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

Algebra ->  Equations -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
T%28t%29+=+38+%2B+43%2Ae%5E%28-0.058t%29


T%2820%29+=+38+%2B+43%2Ae%5E%28-0.058%2A20%29 Replace EVERY copy of 't' with 20. Then use a calculator to evaluate


T%2820%29+=+38+%2B+43%2Ae%5E%28-1.16%29


T%2820%29+=+38+%2B+43%2A0.31348618088248


T%2820%29+=+38+%2B+13.479905777947


T%2820%29+=+51.479905777947 This is approximate


T%2820%29+=+51.5 Round to the nearest tenth


The final answer is approximately 51.5 degrees