Answer :
Certainly! Let's tackle each part individually with detailed explanations:
Part A:
To determine the type of function, we need to analyze the given values for each car year by year.
- Car 1:
- Year 1: \$40,500
- Year 2: \$36,450
- Year 3: \$32,805
For Car 1, the value decrement isn't constant. Instead, it seems to decrease by a fixed percentage each year. This type of behavior is characteristic of an exponential decay function.
- Car 2:
- Year 1: \$42,000
- Year 2: \$39,000
- Year 3: \$36,000
For Car 2, the difference in value between years is \$3,000 each time. This indicates a constant decrease, which is characteristic of a linear decay function.
Part B:
Next, we will establish the functions that describe the value of each car after \(x\) years.
- Car 1:
Since Car 1 is decreasing by 10% annually, we can express it as:
[tex]\[ f_1(x) = 45000 \times (0.9)^x \][/tex]
Here, 45,000 is the initial value, and 0.9 represents the remaining value after a 10% decrease each year.
- Car 2:
Since Car 2 decreases by \$3,000 each year, we can express it as:
[tex]\[ f_2(x) = 45000 - 3000 \times x \][/tex]
Here, 45,000 is the initial value, and the term \(3000 \times x\) represents a linear decrease of \$3,000 per year.
Part C:
To find out which car will have the greater value after 13 years and whether there's a significant difference in their values, we use our functions:
- For Car 1 after 13 years:
[tex]\[ f_1(13) = 45000 \times (0.9)^{13} \approx 11438.40 \][/tex]
- For Car 2 after 13 years:
[tex]\[ f_2(13) = 45000 - 3000 \times 13 = 45000 - 39000 = 6000 \][/tex]
Now, comparing these values, Car 1’s value is approximately \(\[tex]$11,438.40\) and Car 2’s value is \(\$[/tex]6,000\).
To determine the difference:
[tex]\[ \text{Difference} = |11438.40 - 6000| = 5438.40 \][/tex]
Therefore, after 13 years, Car 1 will have a value of approximately \(\[tex]$11,438.40\) and Car 2 will have a value of \$[/tex]6,000. The difference in their values will be around \$5,438.40, which is significant.
Conclusion:
Belinda should consider buying Car 1 since it retains more value after 13 years compared to Car 2. There is indeed a significant difference in their values, with Car 1 being worth approximately \$5,438.40 more than Car 2 after 13 years.
Part A:
To determine the type of function, we need to analyze the given values for each car year by year.
- Car 1:
- Year 1: \$40,500
- Year 2: \$36,450
- Year 3: \$32,805
For Car 1, the value decrement isn't constant. Instead, it seems to decrease by a fixed percentage each year. This type of behavior is characteristic of an exponential decay function.
- Car 2:
- Year 1: \$42,000
- Year 2: \$39,000
- Year 3: \$36,000
For Car 2, the difference in value between years is \$3,000 each time. This indicates a constant decrease, which is characteristic of a linear decay function.
Part B:
Next, we will establish the functions that describe the value of each car after \(x\) years.
- Car 1:
Since Car 1 is decreasing by 10% annually, we can express it as:
[tex]\[ f_1(x) = 45000 \times (0.9)^x \][/tex]
Here, 45,000 is the initial value, and 0.9 represents the remaining value after a 10% decrease each year.
- Car 2:
Since Car 2 decreases by \$3,000 each year, we can express it as:
[tex]\[ f_2(x) = 45000 - 3000 \times x \][/tex]
Here, 45,000 is the initial value, and the term \(3000 \times x\) represents a linear decrease of \$3,000 per year.
Part C:
To find out which car will have the greater value after 13 years and whether there's a significant difference in their values, we use our functions:
- For Car 1 after 13 years:
[tex]\[ f_1(13) = 45000 \times (0.9)^{13} \approx 11438.40 \][/tex]
- For Car 2 after 13 years:
[tex]\[ f_2(13) = 45000 - 3000 \times 13 = 45000 - 39000 = 6000 \][/tex]
Now, comparing these values, Car 1’s value is approximately \(\[tex]$11,438.40\) and Car 2’s value is \(\$[/tex]6,000\).
To determine the difference:
[tex]\[ \text{Difference} = |11438.40 - 6000| = 5438.40 \][/tex]
Therefore, after 13 years, Car 1 will have a value of approximately \(\[tex]$11,438.40\) and Car 2 will have a value of \$[/tex]6,000. The difference in their values will be around \$5,438.40, which is significant.
Conclusion:
Belinda should consider buying Car 1 since it retains more value after 13 years compared to Car 2. There is indeed a significant difference in their values, with Car 1 being worth approximately \$5,438.40 more than Car 2 after 13 years.