David's car was valued at  
$
38
,
000
  in the year  
2007
 .  If it is continuously depreciating in value by  
6
%
  every year, how much will it be worth in  
2017
 ?

Round your answer to the nearest dollar 

To solve this problem, we use the formula for continuous depreciation, which is similar to the formula for continuous compound interest:

\[
V(t) = V_0 e^{-rt}
\]

Where:
- $ V(t) $ is the value of the car after time $ t $,
- $ V_0 $ is the initial value of the car,
- $ r $ is the annual depreciation rate (as a decimal),
- $ t $ is the number of years,
- $ e $ is the base of the natural logarithm (approximately 2.71828).

In this case:
- $ V_0 = 38,000 $,
- $ r = 0.06 $ (since the depreciation rate is 6%),
- $ t = 2017 - 2007 = 10 $ years.

Now, we can plug in these values and calculate $ V(10) $:

\[
V(10) = 38,000 \cdot e^{-0.06 \cdot 10}
\]

Let me compute that for you.The value of David's car in 2017, after continuously depreciating by 6% annually, is approximately \$20,855.

Do you want any further details or have any questions?

Here are five related questions to explore:
1. What would be the value of the car after 5 years instead of 10?
2. How would the car's value change if the depreciation rate was 4% annually?
3. What is the car's value at any point in time if the depreciation continues indefinitely?
4. How much total value has the car lost by 2017?
5. How can we calculate the car's half-life based on this depreciation rate?

**Tip:** Continuous depreciation models are based on exponential decay, which means the car's value decreases faster at the beginning and more slowly over time.

David's car was valued at $38,000 in the year 2007. If it is continuously depreciating in value by 6% every year, how much will it be worth in 2017? Round your answer to the nearest dollar.

Calculate the value of a car worth $38,000 in 2007, depreciating by 6% annually using continuous depreciation. Learn how exponential decay works and use the formula V(t) = V_0 * e^(-rt) to find the car's worth in 2017. Ideal for students in Grades 10-12.

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