SOLUTION: I'm having trouble figuring this one out and would appreciate some help! A can of soda at 81� F. is placed in a refrigerator that maintains a constant temperature of 38�F. The t

Question 1040431: I'm having trouble figuring this one out and would appreciate some help!
A can of soda at 81� F. is placed in a refrigerator that maintains a constant temperature of 38�F. The temperature T of the soda t minutes after it is placed in the refrigerator is given by T(t) = 38 + 43e^� 0.058t
Find the temperature of the soda 20 minutes after it is placed in the refrigerator. (Round to the nearest tenth of a degree.)
So I would write T(20) = 38 + 43e^-.058t, but where do I go from there?
I really appreciate any help, thank you!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Replace EVERY copy of 't' with 20. Then use a calculator to evaluate

This is approximate

Round to the nearest tenth

The final answer is approximately 51.5 degrees
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