anonymous
  • anonymous
Algebra II question drawing below
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1354492999386:dw|
anonymous
  • anonymous
Is this x^(1/12)
anonymous
  • anonymous
cross multiply

Looking for something else?

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

More answers

anonymous
  • anonymous
\[{3\over4}x\times{1\over9}x\]multiply straight across (3 times 1, 4 times 9, x times x):\[{3\over36}x^2\]simplify:\[{1\over12}x^2\]this can also be written as:\[x^2\over12\]:) hope this helps :)
anonymous
  • anonymous
|dw:1354493673382:dw| that's what I meant :)
anonymous
  • anonymous
sry if it looked wrong on the 1st drawing :). pretty sure you just multiply the fractions and it becomes x^1/12 right?
anonymous
  • anonymous
oh im sorry :O okay so you have to add the exponents :) \[\huge x^{3\over4}\times x^{1\over9}\]so find the common denominator:\[\huge x^{27\over36}\times x^{4\over36}\]now add:\[\huge x^{31\over36}\] when multiplying you add the exponents :)
anonymous
  • anonymous
oh okay thx :) was just reviewing something I hadn't gone over in a long time so was making sure if i was doing it right or not :)
anonymous
  • anonymous
no problem Lou :)

Looking for something else?

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