Question Number 197010 by sonukgindia last updated on 06/Sep/23 Answered by nikif99 last updated on 06/Sep/23 $${Triangle}\:{inequalitues}: \\ $$$$\mathrm{3}{x}−{x}<\mathrm{10}\:\Rightarrow\mathrm{2}{x}<\mathrm{10}\:\Rightarrow{x}<\mathrm{5}\:\left(\mathrm{1}\right) \\ $$$${x}+\mathrm{3}{x}>\mathrm{10}\:\Rightarrow\mathrm{4}{x}>\mathrm{10}\:\Rightarrow{x}>\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow{x}\geqslant\mathrm{3}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{max}\:{x}=\mathrm{4} \\ $$$${max}\:{perim}.={x}+\mathrm{3}{x}+\mathrm{10}=\mathrm{26}…
Question Number 197036 by SANOGO last updated on 06/Sep/23 $${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 $$\:{g}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right)^{\mathrm{2}} }×\frac{{d}}{{dx}}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right) \\…
Question Number 197011 by sonukgindia last updated on 06/Sep/23 Answered by MM42 last updated on 06/Sep/23 $${let}\:\:{u}={m}+{ni}\:\:\:\&\:\:\:{v}={k}+{li} \\ $$$$\left({e}^{−\frac{{a}}{\mathrm{2}}} −{i}\right)\left({m}+{ni}\right)+\left({e}^{−\frac{{a}}{\mathrm{2}}} +{i}\right)\left({k}+{li}\right) \\ $$$$\Rightarrow{e}^{−\frac{{a}}{\mathrm{2}}} {m}+{n}+{e}^{−\frac{{a}}{\mathrm{2}}} {k}−{l}=\left({m}+{k}\right){e}^{−\frac{{a}}{\mathrm{2}}}…
Question Number 197005 by MrGHK last updated on 06/Sep/23 Answered by MathematicalUser2357 last updated on 10/Sep/23 $$\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) \\ $$$$=\left(\frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\mathrm{0}}\left(\psi\left(\frac{\cancel{{a}\centerdot\mathrm{0}}+\mathrm{2}}{\mathrm{2}}\right)−\psi\left(\frac{\cancel{{a}\centerdot\mathrm{0}}+\mathrm{1}}{\mathrm{2}}\right)\right)\right)+\centerdot\centerdot\centerdot \\ $$$$\mathrm{Division}\:\mathrm{by}\:\mathrm{0}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}…
Question Number 197032 by bbbbbbbb last updated on 06/Sep/23 $$\left(\mathrm{logtan3}\right)\left(\mathrm{logtan6}\right)\left(\mathrm{logtan9}\right)\left(…\left(\mathrm{logtan87}\right)=?\right. \\ $$$$\boldsymbol{\mathrm{plz}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}} \\ $$ Answered by MM42 last updated on 06/Sep/23 $$\mathrm{0} \\ $$ Terms…
Question Number 197035 by bbbbbbbb last updated on 06/Sep/23 $$\mathrm{tan18}=\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{you}}\:\mathrm{tan72}? \\ $$ Answered by som(math1967) last updated on 06/Sep/23 $${tan}\mathrm{72}={cot}\left(\mathrm{90}−\mathrm{72}\right)={cot}\mathrm{18}=\frac{\mathrm{1}}{{a}} \\ $$ Terms of Service…
Question Number 197027 by sonukgindia last updated on 06/Sep/23 Answered by MM42 last updated on 06/Sep/23 $${c}_{\mathrm{1}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36}\:\:\:\:\&\:\:{c}_{\mathrm{2}} :\:\left({x}−\mathrm{10}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36} \\ $$$${c}_{\mathrm{1}}…
Question Number 196972 by otchereabdullai@gmail.com last updated on 05/Sep/23 Answered by Frix last updated on 05/Sep/23 $${t}=\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{14}{t}^{\mathrm{2}} +\mathrm{9}{t}−\mathrm{14}\right)=\mathrm{0} \\ $$$${t}=−\frac{\mathrm{9}}{\mathrm{28}}\pm\frac{\sqrt{\mathrm{865}}}{\mathrm{28}} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}}…
Question Number 196976 by sonukgindia last updated on 05/Sep/23 Answered by Frix last updated on 05/Sep/23 $$\mathrm{6} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 196973 by sonukgindia last updated on 05/Sep/23 Answered by Frix last updated on 05/Sep/23 $$\mathrm{We}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{the}\:\mathrm{approximate}\:\mathrm{solution}. \\ $$$${x}^{\mathrm{5}} −\mathrm{5}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0}…