find the value of the investments at the end of 5 years for the following compounding methods , semiannual, monthly, daily

Given:
There are given that the total amount is 34900 dollars
Explanation:
According to the question:
We need to find the value of the investment.
So,
To find the value of investments, we need to use the compound interest formula:
So,
From the formula of compound interest:
(a): For annually:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Then,
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ A=34900(1+\frac{0.08}{1})^5 \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34,900(1+\frac{0.08}{1})^{5} \\ A=34900(1+0.08)^5 \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34,900(1+0.08)^{5} \\ A=34900(1.08)^5 \\ A=34900(1.46) \\ A=51279.55 \end{gathered}[/tex]Now,
(b): For the semiannual:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ A=P(1+\frac{r}{2})^{2t} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=P(1+\frac{r}{2})^{2t} \\ A=34900(1+\frac{0.08}{2})^{2(5)} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34900(1+\frac{0.08}{2})^{2(5)} \\ A=34900(1+0.04)^{10} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34900(1+0.04)^{10} \\ A=34900(1.04)^{10} \\ A=34900(1.48) \\ A=51660.5 \end{gathered}[/tex](c): For monthly:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ A=P(1+\frac{r}{12})^{12t} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=P(1+\frac{r}{12})^{12t} \\ A=P(1+\frac{r}{12})^{12(5)} \\ A=P(1+\frac{0.08}{12})^{12(5)} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34900(1+\frac{0.08}{12})^{12(5)} \\ A=34900(1+0.0067)^{60} \\ A=34,900(1.0067)^{60} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34900(1.0067)^{60} \\ A=34900(1.49) \\ A=52099.02 \end{gathered}[/tex]And,
(d): For daily:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ A=P(1+\frac{r}{365})^{365(5)} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=P(1+\frac{r}{365})^{365(5)} \\ A=34900(1+\frac{0.08}{365})^{365(5)} \\ A=34,900(1+0.000219)^{365(5)} \end{gathered}[/tex]Then,
[tex]\begin{gathered} A=34,900(1+0.000219)^{365(5)} \\ A=34,900(1.000219)^{1825} \\ A=34,900(1.49) \\ A=52140.53 \end{gathered}[/tex]Final answer:
Hence, the all values is shown below:
[tex]\begin{gathered} (a).Annual:51279.55 \\ (b).Semiannual:51660.5 \\ (c).Monthly:52099.02 \\ (d).Daily:52140.53 \end{gathered}[/tex]