1. **Stating the problem:** Simplify the expression $$60\sqrt{4^{10}} \cdot \sqrt[6]{4^{18}} : 4^2 = x\sqrt{4^y}$$ and find $x$ and $y$. 2. **Recall the rules:** - The $n$th root of $a^m$ is $a^{\frac{m}{n}}$. - Multiplication of powers with the same base adds exponents. - Division of powers with the same base subtracts exponents. 3. **Rewrite the radicals as exponents:** $$60\sqrt{4^{10}} = 60 \cdot (4^{10})^{\frac{1}{2}} = 60 \cdot 4^{\frac{10}{2}} = 60 \cdot 4^5$$ $$\sqrt[6]{4^{18}} = (4^{18})^{\frac{1}{6}} = 4^{\frac{18}{6}} = 4^3$$ 4. **Multiply the terms:** $$60 \cdot 4^5 \cdot 4^3 = 60 \cdot 4^{5+3} = 60 \cdot 4^8$$ 5. **Divide by $4^2$:** $$\frac{60 \cdot 4^8}{4^2} = 60 \cdot 4^{8-2} = 60 \cdot 4^6$$ 6. **Express in the form $x\sqrt{4^y}$:** Since $\sqrt{4^y} = 4^{\frac{y}{2}}$, we want to write $60 \cdot 4^6$ as $x \cdot 4^{\frac{y}{2}}$. 7. **Choose $y$ so that $\frac{y}{2} = 6$, thus $y = 12$.** 8. **Then $x = 60$.** **Final answer:** $$x = 60, \quad y = 12$$