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Question Number 206794 by MaruMaru last updated on 25/Apr/24 $${help}\:{me}… \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left({t}\right)\mathrm{ln}\left({t}\right)}{{t}}{e}^{−{t}} \:{dt} \\ $$ Answered by Berbere last updated on 25/Apr/24 $${let}\:{f}\left({a}\right)=\int_{\mathrm{0}}…
Question Number 206788 by SANOGO last updated on 25/Apr/24 Answered by A5T last updated on 25/Apr/24 $$\mathrm{1}.\:\overline {\left(\frac{{x}}{{y}}\right)}=\frac{\overset{−} {{x}}}{\overset{−} {{y}}}\Rightarrow\overset{} {\left(\frac{\mathrm{1}}{{z}}\right)}=\frac{\overset{−} {\mathrm{1}}}{\overset{−} {{z}}}=\frac{\mathrm{1}}{\overset{−} {{z}}} \\…
Question Number 206764 by mustafazaheen last updated on 24/Apr/24 Answered by A5T last updated on 24/Apr/24 $${f}\left({g}\left(−\mathrm{3}\right)\right)={f}\left(\sqrt{−\mathrm{3}}\right)={f}\left(\sqrt{\mathrm{3}}{i}\right)=\left(\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} =−\mathrm{3} \\ $$ Commented by JDamian last updated…
Question Number 206746 by sonukgindia last updated on 23/Apr/24 Answered by A5T last updated on 23/Apr/24 $${f}\left(\mathrm{2}\right)+\mathrm{2}{f}\left(−\mathrm{1}\right)=\mathrm{2}…\left({i}\right) \\ $$$${f}\left(−\mathrm{1}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{1}…\left({ii}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}{f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}…\left({iii}\right) \\ $$$$\mathrm{2}×\left({iii}\right)−\left({ii}\right):\:\mathrm{4}{f}\left(\mathrm{2}\right)−{f}\left(−\mathrm{1}\right)=\mathrm{2}…\left({iv}\right) \\ $$$$\mathrm{2}×\left({iv}\right)+\left({i}\right):\:\mathrm{9}{f}\left(\mathrm{2}\right)=\mathrm{6}\Rightarrow{f}\left(\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{3}}…
Question Number 206674 by 073 last updated on 22/Apr/24 Answered by Frix last updated on 22/Apr/24 $$\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}} \mathrm{e}^{−{t}} {dt} \\ $$$$\frac{{d}\Gamma\left({x}\right)}{{dx}}=\Gamma'\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}}…
Question Number 206600 by naka3546 last updated on 20/Apr/24 Answered by aleks041103 last updated on 22/Apr/24 $$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\overset{\mathrm{3}{rd}\:{row}\:+\:\mathrm{1}{st}\:{row}} {\rightarrow}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\overset{\mathrm{1}{st}\:{row}\:+\:\mathrm{2}{nd}\:{row}} {\rightarrow}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}…
Question Number 206629 by aras18 last updated on 20/Apr/24 Answered by Frix last updated on 20/Apr/24 $${a}+{b}−{c}=\mathrm{170} \\ $$$${a}−{b}+{c}=\mathrm{130} \\ $$$$======= \\ $$$$\mathrm{2}{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{300} \\ $$$$\:\:\:{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{150}…
Question Number 206550 by SANOGO last updated on 18/Apr/24 $${calcul}\:\:{Residus}\:{de}\:{f}\:{en}\:{o} \\ $$$${f}\left({z}\right)=\frac{{ze}^{{z}} }{\left({z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Berbere last updated on 18/Apr/24 $${f}\:\:{n}'{a}\:{pas}\:{de}\:{poles}\:{en}\:\mathrm{0};{Res}\left({f},\mathrm{0}\right)=\mathrm{0} \\…