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Category: Algebra

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Question Number 7323 by rohit meena last updated on 23/Aug/16 $${find}\:{the}\:{value}\:{of} \\ $$$$\Delta'\Lambda\:{b}\:{and}\:\left({B}'{U}\:{A}\right)\:\boldsymbol{\Lambda}\:{B} \\ $$$${if}\:{U}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\} \\ $$$${A}=\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\} \\ $$$${B}=\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\Lambda\:{is}\:{a}\:{unoin} \\ $$ Commented by…

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Question Number 138383 by henderson last updated on 12/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{for}}\:{a}_{\mathrm{0}} \:=\:\mathrm{1}\:\boldsymbol{\mathrm{and}}\:\forall\:{n}\:\geqslant\:\mathrm{1},\:{a}_{{n}} \:=\:\frac{\mathrm{1}}{{n}}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\:\:\frac{{a}_{{k}} }{{n}−{k}}\:. \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\forall\:{n}\:\geqslant\:\mathrm{0},\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\mathrm{0}\:\leqslant\:{a}_{{n}} \:\leqslant\:\mathrm{1}. \\ $$ Commented by mitica…

Question-7299

Question Number 7299 by Tawakalitu. last updated on 22/Aug/16 Commented by sou1618 last updated on 22/Aug/16 $${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${x}={y}−\frac{\mathrm{2}}{\mathrm{4}} \\ $$$${f}\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\left({y}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{4}} +\mathrm{2}\left({y}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{3}}…

Question-138329

Question Number 138329 by JulioCesar last updated on 12/Apr/21 Answered by Ñï= last updated on 12/Apr/21 $$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{{dx}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\int\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}} \\ $$$$=\int\frac{{csc}^{\mathrm{2}} \mathrm{2}{xd}}{{csc}^{\mathrm{2}}…

Question-138330

Question Number 138330 by JulioCesar last updated on 12/Apr/21 Answered by Dwaipayan Shikari last updated on 12/Apr/21 $${tanh}^{−\mathrm{1}} {a}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\right) \\ $$$${tanh}^{−\mathrm{1}} {a}+{tanh}^{−\mathrm{1}} {b}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{ab}+{a}+{b}}{\mathrm{1}+{ab}−{a}−{b}}\right)…