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Question-129955

Question Number 129955 by mohammad17 last updated on 21/Jan/21 Answered by Olaf last updated on 21/Jan/21 $$\mathrm{A}\:=\:\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{15}}&{\mathrm{4}}\\{\mathrm{16}}&{−\mathrm{2}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{17}}\\{\mathrm{2}}&{−\mathrm{14}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{A}\right)\:=\:−\mathrm{56460}\:\neq\:\mathrm{0} \\ $$$$\mathrm{A}_{\mathrm{1}} \:=\:\begin{bmatrix}{−\mathrm{2}}&{\mathrm{1}}&{−\mathrm{15}}&{\mathrm{4}}\\{\mathrm{2}}&{−\mathrm{2}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{9}}&{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{17}}\\{−\mathrm{8}}&{−\mathrm{14}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{A}_{\mathrm{1}} \right)\:=\:−\mathrm{14115}…

Question-64384

Question Number 64384 by aliesam last updated on 17/Jul/19 Answered by MJS last updated on 17/Jul/19 $$\left({y}={x}\right)\cap\left({y}=\frac{{x}}{\mathrm{8}}\right)=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\left({y}={x}\right)\cap\left({y}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\right] \\…

Question-129892

Question Number 129892 by mohammad17 last updated on 20/Jan/21 Answered by Ar Brandon last updated on 20/Jan/21 $$\mathrm{GP}\:\mathrm{with}\:\mathrm{U}\left(\mathrm{1}\right)=\frac{\mathrm{11x}}{\mathrm{12}}\:\mathrm{and}\:\mathrm{r}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{S}_{\infty} =\frac{\frac{\mathrm{11}}{\mathrm{12}}\mathrm{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{12}}}=\mathrm{x} \\ $$ Terms of…

Question-129903

Question Number 129903 by 0731619177 last updated on 20/Jan/21 Commented by Dwaipayan Shikari last updated on 20/Jan/21 $$\int{e}^{{x}^{\mathrm{2}} } {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{1}} =\frac{{x}}{\mathrm{1}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{10}}+\frac{{x}^{\mathrm{7}}…

Question-129890

Question Number 129890 by 0731619177 last updated on 20/Jan/21 Answered by Olaf last updated on 20/Jan/21 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{{xx}−\mathrm{33}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{11}\left({x}−\mathrm{3}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{11}} \\…