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Solve for m √m - 7= n + 3 a)m = n^2 + 6n + 2 b)m = n^2 + 6n - 2 c)m = n^2 + 6n + 16
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add 7 to both sides, then square both sides
So b?
no, that is not what you get when you compute \((n+4)^2\) in fact, you get none of those
^ Thank you so much you actually took time solving this on a notepad just for me! :D
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