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Mathematics, 26.06.2020 15:01, andrew412603
In this problem, you will prove that division with remainder is welldefined. That is, you will prove that for any two integers a > 0 and b, there exists a unique remainder after the division of b by a. We will do the proof in two parts: first by proving that there is at most one remainder, and then by proving that there exists at least one remainder. The conclusion will be that there exists a unique remainder. Assume a and b are both integers and a > 0. Define a remainder after the division of b by a to be a value r such that r ≥ 0, r < a, and there exists an integer q for which b = aq + r
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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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In this problem, you will prove that division with remainder is welldefined. That is, you will prove...
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