Question Number 69212 by aliesam last updated on 21/Sep/19 $${use}\:{limit}\:{comparison}\:{test}\:{to}\:{determine} \\ $$$${the}\:{series}\:{converge}\:{or}\:{diverge} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$ Answered by mind is power last…
Question Number 134750 by mnjuly1970 last updated on 06/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:\:{that}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{3}} {sin}\left(\sqrt{\mathrm{2}}\:\pi{n}\right)}\:=\frac{−\mathrm{13}\sqrt{\mathrm{2}}\:\pi^{\mathrm{3}} \:}{\mathrm{720}} \\ $$$$\:\:\:\:\:…{m}.{n}… \\ $$ Terms of Service Privacy Policy…
Question Number 69211 by Mikael last updated on 21/Sep/19 $${To}\:{conserve}\:{rice}\:{from}\:{a}\:{cooperative}\:{there}\:{is} \\ $$$${a}\:{foil}\:{of}\:{area}\:{A}=\mathrm{24}{cm}^{\mathrm{2}} .\:{The}\:{cooperatives} \\ $$$${want}\:{to}\:{build}\:{a}\:{barn}\:{in}\:{the}\:{shape}\:{of}\:{a}\:{parallelopiped} \\ $$$${square}\:{rectangle}\:{without}\:{cover}. \\ $$$${What}\:{should}\:{be}\:{the}\:{symmetrical}\:{dimensions}\:{for} \\ $$$${the}\:{volume}\:{to}\:{be}\:{maximum}? \\ $$ Terms of…
Question Number 134746 by zahaku last updated on 06/Mar/21 $${How}\:{to}\:{calculate} \\ $$$${li}\underset{{x}\rightarrow−\mathrm{3}} {{m}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:\:\:? \\ $$ Answered by mathmax by abdo last updated on 06/Mar/21…
Question Number 134741 by zahaku last updated on 06/Mar/21 $$\underset{\rightarrow−\mathrm{3}} {{lim}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:? \\ $$ Answered by mathmax by abdo last updated on 07/Mar/21 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+\mathrm{3}}{\:\sqrt{\mid\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{6}\mid}}\:\mathrm{changement}\:\mathrm{x}+\mathrm{3}=\mathrm{t}\:\mathrm{give}…
Question Number 69207 by Rio Michael last updated on 21/Sep/19 $${help}\:{please}. \\ $$$$ \\ $$$${A}\:{river}\:{is}\:\mathrm{5}{m}\:{wide}\:{and}\:{flows}\:{at}\:\mathrm{3}.\mathrm{0}{ms}^{−\mathrm{1}} .\:{A}\:{man}\:{can}\:{swim}\:{at}\:\mathrm{2}.\mathrm{0}{ms}^{−\mathrm{1}} \\ $$$${in}\:{still}\:{water}.\:{if}\:{he}\:{sets}\:{off}\:{at}\:{an}\:{angle}\:{of}\:\mathrm{90}°\:{to}\:{the}\:{bank} \\ $$$${calculate} \\ $$$$\left.{a}\right)\:{the}\:{mans}\:{time}\:{and}\:{velocity} \\ $$$$\left.{b}\right)\:{his}\:{distance}\:{downstream}\:{from}\:{the}\:{starting}\:{point}\:{till} \\…
Question Number 134737 by CutieJanab last updated on 06/Mar/21 Answered by floor(10²Eta[1]) last updated on 07/Mar/21 $$\mathrm{if}\:\mathrm{g}=\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)\Rightarrow\mathrm{g}\mid\mathrm{a},\:\mathrm{g}\mid\mathrm{b}\Rightarrow\mathrm{g}\mid\mathrm{ax}+\mathrm{by}\Rightarrow\mathrm{g}\mid\mathrm{d} \\ $$ Commented by CutieJanab last updated on…
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Question Number 69201 by ahmadshah last updated on 21/Sep/19 Answered by Kunal12588 last updated on 21/Sep/19 $${f}\left({x}\right)=\frac{{x}−\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{25}}=\frac{{x}−\mathrm{5}}{\left({x}−\mathrm{5}\right)\left({x}+\mathrm{5}\right)} \\ $$$$\Rightarrow\left({x}−\mathrm{5}\right)\left({x}+\mathrm{5}\right)\neq\mathrm{0} \\ $$$$\Rightarrow{x}\neq\mathrm{5},−\mathrm{5} \\ $$$$\therefore{x}\in\mathbb{R}−\left\{−\mathrm{5},\mathrm{5}\right\} \\…
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