Solve x/3ydiv(3xy)^2/25xy^3

$$\frac { x } { 3 y } \div \frac { ( 3 x y ) ^ { 2 } } { 25 x y ^ { 3 } }$$

Evaluate

$$\frac{25}{27}\approx 0.925925926$$

Solution Steps

Divide \frac{x}{3y} by \frac{\left(3xy\right)^{2}}{25xy^{3}} by multiplying \frac{x}{3y} by the reciprocal of \frac{\left(3xy\right)^{2}}{25xy^{3}}.

$$\frac{x\times 25xy^{3}}{3y\times \left(3xy\right)^{2}}$$

Cancel out y in both numerator and denominator.

$$\frac{25xxy^{2}}{3\times \left(3xy\right)^{2}}$$

Multiply x and x to get x^{2}.

$$\frac{25x^{2}y^{2}}{3\times \left(3xy\right)^{2}}$$

Expand \left(3xy\right)^{2}.

$$\frac{25x^{2}y^{2}}{3\times 3^{2}x^{2}y^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{25x^{2}y^{2}}{3\times 9x^{2}y^{2}}$$

Multiply 3 and 9 to get 27.

$$\frac{25x^{2}y^{2}}{27x^{2}y^{2}}$$

Cancel out x^{2}y^{2} in both numerator and denominator.

$$\frac{25}{27}$$

Factor

$$\frac{5 ^ {2}}{3 ^ {3}} = 0.9259259259259259$$