Question Number 155466 by cortano last updated on 01/Oct/21 $$\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\ldots+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)=? \\ $$ Commented by VIDDD last updated on 01/Oct/21 $${can}\:{u}\:{show}\:{u}\:{solution} \\ $$…
Question Number 155461 by Jonathanwaweh last updated on 30/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 155462 by Jonathanwaweh last updated on 30/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89925 by john santu last updated on 20/Apr/20 $${x}^{\mathrm{2}} \left({yy}''−{y}^{\mathrm{2}} \right)+{xyy}'\:=\:{y}\sqrt{{x}^{\mathrm{2}} \left({y}'\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89922 by student work last updated on 20/Apr/20 Commented by john santu last updated on 20/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\mathrm{5}−\mathrm{2}{x}^{\mathrm{2}} }\:\leqslant\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)\:\leqslant\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{\mathrm{5}+\mathrm{2}{x}^{\mathrm{2}} } \\…
Question Number 24387 by chernoaguero@gmail.com last updated on 17/Nov/17 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{particle}}\:\boldsymbol{\mathrm{move}}\:\boldsymbol{\mathrm{along}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{axis}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}; \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{x}}=\mathrm{6}.\mathrm{00}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{3}.\mathrm{00}\boldsymbol{\mathrm{t}}+\mathrm{2}.\mathrm{00}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right) \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{acceleration}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}} \\ $$$$\boldsymbol{\mathrm{by}}\:\:\boldsymbol{\mathrm{a}}=−\mathrm{8}.\mathrm{00}\boldsymbol{\mathrm{t}}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}^{\mathrm{2}} }\right)\:\boldsymbol{\mathrm{and}},\mathrm{at}\:\mathrm{t}=\mathrm{0},\boldsymbol{\mathrm{its}} \\ $$$$\boldsymbol{\mathrm{velocity}}\:\boldsymbol{\mathrm{is}}\:\mathrm{20}\:\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right).\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{velocities}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{particles}}\:\boldsymbol{\mathrm{match}}, \\…
Question Number 89920 by akash4081 last updated on 20/Apr/20 $${x}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{x}}}\:{and}\:{y}=\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{1}+{y}\:}}\:{find}\:{x}+{y} \\ $$ Commented by john santu last updated on 20/Apr/20 $${x}\:=\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\:\Rightarrow\:{x}\:=\:\frac{−\mathrm{1}\:\pm\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${y}\:=\:\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{\mathrm{2}{y}+\mathrm{3}}\:\Rightarrow\:{y}\:=\:\frac{−\mathrm{1}\:\pm\:\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$$\therefore\:{x}+{y}\:=\:\left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+\left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{4}}\right)…
Question Number 89918 by jagoll last updated on 20/Apr/20 $$\int\:\frac{\mathrm{x}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\mathrm{dx}\: \\ $$ Commented by john santu last updated on 20/Apr/20 $$\left[\:{u}\:=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{du}\:=\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 24379 by j.masanja06@gmail.com last updated on 16/Nov/17 Answered by mrW1 last updated on 17/Nov/17 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{8}{y}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}+\left({y}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}}…
Question Number 155450 by SANOGO last updated on 30/Sep/21 Answered by Kamel last updated on 30/Sep/21 $${L}=\underset{{n}\rightarrow+\infty} {{lim}e}^{\frac{\mathrm{2}}{\mathrm{2}{n}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}{Ln}\left({sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right)} ={e}^{\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left({sin}\left(\pi{x}\right)\right){dx}} ={e}^{−\mathrm{2}{Ln}\left(\mathrm{2}\right)} =\frac{\mathrm{1}}{\mathrm{4}}…