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Let's analyze the problem and solve it step by step.

### Given:
A battery was recharged, stayed fully charged, and then started losing its charge. The table provides the time and the corresponding charge levels:

| Time (h) | 0  | 5  | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55 | 60 |
|----------|----|----|----|----|----|----|----|----|----|----|----|----|----|
| Charge (%) | 92 | 94 | 96 | 98 | 100 | 100 | 100 | 100 | 100 | 95 | 90 | 85 | 80 | 75 |

There are three phases to model:
1. **When the battery is being recharged** (first 15 hours).
2. **When the battery is fully charged** (from 15 to 40 hours).
3. **When the battery is losing its charge** (from 40 to 60 hours).

### 1. Equation for the battery when it is being recharged (0h - 15h)
In the first phase, the battery is charging from 92% to 100%. We can approximate this with a linear equation since the increments are roughly consistent.

At time \( t = 0 \), the charge is 92%.
At time \( t = 15 \), the charge is 100%.

The slope (rate of change of charge with respect to time) is:
\[
m = \frac{100 - 92}{15 - 0} = \frac{8}{15}
\]
Using the point-slope form of the equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept (initial charge at \( t = 0 \)):

\[
\text{Charge} = \frac{8}{15}t + 92
\]
So, the equation for the charging phase is:
\[
\boxed{\text{Charge} = \frac{8}{15}t + 92}
\]

### 2. Equation for the battery when it is fully charged (15h - 40h)
During this phase, the charge remains constant at 100%. So, the equation is simply:
\[
\boxed{\text{Charge} = 100}
\]
This applies for \( 15 \leq t \leq 40 \).

### 3. Equation for the battery when it is losing its charge (40h - 60h)
In the final phase, the battery loses charge linearly, from 100% to 75%.

At time \( t = 40 \), the charge is 100%.
At time \( t = 60 \), the charge is 75%.

The slope is:
\[
m = \frac{75 - 100}{60 - 40} = \frac{-25}{20} = -1.25
\]
Using the point-slope form again, with the point \( (40, 100) \):
\[
\text{Charge} = -1.25(t - 40) + 100
\]
Simplifying:
\[
\text{Charge} = -1.25t + 150
\]
So, the equation for the discharging phase is:
\[
\boxed{\text{Charge} = -1.25t + 150}
\]

### Summary of Equations:
1. **Charging phase (0h - 15h):**
   \[
   \boxed{\text{Charge} = \frac{8}{15}t + 92}
   \]
2. **Fully charged phase (15h - 40h):**
   \[
   \boxed{\text{Charge} = 100}
   \]
3. **Discharging phase (40h - 60h):**
   \[
   \boxed{\text{Charge} = -1.25t + 150}
   \]

Let me know if you'd like further details or have any questions!

### Related Questions:
1. How would the equations change if the charging rate was not constant?
2. Can we model the discharging phase with a non-linear equation?
3. How would the model change if the battery lost charge faster after 60 hours?
4. What is the total energy lost by the battery in the discharging phase?
5. How can we calculate the rate of charge increase during recharging using calculus?

### Tip:
When dealing with real-world data, always check if a linear model fits well or if a more complex (e.g., quadratic or exponential) model might be better for accuracy.

A battery was recharged, remained fully charged, and then slowly lost its charge. Determine the equation for the battery when it is being recharged, fully charged, and losing its charge.

Learn how to determine the equations for a battery's charging, fully charged, and discharging phases using linear equations and rates of change. This problem involves real-world data and applies algebraic concepts like slope and point-slope form.

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