Answer :
Answer:
$27,436
Step-by-step explanation:
Exponential Functions
Recall that an exponential function or equation take the form of [tex]y=a(b)^x[/tex], where a is the initial value, b is the growth/decay rate
(growth: b > 1; decay: 0 < b < 1), and x is the time that passes.
Applying Exponential Functions/Equations
Finding b
The problem mentions "depreciating", which means decreasing. If it depreciates at a rate of 5%, then it decreases the car's value at 5% each year that passes.
The value of b, however, will be 0.95 rather than 0.05.The car's overall value decreases by 5% yearly, meaning the car's value is 95% of its value the previous year.
Think of sales such as "50% off" and "25% off" seen in shops; the larger the percentage, the cheaper the product. A customer wouldn't think too much of a "5% off" sale compared to a "75% off" one, as the price would decrease slightly.
So, b is now identified!
Finding a
The problem mentions that the car was worth $32,000 in 2010 and is asking for its value in 2013. This must mean that the car's value at 2010 is the initial value or a!
Finding x
Repeating what's been said in "Finding a" the problem gives us two measures of time: 2010 and 2013. 2010 is the initial time, and 2013 is the final time.
Since we're finding the value of the car at 2013, where 3 years elapsed between the car's initial value and the value we're trying to find, x must equal 3!
Putting it Together
Using the exponential equation and plugging in the known values, the car's value in 2013 is,
[tex]32,000(0.95)^3=27,436[/tex].
