Solve (1/9-1/18+1/4+1/36)^2

$$( \frac { 1 } { 9 } - \frac { 1 } { 18 } + \frac { 1 } { 4 } + \frac { 1 } { 36 } ) ^ { 2 }$$

Evaluate

$$\frac{1}{9}\approx 0.111111111$$

Solution Steps

Subtract \frac{1}{18} from \frac{1}{9} to get \frac{1}{18}.

$$\left(\frac{1}{18}+\frac{1}{4}+\frac{1}{36}\right)^{2}$$

Add \frac{1}{18} and \frac{1}{4} to get \frac{11}{36}.

$$\left(\frac{11}{36}+\frac{1}{36}\right)^{2}$$

Add \frac{11}{36} and \frac{1}{36} to get \frac{1}{3}.

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

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

$$\frac{1}{9}$$

Factor

$$\frac{1}{3 ^ {2}} = 0.1111111111111111$$