Solve =(-2-0)^2*4/9+(0-0)^2*3/9+(1-0)^2*2/9

$$= ( - 2 - 0 ) ^ { 2 } \times \frac { 4 } { 9 } + ( 0 - 0 ) ^ { 2 } \times \frac { 3 } { 9 } + ( 1 - 0 ) ^ { 2 } \times \frac { 2 } { 9 }$$

Evaluate

$$2$$

Solution Steps

Subtract 0 from -2 to get -2.

$$\left(-2\right)^{2}\times \frac{4}{9}+\left(0-0\right)^{2}\times \frac{3}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Calculate -2 to the power of 2 and get 4.

$$4\times \frac{4}{9}+\left(0-0\right)^{2}\times \frac{3}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Multiply 4 and \frac{4}{9} to get \frac{16}{9}.

$$\frac{16}{9}+\left(0-0\right)^{2}\times \frac{3}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Subtracting 0 from itself leaves 0.

$$\frac{16}{9}+0^{2}\times \frac{3}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Calculate 0 to the power of 2 and get 0.

$$\frac{16}{9}+0\times \frac{3}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$\frac{16}{9}+0\times \frac{1}{3}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Multiply 0 and \frac{1}{3} to get 0.

$$\frac{16}{9}+0+\left(1-0\right)^{2}\times \frac{2}{9}$$

Add \frac{16}{9} and 0 to get \frac{16}{9}.

$$\frac{16}{9}+\left(1-0\right)^{2}\times \frac{2}{9}$$

Subtract 0 from 1 to get 1.

$$\frac{16}{9}+1^{2}\times \frac{2}{9}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{16}{9}+1\times \frac{2}{9}$$

Multiply 1 and \frac{2}{9} to get \frac{2}{9}.

$$\frac{16}{9}+\frac{2}{9}$$

Add \frac{16}{9} and \frac{2}{9} to get 2.

$$2$$

Factor

$$2$$