Question Number 56824 by Gulay last updated on 24/Mar/19 Commented by Gulay last updated on 24/Mar/19 $$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 56818 by Gulay last updated on 24/Mar/19 Commented by Gulay last updated on 24/Mar/19 $$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on…
Question Number 122350 by shaker last updated on 16/Nov/20 Answered by liberty last updated on 16/Nov/20 $$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{{a}} −{a}^{{x}} }{{a}^{{x}} −{a}^{{a}} }\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ax}^{{a}−\mathrm{1}} −{a}^{{x}} .\mathrm{ln}\:{a}}{{a}^{{x}}…
Question Number 122334 by help last updated on 15/Nov/20 Answered by liberty last updated on 16/Nov/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 122329 by mathocean1 last updated on 15/Nov/20 $${find}\:{n}\:\in\:\mathbb{N}\:{such}\:{that} \\ $$$$\mathrm{5}^{\mathrm{2}{n}} +\mathrm{5}^{{n}} \equiv\mathrm{0}\left[\mathrm{13}\right] \\ $$ Answered by 676597498 last updated on 15/Nov/20 $${let}\:{x}=\mathrm{5}^{{n}} \\…
Question Number 187857 by thean last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 23/Feb/23 $$\left.{a}\right){z}^{\mathrm{2}} =−{i}=\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right) \\ $$$${z}=\left[\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{4}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{4}}\right)= \\ $$$$\:\:\:\:{arg}\left({z}\right)=\left(−\frac{\pi}{\mathrm{4}};\frac{\mathrm{3}\pi}{\mathrm{4}}\right)…
Question Number 122319 by help last updated on 15/Nov/20 Commented by help last updated on 15/Nov/20 $${explain}\:{your}\:{answer}\:{pls} \\ $$ Answered by 676597498 last updated on…
Question Number 56772 by Hassen_Timol last updated on 23/Mar/19 $$\mathrm{Given}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\Phi\:}\:\:=\:\:\frac{\Phi}{\:\mathrm{1}\:+\:\Phi\:} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\::\:{without}\:{using}\:{the}\:{exact} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{value}\:{of}\:\Phi… \\ $$$$\:\:\:\:\:\frac{\:\mathrm{1}\:}{\:\Phi\:}\:\:=\:\:\Phi\:−\:\mathrm{1} \\ $$$$ \\ $$$$ \\…
Question Number 122295 by ZiYangLee last updated on 15/Nov/20 Answered by som(math1967) last updated on 15/Nov/20 $$\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}−\mathrm{sinASinB}} \\ $$$$=\frac{\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}}}{\frac{\mathrm{cosAcosB}−\mathrm{sinAsinB}}{\mathrm{cosAcosB}}} \\ $$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$$$\mathrm{now}…
Question Number 122293 by mathocean1 last updated on 15/Nov/20 $${given}\:{f}\left({x}\right)={cos}^{\mathrm{2}} {x} \\ $$$${for}\:{x}\:\in\:\left[−\frac{\pi}{\mathrm{12}};\frac{\pi}{\mathrm{12}\:}\right]\:,\:\mid{f}'\left({x}\right)\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$\left.\mathrm{1}\right)\:{show}\:{that}\:\forall\:{x},\:{y}\:\in\:\left[−\frac{\pi}{\mathrm{12}};\frac{\pi}{\mathrm{12}}\right]\:; \\ $$$$\mid{cos}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {y}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}−{y}\mid \\ $$ Commented by ZiYangLee last…