To create the exponential function that represents the value of the car after x years, we start by determining the value decline.
Part A: The initial value of the car is $40,000, and it decreases by 15% each year. Therefore, it retains 85% of its value each year (100% - 15% = 85%). In decimal form, this is 0.85.
Thus, the exponential function can be defined as:
\[ f(x) = 40000(0.85)^x \]
Part B: Now, we can fill in the table based on the function we have. We will calculate the value of the car for \( x = 0, 1, 3, 5, \) and \( 8 \).
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\( x = 0 \): \[ f(0) = 40000(0.85)^0 = 40000(1) = 40000.00 \]
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\( x = 1 \): \[ f(1) = 40000(0.85)^1 = 40000(0.85) = 34000.00 \]
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\( x = 3 \): \[ f(3) = 40000(0.85)^3 = 40000(0.85 \times 0.85 \times 0.85) = 40000(0.614125) \approx 24565.00 \]
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\( x = 5 \): \[ f(5) = 40000(0.85)^5 = 40000(0.85 \times 0.85 \times 0.85 \times 0.85 \times 0.85) = 40000(0.443705) \approx 17748.20 \]
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\( x = 8 \): \[ f(8) = 40000(0.85)^8 \approx 40000(0.272490) \approx 10899.65 \]
Now, we compile the values into the table:
| x | f(x) | |-----|------------| | 0 | 40000.00 | | 1 | 34000.00 | | 3 | 24565.00 | | 5 | 17748.20 | | 8 | 10899.65 |
So the final rounded amounts for filling out the table are:
- For \( x=0 \): 40000.00
- For \( x=1 \): 34000.00
- For \( x=3 \): 24565.00
- For \( x=5 \): 17748.20
- For \( x=8 \): 10899.65
Note: Ensure you check the calculations for \( f(5) \) and \( f(8) \) since the intermediate decimals can slightly vary depending on rounding at each step. After recalculating:
- \( f(5) = 40000 \times (0.85^5) \approx 40000 \times 0.443705 \approx 17748.20 \)
- \( f(8) = 40000 \times (0.85^8) \approx 40000 \times 0.272490 \approx 10899.60 \)
This means the final answers are confirmed as is.
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