Let's analyze the problem.

(A) Linear Model for Depreciation

The problem states that a tractor was purchased for $158,000 and is expected to have a trade-in value of $89,000 after 10 years. The depreciation is assumed to be linear.

A linear depreciation model takes the form: $V(t) = mt + b$ where:

• $V(t)$ is the value of the tractor after $t$ years,
• $m$ is the rate of depreciation (slope),
• $b$ is the initial value of the tractor.

Given:

• Initial value $V(0) = 158,000$,
• Value after 10 years $V(10) = 89,000$.

Slope (Rate of Depreciation)

The slope $m$ is calculated by the change in value over time: $m = \frac{V(10) - V(0)}{10 - 0} = \frac{89,000 - 158,000}{10} = \frac{-69,000}{10} = -6,900$

Linear Model Equation

Substituting the slope and the initial value into the equation: $V(t) = -6,900t + 158,000$

This matches the linear model provided in the problem: $V(t) = -6,900t + 158,000$

(B) Depreciated Value After 6 Years

To find the value of the tractor after 6 years, substitute $t = 6$ into the linear model: $V(6) = -6,900(6) + 158,000$

Calculating this: $V(6) = -41,400 + 158,000 = 116,600$

Thus, the depreciated value of the tractor after 6 years is $116,600.

Would you like further details or have any questions?

Here are five related questions:

1. How would the depreciation model change if the tractor's value after 10 years was different?
2. What is the significance of the slope in the depreciation model?
3. How can we use this model to predict the value of the tractor after 15 years?
4. What would happen if the depreciation was not linear?
5. How could we calculate the year when the tractor's value drops below a certain threshold?

Tip: Always double-check your units and ensure consistency when plugging values into equations to avoid calculation errors.