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Category: Algebra

2-x-x-4-if-x-Z-2x-x-4-x-4-3-impossible-if-x-Z-x-x-1-2-x-x-3-x-1-1-x-lt-2-x-1-x-1-2-

Question Number 188430 by mnjuly1970 last updated on 01/Mar/23 $$ \\ $$$$\:\:\:\:\mathrm{2}\lfloor\:{x}\:\rfloor\:−\:\lfloor\:−{x}\:\rfloor\:=\mathrm{4} \\ $$$$\:\:\:−−−− \\ $$$$\:\:{if}\:\:{x}\in\mathbb{Z}\:\Rightarrow\:\:\mathrm{2}{x}\:+{x}\:=\:\mathrm{4}\:\Rightarrow\:{x}=\frac{\mathrm{4}}{\mathrm{3}}\:\:,{impossible} \\ $$$$\:\:{if}\:{x}\notin\:\mathbb{Z}\:\overset{\lfloor−{x}\rfloor=−\lfloor{x}\rfloor−\mathrm{1}} {\Rightarrow}\mathrm{2}\lfloor{x}\rfloor+\lfloor{x}\rfloor=\mathrm{3} \\ $$$$\:\:\:\:\:\Rightarrow\:\lfloor\:{x}\:\rfloor=\:\mathrm{1}\:\Rightarrow\:\:\mathrm{1}\leqslant\:{x}\:<\:\mathrm{2}\:\:\:\:\overset{{x}\neq\mathrm{1}} {\Rightarrow}\:{x}\in\:\left(\mathrm{1}\:,\:\mathrm{2}\right)\:\:\:\checkmark \\ $$$$ \\…

1-sin-x-x-

Question Number 122883 by kolos last updated on 20/Nov/20 $$\sum_{\mathrm{1}} ^{\infty} \left(\frac{\mathrm{sin}\:\left(\mathrm{x}\right)}{\mathrm{x}}\right)=? \\ $$ Commented by Dwaipayan Shikari last updated on 20/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sinx}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{1}}…

xf-x-f-x-2-f-2-2-f-8-

Question Number 188407 by HeferH last updated on 01/Mar/23 $${xf}\left({x}\right)\:=\:{f}\left({x}\:+\:\mathrm{2}\right) \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{2} \\ $$$${f}\left(\mathrm{8}\right)\:=\:?\: \\ $$ Answered by horsebrand11 last updated on 01/Mar/23 $$\left({i}\right)\:{x}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right) \\…

Question-57328

Question Number 57328 by ANTARES VY last updated on 02/Apr/19 Answered by mr W last updated on 02/Apr/19 $${let}\:\alpha=\frac{\angle{A}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{r}}{{AO}}=\frac{\mathrm{2}×\mathrm{3}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }}=\frac{\mathrm{3}}{\mathrm{4}}…

Question-122860

Question Number 122860 by greg_ed last updated on 20/Nov/20 Answered by MJS_new last updated on 20/Nov/20 $$\mathrm{both}\:\mathrm{sides}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{squaring}\:\mathrm{allowed} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}\mid{xy}\mid\leqslant\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}\mid{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \mid…

Question-122854

Question Number 122854 by Study last updated on 20/Nov/20 Answered by mathmax by abdo last updated on 20/Nov/20 $$\left(−\mathrm{2}\right)^{\sqrt{\mathrm{2}}} =\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(−\mathrm{2}\right)} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\left(\mathrm{ln}\left(−\mathrm{1}\right)+\mathrm{ln}\left(\mathrm{2}\right)\right)} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\left(\mathrm{i}\pi+\mathrm{ln2}\right)} \\ $$$$=\:\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{i}\sqrt{\mathrm{2}}\pi} \:=\mathrm{e}^{\sqrt{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)}…