Answer:
After dividing [tex]\frac{\sqrt{9x^2}}{\sqrt{18y^2}}[/tex] we get [tex]\frac{\sqrt{2}x}{2y}[/tex]
Hence, Option D is correct
Step-by-step explanation:
Divide [tex]\frac{\sqrt{9x^2}}{\sqrt{18y^2}}[/tex]
We know that:
[tex]\sqrt{9x^2[/tex] = 3x
[tex]\sqrt{18y^2}[/tex] = [tex]\sqrt{2*3*3 y^2}[/tex] = [tex]3\sqrt{2}y[/tex]
[tex]\frac{3x}{3\sqrt{2}y} \\Solving:\\\frac{x}{\sqrt{2}y}[/tex]
Now rationalize denominator
[tex]=\frac{x}{\sqrt{2}y}\\=\frac{x}{\sqrt{2}y} * \frac{\sqrt{2}}{\sqrt{2}} \\=\frac{\sqrt{2}x}{\sqrt{2}*\sqrt{2}y}\\=\frac{\sqrt{2}x}{2y}[/tex]
After dividing [tex]\frac{\sqrt{9x^2}}{\sqrt{18y^2}}[/tex] we get [tex]\frac{\sqrt{2}x}{2y}[/tex]
Hence, Option D is correct
