Question Number 164149 by SANOGO last updated on 14/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 164138 by LEKOUMA last updated on 14/Jan/22 Answered by floor(10²Eta[1]) last updated on 14/Jan/22 $$\mathrm{L}=\mathrm{lim}\left(…\right) \\ $$$$\mathrm{lnL}=\mathrm{lim}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right).\mathrm{ln}\left(\mathrm{lim}\left(\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{5}}{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{lim}\left(\mathrm{ln}\left(\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{5}}{\mathrm{x}^{\mathrm{2}}…
Question Number 33064 by artibunja last updated on 09/Apr/18 Commented by abdo imad last updated on 09/Apr/18 $${we}\:{must}\:{have}\:\mathrm{2}{x}+\mathrm{3}>\mathrm{0}\:{and}\:\mathrm{10}{x}^{\mathrm{2}} \:+\mathrm{13}{x}\:+\mathrm{4}>\mathrm{0} \\ $$$$\Delta\:=\mathrm{13}^{\mathrm{2}} \:−\mathrm{4}\:×\mathrm{10}\:×\mathrm{4}\:=\mathrm{169}\:−\mathrm{160}\:=\mathrm{9}\:>\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{13}\:+\mathrm{3}}{\mathrm{20}}=−\frac{\mathrm{1}}{\mathrm{2}}\:,\:{x}_{\mathrm{2}}…
Question Number 164121 by mkam last updated on 14/Jan/22 $${if}\:{w}\:=\:{f}\left({x},{y}\right)\:{and}\:{x}\:=\:{r}\:{cos}\theta\:,\:{y}\:=\:{rsin}\theta \\ $$$$ \\ $$$${then}\:{prove}\:{that}\:{w}_{{rr}} \:+\:{w}_{\theta\theta} \:=\:\mathrm{0}? \\ $$ Commented by mkam last updated on 14/Jan/22…
Question Number 164117 by SANOGO last updated on 14/Jan/22 $${soit}\:\left({a}_{{n}} \right)_{{n}} {une}\:{suite}\:{define}\:{par} \\ $$$${a}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}−{ln}\left({n}\right)\:\:{montrons}\: \\ $$$${a}_{{n}} {converge} \\ $$ Terms of Service…
Question Number 33043 by mondodotto@gmail.com last updated on 09/Apr/18 $$\:\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{squares}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{large}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{possible}} \\ $$$$\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{drawn}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{rectangular}}\:\boldsymbol{\mathrm{ceiling}}\:\boldsymbol{\mathrm{board}} \\ $$$$\:\boldsymbol{\mathrm{measuring}}\:\mathrm{54}\boldsymbol{\mathrm{cm}}\:\boldsymbol{\mathrm{by}}\:\mathrm{78}\boldsymbol{\mathrm{cm}},\boldsymbol{\mathrm{find}} \\ $$$$\:\left(\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{size}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{squares}} \\ $$$$\:\left(\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{total}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{squares}} \\ $$ Answered by math1967 last updated…
Question Number 98544 by abony1303 last updated on 14/Jun/20 Commented by abony1303 last updated on 14/Jun/20 $$\mathrm{Pls}\:\mathrm{help} \\ $$ Answered by mathmax by abdo last…
Question Number 32999 by artibunja last updated on 09/Apr/18 Commented by prof Abdo imad last updated on 09/Apr/18 $$\Leftrightarrow\:\:\:\left(\:\:\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\right)^{−\mathrm{1}} \:>\mathrm{1}\:\Leftrightarrow\:\left(\:\frac{{ln}\left({x}+\mathrm{1}\right)}{−{ln}\left(\mathrm{3}\right)}\right)^{−\mathrm{1}} \:>\mathrm{1}\:\:{with}\:{x}>−\mathrm{1} \\ $$$$\Leftrightarrow\:−\:\frac{{ln}\mathrm{3}}{{ln}\left({x}+\mathrm{1}\right)}\:>\:\mathrm{1}\:\Leftrightarrow\:\:\frac{{ln}\mathrm{3}}{{ln}\left({x}+\mathrm{1}\right)}\:<−\mathrm{1} \\ $$$$\Leftrightarrow\:\:\frac{{ln}\mathrm{3}}{{ln}\left({x}+\mathrm{1}\right)}\:+\mathrm{1}\:<\mathrm{0}\:\Leftrightarrow\:\:\frac{{ln}\left({x}+\mathrm{1}\right)\:+{ln}\left(\mathrm{3}\right)}{{ln}\left({x}+\mathrm{1}\right)}\:<\mathrm{0}…
Question Number 98528 by DELETED last updated on 14/Jun/20 Commented by john santu last updated on 14/Jun/20 $$\left(\mathrm{3}\right)\mathrm{x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{3} \\ $$$$\frac{\mathrm{34}}{\mathrm{13}}\:+\:\frac{\mathrm{4}}{\mathrm{13}}\:+\:\frac{\mathrm{12}}{\mathrm{13}}\:=\:\frac{\mathrm{50}}{\mathrm{13}}\:=\:\mathrm{3}\:???????? \\ $$ Commented by bobhans…
Question Number 32987 by NECx last updated on 08/Apr/18 $${If}\:\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{2}=\mathrm{4} \\ $$$$\mathrm{3}=\mathrm{10}\: \\ $$$$\mathrm{4}=\mathrm{20} \\ $$$$\mathrm{5}=? \\ $$ Answered by MJS last updated…