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Question-197050

Question Number 197050 by Skabetix last updated on 06/Sep/23 Commented by TheHoneyCat last updated on 07/Sep/23 $$\mathrm{Maintenant},\:\mathrm{2b} \\ $$$${Y}_{{k}} =\frac{\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{{k}} \right)−{n}}{{x}−\mathrm{1}} \\ $$$$=\frac{\left(\underset{{k}=\mathrm{1}}…

Question-197008

Question Number 197008 by tri26112004 last updated on 06/Sep/23 Answered by AST last updated on 06/Sep/23 $$\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}}…

sinx-b-cosx-a-k-find-k-a-b-plz-i-need-dears-

Question Number 197042 by bbbbbbbb last updated on 07/Sep/23 $$\frac{\mathrm{sinx}}{\mathrm{b}}+\frac{\mathrm{cosx}}{\mathrm{a}}=\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\:\boldsymbol{\mathrm{k}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)? \\ $$$$\boldsymbol{\mathrm{plz}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{dears}} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-197010

Question Number 197010 by sonukgindia last updated on 06/Sep/23 Answered by nikif99 last updated on 06/Sep/23 $${Triangle}\:{inequalitues}: \\ $$$$\mathrm{3}{x}−{x}<\mathrm{10}\:\Rightarrow\mathrm{2}{x}<\mathrm{10}\:\Rightarrow{x}<\mathrm{5}\:\left(\mathrm{1}\right) \\ $$$${x}+\mathrm{3}{x}>\mathrm{10}\:\Rightarrow\mathrm{4}{x}>\mathrm{10}\:\Rightarrow{x}>\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow{x}\geqslant\mathrm{3}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{max}\:{x}=\mathrm{4} \\ $$$${max}\:{perim}.={x}+\mathrm{3}{x}+\mathrm{10}=\mathrm{26}…

Question-197011

Question Number 197011 by sonukgindia last updated on 06/Sep/23 Answered by MM42 last updated on 06/Sep/23 $${let}\:\:{u}={m}+{ni}\:\:\:\&\:\:\:{v}={k}+{li} \\ $$$$\left({e}^{−\frac{{a}}{\mathrm{2}}} −{i}\right)\left({m}+{ni}\right)+\left({e}^{−\frac{{a}}{\mathrm{2}}} +{i}\right)\left({k}+{li}\right) \\ $$$$\Rightarrow{e}^{−\frac{{a}}{\mathrm{2}}} {m}+{n}+{e}^{−\frac{{a}}{\mathrm{2}}} {k}−{l}=\left({m}+{k}\right){e}^{−\frac{{a}}{\mathrm{2}}}…

calcule-la-derive-de-g-x-arctan-x-1-2x-3-

Question Number 197036 by SANOGO last updated on 06/Sep/23 $${calcule}\:{la}\:{derive}\:{de}: \\ $$$${g}\left({x}\right)=\:{arctan}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right) \\ $$ Answered by som(math1967) last updated on 06/Sep/23 $$\:{g}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right)^{\mathrm{2}} }×\frac{{d}}{{dx}}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right) \\…

Question-197005

Question Number 197005 by MrGHK last updated on 06/Sep/23 Answered by MathematicalUser2357 last updated on 10/Sep/23 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\left(\psi\left(\frac{{an}+\mathrm{2}}{\mathrm{2}}\right)−\psi\left(\frac{{an}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$=\left(\frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\mathrm{0}}\left(\psi\left(\frac{\cancel{{a}\centerdot\mathrm{0}}+\mathrm{2}}{\mathrm{2}}\right)−\psi\left(\frac{\cancel{{a}\centerdot\mathrm{0}}+\mathrm{1}}{\mathrm{2}}\right)\right)\right)+\centerdot\centerdot\centerdot \\ $$$$\mathrm{Division}\:\mathrm{by}\:\mathrm{0}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}…