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3-sin-2-40-1-cos-2-40-64-sin-2-40-

Question Number 170737 by naka3546 last updated on 30/May/22 $$\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{2}} \mathrm{40}°}\:−\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{40}°}\:+\:\mathrm{64}\:\mathrm{sin}^{\mathrm{2}} \mathrm{40}°\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 30/May/22 $$\frac{\mathrm{3}}{{sin}^{\mathrm{2}} \mathrm{40}}\:+\frac{\mathrm{64}{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}}…

Question-105195

Question Number 105195 by mohammad17 last updated on 26/Jul/20 Answered by Aziztisffola last updated on 26/Jul/20 $$\mathrm{L}\left(\mathrm{2y}''+\mathrm{3y}'−\mathrm{2y}\right)=\mathrm{L}\left(\mathrm{te}^{−\mathrm{2t}} \right) \\ $$$$\mathrm{2s}^{\mathrm{2}} \mathrm{Y}\left(\mathrm{s}\right)−\mathrm{2sy}\left(\mathrm{0}\right)−\mathrm{2y}\left(\mathrm{0}\right)+\mathrm{3sY}\left(\mathrm{s}\right)−\mathrm{3y}\left(\mathrm{0}\right)−\mathrm{2Y}\left(\mathrm{s}\right)=−\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathrm{L}\left(\mathrm{e}^{−\mathrm{2t}} \right)\right) \\ $$$$\left(\mathrm{2s}^{\mathrm{2}} +\mathrm{3s}−\mathrm{2}\right)\mathrm{Y}\left(\mathrm{s}\right)+\mathrm{4}=−\frac{\mathrm{d}}{\mathrm{ds}}\left(\frac{\mathrm{1}}{\mathrm{s}+\mathrm{2}}\right)…

Question-39651

Question Number 39651 by KMA last updated on 09/Jul/18 $$ \\ $$ Answered by MJS last updated on 09/Jul/18 $$−\frac{\mathrm{2}}{\mathrm{25e}^{\sqrt{{x}}} }\left(\mathrm{2}\left(\mathrm{5}\sqrt{{x}}+\mathrm{2}\right)\mathrm{cos}\left(\mathrm{2}\sqrt{{x}}−\mathrm{3}\right)+\left(\sqrt{\mathrm{5}}{x}−\mathrm{3}\right)\mathrm{sin}\left(\mathrm{2}\sqrt{{x}}−\mathrm{3}\right)\right)+{C} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{question}\:\mathrm{LOL} \\ $$…

Question-105155

Question Number 105155 by mohammad17 last updated on 26/Jul/20 Answered by mathmax by abdo last updated on 26/Jul/20 $$\int_{−\infty} ^{+\infty} \:\pi\:\mathrm{e}^{−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}} \mathrm{d}\alpha\:=\pi\:\int_{−\infty} ^{+\infty\:} \:\mathrm{e}^{−\frac{\alpha^{\mathrm{2}}…

if-a-1-i-3-find-a-1-5-

Question Number 170675 by MathsFan last updated on 29/May/22 $$\boldsymbol{\mathrm{if}}\:\:\boldsymbol{\mathrm{a}}=−\mathrm{1}−\boldsymbol{{i}}\sqrt{\mathrm{3}} \\ $$$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$ Answered by Mathspace last updated on 29/May/22 $$\mid{a}\mid=\sqrt{\mathrm{1}+\mathrm{3}}=\mathrm{2}\:\Rightarrow{a}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right) \\ $$$$=\mathrm{2}\:{e}^{{i}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)} \:\Rightarrow{a}^{\frac{\mathrm{1}}{\mathrm{5}}}…

Question-105132

Question Number 105132 by mohammad17 last updated on 26/Jul/20 Answered by Dwaipayan Shikari last updated on 26/Jul/20 $$\left.\mathrm{1}\right)\int\mathrm{5}^{\left({x}^{\mathrm{2}} +\mathrm{6}\right)} \left({x}+\mathrm{3}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{5}^{{u}} {du}\:\:\:\:\:\:\:\:\left\{{u}={x}^{\mathrm{2}} +\mathrm{6}{x}\:\:\:\frac{{du}}{{dx}}=\mathrm{2}{x}+\mathrm{6}\right. \\…