Question Number 119291 by 675480065 last updated on 23/Oct/20 $$\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{x}} \left({x}^{\mathrm{10}} −\mathrm{1}\right)}{{ln}\left({x}\right)}\:{dx}\: \\ $$ Answered by TANMAY PANACEA last updated on 23/Oct/20 $${I}\left({a}\right)=\int_{\mathrm{0}}…
Question Number 119257 by Cristina last updated on 23/Oct/20 $$\mathrm{Isomeri}\:\mathrm{di}\:\mathrm{posizione} \\ $$$$\mathrm{C}_{\mathrm{4}} \mathrm{H}_{\mathrm{8}} \rightarrow\:\mathrm{ciclobutano}\:\mathrm{e}\:\mathrm{metilciclopropano} \\ $$$$\mathrm{C}_{\mathrm{5}} \mathrm{H}_{\mathrm{10}} \rightarrow\:\mathrm{ciclopentano},\:\mathrm{metilciclobutano},\:\mathrm{dimetilciclopropano},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{etilciclo}\:\mathrm{propano} \\ $$$$\mathrm{isomeri}\:\mathrm{configurazionali}\:\mathrm{cis}−\mathrm{trans}\:\mathrm{dell}'\mathrm{1},\mathrm{2}\:\mathrm{dimetilciclobutano} \\ $$$$\mathrm{Guardo}\:\mathrm{il}\:\mathrm{legame}\:\mathrm{C}_{\mathrm{1}} −\mathrm{C}_{\mathrm{2}}…
Question Number 184775 by SEKRET last updated on 11/Jan/23 $$\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}\:+\:\frac{\mathrm{9}}{\mathrm{6}\:+\:\:\frac{\mathrm{25}}{\mathrm{6}\:\:+\:\:\frac{\mathrm{49}}{\mathrm{6}\:+\:\frac{\mathrm{81}}{\mathrm{6}+\:……}\:\:\:\:\:}\:\:\:\:\:}\:\:}\:\:\:\:\:\:\:\:}\:\:=? \\ $$$$ \\ $$$$ \\ $$ Commented by Frix…
Question Number 184773 by Ml last updated on 11/Jan/23 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{ax}+\mathrm{b}}{\:\sqrt{\mathrm{1}+\mathrm{3x}}−\mathrm{2}}=\mathrm{c} \\ $$$$\mathrm{2a}−\mathrm{2b}+\mathrm{3c}=? \\ $$$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\neq\mathrm{0} \\ $$$$\mathrm{pease}\:\mathrm{solution}???? \\ $$ Commented by SEKRET last updated on…
Question Number 184774 by lapache last updated on 11/Jan/23 $${Calcul}\:{the}\:{sum} \\ $$$$\mathrm{1}.\Sigma{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{2}.\Sigma{xarctan}\left({x}\right) \\ $$$$\mathrm{3}.\Sigma{e}^{{x}} {sinx} \\ $$$$\mathrm{4}.\Sigma\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{20}} \\ $$$$\mathrm{5}.\Sigma\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} \:}\:{a}>\mathrm{0} \\…
Question Number 184768 by SANOGO last updated on 11/Jan/23 $$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\:\frac{{x}^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$ Commented by SANOGO last updated on 11/Jan/23 $${merci} \\…
Question Number 119212 by help last updated on 23/Oct/20 Answered by Olaf last updated on 23/Oct/20 $$\forall{k}\in\mathbb{N}^{\ast} ,\:\left({k}−\mathrm{1}\right)^{\mathrm{3}} −\left({k}^{\mathrm{3}} +\mathrm{4}\right)\:=\:−\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}−\mathrm{5}\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\left({k}−\mathrm{1}\right)^{\mathrm{3}} \:>\:{k}^{\mathrm{3}} +\mathrm{4}\:\mathrm{and}\::…
Question Number 119192 by help last updated on 22/Oct/20 Answered by Olaf last updated on 22/Oct/20 $$\forall{k}\in\mathbb{N}^{\ast} ,\:{k}^{\mathrm{2}} +\mathrm{1}\:>\:{k}^{\mathrm{2}} \\ $$$$\forall{k}\in\mathbb{N}^{\ast} ,\:\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \:>\:{k}^{\mathrm{6}} \\…
Question Number 184726 by yaslm last updated on 10/Jan/23 Answered by mr W last updated on 11/Jan/23 $${k}\sqrt{\mathrm{20}}=\mathrm{6000} \\ $$$$\Rightarrow{k}=\frac{\mathrm{6000}}{\:\sqrt{\mathrm{20}}}=\frac{\mathrm{3000}}{\:\sqrt{\mathrm{5}}}\approx\mathrm{1342} \\ $$$${R}=\int_{\mathrm{0}} ^{\mathrm{20}} {k}\sqrt{{x}}{dx}=\frac{\mathrm{2}{k}}{\mathrm{3}}\left(\mathrm{20}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{2}×\mathrm{20}×\mathrm{6000}}{\mathrm{3}}=\mathrm{80}\:\mathrm{000}\:{N}…
Question Number 119173 by help last updated on 22/Oct/20 Answered by Olaf last updated on 22/Oct/20 $$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{2}}−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{3}} \\ $$$$\mathrm{S}\:=\:\left(\frac{\mathrm{1}}{\mathrm{3}}+\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{2}}\right)−\underset{{k}=\mathrm{2}}…