braylon wants to invest his high school earnings for the next 7 years. he deposits $1,500 into an investment…

braylon wants to invest his high school earnings for the next 7 years. he deposits $1,500 into an investment account at the beginning of the first year, $2,000 at the beginning of the second year, $1,870 at the beginning of the third year, and $2,230 at the beginning of the fourth year. use $x = 1 + r$, where $r$ is the interest rate, and the equation $c(x)=1500x^{7}+2000x^{6}+1870x^{5}+2230x^{4}$. the equation represents the relationship between $c(x)$, the value of the investment after 7 years, and its annual interest rate, $r$. find the value of the investment account if the interest rate is 2.85%. (1 point)\n$7,600.00\n$30,314.09\n$8,840.80\n$7,716.96

braylon wants to invest his high school earnings for the next 7 years. he deposits $1,500 into an investment account at the beginning of the first year, $2,000 at the beginning of the second year, $1,870 at the beginning of the third year, and $2,230 at the beginning of the fourth year. use $x = 1 + r$, where $r$ is the interest rate, and the equation $c(x)=1500x^{7}+2000x^{6}+1870x^{5}+2230x^{4}$. the equation represents the relationship between $c(x)$, the value of the investment after 7 years, and its annual interest rate, $r$. find the value of the investment account if the interest rate is 2.85%. (1 point)\n$7,600.00\n$30,314.09\n$8,840.80\n$7,716.96

Answer

Explanation:

Step1: Calculate the value of x

Given $r = 2.85%=0.0285$, then $x=1 + r=1 + 0.0285 = 1.0285$.

Step2: Substitute x into the function C(x)

$C(x)=1500x^{7}+2000x^{6}+1870x^{5}+2230x^{4}$. $C(1.0285)=1500\times(1.0285)^{7}+2000\times(1.0285)^{6}+1870\times(1.0285)^{5}+2230\times(1.0285)^{4}$. First, calculate each term: $(1.0285)^{7}\approx1.2197$, so $1500\times(1.0285)^{7}\approx1500\times1.2197 = 1829.55$. $(1.0285)^{6}\approx1.1860$, so $2000\times(1.0285)^{6}\approx2000\times1.1860 = 2372$. $(1.0285)^{5}\approx1.1531$, so $1870\times(1.0285)^{5}\approx1870\times1.1531=2156.2$. $(1.0285)^{4}\approx1.1210$, so $2230\times(1.0285)^{4}\approx2230\times1.1210 = 2499.83$. Then sum up the terms: $C(1.0285)\approx1829.55 + 2372+2156.2+2499.83=8840.8$.

Answer:

$8,840.80$