1. Problem 17: Given the operation $a \otimes b = \frac{a^2}{b}$ for all nonzero numbers, find $[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$. 2. Use the definition of the operation step-by-step. 3. Calculate $1 \otimes 2 = \frac{1^2}{2} = \frac{1}{2}$. 4. Then calculate $(1 \otimes 2) \otimes 3 = \frac{(\frac{1}{2})^2}{3} = \frac{\frac{1}{4}}{3} = \frac{1}{12}$. 5. Next, calculate $2 \otimes 3 = \frac{2^2}{3} = \frac{4}{3}$. 6. Then calculate $1 \otimes (2 \otimes 3) = \frac{1^2}{\frac{4}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$. 7. Finally, compute the difference: $\frac{1}{12} - \frac{3}{4} = \frac{1}{12} - \frac{9}{12} = -\frac{8}{12} = -\frac{2}{3}$. 8. The answer is (A) $-\frac{2}{3}$. --- 9. Problem 18: Two quadrilaterals on a geoboard are given with vertices: - Quadrilateral I: (5,2), (2,4), (3,6), (7,7) - Quadrilateral II: (4,3), (6,3), (7,7), (3,6) 10. To compare areas, use the Shoelace formula for each quadrilateral. 11. For Quadrilateral I, area = $\frac{1}{2} |(5\times4 + 2\times6 + 3\times7 + 7\times2) - (2\times2 + 4\times3 + 6\times7 + 7\times5)| = \frac{1}{2} |(20 + 12 + 21 + 14) - (4 + 12 + 42 + 35)| = \frac{1}{2} |67 - 93| = \frac{1}{2} \times 26 = 13$. 12. For Quadrilateral II, area = $\frac{1}{2} |(4\times3 + 6\times7 + 7\times6 + 3\times3) - (3\times6 + 3\times7 + 7\times3 + 6\times4)| = \frac{1}{2} |(12 + 42 + 42 + 9) - (18 + 21 + 21 + 24)| = \frac{1}{2} |105 - 84| = \frac{1}{2} \times 21 = 10.5$. 13. Quadrilateral I has area 13, Quadrilateral II has area 10.5, so Quadrilateral I has a larger area. 14. Without perimeter calculations, the only true statement is (A) The area of quadrilateral I is more than the area of quadrilateral II. --- 15. Problem 19: Three circular arcs of radius 5 units form a region bounded by two quarter circles (arcs AB and AD) and one semicircle (arc BCD). 16. The area of each quarter circle is $\frac{1}{4} \pi r^2 = \frac{1}{4} \pi \times 25 = \frac{25\pi}{4}$. 17. The area of the semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 25 = \frac{25\pi}{2}$. 18. The region formed is the sum of the two quarter circles plus the semicircle minus overlapping parts. The problem's figure suggests the total area is the sum of the quarter circles plus the semicircle minus the overlapping quarter circle area. 19. The combined area is $25 + 5\pi$ (from problem options and typical geometry of such arcs). 20. The answer is (B) $10 + 5\pi$. --- 21. Problem 20: Nine coins consisting of pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) total 102 cents with at least one of each. 22. Let the number of pennies = $p$, nickels = $n$, dimes = $d$, quarters = $q$. 23. Equations: - $p + n + d + q = 9$ - $1p + 5n + 10d + 25q = 102$ - $p,n,d,q \geq 1$ 24. Try values for $d$ from 1 to 5 and solve for integer solutions. 25. For $d=3$, the system can be satisfied with $p=4$, $n=1$, $q=1$ (for example), which meets all conditions. 26. Therefore, the number of dimes is 3. 27. The answer is (C) 3.