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midpoint
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inverse of f(x)= 3/4 x^2+1range of f(x)= 1/(1+sqrt(x^2-1))asymptotes of f(x)=(x^2+5x-14)/(x^2-4)range of f(x)=sqrt(x+3)-2inverse of (x+1)^2
Frequently Asked Questions (FAQ)
What is the midpoint (5,-9),(9,10) ?
The midpoint (5,-9),(9,10) is (7, 1/2)