1. Problem 10a: Complete the table assuming linear growth. Given points: $t=0$, $s(0)=24$ and $t=1$, $s(1)=186$. 2. Find the slope $m$ of the linear function $s(t) = mt + b$: $$m = \frac{s(1) - s(0)}{1 - 0} = \frac{186 - 24}{1} = 162$$ 3. Find $b$ (the y-intercept) using $s(0) = 24$: $$s(0) = m \cdot 0 + b = b = 24$$ 4. The linear model is: $$s(t) = 162t + 24$$ 5. Complete the table for $t=0,1,2,3,4$: - $s(0) = 162 \times 0 + 24 = 24$ - $s(1) = 162 \times 1 + 24 = 186$ - $s(2) = 162 \times 2 + 24 = 324 + 24 = 348$ - $s(3) = 162 \times 3 + 24 = 486 + 24 = 510$ - $s(4) = 162 \times 4 + 24 = 648 + 24 = 672$ 6. Problem 10b: Write the linear model: $$s(t) = 162t + 24$$ 7. Problem 10c: Complete the table assuming exponential growth. Given $s(0) = 24$ and $s(1) = 186$, the exponential model is: $$s(t) = a \cdot r^t$$ where $a = s(0) = 24$. 8. Find the growth rate $r$ using $s(1) = 186$: $$186 = 24 \cdot r^1 \Rightarrow r = \frac{186}{24} = 7.75$$ 9. The exponential model is: $$s(t) = 24 \cdot 7.75^t$$ 10. Complete the table for $t=0,1,2,3,4$: - $s(0) = 24 \cdot 7.75^0 = 24$ - $s(1) = 24 \cdot 7.75^1 = 186$ - $s(2) = 24 \cdot 7.75^2 = 24 \cdot 60.0625 = 1441.5$ - $s(3) = 24 \cdot 7.75^3 = 24 \cdot 465.484 = 11171.6$ - $s(4) = 24 \cdot 7.75^4 = 24 \cdot 3608.0 = 86592$ 11. Problem 10d: Write the exponential model: $$s(t) = 24 \cdot 7.75^t$$ 12. Problem 11: Given population model: $$P(x) = 58000 (1.036)^x$$ where $x$ is years. We want population after $m$ months. Since $1$ year = $12$ months, convert months to years: $$x = \frac{m}{12}$$ 13. Substitute into $P(x)$: $$P(m) = 58000 (1.036)^{\frac{m}{12}}$$ 14. The correct choice is A: $$P(m) = 58000 (1.036)^{\frac{m}{12}}$$