Question Number 18872 by mondodotto@gmail.com last updated on 31/Jul/17 Answered by ajfour last updated on 31/Jul/17 $$\mathrm{b}=\mathrm{a}+\frac{\mathrm{1}}{\mathrm{b}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{b}−\frac{\mathrm{1}}{\mathrm{b}}\: \\ $$ Answered by Joel577 last…
Question Number 18866 by mondodotto@gmail.com last updated on 31/Jul/17 Answered by dioph last updated on 31/Jul/17 $$\mathrm{Define}\:{y}\:\equiv\:\mathrm{log}\:{x} \\ $$$$\mathrm{Substitute}\:{x}\:=\:\mathrm{10}^{{y}} \:\left(\mathrm{here}\:\mathrm{I}\:\mathrm{assumed}\right. \\ $$$$\left.\mathrm{base}\:\mathrm{10}\:\mathrm{but}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{any}\:\mathrm{base}\right) \\ $$$$\mathrm{5}^{{y}} \:+\:\left(\mathrm{10}^{{y}}…
Question Number 149914 by 7770 last updated on 08/Aug/21 $$\boldsymbol{{show}}\:\boldsymbol{{that}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{2}+\mathrm{6}\theta^{\mathrm{2}} −\mathrm{2}\boldsymbol{\theta}^{\mathrm{3}} }{\boldsymbol{\theta}^{\mathrm{2}} \left(\boldsymbol{\theta}^{\mathrm{2}} +\mathrm{1}\right)}\right)\boldsymbol{{d}\theta}=\mathrm{1}.\mathrm{606} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18834 by mondodotto@gmail.com last updated on 30/Jul/17 Answered by Tinkutara last updated on 30/Jul/17 $$\mathrm{tan}\:\mathrm{2A}\:=\:\mathrm{cot}\:\left(\mathrm{90}°\:−\:\mathrm{2A}\right)\:=\:\mathrm{cot}\:\left(\mathrm{A}\:−\:\mathrm{18}°\right) \\ $$$$\mathrm{Since}\:\mathrm{A}\:\mathrm{is}\:\mathrm{acute},\:\mathrm{so}\:\mathrm{cot}\:\mathrm{A}\:=\:\mathrm{cot}\:\mathrm{B}\:\Leftrightarrow\:\mathrm{A}\:=\:\mathrm{B} \\ $$$$\mathrm{90}°\:−\:\mathrm{2A}\:=\:\mathrm{A}\:−\:\mathrm{18}° \\ $$$$\mathrm{A}\:=\:\mathrm{36}° \\ $$…
Question Number 18827 by aplus last updated on 30/Jul/17 Commented by aplus last updated on 30/Jul/17 $$\mathrm{help}\:\mathrm{guys} \\ $$ Commented by mrW1 last updated on…
Question Number 149891 by mathdanisur last updated on 08/Aug/21 $$\mathrm{if}\:\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{m};\mathrm{n};\mathrm{p}\in\mathbb{R}^{+} \:\mathrm{then}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{m}\left(\mathrm{x}+\mathrm{y}\right)}{\:\sqrt{\left(\mathrm{n}+\mathrm{2p}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{2nxy}+\left(\mathrm{n}+\mathrm{2p}\right)\mathrm{y}^{\mathrm{2}} }}\:\leqslant\:\frac{\mathrm{3m}}{\:\sqrt{\mathrm{n}+\mathrm{p}}} \\ $$ Answered by dumitrel last updated on 08/Aug/21…
Question Number 18818 by khamizan833@yahoo.com last updated on 30/Jul/17 Answered by daffa22 last updated on 30/Jul/17 $$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{{x}^{\mathrm{2}} −\mathrm{4}} =\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{\mathrm{0}} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{0} \\…
Question Number 18808 by Bruce Lee last updated on 30/Jul/17 $$\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\:=? \\ $$$$\boldsymbol{{help}}\:\boldsymbol{{me}} \\ $$ Answered by khamizan833@yahoo.com last updated on 30/Jul/17 Terms of Service…
Question Number 84333 by M±th+et£s last updated on 11/Mar/20 $$\left.\mathrm{1}\right){find}\:{without}\:{l}'{hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2}\sqrt{{x}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}}{{x}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{the}\:{general}\:{solution}\:{for}\:{tbe}\:{differential}\:{equation} \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\frac{{dy}}{{dx}}\right)=\mathrm{0}\:{is}\:{y}=\frac{{k}−{x}}{\mathrm{1}+{kx}},{k}\:{is}\:{a}\:{constant} \\ $$$${then}\:{find}\:{the}\:{special}\:{solution}\:{if}\:{y}=\frac{\mathrm{2}}{\mathrm{3}\:}\:{when}\:{x}=\mathrm{1} \\ $$…
Question Number 149853 by 0731619 last updated on 07/Aug/21 Answered by mindispower last updated on 08/Aug/21 $$\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{{izcos}\left(\theta\right)} {d}\theta={J}_{\mathrm{0}} \left({z}\right)\:\:{bassel}\:{function} \\ $$$${cos}^{\mathrm{2}} \left(\theta\right)=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}} \\…