Question Number 122162 by naka3546 last updated on 14/Nov/20 $${Find}\:\:{the}\:\:{real}\:\:{solution}\:\:{of}\:\:{equality}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \:−\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$${Please}\:\:{show}\:\:{your}\:\:{workings}\:! \\ $$ Answered by mathmax by abdo last updated…
Question Number 122142 by A8;15: last updated on 14/Nov/20 Answered by ebi last updated on 14/Nov/20 $${y}={mx}+{c} \\ $$$${m}=\frac{{N}\Sigma\left({xy}\right)−\Sigma{x}\Sigma{y}}{{N}\Sigma{x}^{\mathrm{2}} −\left(\Sigma{x}\right)^{\mathrm{2}} } \\ $$$${c}=\frac{\Sigma{y}−{m}\Sigma{x}}{{N}} \\ $$$$\begin{vmatrix}{{x}}\\{−\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{1}}\\{\Sigma{x}=\mathrm{0}}\end{vmatrix}\begin{vmatrix}{{y}}\\{\mathrm{3}}\\{\mathrm{2}}\\{\mathrm{4}}\\{\Sigma{y}=\mathrm{9}}\end{vmatrix}\begin{vmatrix}{{x}^{\mathrm{2}}…
Question Number 187683 by 073 last updated on 20/Feb/23 Answered by anurup last updated on 20/Feb/23 $$\mathrm{I}_{\mathrm{1}} =\int\sqrt{\mathrm{tan}\:{x}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\left(\sqrt{\mathrm{tan}\:{x}}\:+\sqrt{\mathrm{cot}\:{x}}\:\right)+\left(\sqrt{\mathrm{tan}\:{x}}\:−\sqrt{\mathrm{cot}\:{x}}\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\left(\sqrt{\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}}\:+\sqrt{\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}\right)+\:\left(\sqrt{\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}}\:−\sqrt{\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}\:\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\left(\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{sin}\:{x}\mathrm{cos}\:{x}}}\right)+\left(\frac{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{sin}\:{x}\mathrm{cos}\:{x}}}\right)\right\}{dx} \\…
Question Number 187672 by Humble last updated on 20/Feb/23 $$\mathrm{4}^{−\frac{\mathrm{1}}{{x}}} +\mathrm{6}^{−\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{9}^{−\frac{\mathrm{1}}{{x}}} \\ $$$${x}\in\mathbb{R} \\ $$ Answered by SEKRET last updated on 20/Feb/23 $$\:\:\frac{\mathrm{4}^{−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} }{\mathrm{9}^{−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}}…
Question Number 56607 by problem solverd last updated on 19/Mar/19 Commented by Kunal12588 last updated on 19/Mar/19 $${is}\:{answer}=\left(\mathrm{1},\mathrm{0}\right)\:? \\ $$ Commented by MJS last updated…
Question Number 187629 by a.lgnaoui last updated on 19/Feb/23 $${Classer}\:{par}\:{ordre}\:{croissant} \\ $$$$\left({from}\:{min}\:\:{to}\:{max}\right) \\ $$$$\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}};\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}};\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{2}}; \\ $$$$\frac{\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}};\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\mathrm{1}+\sqrt{\mathrm{2}}};\frac{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{1}}\: \\ $$ Commented by a.lgnaoui last updated on 19/Feb/23…
Question Number 122093 by SOMEDAVONG last updated on 14/Nov/20 $$\mathrm{I}.\mathrm{Study}\:\mathrm{coutinuity}\:\mathrm{of}\:\mathrm{function}: \\ $$$$\:\mathrm{1}/.\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\begin{cases}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\:,\mathrm{if}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{1}\:}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:,\mathrm{if}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} >\mathrm{1}}\end{cases} \\ $$$$\left(\boldsymbol{\mathrm{Helpe}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{please}}\right) \\ $$ Terms of Service…
Question Number 56553 by problem solverd last updated on 18/Mar/19 $$\mathrm{solve}\:\mathrm{for}\:{m} \\ $$$${m}^{\mathrm{8}} =\mathrm{3125} \\ $$ Answered by MJS last updated on 18/Mar/19 $${x}^{\mathrm{8}} ={a}…
Question Number 56555 by Hassen_Timol last updated on 18/Mar/19 Commented by Hassen_Timol last updated on 18/Mar/19 $${Can}\:{you},\:{please},\:{answer}\:{me}\:{using}\:{somewhere} \\ $$$${the}\:{derivatives}…\:{in}\:\boldsymbol{{an}}\:\boldsymbol{{easy}}\:\boldsymbol{{way}}… \\ $$$${Thank}\:{you} \\ $$ Commented by…
Question Number 187612 by SEKRET last updated on 19/Feb/23 $$\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}\right)\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{2}\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{f}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\boldsymbol{\mathrm{f}}\::\boldsymbol{\mathrm{R}}\rightarrow\boldsymbol{\mathrm{R}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com