Question Number 192277 by York12 last updated on 13/May/23 $$ \\ $$$${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,…,\:{a}_{{n}} \:{be}\:{real}\:{numbers}\:{such}\:{that}: \\ $$$$\sqrt{{a}_{\mathrm{1}} }\:+\:\sqrt{{a}_{\mathrm{2}} −\mathrm{1}\:\:}\:+\sqrt{{a}_{\mathrm{3}} −\mathrm{2}\:}\:+…+\sqrt{{a}_{{n}} −\left({n}−\mathrm{1}\right)\:}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}}…
Question Number 61202 by aanur last updated on 30/May/19 Commented by aanur last updated on 30/May/19 $${a}=\mathrm{73}\frac{\mathrm{5}}{\mathrm{7}}\:\:\:\:\:{b}=\mathrm{31}\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${sir}\:{help}\:{me}\:{plz} \\ $$ Answered by Kunal12588 last…
Question Number 192274 by York12 last updated on 13/May/23 $${prove}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}} }{\:\sqrt{{n}^{\mathrm{6}} +{k}}}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$ Answered by aleks041103 last…
Question Number 192275 by York12 last updated on 13/May/23 $$\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } +\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \equiv{x}\:{mod}\left(\mathrm{11}\right)\:{where}\:{n}\:{is}\:{any}\:{integer}\:\geq\mathrm{0} \\ $$$$ \\ $$ Answered by BaliramKumar last updated on 14/May/23…
Question Number 126732 by 0731619177 last updated on 23/Dec/20 Answered by Dwaipayan Shikari last updated on 24/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\psi\left({x}\right)+\frac{\mathrm{1}}{{x}}\right]=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)+{x}}+\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)+{x}}\sim−\frac{\mathrm{1}}{{x}}\:\:\left({x}\:{is}\:{very}\:{small}\:{with}\:{respect}\:{to}\:\overset{\infty}…
Question Number 126731 by 0731619177 last updated on 23/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192265 by yaslm last updated on 13/May/23 Answered by Frix last updated on 13/May/23 $${x}={p}+\mathrm{1} \\ $$$${y}={q}+\mathrm{1} \\ $$$$\left({p},\:{q}\right)\:\rightarrow\:\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$$\frac{\left({x}−{y}\right)^{\mathrm{2}} }{{x}−{y}^{\mathrm{2}} }=−\frac{\left({p}−{q}\right)^{\mathrm{2}}…
Question Number 126722 by 0731619177 last updated on 23/Dec/20 Answered by mahdipoor last updated on 23/Dec/20 $${x}=\mathrm{5\begin{cases}{{x}!!!−\mathrm{10}>\mathrm{0}}\\{\mathrm{2}{x}−\mathrm{10}=\mathrm{0}}\end{cases}} \\ $$$${l}\underset{{x}\rightarrow\mathrm{5}^{−} } {{i}m}\:\frac{{x}!!!−\mathrm{10}}{\mathrm{2}{x}−\mathrm{10}}=−\infty \\ $$$${l}\underset{{x}\rightarrow\mathrm{5}^{+} } {{i}m}\:\frac{{x}!!!−\mathrm{10}}{\mathrm{2}{x}−\mathrm{10}}=+\infty…
Question Number 126720 by mathocean1 last updated on 23/Dec/20 $$ \\ $$$${show}\:{that} \\ $$$${cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right) \\ $$ Answered by mathmax by abdo last updated on 23/Dec/20…
Question Number 126712 by help last updated on 23/Dec/20 Commented by liberty last updated on 24/Dec/20 $$\:\left(\bullet\right)\:{x}+\frac{\mathrm{1}}{{x}}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\…