turkey temperatu temperature is 75°F. After 10 minutes, the temperature of the turkey is 172°F, and after 20 minutes, it is 160°F. Use a linear approximation to predict the temperature of the turkey after half an hour (use secant slopes as an approximation for the derivative).
turkey temperatu temperature is 75°F. After 10 minutes, the temperature of the turkey is 172°F, and after 20 minutes, it is 160°F. Use a linear approximation to predict the temperature of the turkey after half an hour (use secant slopes as an approximation for the derivative).
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
A turkey is removed from the oven when its temperature reaches 185°F and is placed on a table in a room where the temperature is 75°F. After 10 minutes, the temperature of the turkey is 172°F, and after 20 minutes, it is 160°F. Use a linear approximation to predict the temperature of the turkey after half an hour (use secant slopes as an approximation for the derivative).
#### Task:
Predict the temperature of the turkey after 30 minutes.
#### Given Data:
- Initial temperature of turkey: **185°F**
- Room temperature: **75°F**
- Temperature of turkey after 10 minutes: **172°F**
- Temperature of turkey after 20 minutes: **160°F**
#### Instructions:
Use a linear approximation to find the temperature after 30 minutes.
#### Calculation Outline:
- Compute the slope (secant slope) between the provided data points.
- Use the linear approximation formula.
Solve for the temperature (T) using:
\[ T(t) = T_0 + (slope \times time) \]
> \( \text{where } T_0 \text{ is the initial temperature and time is the duration.} \)
#### Box for Answer:
\[ \boxed{ \quad \textrm{T}_{\text{30 minutes}} = \quad \quad \quad °F \quad } \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8212daa0-4bcd-44c2-b8bb-69661bb0db17%2F2d89a833-d349-4dda-913d-43baf0d9cd44%2Fd5hv3f7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A turkey is removed from the oven when its temperature reaches 185°F and is placed on a table in a room where the temperature is 75°F. After 10 minutes, the temperature of the turkey is 172°F, and after 20 minutes, it is 160°F. Use a linear approximation to predict the temperature of the turkey after half an hour (use secant slopes as an approximation for the derivative).
#### Task:
Predict the temperature of the turkey after 30 minutes.
#### Given Data:
- Initial temperature of turkey: **185°F**
- Room temperature: **75°F**
- Temperature of turkey after 10 minutes: **172°F**
- Temperature of turkey after 20 minutes: **160°F**
#### Instructions:
Use a linear approximation to find the temperature after 30 minutes.
#### Calculation Outline:
- Compute the slope (secant slope) between the provided data points.
- Use the linear approximation formula.
Solve for the temperature (T) using:
\[ T(t) = T_0 + (slope \times time) \]
> \( \text{where } T_0 \text{ is the initial temperature and time is the duration.} \)
#### Box for Answer:
\[ \boxed{ \quad \textrm{T}_{\text{30 minutes}} = \quad \quad \quad °F \quad } \]
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