Question Number 118896 by bobhans last updated on 20/Oct/20 Answered by benjo_mathlover last updated on 20/Oct/20 $$\:\rightarrow\begin{pmatrix}{\mathrm{2}\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{{ac}^{\mathrm{2}} +{c}}\\{\:\:{c}−\mathrm{1}}\end{pmatrix} \\ $$$$\rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{4}−{b}^{\mathrm{2}} }\:\begin{pmatrix}{\mathrm{2}\:\:\:−{b}}\\{−{b}\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{{ac}^{\mathrm{2}} +{c}}\\{\:\:\:{c}−\mathrm{1}}\end{pmatrix} \\ $$$$\rightarrow\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{4}−{b}^{\mathrm{2}} }\:\begin{pmatrix}{\mathrm{2}{ac}^{\mathrm{2}}…
Question Number 184401 by mr W last updated on 06/Jan/23 $${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \:{with}\:{U}_{\mathrm{0}} =\mathrm{1} \\ $$$${find}\:{U}_{{n}\:} \:{in}\:{terms}\:{of}\:{n}\:\: \\ $$$$ \\ $$$$\left({question}\:{Q}\mathrm{173132}\:{reposted}\right) \\ $$…
Question Number 184384 by Shrinava last updated on 05/Jan/23 Commented by Rasheed.Sindhi last updated on 05/Jan/23 $$\mathrm{0}? \\ $$ Commented by Shrinava last updated on…
Question Number 118849 by mr W last updated on 20/Oct/20 Commented by PRITHWISH SEN 2 last updated on 20/Oct/20 $$\left(\mathrm{a}_{\mathrm{n}} +\mathrm{b}_{\mathrm{n}} \right)=\:\mathrm{3}^{\mathrm{n}} −\mathrm{1} \\ $$…
Question Number 184377 by mathlove last updated on 05/Jan/23 Commented by mathlove last updated on 05/Jan/23 $$ \\ $$Determine the distance between the first two…
Question Number 184351 by mr W last updated on 05/Jan/23 $${f}\left({x}\right)=\frac{{a}^{{x}} +{b}^{{x}} +{c}^{{x}} }{{x}}\:{with}\:{a}+{b}+{c}=\mathrm{0} \\ $$$${prove}\:{f}\left(\mathrm{7}\right)={f}\left(\mathrm{5}\right)×{f}\left(\mathrm{2}\right) \\ $$ Answered by greougoury555 last updated on 05/Jan/23…
Question Number 118790 by mathocean1 last updated on 19/Oct/20 $$\mathrm{Show}\:\mathrm{by}\:\mathrm{recurence}\:\mathrm{that} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}\:} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} ×\mathrm{a}^{\mathrm{k}} ×\mathrm{b}^{\mathrm{n}−\mathrm{k}} \\ $$ Terms of Service Privacy Policy…
Question Number 118780 by mathocean1 last updated on 19/Oct/20 $$\mathrm{z}\:\mathrm{and}\:\mathrm{z}'\:\in\:\mathbb{C}\:. \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\mathrm{1}.\:\:\:\:\:\:\overline {\mathrm{zz}'}=\overset{−} {\mathrm{z}}×\overline {\mathrm{z}'} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\:\overline {\left(\frac{\mathrm{z}}{\mathrm{z}'}\right)}=\frac{\overset{−} {\mathrm{z}}}{\overline {\mathrm{z}'}} \\ $$$$ \\…
Question Number 118775 by obaidullah last updated on 19/Oct/20 $$\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2} \\ $$$${X}=\frac{\mathrm{2}+\mathrm{2}+\mathrm{2}+\mathrm{2}+\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{10}}{\mathrm{5}}=\mathrm{2} \\ $$$${V}=\frac{\left(\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{5}} \\ $$$${Variance}=\mathrm{0} \\ $$ Terms of…
Question Number 118777 by obaidullah last updated on 19/Oct/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}{n}\pi}\:{n}^{{n}} }{{n}^{{n}} ×{e}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{e}}\left(\sqrt{\mathrm{2}{n}\pi}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{e}} \\ $$ Answered by…