Question

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.

Answer 1

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.

Answer 2

Answer:

D(t)=50(0.8)^t

Step-by-step explanation: