Using the quotient property we would have
[tex]\begin{gathered} \sqrt{\frac{75r^9}{8^8}}=\frac{\sqrt{75r^9}}{\sqrt{8^8}} \\ \end{gathered}[/tex]
This follows the rule
[tex]\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/tex]
From the simplified expression as shown above
[tex]\frac{\sqrt{75r^9}}{\sqrt{8^8}}=\frac{\sqrt{3\times25\times r^9}}{\sqrt{(2^3)^8}}[/tex]
Thus;
[tex]\frac{\sqrt{3\times25r^9}}{\sqrt{(2^3)^8}}=\frac{\sqrt{25}\times\sqrt{3}\times\sqrt{r^9}}{\sqrt{2^{^3\times8}}}=\frac{5\sqrt{3r^9}}{\sqrt{2^{24}}}[/tex]
Therefore
[tex]\begin{gathered} Using\text{ fraction index law we could simplify the denominator} \\ \frac{5\sqrt{3r^9}}{2^{\frac{24}{2}}}=\frac{5\sqrt{3r^9}}{2^{12}} \end{gathered}[/tex]
We can not simplify the 3 and the r raised to power of 9 as their power is not even, hence the final answer is given below
[tex]\frac{5\sqrt{3r^9}}{2^{12}}=\frac{5\sqrt{3r^9}}{4096}[/tex]
The final answer is :
[tex]\frac{5\sqrt{3r^9}}{4096}[/tex]
