1. Problem 18: Mutoni spins 2 spinners; one with colors Red (R), Yellow (Y), Blue (B), and the other with Green (G), White (W), Purple (P). 2. a) Draw a tree diagram: The first spinner has 3 outcomes (R, Y, B). Each branches into 3 outcomes of the second spinner (G, W, P). So total outcomes are 3 \times 3 = 9. 3. b) Probability spinners stop at B and G: - Probability first spinner stops at B is $\frac{1}{3}$. - Probability second spinner stops at G is $\frac{1}{3}$. - Since independent, combined probability is $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$. 4. c) Probability spinners do not stop at B and G: - Probability of B and G is $\frac{1}{9}$. - Probability of not (B and G) is $1 - \frac{1}{9} = \frac{8}{9}$. 5. d) Probability first spinner does not stop at R: - Probability first spinner stops at R is $\frac{1}{3}$. - So probability it does not stop at R is $1 - \frac{1}{3} = \frac{2}{3}$. 6. Problem 19: Solve equation $$\frac{\log_4 x^2}{5 + \log_4 x^2} + (\log_4 x)^2 = 0$$ 7. Let $y = \log_4 x$. Then $\log_4 x^2 = 2y$. Rewrite equation: $$\frac{2y}{5 + 2y} + y^2 = 0$$ Multiply both sides by $5 + 2y$: $$2y + y^2(5 + 2y) = 0$$ Expand: $$2y + 5y^2 + 2y^3 = 0$$ Rewrite: $$2y^3 + 5y^2 + 2y = 0$$ Factor out $y$: $$y(2y^2 + 5y + 2) = 0$$ 8. Solve $y=0$ or $2y^2 + 5y + 2=0$. - For $y=0$, $\log_4 x = 0 \Rightarrow x = 4^0 = 1$. - For quadratic, use formula: $$y = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4}$$ 9. Roots: - $y = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2}$ - $y = \frac{-5 - 3}{4} = \frac{-8}{4} = -2$ 10. Convert back to $x$: - For $y = -\frac{1}{2}$, $x = 4^{-1/2} = \frac{1}{\sqrt{4}} = \frac{1}{2}$. - For $y = -2$, $x = 4^{-2} = \frac{1}{16}$. 11. Final solutions for $x$ are: $$x = 1, \frac{1}{2}, \frac{1}{16}$$ 12. Problem 20: Find total area of garden. - Area1 is a square with side length 2m, so area is $2 \times 2 = 4$ m$^2$. - Area2 is a rectangle adjacent to Area1 with height 2m. - Total length of two zones combined is 10m. - Since Area1 is square with length 2m, Area2 length is $10 - 2 = 8$ m. - Area2 area is $8 \times 2 = 16$ m$^2$. 13. Total area of garden is sum of areas: $$4 + 16 = 20 \text{ m}^2$$ Final answers: - 18b) $\frac{1}{9}$ - 18c) $\frac{8}{9}$ - 18d) $\frac{2}{3}$ - 19) $x = 1, \frac{1}{2}, \frac{1}{16}$ - 20) Total area = 20 m$^2$