Cecily bought a car that is depreciating at a rate of 18% each year. The value V of her car after t years can be modeled by the function V(t) = 65000(0.82)t. She wants to track the value of her car each year, for 5 years.

Question

What would be a possible range for this situation?
A) 0 < v ≤ 5
B) All real numbers
C) 0 < v ≤ 65000
D) 30 < v ≤ 65000

Answer 1

To analyze the situation, let's consider the function provided:

$$V(t)=65000(0.82{)}^t$$

This function models the value of Cecily's car over time, where:

• $V(t)$ is the value of the car after $t$ years,
• 65000 is the initial value of the car,
• $0.82$ represents the depreciation rate (100% - 18% = 82%).

As $t$ increases, $V(t)$ will decrease because $(0.82{)}^t$ will become smaller over time, approaching but never reaching zero.

Now, let's find the possible range of values for $V(t)$ over the 5-year period (from $t=0$ to $t=5$):

1. At $t=0$: $$V(0)=65000(0.82{)}^0=65000$$

2. At $t=1$: $$V(1)=65000(0.82{)}^1=65000\times 0.82\approx 53300$$

3. At $t=2$: $$V(2)=65000(0.82{)}^2=65000\times {0.82}^2\approx 43660$$

4. At $t=3$: $$V(3)=65000(0.82{)}^3=65000\times {0.82}^3\approx 35850$$

5. At $t=4$: $$V(4)=65000(0.82{)}^4=65000\times {0.82}^4\approx 29480$$

6. At $t=5$: $$V(5)=65000(0.82{)}^5=65000\times {0.82}^5\approx 24170$$

Over the first five years, the value of the car decreases from 65,000 down to approximately 24,170.

Thus, the range of values for $V(t)$ is:

$$0<V(t)\le 65000$$

This means the car value is always greater than 0 (the car can’t have a negative value) and less than or equal to 65,000.

The correct option for the possible range for this situation is:

C) 0 < v ≤ 65000