1. **Problem 12a:** Find the prime factorization of 576 using exponents. 2. Start by dividing 576 by the smallest prime 2 repeatedly: $$576 \div 2 = 288$$ $$288 \div 2 = 144$$ $$144 \div 2 = 72$$ $$72 \div 2 = 36$$ $$36 \div 2 = 18$$ $$18 \div 2 = 9$$ 3. Now 9 is not divisible by 2, but 9 can be divided by 3: $$9 \div 3 = 3$$ $$3 \div 3 = 1$$ 4. Count the number of times 2 and 3 were factors: - 2 divides 576 six times - 3 divides 576 two times 5. Hence the prime factorization is: $$576 = 2^6 \times 3^2$$ 6. **Problem 12b:** Find $$\sqrt{576}$$. 7. Since $$576 = 2^6 \times 3^2$$, then $$\sqrt{576} = \sqrt{2^6 \times 3^2} = 2^{6/2} \times 3^{2/2} = 2^3 \times 3^1 = 8 \times 3 = 24$$ 8. **Problem 13:** Convert mixed numbers to improper fractions and multiply. - Convert $$5 \tfrac{5}{6}$$ to improper fraction: $$5 \times 6 + 5 = 30 + 5 = 35$$ So, $$5 \tfrac{5}{6} = \frac{35}{6}$$ - Convert $$6 \tfrac{6}{7}$$ to improper fraction: $$6 \times 7 + 6 = 42 + 6 = 48$$ So, $$6 \tfrac{6}{7} = \frac{48}{7}$$ - Multiply the fractions: $$\frac{35}{6} \times \frac{48}{7} = \frac{35 \times 48}{6 \times 7}$$ - Simplify numerator and denominator: $$35 = 5 \times 7, \quad 48 = 6 \times 8$$ So: $$\frac{35 \times 48}{6 \times 7} = \frac{5 \times 7 \times 6 \times 8}{6 \times 7}$$ Canceling 6 and 7: $$= 5 \times 8 = 40$$ 9. **Problem 14a:** What kind of angle is $$\angle ACD$$? - Since ABCF and FCDE are squares, $$\angle ACD$$ is a right angle in square FCDE. - Therefore, $$\angle ACD = 90^\circ$$, a right angle. 10. **Problem 14b:** Name two segments parallel to FC. - FC is a side of the square. - Segments parallel to FC are $$AB$$ and $$DE$$ from the rectangle/squares configuration. 11. **Problem 15a:** Fraction of square CDEF that is shaded (triangle EFA). - Triangle EFA lies partly inside square CDEF. - Since triangle EFA shares side EF with square CDEF and is right triangle formed by EF and FA, its area is half of square CDEF. - Thus, fraction shaded of square CDEF is $$\frac{1}{2}$$. 12. **Problem 15b:** Fraction of square ABCF that is shaded. - Triangle EFA is not inside square ABCF, so shaded fraction is 0. 13. **Problem 15c:** Fraction of rectangle ABDE that is shaded. - Rectangle ABDE consists of both squares ABCF and FCDE. - The shaded area is triangle EFA, half of square FCDE, which is half of one square's area. - Total rectangle area is two squares, so shaded fraction is: $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ 14. **Problem 16a:** If $$AB = 3$$ ft, find perimeter of rectangle ABDE. - Each square side length equals $$AB = 3$$ ft. - Rectangle ABDE consists of two squares side by side, so length $$= 2 \times 3 = 6$$ ft and width $$= 3$$ ft. - Perimeter $$P = 2(\text{length} + \text{width}) = 2(6 + 3) = 2 \times 9 = 18$$ ft. 15. **Problem 16b:** Area of rectangle ABDE. - Area $$= \text{length} \times \text{width} = 6 \times 3 = 18$$ ft². **Final answers:** - 12a: $$576 = 2^6 \times 3^2$$ - 12b: $$\sqrt{576} = 24$$ - 13: Product = $$40$$ - 14a: $$\angle ACD$$ is a right angle - 14b: Segments parallel to FC are $$AB$$ and $$DE$$ - 15a: Fraction shaded of CDEF = $$\frac{1}{2}$$ - 15b: Fraction shaded of ABCF = $$0$$ - 15c: Fraction shaded of ABDE = $$\frac{1}{4}$$ - 16a: Perimeter of ABDE = $$18$$ ft - 16b: Area of ABDE = $$18$$ ft²