Question
Leo's Mom purchased a car for $32,000. The value of her car decreases by 14% annually. You can use a function to describe the value of the car x years since Leo's mom bought it.
Match the below with the correct answer.
(3 points)
Questions:
Is this Linear or Exponential?
What formula would you need to solve this type of equation?
After 2 years what would the formula for depreciation be on Leo's mom's Car?
What is the value of her car after 2 years?
Match the below with the correct answer.
(3 points)
Questions:
Is this Linear or Exponential?
What formula would you need to solve this type of equation?
After 2 years what would the formula for depreciation be on Leo's mom's Car?
What is the value of her car after 2 years?
Answers
GPT-4o mini
Let's address each of your questions regarding the depreciation of Leo's Mom's car.
### 1. Is this Linear or Exponential?
The explanation for the depreciation in the value of the car is exponential because the car's value decreases by a fixed percentage each year (14%). This results in an exponential decay rather than a linear decrease.
**Answer**: Exponential
---
### 2. What formula would you need to solve this type of equation?
The formula for exponential decay can be expressed as:
\[ V(t) = V_0 \times (1 - r)^t \]
Where:
- \( V(t) \) is the value of the car after \( t \) years,
- \( V_0 \) is the initial value ($32,000),
- \( r \) is the rate of depreciation (14% or 0.14), and
- \( t \) is the number of years.
So, the formula specifically for Leo's mom's car would be:
\[ V(t) = 32000 \times (1 - 0.14)^t \]
or simplified,
\[ V(t) = 32000 \times (0.86)^t \]
**Answer**: \( V(t) = 32000 \times (0.86)^t \)
---
### 3. After 2 years what would the formula for depreciation be on Leo's mom's car?
Using the value function we established above, we substitute \( t = 2 \) into the formula:
\[ V(2) = 32000 \times (0.86)^2 \]
Calculating \( (0.86)^2 \):
\[
(0.86)^2 = 0.7396
\]
Thus,
\[
V(2) = 32000 \times 0.7396 \approx 23675.20
\]
So the formula for the value after 2 years is:
\[ V(2) = 32000 \times (0.86)^2 \]
**Answer**: \( V(2) = 32000 \times (0.86)^2 \)
---
### 4. What is the value of her car after 2 years?
Using the formula we derived:
\[
V(2) \approx 32000 \times 0.7396 \approx 23675.20
\]
So, the value of her car after 2 years is approximately $23,675.20.
**Answer**: Approximately $23,675.20 after 2 years.
### 1. Is this Linear or Exponential?
The explanation for the depreciation in the value of the car is exponential because the car's value decreases by a fixed percentage each year (14%). This results in an exponential decay rather than a linear decrease.
**Answer**: Exponential
---
### 2. What formula would you need to solve this type of equation?
The formula for exponential decay can be expressed as:
\[ V(t) = V_0 \times (1 - r)^t \]
Where:
- \( V(t) \) is the value of the car after \( t \) years,
- \( V_0 \) is the initial value ($32,000),
- \( r \) is the rate of depreciation (14% or 0.14), and
- \( t \) is the number of years.
So, the formula specifically for Leo's mom's car would be:
\[ V(t) = 32000 \times (1 - 0.14)^t \]
or simplified,
\[ V(t) = 32000 \times (0.86)^t \]
**Answer**: \( V(t) = 32000 \times (0.86)^t \)
---
### 3. After 2 years what would the formula for depreciation be on Leo's mom's car?
Using the value function we established above, we substitute \( t = 2 \) into the formula:
\[ V(2) = 32000 \times (0.86)^2 \]
Calculating \( (0.86)^2 \):
\[
(0.86)^2 = 0.7396
\]
Thus,
\[
V(2) = 32000 \times 0.7396 \approx 23675.20
\]
So the formula for the value after 2 years is:
\[ V(2) = 32000 \times (0.86)^2 \]
**Answer**: \( V(2) = 32000 \times (0.86)^2 \)
---
### 4. What is the value of her car after 2 years?
Using the formula we derived:
\[
V(2) \approx 32000 \times 0.7396 \approx 23675.20
\]
So, the value of her car after 2 years is approximately $23,675.20.
**Answer**: Approximately $23,675.20 after 2 years.