Question Number 58962 by maxmathsup by imad last updated on 02/May/19 $${you}\:{are}\:{welcome}\:{sir}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124487 by Backer last updated on 03/Dec/20 Commented by Backer last updated on 03/Dec/20 $$\mathrm{Salut}\: \\ $$$$\mathrm{Pouviez}−\mathrm{vous}\:\mathrm{m}'\mathrm{aider}? \\ $$ Answered by Olaf last…
Question Number 124474 by sogol last updated on 03/Dec/20 $$\left(\frac{\mathrm{4}}{−\mathrm{6}+{i}\sqrt{\mathrm{5}}}\right)^{\mathrm{4}} =??? \\ $$$$ \\ $$$${polar}???? \\ $$ Commented by MJS_new last updated on 03/Dec/20 $$\mathrm{no}\:\mathrm{need}\:\mathrm{to}\:\mathrm{convert}\:\mathrm{to}\:\mathrm{polar}…
Question Number 58928 by mr W last updated on 01/May/19 Commented by mr W last updated on 01/May/19 $${Find}\:{area}\:{of}\:{red}. \\ $$ Commented by MJS last…
Question Number 124461 by Dwaipayan Shikari last updated on 03/Dec/20 $$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{7}} } {sin}\left({x}^{\mathrm{7}} \right){dx} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 124459 by Dwaipayan Shikari last updated on 03/Dec/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\right)+… \\ $$ Commented by Dwaipayan Shikari last updated on 03/Dec/20 $${I}\:{have}\:{found}\:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$ Answered…
Question Number 58880 by jayanta10 last updated on 01/May/19 $${t}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} }\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:{t} \\ $$ Commented by gunawan last updated on 01/May/19 $${t}=\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{2}}}×\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}=−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$ Terms of…
Question Number 124347 by Dwaipayan Shikari last updated on 02/Dec/20 Commented by Dwaipayan Shikari last updated on 02/Dec/20 $${What}\:{will}\:\:{be}\:{the}\:{approach}\:{to}\:{this}\:{problem}? \\ $$ Answered by mindispower last…
Question Number 189865 by uchihayahia last updated on 23/Mar/23 $$ \\ $$$$\:{i}\:{know}\:{b}_{{n}} =\mathrm{0},\:{but}\:{a}_{\mathrm{0}} =\frac{\mathrm{4}}{\mathrm{3}}{F}_{\mathrm{0}} \left({according}\:\right. \\ $$$$\left.\:{to}\:{solution}\right)\:{my}\:{answer}\:{is}\:{a}_{\mathrm{0}} =\frac{\mathrm{8}}{\mathrm{3}}{F}_{\mathrm{0}} \\ $$$$\:{where}\:{did}\:{i}\:{did}\:{wrong}\:{how}\:{to} \\ $$$${find}\:{answer}\:\left({Fourier}\:{transformation}\right) \\ $$$$\:{like}\:{picture}\:{below}\:\left({in}\:{the}\:{comment}?\right. \\…
Question Number 124320 by sogol last updated on 02/Dec/20 $${z}+\frac{\mathrm{1}}{{z}}=\mathrm{1} \\ $$$${z}^{\mathrm{3}} =?? \\ $$ Answered by Dwaipayan Shikari last updated on 02/Dec/20 $${z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0}\Rightarrow{z}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\pm\frac{{i}\pi}{\mathrm{3}}}…