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Question-153426

Question Number 153426 by SANOGO last updated on 07/Sep/21 Answered by puissant last updated on 07/Sep/21 $$\forall{x}\in\left[−\mathrm{1};\mathrm{1}\right]\:,\:{f}\left({x}\right)=\frac{\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}.. \\ $$ Commented by SANOGO last updated…

lim-x-a-x-3-2-x-

Question Number 153422 by ZiYangLee last updated on 07/Sep/21 $$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{\mathrm{2}^{{x}} }\:= \\ $$ Answered by puissant last updated on 07/Sep/21 $$=\frac{{a}^{\mathrm{3}} }{\mathrm{2}^{{a}} }\:,\:{a}>\mathrm{0}…

Show-that-S-4pir-2-

Question Number 153409 by SOMEDAVONG last updated on 07/Sep/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{S}=\mathrm{4}\pi\mathrm{r}^{\mathrm{2}} \\ $$ Answered by puissant last updated on 07/Sep/21 $${S}\:{is}\:{derivate}\:{of}\:{V}.\:\:{now},\: \\ $$$${V}=\frac{\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} . \\ $$$$\frac{\partial}{\partial{R}}{V}=\:{S}\:\Rightarrow\:{S}=\frac{\mathrm{4}}{\mathrm{3}}×\mathrm{3}\pi{R}^{\mathrm{2}}…

Question-87860

Question Number 87860 by naka3546 last updated on 06/Apr/20 Commented by MJS last updated on 06/Apr/20 $$\mathrm{again},\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{such} \\ $$$$\mathrm{questions} \\ $$$$\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{3}} −\pi}{\pi^{\mathrm{5}} +\mathrm{e}}} \\ $$$$\mathrm{do}\:\mathrm{you}\:\mathrm{want}\:\mathrm{my}\:\mathrm{equation}?…

given-that-f-x-4x-3-48x-find-the-stationary-point-of-f-x-

Question Number 153395 by Rankut last updated on 07/Sep/21 $${given}\:{that}\:\:{f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{48}{x}.\:{find}\: \\ $$$${the}\:{stationary}\:{point}\:{of}\:{f}\left({x}\right) \\ $$ Answered by puissant last updated on 07/Sep/21 $$\frac{\partial}{\partial{x}}{f}\left({x}\right)=\mathrm{0}\:\Rightarrow\:\mathrm{12}{x}^{\mathrm{2}} −\mathrm{48}=\mathrm{0} \\…

Question-153384

Question Number 153384 by SANOGO last updated on 06/Sep/21 Answered by mindispower last updated on 06/Sep/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\mid=\mathrm{0} \\ $$$$\Rightarrow\epsilon=\frac{\mathrm{1}}{\mathrm{2}},\exists{N}\in\mathbb{N}\:\forall{n}\geqslant{N}\:\:\mid\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\…