Solve 3*(-1/2)^2+1/4/3*(-frac{1{3})^2-3/2}

$$\frac { 3 \cdot ( - \frac { 1 } { 2 } ) ^ { 2 } + \frac { 1 } { 4 } } { 3 \cdot ( - \frac { 1 } { 3 } ) ^ { 2 } - \frac { 3 } { 2 } }$$

Evaluate

$$-\frac{6}{7}\approx -0.857142857$$

Solution Steps

Calculate -\frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{3\times \frac{1}{4}+\frac{1}{4}}{3\left(-\frac{1}{3}\right)^{2}-\frac{3}{2}}$$

Multiply 3 and \frac{1}{4} to get \frac{3}{4}.

$$\frac{\frac{3}{4}+\frac{1}{4}}{3\left(-\frac{1}{3}\right)^{2}-\frac{3}{2}}$$

Since \frac{3}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3+1}{4}}{3\left(-\frac{1}{3}\right)^{2}-\frac{3}{2}}$$

Add 3 and 1 to get 4.

$$\frac{\frac{4}{4}}{3\left(-\frac{1}{3}\right)^{2}-\frac{3}{2}}$$

Divide 4 by 4 to get 1.

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

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

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

Multiply 3 and \frac{1}{9} to get \frac{3}{9}.

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

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

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

Least common multiple of 3 and 2 is 6. Convert \frac{1}{3} and \frac{3}{2} to fractions with denominator 6.

$$\frac{1}{\frac{2}{6}-\frac{9}{6}}$$

Since \frac{2}{6} and \frac{9}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{2-9}{6}}$$

Subtract 9 from 2 to get -7.

$$\frac{1}{-\frac{7}{6}}$$

Divide 1 by -\frac{7}{6} by multiplying 1 by the reciprocal of -\frac{7}{6}.

$$1\left(-\frac{6}{7}\right)$$

Multiply 1 and -\frac{6}{7} to get -\frac{6}{7}.

$$-\frac{6}{7}$$

Factor

$$-\frac{6}{7} = -0.8571428571428571$$