Question Number 149180 by mathdanisur last updated on 03/Aug/21 Answered by Kamel last updated on 03/Aug/21 $${L}=\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{k}}}\right)} =\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}}…
Question Number 149171 by liberty last updated on 03/Aug/21 $$\:\mathrm{2}^{\mathrm{2x}\:} \:+\:\mathrm{4}^{\mathrm{3x}} \:=\:\mathrm{128}\:\Rightarrow\mathrm{x}=? \\ $$ Answered by Ar Brandon last updated on 03/Aug/21 $$\mathrm{2}^{\mathrm{2}{x}} +\mathrm{4}^{\mathrm{3}{x}} =\mathrm{128}\:\:\Rightarrow\:\:\mathrm{4}^{{x}}…
Question Number 149163 by mathdanisur last updated on 03/Aug/21 $${Q}\left[\sqrt{\mathrm{3}}\right]/{ker}\:{f}\:\:{define}\:{the}\:{circular} \\ $$$${element}\:{of}\:{the}\:{unit} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 83621 by jagoll last updated on 04/Mar/20 $$\mid\:{x}+\frac{\mathrm{1}}{{x}}\mid\:<\:\mathrm{4}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented by jagoll last updated on 05/Mar/20 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$ Answered…
Question Number 149155 by gsk2684 last updated on 03/Aug/21 $${find}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{50}} \:{in}\:{the}\: \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{1000}} +\mathrm{2}{x}\left(\mathrm{1}+{x}\right)^{\mathrm{999}} +\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{998}} +…\infty\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 149151 by mathdanisur last updated on 03/Aug/21 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:…\:+\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\right)\:=\:? \\ $$ Answered by Kamel last updated on 03/Aug/21 $${L}=\underset{\boldsymbol{{n}}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:…\:+\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\right)\: \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty} {{lim}}\frac{\mathrm{1}}{\:\sqrt{{n}}}\underset{{k}=\mathrm{1}}…
Question Number 149150 by mathdanisur last updated on 03/Aug/21 $$\int\:{ln}\:\left({cosx}\right)\:{dx}\:=\:? \\ $$ Answered by puissant last updated on 03/Aug/21 $$\mathrm{K}=\mathrm{xln}\left(\mathrm{cosx}\right)+\int\mathrm{xtanxdx} \\ $$$$=\mathrm{xln}\left(\mathrm{cosx}\right)+\int\mathrm{x}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\mathrm{B}_{\mathrm{2n}} \mid\frac{\mathrm{2}^{\mathrm{2n}}…
Question Number 83608 by john santu last updated on 04/Mar/20 $$\sqrt[{\mathrm{4}\:\:}]{\mathrm{17}+\mathrm{x}}\:+\:\sqrt[{\mathrm{4}\:\:}]{\mathrm{17}−\mathrm{x}}\:=\:\mathrm{2}\: \\ $$$$\mathrm{find}\:\mathrm{x}\: \\ $$ Commented by mr W last updated on 04/Mar/20 $${LHS}\geqslant\sqrt[{\mathrm{4}}]{\mathrm{2}×\mathrm{17}}>\sqrt[{\mathrm{4}}]{\mathrm{2}×\mathrm{16}}>\sqrt[{\mathrm{4}}]{\mathrm{16}}=\mathrm{2}={RHS} \\…
Question Number 149141 by mathdanisur last updated on 03/Aug/21 $$\frac{{cos}^{\mathrm{2}} \left(\mathrm{70}\right)\:-\:{sin}^{\mathrm{2}} \left(\mathrm{70}\right)}{{sin}\left(\mathrm{100}\right)\:+\:{sin}\left(\mathrm{20}\right)}\:=\:? \\ $$ Commented by liberty last updated on 03/Aug/21 $$=\frac{\left(\mathrm{cos}\:\mathrm{70}°−\mathrm{sin}\:\mathrm{70}°\right)\left(\mathrm{cos}\:\mathrm{70}°+\mathrm{sin}\:\mathrm{70}°\right)}{\mathrm{sin}\:\mathrm{100}°+\mathrm{sin}\:\mathrm{20}°} \\ $$$$=\frac{\left(\mathrm{sin}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{70}°\right)\left(\mathrm{sin}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{70}°\right)}{\mathrm{2sin}\:\mathrm{60}°\mathrm{cos}\:\mathrm{40}°} \\…
Question Number 149140 by mathdanisur last updated on 03/Aug/21 $$\begin{cases}{\mid{x}\mid\:+\:{y}\:-\:\mathrm{1}\:=\:\mathrm{0}}\\{{x}\:-\:{y}\:-\:\mathrm{1}\:=\:\mathrm{0}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{2}{x}\:-\:\mathrm{3}{y}\:=\:? \\ $$ Commented by liberty last updated on 03/Aug/21 $$\mathrm{y}=\:\mathrm{x}−\mathrm{1} \\ $$$$\Rightarrow\mid\mathrm{x}\mid+\mathrm{x}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{x}<\mathrm{0}\:\Rightarrow−\mathrm{x}+\mathrm{x}−\mathrm{2}=\mathrm{0}\:,\mathrm{no}\:\mathrm{solution} \\…