Question Number 210958 by RojaTaniya last updated on 24/Aug/24 Commented by Ghisom last updated on 26/Aug/24 $$\mathrm{I}\:\mathrm{get} \\ $$$${x}=\frac{\mathrm{36}}{\mathrm{25}}\wedge{y}=\frac{\mathrm{64}}{\mathrm{25}} \\ $$$${x}=−\frac{\mathrm{7}}{\mathrm{16}}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\wedge{y}=\frac{\mathrm{63}}{\mathrm{400}}−\frac{\mathrm{27}}{\mathrm{50}}\mathrm{i} \\ $$$${x}=−\frac{\mathrm{7}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\wedge{y}=\frac{\mathrm{63}}{\mathrm{400}}+\frac{\mathrm{27}}{\mathrm{50}}\mathrm{i} \\ $$…
Question Number 210961 by RojaTaniya last updated on 24/Aug/24 Answered by A5T last updated on 24/Aug/24 $$\frac{{a}−\mathrm{1}+\mathrm{2}}{{a}−\mathrm{1}}+\frac{{b}−\mathrm{1}+\mathrm{2}}{{b}−\mathrm{1}}+\frac{{c}−\mathrm{1}+\mathrm{2}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\mathrm{0};{ab}+{bc}+{ca}=\frac{{m}−\mathrm{1}}{{m}};{abc}=\frac{−{m}−\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}−\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3}}{{abc}−{ab}−{bc}−{ac}+{a}+{b}+{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{{m}−\mathrm{1}+\mathrm{3}{m}}{{m}}}{\frac{−{m}−\mathrm{1}+\mathrm{1}−{m}−{m}}{{m}}}=\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\frac{\mathrm{4}{m}−\mathrm{1}}{−\mathrm{3}{m}}=\frac{\mathrm{7}}{\mathrm{2}}…
Question Number 210922 by RojaTaniya last updated on 22/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 210934 by hardmath last updated on 22/Aug/24 $$\mathrm{If}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{ax}\:−\:\mathrm{18}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}\:+\:\mathrm{2b}}\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{c}}{\mathrm{x}\:+\:\mathrm{5}} \\ $$$$\mathrm{Find}\:\:\:\boldsymbol{\mathrm{a}}\:+\:\boldsymbol{\mathrm{b}}\:+\:\boldsymbol{\mathrm{c}}\:=\:? \\ $$ Answered by A5T last updated on 22/Aug/24 $$\left({x}+\mathrm{5}\right)\left({x}\right)+\mathrm{2}\left({x}+\mathrm{5}\right)+\mathrm{2}{b}−\mathrm{10}\Rightarrow\mathrm{2}{b}−\mathrm{10}=\mathrm{0}\Rightarrow{b}=\mathrm{5} \\…
Question Number 210919 by RojaTaniya last updated on 22/Aug/24 Answered by mr W last updated on 22/Aug/24 $${yes},\:{we}\:{can}\:{design}\:{such}\:{two}\:{dices}. \\ $$$${the}\:{first}\:{one}\:{is}\:{a}\:{normal}\:{die} \\ $$$${with}\:{six}\:{faces}\:{which}\:{have}\:{digit}\: \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\:{respectively}. \\…
Question Number 210906 by hardmath last updated on 21/Aug/24 $$\frac{\mathrm{19x}\:−\:\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:+\:\mathrm{1}}\:\centerdot\:\left(\mathrm{x}\:+\:\frac{\mathrm{19}\:−\:\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:=\:\mathrm{78} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 21/Aug/24 $$\mathrm{Reconstructing}\:\mathrm{your}\:\mathrm{equation}: \\ $$$${x}=\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}}…
Question Number 210895 by RojaTaniya last updated on 21/Aug/24 Answered by Rasheed.Sindhi last updated on 21/Aug/24 $${x}−\sqrt{\frac{\mathrm{10}}{{x}}}\:=\mathrm{11}\wedge\:{x}\in\mathbb{R}\Rightarrow{x}>\mathrm{0} \\ $$$$\: \\ $$$${x}−\sqrt{\frac{\mathrm{10}{x}}{{x}^{\mathrm{2}} }}\:=\mathrm{11} \\ $$$${x}−\frac{\sqrt{\mathrm{10}{x}}}{\mid{x}\mid}=\mathrm{11} \\…
Question Number 210873 by hardmath last updated on 20/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 210874 by peter frank last updated on 20/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 210868 by universe last updated on 20/Aug/24 $$\mathrm{let}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{be}\:\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}>\mathrm{q}>\mathrm{0}\:\mathrm{define} \\ $$$$\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{x}_{\mathrm{n}} \right\}\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{p}+\mathrm{q}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{x}_{\mathrm{1}} −\frac{\mathrm{pq}}{\mathrm{x}_{\mathrm{n}−\mathrm{1}} }\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\:\mathrm{then}\:\mathrm{x}_{\mathrm{n}} \:=\:?? \\ $$ Commented by Ghisom…