Question Number 197001 by josemate19 last updated on 05/Sep/23 $$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{−\pi} {\overset{\:\:\:\pi} {\int}}\frac{{n}!\mathrm{2}^{\mathrm{2}{ncos}\left(\phi\right)} }{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{ne}^{{i}\phi} −{k}\right)}{d}\phi \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 196968 by sonukgindia last updated on 05/Sep/23 Answered by MM42 last updated on 05/Sep/23 $${if}\:\:{x},{y}\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{5}{s}−{p}=\mathrm{10}+\mathrm{7}\sqrt{\mathrm{3}}}\\{{s}^{\mathrm{2}} −\mathrm{2}{p}=\mathrm{7}}\end{cases} \\ $$$$\Rightarrow{s}^{\mathrm{2}} +\mathrm{10}{s}−\mathrm{27}−\mathrm{14}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{2}+\sqrt{\mathrm{3}\:}\Rightarrow{p}=\mathrm{2}\sqrt{\mathrm{3}}\:…
Question Number 196965 by sonukgindia last updated on 05/Sep/23 Answered by qaz last updated on 05/Sep/23 $$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\right)}\overset{{x}\rightarrow{x}^{\mathrm{4}} } {=}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}\right)} \\…
Question Number 196940 by sonukgindia last updated on 04/Sep/23 Answered by MM42 last updated on 04/Sep/23 $${sin}\mathrm{8}{x}=\mathrm{8}{sinxcosxcos}\mathrm{2}{xcos}\mathrm{4}{x} \\ $$$$\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cosxcos}\mathrm{2}{xcos}\mathrm{4}{xdx} \\ $$$${cosxcos}\mathrm{2}{xcos}\mathrm{4}{x}=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{7}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{3}{x}+{cosx}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{7}}{sin}\mathrm{7}{x}+\frac{\mathrm{1}}{\mathrm{5}}{sin}\mathrm{5}{x}+\frac{\mathrm{1}}{\mathrm{3}}{sin}\mathrm{3}{x}+{sinx}\right)\mid_{\mathrm{0}}…
Question Number 196957 by otchereabdullai@gmail.com last updated on 04/Sep/23 Answered by dimentri last updated on 05/Sep/23 $$\:\:\:\:\frac{\mathrm{9}!}{{x}!\:\left(\mathrm{9}−{x}\right)!}\:=\:\mathrm{4}\left(\frac{\mathrm{7}!}{\left({x}−\mathrm{1}\right)!\left(\mathrm{8}−{x}\right)!}\right) \\ $$$$\:\:\:\:\frac{\mathrm{72}}{{x}\left(\mathrm{9}−{x}\right)}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\mathrm{18}=\:\mathrm{9}{x}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{18}=\mathrm{0} \\…
Question Number 196954 by Khalmohmmad last updated on 04/Sep/23 Answered by MM42 last updated on 05/Sep/23 $$\left.\mathrm{1}\right){x}×\mathrm{2}^{{m}} ×\mathrm{5}^{{n}} =\mathrm{24}\Rightarrow{n}=\mathrm{0} \\ $$$$\left.{a}\right){x}×\mathrm{2}^{{m}} =\mathrm{24}×\mathrm{1}\Rightarrow{x}=\mathrm{24\&}{m}=\mathrm{0}\Rightarrow\mathrm{2}{x}+\mathrm{3}{n}−{m}=\mathrm{48} \\ $$$$\left.{b}\right){x}×\mathrm{2}^{{m}} =\mathrm{12}×\mathrm{2}\Rightarrow{x}=\mathrm{12\&}{m}=\mathrm{1}\Rightarrow\mathrm{2}{x}+\mathrm{3}{n}−{m}=\mathrm{23}…
Question Number 196953 by Khalmohmmad last updated on 04/Sep/23 Answered by MM42 last updated on 05/Sep/23 $${p}×\mathrm{2}^{{n}} =\mathrm{15}\Rightarrow{p}=\mathrm{15\&}{n}=\mathrm{0}\Rightarrow\mathrm{2}{p}−\mathrm{3}{n}=\mathrm{30} \\ $$ Terms of Service Privacy Policy…
Question Number 196938 by sonukgindia last updated on 04/Sep/23 Answered by AST last updated on 04/Sep/23 $${xy}+{yz}+{zx}\leqslant\mathrm{147}\left({equality}\:{when}\:{x}={y}={z}=\underset{−} {+}\mathrm{7}\right) \\ $$$$\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\mathrm{3}}\right)\geqslant\left(\frac{{x}+{y}+{z}}{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow\mathrm{3}\left(\mathrm{147}\right)\geqslant\left({x}+{y}+{z}\right)^{\mathrm{2}} \\…
Question Number 196955 by mokys last updated on 04/Sep/23 Commented by mokys last updated on 05/Sep/23 $$????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 196946 by Khalmohmmad last updated on 04/Sep/23 Answered by trisetyo last updated on 04/Sep/23 $$\mathrm{3}^{{x}} +\mathrm{2}^{{x}} =\mathrm{5} \\ $$$$\mathrm{3}^{{x}} =−\left(\mathrm{2}^{{x}} −\mathrm{5}\right)\vee\mathrm{2}^{{x}} =−\left(\mathrm{3}^{{x}} −\mathrm{5}\right)…