1. **State the problem:** Simplify the expression $$\sqrt{75y^7}$$ assuming $$y \geq 0$$. 2. **Recall the property of square roots:** $$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$$ and $$\sqrt{y^{2n}} = y^n$$ for $$y \geq 0$$. 3. **Factor inside the square root:** $$75 = 25 \times 3$$, so $$\sqrt{75y^7} = \sqrt{25 \times 3 \times y^7} = \sqrt{25} \cdot \sqrt{3} \cdot \sqrt{y^7}$$. 4. **Simplify the perfect square:** $$\sqrt{25} = 5$$. 5. **Simplify $$\sqrt{y^7}$$:** Express $$y^7$$ as $$y^{6} \times y^{1}$$ because $$6$$ is even: $$\sqrt{y^7} = \sqrt{y^6 \times y} = \sqrt{y^6} \cdot \sqrt{y}$$. 6. **Simplify $$\sqrt{y^6}$$:** Since $$y^6 = (y^3)^2$$, $$\sqrt{y^6} = y^3$$ (because $$y \geq 0$$). 7. **Combine all parts:** $$5 \cdot \sqrt{3} \cdot y^3 \cdot \sqrt{y} = 5y^3 \sqrt{3y}$$. **Final answer:** $$\boxed{5y^3 \sqrt{3y}}$$