Solve 18*(1/9)^2-20*1/9

$$18 \cdot ( \frac { 1 } { 9 } ) ^ { 2 } - 20 \cdot \frac { 1 } { 9 }$$

Evaluate

$$-2$$

Solution Steps

Calculate \frac{1}{9} to the power of 2 and get \frac{1}{81}.

$$18\times \frac{1}{81}-20\times \frac{1}{9}$$

Multiply 18 and \frac{1}{81} to get \frac{18}{81}.

$$\frac{18}{81}-20\times \frac{1}{9}$$

Reduce the fraction \frac{18}{81} to lowest terms by extracting and canceling out 9.

$$\frac{2}{9}-20\times \frac{1}{9}$$

Multiply 20 and \frac{1}{9} to get \frac{20}{9}.

$$\frac{2}{9}-\frac{20}{9}$$

Since \frac{2}{9} and \frac{20}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-20}{9}$$

Subtract 20 from 2 to get -18.

$$\frac{-18}{9}$$

Divide -18 by 9 to get -2.

$$-2$$

Factor

$$-2$$