anonymous
  • anonymous
how do you do 32y^4z^3 under a sq. root sign?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
32 is 4*4*2 so sqrt 32 is 4*sqrt2 4 is 2*2 so sqrt 4 is 2 sqrt y^4=y^2 so: 2*4*y^2*sqrt(2z^3)
anonymous
  • anonymous
sqrt(2z^3) with this part you cannot do anything
anonymous
  • anonymous
oh but you can

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anonymous
  • anonymous
my guess is you are supposed to write this in "simplest radical form" which means you cannot have a cube under a square root.
anonymous
  • anonymous
\[\sqrt{32y^4z^3}=\sqrt{4^2\times 2 \times y^2 \times y^2 \times z^2 \times z}\] \[=\sqrt{4^2}\times \sqrt{y^2}\times \sqrt{y^2}\times \sqrt{z^2} \times\sqrt{2z}\] \[=4y^2z\sqrt{2z}\]
anonymous
  • anonymous
thanks, still confused a lot but thanks
anonymous
  • anonymous
are you still confused?

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