Question Number 120534 by help last updated on 01/Nov/20 Commented by john santu last updated on 01/Nov/20 $$\rightarrow\begin{cases}{\mathrm{sin}\:{A}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\mathrm{cos}\:{B}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\end{cases} \\ $$$$\rightarrow\begin{cases}{\mathrm{sin}\:\left({A}+{B}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{1}}{\mathrm{6}}}\\{\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}{\mathrm{6}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{tan}\:\left({A}+{B}\right)\:=\frac{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}\:=\:\frac{\left(\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{1}\right)\left(\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)}{\mathrm{5}} \\ $$ Answered…
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}}…