Question Number 133438 by 07711990888 last updated on 22/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133433 by greg_ed last updated on 22/Feb/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}, \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\:\mathrm{2}\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}. \\ $$ Answered by…
Question Number 67898 by ramirez105 last updated on 01/Sep/19 $$ \\ $$$${differential}\:{equation}. \\ $$$${homogenous}. \\ $$$$ \\ $$$${ydx}+\left(\mathrm{2}{x}+\mathrm{3}{y}\right){dy}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service…
Question Number 133432 by Dwaipayan Shikari last updated on 22/Feb/21 $$\frac{{sin}\sqrt{\pi}}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{4}\pi}}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{9}\pi}}{\mathrm{3}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{16}\pi}}{\mathrm{4}^{\mathrm{3}} }+….=\frac{\pi\sqrt{\pi}}{\mathrm{12}}\left(\mathrm{1}−\mathrm{3}\sqrt{\pi}+\mathrm{2}\pi\right) \\ $$ Answered by mindispower last updated on 24/Feb/21 $$\underset{{n}\geqslant\mathrm{0}}…
Question Number 67899 by ramirez105 last updated on 01/Sep/19 $${homogenous}\:{differential}\:{equation}. \\ $$$$ \\ $$$$\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dr}−\mathrm{2}{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$${y}={e} \\ $$$${x}={e} \\ $$ Commented by AnJan_Math_Max…
Question Number 2362 by 123456 last updated on 18/Nov/15 $${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$ Commented by Rasheed Soomro last…
Question Number 2358 by 123456 last updated on 18/Nov/15 $${a}_{{n}+\mathrm{1}} =−\frac{{a}_{{n}} \left(\mid{a}_{{n}} \mid+\mathrm{1}\right)}{\mid{a}_{{n}} \mid},{a}_{\mathrm{1}} =\mathrm{1} \\ $$$${b}_{{n}+\mathrm{1}} ={b}_{{n}} +{a}_{{n}+\mathrm{1}} ,{b}_{\mathrm{1}} ={a}_{\mathrm{1}} \\ $$$${b}_{\mathrm{10}} =? \\…
Question Number 133431 by Algoritm last updated on 22/Feb/21 Answered by benjo_mathlover last updated on 22/Feb/21 $$\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{5xy}−\mathrm{y}=\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{1} \\ $$$$\left(\mathrm{y}−\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} −\left(\mathrm{5y}−\mathrm{3}\right)\mathrm{x}+\mathrm{1}−\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{5y}−\mathrm{3}\pm\sqrt{\left(\mathrm{5y}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{y}\right)}}{\mathrm{2}\left(\mathrm{y}−\mathrm{1}\right)}…
Question Number 2355 by 123456 last updated on 17/Nov/15 $$\frac{{df}}{{d}\zeta}\mathrm{sinh}\:\zeta+\frac{{df}}{{d}\theta}\mathrm{sin}\:\theta+\frac{{df}}{{d}\rho}\rho=\mathrm{0} \\ $$$${f}\left(\rho,\zeta,\theta\right)=?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133424 by liberty last updated on 22/Feb/21 $$\mathrm{Given}\:\mathrm{10}\:\mathrm{white}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{ten}\:\mathrm{black}\:\mathrm{balls} \\ $$$$\mathrm{numbered}\:\mathrm{1},\mathrm{2},…,\mathrm{10}.\:\mathrm{How}\:\mathrm{many}\: \\ $$$$\mathrm{ways}\:\mathrm{can}\:\mathrm{we}\:\mathrm{choose}\:\mathrm{6}\:\mathrm{balls}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{no}\:\mathrm{two}\:\mathrm{chosen}\:\mathrm{balls}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{number} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{two}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{chosen}\:\mathrm{balls}\:\mathrm{have} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{number}? \\ $$ Answered by EDWIN88…