Mathematics
Mathematics, 27.05.2021 20:20, laylay120

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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?
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Mathematics, 20.06.2019 18:04, qdogisbeast7662
You have two cards with a sum of -12 in both hands. a. what two cards could you have? b. you add two more cards to your hand, but the total sum if the cards remains the same, (-12).
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Mathematics, 21.06.2019 15:30, priceisright11401
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Mathematics, 21.06.2019 19:30, youcandoit13
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