1. **State the problem:** Simplify the expressions $$\frac{\sqrt{-48}}{\sqrt{-16}}$$ and $$\sqrt{-6} \cdot \sqrt{-10}$$ with no negative numbers under radicals and no radicals in denominators. 2. **Recall the rule:** For negative numbers under radicals, use $$\sqrt{-a} = i\sqrt{a}$$ where $$i$$ is the imaginary unit. --- ### First expression: $$\frac{\sqrt{-48}}{\sqrt{-16}}$$ 3. Rewrite using imaginary unit: $$\frac{\sqrt{-48}}{\sqrt{-16}} = \frac{i\sqrt{48}}{i\sqrt{16}}$$ 4. Cancel $$i$$ in numerator and denominator: $$\frac{\cancel{i}\sqrt{48}}{\cancel{i}\sqrt{16}} = \frac{\sqrt{48}}{\sqrt{16}}$$ 5. Simplify radicals: $$\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$$ $$\sqrt{16} = 4$$ 6. Substitute back: $$\frac{4\sqrt{3}}{4}$$ 7. Cancel 4: $$\frac{\cancel{4}\sqrt{3}}{\cancel{4}} = \sqrt{3}$$ --- ### Second expression: $$\sqrt{-6} \cdot \sqrt{-10}$$ 8. Rewrite each radical: $$\sqrt{-6} = i\sqrt{6}$$ $$\sqrt{-10} = i\sqrt{10}$$ 9. Multiply: $$i\sqrt{6} \cdot i\sqrt{10} = i^2 \sqrt{6 \cdot 10} = i^2 \sqrt{60}$$ 10. Recall $$i^2 = -1$$: $$-1 \cdot \sqrt{60} = -\sqrt{60}$$ 11. Simplify $$\sqrt{60}$$: $$\sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15}$$ 12. Final answer: $$-2\sqrt{15}$$ --- **Final answers:** $$\frac{\sqrt{-48}}{\sqrt{-16}} = \sqrt{3}$$ $$\sqrt{-6} \cdot \sqrt{-10} = -2\sqrt{15}$$