Solve (7/15)^-2*(7/5)^2quad(sqrt{1}+1/4)

$$( \frac { 7 } { 15 } ) ^ { - 2 } \cdot ( \frac { 7 } { 5 } ) ^ { 2 } \quad ( \sqrt { 1 } + \frac { 1 } { 4 } )$$

Evaluate

$$\frac{45}{4}=11.25$$

Solution Steps

Calculate \frac{7}{15} to the power of -2 and get \frac{225}{49}.

$$\frac{225}{49}\times \left(\frac{7}{5}\right)^{2}\left(\sqrt{1}+\frac{1}{4}\right)$$

Calculate \frac{7}{5} to the power of 2 and get \frac{49}{25}.

$$\frac{225}{49}\times \frac{49}{25}\left(\sqrt{1}+\frac{1}{4}\right)$$

Multiply \frac{225}{49} and \frac{49}{25} to get 9.

$$9\left(\sqrt{1}+\frac{1}{4}\right)$$

Calculate the square root of 1 and get 1.

$$9\left(1+\frac{1}{4}\right)$$

Add 1 and \frac{1}{4} to get \frac{5}{4}.

$$9\times \frac{5}{4}$$

Multiply 9 and \frac{5}{4} to get \frac{45}{4}.

$$\frac{45}{4}$$

Factor

$$\frac{3 ^ {2} \cdot 5}{2 ^ {2}} = 11\frac{1}{4} = 11.25$$