Solve frac{2^{5}+2^{3}-(1/6)^-1}{3+2^3*4}*((-3/2)^2+2^-1)

$$\frac { 2 ^ { 5 } + 2 ^ { 3 } - ( \frac { 1 } { 6 } ) ^ { - 1 } } { 3 + 2 ^ { 3 } \cdot 4 } \cdot ( ( - \frac { 3 } { 2 } ) ^ { 2 } + 2 ^ { - 1 } )$$

Evaluate

$$\frac{187}{70}\approx 2.671428571$$

Solution Steps

Calculate 2 to the power of 5 and get 32.

$$\frac{32+2^{3}-\left(\frac{1}{6}\right)^{-1}}{3+2^{3}\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{32+8-\left(\frac{1}{6}\right)^{-1}}{3+2^{3}\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Add 32 and 8 to get 40.

$$\frac{40-\left(\frac{1}{6}\right)^{-1}}{3+2^{3}\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Calculate \frac{1}{6} to the power of -1 and get 6.

$$\frac{40-6}{3+2^{3}\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Subtract 6 from 40 to get 34.

$$\frac{34}{3+2^{3}\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{34}{3+8\times 4}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Multiply 8 and 4 to get 32.

$$\frac{34}{3+32}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

Add 3 and 32 to get 35.

$$\frac{34}{35}\left(\left(-\frac{3}{2}\right)^{2}+2^{-1}\right)$$

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

$$\frac{34}{35}\left(\frac{9}{4}+2^{-1}\right)$$

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

$$\frac{34}{35}\left(\frac{9}{4}+\frac{1}{2}\right)$$

Add \frac{9}{4} and \frac{1}{2} to get \frac{11}{4}.

$$\frac{34}{35}\times \frac{11}{4}$$

Multiply \frac{34}{35} and \frac{11}{4} to get \frac{187}{70}.

$$\frac{187}{70}$$

Factor

$$\frac{11 \cdot 17}{2 \cdot 5 \cdot 7} = 2\frac{47}{70} = 2.6714285714285713$$