Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Loujoelou

  • 3 years ago

Algebra II question drawing below

  • This Question is Closed
  1. Loujoelou
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354492999386:dw|

  2. Loujoelou
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this x^(1/12)

  3. bay
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cross multiply

  4. yummydum
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[{3\over4}x\times{1\over9}x\]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 :)

  5. Loujoelou
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354493673382:dw| that's what I meant :)

  6. Loujoelou
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sry if it looked wrong on the 1st drawing :). pretty sure you just multiply the fractions and it becomes x^1/12 right?

  7. yummydum
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 :)

  8. Loujoelou
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 :)

  9. yummydum
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no problem Lou :)

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy