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At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures 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.

D(t)=

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is $50^{\circ}$ Celsius. This causes the cake to cool and the temperature difference loses $\frac{1}{5}$ of its value every minute. $\newline$ Write a function that gives the temperature difference in degrees Celsius, $D(t), t$ minutes after the cake was put in the cooler. $\newline$ $D(t)=\square$

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Interpret the exponential function word problem

Full solution

Q. At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is

50^{\circ}

Celsius. This causes the cake to cool and the temperature difference loses

\frac{1}{5}

of its value every minute.

\newline

Write a function that gives the temperature difference in degrees Celsius,

D(t), t

minutes after the cake was put in the cooler.

\newline

D(t)=\square

Identify Initial Difference: Step $1$ : Identify the initial temperature difference and the rate at which it decreases. The initial temperature difference is $50$ degrees Celsius, and it loses $\frac{1}{5}$ of its value every minute.
Determine Decay Function Form: Step $2$ : Determine the form of the function that models exponential decay. The general form of an exponential decay function is $D(t) = D_0 \times (1 - r)^t$ , where $D_0$ is the initial value, $r$ is the decay rate, and $t$ is time in minutes.
Plug in Values: Step $3$ : Plug in the initial temperature difference and the decay rate into the exponential decay function. The initial temperature difference $D_0$ is $50$ degrees Celsius, and the decay rate $r$ is $1/5$ or $0.2$ . Therefore, the function becomes $D(t) = 50 \times (1 - 0.2)^t$ .
Simplify Function: Step $4$ : Simplify the function. The expression $(1 - 0.2)$ is equal to $0.8$ . So, the function simplifies to $D(t) = 50 \times 0.8^t$ .
Write Final Function: Step $5$ : Write the final function. The function that gives the temperature difference in degrees Celsius, $D(t)$ , $t$ minutes after the cake was put in the cooler is $D(t) = 50 \times 0.8^t$ .

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