use $z = 1 + r$, where $r$ is the interest rate paid each year. write a model polynomial, $c(z)$, that…

use $z = 1 + r$, where $r$ is the interest rate paid each year. write a model polynomial, $c(z)$, that represents the final amount of an 8 - year investment if $500 was deposited at the beginning of the first year, $800 was deposited at the beginning of the fourth year, and an additional $370 was deposited at the beginning of the eighth year. (1 point)\n$c(z)=500z + 800z^{5}+370z^{8}$\n$c(z)=500z^{8}+800z^{4}+370$\n$c(z)=500z^{8}+800z^{5}+370z$\n$c(z)=500z^{8}+800z^{4}+370z$

use $z = 1 + r$, where $r$ is the interest rate paid each year. write a model polynomial, $c(z)$, that represents the final amount of an 8 - year investment if $500 was deposited at the beginning of the first year, $800 was deposited at the beginning of the fourth year, and an additional $370 was deposited at the beginning of the eighth year. (1 point)\n$c(z)=500z + 800z^{5}+370z^{8}$\n$c(z)=500z^{8}+800z^{4}+370$\n$c(z)=500z^{8}+800z^{5}+370z$\n$c(z)=500z^{8}+800z^{4}+370z$

Answer

Explanation:

Step1: Analyze first - deposit

The $500$ deposited at the beginning of the first year will earn interest for 8 years. Using the compound - interest formula with $z = 1 + r$, the amount it grows to is $500z^{8}$.

Step2: Analyze second - deposit

The $800$ deposited at the beginning of the fourth year will earn interest for $8 - 4=4$ years. So the amount it grows to is $800z^{4}$.

Step3: Analyze third - deposit

The $370$ deposited at the beginning of the eighth year will earn interest for $8 - 8 = 0$ years (in terms of compounding for this investment period), which is just $370z^{0}=370$.

Step4: Form the polynomial

The polynomial $C(z)$ representing the final amount of the investment is the sum of the amounts from each deposit, so $C(z)=500z^{8}+800z^{4}+370$.

Answer:

$C(z)=500z^{8}+800z^{4}+370$