Question Number 151614 by mathdanisur last updated on 22/Aug/21 $$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\frac{\mathrm{x}\:+\:\mathrm{tan}\left(\mathrm{sin}\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\lambda}\:+\:\mathrm{cos}\left(\boldsymbol{\mathrm{x}}\right)}\:\mathrm{dx}\:\:;\:\:\boldsymbol{\lambda}>\mathrm{1} \\ $$ Answered by ArielVyny last updated on 22/Aug/21 $$\Omega=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{x}+{tan}\left({sinx}\right)}{\lambda+{cos}\left({x}\right)}{dx} \\…
Question Number 20540 by tammi last updated on 28/Aug/17 $$\int\frac{\sqrt{{x}}}{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx} \\ $$ Answered by $@ty@m last updated on 31/Aug/17 $${Let}\:{x}={t}^{\mathrm{6}} \:\:\:\:\left({Remark}:\:{LCM}\:{of}\:\mathrm{3}\:\&\:\mathrm{2}=\mathrm{6}\right) \\ $$$$\Rightarrow{dx}=\mathrm{6}{t}^{\mathrm{5}} {dt}…
Question Number 151609 by mathdanisur last updated on 22/Aug/21 Answered by ghimisi last updated on 22/Aug/21 $$\Leftrightarrow{log}_{{xy}} \left(\mathrm{1}+\sqrt{{xy}}\right)^{\mathrm{2}} \geqslant{log}_{\frac{{x}+{y}}{\mathrm{2}}} \left(\frac{{x}+{y}}{\mathrm{2}}+\mathrm{1}\right)\Leftrightarrow \\ $$$$\Leftrightarrow\frac{{ln}\left(\mathrm{1}+\sqrt{{xy}}\right)}{{ln}\sqrt{{xy}}}\geqslant\frac{{ln}\left(\mathrm{1}+\frac{{x}+{y}}{\mathrm{2}}\right)}{{ln}\frac{{x}+{y}}{\mathrm{2}}}\:\:\left(\bullet\right) \\ $$$${f}\left({t}\right)=\frac{{ln}\left(\mathrm{1}+{t}\right)}{{lnt}},{f}:\left(\mathrm{1};\infty\right)\rightarrow{R} \\…
Question Number 20539 by tammi last updated on 28/Aug/17 $$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{6}}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20538 by tammi last updated on 28/Aug/17 $$\int\frac{{dx}}{{e}^{\mathrm{2}{x}} −\mathrm{3}{e}^{{x}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 151602 by DELETED last updated on 22/Aug/21 Answered by DELETED last updated on 22/Aug/21 $$\left.\mathrm{1}.\mathrm{g}\right).\:\mathrm{Letak}\:\mathrm{persentil}\:\mathrm{ke}−\mathrm{65} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{P}_{\mathrm{i}} =\frac{\mathrm{i}\left(\mathrm{N}+\mathrm{1}\right)}{\mathrm{100}}\:\rightarrow\mathrm{P}_{\mathrm{65}} \:=\frac{\mathrm{65}\left(\mathrm{25}+\mathrm{1}\right)}{\mathrm{100}} \\ $$$$\:\:\:\:=\frac{\mathrm{65}×\mathrm{26}}{\mathrm{100}}=\mathrm{16},\mathrm{9} \\ $$$$\:\:\:\:\mathrm{Nilai}\:\mathrm{persentil}\:\mathrm{ke}\:\mathrm{65}=\mathrm{x}_{\mathrm{16}}…
Question Number 86062 by ar247 last updated on 26/Mar/20 $$\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}}{{x}}{dx} \\ $$ Commented by abdomathmax last updated on 27/Mar/20 $${I}\:=\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}}{{x}}{dx}\:\:{changement}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}={t}\:{give} \\ $$$${x}^{\mathrm{2}}…
Question Number 151596 by mathdanisur last updated on 22/Aug/21 Answered by dumitrel last updated on 22/Aug/21 $$\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)^{\mathrm{2}} \overset{{cbs}} {\leqslant}\mathrm{3}\left({a}+{b}+{c}\right)\Rightarrow\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\leqslant\sqrt{\mathrm{3}\left({a}+{b}+{c}\right)} \\ $$$$\left(\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\right)^{\mathrm{2}} \overset{{cbs}} {\leqslant}\mathrm{3}\left({ab}+{bc}+{ac}\right)\Rightarrow\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\leqslant\sqrt{\mathrm{3}\left({ab}+{bc}+{ac}\right.} \\ $$$$\Rightarrow\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)\left(\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\leqslant\sqrt{\mathrm{3}\left({a}+{b}+{b}\right)\centerdot\mathrm{3}\left({ab}+{bc}+{ac}\right)}=\right.…
Question Number 86063 by john santu last updated on 26/Mar/20 Answered by mr W last updated on 26/Mar/20 $$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\…