Question Number 83672 by jagoll last updated on 05/Mar/20 $${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{51}\: \\ $$$${find}\:{x}\: \\ $$ Answered by john santu last updated on 05/Mar/20 $${x}^{\mathrm{2}}…
Question Number 149192 by Samimsultani last updated on 03/Aug/21 Answered by prakash jain last updated on 03/Aug/21 $$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{{x}} ={u} \\ $$$$\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{{x}} =\frac{\mathrm{1}}{{u}} \\ $$$${u}−\frac{\mathrm{1}}{{u}}=\mathrm{40}\sqrt{\mathrm{6}} \\…
Question Number 149188 by mathdanisur last updated on 03/Aug/21 $${f}\left({x}\right)\:=\:{ax}^{\mathrm{99}} \:+\:{bx}^{\mathrm{77}} \:+\:{cx}^{\mathrm{55}} \:+\:\mathrm{975} \\ $$$${f}\left(−\mathrm{975}\right)\:=\:\mathrm{1000} \\ $$$${find}\:\:\:{f}\left(\mathrm{975}\right)\:=\:? \\ $$ Answered by mr W last updated…
Question Number 149187 by mathdanisur last updated on 03/Aug/21 Commented by mr W last updated on 03/Aug/21 $$=\mathrm{6} \\ $$$${see}\:{Q}\mathrm{74970} \\ $$ Commented by mathdanisur…
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}}…