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anisah.21a

  • one year ago

Find k(a^2+6) given k(x)=sqrt(x-6)

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  1. anonymous
    • one year ago
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    ok

  2. jdoe0001
    • one year ago
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    \(\bf k(a^2+6)?\)

  3. anisah.21a
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    |dw:1444175009862:dw|

  5. anonymous
    • one year ago
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    do you see where this is going

  6. anonymous
    • one year ago
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    sorry wait

  7. anisah.21a
    • one year ago
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    sort of but not really?

  8. anonymous
    • one year ago
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    |dw:1444175116780:dw|

  9. anisah.21a
    • one year ago
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    what is that...?

  10. anonymous
    • one year ago
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    a property for a^2 + 6

  11. jdoe0001
    • one year ago
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    \(\bf k({\color{brown}{ x}})=\sqrt{{\color{brown}{ x}}-6}\qquad {\color{brown}{ x}}=a^2+6\qquad thus \\ \quad \\ k({\color{brown}{ a^2+6}})=\sqrt{{\color{brown}{ a^2\cancel{+6}}}\cancel{-6}}\implies k(a^2+6)=?\)

  12. anisah.21a
    • one year ago
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    what the hell is that lol

  13. anisah.21a
    • one year ago
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    you just gave me the what was originally stated

  14. jdoe0001
    • one year ago
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    |dw:1444176190588:dw|

  15. anisah.21a
    • one year ago
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    your answer is wrong when i submitted it

  16. jdoe0001
    • one year ago
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    yes... anything wrong with that? :)

  17. anisah.21a
    • one year ago
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    was*

  18. jdoe0001
    • one year ago
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    well, what did you get for \(k(a^2+6)\) anyway?

  19. anisah.21a
    • one year ago
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    it is supposed to be a sqrt of something

  20. jdoe0001
    • one year ago
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    so it's, so.. what what the above give you then?

  21. anisah.21a
    • one year ago
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    GIVEN K(X)=SQRTX-6 FIND K(a^2+6)

  22. anisah.21a
    • one year ago
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    so you're basically plugging in a^2+6 into x in sqrt(x-6)

  23. anisah.21a
    • one year ago
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    i am not sure how to go further than that

  24. jdoe0001
    • one year ago
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    yeap, that's all that's being asked, yes

  25. jdoe0001
    • one year ago
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    \(\bf k({\color{brown}{ x}})=\sqrt{{\color{brown}{ x}}-6}\qquad {\color{brown}{ x}}=a^2+6\qquad thus \\ \quad \\ k({\color{brown}{ a^2+6}})=\sqrt{{\color{brown}{ a^2\cancel{+6}}}\cancel{-6}}\implies k(a^2+6)=?\) looks simple enough

  26. anisah.21a
    • one year ago
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    well it is wrong when i submitted it and i have 99 attempts thanks though..

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