Solve 1^2-04^2/2*8*0*4-2*8=

$$\frac{ { 1 }^{ 2 } -0 { 4 }^{ 2 } }{ 2 \cdot 8 \cdot 0 \cdot 4-2 \cdot 8 } =$$

Evaluate

$$-\frac{1}{16}=-0.0625$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$\frac{1-0\times 4^{2}}{2\times 8\times 0\times 4-2\times 8}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{1-0\times 16}{2\times 8\times 0\times 4-2\times 8}$$

Multiply 0 and 16 to get 0.

$$\frac{1-0}{2\times 8\times 0\times 4-2\times 8}$$

Subtract 0 from 1 to get 1.

$$\frac{1}{2\times 8\times 0\times 4-2\times 8}$$

Multiply 2 and 8 to get 16.

$$\frac{1}{16\times 0\times 4-2\times 8}$$

Multiply 16 and 0 to get 0.

$$\frac{1}{0\times 4-2\times 8}$$

Multiply 0 and 4 to get 0.

$$\frac{1}{0-2\times 8}$$

Multiply 2 and 8 to get 16.

$$\frac{1}{0-16}$$

Subtract 16 from 0 to get -16.

$$\frac{1}{-16}$$

Fraction \frac{1}{-16} can be rewritten as -\frac{1}{16} by extracting the negative sign.

$$-\frac{1}{16}$$

Factor

$$-\frac{1}{16} = -0.0625$$