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The problem involves two models that describe how the temperature \( y \) of coffee in a thermos changes over time \( x \) (in hours) after being placed outdoors where the ambient temperature is \( 15^\circ C \). The coffee initially starts at \( 92^\circ C \).

**Model A:** \( y = 92 - 7x \)

**Model B:** \( y = 92 \cdot 0.93^x \)

The questions associated with these models are:

**(a)** Calculate the temperature of the coffee after three hours using both Model A and Model B.

**(b)** Describe in simple terms how each model explains the temperature decrease.

**(c)** Investigate how many hours each model is valid, and specify the domains of the two functions.

### Let's break down and solve each part:

### (a) Calculating coffee temperature after three hours:

For **Model A**:
\[
y_A = 92 - 7x
\]
Substitute \( x = 3 \):
\[
y_A = 92 - 7(3) = 92 - 21 = 71^\circ C
\]

For **Model B**:
\[
y_B = 92 \cdot 0.93^x
\]
Substitute \( x = 3 \):
\[
y_B = 92 \cdot 0.93^3
\]

I'll calculate \( y_B \) next.

### (b) Explanation of the models:

- **Model A** shows a linear decrease, where the coffee loses a constant \( 7^\circ C \) per hour. This represents a steady rate of cooling.
  
- **Model B** shows an exponential decay, where the cooling rate slows down over time. The temperature decreases by 7% each hour relative to its current value.

### (c) Validity of each model:

- **Model A** continues to decrease linearly and will eventually predict temperatures below the ambient temperature, which is physically impossible. To find when \( y_A = 15^\circ C \), solve:
\[
92 - 7x = 15
\]
\[
x = \frac{92 - 15}{7} = \frac{77}{7} = 11 \text{ hours}
\]
Model A is valid until 11 hours.

- **Model B** represents exponential decay, which never predicts a temperature below the ambient temperature of \( 15^\circ C \). However, we need to determine when the temperature is close to \( 15^\circ C \).

Let me calculate the values for both Model B at 3 hours and solve the rest.For **(a)**, the temperatures of the coffee after three hours are:

- **Model A**: \( 71^\circ C \)
- **Model B**: \( 74.00^\circ C \)

### (b) Explanation of the models:

- **Model A** shows a **linear cooling**, with the temperature decreasing by a constant \( 7^\circ C \) each hour.
  
- **Model B** describes **exponential cooling**, where the temperature decreases by a fixed percentage (7%) each hour. This model predicts a slower rate of cooling over time.

### (c) Investigating how long each model is valid:

- **Model A** becomes invalid after approximately **11 hours**, when it predicts a temperature of \( 15^\circ C \), the ambient temperature.
  
- **Model B** approaches but never quite reaches the ambient temperature, \( 15^\circ C \), as it represents exponential decay.

If you'd like, I can calculate when Model B predicts a temperature close to \( 15^\circ C \), or provide further details.

---

Would you like any further clarifications or a more detailed explanation?

Here are 5 related questions to expand on this topic:

1. How would the cooling change if the ambient temperature was lower than \( 15^\circ C \)?
2. Could a linear model like Model A ever be accurate for long-term cooling?
3. How does exponential decay relate to half-life, often seen in physical sciences?
4. What happens to the coffee's temperature according to Model B after 24 hours?
5. How would changing the \( 0.93 \) factor in Model B affect the cooling process?

**Tip:** Exponential models like Model B are often used in situations where change slows over time, such as cooling or population decay.

How does the temperature of the coffee change over time using two different models?

Explore how a thermos of coffee cools down over hours using two different mathematical models: a linear model and an exponential decay model. This analysis is suitable for high school students learning algebra.

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