Answer:
[tex]\displaystyle V_n=V_o\left(0.60\right)^n[/tex]
The car is worth 360 after two years
Step-by-step explanation:
Model For Depretiation
Assume the situation of something that has an initial value Vo and loses its value by a x% every year. The first year it will lose
[tex]\displaystyle \frac{xV_o}{100}[/tex]
and it will have a new value of
[tex]\displaystyle V_1=V_o-x\frac{V_o}{100}=V_o\left(\frac{100-x}{100}\right)[/tex]
Next year it will have a value of
[tex]\displaystyle V_2=V_o\left(\frac{100-x}{100}\right)^2[/tex]
Since x=40%, we can deduct the general formula
[tex]\displaystyle V_n=V_o\left(0.60\right)^n[/tex]
For n=2
[tex]\displaystyle V_2=1000\left(0.60\right)^2[/tex]
[tex]V_2=360[/tex]
