Solve ((23/9div3)^-2/94^2*frac{2/5})^-1

$$( \frac { ( 2 \frac { 3 } { 9 } \div 3 ) ^ { - 2 } } { \frac { 9 } { 4 } ^ { 2 } \cdot \frac { 2 } { 5 } } ) ^ { - 1 }$$

Evaluate

$$\frac{49}{40}=1.225$$

Solution Steps

Express \frac{\frac{2\times 9+3}{9}}{3} as a single fraction.

$$\left(\frac{\left(\frac{2\times 9+3}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Multiply 2 and 9 to get 18.

$$\left(\frac{\left(\frac{18+3}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Add 18 and 3 to get 21.

$$\left(\frac{\left(\frac{21}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Multiply 9 and 3 to get 27.

$$\left(\frac{\left(\frac{21}{27}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Reduce the fraction \frac{21}{27} to lowest terms by extracting and canceling out 3.

$$\left(\frac{\left(\frac{7}{9}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Calculate \frac{7}{9} to the power of -2 and get \frac{81}{49}.

$$\left(\frac{\frac{81}{49}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Calculate \frac{9}{4} to the power of 2 and get \frac{81}{16}.

$$\left(\frac{\frac{81}{49}}{\frac{81}{16}\times \frac{2}{5}}\right)^{-1}$$

Multiply \frac{81}{16} and \frac{2}{5} to get \frac{81}{40}.

$$\left(\frac{\frac{81}{49}}{\frac{81}{40}}\right)^{-1}$$

Divide \frac{81}{49} by \frac{81}{40} by multiplying \frac{81}{49} by the reciprocal of \frac{81}{40}.

$$\left(\frac{81}{49}\times \frac{40}{81}\right)^{-1}$$

Multiply \frac{81}{49} and \frac{40}{81} to get \frac{40}{49}.

$$\left(\frac{40}{49}\right)^{-1}$$

Calculate \frac{40}{49} to the power of -1 and get \frac{49}{40}.

$$\frac{49}{40}$$

Factor

$$\frac{7 ^ {2}}{2 ^ {3} \cdot 5} = 1\frac{9}{40} = 1.225$$