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Mice21 [21]
4 years ago
8

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperature is 50° Celsius. This

causes the cake to cool and the temperature difference loses 1/5 of its value every minute.
Write a function that gives the temperature difference in degrees Celsius, D(t),t minutes after the cake was put in the cooler.
Mathematics
2 answers:
dybincka [34]4 years ago
6 0

Answer:

D(t)=50(0.8)^t

Step-by-step explanation:

We are given that

Initially the difference between the cake's and the cooler's temperature ,a=50 degree Celsius

r=\frac{1}{5}/min

We have to find the function that gives the temperature difference in degrees Celsius D(t).

We know that

D(t)=a(1-r)^t

Substitute the values

D(t)=50(1-\frac{1}{5})^t=50(1-0.2)^t

D(t)=50(0.8)^t

This is required function that gives the temperature difference in degrees Celsius.

lora16 [44]4 years ago
6 0

Answer:

D(t)=50(0.8)^t

Step-by-step explanation:

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