1. **Problem 6a:** A 400 mg dose of Ibuprofen is taken. The half-life is 3 hours, so every 3 hours the amount halves. Find the amount remaining after 12 hours. 2. **Formula:** The amount remaining after time $t$ with half-life $T$ is given by $$M = M_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}$$ where $M_0$ is the initial amount. 3. **Apply to 6a:** Here, $M_0 = 400$ mg, $T = 3$ hours, $t = 12$ hours. $$M = 400 \times \left(\frac{1}{2}\right)^{\frac{12}{3}} = 400 \times \left(\frac{1}{2}\right)^4$$ 4. Calculate $$\left(\frac{1}{2}\right)^4 = \frac{1}{16}$$ 5. So, $$M = 400 \times \frac{1}{16} = 25$$ mg remain after 12 hours. 6. **Problem 6b:** Find amount remaining after 8 hours. 7. Using the same formula with $t=8$ hours: $$M = 400 \times \left(\frac{1}{2}\right)^{\frac{8}{3}}$$ 8. Calculate exponent: $$\frac{8}{3} = 2 + \frac{2}{3}$$ 9. So, $$M = 400 \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^{\frac{2}{3}} = 400 \times \frac{1}{4} \times \left(\frac{1}{2}\right)^{\frac{2}{3}}$$ 10. Calculate $$\left(\frac{1}{2}\right)^{\frac{2}{3}} = e^{\ln(1/2) \times \frac{2}{3}} \approx e^{-0.4621} \approx 0.63$$ 11. So, $$M \approx 400 \times \frac{1}{4} \times 0.63 = 100 \times 0.63 = 63$$ mg (approx.) 12. **Problem 6c:** Write an equation for amount $M$ after $h$ hours. $$M = 400 \times \left(\frac{1}{2}\right)^{\frac{h}{3}}$$ 13. **Problem 7:** Department store shoppers start at 320 and decrease by 2.7% weekly. 14. The decay factor per week is $1 - 0.027 = 0.973$. 15. The function modeling shoppers after $w$ weeks is: $$S(w) = 320 \times 0.973^w$$ 16. **Problem 8a:** Grey seal population in 1997 is 25,400, increasing at 12.8% annually. 17. Find population in 2000 (3 years later). 18. Growth factor is $1 + 0.128 = 1.128$. 19. Population after 3 years: $$P = 25400 \times 1.128^3$$ 20. Calculate: $$1.128^3 \approx 1.128 \times 1.128 \times 1.128 = 1.433$$ 21. So, $$P \approx 25400 \times 1.433 = 36398$$ seals (approx.) 22. **Problem 8b:** Population in 2022 (25 years after 1997). 23. Calculate: $$P = 25400 \times 1.128^{25}$$ 24. Using approximation: $$1.128^{25} \approx e^{25 \times \ln(1.128)} = e^{25 \times 0.120} = e^{3.0} \approx 20.09$$ 25. So, $$P \approx 25400 \times 20.09 = 510286$$ seals (approx.) 26. **Problem 9:** Given table for $g(x)$: | x | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | g(x) | 25 | 20 | 16 | 12.8 | 10.24 | 27. This is exponential decay. Find ratio between consecutive terms: $$\frac{20}{25} = 0.8, \quad \frac{16}{20} = 0.8, \quad \frac{12.8}{16} = 0.8, \quad \frac{10.24}{12.8} = 0.8$$ 28. So common ratio $r = 0.8$. 29. General form: $$g(x) = a \times r^{x-1}$$ 30. Use $x=1$, $g(1)=25$: $$25 = a \times 0.8^{0} = a \times 1 \Rightarrow a = 25$$ 31. Final equation: $$g(x) = 25 \times 0.8^{x-1}$$