A car was valued at $42,000 in the year 1990. The value depreciated to $13,000 by the year 2005.

A) What was the annual rate of change between 1990 and 2005?
 = 
 Round the rate of decrease to 4 decimal places.

B) What is the correct answer to part A written in percentage form?
 = 
 %.

C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2009 ?
value = $ 
 Round to the nearest 50 dollars.

To solve this problem, we'll first calculate the annual rate of depreciation and then use that to answer each part.

### A) What was the annual rate of change between 1990 and 2005?
The formula for calculating the annual rate of change in a value that changes exponentially over time is given by:

\[
V(t) = V_0 (1 - r)^t
\]
where:
- $V(t)$ is the value at time $t$,
- $V_0$ is the initial value,
- $r$ is the annual rate of depreciation, and
- $t$ is the time in years.

In this case:
- $V_0 = 42,000$ (initial value in 1990),
- $V(t) = 13,000$ (value in 2005),
- $t = 2005 - 1990 = 15$ years.

We need to find $r$.

#### Step 1: Rearranging the formula
\[
\frac{V(t)}{V_0} = (1 - r)^t
\]

Substitute the known values:
\[
\frac{13,000}{42,000} = (1 - r)^{15}
\]

\[
0.3095 = (1 - r)^{15}
\]

#### Step 2: Solving for $r$
Take the 15th root of both sides:
\[
(1 - r) = \left(0.3095\right)^{\frac{1}{15}}
\]

Now, subtract from 1 to get $r$:
\[
r = 1 - \left(0.3095\right)^{\frac{1}{15}}
\]

Let me calculate this.### A) The annual rate of change between 1990 and 2005 is approximately 0.0752 (or 7.52% when written as a percentage).

### B) What is the correct answer to part A written in percentage form?

The annual rate of decrease is:

\[
7.52\%
\]

### C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2009?

To find the car's value in 2009, we will use the formula:

\[
V(t) = V_0 (1 - r)^t
\]

Where:
- $V_0 = 13,000$ (the value in 2005),
- $r = 0.0752$ (the rate of depreciation),
- $t = 2009 - 2005 = 4$ years.

Let me calculate the value for 2009.### C) The value of the car in 2009, assuming it continues to drop by the same percentage, is approximately **$9,500** (rounded to the nearest 50 dollars).

---

Do you want more details on any of the calculations, or do you have any questions?

Here are 5 related questions:
1. How would the depreciation change if the car's value dropped faster over the same period?
2. Can you calculate the value of the car in 2010 using the same method?
3. How would you calculate the annual rate of increase instead of decrease?
4. What happens to the car's value if depreciation accelerates after 2005?
5. How would the result change if the initial car value in 1990 was $50,000?

**Tip:** Always remember to express the depreciation rate in decimal form when using it in formulas. For example, 7.52% becomes 0.0752.

A car was valued at $42,000 in the year 1990. The value depreciated to $13,000 by the year 2005.

A) What was the annual rate of change between 1990 and 2005? Round the rate of decrease to 4 decimal places.

B) What is the correct answer to part A written in percentage form?

C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2009? Round to the nearest 50 dollars.

Learn how to calculate the annual rate of depreciation for a car that dropped in value from $42,000 in 1990 to $13,000 by 2005. This problem involves exponential decay and percentage decrease. Also, predict the car's value in 2009 using the same rate of depreciation. Suitable for Grades 10-12.

Math Problem Statement

A) What was the annual rate of change between 1990 and 2005?

B) What is the correct answer to part A written in percentage form?

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