Question Number 37231 by abdo.msup.com last updated on 11/Jun/18 $${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}\:\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{p}_{{c}} \left({A}\right)\:{the}\:{caracteristic}\: \\ $$$${polunom}\:{of}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{3}\right)\:{calcypulate}\:{e}^{{tA}} \:\:\:\:\:{t}\in\:{R}\:\: \\ $$…
Question Number 37232 by abdo.msup.com last updated on 11/Jun/18 $${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{3}\:\right) \\ $$$${calculate}\:{A}^{{n}} \:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 37230 by abdo.msup.com last updated on 11/Jun/18 $${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}}\end{pmatrix} \\ $$$${calculate}\:\:{A}^{{n}} \: \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:\:,\:{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:{e}^{{iA}} ,\:{e}^{−{iA}} \:\:\:{and}\:{e}^{{iA}} \:+{e}^{−{iA}} \:\:. \\ $$ Terms…
Question Number 37228 by abdo.msup.com last updated on 11/Jun/18 $${E}\:{id}\:{k}\:{vectorial}\:{space}\:{and}\:{f}\in{L}\left({E}\right) \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{if}\:{f}\:{is}\:{nilpotent}\:{with}\:{indice} \\ $$$${p}\geqslant\mathrm{1}\:,{I}\:−{f}\:{is}\:{bijective}\:{and} \\ $$$$\left({I}−{f}\right)^{−\mathrm{1}} =\sum_{{i}=\mathrm{0}} ^{{p}−\mathrm{1}} {f}^{{i}} \\ $$$$\left.\mathrm{2}\right){let}\:{E}={R}_{{n}} \left[{x}\right]\:{and}\:{f}\in{L}\left({E}\right)\:/ \\ $$$${f}\left({p}\right)\:={p}−{p}^{'} \:\:{prove}\:{that}\:{f}\:{is}\:{inversible}…
Question Number 168259 by peter frank last updated on 07/Apr/22 $$\mathrm{If}\:\mathrm{the}\:\mathrm{odds}\:\mathrm{in}\:\mathrm{favour}\:\mathrm{of}\:\mathrm{an}\:\mathrm{event} \\ $$$$\mathrm{be}\:\frac{\mathrm{1}}{\mathrm{3}}.\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\: \\ $$$$\mathrm{an}\:\mathrm{occurrence}\:\mathrm{of}\:\mathrm{the}\:\mathrm{event}. \\ $$ Answered by som(math1967) last updated on 07/Apr/22 $${if}\:{the}\:{probability}\:{of}\:{occurance}\:…
Question Number 168248 by mathlove last updated on 07/Apr/22 Commented by MJS_new last updated on 07/Apr/22 $$\mathrm{6}^{\mathrm{4}} >\mathrm{405}\:\Rightarrow\:{x}<\mathrm{4} \\ $$$$\mathrm{trying}\:{x}=\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\Rightarrow \\ $$$${x}=\mathrm{3} \\ $$ Terms…
Question Number 37166 by nishant last updated on 09/Jun/18 $${if}\:\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1} \\ $$$${can}\:{be}\:{resolved}\:\:{into}\:\:{two}\:\:{linear} \\ $$$${factors},\:\:{prove}\:\:{that}\:\:'\alpha'\:\:{is}\:{a}\:{root}\: \\ $$$${of}\:{the}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0} \\ $$ Commented by prakash…
Question Number 168229 by mathlove last updated on 06/Apr/22 Answered by mr W last updated on 06/Apr/22 $${x}=\mathrm{2}: \\ $$$${P}\left(\mathrm{0}\right)−{P}\left(\mathrm{0}\right)=\mathrm{4}{a}−\mathrm{2}{b}+\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}−{b}+\mathrm{6}=\mathrm{0} \\ $$$${x}=\mathrm{4}: \\…
Question Number 37139 by nishant last updated on 09/Jun/18 $${if}\:\alpha\:,\:\beta\:\:{are}\:{the}\:{roots}\:{of}\:{the}\:{quadratic} \\ $$$${equation}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\mathrm{0}\:{then}\:\:{find} \\ $$$${the}\:{quadratic}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\:\alpha^{\mathrm{2}\:\:\:} ,\:\beta^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$ Answered…
Question Number 168210 by mathlove last updated on 06/Apr/22 Answered by mr W last updated on 06/Apr/22 $${y}={x}^{\mathrm{3}} +{ex} \\ $$$${x}^{\mathrm{3}} +{ex}−{y}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{{e}^{\mathrm{3}} }{\mathrm{27}}+\frac{{y}^{\mathrm{2}}…