Question Number 82066 by john santu last updated on 18/Feb/20 $$\mathrm{log}_{\mathrm{3}+\mathrm{2}{x}−{x}^{\mathrm{2}} } \:\left(\frac{\mathrm{sin}\:{x}+\sqrt{\mathrm{3}}\mathrm{cos}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}{x}−{x}^{\mathrm{2}} \right)}\: \\ $$ Commented by john santu last updated on 18/Feb/20…
Question Number 82067 by john santu last updated on 18/Feb/20 $$\begin{cases}{\mid{x}\mid\:−\sqrt[{\mathrm{3}\:}]{{y}+\mathrm{3}\:}\:=\:\mathrm{1}}\\{\left(−{x}\sqrt{−{x}}\right)^{\mathrm{2}} \:=\:{y}\:+\mathrm{10}}\end{cases} \\ $$$${find}\:{solution} \\ $$ Commented by john santu last updated on 18/Feb/20 Commented…
Question Number 147603 by puissant last updated on 22/Jul/21 Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\mathrm{Soit}\:\mathrm{G}\:\mathrm{le}\:\mathrm{nombre}\:\mathrm{de}\:\mathrm{garcons}. \\ $$$$\mathrm{Soit}\:\mathrm{F}\:\mathrm{le}\:\mathrm{nombre}\:\mathrm{de}\:\mathrm{filles}. \\ $$$$\mathrm{G}+\mathrm{F}\:=\:\mathrm{10}\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{Chaque}\:\mathrm{garcon}\:\mathrm{a}\:\mathrm{G}−\mathrm{1}\:\mathrm{freres}\:\mathrm{F}\:\mathrm{soeurs}…
Question Number 147602 by puissant last updated on 22/Jul/21 Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\mathrm{Soit}\:\mathrm{E}\:\mathrm{le}\:\mathrm{nombre}\:\mathrm{d}'\mathrm{eleves}. \\ $$$$\mathrm{Soit}\:\mathrm{T}\:\mathrm{le}\:\mathrm{nombre}\:\mathrm{de}\:\mathrm{tables}. \\ $$$$\begin{cases}{\mathrm{E}\:=\:\mathrm{2T}+\mathrm{11}\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{T}\:=\:\frac{\mathrm{E}}{\mathrm{3}}+\mathrm{7}\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$ \\ $$$$\mathrm{En}\:\mathrm{injectant}\:\left(\mathrm{2}\right)\:\mathrm{dans}\:\left(\mathrm{1}\right)\::…
Question Number 16528 by I’m a gamer last updated on 23/Jun/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Jun/17 $$\mathrm{3}\left({a}+{b}+{c}\right)−\frac{{a}}{{b}}−\frac{{b}}{{c}}−\frac{{c}}{{a}}\leqslant\mathrm{6} \\ $$$$\mathrm{3}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}\sqrt{{abc}}}=\mathrm{3}\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \Rightarrow{abc}\leqslant\mathrm{1} \\ $$$${a}+{b}+{c}\leqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\leqslant\mathrm{3} \\…
Question Number 147593 by 0731619 last updated on 22/Jul/21 Answered by gsk2684 last updated on 22/Jul/21 $${put}\:\mathrm{sin}\:{y}\:=\:{t}\Rightarrow\mathrm{cos}\:{y}\:{dy}\:=\:{dt} \\ $$$$\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}−\mathrm{6}}{dt}=\int\frac{\mathrm{1}}{\left({t}−\mathrm{2}\right)\left({t}+\mathrm{3}\right)}{dt} \\ $$$$=\int\left(\frac{\frac{\mathrm{1}}{\mathrm{5}}}{{t}−\mathrm{2}}+\frac{−\frac{\mathrm{1}}{\mathrm{5}}}{{t}+\mathrm{3}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{ln}\:\left({t}−\mathrm{2}\right)−\mathrm{ln}\:\left({t}+\mathrm{3}\right)\right)+{c} \\…
Question Number 82059 by jagoll last updated on 18/Feb/20 $${what}\:{is}\:{derivative}\:{of}\:\:{h}\:=\:\sqrt{{ln}\left({x}\right)} \\ $$$${by}\:{first}\:{principle}\:{method}\: \\ $$ Answered by Henri Boucatchou last updated on 18/Feb/20 $$\frac{\mathrm{dh}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{\mathrm{d}\sqrt{\mathrm{lnx}}}{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{dlnx}}{\mathrm{dx}}}{\mathrm{2}\sqrt{\mathrm{lnx}}}…
Question Number 82056 by mathocean1 last updated on 17/Feb/20 $${g}\left({M}\right)=\mathrm{2}{M}\overset{\rightarrow} {{B}}.{M}\overset{\rightarrow} {{C}}+{M}\overset{\rightarrow} {{C}}.{M}\overset{\rightarrow} {{A}}+{M}\overset{\rightarrow} {{A}}.{M}\overset{\rightarrow} {{B}} \\ $$$${g}\left({G}\right)=\mathrm{4}{MA}^{\mathrm{2}} +\mathrm{3}{M}\overset{\rightarrow} {{A}}\left({A}\overset{\rightarrow} {{B}}+{A}\overset{\rightarrow} {{C}}\right) \\ $$$$ \\…
Question Number 82057 by ajfour last updated on 17/Feb/20 Commented by ajfour last updated on 17/Feb/20 $${The}\:{plane}\:{diving}\:{at}\:{speed}\:{u}\:{drops} \\ $$$${a}\:{ball}\:{which}\:{upon}\:{one}\:{bounce} \\ $$$${collides}\:{with}\:{the}\:{plane}\:{when} \\ $$$${it}\:{reaches}\:{its}\:{maximum}\:{height}. \\ $$$${Find}\:{the}\:{declination}\:{of}\:\bar…
Question Number 16519 by tawa tawa last updated on 23/Jun/17 $$\mathrm{Solve}: \\ $$$$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$ Answered by ajfour last updated on 23/Jun/17 Commented by ajfour…