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Mathematics, 04.03.2021 01:00 Nicolas9979

What the Solution for the equation above


What the Solution for the equation above

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Mathematics, 21.06.2019 16:30, angeline310
Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane
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Mathematics, 22.06.2019 01:10, abadir2008
Given: ae ≅ ce ; de ≅ be prove: abcd is a parallelogram. we have that ab || dc. by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by. so, ∠cad ≅ ∠ by cpctc. therefore, ad || bc by the converse of the theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.
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