1. **Problem statement:** Simplify the expressions $$\sqrt{-3} \cdot \sqrt{-14}$$ and $$\frac{\sqrt{-42}}{\sqrt{7}}$$ with no negative numbers under radicals and no radicals in the denominator. 2. **Recall:** For negative numbers under square roots, use $$\sqrt{-a} = i\sqrt{a}$$ where $$i = \sqrt{-1}$$. 3. **First expression:** $$\sqrt{-3} \cdot \sqrt{-14} = (i\sqrt{3})(i\sqrt{14}) = i^2 \sqrt{3 \cdot 14} = -1 \cdot \sqrt{42} = -\sqrt{42}$$. 4. **Second expression:** $$\frac{\sqrt{-42}}{\sqrt{7}} = \frac{i\sqrt{42}}{\sqrt{7}} = i \frac{\sqrt{42}}{\sqrt{7}} = i \sqrt{\frac{42}{7}} = i \sqrt{6}$$. 5. **Final answers:** - $$\sqrt{-3} \cdot \sqrt{-14} = -\sqrt{42}$$ - $$\frac{\sqrt{-42}}{\sqrt{7}} = i \sqrt{6}$$ These expressions have no negative numbers under radicals and no radicals in the denominator.