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

D(t)=50(0.8)^t

Explanation:

Answer 2

`D(t)=50(0.8)^t`

Explanation:

We are given that

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

`r=(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-(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.