Question Number 148189 by abdurehime last updated on 25/Jul/21 Commented by abdurehime last updated on 25/Jul/21 $$\mathrm{help}\:\mathrm{me} \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 82654 by jagoll last updated on 23/Feb/20 $$\boldsymbol{\mathrm{I}}\mathrm{f}\:\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{find}\:\mathrm{sin}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{x}\:? \\ $$ Commented by jagoll last updated on 23/Feb/20 Answered by mr W…
Question Number 17119 by gourav~ last updated on 01/Jul/17 Commented by prakash jain last updated on 01/Jul/17 $$\frac{\mathrm{sin}\:\left({A}+\mathrm{3}{B}\right)+\mathrm{sin}\:\left(\mathrm{3}{A}+{B}\right)}{\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}} \\ $$$$=\frac{\mathrm{2sin}\:\left(\frac{\mathrm{4}{A}+\mathrm{4}{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}{B}−\mathrm{2}{A}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\…
Question Number 17117 by tawa tawa last updated on 01/Jul/17 $$\mathrm{A}\:\mathrm{clock}\:\mathrm{has}\:\mathrm{a}\:\mathrm{pendulum}\:\mathrm{made}\:\mathrm{of}\:\mathrm{iron}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}.\mathrm{5m}, \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{clock}\:\mathrm{keeps}\:\mathrm{accurate}\:\mathrm{time}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}.\:\mathrm{By}\:\mathrm{how}\:\mathrm{much}\:\mathrm{time}\:\mathrm{will}\:\mathrm{it}\:\mathrm{be}\:\mathrm{late} \\ $$$$\mathrm{running}\:\mathrm{at}\:\mathrm{a}\:\mathrm{temperature}\:\mathrm{30}°\mathrm{C}\:\mathrm{for}\:\mathrm{1}\:\mathrm{day}.\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of} \\ $$$$\mathrm{iron}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{2}\:×\:\mathrm{10}^{−\mathrm{5}} \mathrm{per}\:\mathrm{k}. \\ $$ Answered by ajfour last updated…
Question Number 82647 by Learner-123 last updated on 23/Feb/20 $${In}\:{the}\:{rectangular}\:{region},\:−\mathrm{2}<{x}<\mathrm{2}, \\ $$$$−\mathrm{3}<{y}<\mathrm{3},\:{the}\:{surface}\:{charge}\:{density} \\ $$$${is}\:{given}\:{as}\:\rho_{{s}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} .\:{If}\:{no}\:{other} \\ $$$${charge}\:{is}\:{present},\:{find}\:{E}\:{at}\:{P}\left(\mathrm{0},\mathrm{0},\mathrm{1}\right). \\ $$ Commented by Learner-123…
Question Number 82644 by Power last updated on 23/Feb/20 Commented by Power last updated on 23/Feb/20 $$\mathrm{sorry}\:\:\mathrm{it}\:\mathrm{worked} \\ $$ Commented by john santu last updated…
Question Number 82638 by jagoll last updated on 23/Feb/20 $${given}\:{x}=\:\mathrm{cos}\:^{\mathrm{3}} {x} \\ $$$${what}\:{is}\:{x}\:? \\ $$ Commented by mr W last updated on 23/Feb/20 $${if}\:{you}\:{want}\:{to}\:{calculate}\:{it}\:{by}\:{hand}\:{by} \\…
Question Number 82639 by TawaTawa last updated on 23/Feb/20 Commented by TawaTawa last updated on 23/Feb/20 $$\mathrm{Please}\:\mathrm{help}. \\ $$ Commented by jagoll last updated on…
Question Number 17102 by tawa tawa last updated on 30/Jun/17 $$\mathrm{compute}:\:\:\:\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2k}\:+\:\mathrm{1}}{\mathrm{2}^{\mathrm{2}\left(\mathrm{k}\:+\:\mathrm{1}\right)} } \\ $$ Commented by prakash jain last updated on 01/Jul/17 $$\mathrm{S}=\:\underset{\mathrm{k}\:=\:\mathrm{0}}…
Question Number 148174 by tabata last updated on 25/Jul/21 $${find}\:{the}\:{residue}\:{of}\:\:{f}\left({z}\right)=\frac{{sin}\left({z}\right)}{{cos}\left({z}^{\mathrm{3}} \right)−\mathrm{1}} \\ $$ Answered by mathmax by abdo last updated on 25/Jul/21 $$\mathrm{cosu}\sim\mathrm{1}−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{cos}\left(\mathrm{z}^{\mathrm{3}} \right)\sim\mathrm{1}−\frac{\mathrm{z}^{\mathrm{6}}…