Question Number 130478 by liberty last updated on 26/Jan/21 $$\:{Integrate}\:{the}\:{function}\:{f}\left({x},{y}\right)={xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$${over}\:{the}\:{domain}\:{R}=\left\{−\mathrm{3}\leqslant{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \leqslant\mathrm{3},\:\mathrm{1}\leqslant{y}\leqslant\mathrm{4}\:\right\} \\ $$ Answered by EDWIN88 last updated on 26/Jan/21…
Question Number 130476 by shaker last updated on 26/Jan/21 Answered by TheSupreme last updated on 26/Jan/21 $$\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\mathrm{1} \\ $$$${f}\left({x}\right)=\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} \\ $$$${f}'\left({x}\right)<\mathrm{0}\:\forall{x}\in\mathbb{R} \\…
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Question Number 130474 by EDWIN88 last updated on 26/Jan/21 $$\mathcal{E}\:=\:\int_{\:\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\frac{\mathrm{dy}}{\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{5}/\mathrm{2}} }\: \\ $$ Answered by liberty last updated on 26/Jan/21 Terms of…
Question Number 130472 by shaker last updated on 26/Jan/21 Answered by mathmax by abdo last updated on 26/Jan/21 $$\mathrm{let}\:\mathrm{w}\left(\mathrm{x}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{2k}−\mathrm{1}\right)\mathrm{x}^{\mathrm{k}} \:\Rightarrow\mathrm{w}\left(\mathrm{x}\right)=\mathrm{2}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{kx}^{\mathrm{k}} −\sum_{\mathrm{k}=\mathrm{1}}…
Question Number 130473 by liberty last updated on 26/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 130469 by liberty last updated on 26/Jan/21 $$\:{P}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$ Answered by EDWIN88 last updated on 26/Jan/21 $$\mathrm{2psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{4psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{8psin}\:\frac{\pi}{\mathrm{15}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\…
Question Number 130465 by naka3546 last updated on 25/Jan/21 $${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{4}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:+\:\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)} \\ $$ Answered by Dwaipayan Shikari last updated on 26/Jan/21 $$\frac{\mathrm{4}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\pi}{\mathrm{14}}\right){sin}\left(\frac{\pi}{\mathrm{7}}\right)+{sin}\left(\frac{\pi}{\mathrm{7}}\right)}{{cos}\frac{\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}} \\ $$$$=\frac{\mathrm{2}{cos}\frac{\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}−\mathrm{2}{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}+{sin}\frac{\pi}{\mathrm{14}}}{{cos}\frac{\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}}…
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Question Number 130455 by ayoubbacmath0 last updated on 25/Jan/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\mid\mathrm{e}^{\mathrm{x}} −\mathrm{2}\mid \\ $$$$\begin{cases}{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\:\:\:\:\:\:\mathrm{x}\in\right]\mathrm{ln2};+\infty\left[\right.}\\{\left.\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\:\:\:\mathrm{x}\in\right]−\infty;\mathrm{ln2}\left[\right.}\end{cases} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{2}\right)\right)=+\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{ln}\left(−\mathrm{e}^{\mathrm{x}} +\mathrm{2}\right)\right)=−\infty…