Suppose that $\$ 84,000$ is invested at $4 \frac{1}{2} \%$ interest, compounded quarterly.
a) Find the function for the amount to which the investment grows after $t$ years.
b) Find the amount of money in the account at $\mathrm{t}=0,2,5$, and 10 years.
a) The function for the amount to which the investment grows after $t$ years is $A(t)=$ (Simplify your answer. Type an expression using $t$ as the variable.)
b) Find the amount of money in the account at $t=0,2,5$, and 10 years.
The amount of money in the account at $t=0$ years is $\$ \square$. (Round to the nearest dollar as needed.)

The amount of money in the account at $t=2$ years is $\$ \square$. (Round to the nearest dollar as needed.)

The amount of money in the account at $t=5$ years is $\$ \square$. (Round to the nearest dollar as needed.)

The amount of money in the account at $t=10$ years is $\$ \square$. (Round to the nearest dollar as needed.)

Step 1 ：Given that the principal amount (P) is $84,000, the annual interest rate (r) is 4.5% or 0.045 in decimal form, and the interest is compounded quarterly, so n = 4.

Step 2 ：Using the compound interest formula, which is \(A = P(1 + \frac{r}{n})^{nt}\), we can find the amount of money (A) in the account after t years.

Step 3 ：Substitute the given values into the formula, we get \(A(t) = 84000(1 + \frac{0.045}{4})^{4t}\).

Step 4 ：For t = 0 years, substitute 0 into the formula, we get \(A(0) = 84000(1 + \frac{0.045}{4})^{4*0} = 84000\).

Step 5 ：For t = 2 years, substitute 2 into the formula, we get \(A(2) = 84000(1 + \frac{0.045}{4})^{4*2} = 91864\).

Step 6 ：For t = 5 years, substitute 5 into the formula, we get \(A(5) = 84000(1 + \frac{0.045}{4})^{4*5} = 105063\).

Step 7 ：For t = 10 years, substitute 10 into the formula, we get \(A(10) = 84000(1 + \frac{0.045}{4})^{4*10} = 131408\).

Step 8 ：So, the amount of money in the account at t = 0 years is \(\boxed{84000}\), at t = 2 years is \(\boxed{91864}\), at t = 5 years is \(\boxed{105063}\), and at t = 10 years is \(\boxed{131408}\).