Answers: 1
Mathematics, 20.06.2019 18:04, Uc34758
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?
Answers: 2
Mathematics, 21.06.2019 22:00, Jasten
Set $r$ is a set of rectangles such that (1) only the grid points shown here are used as vertices, (2) all sides are vertical or horizontal and (3) no two rectangles in the set are congruent. if $r$ contains the maximum possible number of rectangles given these conditions, what fraction of the rectangles in set $r$ are squares? express your answer as a common fraction.
Answers: 1
Use order of operations to check your 5 = 28 74...
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