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Answer :
We start with the temperature function for the lasagna:
[tex]$$
T(t) = e^{(4.86753-t)} + 70.
$$[/tex]
Step 1. Differentiate the function
To find the rate of change of temperature, we differentiate [tex]$T(t)$[/tex] with respect to time [tex]$t$[/tex]. Since the derivative of a constant is zero, we only need to differentiate the exponential part. Applying the chain rule to [tex]$e^{(4.86753-t)}$[/tex], we have:
[tex]$$
\frac{d}{dt}\left(e^{(4.86753-t)}\right) = e^{(4.86753-t)} \cdot \frac{d}{dt}(4.86753-t).
$$[/tex]
Noting that
[tex]$$
\frac{d}{dt}(4.86753-t) = -1,
$$[/tex]
the derivative of the temperature function is:
[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]
Step 2. Evaluate at [tex]$t = 2$[/tex] hours
Substitute [tex]$t = 2$[/tex] into the derivative:
[tex]$$
T'(2) = -e^{(4.86753-2)} = -e^{2.86753}.
$$[/tex]
Step 3. Compute the value
Evaluating the exponential term,
[tex]$$
e^{2.86753} \approx 17.59,
$$[/tex]
we obtain:
[tex]$$
T'(2) \approx -17.59 \text{ degrees per hour}.
$$[/tex]
Conclusion
The lasagna is cooling at approximately [tex]$-17.59$[/tex] degrees per hour exactly 2 hours after it comes out of the oven. This corresponds to option C:
[tex]$$\boxed{-17.59 \text{ degrees/hour}}.$$[/tex]
[tex]$$
T(t) = e^{(4.86753-t)} + 70.
$$[/tex]
Step 1. Differentiate the function
To find the rate of change of temperature, we differentiate [tex]$T(t)$[/tex] with respect to time [tex]$t$[/tex]. Since the derivative of a constant is zero, we only need to differentiate the exponential part. Applying the chain rule to [tex]$e^{(4.86753-t)}$[/tex], we have:
[tex]$$
\frac{d}{dt}\left(e^{(4.86753-t)}\right) = e^{(4.86753-t)} \cdot \frac{d}{dt}(4.86753-t).
$$[/tex]
Noting that
[tex]$$
\frac{d}{dt}(4.86753-t) = -1,
$$[/tex]
the derivative of the temperature function is:
[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]
Step 2. Evaluate at [tex]$t = 2$[/tex] hours
Substitute [tex]$t = 2$[/tex] into the derivative:
[tex]$$
T'(2) = -e^{(4.86753-2)} = -e^{2.86753}.
$$[/tex]
Step 3. Compute the value
Evaluating the exponential term,
[tex]$$
e^{2.86753} \approx 17.59,
$$[/tex]
we obtain:
[tex]$$
T'(2) \approx -17.59 \text{ degrees per hour}.
$$[/tex]
Conclusion
The lasagna is cooling at approximately [tex]$-17.59$[/tex] degrees per hour exactly 2 hours after it comes out of the oven. This corresponds to option C:
[tex]$$\boxed{-17.59 \text{ degrees/hour}}.$$[/tex]
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