The problem asks for a linear model for the depreciated value $V$ of the tractor, where $t$ represents the number of years after the tractor was purchased.

Step 1: Identify key points.

We are given:

• The tractor is bought for $146,000, meaning at $t = 0$, the value $V(0) = 146,000$.
• After 10 years, the tractor's trade-in value is $79,000, meaning at $t = 10$, $V(10) = 79,000$.

These give us two points: $(0, 146000)$ and $(10, 79000)$.

Step 2: Find the slope of the linear model.

The slope $m$ of the linear equation $V = mt + b$ can be calculated using the slope formula: $m = \frac{V(10) - V(0)}{10 - 0} = \frac{79000 - 146000}{10 - 0} = \frac{-67000}{10} = -6700$

Step 3: Use the slope-intercept form.

The general equation of a line is $V = mt + b$, where $m$ is the slope and $b$ is the y-intercept. From the given data, we know $b = 146,000$, as the initial value of the tractor at $t = 0$.

So the equation is: $V = -6700t + 146000$

Final answer:

$V(t) = -6700t + 146000$

This equation represents the depreciated value of the tractor $t$ years after it was purchased.

Would you like further details on how this model works, or do you have any specific questions?

5 Related Questions:

1. How would the equation change if the tractor's initial value was higher or lower?
2. What happens to the value of the tractor after 15 years?
3. How would you calculate the tractor's value at $t = 5$ years using the linear model?
4. Can you use this method for non-linear depreciation models?
5. What if the tractor's trade-in value was lower than expected after 10 years?

Tip:

Always verify the accuracy of your slope by double-checking the difference in value over time when working with linear depreciation models.