Question Number 99300 by mr W last updated on 20/Jun/20 $$\mathrm{Find}\:\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{{n}}\right)!}=? \\ $$ Answered by maths mind last updated on 20/Jun/20 $${Z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\Rightarrow{z}\in\left\{\mathrm{1},{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}}…
Question Number 164826 by mnjuly1970 last updated on 22/Jan/22 $$ \\ $$$$\:\:\:\:\:\:{if}\:\:\:\:{h}\left({x}\right)\:=\:{x}\:−\:\lfloor\frac{\mathrm{1}}{{x}}\rfloor\:\:,\:\:{x}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{then}\:\:\:{h}^{\:−\mathrm{1}} \left(\:{x}\:\right)\:=\:? \\ $$$$ \\ $$ Answered by TheSupreme last updated on…
Question Number 33743 by prof Abdo imad last updated on 23/Apr/18 $${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)….\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right)\:{with}\:{n}\:{integr} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$ Commented by caravan…
Question Number 99256 by student work last updated on 19/Jun/20 $$\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:…..} } } =?\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by mr W last updated on 19/Jun/20 $$\left(\sqrt{\mathrm{2}}\right)^{{x}}…
Question Number 164795 by HongKing last updated on 22/Jan/22 Answered by puissant last updated on 22/Jan/22 $$\Omega=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{sinhx}}−\frac{\mathrm{1}}{{xcoshx}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}}−\frac{\mathrm{1}}{{x}\frac{{e}^{{x}}…
Question Number 99257 by peter frank last updated on 19/Jun/20 Answered by Ar Brandon last updated on 19/Jun/20 $$\mathrm{1a}\backslash\mathrm{log}_{\mathrm{ab}} \mathrm{x}=\frac{\mathrm{log}_{\mathrm{x}} \mathrm{x}}{\mathrm{log}_{\mathrm{x}} \mathrm{ab}}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{a}+\mathrm{log}_{\mathrm{x}} \mathrm{b}}=\frac{\mathrm{1}}{\frac{\mathrm{log}_{\mathrm{a}} \mathrm{a}}{\mathrm{log}_{\mathrm{a}}…
Question Number 164788 by cortano1 last updated on 22/Jan/22 Answered by leonhard77 last updated on 22/Jan/22 $$\:\Rightarrow\mathrm{tan}\:\mathrm{30}°\:=\:\frac{\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{28}.\mathrm{tan}\:\mathrm{2}°°} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{28}°.\mathrm{tan}\:\mathrm{2}°} \\ $$$$\mathrm{1}−\mathrm{tan}\:\mathrm{28}°.\mathrm{tan}\:\mathrm{2}°=\sqrt{\mathrm{3}}\:\left(\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°\right) \\ $$$$ \\ $$$$\mathrm{tan}\:\mathrm{28}°=\frac{\mathrm{tan}\:\mathrm{26}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{26}°.\mathrm{tan}\:\mathrm{2}°}…
Question Number 164787 by cortano1 last updated on 22/Jan/22 Answered by mr W last updated on 22/Jan/22 $${xy}\left[\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}\right]=\frac{\mathrm{10}\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{9}} \\ $$$${xy}\left[\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)\right]=\frac{\mathrm{2}\left({x}+{y}\right)^{\mathrm{3}} }{\mathrm{3}} \\…
Question Number 164786 by mathlove last updated on 21/Jan/22 Answered by MJS_new last updated on 22/Jan/22 $$\frac{{A}}{{B}}=\frac{{x}+\mathrm{3}}{{x}^{\mathrm{6}} +\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 33702 by math khazana by abdo last updated on 22/Apr/18 $${let}\:{p}\left({x}\right)\:={a}_{\mathrm{0}} \:+{a}_{\mathrm{1}} {x}\:+{a}_{\mathrm{2}} {x}^{\mathrm{2}} \:+…{a}_{{n}} {x}^{{n}} \\ $$$${prove}\:{that}\:\:{a}_{{k}} =\:\frac{{p}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:\:\forall\:{k}\:\in\left[\left[\mathrm{0},{n}\right]\right]\:. \\ $$ Commented…