Here are the coordinates for eight points on the coordinate plane (q > p > 0): a (0, 0), b (p, 0), c (q, 0), d (p + q, 0), e (0, q), f (p, q), g (q, q), and h (p + q, q). which four points, if any, are on the vertices of a square?

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

In which figure is de bc ? a. figure 1 b. figure 2 c. figure 3 d. figure 4

Let f(x)=2x and g(x)=2x. graph the functions on the same coordinate plane. what are the solutions to the equation f(x)=g(x) ?me asap