anonymous
  • anonymous
JUST NEED CHECKED!! Suppose that x* means 5x^2 - x then what does (y+3)* mean? Do you plug in (y+3) for x, expand the squared, and the (y+3)'s cross out to 5(y+3)? (y+3)* = 5y+15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Hero
  • Hero
Interesting...
Hero
  • Hero
x* = 5x^2 - x y* = y+3 just a guess
Mertsj
  • Mertsj
My thought would be that x* is like a function. f(x) = 5x^2-x x* = 5x^2-x (y+3)* = 5(y+3)^2-(y+3)

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anonymous
  • anonymous
I think so too. But wouldn't you expand it after that? 5(y+3)(y+3)-(y+3) Then the (y+3)s cross out leaving just 5(y+3)?
Hero
  • Hero
Yeah, that probably is right. the y variable threw me off
Mertsj
  • Mertsj
Yes. I think you should go ahead and square and then collect like terms.
anonymous
  • anonymous
Okay, I think my final answer is 5y+15, since (y+3) are like terms.
Mertsj
  • Mertsj
\[5(y+3)^{2}-(y+3)=5(y^2+6y+9)-y-3=5y^2+30y+45-y-3\]
Mertsj
  • Mertsj
\[5y^2+29y+42\]
anonymous
  • anonymous
So the last (y+3) doesn't cross out anything?
anonymous
  • anonymous
Doesn't take a y out of the squared term?
Mertsj
  • Mertsj
Remember the "powers first" rule in order of operations.
Mertsj
  • Mertsj
This could be the approach you favor:
Mertsj
  • Mertsj
\[5(y+3)^2-(y+3)=(y+3)[5(y+3)-1]=(y+3)[5y+15-1]=\]
Mertsj
  • Mertsj
\[(y+3)(5y+14)=5y^2+14y+15y+42=5y^2+29y+42\]
anonymous
  • anonymous
Thank you so much! You really helped. :)
Mertsj
  • Mertsj
yw

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