anonymous
  • anonymous
What is sqrt(38)/2sqrt(6)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
to rationalize the denominator, you need to multiply top and bottom by sqrt(6)
anonymous
  • anonymous
???????????????????????????????????????????

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jim_thompson5910
  • jim_thompson5910
where are you stuck
jim_thompson5910
  • jim_thompson5910
no offense, putting up a million question marks doesn't help at all
anonymous
  • anonymous
srry. I was taught that you don't do anything with the sqrts ?
jim_thompson5910
  • jim_thompson5910
the instructions are to rationalize the denominator no?
anonymous
  • anonymous
it says "What is the exact value of..."
jim_thompson5910
  • jim_thompson5910
and the full expression given is \[\large \frac{\sqrt{38}}{2\sqrt{6}}\] right?
anonymous
  • anonymous
yesh
anonymous
  • anonymous
7.54983443527
jim_thompson5910
  • jim_thompson5910
ok you multiply top and bottom by sqrt(6) like this \[\large \frac{\sqrt{38}}{2\sqrt{6}}\] \[\large \frac{\sqrt{38}*\sqrt{6}}{2\sqrt{6}*\sqrt{6}}\] \[\large \frac{\sqrt{38*6}}{2\sqrt{6*6}}\] \[\large \frac{\sqrt{228}}{2\sqrt{36}}\] \[\large \frac{\sqrt{228}}{2*6}\] \[\large \frac{\sqrt{228}}{12}\] \[\large \frac{\sqrt{4*57}}{12}\] \[\large \frac{\sqrt{4}*\sqrt{57}}{12}\] \[\large \frac{2*\sqrt{57}}{12}\] \[\large \frac{\sqrt{57}}{6}\]
anonymous
  • anonymous
^
anonymous
  • anonymous
i think i did something wrong no ?
anonymous
  • anonymous
ok so that is it? Wow...
anonymous
  • anonymous
THANKYOU!!!!!!!!!!!!!!!
anonymous
  • anonymous
no problem for nothing lol
jim_thompson5910
  • jim_thompson5910
DUDE..IM..A..DUCK you provided the approximate answer they want the exact answer (in terms of radicals)

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