Question Number 171402 by mnjuly1970 last updated on 14/Jun/22 $$\:\:{solve}… \\ $$$$\:\:\:\lfloor{x}^{\:\mathrm{2}} \rfloor+\lfloor{x}\rfloor^{\:\mathrm{2}} =\:{x}^{\:\mathrm{2}} +{x}\:+\mathrm{1} \\ $$$$\:\:\:\:\:\:{x}=? \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 171388 by infinityaction last updated on 15/Jun/22 $${a},{b},\:{and}\:{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${such}\:{that} \\ $$$${a}^{\mathrm{2}} +{ab}+\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\:=\:\mathrm{25}\:\:,\:\:\:\:\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\:+\:{c}^{\mathrm{2}} \:=\mathrm{9} \\ $$$${and}\:{c}^{\mathrm{2}} +{ca}+{a}^{\mathrm{2}} \:=\:\mathrm{16} \\ $$$${find}\:\:{the}\:\:{value}\:{of}\:\:{ab}+\mathrm{2}{bc}+\mathrm{3}{ac} \\…
Question Number 105851 by 1549442205PVT last updated on 01/Aug/20 $$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{students}\:\mathrm{in}\:\mathrm{the}\:\mathrm{class}\:\mathrm{was} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{theater}\:\mathrm{exactly}\:\mathrm{two}\:\mathrm{times}\:\mathrm{during}\: \\ $$$$\:\mathrm{the}\:\mathrm{winter}\:\mathrm{holidays},\mathrm{while} \\ $$$$\mathrm{performances}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{were}\:\mathrm{seen}\:\mathrm{by} \\ $$$$\mathrm{25},\mathrm{12}\:\mathrm{and}\:\mathrm{23}\:\mathrm{students},\mathrm{respectively} \\ $$$$.\mathrm{How}\:\mathrm{many}\:\mathrm{students}\:\mathrm{are}\:\mathrm{there}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{class}?\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them}\:\mathrm{saw}\:\mathrm{performances} \\ $$$$\:\mathrm{A}\:\mathrm{and}\:\mathrm{B},\mathrm{B}\:\mathrm{and}\:\mathrm{C},\mathrm{C}\:\mathrm{and}\:\mathrm{A}? \\…
Question Number 40306 by behi83417@gmail.com last updated on 19/Jul/18 Commented by MJS last updated on 20/Jul/18 $$\mathrm{can}\:\mathrm{we}\:\mathrm{show}\:\mathrm{the}\:\mathrm{following}? \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant\frac{{x}^{\mathrm{2}} +{yz}}{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} +{xz}}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}}…
Question Number 171360 by Shrinava last updated on 13/Jun/22 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{log}_{\mathrm{9}} \:\mathrm{x}\:\:\centerdot\:\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{7}\:−\:\mathrm{x}\right)\:=\:\mathrm{1} \\ $$ Answered by MJS_new last updated on 13/Jun/22 $$\frac{\mathrm{ln}\:{x}}{\mathrm{2ln}\:\mathrm{3}}×\frac{\mathrm{ln}\:\left(\mathrm{7}−{x}\right)}{\mathrm{ln}\:\mathrm{2}}=\mathrm{1} \\…
Question Number 105822 by bemath last updated on 01/Aug/20 $$\mathcal{G}{iven}\:{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\mathrm{3}{x}+\frac{\mathrm{3}}{\mathrm{2}{x}} \\ $$$${find}\:{f}\left(\mathrm{2}\right)\: \\ $$ Answered by bobhans last updated on 01/Aug/20 $$\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}}…
Question Number 171346 by Mr.D.N. last updated on 13/Jun/22 $$\:\:\mathrm{solve}: \\ $$$$\:\:\:\frac{\mathrm{m}+\left(\mathrm{mn}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{m}^{\mathrm{2}} \mathrm{n}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{m}−\mathrm{n}}\:×\:\frac{\left(\mathrm{m}^{\mathrm{1}/\mathrm{3}} −\mathrm{n}^{\mathrm{1}/\mathrm{3}} \right)}{\mathrm{m}^{\mathrm{1}/\mathrm{3}} } \\ $$ Commented by infinityaction last…
Question Number 171342 by mathlove last updated on 13/Jun/22 Commented by infinityaction last updated on 13/Jun/22 $$\mathrm{3} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 171318 by sciencestudent last updated on 12/Jun/22 $${if}\:{f}\left({x}\right)={x}^{\frac{\mathrm{13}}{\mathrm{5}}} \:{and}\:{g}\left({x}\right)=\sqrt[{\mathrm{5}}]{{x}^{\mathrm{13}} }\: \\ $$$${Which}\:{one}\:{is}\:{correct}? \\ $$$$\left.\mathrm{1}\left.\right)\:{f}\left({x}\right)={g}\left({x}\right)\:\:\:\:\mathrm{2}\right){f}\left({x}\right)\neq{g}\left({x}\right) \\ $$ Commented by mr W last updated on…
Question Number 171311 by mathlove last updated on 12/Jun/22 $$\mathrm{2}^{{x}^{−{i}^{\mathrm{2}} } } +\mathrm{2}^{{x}^{\mathrm{2}} } =\mathrm{2}^{{x}^{\mathrm{3}} −{i}^{\mathrm{2}} } \\ $$$${x}=? \\ $$ Terms of Service Privacy…