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Question-212181

Question Number 212181 by RojaTaniya last updated on 05/Oct/24 Answered by A5T last updated on 05/Oct/24 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{5} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{2} \\ $$$${a}^{\mathrm{3}}…

f-x-arctan-1-x-1-x-ask-f-2023-0-

Question Number 212177 by MrGaster last updated on 05/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ask}:{f}^{\mathrm{2023}} \left(\mathrm{0}\right) \\ $$ Answered by a.lgnaoui last updated on 05/Oct/24 $$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\Rightarrow\:\:\:\mathrm{f}^{\mathrm{2023}} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2023}} }{\mathrm{4}^{\mathrm{2023}}…

Question-212159

Question Number 212159 by boblosh last updated on 04/Oct/24 Answered by Rasheed.Sindhi last updated on 04/Oct/24 $$\begin{bmatrix}{\:\:\:\:\:\mathrm{2}}&{\:\:\:\:\mathrm{3}}\\{−\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}+{a}\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{4}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}{a}+\mathrm{2}}&{{a}+\mathrm{3}}\\{\mathrm{0}{a}−\mathrm{1}}&{\mathrm{4}{a}−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\mathrm{2}{a}+\mathrm{2}=\mathrm{8}\Rightarrow{a}=\mathrm{3} \\ $$$${c}=−\mathrm{1} \\ $$$${b}={a}+\mathrm{3}=\mathrm{3}+\mathrm{3}=\mathrm{6}…

Question-212152

Question Number 212152 by RojaTaniya last updated on 04/Oct/24 Answered by som(math1967) last updated on 04/Oct/24 $$\:{x}^{\mathrm{3}} +{ax}+{b}=\left({x}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right) \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\:{a}+{b}=−\mathrm{1}\:….{case}\mathrm{1} \\ $$$${put}\:{x}=\mathrm{2} \\…

lim-0-2-e-x-1-2-x-dx-

Question Number 212169 by universe last updated on 04/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\underset{\lambda\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\:\int_{\lambda} ^{\mathrm{2}\lambda} \:\frac{{e}^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } }{{x}}{dx}\:=\:? \\ $$ Answered by MrGaster last updated on 03/Nov/24…