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anonymous

  • 4 years ago

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

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  1. Hero
    • 4 years ago
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    Interesting...

  2. Hero
    • 4 years ago
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    x* = 5x^2 - x y* = y+3 just a guess

  3. Mertsj
    • 4 years ago
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    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)

  4. anonymous
    • 4 years ago
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    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)?

  5. Hero
    • 4 years ago
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    Yeah, that probably is right. the y variable threw me off

  6. Mertsj
    • 4 years ago
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    Yes. I think you should go ahead and square and then collect like terms.

  7. anonymous
    • 4 years ago
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    Okay, I think my final answer is 5y+15, since (y+3) are like terms.

  8. Mertsj
    • 4 years ago
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    \[5(y+3)^{2}-(y+3)=5(y^2+6y+9)-y-3=5y^2+30y+45-y-3\]

  9. Mertsj
    • 4 years ago
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    \[5y^2+29y+42\]

  10. anonymous
    • 4 years ago
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    So the last (y+3) doesn't cross out anything?

  11. anonymous
    • 4 years ago
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    Doesn't take a y out of the squared term?

  12. Mertsj
    • 4 years ago
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    Remember the "powers first" rule in order of operations.

  13. Mertsj
    • 4 years ago
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    This could be the approach you favor:

  14. Mertsj
    • 4 years ago
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    \[5(y+3)^2-(y+3)=(y+3)[5(y+3)-1]=(y+3)[5y+15-1]=\]

  15. Mertsj
    • 4 years ago
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    \[(y+3)(5y+14)=5y^2+14y+15y+42=5y^2+29y+42\]

  16. anonymous
    • 4 years ago
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    Thank you so much! You really helped. :)

  17. Mertsj
    • 4 years ago
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    yw

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