Question Number 105181 by mohammad17 last updated on 26/Jul/20 $${graph}\:{the}\:{function}\:{x}^{\mathrm{2}} +\left({y}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Answered by bemath last updated on 26/Jul/20 Commented by mr…
Question Number 39638 by jasno91 last updated on 09/Jul/18 Answered by $@ty@m last updated on 09/Jul/18 $${P}=\mathrm{2}\left(\mathrm{30}+\mathrm{24}\right) \\ $$$$\:\:\:\:=\mathrm{118}{m} \\ $$ Terms of Service Privacy…
Question Number 105167 by mohammad17 last updated on 26/Jul/20 Commented by bemath last updated on 26/Jul/20 Answered by Aziztisffola last updated on 26/Jul/20 $$\int_{−\infty} ^{\:\infty}…
Question Number 105155 by mohammad17 last updated on 26/Jul/20 Answered by mathmax by abdo last updated on 26/Jul/20 $$\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}}…
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 $$\mid{a}\mid=\sqrt{\mathrm{1}+\mathrm{3}}=\mathrm{2}\:\Rightarrow{a}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right) \\ $$$$=\mathrm{2}\:{e}^{{i}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)} \:\Rightarrow{a}^{\frac{\mathrm{1}}{\mathrm{5}}}…
Question Number 105132 by mohammad17 last updated on 26/Jul/20 Answered by Dwaipayan Shikari last updated on 26/Jul/20 $$\left.\mathrm{1}\right)\int\mathrm{5}^{\left({x}^{\mathrm{2}} +\mathrm{6}\right)} \left({x}+\mathrm{3}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{5}^{{u}} {du}\:\:\:\:\:\:\:\:\left\{{u}={x}^{\mathrm{2}} +\mathrm{6}{x}\:\:\:\frac{{du}}{{dx}}=\mathrm{2}{x}+\mathrm{6}\right. \\…
Question Number 170646 by Gbenga last updated on 28/May/22 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{can}}\:\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{done}} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 39529 by Rio Mike last updated on 07/Jul/18 Commented by Rio Mike last updated on 07/Jul/18 $${i}\:{need}\:{help}\:{guys}.{can}\:{someone}\:{pleasd} \\ $$$${explain}\:{to}\:{me}\:\:{when}\:{to}\:{use}\:{each} \\ $$$${of}\:{the}\:{above}\:{quadrants}\:{and}\:{how}?\: \\ $$$${in}\:{trigonometry}\:{i}\:{can}\:{solve}\:{the}\:{equation}…
Question Number 170572 by daus last updated on 27/May/22 Answered by mr W last updated on 27/May/22 $$\frac{{k}+\mathrm{2}}{{k}!+\left({k}+\mathrm{1}\right)!+\left({k}+\mathrm{2}\right)!} \\ $$$$=\frac{{k}+\mathrm{2}}{{k}!\left({k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{{k}!\left({k}+\mathrm{2}\right)} \\ $$$$=\frac{{k}+\mathrm{1}}{\left({k}+\mathrm{2}\right)!}…