Respuesta :

kudzordzifrancis

Answer:

B. [tex]\frac{x\sqrt{2}}{2y}[/tex]

Step-by-step explanation:

We want to divide [tex]\sqrt{9x^2}[/tex] by[tex]\sqrt{18y^2}[/tex].

This becomes:

[tex]\frac{\sqrt{9x^2}}{\sqrt{18y^2}}[/tex]

[tex]\frac{\sqrt{(3x)^2}}{\sqrt{2(3y)^2}}[/tex]

We remove the perfect squares to obtain

[tex]\frac{3x}{3y\sqrt{2}}[/tex]

Cancel out the common factors to get;

[tex]\frac{x}{y\sqrt{2}}[/tex]

Rationalize the denominator to get:

[tex]\frac{x}{y\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}[/tex]

[tex]\frac{x\sqrt{2}}{2y}[/tex]

The correct answer is B

gmany

Answer:

[tex]\large\boxed{B.\ \dfrac{x\sqrt2}{2y}}[/tex]

Step-by-step explanation:

[tex]\sqrt{9x^2}:\sqrt{18y^2}=\dfrac{\sqrt{9x^2}}{\sqrt{18y^2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt9\cdot\sqrt{x^2}}{\sqrt{18}\cdot\sqrt{y^2}}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\dfrac{3\cdot x}{\sqrt{9\cdot2}\cdot y}=\dfrac{3x}{\sqrt9\cdot\sqrt2\cdot y}=\dfrac{3x}{3y\sqrt2}\qquad\text{cancel 3}\\\\=\dfrac{x}{y\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\qquad\text{use}\ \sqrt{a}\cdot\sqrt{a}=a\\\\=\dfrac{x\sqrt2}{2y}[/tex]

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