1. **Problem Statement:** Briac Steel deposits 44004400 at the beginning of every month into a fund with an 8% annual interest rate compounded semi-annually. We want to find the total amount in the fund after 10 years. 2. **Formula Used:** Since deposits are made monthly but interest is compounded semi-annually, we treat this as an annuity with periodic deposits and compound interest. The future value of an annuity with payments at the beginning of each period (annuity due) is: $$FV = P \times \frac{(1 + r)^n - 1}{r} \times (1 + r)$$ where: - $P$ = payment per period - $r$ = interest rate per period - $n$ = total number of periods 3. **Important Rules:** - Interest rate per period must match the compounding period. - Since interest is compounded semi-annually, each period is 6 months. - Payments are monthly, so we need to adjust the formula accordingly. 4. **Adjusting for Monthly Payments and Semi-Annual Compounding:** - Number of years = 10 - Number of months = $10 \times 12 = 120$ - Number of semi-annual periods = $10 \times 2 = 20$ - Interest rate per semi-annual period = $\frac{8}{100} \div 2 = 0.04$ Since payments are monthly but compounding is semi-annual, we use the formula for future value of an annuity with monthly payments and semi-annual compounding by converting the interest rate to an effective monthly rate: $$r_{monthly} = (1 + 0.04)^{\frac{1}{6}} - 1$$ Calculate $r_{monthly}$: $$r_{monthly} = (1.04)^{\frac{1}{6}} - 1 = 0.006542$$ 5. **Calculate total number of payments:** $$n = 120$$ 6. **Calculate future value of annuity due:** $$FV = 44004400 \times \frac{(1 + 0.006542)^{120} - 1}{0.006542} \times (1 + 0.006542)$$ Calculate $(1 + 0.006542)^{120}$: $$ (1.006542)^{120} = 2.219640$$ Calculate numerator: $$2.219640 - 1 = 1.219640$$ Calculate fraction: $$\frac{1.219640}{0.006542} = 186.423$$ Multiply by $(1 + 0.006542)$: $$186.423 \times 1.006542 = 187.646$$ Finally, calculate $FV$: $$44004400 \times 187.646 = 8,255,927,000.00$$ 7. **Answer:** The fund will be worth approximately 8255927000.00 after 10 years.