Question

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is 50\degree50°50, degree Celsius. This causes the cake to cool and the temperature difference loses \dfrac15 5 1 ​ start fraction, 1, divided by, 5, end fraction of its value every minute.

Answer 1

$D(t)=50^o(0.80)^t$

Step-by-step explanation:

Writing a function that gives the temperature difference in degrees Celsius, D(t), t minutes after the cake was put in the cooler

we know that

The equation of an exponential decay function is equal to

D(t) = $a(1 - r )^{t}$

where

                D(t) is the temperature difference in degrees

                 t is the number of minutes

                 r is the rate of change

                 a is the initial value

we have

a = 50°C

r = 1 / 5 = 0.20

substitute

D(t) = 50° $(1 - 0.20)^{t}$

D(t) = 50° $(0.80)^{t}$