1. We are given the half-life of mercury-210 as 600 seconds and an initial mass of 2500 grams. 2. We want to find the remaining mass after 17 minutes. 3. Convert 17 minutes to seconds: $$17 \times 60 = 1020 \text{ seconds}$$. 4. Use the half-life decay formula: $$m = m_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}$$ where $m_0 = 2500$, $t = 1020$, and $T=600$. 5. Calculate the exponent: $$\frac{t}{T} = \frac{1020}{600} = 1.7$$ 6. Calculate the remaining mass: $$m = 2500 \times \left(\frac{1}{2}\right)^{1.7} = 2500 \times 2^{-1.7}$$ 7. Use logs to compute $2^{-1.7}$: $$2^{-1.7} = e^{-1.7 \ln 2} \approx e^{-1.7 \times 0.6931} = e^{-1.1783} \approx 0.307$$ 8. Multiply to find remaining mass: $$m \approx 2500 \times 0.307 = 767.5 \text{ grams}$$ --- 9. For Balthazar's investment: Initial amount $P=1800$, interest rate per year $r=3.98\% = 0.0398$, compounded triannually means $n=3$ times per year, and $t=14$ years. 10. Compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ 11. Calculate: $$\frac{r}{n} = \frac{0.0398}{3} \approx 0.013267$$ $$nt = 3 \times 14 = 42$$ 12. Calculate the total amount: $$A = 1800 \times (1.013267)^{42}$$ 13. Calculate the exponent: $$\ln(A/1800) = 42 \times \ln(1.013267) \approx 42 \times 0.01318 = 0.5536$$ 14. Thus: $$A = 1800 \times e^{0.5536} \approx 1800 \times 1.739 = 3130.3$$ **Final answers:** - Mercury-210 remaining mass after 17 minutes is approximately **767.5 grams**. - Investment value after 14 years compounded triannually is approximately **3130.3**.