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anonymous
 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 x3
I cannot get the 1/4 help please!
anonymous
 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 x3 I cannot get the 1/4 help please!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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 \(\frac{ 1 }{ 4 }x^2\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once you've done that, you can now easily solve for y.
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