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Sagot :
Certainly! Let's address each part of the question step-by-step.
### Part A (2 points)
What type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? Explain your answer.
1. Car 1: The values for Car 1 are 32,000, 26,000, and 20,000 for the first three years. Notice how each year the value decreases by a constant amount:
- From Year 1 to Year 2: [tex]\( 32,000 - 26,000 = 6,000 \)[/tex]
- From Year 2 to Year 3: [tex]\( 26,000 - 20,000 = 6,000 \)[/tex]
The decrease is constant by [tex]\( \$6,000 \)[/tex] each year. This indicates a linear depreciation.
2. Car 2: The values for Car 2 are 32,300, 27,455, and 23,336.75 for the first three years. Notice how the amount of decrease every year is not constant:
- From Year 1 to Year 2: [tex]\( 32,300 - 27,455 = 4,845 \)[/tex]
- From Year 2 to Year 3: [tex]\( 27,455 - 23,336.75 = 4,118.25 \)[/tex]
The value of the car decreases by a proportion rather than a constant amount every year. This indicates an exponential depreciation.
### Part B (4 points)
Write one function for each car to describe the value of the car [tex]\( f(x) \)[/tex], in dollars, after [tex]\( x \)[/tex] years.
1. Car 1 (Linear depreciation):
We know the initial value of Car 1 is [tex]\( \$38,000 \)[/tex] and it loses [tex]\( \$6,000 \)[/tex] every year. The function describing the value of Car 1 after [tex]\( x \)[/tex] years can be written as:
[tex]\[ f_1(x) = 38000 - 6000x \][/tex]
2. Car 2 (Exponential depreciation):
The initial value of Car 2 is [tex]\( \$38,000 \)[/tex]. To model the exponential depreciation, we need to calculate the rate which would make [tex]\( 38,000 \)[/tex] decrease to [tex]\( 23,336.75 \)[/tex] over 3 years. Using the given values, the depreciation rate can be found as:
[tex]\[ r = \left(\frac{23336.75}{38000}\right)^{\frac{1}{3}} \][/tex]
The function describing the value of Car 2 after [tex]\( x \)[/tex] years using this rate [tex]\( r \)[/tex] can be written as:
[tex]\[ f_2(x) = 38000 \times r^x \][/tex]
### Part C (4 points)
Belinda wants to purchase a car that would have the greatest value in five years. Will there be any significant difference in the value of either car after five years? Explain your answer, and show the value of each car after five years.
1. Calculating the value of Car 1 after 5 years:
[tex]\[ f_1(5) = 38000 - 6000 \times 5 = 38000 - 30000 = 8000 \][/tex]
The value of Car 1 after 5 years will be [tex]\( \$8,000 \)[/tex].
2. Calculating the value of Car 2 after 5 years:
Using the calculated rate [tex]\( r \approx 0.85 \)[/tex] from part B:
[tex]\[ f_2(5) = 38000 \times 0.85^5 \approx 16860.80 \][/tex]
The value of Car 2 after 5 years will be approximately [tex]\( \$16,860.80 \)[/tex].
3. Determining the greater value and significance of the difference:
Comparing the two values after 5 years, Car 2 has a significantly higher value than Car 1. Car 1 is valued at [tex]\( \$8,000 \)[/tex] while Car 2 is valued at approximately [tex]\( \$16,860.80 \)[/tex]. The difference is:
[tex]\[ 16860.80 - 8000 = 8860.80 \][/tex]
This is a significant difference since [tex]\( \$8,860.80 \)[/tex] is a substantial amount. Therefore, Belinda should opt to purchase Car 2, as it retains a significantly higher value after five years.
### Summary:
- Car 1: Linear depreciation; [tex]\( f_1(x) = 38000 - 6000x \)[/tex]
- Car 2: Exponential depreciation; [tex]\( f_2(x) = 38000 \times r^x \)[/tex]
- After 5 years:
- Car 1 Value: [tex]\( \$8,000 \)[/tex]
- Car 2 Value: [tex]\( \$16,860.80 \)[/tex]
- Significant difference: Yes, Car 2 has a significantly higher value after 5 years.
### Part A (2 points)
What type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? Explain your answer.
1. Car 1: The values for Car 1 are 32,000, 26,000, and 20,000 for the first three years. Notice how each year the value decreases by a constant amount:
- From Year 1 to Year 2: [tex]\( 32,000 - 26,000 = 6,000 \)[/tex]
- From Year 2 to Year 3: [tex]\( 26,000 - 20,000 = 6,000 \)[/tex]
The decrease is constant by [tex]\( \$6,000 \)[/tex] each year. This indicates a linear depreciation.
2. Car 2: The values for Car 2 are 32,300, 27,455, and 23,336.75 for the first three years. Notice how the amount of decrease every year is not constant:
- From Year 1 to Year 2: [tex]\( 32,300 - 27,455 = 4,845 \)[/tex]
- From Year 2 to Year 3: [tex]\( 27,455 - 23,336.75 = 4,118.25 \)[/tex]
The value of the car decreases by a proportion rather than a constant amount every year. This indicates an exponential depreciation.
### Part B (4 points)
Write one function for each car to describe the value of the car [tex]\( f(x) \)[/tex], in dollars, after [tex]\( x \)[/tex] years.
1. Car 1 (Linear depreciation):
We know the initial value of Car 1 is [tex]\( \$38,000 \)[/tex] and it loses [tex]\( \$6,000 \)[/tex] every year. The function describing the value of Car 1 after [tex]\( x \)[/tex] years can be written as:
[tex]\[ f_1(x) = 38000 - 6000x \][/tex]
2. Car 2 (Exponential depreciation):
The initial value of Car 2 is [tex]\( \$38,000 \)[/tex]. To model the exponential depreciation, we need to calculate the rate which would make [tex]\( 38,000 \)[/tex] decrease to [tex]\( 23,336.75 \)[/tex] over 3 years. Using the given values, the depreciation rate can be found as:
[tex]\[ r = \left(\frac{23336.75}{38000}\right)^{\frac{1}{3}} \][/tex]
The function describing the value of Car 2 after [tex]\( x \)[/tex] years using this rate [tex]\( r \)[/tex] can be written as:
[tex]\[ f_2(x) = 38000 \times r^x \][/tex]
### Part C (4 points)
Belinda wants to purchase a car that would have the greatest value in five years. Will there be any significant difference in the value of either car after five years? Explain your answer, and show the value of each car after five years.
1. Calculating the value of Car 1 after 5 years:
[tex]\[ f_1(5) = 38000 - 6000 \times 5 = 38000 - 30000 = 8000 \][/tex]
The value of Car 1 after 5 years will be [tex]\( \$8,000 \)[/tex].
2. Calculating the value of Car 2 after 5 years:
Using the calculated rate [tex]\( r \approx 0.85 \)[/tex] from part B:
[tex]\[ f_2(5) = 38000 \times 0.85^5 \approx 16860.80 \][/tex]
The value of Car 2 after 5 years will be approximately [tex]\( \$16,860.80 \)[/tex].
3. Determining the greater value and significance of the difference:
Comparing the two values after 5 years, Car 2 has a significantly higher value than Car 1. Car 1 is valued at [tex]\( \$8,000 \)[/tex] while Car 2 is valued at approximately [tex]\( \$16,860.80 \)[/tex]. The difference is:
[tex]\[ 16860.80 - 8000 = 8860.80 \][/tex]
This is a significant difference since [tex]\( \$8,860.80 \)[/tex] is a substantial amount. Therefore, Belinda should opt to purchase Car 2, as it retains a significantly higher value after five years.
### Summary:
- Car 1: Linear depreciation; [tex]\( f_1(x) = 38000 - 6000x \)[/tex]
- Car 2: Exponential depreciation; [tex]\( f_2(x) = 38000 \times r^x \)[/tex]
- After 5 years:
- Car 1 Value: [tex]\( \$8,000 \)[/tex]
- Car 2 Value: [tex]\( \$16,860.80 \)[/tex]
- Significant difference: Yes, Car 2 has a significantly higher value after 5 years.
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