Question Number 173697 by KONE last updated on 16/Jul/22 Answered by KONE last updated on 16/Jul/22 $${besoin}\:{d}\:{aide}\:{svp} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 42627 by mondodotto@gmail.com last updated on 29/Aug/18 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{log}}\left(\boldsymbol{{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{log}}\mathrm{5}−\boldsymbol{\mathrm{log}}\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Answered by $@ty@m last updated on 30/Aug/18 $$\mathrm{log}\:\left({x}−\mathrm{3}\right)+\mathrm{3log}\:\mathrm{5}−\mathrm{3log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{log}\:\mathrm{1} \\…
Question Number 108162 by mathdave last updated on 15/Aug/20 Answered by JDamian last updated on 15/Aug/20 $${k}={m}−\mathrm{4}\:\:\Rightarrow\:\:{m}={k}+\mathrm{4} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{k}+\mathrm{4}} }{\left({k}+\mathrm{4}\right)!} \\…
Question Number 173696 by KONE last updated on 16/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108158 by Tinku Tara last updated on 15/Aug/20 $$\mathrm{test}\:\mathrm{for}\:\mathrm{edit}\:\mathrm{post}. \\ $$$${edited} \\ $$ Commented by Tinku Tara last updated on 15/Aug/20 $$\mathrm{comment}\:{edited} \\…
Question Number 173695 by KONE last updated on 16/Jul/22 Answered by a.lgnaoui last updated on 19/Jul/22 $${l}\:{image}\:{n}\:{est}\:{pas}\:{claire} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 173677 by naka3546 last updated on 16/Jul/22 $$\mathrm{Find}\:\:\mathrm{tangent}\:\:\mathrm{equation}\:\:\mathrm{of}\:\: \\ $$$$\mathrm{two}\:\:\mathrm{circles}\:: \\ $$$${L}_{\mathrm{1}} \::\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} \:+\:\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$${L}_{\mathrm{2}} \::\:\left({x}+\mathrm{4}\right)^{\mathrm{2}} \:+\:\left({y}−\mathrm{11}\right)^{\mathrm{2}} \:=\:\mathrm{144} \\ $$ Commented…
Question Number 108143 by abony1303 last updated on 15/Aug/20 $$\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}=\mathrm{sin}^{\mathrm{2}} \mathrm{3x} \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{ASAP} \\ $$ Answered by ajfour last updated on…
Question Number 108131 by mathdave last updated on 14/Aug/20 Commented by Her_Majesty last updated on 14/Aug/20 $${just}\:{a}\:{try} \\ $$$${for}\:{x}\rightarrow+\infty\:{we}\:{have}\:\frac{{x}+\sqrt{{x}}}{{x}}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\rightarrow\mathrm{1} \\ $$$$\Rightarrow\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}\sim\sqrt{{x}+\mathrm{1}} \\ $$$$\Rightarrow\:{the}\:{limit}=\mathrm{1} \\ $$$${or}:\:\sqrt{{x}+\sqrt{{x}}}\sim\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}}\:{when}\:{x}\rightarrow+\infty…
Question Number 173661 by aliibrahim1 last updated on 15/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com