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anonymous

  • 5 years ago

how would you solve (sqrt(x+2)) + (sqrt(3-x))

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  1. anonymous
    • 5 years ago
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    \[\sqrt{x+2}+\sqrt{3-x}=0 ?\]

  2. anonymous
    • 5 years ago
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    \[\sqrt{a} ~~~\,a \ge0\]

  3. anonymous
    • 5 years ago
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    \[x+2\ge0 \,and \,~3-x \ge0\]

  4. anonymous
    • 5 years ago
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    f(x) = \[\sqrt{x+2}\] ; g(x) = \[\sqrt{3-x}\] find (f+g)(x) how would I do that?

  5. anonymous
    • 5 years ago
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    \[x \ge2~\, and~\,x \le3\]

  6. anonymous
    • 5 years ago
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    ohh ok

  7. anonymous
    • 5 years ago
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    (f+g)(x)=f(x)+g(x) \[(f+g)(x)=\sqrt{x+2}+\sqrt{3-x}\] this is the answer you cant do anything

  8. anonymous
    • 5 years ago
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    what if it was (f * g) ? could you do something then?

  9. anonymous
    • 5 years ago
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    since inside of the roots are different you cannot make addition

  10. anonymous
    • 5 years ago
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    is it multiply sign?

  11. anonymous
    • 5 years ago
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    yeah

  12. anonymous
    • 5 years ago
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    if it was (f*g)(x), you can multiply the expression inside the radical signs since they have common indices.

  13. anonymous
    • 5 years ago
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    (f*g)(x)=f(x)*g(x) \[(f*g)(x)=(\sqrt{x+2})(\sqrt{3-x})\] \[\sqrt{(x+2)(3-x)}\]\[\sqrt{-x^2+x-6}\]

  14. anonymous
    • 5 years ago
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    it should be +6 not -6

  15. anonymous
    • 5 years ago
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    o okay that makes sense thank you

  16. anonymous
    • 5 years ago
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    yw

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