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 $$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\Rightarrow\:\:\:\mathrm{f}^{\mathrm{2023}} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2023}} }{\mathrm{4}^{\mathrm{2023}}…
Question Number 212179 by cherokeesay last updated on 05/Oct/24 Answered by cemoosky last updated on 05/Oct/24 $${x}\:=\:\mathrm{6}.\mathrm{6} \\ $$ Answered by mr W last updated…
Question Number 212159 by boblosh last updated on 04/Oct/24 Answered by Rasheed.Sindhi last updated on 04/Oct/24 $$\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} \\ $$$$\begin{bmatrix}{\mathrm{2}{a}+\mathrm{2}}&{{a}+\mathrm{3}}\\{\mathrm{0}{a}−\mathrm{1}}&{\mathrm{4}{a}−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\mathrm{2}{a}+\mathrm{2}=\mathrm{8}\Rightarrow{a}=\mathrm{3} \\ $$$${c}=−\mathrm{1} \\ $$$${b}={a}+\mathrm{3}=\mathrm{3}+\mathrm{3}=\mathrm{6}…
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Question Number 212152 by RojaTaniya last updated on 04/Oct/24 Answered by som(math1967) last updated on 04/Oct/24 $$\:{x}^{\mathrm{3}} +{ax}+{b}=\left({x}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right) \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\:{a}+{b}=−\mathrm{1}\:….{case}\mathrm{1} \\ $$$${put}\:{x}=\mathrm{2} \\…
Question Number 212169 by universe last updated on 04/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\underset{\lambda\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\:\int_{\lambda} ^{\mathrm{2}\lambda} \:\frac{{e}^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } }{{x}}{dx}\:=\:? \\ $$ Answered by MrGaster last updated on 03/Nov/24…
Question Number 212170 by ajfour last updated on 04/Oct/24 Commented by ajfour last updated on 08/Oct/24 Commented by ajfour last updated on 08/Oct/24 $${I}\:{suspect}\:{it}\:{might}\:{happen}\:{this}\:{way}. \\…
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Question Number 212164 by mnjuly1970 last updated on 04/Oct/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:−\:\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\right)\:=\:? \\ $$$$\: \\ $$$$ \\ $$ Answered by…
Question Number 212160 by boblosh last updated on 04/Oct/24 Answered by A5T last updated on 04/Oct/24 $$\frac{{x}}{\mathrm{2}}+\mathrm{4}{y}=\mathrm{4}\Rightarrow{x}+\mathrm{8}{y}=\mathrm{8}…\left({i}\right) \\ $$$$\frac{{x}}{\mathrm{4}}−\frac{\mathrm{2}{y}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\mathrm{3}{x}−\mathrm{8}{y}=\mathrm{8}…\left({ii}\right) \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}\\{\mathrm{8}}\end{bmatrix} \\ $$$${By}\:{Cramer}'{s}\:{rule}:\:{x}=\frac{\begin{vmatrix}{\mathrm{8}}&{\mathrm{8}}\\{\mathrm{8}}&{−\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{128}}{−\mathrm{32}}=\mathrm{4} \\ $$$${and}\:{y}=\frac{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{16}}{−\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{2}}…