Question Number 120531 by A8;15: last updated on 01/Nov/20 Commented by Dwaipayan Shikari last updated on 01/Nov/20 $$\int{e}^{{x}^{{x}} } {dx}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({x}^{{x}} \right)^{{n}} }{{n}!}{dx} \\…
Question Number 120526 by help last updated on 01/Nov/20 Commented by help last updated on 01/Nov/20 $${The}\:{thick}\:{question}…{i}\:{need}\:{help}\:{pls} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120524 by SOMEDAVONG last updated on 01/Nov/20 $$\mathrm{1}/.\mathrm{Given}\:\mathrm{E}=\mathrm{C}\left(\left[\mathrm{0},\mathrm{1}\right],\mathbb{R}\right)\:\mathrm{are}\:\mathrm{mapping}\:\mathrm{set}\:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}. \\ $$$$\left(\mathrm{a}\right).\:\mathrm{Show}\:\mathrm{that}\:\mathrm{E}\:\mathrm{is}\:\mathrm{vector}\:\mathrm{space}. \\ $$$$\left(\mathrm{b}\right).\mathrm{For}\:\:\mathrm{f}\in\mathrm{E}\:,\mathrm{let}\:\mid\mid\mathrm{f}\mid\mid_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mid\mathrm{f}\left(\mathrm{x}\right)\mid\mathrm{dx}\:. \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{E},\mid\mid.\mid\mid_{\mathrm{1}} \right)\:\mathrm{are}\:\mathrm{normed}\:\mathrm{space}. \\ $$$$\:\:\left(\mathrm{Helpe}\:\mathrm{me}\:\mathrm{please}\right) \\ $$ Terms…
Question Number 186058 by 123564 last updated on 31/Jan/23 $${Q}\mathrm{185257} \\ $$ Commented by mr W last updated on 31/Jan/23 $${what}'{s}\:{your}\:{problem}?\:{the}\:{question}\:{was} \\ $$$${answered}\:{completely}. \\ $$…
Question Number 186051 by Tom last updated on 31/Jan/23 $$\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}\mid\mathrm{5}^{{x}} −\mathrm{5}^{−{x}} \mid{dx} \\ $$ Answered by Frix last updated on 31/Jan/23 $$=\underset{−\mathrm{1}} {\overset{\mathrm{0}}…
Question Number 54975 by naka3546 last updated on 15/Feb/19 $${f}\left({x}\right)\:\:=\:\:{x}^{\mathrm{3}} \:−\:\mathrm{2}{x}\:+\:\mathrm{4} \\ $$$${f}\:^{−\mathrm{1}} \left({x}\right)\:\:=\:\:? \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 54944 by naka3546 last updated on 15/Feb/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{{x}}{\mathrm{3}^{{x}} \:−\:\mathrm{1}} \\ $$ Commented by Abdo msup. last updated on 15/Feb/19 $${let}\:{A}\left({x}\right)=\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}}\:\Rightarrow{A}\left({x}\right)=\frac{{x}}{{e}^{{xln}\left(\mathrm{3}\right)} −\mathrm{1}}\:{but}…
Question Number 120477 by help last updated on 31/Oct/20 Answered by Dwaipayan Shikari last updated on 31/Oct/20 $${f}\left({x}\right)−{f}\left({x}−\mathrm{2}\right)=\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{7}\:\:\:,\:\:{f}\left(\mathrm{1}\right)=\mathrm{4} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:\:\:\:\:\: \\…
Question Number 120468 by help last updated on 31/Oct/20 Commented by bemath last updated on 01/Nov/20 $$\mathrm{cot}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{cot}\:\left(\theta−\mathrm{30}°\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{tan}\:\left(\theta−\mathrm{30}°\right)}\:= \\ $$$$\frac{\mathrm{1}}{\left(\frac{\mathrm{tan}\:\theta+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\:.\:\frac{\mathrm{tan}\:\theta−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\right)}\:= \\ $$$$\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{3}}}\:=\:\frac{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}}…
Question Number 120467 by help last updated on 31/Oct/20 Answered by mathmax by abdo last updated on 31/Oct/20 $$\mathrm{tan}\left(\theta+\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{tan}\theta\:+\mathrm{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}−\mathrm{tan}\theta\:\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\mathrm{tan}\theta\:+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\theta} \\ $$$$\mathrm{tan}\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=\mathrm{tan}\left(\theta+\pi−\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{tan}\theta−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\theta} \\ $$$$\mathrm{e}\:\Rightarrow\mathrm{tan}\theta\:+\frac{\mathrm{tan}\theta\:+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\theta}\:+\frac{\mathrm{tan}\theta−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tant}}\:=\mathrm{3}\:\:\:\mathrm{let}\:\mathrm{tan}\theta\:=\mathrm{x}\:\mathrm{so}\:\mathrm{e}\:\Rightarrow \\ $$$$\mathrm{x}\:+\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{x}}+\frac{\mathrm{x}−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{x}}\:=\mathrm{3}\:\Rightarrow\mathrm{x}\:+\frac{\left(\mathrm{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{x}\right)+\left(\mathrm{x}−\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{x}\right)}{\mathrm{1}−\mathrm{3x}^{\mathrm{2}}…