Solve {left(frac{5sqrt{41}}{41}right)}^2

$${ \left( \frac{ 5 \sqrt{ 41 } }{ 41 } \right) }^{ 2 }$$

Evaluate

$$\frac{25}{41}\approx 0.609756098$$

Solution Steps

To raise \frac{5\sqrt{41}}{41} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(5\sqrt{41}\right)^{2}}{41^{2}}$$

Expand \left(5\sqrt{41}\right)^{2}.

$$\frac{5^{2}\left(\sqrt{41}\right)^{2}}{41^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{25\left(\sqrt{41}\right)^{2}}{41^{2}}$$

The square of \sqrt{41} is 41.

$$\frac{25\times 41}{41^{2}}$$

Multiply 25 and 41 to get 1025.

$$\frac{1025}{41^{2}}$$

Calculate 41 to the power of 2 and get 1681.

$$\frac{1025}{1681}$$

Reduce the fraction \frac{1025}{1681} to lowest terms by extracting and canceling out 41.

$$\frac{25}{41}$$

Factor

$$\frac{5 ^ {2}}{41} = 0.6097560975609756$$