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Johan14th

  • one year ago

Help with inverse equation h(x)= 2 sqrt. (x+3) x= 2 sqrt. (y+3) x^2= 2 sqrt. (y+3)^2 x^2= 2y+6 -6. -6 x^2 -6= 2y ÷2. ÷2 (x^2 -6)/2= y ?????? The answer should be 1/4 x-3 I cannot get the 1/4 help please!

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  1. anonymous
    • one year ago
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    You need to get rid of the radical sign \(\sqrt{y+3}\), so you must square it ( \((\sqrt{y+3})^2\)) to get \(y+3\). But of course, you must square it on both sides! So in the left side, you'll get \(\frac{ x^2 }{ 4 }\) or \(\frac{ 1 }{ 4 }x^2\).

  2. anonymous
    • one year ago
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    Once you've done that, you can now easily solve for y.

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