Question Number 180159 by Mastermind last updated on 08/Nov/22 $$\mathrm{Express}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{ze}^{\mathrm{iz}} \:\mathrm{in}\:\mathrm{polar} \\ $$$$\mathrm{form}\:\mathrm{and}\:\mathrm{separate}\:\mathrm{it}\:\mathrm{into}\:\mathrm{Real}\:\mathrm{and}\: \\ $$$$\mathrm{Imaginary}\:\mathrm{part}. \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$ Answered by Frix last…
Question Number 114615 by I want to learn more last updated on 19/Sep/20 Answered by Olaf last updated on 19/Sep/20 $${t}\:\mathrm{is}\:\mathrm{the}\:\mathrm{scheduled}\:\mathrm{time} \\ $$$${x}\:=\:{v}_{\mathrm{1}} {t}_{\mathrm{1}} \:=\:\mathrm{45}\left({t}+\frac{\mathrm{1}}{\mathrm{20}}\right)\:\left(\mathrm{1}\right)…
Question Number 180145 by ali009 last updated on 07/Nov/22 $${determine} \\ $$$$\left.\mathrm{1}\right)\mathcal{L}^{−} \left[\frac{\mathrm{4}{s}^{\mathrm{2}} −\mathrm{17}{s}−\mathrm{24}}{{s}\left({s}+\mathrm{3}\right)\left({s}−\mathrm{4}\right)}\right] \\ $$$$\left.\mathrm{2}\right)\mathcal{L}^{−} \left[\frac{\mathrm{5}{s}^{\mathrm{2}} −\mathrm{4}{s}−\mathrm{7}}{\left({s}−\mathrm{3}\right)\left({s}^{\mathrm{2}} +\mathrm{4}\right)}\right] \\ $$ Answered by Ar Brandon…
Question Number 180101 by Acem last updated on 07/Nov/22 $${Hello}\:{mr}.\:{Tinku}\:{Tara} \\ $$$$ \\ $$$${Please},\:{in}\:{the}\:{comments}\:{part},\:{putting}\:{the}\:{name} \\ $$$$\:{of}\:{the}\:{member}\:{whom}\:{the}\:{comment}\:{is}\:{for}\:{him} \\ $$$$ \\ $$$${Commentd}\:{by}\:{Acem}\:{on}\:{Mr}.\:{w}\:\:{as}\:{an}\:{example} \\ $$$$ \\ $$$${Thank}\:{you}! \\…
Question Number 180103 by Mastermind last updated on 07/Nov/22 $${Solve}\:\mathrm{2}{x}^{\mathrm{2}} =\mathrm{8} \\ $$ Commented by Frix last updated on 07/Nov/22 $$\mathrm{2}{x}^{\mathrm{2}} =\mathrm{8} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}…
Question Number 49028 by 123 45 polytechnicien last updated on 01/Dec/18 $${n}^{\mathrm{2}} −\mathrm{1}\:{est}\:{olympique} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 114561 by Akeyz last updated on 19/Sep/20 Answered by bemath last updated on 19/Sep/20 $$\mathrm{log}\:_{\left(\mathrm{log}\:_{\pi} {x}\right)} \:\left(\mathrm{log}\:_{{x}} \pi\right)\:=\:−\mathrm{1}=\mathrm{log}\:_{\left(\mathrm{log}\:_{\pi} {x}\right)} \left(\mathrm{log}\:_{\pi} {x}\right)^{−\mathrm{1}} \\ $$$$\mathrm{log}\:_{{x}}…
Question Number 49027 by 123 45 polytechnicien last updated on 01/Dec/18 $${omoxnn}\:{dit}\:{qu}\:{un}\:{entier}\:{k}\:{est}\:{olympique}\:{s}\:{il}\:{existe}\:\mathrm{4}\:{entiers}\:{a}\:{b}\:{c}\:{et}\:{d}\:{tous}\:{premiers}\:{avec}\:{k}\:{tel}\:{que}\:{k}\:{divise}\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} .{Soit}\:{n}\:{un}\:{entier}\:{naturel}\:{quelconque} \\ $$$${montrer}\:{que}\:{n}^{\mathrm{2}} −\mathrm{1}\:{divise}\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} \\ $$ Terms…
Question Number 48919 by vajpaithegrate@gmail.com last updated on 30/Nov/18 $$\mathrm{if}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{lengrhs}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{segment} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{focal}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{y}^{\mathrm{2}} =\mathrm{4ax}. \\ $$$$\mathrm{where}\:\mathrm{p},\mathrm{q}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{5}+^{} \sqrt{}\mathrm{2}\:\right)\mathrm{x}^{\mathrm{2}} −\left(\mathrm{4}+\sqrt{\mathrm{5}}\right)\mathrm{x}+\left(\mathrm{4}+\sqrt{}\mathrm{5}\right)=\mathrm{0}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{semi}\:\mathrm{letusrectum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\mathrm{is} \\ $$$$\mathrm{ans}:\mathrm{2}…
Question Number 48918 by vajpaithegrate@gmail.com last updated on 30/Nov/18 $$\mathrm{if}\:\mathrm{r}_{\mathrm{1}} \mathrm{andr}_{\mathrm{2}\:} \mathrm{are}\:\mathrm{the}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{smallest}\:\mathrm{and}\: \\ $$$$\mathrm{largest}\:\mathrm{circles}\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\: \\ $$$$\left(\mathrm{5},\mathrm{6}\right)\mathrm{and}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{r}_{\mathrm{1}} \mathrm{r}_{\mathrm{2}} = \\ $$$$\mathrm{ans}:\frac{\mathrm{41}}{\mathrm{4}} \\…