Here's the question you clicked on:
Loujoelou
Algebra II question drawing below
|dw:1354492999386:dw|
\[{3\over4}x\times{1\over9}x\]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 :)
|dw:1354493673382:dw| that's what I meant :)
sry if it looked wrong on the 1st drawing :). pretty sure you just multiply the fractions and it becomes x^1/12 right?
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 :)
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 :)