1. **Problem:** Given the accumulation function $a(t) = 0.1t^2 + 1$, an initial investment of 1000 at time 0, and an additional investment of $X$ at time 6, find $X$ if the total accumulated value at time 8 is 18000. 2. **Formula:** The accumulated value at time $t$ for an investment made at time $s$ is $\text{Investment} \times \frac{a(t)}{a(s)}$. 3. **Step 1:** Calculate the accumulation factors: - $a(0) = 0.1(0)^2 + 1 = 1$ - $a(6) = 0.1(6)^2 + 1 = 0.1 \times 36 + 1 = 4.6$ - $a(8) = 0.1(8)^2 + 1 = 0.1 \times 64 + 1 = 7.4$ 4. **Step 2:** Calculate the accumulated value of the initial investment at time 8: $$1000 \times \frac{a(8)}{a(0)} = 1000 \times \frac{7.4}{1} = 7400$$ 5. **Step 3:** Calculate the accumulated value of the additional investment $X$ at time 8: $$X \times \frac{a(8)}{a(6)} = X \times \frac{7.4}{4.6}$$ 6. **Step 4:** Set up the equation for total accumulated value at time 8: $$7400 + X \times \frac{7.4}{4.6} = 18000$$ 7. **Step 5:** Solve for $X$: $$X \times \frac{7.4}{4.6} = 18000 - 7400 = 10600$$ $$X = 10600 \times \frac{4.6}{7.4}$$ $$X = 10600 \times \frac{\cancel{4.6}}{\cancel{7.4}} \approx 10600 \times 0.6216 = 6585.12$$ \boxed{X = 6585.12} --- 8. **Problem:** If 64000 grows to 128000 in 4 years at a constant effective annual interest rate $i$, find the amount 100000 will grow to in 3 years at the same rate. 9. **Step 1:** Find the effective annual interest rate $i$: $$128000 = 64000 (1 + i)^4$$ $$2 = (1 + i)^4$$ $$1 + i = \sqrt[4]{2} = 2^{\frac{1}{4}} \approx 1.1892$$ $$i = 0.1892 = 18.92\%$$ 10. **Step 2:** Calculate the amount 100000 will grow to in 3 years: $$100000 (1 + i)^3 = 100000 \times 1.1892^3$$ $$= 100000 \times 1.6818 = 168180$$ \boxed{168180}