Solve 3(-2-6)+4+1/(-3)^3+(-2)^4

$$\frac { 3 ( - 2 - 6 ) + 4 + 1 } { ( - 3 ) ^ { 3 } + ( - 2 ) ^ { 4 } }$$

Evaluate

$$\frac{19}{11}\approx 1.727272727$$

Solution Steps

Subtract 6 from -2 to get -8.

$$\frac{3\left(-8\right)+4+1}{\left(-3\right)^{3}+\left(-2\right)^{4}}$$

Multiply 3 and -8 to get -24.

$$\frac{-24+4+1}{\left(-3\right)^{3}+\left(-2\right)^{4}}$$

Add -24 and 4 to get -20.

$$\frac{-20+1}{\left(-3\right)^{3}+\left(-2\right)^{4}}$$

Add -20 and 1 to get -19.

$$\frac{-19}{\left(-3\right)^{3}+\left(-2\right)^{4}}$$

Calculate -3 to the power of 3 and get -27.

$$\frac{-19}{-27+\left(-2\right)^{4}}$$

Calculate -2 to the power of 4 and get 16.

$$\frac{-19}{-27+16}$$

Add -27 and 16 to get -11.

$$\frac{-19}{-11}$$

Fraction \frac{-19}{-11} can be simplified to \frac{19}{11} by removing the negative sign from both the numerator and the denominator.

$$\frac{19}{11}$$

Factor

$$\frac{19}{11} = 1\frac{8}{11} = 1.7272727272727273$$