Math Problem Statement

Find the average rate of change of the temperature of hot chocolate between t = 0 and t = 20. Use proper units in your answer.

Solution

The problem asks to find the average rate of change of the temperature function H(t)H(t) between t=0t = 0 and t=20t = 20, where the temperature H(t)H(t) is measured in degrees Fahrenheit, and the corresponding values for tt and H(t)H(t) are provided in the table.

Table of values:

\hline t \ (\text{min}) & H(t) \ (\text{°F}) \\ \hline 0 & 170 \\ 2 & 137 \\ 5 & 107 \\ 10 & 83 \\ 13 & 77 \\ 20 & 72 \\ \hline \end{array}$$ ### Average Rate of Change Formula: The average rate of change of a function between two points $$t_1$$ and $$t_2$$ is given by: $$\frac{H(t_2) - H(t_1)}{t_2 - t_1}$$ Here, $$t_1 = 0$$, $$t_2 = 20$$, $$H(t_1) = H(0) = 170$$, and $$H(t_2) = H(20) = 72$$. ### Calculation: $$\frac{H(20) - H(0)}{20 - 0} = \frac{72 - 170}{20} = \frac{-98}{20} = -4.9 \ \text{°F/min}$$ ### Conclusion: The average rate of change of the temperature between $$t = 0$$ and $$t = 20$$ is $$-4.9 \ \text{°F/min}$$. Would you like a deeper explanation or have any questions? #### Related Questions: 1. What does a negative average rate of change represent in this context? 2. How does the rate of change compare between smaller intervals, such as $$t = 0$$ to $$t = 2$$? 3. How can we approximate the instantaneous rate of change using this table? 4. What would happen to the rate of change if we extended the time interval beyond 20 minutes? 5. How is the average rate of change related to the slope of the line connecting two points on the graph of $$H(t)$$? **Tip:** The average rate of change gives an overall sense of how a quantity changes over an interval, while the instantaneous rate of change focuses on behavior at specific points.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rate of Change
Temperature

Formulas

Average rate of change formula: (H(t2) - H(t1)) / (t2 - t1)

Theorems

Basic Algebraic Rate of Change

Suitable Grade Level

Grades 8-10