Solve (3/4)^2*8/9+frac{sqrt{1}}{144}

$$( \frac { 3 } { 4 } ) ^ { 2 } \cdot \frac { 8 } { 9 } + \frac { \sqrt { 1 } } { 144 }$$

Evaluate

$$\frac{73}{144}\approx 0.506944444$$

Solution Steps

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{9}{16}\times \frac{8}{9}+\frac{\sqrt{1}}{144}$$

Multiply \frac{9}{16} times \frac{8}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 8}{16\times 9}+\frac{\sqrt{1}}{144}$$

Cancel out 9 in both numerator and denominator.

$$\frac{8}{16}+\frac{\sqrt{1}}{144}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$\frac{1}{2}+\frac{\sqrt{1}}{144}$$

Calculate the square root of 1 and get 1.

$$\frac{1}{2}+\frac{1}{144}$$

Least common multiple of 2 and 144 is 144. Convert \frac{1}{2} and \frac{1}{144} to fractions with denominator 144.

$$\frac{72}{144}+\frac{1}{144}$$

Since \frac{72}{144} and \frac{1}{144} have the same denominator, add them by adding their numerators.

$$\frac{72+1}{144}$$

Add 72 and 1 to get 73.

$$\frac{73}{144}$$

Factor

$$\frac{73}{2 ^ {4} \cdot 3 ^ {2}} = 0.5069444444444444$$