Answer:
1. It's value in 2009 was $16,462.575
2. It's value in 2020 will be $3,558.
Step-by-step explanation:
The equation for the value of car after t years has the following format.
[tex]V(t) = V(0)(1-r)^{t}[/tex]
In which V(t) is the value after t years, V(0) is the initial value and r is the ratio of depreciation.
In this problem, we have that:
[tex]V(0) = 25000, r = 0.13[/tex]
So
[tex]V(t) = V(0)(1-r)^{t}[/tex]
[tex]V(t) = 25000(1-0.13)^{t}[/tex]
[tex]V(t) = 25000(0.87)^{t}[/tex]
1. What was its value in 2009
2009 is 3 years after 2006, so this is V(3)
[tex]V(t) = 25000(0.87)^{t}[/tex]
[tex]V(3) = 25000(0.87)^{3}[/tex]
[tex]V(3) = 16462.575[/tex]
It's value in 2009 was $16,462.575
2.What will be its value in 2020?
2020 is 14 years after 2006, so this is V14)
[tex]V(t) = 25000(0.87)^{t}[/tex]
[tex]V(14) = 25000(0.87)^{14}[/tex]
[tex]V(14) = 3558[/tex]
It's value in 2020 will be $3,558.
