Solve (4div|-4|-1-1/3)^2+(pi-2)^0-(-1)^2010

$$( 4 \div | - 4 | - 1 - \frac { 1 } { 3 } ) ^ { 2 } + ( \pi - 2 ) ^ { 0 } - ( - 1 ) ^ { 2010 }$$

Evaluate

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

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -4 is 4.

$$\left(\frac{4}{4}-1-\frac{1}{3}\right)^{2}+\left(\pi -2\right)^{0}-\left(-1\right)^{2010}$$

Divide 4 by 4 to get 1.

$$\left(1-1-\frac{1}{3}\right)^{2}+\left(\pi -2\right)^{0}-\left(-1\right)^{2010}$$

Subtract 1 from 1 to get 0.

$$\left(-\frac{1}{3}\right)^{2}+\left(\pi -2\right)^{0}-\left(-1\right)^{2010}$$

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

$$\frac{1}{9}+\left(\pi -2\right)^{0}-\left(-1\right)^{2010}$$

Calculate \pi -2 to the power of 0 and get 1.

$$\frac{1}{9}+1-\left(-1\right)^{2010}$$

Add \frac{1}{9} and 1 to get \frac{10}{9}.

$$\frac{10}{9}-\left(-1\right)^{2010}$$

Calculate -1 to the power of 2010 and get 1.

$$\frac{10}{9}-1$$

Subtract 1 from \frac{10}{9} to get \frac{1}{9}.

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

Factor

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