Question Number 163958 by HongKing last updated on 12/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 98416 by 175 last updated on 13/Jun/20 Answered by Farruxjano last updated on 13/Jun/20 $$\boldsymbol{{i}}^{\mathrm{5}\boldsymbol{{n}}+\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{5}\boldsymbol{{n}}+\mathrm{1}=\mathrm{2}\boldsymbol{{k}}\:\left(\boldsymbol{{k}}\in\boldsymbol{{Z}}\right),\:\Rightarrow\:\boldsymbol{{n}}=\mathrm{2}\boldsymbol{{t}}+\mathrm{1}\:\left(\boldsymbol{{t}}\in\boldsymbol{{Z}}\right) \\ $$ Answered by Ramajunan last updated…
Question Number 163942 by Rasheed.Sindhi last updated on 12/Jan/22 $${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left[\:{f}\left({x}\right)\:\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by mr W last updated on 12/Jan/22 $${x}\:\in\:{R} \\…
Question Number 32868 by Tinkutara last updated on 04/Apr/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163905 by HongKing last updated on 11/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163906 by HongKing last updated on 11/Jan/22 $$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}\:+\:\mathrm{6}\:=\:\mathrm{8y}}\\{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{y}\:+\:\mathrm{6}\:=\:\mathrm{8z}}\\{\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\:+\:\mathrm{6}\:=\:\mathrm{8x}}\end{cases}\:\:\:\Rightarrow\:\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:=\:? \\ $$ Answered by ajfour last updated on 11/Jan/22 $${let}\:\:{from}\:{symmetry} \\ $$$${x}={y}={z}\:\:\Rightarrow\:\:{x}^{\mathrm{3}}…
Question Number 32832 by Rasheed.Sindhi last updated on 03/Apr/18 $$\mathrm{Determine}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{1001n}+\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{perfect}\:\mathrm{cube}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163897 by HongKing last updated on 11/Jan/22 $$\mathrm{if}\:\:\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{x}\:\neq\:\mathrm{1} \\ $$$$\mathrm{simplificar}\:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} \\ $$ Answered by mr W last updated on 11/Jan/22…
Question Number 163902 by HongKing last updated on 11/Jan/22 $$\sqrt{\mathrm{x}!^{\boldsymbol{\mathrm{x}}!} }\:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}!} \:\:=\:\mathrm{x}!^{\mathrm{3}} \:\:+\:\:\mathrm{10x}!\:\:+\:\:\mathrm{4} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by MJS_new last updated on 11/Jan/22 $${x}!\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{try}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}……
Question Number 163899 by HongKing last updated on 11/Jan/22 $$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\:\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{1}\:-\:\mathrm{x}}}\:\mathrm{dx} \\ $$ Answered by Kamel last updated on 11/Jan/22 $$\Omega\left({a}\right)=\pi{cos}\left(\frac{{a}}{\mathrm{2}}\right){J}_{\mathrm{0}} \left(\frac{{a}}{\mathrm{2}}\right) \\ $$…