anonymous
  • anonymous
Can someone please Help me with college algebra 2 ?
Mathematics
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anonymous
  • anonymous
Can someone please Help me with college algebra 2 ?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ok
anonymous
  • anonymous
\[\sqrt[4]{32y^13}\]
anonymous
  • anonymous
32 = 2 x 2 x 2 x 2

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anonymous
  • anonymous
i left out one 2
anonymous
  • anonymous
its y raised to the 13th
anonymous
  • anonymous
im guessing im factoring it?
anonymous
  • anonymous
are you trying to solve for y?
anonymous
  • anonymous
no i think you are trying to simplify the expression
anonymous
  • anonymous
i was using prime factorization
anonymous
  • anonymous
right simplify
anonymous
  • anonymous
so you have 32 factored to 2^5, so take away 16 from inside the radical and leave in one 2 as for y, you have y^3 outside the radical and one y inside, since (y^3)^4 * y = y^13
anonymous
  • anonymous
im pretty sure it is 2y^3*radical4(2y)
anonymous
  • anonymous
so its 32^(1/4)*y^(13/4)*3^(1/4)
anonymous
  • anonymous
its a 4 outside the radical and 32y^13 on the inside
anonymous
  • anonymous
Factor completely: 6b^2+4a-8b-3ab
anonymous
  • anonymous
wait, ok your previous expression was messed up, can you re type it?

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