Question Number 126536 by mathocean1 last updated on 21/Dec/20 $${show}\:{by}\:{recurrence}\:{that} \\ $$$$\forall\:{integer}\:{p}\geqslant\mathrm{4},\:\left({p}+\mathrm{1}\right)^{\mathrm{3}} \leqslant\mathrm{3}{p}^{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 126537 by mathocean1 last updated on 21/Dec/20 $${show}\:{by}\:{recurrence}\:{that}\:\forall \\ $$$${integer}\:{p}\geqslant\mathrm{0},\:\mathrm{3}^{{p}} \geqslant{p}^{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126534 by mohammad17 last updated on 21/Dec/20 Commented by mohammad17 last updated on 21/Dec/20 $$?? \\ $$ Answered by mr W last updated…
Question Number 126529 by Study last updated on 21/Dec/20 Commented by Olaf last updated on 22/Dec/20 $$\mathrm{Why}\:{x}^{\frac{\mathrm{1}}{\mathrm{ln}{x}}} \:\mathrm{is}\:\mathrm{not}\:{e}^{\mathrm{ln}\left[{x}^{\frac{\mathrm{1}}{\mathrm{ln}{x}}} \right]} \:=\:{e}^{\frac{\mathrm{1}}{\mathrm{ln}{x}}\mathrm{ln}{x}} \:=\:{e}\:\mathrm{sir}\:? \\ $$$$\mathrm{Why}\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{not}\:{e}.{x}+\mathrm{C}\:? \\ $$…
Question Number 126524 by rs4089 last updated on 21/Dec/20 Answered by Olaf last updated on 21/Dec/20 $$\mathrm{Let}\:{u}\:=\:{e}^{{x}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}.\frac{{du}}{{dx}}\:=\:{e}^{{x}} \frac{{dy}}{{du}}\:=\:{u}\frac{{dy}}{{du}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\frac{{du}}{{dx}}\right)\:=\:\frac{{d}}{{du}}\left({u}\frac{{du}}{{du}}\right)\frac{{du}}{{dx}} \\…
Question Number 192062 by mehdee42 last updated on 07/May/23 $${prove}\:{it}\::\: \\ $$$$\:\:\:{times\_n}\:\:\:;\:\:\:\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+…+\sqrt{\mathrm{4}}}}\:\:}\:<\:\mathrm{3} \\ $$ Commented by ajfour last updated on 08/May/23 $${but}\:\mathrm{4}>\:\mathrm{3}\:\:\:\: \\ $$$${how}\:{can}\:{this}\:{be}… \\…
Question Number 192057 by Shrinava last updated on 07/May/23 $$\mathrm{Simplify}: \\ $$$$\frac{\sqrt{\mathrm{a}}\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{2}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{3}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{4}} }}{\left(\sqrt{\mathrm{a}}\:\:+\:\:\mathrm{1}\right)\centerdot\left(\mathrm{a}\:\:+\:\:\mathrm{1}\right)} \\ $$ Answered by mehdee42 last updated on 07/May/23 $$\:\:\:\frac{\sqrt{{a}}+{a}+{a}\sqrt{{a}}+{a}^{\mathrm{2}}…
Question Number 60987 by ajfour last updated on 28/May/19 Commented by ajfour last updated on 28/May/19 $${Find}\:{the}\:{illuminated}\:{area}\:{of} \\ $$$${the}\:{inner}\:{curved}\:{surface}\:{of}\: \\ $$$${shown},\:{hollow}\:{open}\:{cylinder}. \\ $$$$\:\:\:\:\:\:\left[{where}\:\:{a}>\mathrm{2}{R}\left(\frac{{H}}{{h}}−\mathrm{1}\right)\right]\:\:\:\:\: \\ $$…
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Question Number 126520 by joki last updated on 21/Dec/20 $$\int\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}\:} \:\:}}\mathrm{dx} \\ $$ Answered by liberty last updated on 21/Dec/20 $$−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{d}\left(\mathrm{1}−{x}^{\mathrm{2}}…