Question Number 87089 by TawaTawa1 last updated on 02/Apr/20 Commented by TawaTawa1 last updated on 02/Apr/20 $$\mathrm{That}\:\mathrm{is}\:\:\mathrm{4}\:\mathrm{in} \\ $$ Commented by MJS last updated on…
Question Number 87086 by Chi Mes Try last updated on 02/Apr/20 Answered by mind is power last updated on 03/Apr/20 $$=\int_{\mathrm{0}} ^{+\infty} \mathrm{2}\frac{{dt}}{\left({t}^{\mathrm{4}} +\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 152620 by Tawa11 last updated on 30/Aug/21 Commented by Tawa11 last updated on 30/Aug/21 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 152617 by eman_64 last updated on 30/Aug/21 Commented by mr W last updated on 30/Aug/21 $${L}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−{x}\right)}{−\mathrm{4cos}\:{x}}=\frac{\mathrm{1}}{\mathrm{4}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$ Terms of Service Privacy…
Question Number 21545 by tawa tawa last updated on 27/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21541 by tawa tawa last updated on 26/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87074 by jagoll last updated on 02/Apr/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{t}^{\mathrm{3}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{expanssion}\:\left\{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{6}} }{\mathrm{1}−\mathrm{t}}\right\}^{\mathrm{3}} \: \\ $$ Commented by mr W last updated on 02/Apr/20 $${t}^{\mathrm{2}}…
Question Number 152608 by ZiYangLee last updated on 30/Aug/21 $$\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{substitution}\:{x}=\mathrm{cos}\:\mathrm{2}\theta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx}\:=\:−\mathrm{sin}\:\mathrm{2}\theta−\mathrm{2}\theta+{C} \\ $$ Answered by Olaf_Thorendsen last updated on 30/Aug/21 $$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\mathrm{F}\left(\theta\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+\mathrm{cos2}\theta}{\mathrm{1}−\mathrm{cos2}\theta}}\:\left(−\mathrm{2sin2}\theta{d}\theta\right) \\…
Question Number 21535 by Tinkutara last updated on 26/Sep/17 $$\mathrm{Which}\:\mathrm{forces}\:\mathrm{of}\:\mathrm{attraction}\:\mathrm{are}\:\mathrm{responsible} \\ $$$$\mathrm{for}\:\mathrm{liquefaction}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} ? \\ $$$$\left({a}\right)\:\mathrm{Coulombic}\:\mathrm{forces} \\ $$$$\left({b}\right)\:\mathrm{Dipole}\:\mathrm{forces} \\ $$$$\left({c}\right)\:\mathrm{Hydrogen}\:\mathrm{bonding} \\ $$$$\left({d}\right)\:\mathrm{Van}\:\mathrm{der}\:\mathrm{Waal}'\mathrm{s}\:\mathrm{forces}. \\ $$ Terms of…
Question Number 87069 by jagoll last updated on 02/Apr/20 $$\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}}\:=\:\mathrm{sec}\:\mathrm{2x}−\mathrm{tan}\:\mathrm{2x} \\ $$$$\mathrm{prove}\:\mathrm{it}\: \\ $$ Commented by jagoll last updated on 02/Apr/20 $$\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}\:=\:\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:…