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What is the inverse of y = 16x^2 + 1
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What is the inverse of y = 16x^2 + 1

User Shaboboo
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x = 16y^2 + 1
x - 1 = 16y^2
(x - 1)/16 = y^ 2
( (x-1) / 16 ) ^1/2 = y
( (x-1) ^ 1/2 ) / 4 = y
User Ckg
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