anonymous
  • anonymous
Simplify 6 4*√4 - 7 *3√4 + 2 *3√4 - 3 *3√4.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
64*√4 - 7 *3√4 + 2 *3√4 - 3 *3√4. or 6*4*√4 - 7 *3√4 + 2 *3√4 - 3 *3√4 ?
anonymous
  • anonymous
lol yes i find this confusin also lol
anonymous
  • anonymous
the 6 and the 4 are separate. it's 6 4*√4 - 7 *3√4 + 2 *3√4 - 3 *3√4. or 6*4*√4 - 7 *3√4 + 2 *3√4 - 3 *3√4 by the asteric i mean that the number before it is a small raised number.

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anonymous
  • anonymous
so the numers asterick are cube/4th roots yada yada?
anonymous
  • anonymous
yup.
anonymous
  • anonymous
so in other words it is this \[6\sqrt[4]{4}\]
anonymous
  • anonymous
as the first part?
anonymous
  • anonymous
exactly!
anonymous
  • anonymous
alright i'll equate it out for nick so he can make sure i'm sure he's got hte right answer \[6\sqrt[4]{4}-8\sqrt[3]{4}+2\sqrt[3]{4}-3\sqrt[3]{4}\]
anonymous
  • anonymous
6*4√4 - 7 *3√4 + 2 *3√4 - 3 *3√4 3√4(6*√4-7+2-3) 3√4(6*√4-8)
anonymous
  • anonymous
is that correct?
anonymous
  • anonymous
yes. (:
anonymous
  • anonymous
nick's answer is right or the way i showed it/
anonymous
  • anonymous
out u wrote 8 instead of 7
anonymous
  • anonymous
and 95 girl i am right i just simplified it
anonymous
  • anonymous
i took 3√4 common
anonymous
  • anonymous
out u do it too y will get the same one wht i got
anonymous
  • anonymous
it's correct it just looks so weird when you use a check to show sqroot i see it as all under the root
anonymous
  • anonymous
its correct so i deserve a medal too bro
anonymous
  • anonymous
lol

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