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 $$ \\ $$$$\:\:\:\:\:\:\:..{calculus}.. \\ $$$$\:\:{x},{y},{z}\:\in\mathbb{R}^{+} \:\:{and}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\mathrm{1} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:\:\:\:\: \\ $$$$\:\:\:\:{min}_{{x},{y},{z}\in\mathbb{R}^{+\:\:\:\:} }…
Question Number 118011 by mmmmmm1 last updated on 14/Oct/20 Answered by mr W last updated on 14/Oct/20 $$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{0}−{h}\right)^{\mathrm{2}} +\left(\mathrm{0}−{k}\right)^{\mathrm{2}}…
Question Number 118002 by mmmmmm1 last updated on 14/Oct/20 Answered by TANMAY PANACEA last updated on 14/Oct/20 $${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}+\frac{\mathrm{1}}{{x}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${now}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:=\mathrm{6}…
Question Number 52450 by maxmathsup by imad last updated on 07/Jan/19 $${let}\:{P}\left({x}\right)=\left(\mathrm{1}+{ix}\:−{x}^{\mathrm{2}} \right)^{{n}} −\mathrm{1} \\ $$$${find}\:{roots}\:{of}\:\:{P}\left({x}\right)\:{and}\:{factorize}\:{P}\left({x}\right){inside}\:{C}\left({x}\right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com