Question Number 10414 by amir last updated on 07/Feb/17 Commented by mrW1 last updated on 08/Feb/17 $${the}\:{angle}\:{between}\:{tangent}\:{lines}\:{is} \\ $$$${not}\:{a}\:{constant}! \\ $$ Commented by amir last…
Question Number 10408 by konen last updated on 07/Feb/17 Answered by mrW1 last updated on 08/Feb/17 $${y}^{\mathrm{2}} ={e}^{{x}+{y}} \\ $$$$\Rightarrow\mathrm{2ln}\:{y}={x}+{y} \\ $$$$\Rightarrow{x}=\mathrm{2ln}\:{y}−{y} \\ $$$$\frac{{dx}}{{dy}}=\frac{\mathrm{2}}{{y}}−\mathrm{1}=\frac{\mathrm{2}−{y}}{{y}} \\…
Question Number 75932 by Rio Michael last updated on 21/Dec/19 $${solve}\:{for}\:{x}\:{the}\:{following} \\ $$$${a}.\:\mid{x}\mid\:+\:\mathrm{3}{x}\:−\mathrm{4}\:=\mathrm{0} \\ $$$${b}.\:\:\mid{x}\mid−\mathrm{1}\:=\:\mathrm{0} \\ $$$${c}.\:{x}^{\mathrm{2}} +\mathrm{3}\mid{x}\mid\:+\mathrm{2}\:=\mathrm{0} \\ $$$$ \\ $$ Commented by mathmax…
Question Number 10398 by amir last updated on 07/Feb/17 Answered by arge last updated on 07/Feb/17 $${por}\:{l}'{hopital}, \\ $$$$ \\ $$$${y}=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}} \\ $$$$…
Question Number 75933 by Rio Michael last updated on 21/Dec/19 $${solve}\:{the}\:{inequality} \\ $$$${a}.\:\:{ln}\left(\mathrm{2}{x}−{e}\right)\:>\mathrm{1} \\ $$$${b}.\:\left({lnx}\right)^{\mathrm{2}} −{lnx}−\mathrm{6}<\mathrm{0} \\ $$$${c}.\:\mid{x}\mid\:+\:\mid{x}+\mathrm{2}\mid\:\geqslant\:\mathrm{2} \\ $$$${d}.\:\mid\mathrm{2}{x}−\mathrm{5}\mid\:+\:\mid{x}\:+\mathrm{2}\mid\:>\:\mathrm{7} \\ $$ Commented by mathmax…
Question Number 10397 by amir last updated on 07/Feb/17 Commented by mrW1 last updated on 07/Feb/17 $$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{maximum}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{premetre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{trangle},\:\mathrm{but}\:\mathrm{a}\:\mathrm{minimum}. \\ $$$$ \\ $$$$\mathrm{Due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{seen} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{premetre}\:\mathrm{occurs}…
Question Number 10387 by FilupSmith last updated on 06/Feb/17 $$\mathrm{6}\:\mathrm{people}\:{a},\:{b},\:{c},\:{d},\:{e},\:\mathrm{and}\:{f}\:\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line}. \\ $$$$\: \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{they}\:\mathrm{can}\:\mathrm{stand}\:\mathrm{arranged} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{6}! \\ $$$$\: \\ $$$$\mathrm{If}\:\mathrm{two}\:\mathrm{people}\:\mathrm{have}\:\mathrm{to}\:\mathrm{stand}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}, \\ $$$$\mathrm{but}\:\mathrm{everyone}\:\mathrm{else}\:\mathrm{do}\:\mathrm{not}\:\mathrm{matter},\:\mathrm{how}\:\mathrm{many}\:\mathrm{combinations} \\ $$$$\mathrm{combinations}\:\mathrm{are}\:\mathrm{there}? \\…
Question Number 75918 by Master last updated on 21/Dec/19 Answered by Kunal12588 last updated on 21/Dec/19 $${ab}=\mathrm{21} \\ $$$${a}+{c}=\mathrm{10} \\ $$$${c}−{d}=\mathrm{0} \\ $$$${bd}=\mathrm{4} \\ $$$$\Rightarrow…
Question Number 141454 by loveineq last updated on 19/May/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{\left({a}+{b}+\mathrm{4}\right)^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} }{\left({b}+{c}+\mathrm{4}\right)^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} +{ca}+{a}^{\mathrm{2}} }{\left({c}+{a}+\mathrm{4}\right)^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$…
Question Number 75916 by ajfour last updated on 21/Dec/19 $${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{36}=\mathrm{0} \\ $$$${prove}\:{that}\:{x}=\mathrm{2},\mathrm{3},−\mathrm{6}. \\ $$ Commented by TawaTawa last updated on 21/Dec/19 $$\mathrm{Sir},\:\mathrm{please}\:\mathrm{solve}\:\mathrm{it}.\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{your}\:\mathrm{approach}. \\…