anonymous
  • anonymous
Simplify the expression four root of order three of x times x to the power of start fraction three over two end fraction end power times y to the power of six end power times y to the power of start fraction one over two end fraction end power and rewrite it in radical form.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
misty1212
  • misty1212
\[\huge \sqrt[4]{x^3}\times x^{\frac{3}{2}}\]?
misty1212
  • misty1212
and \[\huge y^6\times y^{\frac{1}{2}}\]?
misty1212
  • misty1212
\[y^6\times y^{\frac{1}{2}}=y^{6+\frac{1}{2}}=y^{\frac{13}{2}}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[4\sqrt[3]{x}*x^\frac{ 3 }{ 2 }*y^6*y ^\frac{ 1 }{ 2 }\]
misty1212
  • misty1212
\[4x^{\frac{1}{3}+\frac{3}{2}}=4x^{\frac{11}{6}}\]
misty1212
  • misty1212
so \[4\sqrt[6]{x^{11}}\times \sqrt{y^{13}}\]
anonymous
  • anonymous
i got \[4\sqrt[11]{x^6} \sqrt[13]{y^2}\]
anonymous
  • anonymous
is that correct?

Looking for something else?

Not the answer you are looking for? Search for more explanations.