Solve (3)|2^3-3^2|-(-2/3)div(-4/9)

$$( 3 ) | 2 ^ { 3 } - 3 ^ { 2 } | - ( - \frac { 2 } { 3 } ) \div ( - \frac { 4 } { 9 } )$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$3|8-3^{2}|-\frac{-\frac{2}{3}}{-\frac{4}{9}}$$

Calculate 3 to the power of 2 and get 9.

$$3|8-9|-\frac{-\frac{2}{3}}{-\frac{4}{9}}$$

Subtract 9 from 8 to get -1.

$$3|-1|-\frac{-\frac{2}{3}}{-\frac{4}{9}}$$

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

$$3\times 1-\frac{-\frac{2}{3}}{-\frac{4}{9}}$$

Multiply 3 and 1 to get 3.

$$3-\frac{-\frac{2}{3}}{-\frac{4}{9}}$$

Divide -\frac{2}{3} by -\frac{4}{9} by multiplying -\frac{2}{3} by the reciprocal of -\frac{4}{9}.

$$3-\left(-\frac{2}{3}\left(-\frac{9}{4}\right)\right)$$

Multiply -\frac{2}{3} times -\frac{9}{4} by multiplying numerator times numerator and denominator times denominator.

$$3-\frac{-2\left(-9\right)}{3\times 4}$$

Do the multiplications in the fraction \frac{-2\left(-9\right)}{3\times 4}.

$$3-\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

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

Convert 3 to fraction \frac{6}{2}.

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

Since \frac{6}{2} and \frac{3}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 3 from 6 to get 3.

$$\frac{3}{2}$$

Factor

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$