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Author: Tinku Tara

If-x-1-17-2-Find-the-value-of-x-3-2x-2-7x-1-x-2-x-1-decimal-point-

Question Number 17270 by VEGAMIND last updated on 03/Jul/17 $$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{2}}.\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}} \\ $$$$\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{7}\boldsymbol{\mathrm{x}}−\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}+\mathrm{1}}\:\:\boldsymbol{\mathrm{decimal}}\:\boldsymbol{\mathrm{point}}. \\ $$ Commented by 18±1 last updated on 03/Jul/17 $$.\mathrm{2}\boldsymbol{{x}}=\mathrm{1}+\sqrt{\mathrm{17}}…

arccos-cos-9-

Question Number 148339 by mathdanisur last updated on 27/Jul/21 $${arccos}\:\left({cos}\:\mathrm{9}\right)\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 27/Jul/21 $$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos9}\right)\right) \\ $$$$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos}\left(\mathrm{9}−\mathrm{2}\pi\right)\right)\right),\:\mathrm{9}−\mathrm{2}\pi\:\in\left[\mathrm{0},\pi\right] \\ $$$${x}\:=\:\mathrm{9}−\mathrm{2}\pi \\…

lim-x-0-1-1-x-2-cos-2x-x-2-

Question Number 82800 by jagoll last updated on 24/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} } \\ $$ Commented by mathmax by abdo last updated on 24/Feb/20 $${let}\:{f}\left({x}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…

x-1-3-1-x-15-Find-the-limit-that-does-not-inclued-the-variable-x-in-the-opening-of-the-binomial-

Question Number 148333 by mathdanisur last updated on 27/Jul/21 $$\left(\sqrt[{\mathrm{3}}]{{x}}\:+\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{15}} \\ $$$${Find}\:{the}\:{limit}\:{that}\:{does}\:{not}\:{inclued} \\ $$$${the}\:{variable}\:\boldsymbol{{x}}\:{in}\:{the}\:{opening}\:{of}\:{the} \\ $$$${binomial}. \\ $$ Answered by qaz last updated on 27/Jul/21…

Question-148334

Question Number 148334 by Sravanth last updated on 27/Jul/21 Answered by Rasheed.Sindhi last updated on 27/Jul/21 $$\mathrm{3}^{\mathrm{6}} ×\left(\mathrm{2}^{−\mathrm{2}} ×\mathrm{3}^{\mathrm{5}×−\mathrm{2}} \right)×\left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{6}} ×\mathrm{2}^{−\mathrm{2}} ×\mathrm{3}^{−\mathrm{10}}…

Question-148330

Question Number 148330 by 0731619 last updated on 27/Jul/21 Answered by qaz last updated on 27/Jul/21 $$\mathrm{y}_{\left(\mathrm{n}\right)} =\mathrm{2}^{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}}} \\ $$$$\mathrm{y}_{\left(\mathrm{n}\right)} '=\mathrm{2}^{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}}} \centerdot\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4n}}\right)\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{2n}−\mathrm{4}}{\mathrm{n}^{\mathrm{2}}…