Question Number 77132 by peter frank last updated on 03/Jan/20 Commented by kaivan.ahmadi last updated on 03/Jan/20 $${z}=\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{4}\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{6}{i}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}{i}=\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}{i}=\mathrm{4}\left(\sqrt{\mathrm{3}}+{i}\right)=\mathrm{8}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{8}{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\Rightarrow\frac{{z}}{\mathrm{8}}+{i}\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{3}}…
Question Number 11597 by tawa last updated on 28/Mar/17 $$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}^{\sqrt{\mathrm{x}}} } ,\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$ Answered by sma3l2996 last updated on 28/Mar/17 $${y}={x}^{{e}^{\sqrt{{x}}{ln}\left({x}\right)} } ={e}^{{e}^{\sqrt{{x}}{ln}\left({x}\right)} {ln}\left({x}\right)}…
Question Number 77131 by peter frank last updated on 03/Jan/20 Answered by mr W last updated on 03/Jan/20 $${perpendicular}\:{tangents}\:{from}\:{P}\left({u},{v}\right): \\ $$$${y}={v}+{m}\left({x}−{u}\right)\:\Rightarrow{mx}−{y}+\left({v}−{mu}\right) \\ $$$${y}={v}−\frac{\mathrm{1}}{{m}}\left({x}−{u}\right)\:\Rightarrow{x}+{my}−\left({mv}+{u}\right) \\ $$$${from}\:{Q}\mathrm{77127}\:{we}\:{have}:…
Question Number 11594 by JAZAR last updated on 28/Mar/17 $${please}\:{how}\:{can}\:{demonstred}\: \\ $$$${sin}\left(\mathrm{2}{x}\underset{} {\right)}/{cos}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right) \\ $$ Answered by mrW1 last updated on 28/Mar/17 $${that}'{s}\:{not}\:{true}! \\ $$$$\mathrm{sin}\:\left(\mathrm{2}{x}\right)/\mathrm{cos}\:\left(\mathrm{2}{x}\right)=\mathrm{tan}\:\left(\mathrm{2}{x}\right)\neq\mathrm{2sin}\:\left(\mathrm{2}{x}\right)…
Question Number 142667 by mnjuly1970 last updated on 03/Jun/21 Answered by mindispower last updated on 03/Jun/21 $${T}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left(\mathrm{1}+{x}\right){dx}=\frac{{ln}\left(\mathrm{2}\right)}{{n}+\mathrm{1}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}.\frac{{dx}}{\mathrm{1}+{x}} \\ $$$$\mathrm{0}\leqslant\int_{\mathrm{0}}…
Question Number 77128 by peter frank last updated on 03/Jan/20 $${Find}\:{the}\:{value}\:{of}\:{constant} \\ $$$$“{a}''\:{such}\:{that}\:{axe}^{−{x}\:} {is} \\ $$$${a}\:{solution}\:{of}\:{Differential} \\ $$$${equation} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}{y}=\mathrm{2}{e}^{−{x}} \\ $$$${solve}\:{D}.{E}\:{for}\:\:{which} \\…
Question Number 77129 by mathocean1 last updated on 03/Jan/20 $$\mathrm{solve}\:\mathrm{in}\:\mathrm{R} \\ $$$$\mid\mathrm{tan2}{x}\mid−\sqrt{\mathrm{3}}\geqslant\mathrm{0} \\ $$ Answered by jagoll last updated on 03/Jan/20 Commented by mathocean1 last…
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 11591 by agni5 last updated on 28/Mar/17 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intercept}\:\mathrm{of}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1}\:\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{x}-\mathrm{axis}\:? \\ $$ Answered by ajfour last updated on 28/Mar/17…
Question Number 77127 by peter frank last updated on 03/Jan/20 $${Prove}\:{that}\:{line}\:{lx}+{my}+{n}=\mathrm{0} \\ $$$${is}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}\:} }=\mathrm{1}\:{if}\:{a}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$…