anonymous
  • anonymous
can someone help me with this equation: 3x^(3/2)(9 sqrt of x - 8/ sq rt of x)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sqrt{3}^{3}*(9*\sqrt{x}-8/\sqrt{x})\]
anonymous
  • anonymous
i'm sorry, i'm a little confused. is that the answer or are you asking me that's what it looks like?
anonymous
  • anonymous
3x^(3/2)(9 sqrt of x - 8/ sq rt of x)? is the prob like this? 3x^(3/2) * (9 sqrt of x - 8) 3x^(3/2) * (9 sqrt of x - 8) = -------------------------= ---------------------- ( sq rt of x) x^1/2 3(9)x^(3/2) * ( x - 8)^1/2 = -------------------- x^1/2 = 27x * ( x - 8)^1/2

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anonymous
  • anonymous
the 3x raised to the 3/2 is one part while 9 sqrt of x -(8/ sq rt of x) is all in one part.
anonymous
  • anonymous
I was trying to get rid of the radical beneath 8 and ended up with 8 sqrt of x all over x
anonymous
  • anonymous
look at my solution that 9 and 3 was multiplied and hence 9(3) (x^3/2 )(x^-1/2)=27x
anonymous
  • anonymous
oh! do you also multiply with the 8?
anonymous
  • anonymous
my solution was based on assumtion that 9 sqrt of x - 8 that you mean it is 9 (x-8)^1/2
anonymous
  • anonymous
no, the 8 is the numerator over the denominator sq rt of x
anonymous
  • anonymous
yes i know that,,, the denominator was taken care of the numerator x^3/2
anonymous
  • anonymous
3(9)(x^3/2 )(x^-1/2)=27x
anonymous
  • anonymous
that one is separate from the 9 multiplied by the sq rt of x. there are two square roots of x in this equation. one multiplied by the 9 and the one divided by the 8 which is why I assumed you multiplied the 3 with the 8 as well?
anonymous
  • anonymous
to get 24x right?
anonymous
  • anonymous
its 3(9)=27
anonymous
  • anonymous
kk i think i got it. sorry about the confusion. and thanks for your help.
anonymous
  • anonymous
3(9)x^(3/2) ------------- = 27 (x^3/2)(x^-1/2) =27x^2/2 = 27x x^1/2 then you multipled that to sqrt(x-8)
anonymous
  • anonymous
kk. it just looks really confusing when first looked at. thanks
anonymous
  • anonymous
ok yw,,,,, good luck now...

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