Question Number 83268 by M±th+et£s last updated on 29/Feb/20 Answered by mr W last updated on 29/Feb/20 Commented by mr W last updated on 29/Feb/20…
Question Number 148806 by Nehayadav last updated on 31/Jul/21 Commented by mr W last updated on 07/Aug/21 $${answer}\:{see}\:{Q}\mathrm{149796} \\ $$ Terms of Service Privacy Policy…
Question Number 83266 by 09658867628 last updated on 29/Feb/20 $$\int\mathrm{sin}^{\mathrm{10}} \Theta\mathrm{cos}\:\Theta{d}\Theta\: \\ $$ Commented by Tony Lin last updated on 29/Feb/20 $${let}\:{sin}\theta={t},\:{dt}={cos}\theta{d}\theta \\ $$$$\int{t}^{\mathrm{10}} {dt}…
Question Number 83264 by peter frank last updated on 29/Feb/20 Commented by peter frank last updated on 29/Feb/20 $$\mathrm{7},\mathrm{12},\mathrm{13} \\ $$ Commented by mr W…
Question Number 17729 by Tinkutara last updated on 09/Jul/17 $$\mathrm{A}\:\mathrm{lotus}\:\mathrm{plant}\:\mathrm{in}\:\mathrm{a}\:\mathrm{pool}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cubit}\:\mathrm{above}\:\mathrm{water}\:\mathrm{level}.\:\mathrm{When} \\ $$$$\mathrm{propelled}\:\mathrm{by}\:\mathrm{air},\:\mathrm{the}\:\mathrm{lotus}\:\mathrm{sinks}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{pool}\:\mathrm{2}\:\mathrm{cubits}\:\mathrm{away}\:\mathrm{from}\:\mathrm{its}\:\mathrm{position}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{depth}\:\mathrm{of}\:\mathrm{water}\:\mathrm{in}\:\mathrm{the}\:\mathrm{pool}. \\ $$ Commented by alex041103 last updated…
Question Number 148797 by kameda last updated on 31/Jul/21 Answered by puissant last updated on 31/Jul/21 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{sin}\alpha=\frac{\mathrm{AB}}{\mathrm{OA}}\:\Rightarrow\:\mathrm{sin}\alpha=\frac{\mathrm{AB}}{\mathrm{1}}\:\mathrm{car}\:\mathrm{c}'\mathrm{est}\:\mathrm{un} \\ $$$$\mathrm{cercle}\:\mathrm{de}\:\mathrm{rayon}\:\mathrm{1}.. \\ $$$$\Rightarrow\mathrm{AB}=\mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{car}\:\alpha\:\mathrm{est}\:\mathrm{represent}\acute {\mathrm{e}}\:\mathrm{par}\:\mathrm{x}. \\…
Question Number 83262 by peter frank last updated on 29/Feb/20 Answered by mr W last updated on 29/Feb/20 $${curve}\:\mathrm{1}:\:{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{8}\:\:\:…\left({i}\right) \\ $$$${curve}\:\mathrm{2}:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} =\mathrm{4}\:\:\:…\left({ii}\right)…
Question Number 148798 by abdurehime last updated on 31/Jul/21 Answered by puissant last updated on 31/Jul/21 $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\int\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\int\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}}\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\sqrt{\mathrm{cotan}\left(\mathrm{x}\right)}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{tan}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\mathrm{dx}…
Question Number 83261 by peter frank last updated on 29/Feb/20 Commented by peter frank last updated on 29/Feb/20 $$\mathrm{6},\mathrm{7} \\ $$ Terms of Service Privacy…
Question Number 148792 by Snail last updated on 31/Jul/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\lfloor{tan}^{\mathrm{2}} {x}\rfloor..{evaluate}\:{the}\:{limit}\:{using}\:{squeeze} \\ $$$${theorem}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com