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Algebra 2 √x = 1/9
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What is the opposite operation of a square root?\[\sqrt{x}^2 = x\]
did you get it?
\[\sqrt{x} = \frac{ 1 }{9} \] \[x = \frac{ 1 }{ 81}\]
The square root of a number \(x\) is a number \(a\) which when multiplied together with itself gives you \(x\)\[\sqrt{x} = a\]\[a*a = x\]So if \[\sqrt{x} = \frac{1}9\]\[x=\frac{1}{9}*\frac{1}{9} = \frac{1}{81}\]
gracias a todos☺
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