1. Stating the problem: Simplify the square root expressions and products for questions 35 to 49. 2. Recall the rule: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$. 3. Simplify each: 35. $\sqrt{\frac{27}{36}} = \frac{\sqrt{27}}{\sqrt{36}} = \frac{\sqrt{9 \cdot 3}}{6} = \frac{3\sqrt{3}}{6} = \frac{\cancel{3}\sqrt{3}}{\cancel{6}2} = \frac{\sqrt{3}}{2}$ 36. $\sqrt{\frac{49}{4}} = \frac{\sqrt{49}}{\sqrt{4}} = \frac{7}{2}$ 37. $\sqrt{\frac{36}{9}} = \frac{\sqrt{36}}{\sqrt{9}} = \frac{6}{3} = \frac{\cancel{6}}{\cancel{3}2} = 2$ 38. $\sqrt{\frac{9}{49}} = \frac{\sqrt{9}}{\sqrt{49}} = \frac{3}{7}$ 39. $\sqrt{\frac{48}{81}} = \frac{\sqrt{48}}{\sqrt{81}} = \frac{\sqrt{16 \cdot 3}}{9} = \frac{4\sqrt{3}}{9}$ 40. $\sqrt{\frac{64}{16}} = \frac{\sqrt{64}}{\sqrt{16}} = \frac{8}{4} = 2$ 41. $\sqrt{\frac{120}{4}} = \frac{\sqrt{120}}{2} = \frac{\sqrt{4 \cdot 30}}{2} = \frac{2\sqrt{30}}{2} = \cancel{\frac{\cancel{2}\sqrt{30}}{\cancel{2}}} = \sqrt{30}$ 42. $\frac{1}{2} \sqrt{32} \cdot \sqrt{2} = \frac{1}{2} \sqrt{32 \cdot 2} = \frac{1}{2} \sqrt{64} = \frac{1}{2} \cdot 8 = 4$ 43. $3 \sqrt{63} \cdot \sqrt{4} = 3 \sqrt{63 \cdot 4} = 3 \sqrt{252} = 3 \sqrt{36 \cdot 7} = 3 \cdot 6 \sqrt{7} = 18 \sqrt{7}$ 44. $\sqrt{9} \cdot 4 \sqrt{25} = 3 \cdot 4 \cdot 5 = 60$ 45. $-2 \sqrt{27} \cdot \sqrt{3} = -2 \sqrt{27 \cdot 3} = -2 \sqrt{81} = -2 \cdot 9 = -18$ 46. $\sqrt{7} \cdot \sqrt{\frac{18}{\sqrt{2}}} = \sqrt{7} \cdot \sqrt{\frac{18}{\sqrt{2}}} = \sqrt{7} \cdot \frac{\sqrt{18}}{\sqrt[4]{4}}$ (simplify $\sqrt{\frac{18}{\sqrt{2}}}$ by rationalizing denominator) $\sqrt{\frac{18}{\sqrt{2}}} = \sqrt{18} \cdot \frac{1}{\sqrt[4]{4}} = 3\sqrt{2} \cdot \frac{1}{\sqrt[4]{4}}$ (approximate or leave as is) So the expression is $\sqrt{7} \cdot 3 \sqrt{2} / \sqrt[4]{4} = 3 \sqrt{14} / \sqrt[4]{4}$ 47. $-\sqrt{4} \cdot \sqrt{\frac{81}{36}} = -2 \cdot \frac{\sqrt{81}}{\sqrt{36}} = -2 \cdot \frac{9}{6} = -2 \cdot \frac{\cancel{9}}{\cancel{6}1.5} = -3$ 48. $\frac{\sqrt{10} \cdot \sqrt{16}}{\sqrt{5}} = \frac{\sqrt{160}}{\sqrt{5}} = \sqrt{\frac{160}{5}} = \sqrt{32} = \sqrt{16 \cdot 2} = 4 \sqrt{2}$ 49. $\frac{-2 \sqrt{20}}{\sqrt{100}} = -2 \cdot \frac{\sqrt{20}}{10} = -2 \cdot \frac{\sqrt{4 \cdot 5}}{10} = -2 \cdot \frac{2 \sqrt{5}}{10} = -2 \cdot \frac{\cancel{2} \sqrt{5}}{\cancel{10}5} = - \frac{2 \sqrt{5}}{5}$ Final answers: 35. $\frac{\sqrt{3}}{2}$ 36. $\frac{7}{2}$ 37. $2$ 38. $\frac{3}{7}$ 39. $\frac{4 \sqrt{3}}{9}$ 40. $2$ 41. $\sqrt{30}$ 42. $4$ 43. $18 \sqrt{7}$ 44. $60$ 45. $-18$ 46. $\frac{3 \sqrt{14}}{\sqrt[4]{4}}$ 47. $-3$ 48. $4 \sqrt{2}$ 49. $- \frac{2 \sqrt{5}}{5}$