Question Number 206093 by mr W last updated on 06/Apr/24 Answered by A5T last updated on 07/Apr/24 Commented by A5T last updated on 07/Apr/24 $${S}_{\mathrm{4}}…
Question Number 206095 by RoseAli last updated on 06/Apr/24 Answered by Frix last updated on 07/Apr/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}−\mathrm{tan}\:{x}}\:\:\overset{\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} {=}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[{x}−\mathrm{sin}\:{x}\right]}{\frac{{d}}{{dx}}\left[{x}−\mathrm{tan}\:{x}\right]}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{−\mathrm{tan}^{\mathrm{2}} \:{x}}…
Question Number 206063 by MATHEMATICSAM last updated on 06/Apr/24 $$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$ Answered by mr W last updated on 06/Apr/24…
Question Number 206082 by hardmath last updated on 06/Apr/24 $$\mathrm{If}\:\:\:\mathrm{cos}\boldsymbol{\alpha}\:=\:\mathrm{sin}\boldsymbol{\alpha}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{sin2}\boldsymbol{\alpha}\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 29/Apr/24 $$\mathrm{cos}\alpha\:=\:\mathrm{sin}\alpha\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\mathrm{cos}\alpha\:−\:\mathrm{sin}\alpha\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\…
Question Number 206045 by mr W last updated on 05/Apr/24 Commented by mr W last updated on 05/Apr/24 $${a}\:{special}\:{case}\:{of}\:{Q}\mathrm{206053} \\ $$ Answered by HeferH24 last…
Question Number 206047 by universe last updated on 05/Apr/24 Answered by aleks041103 last updated on 06/Apr/24 $${y}'={y}+{const}. \\ $$$$\Rightarrow{y}={c}_{\mathrm{2}} {e}^{{x}} −{c}_{\mathrm{1}} \\ $$$$\Rightarrow{y}'={y}+{c}_{\mathrm{1}} \\ $$$$\Rightarrow{c}_{\mathrm{1}}…
Question Number 206036 by Davidtim last updated on 05/Apr/24 $${is}\:{it}\:{a}\:{polynomial}? \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\frac{\mathrm{2}}{{x}^{−\mathrm{2}} } \\ $$ Answered by Frix last updated on 05/Apr/24 $$\frac{\mathrm{2}}{{x}^{−\mathrm{2}} }=\mathrm{2}{x}^{\mathrm{2}}…
Question Number 206037 by Davidtim last updated on 05/Apr/24 $${is}\:{it}\:{a}\:{polynomial}? \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{2}} }+\mathrm{10} \\ $$ Answered by Frix last updated on 05/Apr/24 $${f}\left({x}\right)={x}^{\mathrm{3}}…
Question Number 206038 by BaliramKumar last updated on 05/Apr/24 $$\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by mr W last updated on 05/Apr/24 $${x}+\mathrm{1}=\sqrt{\mathrm{1}+{x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$$\Rightarrow\mathrm{2024}=\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}} \\ $$…
Question Number 206024 by MaruMaru last updated on 05/Apr/24 $${proove} \\ $$$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$ Answered by Tinku Tara last updated on 05/Apr/24 $${e}^{{i}\pi} ={cos}\pi+{isin}\pi=−\mathrm{1}…