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Question-196635

Question Number 196635 by sonukgindia last updated on 28/Aug/23 Answered by sniper237 last updated on 28/Aug/23 $$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {f}\left({n}\right)\:;\:{f}\left({z}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{z}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\overset{{Residu}\:{theo}\:{applicat\%}} {=}\:{Res}\left(\frac{\pi{f}\left({z}\right)}{{sin}\left(\pi{z}\right)};\:−\mathrm{1}/\mathrm{2}\right)\:\:\: \\…

A-pilot-flies-his-plane-directly-from-a-point-A-to-a-point-B-a-distance-of-450Km-The-bearing-of-B-from-A-is-030-A-wind-of-80Km-hr-is-blowing-from-the-east-Given-that-the-plane-can-travel-at-320Km

Question Number 196657 by Tawa11 last updated on 28/Aug/23 $$\mathrm{A}\:\mathrm{pilot}\:\mathrm{flies}\:\mathrm{his}\:\mathrm{plane}\:\mathrm{directly}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{A}\:\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:\mathrm{B}, \\ $$$$\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{450Km}.\:\mathrm{The}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{B}\:\mathrm{from}\:\mathrm{A}\:\mathrm{is}\:\mathrm{030}°.\:\mathrm{A} \\ $$$$\mathrm{wind}\:\mathrm{of}\:\mathrm{80Km}/\mathrm{hr}\:\mathrm{is}\:\mathrm{blowing}\:\mathrm{from}\:\mathrm{the}\:\mathrm{east}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{plane}\:\mathrm{can}\:\mathrm{travel}\:\mathrm{at}\:\mathrm{320Km}/\mathrm{hr}\:\mathrm{in}\:\mathrm{still}\:\mathrm{air}.\:\mathrm{Find}\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{bearing}\:\mathrm{on}\:\mathrm{which}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{must}\:\mathrm{be}\:\mathrm{steered}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{fly}\:\mathrm{from}\:\mathrm{A}\:\mathrm{to}\:\mathrm{B}. \\ $$ Terms of Service…

find-f-x-if-f-x-x-2-1-x-x-1-

Question Number 196621 by universe last updated on 28/Aug/23 $$\mathrm{find}\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:\mathrm{if} \\ $$$$\:\mathrm{f}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\:=\:\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}} \\ $$ Answered by MM42 last updated on 28/Aug/23 $${x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}={u}\Rightarrow{x}=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}}…