Solve (3/2)^2*(-3)^2+(-3)^2*(-11/8)

$$( \frac { 3 } { 2 } ) ^ { 2 } \times ( - 3 ) ^ { 2 } + ( - 3 ) ^ { 2 } \times ( - \frac { 11 } { 8 } )$$

Evaluate

$$\frac{63}{8}=7.875$$

Solution Steps

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$\frac{9}{4}\left(-3\right)^{2}+\left(-3\right)^{2}\left(-\frac{11}{8}\right)$$

Calculate -3 to the power of 2 and get 9.

$$\frac{9}{4}\times 9+\left(-3\right)^{2}\left(-\frac{11}{8}\right)$$

Express \frac{9}{4}\times 9 as a single fraction.

$$\frac{9\times 9}{4}+\left(-3\right)^{2}\left(-\frac{11}{8}\right)$$

Multiply 9 and 9 to get 81.

$$\frac{81}{4}+\left(-3\right)^{2}\left(-\frac{11}{8}\right)$$

Calculate -3 to the power of 2 and get 9.

$$\frac{81}{4}+9\left(-\frac{11}{8}\right)$$

Express 9\left(-\frac{11}{8}\right) as a single fraction.

$$\frac{81}{4}+\frac{9\left(-11\right)}{8}$$

Multiply 9 and -11 to get -99.

$$\frac{81}{4}+\frac{-99}{8}$$

Fraction \frac{-99}{8} can be rewritten as -\frac{99}{8} by extracting the negative sign.

$$\frac{81}{4}-\frac{99}{8}$$

Least common multiple of 4 and 8 is 8. Convert \frac{81}{4} and \frac{99}{8} to fractions with denominator 8.

$$\frac{162}{8}-\frac{99}{8}$$

Since \frac{162}{8} and \frac{99}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{162-99}{8}$$

Subtract 99 from 162 to get 63.

$$\frac{63}{8}$$

Factor

$$\frac{3 ^ {2} \cdot 7}{2 ^ {3}} = 7\frac{7}{8} = 7.875$$