Solve (1/9)^2-19*1/9

$$( \frac { 1 } { 9 } ) ^ { 2 } - 19 \cdot \frac { 1 } { 9 }$$

Evaluate

$$-\frac{170}{81}\approx -2.098765432$$

Solution Steps

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

$$\frac{1}{81}-19\times \frac{1}{9}$$

Multiply 19 and \frac{1}{9} to get \frac{19}{9}.

$$\frac{1}{81}-\frac{19}{9}$$

Least common multiple of 81 and 9 is 81. Convert \frac{1}{81} and \frac{19}{9} to fractions with denominator 81.

$$\frac{1}{81}-\frac{171}{81}$$

Since \frac{1}{81} and \frac{171}{81} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-171}{81}$$

Subtract 171 from 1 to get -170.

$$-\frac{170}{81}$$

Factor

$$-\frac{170}{81} = -2\frac{8}{81} = -2.0987654320987654$$