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Sagot :
### Part A: Determine the Type of Function
To determine whether the investment values follow a linear or exponential function, we analyze how the values change over the years:
Option 1:
- From Year 1 to Year 2: [tex]\( \$1300 \to \$1690 \)[/tex]
- Approximate growth factor: [tex]\( \frac{1690}{1300} \approx 1.3 \)[/tex]
- From Year 2 to Year 3: [tex]\( \$1690 \to \$2197 \)[/tex]
- Approximate growth factor: [tex]\( \frac{2197}{1690} \approx 1.3 \)[/tex]
Since the investment values increase by approximately multiplying by a constant factor (1.3), Option 1 can be described using an exponential function.
Option 2:
- From Year 1 to Year 2: [tex]\( \$1300 \to \$1600 \)[/tex]
- Difference: [tex]\(1600 - 1300 = 300\)[/tex]
- From Year 2 to Year 3: [tex]\( \$1600 \to \$1900 \)[/tex]
- Difference: [tex]\(1900 - 1600 = 300\)[/tex]
Since the investment values increase by a constant difference of \[tex]$300, Option 2 can be described using a linear function. ### Part B: Write the Functions Based on the observations from Part A, we can define the functions for each option: Option 1 (Exponential): The formula for an exponential function is \( f(n) = \text{initial amount} \times (\text{growth rate})^n \). - Initial amount: \$[/tex]1000
- Growth rate: 1.3
[tex]\[ f_1(n) = 1000 \times (1.3)^n \][/tex]
Option 2 (Linear):
The formula for a linear function is [tex]\( f(n) = \text{initial amount} + (\text{annual increase} \times n) \)[/tex].
- Initial amount: \[tex]$1000 - Annual increase: \$[/tex]300
[tex]\[ f_2(n) = 1000 + (300 \times n) \][/tex]
### Part C: Calculate the Investment Value After 20 Years
To find the investment values after 20 years for both options, we substitute [tex]\( n = 20 \)[/tex] into the respective functions:
Option 1 (Exponential):
[tex]\[ f_1(20) = 1000 \times (1.3)^{20} \][/tex]
The investment value after 20 years is approximately:
[tex]\[ 190,049.64 \][/tex]
Option 2 (Linear):
[tex]\[ f_2(20) = 1000 + (300 \times 20) \][/tex]
The investment value after 20 years is:
[tex]\[ 1000 + 6000 = 7000 \][/tex]
### Determine the Difference:
To find if there is a significant difference, we calculate the difference between the two investment values:
[tex]\[ \text{Difference} = 190,049.64 - 7000 = 183,049.64 \][/tex]
### Conclusion:
Belinda's investment value after 20 years using Option 1 (exponential growth) will be \[tex]$190,049.64, and using Option 2 (linear growth) will be \$[/tex]7000. There is a significant difference of \$183,049.64 between the two options. Therefore, Belinda should choose Option 1 to maximize her investment's growth over 20 years.
To determine whether the investment values follow a linear or exponential function, we analyze how the values change over the years:
Option 1:
- From Year 1 to Year 2: [tex]\( \$1300 \to \$1690 \)[/tex]
- Approximate growth factor: [tex]\( \frac{1690}{1300} \approx 1.3 \)[/tex]
- From Year 2 to Year 3: [tex]\( \$1690 \to \$2197 \)[/tex]
- Approximate growth factor: [tex]\( \frac{2197}{1690} \approx 1.3 \)[/tex]
Since the investment values increase by approximately multiplying by a constant factor (1.3), Option 1 can be described using an exponential function.
Option 2:
- From Year 1 to Year 2: [tex]\( \$1300 \to \$1600 \)[/tex]
- Difference: [tex]\(1600 - 1300 = 300\)[/tex]
- From Year 2 to Year 3: [tex]\( \$1600 \to \$1900 \)[/tex]
- Difference: [tex]\(1900 - 1600 = 300\)[/tex]
Since the investment values increase by a constant difference of \[tex]$300, Option 2 can be described using a linear function. ### Part B: Write the Functions Based on the observations from Part A, we can define the functions for each option: Option 1 (Exponential): The formula for an exponential function is \( f(n) = \text{initial amount} \times (\text{growth rate})^n \). - Initial amount: \$[/tex]1000
- Growth rate: 1.3
[tex]\[ f_1(n) = 1000 \times (1.3)^n \][/tex]
Option 2 (Linear):
The formula for a linear function is [tex]\( f(n) = \text{initial amount} + (\text{annual increase} \times n) \)[/tex].
- Initial amount: \[tex]$1000 - Annual increase: \$[/tex]300
[tex]\[ f_2(n) = 1000 + (300 \times n) \][/tex]
### Part C: Calculate the Investment Value After 20 Years
To find the investment values after 20 years for both options, we substitute [tex]\( n = 20 \)[/tex] into the respective functions:
Option 1 (Exponential):
[tex]\[ f_1(20) = 1000 \times (1.3)^{20} \][/tex]
The investment value after 20 years is approximately:
[tex]\[ 190,049.64 \][/tex]
Option 2 (Linear):
[tex]\[ f_2(20) = 1000 + (300 \times 20) \][/tex]
The investment value after 20 years is:
[tex]\[ 1000 + 6000 = 7000 \][/tex]
### Determine the Difference:
To find if there is a significant difference, we calculate the difference between the two investment values:
[tex]\[ \text{Difference} = 190,049.64 - 7000 = 183,049.64 \][/tex]
### Conclusion:
Belinda's investment value after 20 years using Option 1 (exponential growth) will be \[tex]$190,049.64, and using Option 2 (linear growth) will be \$[/tex]7000. There is a significant difference of \$183,049.64 between the two options. Therefore, Belinda should choose Option 1 to maximize her investment's growth over 20 years.
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