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dtan5457

  • one year ago

Suppose h(2z+3)=(5-2z)/(4+z) find h(-9) find h^-1(4) find h(z)

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  1. dtan5457
    • one year ago
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    Anyone know what type of math this called? I never seen this

  2. dtan5457
    • one year ago
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    @DanJS @dan815 @Loser66

  3. FireKat97
    • one year ago
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    do you by any chance have answers to these?

  4. dtan5457
    • one year ago
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    no

  5. Loser66
    • one year ago
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    I don't know whether it works or not, just give it a try; Let t = 2z +3, then \(z = \dfrac{t-3}{2}\) So \(h(2z+3) = h (t) =\dfrac{5-(t-3)}{4+\dfrac{t-3}{2}}= \dfrac{16-2t}{5-t} \)

  6. Loser66
    • one year ago
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    So as h(z) , that is \(h(z) = \dfrac{16-2z}{5-z}\)

  7. Loser66
    • one year ago
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    From it, we can find h(-9), \(h^{-1} (4)\) is just let it =4, solve for z

  8. anonymous
    • one year ago
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    If you're looking for more information on this type of problem, it's called a functional equation. It may or may not be a part of functional analysis, I'm not sure.

  9. dtan5457
    • one year ago
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    its alright, i figured these 3 out.

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