Question Number 181747 by SANOGO last updated on 29/Nov/22 $${calcul}\:{la}\:{somme} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{+{oo}} {\sum}}\frac{{x}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Frix last updated on 01/Dec/22…
Question Number 181741 by yaslm last updated on 29/Nov/22 Answered by cortano1 last updated on 30/Nov/22 $$\:\Rightarrow\mathrm{11}\:\mathrm{sin}\:\theta\:+\:\mathrm{60}\:\mathrm{cos}\:\theta\:=\:\mathrm{61}\:;\:\theta\:\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{11}\:\mathrm{tan}\:\theta\:+\mathrm{60}\:=\:\mathrm{61}\:\mathrm{sec}\:\theta\: \\ $$$$\Rightarrow\left(\mathrm{11}\:\mathrm{tan}\:\theta+\mathrm{60}\right)^{\mathrm{2}} =\:\mathrm{3721}\:\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\mathrm{121}\:\mathrm{tan}\:^{\mathrm{2}}…
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Question Number 181733 by SANOGO last updated on 30/Nov/22 $$\left.\mathrm{1}\right){calcul}\:{n}−+{oo}:\:{sn}=\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\:\frac{\mathrm{1}}{{k}} \\ $$$$\left.\mathrm{2}\right){find}:\:\underset{{k}=\mathrm{1}} {\overset{+{oo}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 116184 by ZiYangLee last updated on 01/Oct/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{6}-\mathrm{digits}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{which}\:\mathrm{are} \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1}\:\mathrm{to}\:\mathrm{9}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the}\:\mathrm{number}\:\mathrm{appears}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{twice}?\:\left[\mathrm{For}\:\mathrm{instance}:\:\mathrm{121233},\mathrm{122221},\mathrm{777777}\:\mathrm{and}\:\mathrm{etc}.\right] \\ $$ Answered by mr W last updated on…
Question Number 116149 by A8;15: last updated on 01/Oct/20 Commented by A8;15: last updated on 01/Oct/20 please help Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 181676 by yaslm last updated on 28/Nov/22 Answered by floor(10²Eta[1]) last updated on 28/Nov/22 $$\mathrm{def}\:\mathrm{of}\:\mathrm{square}\:\mathrm{root}: \\ $$$$\sqrt{\mathrm{b}}=\mathrm{a}\Leftrightarrow\mathrm{b}=\mathrm{a}^{\mathrm{2}} \wedge\mathrm{a}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\sqrt{\mathrm{x}}−\mathrm{7}=\sqrt{\mathrm{x}−\mathrm{7}}\Leftrightarrow\mathrm{x}−\mathrm{7}=\left(\sqrt{\mathrm{x}}−\mathrm{7}\right)^{\mathrm{2}} \wedge\sqrt{\mathrm{x}}\geqslant\mathrm{7}…
Question Number 116114 by Khalmohmmad last updated on 01/Oct/20 Answered by Dwaipayan Shikari last updated on 01/Oct/20 $$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}+\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}…
Question Number 50560 by Gulay last updated on 17/Dec/18 $$\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}−\mathrm{4y}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{3}\:\:\:\:\:\:\mathrm{y}=−\mathrm{1} \\ $$$$\mathrm{help}\:\mathrm{md}\:\mathrm{plz}\:\mathrm{sir} \\ $$ Answered by ajfour last updated on 17/Dec/18 $$\frac{\mathrm{7}{x}}{\mathrm{3}}−\mathrm{4}{y}\:=\:\frac{\mathrm{7}×\mathrm{3}}{\mathrm{3}}−\mathrm{4}\left(−\mathrm{1}\right)\:=\:\mathrm{7}+\mathrm{4}\:=\:\mathrm{11}\:. \\ $$ Commented…
Question Number 50558 by Gulay last updated on 17/Dec/18 $$\mathrm{thanks}\:\mathrm{sir} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com