expression you used is misleading and not the same as 
.![\sqrt[n]s](/tpl/images/0498/5709/40ae7.png)
and 
are equivalent in this context:![\dfrac{\sqrt[6]s\sqrt{s^9}}{\sqrt{16s^{18}}}=\dfrac{s^{1/6}s^{9/2}}{\sqrt{16}s^{18/2}}=\dfrac{s^{1/6+9/2}}{4s^9}=\dfrac{s^{14/3}}{4s^9}=\dfrac{s^{-13/3}}4=\dfrac4{s^{13/3}}=\dfrac4{\sqrt[3]{s^{13}}}](/tpl/images/0498/5709/42714.png)
sixth root of s
times
square root of s to the 9th
divided by
square root of 16 t...
Mathematics, 04.02.2020 01:01 kefernanda40
sixth root of s
times
square root of s to the 9th
divided by
square root of 16 times s to the 18th
Answers: 3
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