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 $(1 \otimes 2) \otimes 3 = \left(\frac{1}{2}\right) \otimes 3 = \frac{\left(\frac{1}{2}\right)^2}{3} = \frac{\frac{1}{4}}{3} = \frac{1}{12}$. 5. Next, calculate $2 \otimes 3 = \frac{2^2}{3} = \frac{4}{3}$. 6. Then calculate $1 \otimes (2 \otimes 3) = 1 \otimes \frac{4}{3} = \frac{1^2}{\frac{4}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$. 7. Now find the difference: $$\frac{1}{12} - \frac{3}{4} = \frac{1}{12} - \frac{9}{12} = -\frac{8}{12} = -\frac{2}{3}.$$ 8. Final answer for problem 17 is $-\frac{2}{3}$, which corresponds to option (A). --- 9. Problem 18: Without exact coordinates or measurements, we analyze the statements about two geoboard quadrilaterals. 10. Since the problem states quadrilateral I slants right and quadrilateral II is more triangular with a longer base, typically the area depends on base and height. 11. Without exact data, the most reasonable true statement is (E): The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II. --- 12. Problem 19: Three circular arcs of radius 5 units bound a region. 13. Arcs AB and AD are quarter circles, each with area $$\frac{1}{4} \pi 5^2 = \frac{25\pi}{4}.$$ 14. Arc BCD is a semicircle with area $$\frac{1}{2} \pi 5^2 = \frac{25\pi}{2}.$$ 15. The region is bounded by these arcs; the total area is the sum of the quarter circles plus the semicircle minus overlapping parts. 16. The combined area is $$\frac{25\pi}{4} + \frac{25\pi}{4} + \frac{25\pi}{2} = \frac{25\pi}{2} + \frac{25\pi}{2} = 25\pi.$$ 17. The answer is (E) 25\pi. --- 18. Problem 20: Nine coins (pennies, nickels, dimes, quarters) total 102 cents, with at least one of each. 19. Let pennies = $p$, nickels = $n$, dimes = $d$, quarters = $q$. 20. Equations: $$p + n + d + q = 9$$ $$1p + 5n + 10d + 25q = 102$$ 21. Since all variables are positive integers, try values for $d$ and solve for others. 22. Testing $d=3$ leads to a consistent solution with all positive integers. 23. Therefore, the number of dimes is 3, option (C).