Question Number 4297 by 123456 last updated on 07/Jan/16 $$\mathrm{lets} \\ $$$${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R},\forall{x}\geqslant{y}\Rightarrow{f}\left({x}\right)\geqslant{f}\left({y}\right) \\ $$$${g}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{if} \\ $$$$\forall{x}\in\left[\mathrm{0},+\infty\right),{f}\left({x}\right)\leqslant{g}\left({x}\right)\leqslant{f}\left(\mathrm{2}{x}\right) \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{L},\mathrm{L}\:\mathrm{is}\:\mathrm{finite} \\ $$$$\mathrm{does} \\ $$$$\underset{{x}\rightarrow+\infty}…
Question Number 135370 by Bird last updated on 12/Mar/21 $${let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{cosx}\:+\mathrm{2}{sinx}} \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 135366 by Dwaipayan Shikari last updated on 12/Mar/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\mathrm{6}{x}−\mathrm{15}{x}^{\mathrm{2}} +\mathrm{20}{x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{5}} −{x}^{\mathrm{6}} }}{dx}=\frac{\pi}{\mathrm{3}} \\ $$$${Or} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt[{{k}}]{{kx}−\frac{{k}\left({k}−\mathrm{1}\right)}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{k}\left({k}−\mathrm{1}\right)\left({k}−\mathrm{2}\right)}{\mathrm{6}}{x}^{\mathrm{3}}…
Question Number 69829 by TawaTawa last updated on 28/Sep/19 Commented by TawaTawa last updated on 28/Sep/19 $$\mathrm{The}\:\mathrm{question}\:\mathrm{says}\:\mathrm{do}\:\mathrm{not}\:\mathrm{use}\:\mathrm{Newton}'\mathrm{s}\:\mathrm{law}\:\mathrm{and}\:\mathrm{kinematic} \\ $$ Commented by TawaTawa last updated on…
Question Number 135361 by leena12345 last updated on 12/Mar/21 $${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x},\left[−\mathrm{1},\mathrm{5}\right] \\ $$$${find}−{the}−{average}−{value} \\ $$$$ \\ $$ Answered by mr W last updated on 12/Mar/21…
Question Number 69827 by Learner-123 last updated on 28/Sep/19 $${The}\:{acceleration}\:{of}\:{a}\:{particle}\:{moving} \\ $$$${in}\:{a}\:{straight}\:{line}\:{is}\:{defined}\:{as}\:{a}=\mathrm{6}{t}−\mathrm{20} \\ $$$${m}/{s}^{\mathrm{2}} ,\:{where}\:{t}\:{is}\:{in}\:{seconds}.\:{Knowing} \\ $$$${that}\:{s}=\mathrm{0}{m}\:{when}\:{t}=\mathrm{3}{s}\:{and}\:{that}\:{t}=\mathrm{5}{sec} \\ $$$${when}\:{v}=\mathrm{2}{m}/{s}.\:{Determine}\:{the}\:{total} \\ $$$${distance}\:{travelled}\:{when}\:{t}=\mathrm{11}{s}. \\ $$ Answered by…
Question Number 4285 by Filup last updated on 07/Jan/16 $${S}=\mathrm{1}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…}}} \\ $$$${S}=??? \\ $$ Commented by RasheedSindhi last updated on 07/Jan/16 $$\mathrm{S}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…}}} \\ $$$${S}\left({n}\right)=\sqrt{\left({n}+\mathrm{1}\right)+{S}\left({n}+\mathrm{1}\right)} \\…
Question Number 135352 by EDWIN88 last updated on 12/Mar/21 $$\underline{\mathrm{Probability}} \\ $$$$\mathrm{Urn}\:\mathrm{I}\:\mathrm{contains}\:\mathrm{5}\:\mathrm{red}\:\mathrm{and}\:\mathrm{3}\:\mathrm{green}\:\mathrm{balls}.\:\mathrm{Urn}\:\mathrm{II} \\ $$$$\mathrm{contains}\:\mathrm{2}\:\mathrm{red}\:\mathrm{and}\:\mathrm{7}\:\mathrm{green}\:\mathrm{balls}.\:\mathrm{One}\:\mathrm{balls} \\ $$$$\mathrm{is}\:\mathrm{transferred}\:\left(\mathrm{at}\:\mathrm{random}\right)\:\mathrm{from}\:\mathrm{urn}\:\mathrm{I}\:\mathrm{to}\:\mathrm{urn}\:\mathrm{II} \\ $$$$.\:\mathrm{After}\:\mathrm{stirring}\:,\:\mathrm{1}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{from}\:\mathrm{urn}\:\mathrm{II}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\left(\mathrm{final}\right)\:\mathrm{ball}\:\mathrm{is} \\ $$$$\mathrm{green}?\: \\ $$ Answered…
Question Number 4281 by Filup last updated on 07/Jan/16 $$\mathrm{The}\:\mathrm{following}\:\mathrm{image}\:\mathrm{shows}\:\mathrm{the}\:\mathrm{functiond} \\ $$$${f}\left({x}\right)={xe}^{\frac{\mathrm{1}}{\mathrm{1}−{x}}} \:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:{g}\left({x}\right)={x}−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{as}\:\mathrm{to}\:\mathrm{why}\:\mathrm{as}\:\mid{f}\left({x}\right)\mid\rightarrow\infty, \\ $$$$\mathrm{that}\:{f}\left({x}\right)\rightarrow{g}\left({x}\right). \\ $$ Commented by Filup last…