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midpoint
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line m=-1,(-5,1)inverse of f(x)=x^2+4,x>= 0range of f(x)=((6x^2-5))/((2x^2+6))domain of ((1))/((-10(\frac{(1)){(-5x-6)})+3)}inverse of 3x^2
Frequently Asked Questions (FAQ)
What is the midpoint (1,-1),(5,3) ?
The midpoint (1,-1),(5,3) is (3,1)