1. **State the problem:** We need to determine which of the given equations is true for all positive values of $x$ and $y$. 2. **List the equations:** - (a) $\sqrt{x} + \sqrt{y} = \sqrt{x + y}$ - (b) $\sqrt{x^4 y^{16}} = x^2 y^4$ - (c) $(x \sqrt{y})(y \sqrt{x}) = x^2 y^2$ - (d) $y \sqrt{x} + y \sqrt{x} = \sqrt{4 x y^2}$ - (e) $(xy)(yy) = (xy)^{2y}$ 3. **Check each equation:** **(a)** $\sqrt{x} + \sqrt{y} = \sqrt{x + y}$ This is generally false because $\sqrt{a} + \sqrt{b} \neq \sqrt{a + b}$ for positive $a,b$. **(b)** $\sqrt{x^4 y^{16}} = x^2 y^4$ Recall $\sqrt{a} = a^{1/2}$, so: $$\sqrt{x^4 y^{16}} = (x^4 y^{16})^{1/2} = x^{4 \times \frac{1}{2}} y^{16 \times \frac{1}{2}} = x^2 y^8$$ Right side is $x^2 y^4$, so this is false. **(c)** $(x \sqrt{y})(y \sqrt{x}) = x^2 y^2$ Rewrite: $$x \sqrt{y} = x y^{1/2}, \quad y \sqrt{x} = y x^{1/2}$$ Multiply: $$x y^{1/2} \times y x^{1/2} = x^{1 + \frac{1}{2}} y^{1 + \frac{1}{2}} = x^{\frac{3}{2}} y^{\frac{3}{2}}$$ Right side is $x^2 y^2$, so false. **(d)** $y \sqrt{x} + y \sqrt{x} = \sqrt{4 x y^2}$ Left side: $$y \sqrt{x} + y \sqrt{x} = 2 y \sqrt{x} = 2 y x^{1/2}$$ Right side: $$\sqrt{4 x y^2} = (4 x y^2)^{1/2} = 2 x^{1/2} y$$ Both sides equal $2 y x^{1/2}$, so this is true. **(e)** $(xy)(yy) = (xy)^{2y}$ Left side: $$(xy)(yy) = x y \times y y = x y^3$$ Right side: $$(xy)^{2y} = x^{2y} y^{2y}$$ These are not equal for all positive $x,y$, so false. 4. **Conclusion:** Only equation (d) is true for all positive $x$ and $y$. **Final answer:** $$y \sqrt{x} + y \sqrt{x} = \sqrt{4 x y^2}$$