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Question Number 20867 by Tinkutara last updated on 05/Sep/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{irrational}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\left({x}\:−\:\mathrm{1}\right)\left({x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{21}\:\mathrm{is} \\ $$ Answered by alex041103 last updated on 09/Sep/17 $$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left(\mathrm{3}{x}−\mathrm{2}\right)\left(\mathrm{3}{x}+\mathrm{1}\right)= \\…
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Question Number 151911 by mathdanisur last updated on 24/Aug/21 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{digit}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\mathrm{prime} \\ $$$$\mathrm{numbers}\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\:\mathrm{such}\:\mathrm{that}\:\boldsymbol{\mathrm{x}}<\boldsymbol{\mathrm{y}},\:\boldsymbol{\mathrm{z}}<\mathrm{1000} \\ $$$$\mathrm{and}\:\:\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}\boldsymbol{\mathrm{a}}} \:=\:\mathrm{z} \\ $$ Commented by Rasheed.Sindhi last updated on 24/Aug/21 $$\mathrm{2}\boldsymbol{\mathrm{a}}\:{means}\:\mathrm{20},\mathrm{21},\mathrm{22},…,\mathrm{29}?\:{Or}\:\mathrm{2}\boldsymbol{\mathrm{a}}=\mathrm{2}×\boldsymbol{\mathrm{a}}?…
Question Number 151907 by mathdanisur last updated on 24/Aug/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{2}\boldsymbol{\mathrm{k}}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{diagonals} \\ $$$$\mathrm{is}\:\mathrm{4}\boldsymbol{\mathrm{k}}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{orthodiagonal}\:\mathrm{one}. \\ $$ Commented by mr W last…
Question Number 151905 by mathdanisur last updated on 24/Aug/21 Answered by mindispower last updated on 24/Aug/21 $$\mathrm{8}{x}^{{x}} {y}^{{y}} {z}^{{z}} \mathrm{2}^{{x}+{y}+{z}} =\mathrm{2}\left(\mathrm{2}{x}\right)^{{x}} .\mathrm{2}\left(\mathrm{2}{y}\right)^{{y}} .\left(\mathrm{2}{z}\right)^{{z}_{} } \\…
Question Number 20823 by ANTARES_VY last updated on 04/Sep/17 Answered by ajfour last updated on 04/Sep/17 $$\mathrm{2}{y}+\mathrm{5}\left(\frac{{x}−\mathrm{9}}{\mathrm{2}}\right)=\mathrm{5} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\mathrm{4}{y}+\mathrm{5}{x}=\mathrm{55}\:\:\:\:…\left({i}\right) \\ $$$$\left.\:\:\:{and}\:\:\:\mathrm{4}{x}−{y}=\mathrm{11}\:\:\:\:\right]×\mathrm{4} \\ $$$$\:\:\:\:\:\:\:{so}\:\:\:\:\:\begin{cases}{\mathrm{16}{x}−\mathrm{4}{y}\:=\mathrm{44}}\\{\mathrm{4}{y}+\mathrm{5}{x}\:=\mathrm{55}}\end{cases} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\mathrm{21}{x}=\mathrm{99}\:\:\:\:…
Question Number 151889 by mathdanisur last updated on 23/Aug/21 $$\mathrm{Compare}: \\ $$$$\sqrt[{\mathrm{2020}}]{\left(\mathrm{2020}!\right)^{\mathrm{3}} }\:\:\:\mathrm{and}\:\:\:\mathrm{505}\centerdot\mathrm{2021}^{\mathrm{2}} \\ $$ Answered by MJS_new last updated on 24/Aug/21 $$\forall{n}\in\mathbb{N}\mid{n}>\mathrm{1}:\left({n}!\right)^{\frac{\mathrm{3}}{{n}}} <\frac{{n}}{\mathrm{4}}×\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\…