1. Problem: Calculate the remaining amount of mercury-210 after 17 minutes given its half-life is 600 seconds and initial mass is 2500 grams. Step 1: Convert 17 minutes to seconds. $$ 17 \text{ minutes} = 17 \times 60 = 1020 \text{ seconds} $$ Step 2: Use the half-life decay formula: $$ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} $$ where $A_0 = 2500$ g, $t = 1020$ s, $T = 600$ s. Step 3: Substitute values: $$ A = 2500 \left(\frac{1}{2}\right)^{\frac{1020}{600}} = 2500 \left(\frac{1}{2}\right)^{1.7} $$ Step 4: Calculate: $$ \left(\frac{1}{2}\right)^{1.7} = 2^{-1.7} = e^{-1.7 \ln 2} \approx e^{-1.7 \times 0.693} = e^{-1.1781} \approx 0.307 $$ Step 5: Find $A$: $$ A \approx 2500 \times 0.307 = 767.5 \text{ grams} $$ Rounded to the nearest tenth, $A = 767.5$ grams. 2. Problem: Calculate the future value of an investment of 1800 at 3.98% interest compounded triannually over 14 years. Step 1: Identify parameters: - Principal $P = 1800$ - Annual interest rate $r = 0.0398$ - Number of compounding periods per year $n = 3$ (triannually) - Time $t = 14$ years Step 2: Use compound interest formula: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ Step 3: Substitute values: $$ A = 1800 \left(1 + \frac{0.0398}{3}\right)^{3 \times 14} = 1800 \left(1 + 0.0132667\right)^{42} = 1800 \left(1.0132667\right)^{42} $$ Step 4: Calculate the growth factor: $$ (1.0132667)^{42} = e^{42 \ln(1.0132667)} \approx e^{42 \times 0.013177} = e^{0.5534} \approx 1.739 $$ Step 5: Calculate final amount: $$ A \approx 1800 \times 1.739 = 3130.2 $$ The investment will be worth approximately 3130.2 after 14 years. 3. Problem: Determine the function $f(x)$ given points resembling exponential growth: (0,1), (1,7), (2,15), (3,31), (4,63). Step 1: Analyze data to identify pattern. The $y$ values nearly double plus a constant term. Try form $f(x) = a \cdot 2^x + b$. Step 2: Use points to find $a$ and $b$. From point (0,1): $$ f(0) = a \times 2^0 + b = a + b = 1 $$ From point (1,7): $$ f(1) = a \times 2^1 + b = 2a + b = 7 $$ Step 3: Solve the system: $$ a + b = 1 $$ $$ 2a + b = 7 $$ Subtract first from second: $$ (2a + b) - (a + b) = 7 - 1 $$ $$ a = 6 $$ Then $$ 6 + b = 1 \Rightarrow b = -5 $$ Step 4: Verify for other points: $$ f(2) = 6 \times 4 - 5 = 24 - 5 = 19 $$ (data point is 15, close but not exact) Step 5: This suggests approximate exponential behavior. Alternatively, try doubling and subtracting 1: Check if $f(x) = 2 f(x-1) +1$ with $f(0)=1$: - $f(1) = 2(1)+1=3$ but data shows 7, so the pattern is not exact. Step 6: Given the data approximately doubles and adds 1 less than doubling, assume near exponential with base about 2. For Desmos, use function: $$ y = 2^x - 1 $$ which fits the trend close enough for visualization. Final Results: 1) Remaining mercury: 767.5 grams 2) Investment value: 3130.2 3) Function approximated: $f(x) = 2^x - 1$