Question Number 191342 by mnjuly1970 last updated on 23/Apr/23 $$ \\ $$$$\:\:\:\:{calculate}… \\ $$$$\:\Omega\:=\left\{\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left(\:\:\mathrm{1}\:+\:\sqrt{\mathrm{3}}\:{sin}\left({x}\right)\:+\:{cos}\left({x}\right)\:\right)^{\:{n}} {dx}\right\}^{\frac{\mathrm{1}}{{n}}} =\:? \\ $$$$ \\ $$$$ \\ $$ Answered…
Question Number 125800 by TITA last updated on 13/Dec/20 Commented by TITA last updated on 13/Dec/20 $$\mathrm{prove} \\ $$ Commented by liberty last updated on…
Question Number 60264 by maxmathsup by imad last updated on 19/May/19 $${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\mathrm{3}\:\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dx}\:\:{with}\:{t}>\mathrm{0} \\ $$$$\mathrm{1}.\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\mathrm{2}.\:{find}\:{also}\:{g}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{3}\left[{x}^{\mathrm{2}} \right]}…
Question Number 60263 by maxmathsup by imad last updated on 19/May/19 $${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{n}\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{n}\:{U}_{{n}} \\…
Question Number 191335 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $${radius}\:{of}\:{big}\:{quatercircle}\:={a} \\ $$$${radius}\:{of}\:{big}\:{semicircle}\:={b} \\ $$$${radius}\:{of}\:{small}\:{quatercircle}\:={c} \\ $$$${b}^{\mathrm{2}} +\left(\mathrm{2}{b}\right)^{\mathrm{2}}…
Question Number 60257 by rahul 19 last updated on 19/May/19 $${The}\:{maximum}\:{value}\:{of}\:{Z}=\mathrm{4}{x}+\mathrm{2}{y} \\ $$$${subject}\:{to}\:{constraints}\:\mathrm{2}{x}+\mathrm{3}{y}\leqslant\mathrm{18}\:, \\ $$$${x}+{y}\geqslant\mathrm{10}\:{and}\:{x},{y}\geqslant\mathrm{0}\:{is}\:? \\ $$ Commented by rahul 19 last updated on 19/May/19…
Question Number 60256 by rahul 19 last updated on 19/May/19 Answered by mr W last updated on 19/May/19 $$\mathrm{2}{x}+\mathrm{4}{y}=\mathrm{2}\left({x}+{y}\right)+\mathrm{2}{y}\leqslant\mathrm{8}+\mathrm{2}{y} \\ $$$$\leqslant\mathrm{8}+\mathrm{2}\left(\mathrm{4}−{x}\right)=\mathrm{16}−\mathrm{2}{x}\leqslant\mathrm{16}\:! \\ $$$$ \\ $$$${z}=\mathrm{2}{x}+\mathrm{5}{y}=\mathrm{2}\left({x}+{y}\right)+\mathrm{3}{y}\leqslant\mathrm{8}+\mathrm{3}{y}…
Question Number 191325 by Mingma last updated on 23/Apr/23 Answered by HeferH last updated on 23/Apr/23 $$\mathrm{16}\:=\:\mathrm{2}×\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{Yellow}\:=\:\mathrm{8}\: \\ $$ Commented by Mingma last…
Question Number 125790 by mohammad17 last updated on 13/Dec/20 Answered by liberty last updated on 13/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}{\mathrm{sin}\:\pi{x}}\:=\:\mathrm{4}×\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\left[\:\frac{{x}−\mathrm{2}}{\mathrm{sin}\:\pi{x}}\:\right] \\ $$$$\:\left[\:{let}\:{x}=\mathrm{2}+{t}\:\wedge\:{t}\rightarrow\mathrm{0}\:\right]\: \\ $$$$=\:\mathrm{4}×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{\mathrm{sin}\:\pi\left({t}+\mathrm{2}\right)}\:=\:\mathrm{4}\:×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{\mathrm{sin}\:\left(\mathrm{2}\pi+\pi{t}\right)}…
Question Number 60255 by ANTARES VY last updated on 19/May/19 $$\boldsymbol{\mathrm{b}}=\sqrt[{\mathrm{3}}]{\frac{\boldsymbol{\mathrm{kT}}}{\boldsymbol{\mathrm{P}}}}.\:\boldsymbol{\mathrm{distance}}\:\:\boldsymbol{\mathrm{molekular}} \\ $$$$\boldsymbol{\mathrm{prove}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com