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### **Problem 12**: Celsius and Fahrenheit Temperature Range

The relationship between Celsius and Fahrenheit is given by the equation:

\[
C = \frac{5}{9}(F - 32)
\]

We need to determine the Fahrenheit temperature range for a comfortable house temperature between 18°C and 22°C.

#### a) **Double Linear Inequality**

The temperature range in Celsius is:

\[
18 \leq C \leq 22
\]

Substituting $ C = \frac{5}{9}(F - 32) $ into this inequality:

\[
18 \leq \frac{5}{9}(F - 32) \leq 22
\]

#### b) **Solving the Inequality**

To solve this inequality, we follow these steps:

1. **Multiply the entire inequality by 9/5** to eliminate the fraction:

\[
18 \times \frac{9}{5} \leq F - 32 \leq 22 \times \frac{9}{5}
\]

This gives:

\[
32.4 \leq F - 32 \leq 39.6
\]

2. **Add 32 to all parts** of the inequality to isolate $ F $:

\[
32.4 + 32 \leq F \leq 39.6 + 32
\]

This simplifies to:

\[
64.4^\circ F \leq F \leq 71.6^\circ F
\]

So, the temperature range in Fahrenheit is approximately **64.4°F to 71.6°F**.

---

### **Problem 13**: Long-Distance Call for Charity

We need to find out how long a volunteer can talk if the total cost must not exceed $2.00. The cost structure is as follows:
- $0.50 for the first 3 minutes.
- $0.10 for each additional minute or part thereof.

Let $ t $ represent the total time in minutes.

- The first 3 minutes cost $0.50.
- For $ t > 3 $, the additional time is $ t - 3 $, and each minute costs $0.10.

The total cost should be at most $2.00, so we set up the inequality:

\[
0.50 + 0.10(t - 3) \leq 2.00
\]

Now, solve the inequality:

1. **Subtract 0.50** from both sides:

\[
0.10(t - 3) \leq 1.50
\]

2. **Divide by 0.10**:

\[
t - 3 \leq 15
\]

3. **Add 3** to both sides:

\[
t \leq 18
\]

Thus, the volunteer can talk for at most **18 minutes**.

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### Do you have any questions or would you like more details on these solutions?

Here are 5 related questions you might consider:
1. What would the temperature range be in Fahrenheit if the range in Celsius were 20°C to 25°C?
2. How would the solution change if the cost structure for the call had a higher rate for the first 3 minutes?
3. Can you rewrite the Celsius-Fahrenheit conversion formula to solve for Fahrenheit in terms of Celsius?
4. How would you solve a similar inequality if the temperature range was given in Fahrenheit, and you had to convert it to Celsius?
5. What if a different charity campaign limited the total cost of the calls to $1.50, how would that change the maximum call time?

**Tip**: When solving inequalities, remember to reverse the inequality sign if you multiply or divide by a negative number!

12. The relationship between Celsius and Fahrenheit is represented by C = 5/9(F - 32). In order to be comfortable, but also economical, the temperature in your house should be between 18°C and 22°C. a) Write this statement as a double linear inequality. b) Solve the inequality to determine the temperature range in degrees Fahrenheit. 13. Some volunteers are making long distance phone calls to raise money for a charity. The calls are billed at the rate of $0.50 for the first 3 minutes and $0.10/min for each additional minute or part thereof. If each call cannot cost more than $2.00, how long can each volunteer talk to a prospective donor?

Learn how to solve a double linear inequality for the conversion between Celsius and Fahrenheit, and determine the maximum length of a charity call given a cost limit. This solution explains the temperature range in Fahrenheit based on 18°C to 22°C and uses a simple cost formula to find out how long volunteers can talk without exceeding $2.00.

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