High School

The price of a car was $20,000 and it has depreciated by 8% yearly. Which equation will help find the value of the car after a certain period of time? What is the value of the car 5 years later?

A. [tex]y = 20000(1 - 0.08)^5[/tex]; $13,181.63
B. [tex]y = 20000(1 + 0.08)^5[/tex]; $29,386.56
C. [tex]y = 20000(1 + 0.05)^5[/tex]; $29,549.11
D. [tex]y = 20000(1 - 0.05)^5[/tex]; $13,268.41

Answer :

Answer:

O [tex]\sf y = 20000(1 - 08) ^5[/tex]

O $13181.63

explanation:

[tex]\sf compound \ interest = P(1 + \dfrac{r}{100})^{n}[/tex]

insert the following given values

[tex]\sf \hookrightarrow 20000(1 - \frac{8}{100} )^5[/tex]

[tex]\sf \hookrightarrow 20000(1 - 0.08)^5[/tex]

[tex]\sf \hookrightarrow 13181.63[/tex]

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