Question Number 107428 by abony1303 last updated on 10/Aug/20 $$\mathrm{The}\:\mathrm{polynomial}\:{P}\left({x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{ax}^{\mathrm{2}} −\mathrm{4x}+\mathrm{b}, \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{constants}.\:\mathrm{Given}\:\mathrm{that} \\ $$$$\mathrm{x}−\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:{P}\left({x}\right)\:\mathrm{and}\:\mathrm{that}\:\mathrm{a}\:\mathrm{remainder} \\ $$$$\mathrm{of}\:\mathrm{6}\:\mathrm{is}\:\mathrm{obtained}\:\mathrm{when}\:{P}\left({x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right),\:\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}. \\ $$ Commented by abony1303…
Question Number 107419 by DeepakMahato last updated on 10/Aug/20 $${Factorize}:−\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}\boldsymbol{{x}}+\mathrm{3} \\ $$$$ \\ $$$$ \\ $$ Answered by nimnim last updated on 10/Aug/20 $${x}^{\mathrm{2}}…
Question Number 41878 by lucha116 last updated on 14/Aug/18 $$\sqrt{\mathrm{3}}{sin}\mathrm{3}{x}−{cos}\mathrm{3}{x}+\mathrm{2}{sin}\frac{\mathrm{9}{x}}{\mathrm{4}}=\mathrm{2} \\ $$ Answered by MJS last updated on 15/Aug/18 $${f}\left({x}\right)=\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{3}{x}\:−\mathrm{cos}\:\mathrm{3}{x}\:+\mathrm{2sin}\:\frac{\mathrm{9}{x}}{\mathrm{4}}\:−\mathrm{2} \\ $$$$\mathrm{found}\:\mathrm{by}\:\mathrm{approximation}: \\ $$$$\mathrm{max}\left({f}\left({x}\right)\right)=\begin{pmatrix}{\frac{\mathrm{2}\pi}{\mathrm{9}}+\frac{\mathrm{8}\pi}{\mathrm{3}}{n}}\\{\mathrm{2}}\end{pmatrix}\:\mathrm{with}\:{n}\in\mathbb{Z} \\…
Question Number 172910 by DAVONG last updated on 03/Jul/22 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{e}−\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}} }{\mathrm{x}−\mathrm{1}}=? \\ $$ Answered by FongXD last updated on 03/Jul/22 $$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{e}−\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{lnx}} }{\mathrm{x}−\mathrm{1}}=−\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{e}\left(\mathrm{e}^{\frac{\mathrm{lnx}}{\mathrm{x}−\mathrm{1}}−\mathrm{1}}…
Question Number 172889 by Gbenga last updated on 02/Jul/22 $$\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{tanh}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{x}}\right)\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{laplace}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107328 by mathdave last updated on 10/Aug/20 Commented by mnjuly1970 last updated on 10/Aug/20 Commented by mnjuly1970 last updated on 10/Aug/20 Commented by…
Question Number 107315 by Algoritm last updated on 10/Aug/20 Commented by bemath last updated on 10/Aug/20 $$\frac{\mathrm{1}}{\mathrm{2020}−\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}}=\:{a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$\frac{\mathrm{2020}+\sqrt{\mathrm{2020}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2020}^{\mathrm{2}} −\left(\mathrm{2020}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:{a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$\mathrm{2020}+\sqrt{\mathrm{2020}^{\mathrm{2}}…
Question Number 172839 by DAVONG last updated on 02/Jul/22 $$\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{lnx}}\mathrm{dx}=? \\ $$ Answered by puissant last updated on 02/Jul/22 $${I}=\int\frac{\mathrm{1}}{{lnx}}{dx}\:;\:\:{x}={e}^{{u}} \:\Rightarrow\:{dx}={e}^{{u}} {du} \\ $$$${I}=\int\frac{{e}^{{u}} }{{u}}{du}\:=\:\int\underset{{n}\geqslant\mathrm{0}}…
Question Number 172838 by DAVONG last updated on 02/Jul/22 $$\mathrm{1}.\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{x}−\mathrm{1}−\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}\:=\:? \\ $$ Commented by cortano1 last updated on 02/Jul/22 $$\:\:=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}−\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} }\: \\…
Question Number 107275 by mathdave last updated on 09/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com