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Question Number 200833 by Mastermind last updated on 24/Nov/23 $$\mathrm{Advance}\:\mathrm{question} \\ $$$$ \\ $$$$\mathrm{A}\:\mathrm{phone}\:\mathrm{mistakenly}\:\mathrm{got}\:\mathrm{locked}\:\mathrm{with}\:\mathrm{the}\: \\ $$$$\mathrm{pattern}\:\mathrm{of}\:\:\mathrm{3}\:×\:\mathrm{3}\:\mathrm{form},\:\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{attempts} \\ $$$$\mathrm{person}\:\mathrm{can}\:\mathrm{try}\:\mathrm{before}\:\mathrm{he}\:\mathrm{can}\:\mathrm{eventually}\:\mathrm{get} \\ $$$$\mathrm{it}\:\mathrm{right}\:? \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\…

Question-199391

Question Number 199391 by Calculusboy last updated on 03/Nov/23 Answered by AST last updated on 03/Nov/23 $$\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{10}\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{10} \\ $$$$\Rightarrow\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{4}\:{or}\:\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{4},\mathrm{2}\right);\left(\mathrm{2},\mathrm{6}\right) \\…

Question-199392

Question Number 199392 by Calculusboy last updated on 03/Nov/23 Answered by mr W last updated on 03/Nov/23 $$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{53} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{13} \\…