Solve i+m/m:(-1)^2+(sqrt{6})^2-(-9)+(-6)div2

$$i + \frac { m } { m } : ( - 1 ) ^ { 2 } + ( \sqrt { 6 } ) ^ { 2 } - ( - 9 ) + ( - 6 ) \div 2$$

Evaluate

$$13+i$$

Solution Steps

Divide m by m to get 1.

$$i+\frac{1}{\left(-1\right)^{2}}+\left(\sqrt{6}\right)^{2}-\left(-9\right)+\frac{-6}{2}$$

Calculate -1 to the power of 2 and get 1.

$$i+\frac{1}{1}+\left(\sqrt{6}\right)^{2}-\left(-9\right)+\frac{-6}{2}$$

Anything divided by one gives itself.

$$i+1+\left(\sqrt{6}\right)^{2}-\left(-9\right)+\frac{-6}{2}$$

The square of \sqrt{6} is 6.

$$i+1+6-\left(-9\right)+\frac{-6}{2}$$

Do the additions in i+1+6.

$$7+i-\left(-9\right)+\frac{-6}{2}$$

The opposite of -9 is 9.

$$7+i+9+\frac{-6}{2}$$

Add 7+i and 9 to get 16+i.

$$16+i+\frac{-6}{2}$$

Divide -6 by 2 to get -3.

$$16+i-3$$

Subtract 3 from 16+i to get 13+i.

$$13+i$$