Solve 6*(1/3)^2-17*1/3

$$6 \cdot ( \frac { 1 } { 3 } ) ^ { 2 } - 17 \cdot \frac { 1 } { 3 }$$

Evaluate

$$-5$$

Solution Steps

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$6\times \frac{1}{9}-17\times \frac{1}{3}$$

Multiply 6 and \frac{1}{9} to get \frac{6}{9}.

$$\frac{6}{9}-17\times \frac{1}{3}$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$\frac{2}{3}-17\times \frac{1}{3}$$

Multiply 17 and \frac{1}{3} to get \frac{17}{3}.

$$\frac{2}{3}-\frac{17}{3}$$

Since \frac{2}{3} and \frac{17}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-17}{3}$$

Subtract 17 from 2 to get -15.

$$\frac{-15}{3}$$

Divide -15 by 3 to get -5.

$$-5$$

Factor

$$-5$$