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Karen purchased a car for $40,000 by taking out a loan that will take 7 years to pay off. Given the car depreciates in value by...Question
Karen purchased a car for $40,000 by taking out a loan that will take 7 years to pay off. Given the car depreciates in value by 11% each year, the situation can be modeled by the equation v(t)=40,000(0.89)^t
, where v is the value of the car after t years. If Karen wants to know how much the car will be worth in 7 years, which of the following options would be an appropriate range?
Option #1: 0<t≤40,000
Option #2: 0<v≤40,000
Option #3: 0<t≤7
Option #4: 0<v≤7
(1 point)
Option #
, where v is the value of the car after t years. If Karen wants to know how much the car will be worth in 7 years, which of the following options would be an appropriate range?
Option #1: 0<t≤40,000
Option #2: 0<v≤40,000
Option #3: 0<t≤7
Option #4: 0<v≤7
(1 point)
Option #
Answers
Answer
its option 2
Answered by
GPT-4o mini
To determine the appropriate range for the value of the car after 7 years, we can analyze the equation provided:
\[ v(t) = 40,000(0.89)^t \]
Given that \( t \) represents time in years, if we want to find the value of the car after 7 years, we plug in \( t = 7 \):
\[ v(7) = 40,000(0.89)^7 \]
Calculating \( (0.89)^7 \):
\[ (0.89)^7 \approx 0.5132 \]
Now, substituting back:
\[ v(7) \approx 40,000 \times 0.5132 \approx 20,528 \]
So, the value of the car after 7 years is approximately $20,528.
Now, let's evaluate the options given in the context of this result:
1. **Option #1: \( 0 < t \leq 40,000 \)** - This doesn't make sense as \( t \) represents years, which shouldn't exceed the number of years and it doesn't provide any specific range.
2. **Option #2: \( 0 < v \leq 40,000 \)** - This option considers the value of the car, which will indeed still be less than $40,000, but it does not properly describe the specific situation at t=7.
3. **Option #3: \( 0 < t \leq 7 \)** - This option is relevant because it correctly confines \( t \) to the time frame of interest (the 7 years of loan payment).
4. **Option #4: \( 0 < v \leq 7 \)** - This does not make sense in this context as the car's value will be much greater than 7.
Based on this analysis, the most appropriate option that relates to the scenario is:
**Option #3: \( 0 < t \leq 7 \)**.
\[ v(t) = 40,000(0.89)^t \]
Given that \( t \) represents time in years, if we want to find the value of the car after 7 years, we plug in \( t = 7 \):
\[ v(7) = 40,000(0.89)^7 \]
Calculating \( (0.89)^7 \):
\[ (0.89)^7 \approx 0.5132 \]
Now, substituting back:
\[ v(7) \approx 40,000 \times 0.5132 \approx 20,528 \]
So, the value of the car after 7 years is approximately $20,528.
Now, let's evaluate the options given in the context of this result:
1. **Option #1: \( 0 < t \leq 40,000 \)** - This doesn't make sense as \( t \) represents years, which shouldn't exceed the number of years and it doesn't provide any specific range.
2. **Option #2: \( 0 < v \leq 40,000 \)** - This option considers the value of the car, which will indeed still be less than $40,000, but it does not properly describe the specific situation at t=7.
3. **Option #3: \( 0 < t \leq 7 \)** - This option is relevant because it correctly confines \( t \) to the time frame of interest (the 7 years of loan payment).
4. **Option #4: \( 0 < v \leq 7 \)** - This does not make sense in this context as the car's value will be much greater than 7.
Based on this analysis, the most appropriate option that relates to the scenario is:
**Option #3: \( 0 < t \leq 7 \)**.
Answered by
GPT-4o mini
I apologize for the confusion. Let's clarify the reasoning regarding Option #2 in relation to the value of the car.
The equation for the depreciation of the car is given as:
\[ v(t) = 40,000(0.89)^t \]
By substituting \( t = 7 \):
\[ v(7) = 40,000(0.89)^7 \]
Calculating \( (0.89)^7 \):
\[ (0.89)^7 \approx 0.5132 \]
Now we calculate \( v(7) \):
\[ v(7) \approx 40,000 \times 0.5132 \approx 20,528 \]
From the calculation, the value of the car after 7 years is approximately $20,528. This value is indeed within the range \( 0 < v \leq 40,000 \).
### Evaluating Option #2:
- **Option #2: \( 0 < v \leq 40,000 \)** - This option appropriately describes the situation because it states that the value of the car will always be greater than 0 and will not exceed the purchase price of $40,000.
So, you are correct. The appropriate choice that encapsulates the scenario is indeed:
**Option #2: \( 0 < v \leq 40,000 \)**.
Thank you for your patience and for pointing out the need for clarification!
The equation for the depreciation of the car is given as:
\[ v(t) = 40,000(0.89)^t \]
By substituting \( t = 7 \):
\[ v(7) = 40,000(0.89)^7 \]
Calculating \( (0.89)^7 \):
\[ (0.89)^7 \approx 0.5132 \]
Now we calculate \( v(7) \):
\[ v(7) \approx 40,000 \times 0.5132 \approx 20,528 \]
From the calculation, the value of the car after 7 years is approximately $20,528. This value is indeed within the range \( 0 < v \leq 40,000 \).
### Evaluating Option #2:
- **Option #2: \( 0 < v \leq 40,000 \)** - This option appropriately describes the situation because it states that the value of the car will always be greater than 0 and will not exceed the purchase price of $40,000.
So, you are correct. The appropriate choice that encapsulates the scenario is indeed:
**Option #2: \( 0 < v \leq 40,000 \)**.
Thank you for your patience and for pointing out the need for clarification!