Question Number 180569 by SAMIRA last updated on 13/Nov/22 $$\sqrt{\boldsymbol{{x}}+\mathrm{4}}\:−\:\sqrt{\boldsymbol{{x}}−\mathrm{1}}\:>\:\sqrt{\mathrm{4}\boldsymbol{{x}}+\mathrm{5}} \\ $$ Commented by Frix last updated on 14/Nov/22 $$\mathrm{false}. \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{for}\:{x}\geqslant\mathrm{0}:\:\sqrt{{x}+\mathrm{4}}<\sqrt{\mathrm{4}{x}+\mathrm{5}} \\…
Question Number 180565 by depressiveshrek last updated on 13/Nov/22 $$\left(\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\mathrm{3}{y}+\sqrt{\mathrm{9}{y}^{\mathrm{2}} +\mathrm{1}}\right)=\mathrm{1} \\ $$$$\left(\mathrm{4}{x}+\mathrm{6}{y}\right)^{\mathrm{3}} =? \\ $$$${Show}\:{full}\:{solution} \\ $$ Commented by Frix last updated on…
Question Number 115023 by bobhans last updated on 23/Sep/20 $${solve}\:\begin{cases}{\mathrm{7}{x}−\mathrm{5}{y}+\mathrm{3}{z}=\mathrm{6}}\\{\mathrm{2}{x}+\mathrm{4}{y}−\mathrm{5}{z}=−\mathrm{5}}\\{\mathrm{9}{x}−\mathrm{8}{y}+\mathrm{2}{z}=−\mathrm{1}}\end{cases}.\:{Find}\:{x}+{y}+{z}\: \\ $$ Answered by bemath last updated on 23/Sep/20 Answered by PRITHWISH SEN 2 last…
Question Number 115018 by bemath last updated on 23/Sep/20 $${If}\:\mathrm{9}^{{x}} +\mathrm{9}^{−{x}} \:=\:\mathrm{3}^{\mathrm{2}+{x}} +\mathrm{3}^{\mathrm{2}−{x}} \:−\mathrm{20},\:{then}\: \\ $$$$\mathrm{27}^{{x}} +\mathrm{27}^{−{x}} \:=? \\ $$ Answered by bobhans last updated…
Question Number 180550 by a.lgnaoui last updated on 13/Nov/22 $${Resoudre}\: \\ $$$${af}^{'} \left({x}\right)+\frac{{b}}{{f}\left({x}\right)}+{c}=\mathrm{0}\:\:\:\:\left({a},{b},{c}\right)\in\mathbb{R}^{\mathrm{3}} \\ $$ Answered by mr W last updated on 13/Nov/22 $${a}\frac{{dy}}{{dx}}=−\frac{{b}+{cy}}{{y}} \\…
Question Number 180542 by a.lgnaoui last updated on 13/Nov/22 $${Resoudre}\:{dans}\:\mathbb{R} \\ $$$$\left.\mathrm{1}\right) \\ $$$${a}+{b}+{c}=\mathrm{2} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{6} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)…
Question Number 114956 by I want to learn more last updated on 22/Sep/20 Commented by I want to learn more last updated on 22/Sep/20 $$\mathrm{really}?.\:\mathrm{But}\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{sir}.…
Question Number 180490 by Shrinava last updated on 12/Nov/22 Answered by a.lgnaoui last updated on 13/Nov/22 $${m}=\mathrm{60\%}{M}={Masse}\left({final}\right) \\ $$$${m}+{x}\:\:\:\:\:{x}={masse}\:{sugar} \\ $$$$\mathrm{76\%}\left(\mathrm{60\%}{M}+{x}\right)+{x}=\mathrm{100\%}{M} \\ $$$$\mathrm{76}\left(\mathrm{60\%}{M}+{x}\right)=\mathrm{100}{M} \\ $$$$\mathrm{76}×\mathrm{60\%}{M}+\mathrm{76}{x}+{x}=\mathrm{100}{M}…
Question Number 180483 by Shrinava last updated on 12/Nov/22 Commented by MJS_new last updated on 12/Nov/22 $$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an}\:\mathrm{infinite}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{sequences}\:{a}_{{n}} \:\mathrm{with} \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}\wedge{a}_{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{5}}\wedge{a}_{\mathrm{3}} =\frac{\mathrm{24}}{\mathrm{25}}\wedge{a}_{\mathrm{4}}…
Question Number 114917 by bemath last updated on 22/Sep/20 Commented by bemath last updated on 22/Sep/20 $${anyone}\:{can}\:{translate}\:{to}\:{english}? \\ $$ Commented by Olaf last updated on…