Question Number 149503 by mathdanisur last updated on 05/Aug/21 $$\mathrm{Calcular}: \\ $$$$\mathrm{3sin}\centerdot\left(\mathrm{150}\:-\:\boldsymbol{\alpha}\right)\:-\:\mathrm{2cos}\centerdot\left(\mathrm{60}\:-\:\boldsymbol{\alpha}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by MJS_new last updated on 05/Aug/21 $$\Leftrightarrow \\ $$$$\mathrm{3sin}\:\left(\alpha+\mathrm{30}\right)\:−\mathrm{2sin}\:\left(\alpha+\mathrm{30}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 149481 by mathdanisur last updated on 05/Aug/21 Answered by mr W last updated on 05/Aug/21 $${say}\:{AB}={CK}={x},\:{BM}={BK}=\mathrm{1} \\ $$$${AC}={x}−\mathrm{1}+{x}=\mathrm{2}{x}−\mathrm{1} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 149478 by mathdanisur last updated on 05/Aug/21 $${Solve}\:{for}\:{real}\:{numbers}: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}{sin}\boldsymbol{{y}}−\mathrm{4}{cos}\boldsymbol{{y}}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$ Commented by iloveisrael last updated on 06/Aug/21 $$\Delta\geqslant\mathrm{0} \\ $$$$\mathrm{3sin}\:\mathrm{y}+\mathrm{4cosy}−\mathrm{6}\geqslant−\mathrm{1}…
Question Number 149474 by mathdanisur last updated on 05/Aug/21 $${x};{y};{z}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:{x}+{y}+{z}=\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{12}}]{{x}^{\mathrm{5}} }\:+\:\sqrt[{\mathrm{12}}]{{y}^{\mathrm{5}} }\:+\:\sqrt[{\mathrm{12}}]{{z}^{\mathrm{5}} }\:\geqslant\:{xy}\:+\:{yz}\:+\:{zx} \\ $$ Commented by dumitrel last updated on 06/Aug/21 $${hint}\:?…
Question Number 83941 by john santu last updated on 08/Mar/20 $$\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…