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Author: Tinku Tara

629-x-1-4-77-x-1-4-8-Find-x-

Question Number 218138 by hardmath last updated on 30/Mar/25 $$\sqrt[{\mathrm{4}}]{\mathrm{629}\:−\:\mathrm{x}}\:\:+\:\:\sqrt[{\mathrm{4}}]{\mathrm{77}\:\:+\:\:\mathrm{x}}\:\:=\:\:\mathrm{8} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Ghisom last updated on 30/Mar/25 $$\mathrm{easy}\:\mathrm{to}\:\mathrm{see}: \\ $$$$\mathrm{629}\:\mathrm{is}\:\mathrm{almost}\:\mathrm{5}^{\mathrm{4}} ,\:\mathrm{77}\:\mathrm{is}\:\mathrm{almost}\:\mathrm{3}^{\mathrm{4}}…

determinant-1-0-0-0-1-1-1-0-0-0-0-1-1-0-0-0-0-1-1-0-0-0-0-1-1-1-0-0-0-1-

Question Number 218119 by MrGaster last updated on 30/Mar/25 $$\begin{vmatrix}{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}\end{vmatrix}=? \\ $$$$ \\ $$ Answered by SdC355 last updated on 30/Mar/25 $$\mathrm{Holy}\:\mathrm{shit}….\mathrm{what}\:\mathrm{is}\:\mathrm{that}\:\mathrm{Lol} \\ $$$$\mathrm{But}\:\mathrm{that}\:\mathrm{det}\left\{\mathrm{A}\right\}\:\mathrm{is}\:\varepsilon^{\mathrm{6}} −\mathrm{6}\varepsilon^{\mathrm{4}}…

determinant-determinant-8044-determinant-

Question Number 218128 by mr W last updated on 30/Mar/25 $$\begin{array}{|c|}{?}&\hline{?}&\hline{?}&\hline{?}\\\hline\end{array}×\begin{array}{|c|}{?}\\\hline\end{array}=\mathrm{8044}\begin{array}{|c|}{?}\\\hline\end{array} \\ $$ Answered by A5T last updated on 30/Mar/25 $$\frac{\mathrm{8044}?}{\mathrm{8}}>\frac{\mathrm{80000}}{\mathrm{8}}=\mathrm{10000}\: \\ $$$$\Rightarrow\:\mathrm{8044}?\:\mathrm{should}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{9} \\ $$$$\Rightarrow\mathrm{9}\:\mid\:\mathrm{8}+\mathrm{0}+\mathrm{4}+\mathrm{4}+?\Rightarrow\:?=\mathrm{2}…

how-many-different-words-can-be-formed-from-the-word-MATHEMATICS-note-here-a-word-should-have-at-least-two-letters-but-mustn-t-have-a-meaning-

Question Number 218129 by mr W last updated on 30/Mar/25 $${how}\:{many}\:{different}\:{words}\:{can}\:{be} \\ $$$${formed}\:{from}\:{the}\:{word}\: \\ $$$$\boldsymbol{\mathrm{MATHEMATICS}}? \\ $$$${note}:\:\:{here}\:{a}\:{word}\:{should}\:{have}\:{at}\: \\ $$$${least}\:{two}\:{letters},\:{but}\:{mustn}'{t}\:{have}\:{a} \\ $$$${meaning}. \\ $$ Answered by…

x-1-x-3-x-5-1-x-5-

Question Number 218103 by ArshadS last updated on 29/Mar/25 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:,\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$ Answered by Rasheed.Sindhi last updated on 29/Mar/25 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:,\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\…

Solve-for-x-2x-3-x-2-x-2-

Question Number 218093 by ArshadS last updated on 29/Mar/25 $${Solve}\:{for}\:{x} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}\:−\sqrt{{x}−\mathrm{2}}\:=\sqrt{{x}+\mathrm{2}}\: \\ $$ Answered by vnm last updated on 29/Mar/25 $$\mathrm{2}{x}+\mathrm{3}=\left(\sqrt{{x}+\mathrm{2}}+\sqrt{{x}−\mathrm{2}}\right)^{\mathrm{2}} = \\ $$$${x}+\mathrm{2}+{x}−\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}…