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Category: Algebra

If-x-y-1-2-which-of-the-following-cannot-be-xy-

Question Number 150006 by mathdanisur last updated on 08/Aug/21 $$\mathrm{If}\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\:\mathrm{xy}.? \\ $$ Answered by dumitrel last updated on 08/Aug/21 $${xy}={k}\Rightarrow{x}+\frac{{k}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2}{k}=\mathrm{0}\Rightarrow\bigtriangleup\geqslant\mathrm{0}\Rightarrow \\ $$$$\mathrm{1}−\mathrm{16}{k}\geqslant\mathrm{0}\Rightarrow{k}\in\left(−\infty;\frac{\mathrm{1}}{\mathrm{16}}\right]\Rightarrow{k}\notin\left(\frac{\mathrm{1}}{\mathrm{16}};\infty\right)…

x-2-x-x-6-1-dx-

Question Number 150007 by mathdanisur last updated on 08/Aug/21 $$\int\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}}{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 08/Aug/21 $${I}=\int\frac{{x}^{\mathrm{2}} +{x}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}=\int\frac{{x}^{\mathrm{2}}…

lim-x-0-cos-x-1-x-

Question Number 149958 by mathdanisur last updated on 08/Aug/21 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{x}}}]{\mathrm{cos}\sqrt{\mathrm{x}}}\:=\:? \\ $$ Commented by dumitrel last updated on 08/Aug/21 $${e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ Commented by…

x-2-2-x-2-4-x-2-6-x-2-2020-1-x-

Question Number 84404 by Wepa last updated on 12/Mar/20 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{6}\right)…\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2020}\right)=\mathrm{1} \\ $$$$\mathrm{x}=? \\ $$ Commented by TANMAY PANACEA last updated on…

Question-149932

Question Number 149932 by mathdanisur last updated on 08/Aug/21 Answered by mr W last updated on 08/Aug/21 $${say}\:{AB}={CD}=\mathrm{1} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{5}{x}}=\frac{{BC}}{\mathrm{sin}\:\mathrm{9}{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{9}{x}}{\mathrm{sin}\:\mathrm{4}{x}} \\…