is minutes after it is placed in the refrigerator. So now you could just solve for T(14)
\[T(14) =39+43e ^{-0.058(14)}\] then\[T(14) = 39 + 19.09067729\]so T(14) = 58.09 degrees F

So the second part they give you T(t) = 50 degrees F since T(t) is the temperature of the soda. Now by what the way I showed you from the first part, can you do the second part? You'll have\[50 = 39+43e ^{-0.058t}\] you'll just have to solve for t

Answer 2

@hwtrain thanks!!