If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given number of years.

Answer:Part a) [tex]\$17,958.56[/tex]  Part b) [tex]\$32,251.00[/tex]  Part c) [tex]\$57,918.16[/tex] Step-by-step explanation:The complete question isIf $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)a)6 yearsb)12 yearsc)18 yearswe know that    The compound interest formula is equal to  [tex]A=P(1+\frac{r}{n})^{nt}[/tex]  where  A is the Final Investment Value  P is the Principal amount of money to be invested  r is the rate of interest  in decimalt is Number of Time Periods  n is the number of times interest is compounded per yearPart a) 6 yearswe have  [tex]t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]  substitute in the formula above  [tex]A=10,000(1+\frac{0.10}{2})^{2*6}[/tex]  [tex]A=10,000(1.05)^{12}[/tex]  [tex]A=\$17,958.56[/tex]  Part b) 12 yearswe have  [tex]t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]  substitute in the formula above  [tex]A=10,000(1+\frac{0.10}{2})^{2*12}[/tex]  [tex]A=10,000(1.05)^{24}[/tex]  [tex]A=\$32,251.00[/tex]  Part c) 18 yearswe have  [tex]t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]  substitute in the formula above  [tex]A=10,000(1+\frac{0.10}{2})^{2*18}[/tex]  [tex]A=10,000(1.05)^{36}[/tex]  [tex]A=\$57,918.16[/tex]