Solve {left(16/3-3/frac{1{2}-4}+1right)}^2

$${ \left( \frac{ \frac{ 16 }{ 3 } -3 }{ \frac{ 1 }{ 2 } -4 } +1 \right) }^{ 2 }$$

Evaluate

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

Solution Steps

Subtract 3 from \frac{16}{3} to get \frac{7}{3}.

$$\left(\frac{\frac{7}{3}}{\frac{1}{2}-4}+1\right)^{2}$$

Subtract 4 from \frac{1}{2} to get -\frac{7}{2}.

$$\left(\frac{\frac{7}{3}}{-\frac{7}{2}}+1\right)^{2}$$

Divide \frac{7}{3} by -\frac{7}{2} by multiplying \frac{7}{3} by the reciprocal of -\frac{7}{2}.

$$\left(\frac{7}{3}\left(-\frac{2}{7}\right)+1\right)^{2}$$

Multiply \frac{7}{3} and -\frac{2}{7} to get -\frac{2}{3}.

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

Add -\frac{2}{3} and 1 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

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