College

The function [tex]f(t) = 349.2(0.98)^t[/tex] models the relationship between [tex]t[/tex], the time an oven spends cooling, and the temperature of the oven.

**Oven Cooling Time Table:**

[tex]
\[
\begin{array}{|c|c|}
\hline
\text{Time (minutes) } t & \text{Oven temperature (degrees Fahrenheit) } f(t) \\
\hline
5 & 315 \\
\hline
10 & 285 \\
\hline
15 & 260 \\
\hline
20 & 235 \\
\hline
25 & 210 \\
\hline
\end{array}
\]
[/tex]

For which temperature will the model predict the time spent cooling?

A. 0
B. 100
C. 300
D. 400

Answer :

To determine the time it takes for the oven to cool to a specific temperature using the function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex], we need to find the value of [tex]\( t \)[/tex] when the oven reaches a temperature of 100 degrees Fahrenheit.

Here’s how we can find that:

1. Set up the equation:
You want to find [tex]\( t \)[/tex] when the temperature [tex]\( f(t) \)[/tex] is 100. So, you set up the equation:
[tex]\[
f(t) = 349.2 \times (0.98)^t = 100
\][/tex]

2. Solve for [tex]\( t \)[/tex]:
To solve for [tex]\( t \)[/tex], first divide both sides of the equation by 349.2:
[tex]\[
(0.98)^t = \frac{100}{349.2}
\][/tex]

3. Use logarithms to solve for [tex]\( t \)[/tex]:
Take the natural logarithm on both sides to bring down the exponent:
[tex]\[
\ln((0.98)^t) = \ln\left(\frac{100}{349.2}\right)
\][/tex]

4. Apply the power rule for logarithms:
Using the property of logarithms [tex]\(\ln(a^b) = b \ln(a)\)[/tex], the equation becomes:
[tex]\[
t \cdot \ln(0.98) = \ln\left(\frac{100}{349.2}\right)
\][/tex]

5. Solve for [tex]\( t \)[/tex]:
Isolate [tex]\( t \)[/tex] by dividing both sides by [tex]\(\ln(0.98)\)[/tex]:
[tex]\[
t = \frac{\ln\left(\frac{100}{349.2}\right)}{\ln(0.98)}
\][/tex]

6. Calculate the value of [tex]\( t \)[/tex]:
After performing the calculations, you find that [tex]\( t \)[/tex] approximately equals 62 minutes.

Therefore, the time it will take for the oven to cool to 100 degrees Fahrenheit is approximately 62 minutes.