Question Number 219761 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by mr W last updated on 01/May/25 $$\mathrm{1}+{x}+{x}^{\mathrm{2}}…
Question Number 219697 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 2 2…
Question Number 219763 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by…
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Question Number 219733 by Spillover last updated on 01/May/25 Answered by SdC355 last updated on 01/May/25 $$\mid\mathrm{g}\left({x}\right)−\mathrm{2}\mid\leq\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$−\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\leq\mathrm{g}\left({x}\right)\leq\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\leq\underset{{x}\rightarrow\mathrm{1}}…
Question Number 219730 by SdC355 last updated on 01/May/25 $${f}\left({w}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\theta\left({s}−\mathrm{1}\right)}{{s}\left({s}−\mathrm{1}\right)^{\alpha} }{e}^{−{sw}} \:\mathrm{d}{s} \\ $$$$\hat {\theta}\left({s}\right)=\begin{cases}{\mathrm{0}\:\:{s}<\mathrm{0}}\\{\mathrm{1}\:\:{s}>\mathrm{0}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 219731 by Nescio last updated on 01/May/25 Answered by breniam last updated on 03/May/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }×\frac{{x}}{\mathrm{log}\left(\mathrm{1}+\mathrm{2}{x}\right)}×\mathrm{tan}\left({x}\right)={L} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }=\mathrm{1}…
Question Number 219724 by Nicholas666 last updated on 01/May/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}; \\ $$$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$ Answered by nothing48 last updated…
Question Number 219789 by Firhad last updated on 01/May/25 Commented by Firhad last updated on 02/May/25 $${can}\:{you}\:{help}\:{me} \\ $$ Commented by mr W last updated…