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midpoint
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inverse of (x^2)/(x-2)distance (0,0),(6,8)asymptotes of f(x)=(7x^2+5x-2)/(2x^2-18)y=x^2-2xdomain of f(x)= 5/(1-x^2)
Frequently Asked Questions (FAQ)
What is the midpoint (-1,-2),(1,2) ?
The midpoint (-1,-2),(1,2) is (0,0)