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
Question Number 52449 by maxmathsup by imad last updated on 07/Jan/19 $${let}\:{j}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{and}\:{P}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} −\left(\mathrm{1}−{jx}\right)^{{n}} \:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){factorize}\:{P}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right)\:\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {P}\left({x}\right){dx}. \\ $$$$\left.\mathrm{4}\right)\:{decompose}\:{inside}\:{C}\left({x}\right)\:{the}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{P}\left({x}\right)}…
Question Number 117984 by Ar Brandon last updated on 14/Oct/20 $$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{0}\neq{x}\in\mathbb{R}\:\mathrm{and}\:\mathrm{if}\:{f}\left(\mathrm{2}\right)=\mathrm{9},\:\mathrm{then}\:\mathrm{f}\left(\mathrm{6}\right)\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{216}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{217}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{126}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{127} \\ $$ Answered by Ar Brandon last updated…
Question Number 117972 by Ar Brandon last updated on 14/Oct/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{surjections}\:\mathrm{of}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\}\:\mathrm{onto}\:\left\{\mathrm{x},\mathrm{y}\right\}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{16}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{14}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{6} \\ $$ Answered by Lordose last updated on 14/Oct/20 $$\mathrm{N}=\:\mathrm{2}^{\mathrm{n}\left(\mathrm{A}\right)−\mathrm{1}} =\:\mathrm{2}^{\mathrm{3}} =\mathrm{8}…
Question Number 52421 by ahmetcu07@hotmail.com last updated on 07/Jan/19 Commented by ahmetcu07@hotmail.com last updated on 07/Jan/19 $${help}\:{me}\:{please} \\ $$ Commented by MJS last updated on…