A student imagined one number.2 is written to the right side of the number and 14 is added to the obtained number.3 is written to the right side of the obtained number and 52 is added to the newly obtained number. The result of dividing the final number by 60 is the quotient that is for 6 greater than the initial number and the remainder is a two-digit number with both digits the same as the initial number. Find the initial number.

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?

(05.08a)triangle abc is transformed to similar triangle a′b′c′ below: a coordinate plane is shown. triangle abc has vertices a at 2 comma 6, b at 2 comma 4, and c at 4 comma 4. triangle a prime b prime c prime has vertices a prime at 1 comma 3, b prime at 1 comma 2, and c prime at 2 comma 2. what is the scale factor of dilation? 1 over 2 1 over 3 1 over 4 1 over 5

What statement is true? 2x^2-9x+2=-1