Answer :
To solve this problem, we need to determine which function best models the value of the car, where [tex]\( n \)[/tex] represents the number of years after 2023.
The original function given is [tex]\( f(t) = 20000(0.88)^t \)[/tex], which models the car's value [tex]\( t \)[/tex] years after 2020. Since [tex]\( n \)[/tex] represents the years after 2023, we have the relationship that [tex]\( t = n + 3 \)[/tex] because 2023 is 3 years after 2020.
Let's substitute [tex]\( t = n + 3 \)[/tex] into the original function to model the value of the car [tex]\( n \)[/tex] years after 2023:
1. Start with the expression for [tex]\( t \)[/tex]: [tex]\( t = n + 3 \)[/tex].
2. Substitute into the original function:
[tex]\[
f(t) = 20000(0.88)^t = 20000(0.88)^{n+3}
\][/tex]
3. Use the property of exponents: [tex]\( (a^{b+c} = a^b \cdot a^c) \)[/tex], so:
[tex]\[
f(t) = 20000 \cdot (0.88)^3 \cdot (0.88)^n
\][/tex]
The expression [tex]\( 20000 \cdot (0.88)^3 \cdot (0.88)^n \)[/tex] can be simplified to:
[tex]\[
20000(0.88)^3(0.88)^n
\][/tex]
Among the answer choices, this matches option C:
[tex]\[
v(n) = 20000(0.88)^3(0.88)^n
\][/tex]
Therefore, the function [tex]\( v(n) = 20000(0.88)^3(0.88)^n \)[/tex] is the one that best models the value of the car where [tex]\( n \)[/tex] represents the number of years after 2023. So, the correct answer is option C.
The original function given is [tex]\( f(t) = 20000(0.88)^t \)[/tex], which models the car's value [tex]\( t \)[/tex] years after 2020. Since [tex]\( n \)[/tex] represents the years after 2023, we have the relationship that [tex]\( t = n + 3 \)[/tex] because 2023 is 3 years after 2020.
Let's substitute [tex]\( t = n + 3 \)[/tex] into the original function to model the value of the car [tex]\( n \)[/tex] years after 2023:
1. Start with the expression for [tex]\( t \)[/tex]: [tex]\( t = n + 3 \)[/tex].
2. Substitute into the original function:
[tex]\[
f(t) = 20000(0.88)^t = 20000(0.88)^{n+3}
\][/tex]
3. Use the property of exponents: [tex]\( (a^{b+c} = a^b \cdot a^c) \)[/tex], so:
[tex]\[
f(t) = 20000 \cdot (0.88)^3 \cdot (0.88)^n
\][/tex]
The expression [tex]\( 20000 \cdot (0.88)^3 \cdot (0.88)^n \)[/tex] can be simplified to:
[tex]\[
20000(0.88)^3(0.88)^n
\][/tex]
Among the answer choices, this matches option C:
[tex]\[
v(n) = 20000(0.88)^3(0.88)^n
\][/tex]
Therefore, the function [tex]\( v(n) = 20000(0.88)^3(0.88)^n \)[/tex] is the one that best models the value of the car where [tex]\( n \)[/tex] represents the number of years after 2023. So, the correct answer is option C.
