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Category: Algebra

Find-1-lim-n-5n-25-3n-15-1-5n-2-lim-x-x-3-3x-2-1-1-3-x-3-3x-2-1-1-3-3-lim-x-0-cos-4x-3-1-sin-6-2x-

Question Number 199451 by hardmath last updated on 03/Nov/23 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{5}\boldsymbol{\mathrm{n}}}]{\frac{\mathrm{5n}\:−\:\mathrm{25}}{\mathrm{3n}\:+\:\mathrm{15}}} \\ $$$$\mathrm{2}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:\right) \\ $$$$\mathrm{3}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4x}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{sin}^{\mathrm{6}} \:\mathrm{2x}}…

without-using-calculator-what-is-larger-log-2-3-or-log-3-5-

Question Number 199432 by mr W last updated on 03/Nov/23 $${without}\:{using}\:{calculator}: \\ $$$${what}\:{is}\:{larger}?\:\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\:{or}\:\mathrm{log}_{\mathrm{3}} \:\mathrm{5}? \\ $$ Answered by witcher3 last updated on 03/Nov/23 $$\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{ln}\left(\mathrm{2}\right)},\frac{\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{ln}\left(\mathrm{3}\right)}…

Solve-log-3-p-log-r-8-5-r-p-11-find-r-amp-p-

Question Number 199458 by Calculusboy last updated on 03/Nov/23 $$\boldsymbol{{Solve}}:\:\boldsymbol{{log}}_{\mathrm{3}} \boldsymbol{{p}}\:+\:\boldsymbol{{log}}_{\boldsymbol{{r}}} \mathrm{8}\:=\mathrm{5} \\ $$$$\boldsymbol{{r}}+\boldsymbol{{p}}=\mathrm{11}.\:\:\boldsymbol{{find}}\:\boldsymbol{{r\&p}} \\ $$ Answered by Frix last updated on 04/Nov/23 $$\mathrm{Obviously}\:{p}=\mathrm{9}\wedge{r}=\mathrm{2} \\…

Question-199381

Question Number 199381 by Mingma last updated on 02/Nov/23 Answered by witcher3 last updated on 02/Nov/23 $$\left(\mathrm{202}\right)=\mathrm{2}.\mathrm{101} \\ $$$$\frac{\left(\mathrm{201}\right)!}{\mathrm{k}}\equiv\mathrm{0}\left[\mathrm{202}\right]\mathrm{0},\forall\mathrm{k}\in\left\{\mathrm{1},……\mathrm{201}\right)−\left\{\mathrm{2},\mathrm{101}\right) \\ $$$$\mathrm{J}\equiv\frac{\mathrm{201}!}{\mathrm{101}}+\frac{\mathrm{201}!}{\mathrm{2}}\left[\mathrm{202}\right] \\ $$$$\frac{\mathrm{201}!}{\mathrm{2}}=\mathrm{202}.\mathrm{3}.\mathrm{2}.\underset{\mathrm{k}=\mathrm{5},\mathrm{k}\neq\mathrm{101}} {\overset{\mathrm{201}} {\prod}}\mathrm{k}\equiv\mathrm{0}\left[\mathrm{202}\right]…

Question-199355

Question Number 199355 by hardmath last updated on 01/Nov/23 Answered by MathematicalUser2357 last updated on 04/Nov/23 $$\Omega\approx\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}!}{{n}^{{n}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}{n}} }??? \\ $$ Terms…