Let's go through each part of the question in detail:

Part 1: Linear Model for Depreciation

The tractor's initial value is $158,000, and after 10 years, it will have a trade-in value of $85,000. To find a linear model, we need to determine the rate of depreciation per year.

The depreciation is calculated as: $\text{Depreciation per year} = \frac{158,000 - 85,000}{10} = 7,300 \, \text{dollars per year}.$ So, the value of the tractor decreases by $7,300 each year.

The linear model for the value $V$ of the tractor after $t$ years is given by: $V = -7,300t + 158,000.$ This is the linear model where $V$ is the value after $t$ years, with a negative slope representing the annual depreciation.

Part 2: Depreciated Value After 6 Years

To find the depreciated value after 6 years, substitute $t = 6$ into the equation: $V = -7,300(6) + 158,000 = -43,800 + 158,000 = 114,200.$ So, after 6 years, the tractor's value will be $114,200.

Part 3: When Will the Value Fall Below $70,000?

We need to determine when the depreciated value falls below $70,000. Set the equation equal to 70,000 and solve for $t$: $70,000 = -7,300t + 158,000.$ Solving for $t$: $7,300t = 158,000 - 70,000 = 88,000,$ $t = \frac{88,000}{7,300} \approx 12.05.$ So, the value will fall below $70,000 during the 13th year (since we round up to the nearest integer).

Part 4: Graph of Depreciation

The graph of $V = -7,300t + 158,000$ is a straight line that starts at $158,000 when $t = 0$ and decreases by $7,300 per year. When $t = 10$, the value is $85,000.

Correct Graph:

The line should start at $158,000 at $t = 0$ and decrease to $85,000 at $t = 10$. It should continue to fall as $t$ increases.

The correct graph is B: The line falls from left to right, passing through the points (0, 158,000) and (5, 121,500).

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Would you like more details or have any further questions?

5 Related Questions:

1. How is the slope of a linear depreciation model determined?
2. What would the value of the tractor be after 8 years?
3. How does the depreciation rate affect the graph of the value over time?
4. What happens to the model after the 10th year if the depreciation stops at $85,000?
5. Can this model be applied to other assets? Why or why not?

Tip:

When solving problems involving linear depreciation, remember that the slope of the line (rate of depreciation) is key to understanding how the value changes over time.