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Calc question using e?
Coffee question
T = 80+85e^-0.045t
T= degrees farenheit
t= time (minutes)
a. what was the temperature of the coffee when it was poured
b. When will the coffee be 120 degrees farenheit
2 Answers
- cidyahLv 75 years ago
a)
substitute t=0
T = 80+85 e^0
T = 80+85 = 165 °F
b)
solve:
120 = 80 + 85 e^(-0.045 t)
120-80 = 85 e^(-0.045t)
85 e^(-0.045t) = 40
e^(-0.045t) = 40/85
-0.045 t = ln (40/85)
t = -ln(40/85) / 0.045
t = 16.75 (time at which coffee will be 120 °F
- BrainardLv 75 years ago
a. what was the temperature of the coffee when it was poured
At t = 0
T = 80 + 85e^0
= 80 + 85
= 165 ° F
b. When will the coffee be 120 degrees farenheit
When T = 120
120 = 80 + 85e^(-0.045t)
40 = 85e^(-0.045t)
40e^(0.045t) = 85
e^(0.045t) = 85/40
= 2.125
Taking logs to base e
0.045t = ln(2.125)
t = ln(2.125)/0.045
= 16.7504845
= 16.75 minutes
