Solve =|19-4/sqrt{(19+2)^2+2(0)^2}

$$= \frac { | 19 - 4 } { \sqrt { ( 19 + 2 ) ^ { 2 } + 2 ( 0 ) ^ { 2 } } }$$

Evaluate

$$\frac{5}{7}\approx 0.714285714$$

Solution Steps

Subtract 4 from 19 to get 15.

$$\frac{|15|}{\sqrt{\left(19+2\right)^{2}+2\times 0^{2}}}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 15 is 15.

$$\frac{15}{\sqrt{\left(19+2\right)^{2}+2\times 0^{2}}}$$

Add 19 and 2 to get 21.

$$\frac{15}{\sqrt{21^{2}+2\times 0^{2}}}$$

Calculate 21 to the power of 2 and get 441.

$$\frac{15}{\sqrt{441+2\times 0^{2}}}$$

Calculate 0 to the power of 2 and get 0.

$$\frac{15}{\sqrt{441+2\times 0}}$$

Multiply 2 and 0 to get 0.

$$\frac{15}{\sqrt{441+0}}$$

Add 441 and 0 to get 441.

$$\frac{15}{\sqrt{441}}$$

Calculate the square root of 441 and get 21.

$$\frac{15}{21}$$

Reduce the fraction \frac{15}{21} to lowest terms by extracting and canceling out 3.

$$\frac{5}{7}$$

Factor

$$\frac{5}{7} = 0.7142857142857143$$