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 Number 64356 by udu last updated on 17/Jul/19 $$ \\ $$ Answered by MJS last updated on 17/Jul/19 $$\mathrm{42} \\ $$ Terms of Service…
Question Number 129890 by 0731619177 last updated on 20/Jan/21 Answered by Olaf last updated on 20/Jan/21 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{{xx}−\mathrm{33}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{11}\left({x}−\mathrm{3}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{11}} \\…
Question Number 64355 by turbo msup by abdo last updated on 17/Jul/19 $${let}\:{F}\left({x}\right)=\int_{{u}\left({x}\right)} ^{{v}\left({x}\left\{\right.\right.} {f}\left({x},{t}\right){dt} \\ $$$${how}\:{to}\:{calculate}\:\:\frac{{dF}}{{dx}}\left({x}\right)? \\ $$ Commented by MJS last updated on…
Question Number 64354 by Rio Michael last updated on 17/Jul/19 $${some}\:{one}\:{write}\:{the}\:{statement} \\ $$$$\:{a}\:\equiv−{a}\left({mod}\:{m}\right)\: \\ $$$${show}\:{that}\:{this}\:{statement}\:{is}\:{not}\:{generally}\:{true}.!\:{giving}\:{a}\:{counter} \\ $$$${example} \\ $$ Answered by MJS last updated on…
Question Number 129889 by 0731619177 last updated on 20/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129886 by zarawan last updated on 23/Jan/21 $${Find}\:{Null}\:{space}\:{of}\:{the}\:{following}\:{matrix}\:{and}\:{also}\:{find}\:{basis}\:{for}\:{the}\:{null}\:{space}. \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}&{\mathrm{1}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129887 by zarawan last updated on 20/Jan/21 $${Is}\:{the}\:{vector}\:\begin{bmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{bmatrix}{an}\:{eigen}\:{vector}\:{of}\:\begin{bmatrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{7}}\\{\mathrm{5}}&{\mathrm{6}}&{\mathrm{5}\:}\end{bmatrix}?\:{if}\:{sp}\:{find}\:{the}\:{corresponding}\:{eigen}\:{value}? \\ $$ Answered by Olaf last updated on 20/Jan/21 $$\mathrm{AX}\:=\:\begin{bmatrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{7}}\\{\mathrm{5}}&{\mathrm{6}}&{\mathrm{5}}\end{bmatrix}\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix} \\ $$$$=\:−\mathrm{2}\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:=\:−\mathrm{2X} \\ $$$$\exists\lambda\backslash\:\mathrm{AX}\:=\:\lambda\mathrm{X} \\…
Question Number 64350 by Rio Michael last updated on 16/Jul/19 $${Given}\:{that}\:{f}\left({x}\right)\:=\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{2}} }&{{x}^{\mathrm{3}} }\\{\mathrm{1}}&{\mathrm{2}{x}}&{\mathrm{3}{x}^{\mathrm{2}} }\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{6}{x}}\end{vmatrix},\:{find}\:{f}\:'\:\left({x}\right) \\ $$ Commented by Tony Lin last updated on 17/Jul/19 $${f}\left({x}\right)=\mathrm{12}{x}^{\mathrm{3}}…
Question Number 64348 by Rio Michael last updated on 16/Jul/19 $${the}\:{vectors}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{are}\:{such}\:{that}\:\mid\boldsymbol{{a}}\mid\:=\mathrm{3}\:,\:\mid\boldsymbol{{b}}\mid=\mathrm{5}\:{and}\:\boldsymbol{{a}}.\boldsymbol{{b}}=−\mathrm{14} \\ $$$${find}\:\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$ Answered by Tanmay chaudhury last updated on 17/Jul/19 $$\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} =\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)…