Question Number 161282 by byaw last updated on 15/Dec/21 Answered by alephzero last updated on 16/Dec/21 $$\boldsymbol{{First}}\:\boldsymbol{{method}}. \\ $$$$\frac{{u}}{\mathrm{2}}\:\rightarrow\:{u}\:\rightarrow\:\mathrm{3}.\mathrm{14}{u} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\rightarrow\:\mathrm{6}\:\rightarrow\:\mathrm{3}.\mathrm{14}{u}\:=\:? \\ $$$${u}\:=\:\mathrm{6}\:\Rightarrow\:\mathrm{3}.\mathrm{14}\left(\mathrm{6}\right)\:=\:\mathrm{18}.\mathrm{84}\: \\ $$$$\left(\mathrm{2}\right)\:\frac{{u}}{\mathrm{2}}\:\rightarrow\:\mathrm{10}\:\rightarrow\:\mathrm{50}.\mathrm{24}…
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Question Number 161156 by LEKOUMA last updated on 13/Dec/21 $${Calculate} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{ln}\:\left(\mathrm{1}+{e}^{−{x}} \right)\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}}{\mathrm{2}+\mathrm{sin}\:\frac{\mathrm{1}}{{x}}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{a}^{{x}} +{b}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered…
Question Number 95604 by 174 last updated on 26/May/20 Commented by MJS last updated on 26/May/20 $${x}^{\mathrm{5}} +\mathrm{1}=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right) \\ $$$$\mathrm{decompose}\:\mathrm{and}\:\mathrm{solve}. \\ $$ Commented…
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Question Number 30017 by jonah last updated on 15/Feb/18 $$\mathrm{find}\:\mathrm{the}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{given}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation} \\ $$$$\mathrm{1}.\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{25}}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1} \\ $$$$\mathrm{2}.\mathrm{x}^{\mathrm{2}} +\mathrm{9y}^{\mathrm{2}} =\mathrm{9} \\ $$ Answered by ajfour last updated…
Question Number 95473 by aurpeyz last updated on 25/May/20 Answered by EmericGent last updated on 25/May/20 $${regroup}\:{every}\:{two}\:{terms} \\ $$$$\mathrm{1}-\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{then}\:\forall{n}\:\in\:\mathbb{N}\ast \\ $$$$\frac{\mathrm{1}}{{n}}\:−\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:=\:\frac{{n}+\mathrm{1}−{n}}{{n}\left({n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${since}\:{n}\left({n}+\mathrm{1}\right)\:{is}\:{a}\:{non}\:{zero}\:{positive}\:…
Question Number 160993 by gbanda95 last updated on 10/Dec/21 Answered by mathmax by abdo last updated on 10/Dec/21 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{x}^{\mathrm{3}} \mathrm{sinx}\:\mathrm{dx}\:=\mathrm{I}_{\mathrm{3}} =\mathrm{3}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{3}−\mathrm{1}\right)\mathrm{I}_{\mathrm{1}} =\frac{\mathrm{3}\pi^{\mathrm{2}}…
Question Number 95420 by bobhans last updated on 25/May/20 $$\mathrm{tinkutara}\:\mathrm{admint} \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{update}\:\mathrm{to}\:\mathrm{version}\:\mathrm{2}.\mathrm{074} \\ $$ Commented by Tinku Tara last updated on 25/May/20 Version uploaded so far on playstore is 2.073. 2.076 is only available on www.tinkutara.com and contains fixes for problem reported by MrW. Commented by…
Question Number 29880 by tawa tawa last updated on 13/Feb/18 Commented by ajfour last updated on 13/Feb/18 Commented by ajfour last updated on 13/Feb/18 $${F}\mathrm{cos}\:\alpha−{f}−{mg}\mathrm{sin}\:\alpha={ma}…