Question Number 118181 by mathocean1 last updated on 15/Oct/20 $${find}\:{all}\:{numbers}\:>\mathrm{1}\:{from}\:\mathbb{N}\:{which} \\ $$$${their}\:{cube}\:{are}\:<\mathrm{18360} \\ $$ Commented by mathocean1 last updated on 15/Oct/20 $${thanks} \\ $$ Answered…
Question Number 118172 by I want to learn more last updated on 15/Oct/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rhombus}\:\mathrm{with}\:\mathrm{side}\:\:\mathrm{8}\:\mathrm{cm} \\ $$ Answered by MJS_new last updated on 15/Oct/20 $$\mathrm{more}\:\mathrm{information}\:\mathrm{needed} \\…
Question Number 183668 by mnjuly1970 last updated on 28/Dec/22 $$ \\ $$$$\:\:\:\:\:\:{x}\:,\:{y}\:,\:{z}\:\in\mathbb{R}: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{If}\:\:\:\:\begin{cases}{\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{y}+{z}}\:=\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{{y}}\:+\frac{\mathrm{1}}{{x}+{z}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{1}}{{z}_{\:} }\:+\frac{\mathrm{1}}{{x}+{y}}\:=\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{x}\:,\:{y}\:,\:{z}\:=? \\ $$$$ \\ $$ Commented by…
Question Number 183669 by mnjuly1970 last updated on 28/Dec/22 $$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\mathrm{1}\:+\:{x}\:+\:{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:{min}_{\:{f}} \:=\:? \\ $$ Answered by manolex last updated on 28/Dec/22…
Question Number 183660 by mr W last updated on 28/Dec/22 $${what}\:{is}\:{larger}, \\ $$$$\mathrm{2022}^{\mathrm{2022}} \:{or}\:\mathrm{2021}^{\mathrm{2023}} \:? \\ $$ Answered by Ar Brandon last updated on 28/Dec/22…
Question Number 52573 by Necxx last updated on 09/Jan/19 $${Find}\:{the}\:{relation}\:{between}\:{p}\:{q}\:{and}\:{r} \\ $$$${if}\:{one}\:{of}\:{the}\:{root}\:{of}\:{the}\:{equation} \\ $$$${px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0}\:{is}\:{double}\:{the}\:{other}. \\ $$ Answered by mr W last updated on 09/Jan/19…
Question Number 52572 by Necxx last updated on 09/Jan/19 $${In}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}.{One} \\ $$$${root}\:{is}\:{the}\:{square}\:{of}\:{the}\:{other}. \\ $$$${Show}\:{that}\:{c}\left({a}−{b}\right)^{\mathrm{3}} ={a}\left({c}−{b}\right)^{\mathrm{3}} \\ $$$$ \\ $$ Answered by math1967 last updated…
Question Number 183592 by greougoury555 last updated on 27/Dec/22 $$\:\:\:\:\:\begin{cases}{\sqrt{\frac{{x}}{{y}}}\:+\sqrt{\frac{{y}}{{z}}}\:+\sqrt{\frac{{z}}{{x}}}\:=\:\mathrm{3}}\\{\sqrt{\frac{{y}}{{x}}}\:+\sqrt{\frac{{z}}{{y}}}\:+\sqrt{\frac{{x}}{{z}}}\:=\:\mathrm{3}}\\{\sqrt{{xyz}}\:=\:\mathrm{1}}\end{cases} \\ $$$$\: \\ $$$$ \\ $$ Answered by mr W last updated on 27/Dec/22 $${a}=\sqrt{\frac{{x}}{{y}}},\:{b}=\sqrt{\frac{{y}}{{z}}},\:{c}=\sqrt{\frac{{z}}{{x}}}…
Question Number 183556 by Shrinava last updated on 26/Dec/22 Answered by Frix last updated on 26/Dec/22 $${f}^{−\mathrm{1}} \left({x}\right)=\frac{{bx}+\mathrm{8}}{\left({a}−\mathrm{2}{b}\right){x}+{a}+\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{8}\right)=\frac{\mathrm{8}\left({b}+\mathrm{1}\right)}{\mathrm{9}{a}−\mathrm{2}\left(\mathrm{8}{b}−\mathrm{1}\right)} \\ $$ Terms of…
Question Number 118023 by 1549442205PVT last updated on 14/Oct/20 $$ \\ $$$$\:\:\:\:\:\:\:..{calculus}.. \\ $$$$\:\:{x},{y},{z}\:\in\mathbb{R}^{+} \:\:{and}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\mathrm{1} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:\:\:\:\: \\ $$$$\:\:\:\:{min}_{{x},{y},{z}\in\mathbb{R}^{+\:\:\:\:} }…