A farmer buys a new tractor for ​$158 comma 000 and assumes that it will have a​ trade-in value of ​$85 comma 000 after 10 years. The farmer uses a constant rate of depreciation to determine the annual value of the tractor.
Question content area bottom
Part 1
​(A) Find a linear model for the depreciated value V of the tractor t years after it was purchased.
Vequals

negative 7,300 t plus 158,000
​(Type your answer in​ slope-intercept form.)
Part 2
​(B) What is the depreciated value of the tractor after 6 ​years?
The depreciated value of the tractor after 6 years is ​$

114,200.
Part 3
​(C) When will the depreciated value fall below ​$70 comma 000​?
The depreciated value will fall below ​$70 comma 000 during the

13th year.
​(Round up to the nearest​ integer.)
Part 4
​(D) Graph V for 0less than or equalstless than or equals20. Choose the correct graph below.
A.
0
10
20
0
170,000
t
V
85,000

A coordinate system has a horizontal t-axis from 0 to 20 in increments of 1 and a vertical V-axis is labeled from 0 to 170,000 in increments of 10000. A line rises from left to right and passes through the points (0, 58000) and (5, 94500).
B.
0
10
20
0
170,000
t
V
85,000

A coordinate system has a horizontal t-axis from 0 to 20 in increments of 1 and a vertical V-axis is labeled from 0 to 170,000 in increments of 10000. A line falls from left to right and passes through the points (0, 158000) and (5, 121500).
C.
0
10
20
0
170,000
t
V
85,000

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).

---

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.

A farmer buys a new tractor for $158,000 and assumes that it will have a trade-in value of $85,000 after 10 years. The farmer uses a constant rate of depreciation to determine the annual value of the tractor. Find a linear model for the depreciated value V of the tractor t years after it was purchased, calculate the value after 6 years, determine when the value falls below $70,000, and select the correct graph for depreciation.

Explore how to create a linear depreciation model for a tractor purchased for $158,000 with a trade-in value of $85,000 after 10 years. Learn how to calculate its value after 6 years and predict when it will fall below $70,000 using algebra and linear equations. Suitable for high school students.

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