Solve sqrt{(6/7)^2+(-2/7)^2+(3/7)^2}

$$\sqrt { ( \frac { 6 } { 7 } ) ^ { 2 } + ( - \frac { 2 } { 7 } ) ^ { 2 } + ( \frac { 3 } { 7 } ) ^ { 2 } }$$

Evaluate

$$1$$

Solution Steps

Calculate \frac{6}{7} to the power of 2 and get \frac{36}{49}.

$$\sqrt{\frac{36}{49}+\left(-\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}}$$

Calculate -\frac{2}{7} to the power of 2 and get \frac{4}{49}.

$$\sqrt{\frac{36}{49}+\frac{4}{49}+\left(\frac{3}{7}\right)^{2}}$$

Since \frac{36}{49} and \frac{4}{49} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{36+4}{49}+\left(\frac{3}{7}\right)^{2}}$$

Add 36 and 4 to get 40.

$$\sqrt{\frac{40}{49}+\left(\frac{3}{7}\right)^{2}}$$

Calculate \frac{3}{7} to the power of 2 and get \frac{9}{49}.

$$\sqrt{\frac{40}{49}+\frac{9}{49}}$$

Since \frac{40}{49} and \frac{9}{49} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{40+9}{49}}$$

Add 40 and 9 to get 49.

$$\sqrt{\frac{49}{49}}$$

Divide 49 by 49 to get 1.

$$\sqrt{1}$$

Calculate the square root of 1 and get 1.

$$1$$

Factor

$$1$$