Solve |-1/2|*(-6)+(-2/3)*3^2

$$| - \frac { 1 } { 2 } | \times ( - 6 ) + ( - \frac { 2 } { 3 } ) \times 3 ^ { 2 }$$

Evaluate

$$-9$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{1}{2} is \frac{1}{2}.

$$\frac{1}{2}\left(-6\right)-\frac{2}{3}\times 3^{2}$$

Multiply \frac{1}{2} and -6 to get \frac{-6}{2}.

$$\frac{-6}{2}-\frac{2}{3}\times 3^{2}$$

Divide -6 by 2 to get -3.

$$-3-\frac{2}{3}\times 3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$-3-\frac{2}{3}\times 9$$

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

$$-3+\frac{-2\times 9}{3}$$

Multiply -2 and 9 to get -18.

$$-3+\frac{-18}{3}$$

Divide -18 by 3 to get -6.

$$-3-6$$

Subtract 6 from -3 to get -9.

$$-9$$

Factor

$$-9$$