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anonymous

  • 4 years ago

x^2+y^2=20 y=x^2 A. {(4, -2), (4, 2) B. {(-2, 4), (2, 4) C.(2, 4) D.(3, 9)

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  1. anonymous
    • 4 years ago
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    X^2+Y^2=20

  2. anonymous
    • 4 years ago
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    Y=X^2

  3. anonymous
    • 4 years ago
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    i think for this one easiest to check answer and see which ones work

  4. anonymous
    • 4 years ago
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    i DONT KNOW HOW TO DO IT

  5. anonymous
    • 4 years ago
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    \( x^2+y^2=20 \) and \( y=x^2 \) \( \implies x^4+x^2=20 \)

  6. perl
    • 4 years ago
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    y^2 + y - 20 = 0 after substitution , (y+5)(y-4) = 0

  7. anonymous
    • 4 years ago
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    \[ (-2+x) (2+x) \left(5+x^2\right) =0 \]

  8. lgbasallote
    • 4 years ago
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    y + y^2 = 20 y^2 + y + 1/4= 20 +1/4 (y+1/2)^2 = 81/4 y + 1/2 = 9/2 y = 4 knowing that... 4 = x^2 x = +/- 2 so it's {(-2,4)(2,4)

  9. anonymous
    • 4 years ago
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    for example, try (2,4) \[2^2=4\] so second equation is satisfied, but \[2^2+4^2=4+16=20\] so first equation is satisfied as well

  10. anonymous
    • 4 years ago
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    Yes substitution is the best choice here.

  11. anonymous
    • 4 years ago
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    now try \[(-2)^2=4\] true and \[(-2)^2+4^2=20\] also true

  12. perl
    • 4 years ago
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    { (sqrt 5, 5) ( -sqrt 5 , 5 ) ( 2, 4) (-2, 4) }

  13. anonymous
    • 4 years ago
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    Analytically it would be messy.

  14. anonymous
    • 4 years ago
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    (3,9) does not work because although \[3^2=9\] is true \[3^2+9^2\neq 20\] so that one is out

  15. perl
    • 4 years ago
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    the 5's dont work, scratch that

  16. anonymous
    • 4 years ago
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    and (4,-2) does not work because \[4^2\neq -2\]

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