lgbasallote
  • lgbasallote
Lgbariddle!! is \[\huge a^{\frac mn} = \sqrt[n]{a^m} = (\sqrt[n] a)^m\]
Mathematics
katieb
  • katieb
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
yup
anonymous
  • anonymous
Can be proved
lgbasallote
  • lgbasallote
so tell me...what happens when a is negative?

Looking for something else?

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

More answers

lgbasallote
  • lgbasallote
for example \[\huge (-1)^{2/4}\]
lgbasallote
  • lgbasallote
\[\huge \sqrt[4]{-1} \implies \text{imaginary}\] \[\huge \sqrt[4] {(-1)^2} \implies 1\]
anonymous
  • anonymous
you're right, again
lgbasallote
  • lgbasallote
that's where your algebra is now shaken :troll:
lgbasallote
  • lgbasallote
this is the part where i advertise this: http://openstudy.com/updates/4fdaecf3e4b0f2662fd13fd6

Looking for something else?

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