Given the function f(x)= 14x^2 + 5x - 1 answer the question if you could change the 14 to any other number, but the rest has to stay the same, what value would it have to be so that there is only 1 real root for the function? show or explain.

This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 () by changing the 14 to 0. That way we'll left with  which only has one real root:

Skew lines(segments) are lines(segments) in 3 dimensions that do not intersect the given segment and are not parallel to it.  ae, bf, eg, and fh are your skew segments.  i hope that : )

The only number less than 8 that can equal 12 would be 6, 36 is the dividend