Question Number 83941 by john santu last updated on 08/Mar/20 $$\mathrm{If}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{4}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}\:}\:=\:\mathrm{x}\: \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{6}\:=\:? \\ $$ Answered by $@ty@m123 last updated on 08/Mar/20 $$\:^{\mathrm{3}}…
Question Number 149469 by mathdanisur last updated on 05/Aug/21 $$\mathrm{Compare}: \\ $$$$\mathrm{1900}^{\frac{\mathrm{3}}{\mathrm{4}}} \:+\:\:\mathrm{99}^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\:\boldsymbol{\mathrm{and}}\:\:\:\mathrm{1999}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$ Answered by mindispower last updated on 05/Aug/21 $${f}\left({x}\right)={x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}}…
Question Number 149468 by mathdanisur last updated on 05/Aug/21 $${sin}\boldsymbol{\alpha}\:\centerdot\:{cos}\boldsymbol{\alpha}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:\Rightarrow\:\mathrm{3}\mid{sin}\boldsymbol{\alpha}\:-\:{cos}\boldsymbol{\alpha}\mid=? \\ $$ Answered by Olaf_Thorendsen last updated on 05/Aug/21 $$\mathrm{Let}\:{x}\:=\:\mathrm{3}\mid\mathrm{sin}\alpha−\mathrm{cos}\alpha\mid \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{9}\left(\mathrm{sin}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \alpha−\mathrm{2sin}\alpha.\mathrm{cos}\alpha\right)…
Question Number 83931 by john santu last updated on 08/Mar/20 $$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{1}}+\sqrt[{\mathrm{4}\:}]{\mathrm{2}}\right)}\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\left(\sqrt[{\:\mathrm{4}}]{\mathrm{2}}+\sqrt[{\mathrm{4}\:}]{\mathrm{3}}\right)}\:+ \\ $$$$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{3}}+\sqrt[{\mathrm{4}\:}]{\mathrm{4}}\right)}\:+\:…\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{255}}+\sqrt{\mathrm{256}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{255}}+\sqrt[{\mathrm{4}\:}]{\mathrm{256}}\right)} \\ $$$$=\:…\: \\ $$ Commented by john santu last updated on 08/Mar/20…
Question Number 18392 by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 $${solve}\:{for}\:{x},{y},{z}. \\ $$ Answered by mrW1 last updated on…
Question Number 18388 by tawa tawa last updated on 19/Jul/17 $$\mathrm{Solve}\:\mathrm{simultaneously}.\: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}\:\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{y}\:=\:\mathrm{15}\:\:\:\:……\:\left(\mathrm{ii}\right) \\ $$ Answered by mrW1 last updated on 19/Jul/17…
Question Number 18386 by Tinkutara last updated on 19/Jul/17 $$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{rational}\:\mathrm{number}. \\ $$ Commented by mrW1 last updated on 19/Jul/17 $$\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:=\mathrm{1} \\ $$ Answered…
Question Number 149458 by mathdanisur last updated on 05/Aug/21 Commented by EDWIN88 last updated on 06/Aug/21 $${n}\left({S}\right)={C}_{\mathrm{2}} ^{\mathrm{7}} =\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}×\mathrm{1}}=\mathrm{21} \\ $$$${n}\left({A}\right)=\:\mathrm{18} \\ $$$${A}=\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{1},\mathrm{4}\right),\left(\mathrm{1},\mathrm{6}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right),\right. \\ $$$$\:\:\:\:\:\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{2},\mathrm{6}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{4},\mathrm{5}\right),\left(\mathrm{4},\mathrm{6}\right),…
Question Number 149454 by mnjuly1970 last updated on 05/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 83910 by redmiiuser last updated on 07/Mar/20 $$\mathrm{find}\:\mathrm{all}\:\mathrm{6}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{only}\:\mathrm{palindrome}\:\mathrm{but}\:\mathrm{also}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{495}. \\ $$ Commented by redmiiuser last updated on 08/Mar/20 $$\mathrm{CAN}\:\mathrm{ANYONE}\: \\ $$$$\mathrm{ANSWER}\:\mathrm{THIS} \\…