Question Number 197706 by universe last updated on 26/Sep/23 Answered by witcher3 last updated on 27/Sep/23 $$\left(\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{a}+\mathrm{b}}{\mathrm{4}}\geqslant\frac{\sqrt{\mathrm{ab}}}{\mathrm{2}}.\mathrm{Am}−\mathrm{GM} \\ $$$$\left.\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\leqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right.}\right) \\…
Question Number 197670 by mr W last updated on 26/Sep/23 Commented by mr W last updated on 26/Sep/23 $${unsolved}\:{old}\:{question}\:{Q}#\mathrm{197017} \\ $$ Commented by ajfour last…
Question Number 197660 by pticantor last updated on 25/Sep/23 Answered by witcher3 last updated on 25/Sep/23 $$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}+…\mathrm{t}^{\mathrm{n}} }\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dt}=\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty}…
Question Number 197661 by pticantor last updated on 25/Sep/23 $$\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}+….+{t}^{{n}} }=?\: \\ $$ Commented by JDamian last updated on 25/Sep/23 $$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{again}\:\mathrm{the}\:\mathrm{same}\:\mathrm{topic} \\…
Question Number 197659 by pticantor last updated on 25/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197644 by gungun last updated on 25/Sep/23 Commented by gungun last updated on 25/Sep/23 $${find}\:{the}\:{pink}\:{area} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 197640 by Mastermind last updated on 25/Sep/23 $${Water}\:{is}\:{leaking}\:{from}\:{a}\:{hemispheric} \\ $$$${bowl}\:{of}\:{radius}\:\mathrm{20}{cm}\:{at}\:{the}\:{rate}\:{of}\: \\ $$$$\mathrm{0}.\mathrm{5}{cm}^{\mathrm{3}} /{s}.\:{Find}\:{the}\:{rate}\:{at}\:{which}\:{surface} \\ $$$${area}\:{of}\:{the}\:{water}\:{decreasing}\:{when}\:{the} \\ $$$${water}\:{level}\:{is}\:{halfway}\:{from}\:{the}\:{top}. \\ $$$$ \\ $$$${Thank}\:{you} \\ $$…
Question Number 197636 by universe last updated on 25/Sep/23 Answered by mr W last updated on 25/Sep/23 $${s}={x}+{y} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{8}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({s}−{x}\right)^{\mathrm{2}}…
Question Number 197637 by SLVR last updated on 25/Sep/23 Commented by SLVR last updated on 25/Sep/23 $${kindly}\:{help}\:{me}\:{sir} \\ $$ Commented by SLVR last updated on…
Question Number 197639 by mokys last updated on 25/Sep/23 $${show}\:{that}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+{k}−\mathrm{1}\right)!}{{k}!\left({n}−\mathrm{1}\right)!}\:{z}^{{k}} \\ $$ Commented by mr W last updated on 25/Sep/23 $$\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }=\underset{{k}=\mathrm{0}}…