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 \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{9}\left(\mathrm{sin}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \alpha−\mathrm{2sin}\alpha.\mathrm{cos}\alpha\right)…
Question Number 83935 by Roland Mbunwe last updated on 08/Mar/20 $${If}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:{and}\:{higher}\:{powers}\:{of}\:{x}\:{are}\:{neglected},\:{show}\:{that} \\ $$$$\sqrt{\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\mathrm{1}−{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149471 by ArielVyny last updated on 05/Aug/21 $${on}\:{distribue}\:{au}\:{hasard}\:\mathrm{8}\:{boules}\:{b}_{\mathrm{1}} …{b}_{\mathrm{8}} \\ $$$${dans}\:\mathrm{6}\:{tiroirs}\:{t}_{\mathrm{1}} …{t}_{\mathrm{6}} .{soit}\:{A}_{{i}} \:{l}'{evenement} \\ $$$$“{le}\:{tiroir}\:{t}_{{i}} \:{est}\:{vide}''\:{les}\:{evenements}\:{A}_{\mathrm{1}} {et} \\ $$$${A}_{\mathrm{2}} {sont}-{ils}\:{independants}\:? \\ $$…
Question Number 18397 by geovane10math last updated on 20/Jul/17 $${Prove}\::\:\forall{n}\:\in\:\mathbb{N},\:{n}\:\geqslant\:\mathrm{2},\:{so}\:\exists\:{x},{y},{z}\:\mid\: \\ $$$${x},{y},{z}\:\in\:\mathbb{N}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{{n}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${Example}: \\ $$$${choose}\:{n}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${If}\:{x}\:=\:\mathrm{1}\:\:,\:\:\:{y}\:=\:\mathrm{2}\:\:{and}\:\:\:{z}\:=\:\mathrm{2},\:{the}\:{equa}- \\ $$$${tion}\:{is}\:{correct}! \\…
Question Number 18396 by Joel577 last updated on 20/Jul/17 $$\int\:\sqrt{\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({x}\:+\:\mathrm{1}\right)^{\mathrm{2}} }}\:\:{dx} \\ $$ Answered by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 $$\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\left({x}^{\mathrm{2}} +{x}\right)^{\mathrm{2}}…
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 18394 by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149467 by fotosy2k last updated on 05/Aug/21 Answered by mindispower last updated on 05/Aug/21 $${ln}\left({a}_{{n}} \right)={nln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$$${ln}\left(\mathrm{1}+{x}\right)\leqslant{x} \\ $$$$\Rightarrow\mathrm{0}\leqslant{ln}\left({a}_{{n}} \right)\leqslant{n}.\frac{\mathrm{1}}{{n}}\Rightarrow\mathrm{0}\leqslant{ln}\left({a}_{{n}} \right)\leqslant\mathrm{1} \\…
Question Number 18392 by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17 $${solve}\:{for}\:{x},{y},{z}. \\ $$ Answered by mrW1 last updated on…