Menu Close

Category: None

Question-194560

Question Number 194560 by Denis last updated on 09/Jul/23 Answered by MM42 last updated on 10/Jul/23 $$\Delta{XAB}\:\:{is}\:\:{equilatral}\:{triangle} \\ $$$$\angle\:{XAD}=\angle{XBC}=\mathrm{30}^{\mathrm{0}} \: \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}}…

Gyanashram-classes-shivkund-Munger-by-Bittu-sir-12-th-physics-test-1-2-

Question Number 194506 by Biteshwar last updated on 08/Jul/23 $$\:\:\:\:\:\:\:\:\:\:\mathrm{Gyanashram}\:\mathrm{classes} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{shivkund}\left(\mathrm{Munger}\right)\:\:\:\:\mathrm{by}−\mathrm{Bittu}\:\mathrm{sir} \\ $$$$\:\:\:\mathrm{12}\:\:\mathrm{th}\:\mathrm{physics}\:\mathrm{test} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}. \: \: \: \: \: \: \: \: \\…

Question-194436

Question Number 194436 by sonukgindia last updated on 06/Jul/23 Answered by Frix last updated on 06/Jul/23 $$\mathrm{3}\:?\:\sqrt{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\mathrm{3}−\sqrt{\mathrm{3}}\:?\:\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\mathrm{54}−\mathrm{30}\sqrt{\mathrm{3}}\:?\:\mathrm{2} \\ $$$$\mathrm{52}\:?\:\mathrm{30}\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{26}}{\mathrm{15}}\:?\:\sqrt{\mathrm{3}}…

calcul-e-2ln-1-u-e-2ln-1-u-

Question Number 194434 by SANOGO last updated on 06/Jul/23 $${calcul} \\ $$$${e}^{\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)\:} −{e}^{−\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)} \:=? \\ $$ Answered by aba last updated on 06/Jul/23 $$\mathrm{e}^{\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} −\mathrm{e}^{−\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)}…

Question-194363

Question Number 194363 by sonukgindia last updated on 05/Jul/23 Commented by Frix last updated on 05/Jul/23 $$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}. \\ $$$$\sqrt{{z}_{\mathrm{1}} +\sqrt{{z}_{\mathrm{2}} }}+\sqrt{{z}_{\mathrm{1}} −\sqrt{{z}_{\mathrm{2}} }}={r} \\ $$$$\mathrm{Solving}\:\mathrm{by}\:\mathrm{2}\:\mathrm{times}\:\left[\mathrm{squaring}\:\left(\mathrm{introduces}\right.\right.…