Question Number 201332 by Rodier97 last updated on 04/Dec/23 $$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:{calcul}\:: \\ $$$$ \\ $$$$\:\:\:\:\mathrm{1}.\:\:\:\:{lim}_{{n}\rightarrow+\infty\:} \frac{{sin}\left({n}^{\mathrm{2}} \right)−{cos}\left({n}^{\mathrm{3}} \right)}{{n}}\: \\ $$$$…
Question Number 201329 by MrGHK last updated on 04/Dec/23 Answered by witcher3 last updated on 04/Dec/23 $$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{m}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{m}+\mathrm{2n}+\mathrm{1}} \mathrm{dx} \\…
Question Number 201325 by sonukgindia last updated on 04/Dec/23 Commented by mr W last updated on 05/Dec/23 $${do}\:{you}\:{have}\:{the}\:{right}\:{answer}? \\ $$ Answered by mr W last…
Question Number 201322 by cherokeesay last updated on 04/Dec/23 Answered by AST last updated on 05/Dec/23 $${Let}\:\angle{EDC}=\beta\Rightarrow\angle{ABD}=\mathrm{2}\beta−\mathrm{90} \\ $$$${EC}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cos}\beta…\left({i}\right) \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{BD}\boldsymbol{{D}}{Ccos}\left(\mathrm{2}\beta\right)…\left({ii}\right) \\…
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Question Number 201302 by hardmath last updated on 03/Dec/23 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{−\mathrm{2}} \:=\:\mathrm{4}}\\{\mathrm{x}^{−\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{5}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:? \\ $$ Answered by witcher3 last updated on 03/Dec/23 $$\left(\mathrm{1}\right)\ast\left(\mathrm{2}\right)\Leftrightarrow\mathrm{2}+\left(\mathrm{xy}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\left(\mathrm{xy}\right)^{\mathrm{2}}…
Question Number 201298 by ajfour last updated on 03/Dec/23 Commented by ajfour last updated on 03/Dec/23 $${Find}\:{IJ}\:\:\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$ Commented by mr W last updated…
Question Number 201292 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\underset{−{a}} {\overset{{a}} {\int}}\frac{{cos}\left({x}\right){dx}}{\mathrm{1}+{e}^{\pi/{x}} }=\underset{{a}} {\overset{−{a}} {\int}}\frac{{cos}\left(−{x}\right){d}\left(−{x}\right)}{\mathrm{1}+{e}^{\pi/\left(−{x}\right)} }= \\ $$$$=\int_{−{a}}…
Question Number 201293 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}}…