Question Number 205794 by mnjuly1970 last updated on 30/Mar/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{A}\:=\:\left\{\:\frac{{k}}{\mathrm{2}^{{n}} }\:\mid\:\mathrm{1}\leqslant\:{k}\:\leqslant\:\mathrm{2}^{{n}} \:,\:{n}\in\mathbb{N}\:\right\} \\ $$$$\:\:\:\:\:{find}\:.\:\:\overset{\:−} {\mathrm{A}}\:=\:? \\ $$$$ \\ $$ Commented by…
Question Number 205749 by mokys last updated on 29/Mar/24 $${whats}\:{the}\:{suficient}\:{condition}\:{to}\:{became}\:{the}\:{question}\: \\ $$$$ \\ $$$$\sigma^{\mathrm{2}} \left(\mathrm{1}−{a}_{{i}} \right)\left[\lambda_{{i}} \left(\mathrm{1}+{a}_{{i}} \right)−\left(\mathrm{1}−{a}_{{i}} \right)\left(\mathrm{1}−{d}_{{i}} \right)+\left(\lambda_{{i}} +{k}\right)\right]\:<\:\mathrm{0}\: \\ $$ Terms of…
Question Number 205750 by MetaLahor1999 last updated on 29/Mar/24 $$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}=? \\ $$ Commented by mokys last updated on 29/Mar/24 $$=\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \:\frac{−\mathrm{2}{e}^{−\mathrm{2}{x}}…
Question Number 205746 by Satyam1234 last updated on 29/Mar/24 Answered by A5T last updated on 29/Mar/24 $${C}\neq{J}+\mathrm{2}\Rightarrow{C}=\mathrm{3}{F} \\ $$$${A}\geqslant\mathrm{1};{A}=\mathrm{1}\Rightarrow{B}=\mathrm{3}\Rightarrow{I}=\mathrm{7}\Rightarrow{H}=\mathrm{4}\Rightarrow{A}=\mathrm{8}\:{X} \\ $$$${A}=\mathrm{2}\Rightarrow{B}=\mathrm{6}\Rightarrow{E}=\mathrm{2}\:{X} \\ $$$${A}=\mathrm{3}\Rightarrow{B}=\mathrm{9}\:{X}\left(\mathrm{6}\Rightarrow{B}<\mathrm{5}\:{or}\:{A}\geqslant\mathrm{4}\right)\:{or}\:{G}=\mathrm{0} \\ $$$$\:\left({G}=\mathrm{0}\Rightarrow{I}=\mathrm{3}\Rightarrow{D}=\mathrm{6}\Rightarrow{E}=\mathrm{18}\right){X}…
Question Number 205725 by NasaSara last updated on 28/Mar/24 Answered by Berbere last updated on 29/Mar/24 $$\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}} \left[{xy}\right]{dydx}\overset{{xy}={t}} {=}\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}{x}}…
Question Number 205726 by hardmath last updated on 28/Mar/24 $$ \\ $$101 is chosen arbitrarily from the numbers 1,2,3,…,199,200. Prove that two of these selected…
Question Number 205727 by mathzup last updated on 28/Mar/24 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{e}^{{x}} −{cosx}}{{x}^{\mathrm{2}} } \\ $$ Commented by lepuissantcedricjunior last updated on 29/Mar/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\boldsymbol{{cosx}}}{\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 205716 by universe last updated on 28/Mar/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$ Answered by MM42 last updated on 28/Mar/24 $$\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)…\left(\mathrm{4}{n}−\mathrm{1}\right)}<{A}<\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)…\left(\mathrm{4}{n}+\mathrm{2}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right){A}}<{A}<\frac{\mathrm{4}{n}+\mathrm{2}}{\left(\mathrm{2}{n}\right){A}} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}<{A}^{\mathrm{2}}…
Question Number 205733 by hardmath last updated on 28/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$ Answered by…
Question Number 205734 by mr W last updated on 28/Mar/24 Answered by A5T last updated on 28/Mar/24 Commented by A5T last updated on 28/Mar/24 $$\frac{\mathrm{48}×\mathrm{144}×\mathrm{13}}{\mathrm{4}{R}}=\frac{\mathrm{48}×\mathrm{36}}{\mathrm{2}}\Rightarrow{R}=\mathrm{26}…