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Answer :
Certainly! To determine the constant temperature at which the object's temperature will eventually stabilize, we need to analyze the given cooling function:
[tex]\[
\pi(t) = 249 e^{0.035 t} + 82
\][/tex]
In this function, [tex]\(\pi(t)\)[/tex] represents the temperature of the object at time [tex]\(t\)[/tex]. Let's break down the components of this function:
1. Exponential Decay Term: The term [tex]\(249 e^{0.035 t}\)[/tex] involves the exponential function [tex]\(e^{0.035 t}\)[/tex]. As time [tex]\(t\)[/tex] increases, the value of this exponential function, [tex]\(e^{0.035 t}\)[/tex], gets very small because of the negative exponent. Therefore, over a long period, the entire term [tex]\(249 e^{0.035 t}\)[/tex] approaches zero.
2. Constant Term: The constant 82 in the equation represents a constant value that is added to the decaying term. This constant value is crucial in determining the temperature at which the object will stabilize.
As the exponential term diminishes to zero, the temperature function [tex]\(\pi(t)\)[/tex] becomes:
[tex]\[
\pi(t) \approx 249 \times 0 + 82 = 82
\][/tex]
Therefore, as time goes to infinity, the object’s temperature approaches and stabilizes at [tex]\(82^\circ F\)[/tex].
The answer is [tex]\(82^\circ F\)[/tex].
[tex]\[
\pi(t) = 249 e^{0.035 t} + 82
\][/tex]
In this function, [tex]\(\pi(t)\)[/tex] represents the temperature of the object at time [tex]\(t\)[/tex]. Let's break down the components of this function:
1. Exponential Decay Term: The term [tex]\(249 e^{0.035 t}\)[/tex] involves the exponential function [tex]\(e^{0.035 t}\)[/tex]. As time [tex]\(t\)[/tex] increases, the value of this exponential function, [tex]\(e^{0.035 t}\)[/tex], gets very small because of the negative exponent. Therefore, over a long period, the entire term [tex]\(249 e^{0.035 t}\)[/tex] approaches zero.
2. Constant Term: The constant 82 in the equation represents a constant value that is added to the decaying term. This constant value is crucial in determining the temperature at which the object will stabilize.
As the exponential term diminishes to zero, the temperature function [tex]\(\pi(t)\)[/tex] becomes:
[tex]\[
\pi(t) \approx 249 \times 0 + 82 = 82
\][/tex]
Therefore, as time goes to infinity, the object’s temperature approaches and stabilizes at [tex]\(82^\circ F\)[/tex].
The answer is [tex]\(82^\circ F\)[/tex].
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