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a-b-c-d-N-a-b-c-d-is-quadruple-of-a-b-c-d-such-that-b-a-2-1-c-b-2-1-d-c-2-1-a-b-c-d-is-odd-number-s-k-is-

Question Number 84206 by naka3546 last updated on 10/Mar/20 $${a},\:{b},\:{c},\:{d}\:\:\in\:\:\mathbb{N} \\ $$$$\left({a},\:{b},\:{c},\:{d}\right)\:\:{is}\:\:{quadruple}\:\:{of}\:\:{a},\:{b},\:{c},\:{d}\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:{b}\:\:=\:\:{a}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{c}\:\:=\:\:{b}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{d}\:\:=\:\:{c}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\tau\left({a}\right)\:+\:\tau\left({b}\right)\:+\:\tau\left({c}\right)\:+\:\tau\left({d}\right)\:\:{is}\:\:\:{odd}\:\:{number}\left({s}\right)\:. \\ $$$$\tau\left({k}\right)\:\:{is}\:\:{the}\:\:{number}\:\:{of}\:\:{positive}\:\:{divisor}\:\:{of}\:\:\:{natural}\:\:{number}\:\:{k}\:.…

Question-84152

Question Number 84152 by naka3546 last updated on 09/Mar/20 Commented by MJS last updated on 09/Mar/20 $$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\mathrm{for}\:\mathrm{which}\:\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{there} \\ $$$$\mathrm{is}\:\mathrm{such}\:\mathrm{an}\:{x}\:\mathrm{at}\:\mathrm{all}… \\ $$$$\mathrm{it}\:\mathrm{works}\:\mathrm{for}\:{n}=\mathrm{3}\:\mathrm{but}\:\mathrm{not}\:\mathrm{for}\:{n}=\mathrm{2}\:\mathrm{and}\:{n}=\mathrm{4} \\ $$$$… \\ $$…

5-x-1-x-4-

Question Number 84123 by Roland Mbunwe last updated on 09/Mar/20 $$\int\frac{\mathrm{5}−\boldsymbol{{x}}}{\mathrm{1}+\sqrt{\left(\boldsymbol{{x}}−\mathrm{4}\right)}} \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20 $${I}=\int\:\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{4}}={t}\:\Rightarrow{x}−\mathrm{4}={t}^{\mathrm{2}} \\ $$$${I}\:=\int\:\frac{\mathrm{5}−\left(\mathrm{4}+{t}^{\mathrm{2}}…

Question-149628

Question Number 149628 by DELETED last updated on 06/Aug/21 Answered by DELETED last updated on 06/Aug/21 $$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }\:=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\:+\:\frac{\mathrm{2}}{\mathrm{6}}\:=\frac{\mathrm{3}}{\mathrm{6}}\:\rightarrow\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2}\Omega \\ $$$$\mathrm{I}_{\mathrm{p}}…

Question-18556

Question Number 18556 by aplus last updated on 24/Jul/17 Answered by Tinkutara last updated on 25/Jul/17 $$\mathrm{V}\:=\:\pi{r}^{\mathrm{2}} {h} \\ $$$$\mathrm{Total}\:\mathrm{surface}\:\mathrm{area},\:{A}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:+\:\mathrm{2}\pi{rh} \\ $$$$=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:+\:\mathrm{2}\frac{\pi{r}^{\mathrm{2}} {h}}{{r}}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}}…