To model the situation with an equation, we'll denote:
- \( x \) as the time in minutes,
- \( y \) as the gallons of gas in the boat.
From the points provided in your question, we can first identify the changes:
- From (4, 50) to (10, 80):
- The time increases from 4 minutes to 10 minutes, which is a change of \( 10 - 4 = 6 \) minutes.
- The gallons of gas increase from 50 gallons to 80 gallons, which is a change of \( 80 - 50 = 30 \) gallons.
The rate of increase in gallons of gas per minute can be calculated as: \[ \text{Rate} = \frac{\text{Change in gallons}}{\text{Change in time}} = \frac{30 \text{ gallons}}{6 \text{ minutes}} = 5 \text{ gallons per minute}. \]
Next, we can establish the equation for the gallons of gas in the boat, starting from the initial point (4, 50):
- At \( x = 4 \), \( y = 50 \).
We can express \( y \) as a function of \( x \): \[ y = 5x + b \] To find \( b \), we can plug in one of the points. Using \( (4, 50) \): \[ 50 = 5(4) + b \ 50 = 20 + b \ b = 30. \]
Now we can write the equation: \[ y = 5x + 30. \]
Predictions
Using the equation, we can predict the gallons of gas at any time \( x \).
Here are the meanings of the specific points:
- The point \( (4, 50) \) means that after 4 minutes, there are 50 gallons of gas in the boat.
- The point \( (10, 80) \) means that after 10 minutes, there are 80 gallons of gas in the boat.
These points directly relate to the linear relationship and confirm that Leo fills the gas at a constant rate of 5 gallons per minute.