1. **State the problem:** Simplify the expressions $$\frac{\sqrt{-100}}{\sqrt{-2}}$$ and $$\sqrt{-15} \cdot \sqrt{5}$$ with no negative numbers under radicals and no radicals in the denominator. 2. **Recall important rules:** - For negative numbers under radicals, use $$\sqrt{-a} = i\sqrt{a}$$ where $$i$$ is the imaginary unit. - For division and multiplication of radicals, $$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$ and $$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$$. - Rationalize denominators by multiplying numerator and denominator by the conjugate or appropriate radical. 3. **Simplify the first expression:** $$\frac{\sqrt{-100}}{\sqrt{-2}} = \frac{i\sqrt{100}}{i\sqrt{2}} = \frac{i \cdot 10}{i \cdot \sqrt{2}}$$ Cancel $$i$$: $$\frac{\cancel{i} \cdot 10}{\cancel{i} \cdot \sqrt{2}} = \frac{10}{\sqrt{2}}$$ Rationalize denominator: $$\frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2}$$ Simplify fraction: $$\frac{\cancel{10}^{5}\sqrt{2}}{\cancel{2}^1} = 5\sqrt{2}$$ 4. **Simplify the second expression:** $$\sqrt{-15} \cdot \sqrt{5} = i\sqrt{15} \cdot \sqrt{5} = i \sqrt{15 \cdot 5} = i \sqrt{75}$$ Simplify $$\sqrt{75}$$: $$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$$ So, $$i \sqrt{75} = i \cdot 5 \sqrt{3} = 5i\sqrt{3}$$ **Final answers:** $$\frac{\sqrt{-100}}{\sqrt{-2}} = 5\sqrt{2}$$ $$\sqrt{-15} \cdot \sqrt{5} = 5i\sqrt{3}$$