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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} \\ $$$$\: \\ $$$${x}−\sqrt{\frac{\mathrm{10}{x}}{{x}^{\mathrm{2}} }}\:=\mathrm{11} \\ $$$${x}−\frac{\sqrt{\mathrm{10}{x}}}{\mid{x}\mid}=\mathrm{11} \\…
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Question Number 210868 by universe last updated on 20/Aug/24 $$\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…
Question Number 210871 by hardmath last updated on 20/Aug/24 $$\mathrm{7}\left(\mathrm{x}−\mathrm{2}\right)\left(\sqrt{\mathrm{10}\:+\:\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{x}}\right)−\mathrm{15x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by Ghisom last updated on 20/Aug/24 $$\mathrm{obviously}\:{x}=−\mathrm{1} \\ $$ Terms…
Question Number 210870 by hardmath last updated on 20/Aug/24 $$\mathrm{8x}^{\mathrm{2}} \:−\:\mathrm{13x}\:+\:\mathrm{11}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{x}}\right)\:\sqrt[{\mathrm{3}}]{\mathrm{3x}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by Ghisom last updated on 20/Aug/24 $$\mathrm{obviously}\:{x}=\mathrm{1} \\…