Question 132649: A hot cup of coffee is left outside where it cools exponentially. The temperature, T, of the coffee in degrees Celsius after t minutes is modeled by the function T=66(b)^(x/2)+17. The temperature of the coffee after 9 minutes is 56 degrees Celsius. What is the rate of cooling?
I think the answer is 3% every two minutes????
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A hot cup of coffee is left outside where it cools exponentially. The temperature, T, of the coffee in degrees Celsius after t minutes is modeled by the function T=66(b)^(x/2)+17. The temperature of the coffee after 9 minutes is 56 degrees Celsius. What is the rate of cooling?
I think the answer is 3% every two minutes????
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T=66(b)^(x/2)+17
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56 = 66*b^(9/2)+17
39 = 66*b^(9/2)
13/22 = b^(9/2)
Raise both sides to the (2/9) to solve for b:
b = 0.889666....
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Formula:
T(x) = 66*(0.889666)^(x/2) + 17
T(0) = 66*(1)+17
T(0) = 83 degrees
T(9) = 56 degrees
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Average = (56-83)/9 = -3 degrees per minute
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Cheers,
Stan H.
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