Question Number 11823 by Joel576 last updated on 01/Apr/17 $$\mathrm{3}\:+\:\mathrm{2}\left(\mathrm{3}^{\mathrm{2}} \right)\:+\:\mathrm{3}\left(\mathrm{3}^{\mathrm{3}} \right)\:+\:…\:+\:\mathrm{10}\left(\mathrm{3}^{\mathrm{10}} \right)\:=\:? \\ $$ Answered by ajfour last updated on 02/Apr/17 $${S}=\:\mathrm{3}+\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{3}\right)^{\mathrm{3}} +…+\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{10}}…
Question Number 77314 by aliesam last updated on 05/Jan/20 Commented by mind is power last updated on 05/Jan/20 $$\mathrm{first}\:\Psi\left(\frac{−\mathrm{2}}{\varphi}\right)\mathrm{is}\:\mathrm{psi}\:\mathrm{function}? \\ $$ Commented by aliesam last…
Question Number 77285 by jagoll last updated on 05/Jan/20 $$\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{Fourier}\:\mathrm{series}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}\:,\:\mathrm{0}\:<\:\mathrm{x}<\frac{\mathrm{1}}{\mathrm{8}} \\ $$ Answered by john santu last updated on 05/Jan/20 $$\mathrm{because}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}\: \\ $$$$,\:\mathrm{we}\:\mathrm{used}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{series}.…
Question Number 11743 by saa last updated on 30/Mar/17 $$\mathrm{9}/\mathrm{3}\left(\mathrm{6}×\mathrm{4}/\mathrm{8}\right)=? \\ $$ Answered by Joel576 last updated on 31/Mar/17 $$\frac{\mathrm{9}}{\mathrm{3}\left(\mathrm{6}\:.\:\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{9}}{\mathrm{3}\:.\:\mathrm{3}}\:=\:\mathrm{1} \\ $$ Terms of Service…
Question Number 11732 by agni5 last updated on 30/Mar/17 $$\mathrm{0}.\mathrm{16}\:\boldsymbol{\div}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{5}}\:\boldsymbol{\div}\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$ Answered by ajfour last updated on 30/Mar/17 $$\mathrm{8} \\ $$ Terms of Service…
Question Number 77132 by peter frank last updated on 03/Jan/20 Commented by kaivan.ahmadi last updated on 03/Jan/20 $${z}=\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{4}\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{6}{i}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}{i}=\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}{i}=\mathrm{4}\left(\sqrt{\mathrm{3}}+{i}\right)=\mathrm{8}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{8}{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\Rightarrow\frac{{z}}{\mathrm{8}}+{i}\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{3}}…
Question Number 77131 by peter frank last updated on 03/Jan/20 Answered by mr W last updated on 03/Jan/20 $${perpendicular}\:{tangents}\:{from}\:{P}\left({u},{v}\right): \\ $$$${y}={v}+{m}\left({x}−{u}\right)\:\Rightarrow{mx}−{y}+\left({v}−{mu}\right) \\ $$$${y}={v}−\frac{\mathrm{1}}{{m}}\left({x}−{u}\right)\:\Rightarrow{x}+{my}−\left({mv}+{u}\right) \\ $$$${from}\:{Q}\mathrm{77127}\:{we}\:{have}:…
Question Number 77128 by peter frank last updated on 03/Jan/20 $${Find}\:{the}\:{value}\:{of}\:{constant} \\ $$$$“{a}''\:{such}\:{that}\:{axe}^{−{x}\:} {is} \\ $$$${a}\:{solution}\:{of}\:{Differential} \\ $$$${equation} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}{y}=\mathrm{2}{e}^{−{x}} \\ $$$${solve}\:{D}.{E}\:{for}\:\:{which} \\…
Question Number 77126 by peter frank last updated on 03/Jan/20 $$\left.\mathrm{1}\right){Express}\:\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\:\:{in} \\ $$$${partial}\:{fraction} \\ $$$$\left.\mathrm{2}\right)\:{Solve} \\ $$$${xdy}+{ydy}−\left(\frac{{xdx}−{ydy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$ \\ $$ Answered…
Question Number 77127 by peter frank last updated on 03/Jan/20 $${Prove}\:{that}\:{line}\:{lx}+{my}+{n}=\mathrm{0} \\ $$$${is}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}\:} }=\mathrm{1}\:{if}\:{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$…