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Given-that-e-i-npi-n-N-show-that-1-2-1-2i-sin-2-1-n-2-n-cos-n-1-2-e-1-2-in-

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Question Number 124595 by physicstutes last updated on 04/Dec/20 $$\mathrm{Given}\:\mathrm{that}\:\:\omega\:=\:{e}^{{i}\theta} ,\:\theta\neq\:{n}\pi,\:{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\left(\mathrm{1}\right)\:\frac{\omega^{\mathrm{2}} −\mathrm{1}}{\omega}\:=\:\mathrm{2}{i}\:\mathrm{sin}\:\theta \\ $$$$\:\left(\mathrm{2}\right)\:\left(\mathrm{1}\:+\:\omega\right)^{{n}} \:=\:\mathrm{2}^{{n}} \mathrm{cos}^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}\left({in}\theta\right)} \\ $$ Answered by…

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Evaluate-A-x-y-2-dxdy-over-the-area-bounded-by-the-ellipse-x-2-a-2-y-2-b-2-1-Anybody-

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Question Number 190052 by Mastermind last updated on 26/Mar/23 $$\mathrm{Evaluate}\:\int\int_{\mathrm{A}} \left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \mathrm{dxdy}\:\mathrm{over}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ellipse}\: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$ \\ $$$$ \\…

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Question-58974

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Question Number 58974 by Tawa1 last updated on 02/May/19 Answered by MJS last updated on 03/May/19 $$\frac{\mathrm{3}.\mathrm{86km}}{{x}\frac{\mathrm{km}}{\mathrm{hr}}}+\frac{\mathrm{180}.\mathrm{2km}}{\mathrm{10}{x}\frac{\mathrm{km}}{\mathrm{hr}}}+\frac{\mathrm{42}.\mathrm{2km}}{\mathrm{5}{x}\frac{\mathrm{km}}{\mathrm{hr}}}=\mathrm{16}.\mathrm{2hr} \\ $$$$\frac{\mathrm{3}.\mathrm{86}}{{x}}\mathrm{hr}+\frac{\mathrm{180}.\mathrm{2}}{\mathrm{10}{x}}\mathrm{hr}+\frac{\mathrm{42}.\mathrm{2}}{\mathrm{5}{x}}\mathrm{hr}=\mathrm{16}.\mathrm{2hr} \\ $$$$\frac{\mathrm{3}.\mathrm{86}}{{x}}+\frac{\mathrm{18}.\mathrm{02}}{{x}}+\frac{\mathrm{8}.\mathrm{44}}{{x}}=\mathrm{16}.\mathrm{2} \\ $$$$\mathrm{3}.\mathrm{86}+\mathrm{18}.\mathrm{02}+\mathrm{8}.\mathrm{44}=\mathrm{16}.\mathrm{2}{x} \\ $$$${x}=\frac{\mathrm{30}.\mathrm{32}}{\mathrm{16}.\mathrm{2}}=\frac{\mathrm{758}}{\mathrm{405}}\approx\mathrm{1}.\mathrm{87}…

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