Question Number 196540 by Matica last updated on 27/Aug/23 $$\:\:{using}\:{definition}\:{of}\:{limit},\:{prove}\: \\ $$$$\:\:\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}}{{x}+\mathrm{1}}\:=\:\mathrm{1} \\ $$ Answered by ERLY last updated on 27/Aug/23 $${tu}\:{appliques}\:{le}\:{monome}\:{de}\:{plus}\:{haut}\:{degre}\:{tu}\:{aura}\:{lim}\:{x}/{x}\:{ce}\:{qui}\:{donne}\:\mathrm{1}\:\:\:\:\:\:{Erly}\:{rolvinst} \\ $$…
Question Number 196539 by qaz last updated on 27/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left({i}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{i}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left({i},{j}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{j},\mathrm{6}+\mathrm{1}−{i}\right) \\…
Question Number 196595 by sonukgindia last updated on 27/Aug/23 Answered by Rasheed.Sindhi last updated on 28/Aug/23 $${f}\left({x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}}+\frac{{ax}^{\mathrm{2}} −{bx}+{c}}{−{dx}+{e}} \\ $$$$=\frac{\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\left({e}−{dx}\right)+\left({ax}^{\mathrm{2}}…
Question Number 196594 by sonukgindia last updated on 27/Aug/23 Answered by JDamian last updated on 27/Aug/23 $${really}? \\ $$ Answered by sniper237 last updated on…
Question Number 196576 by sonukgindia last updated on 27/Aug/23 Commented by Frix last updated on 27/Aug/23 $$\mathrm{Infinite}\:\mathrm{solutions}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R} \\ $$$$\mathrm{No}\:\mathrm{solution}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{Z} \\ $$ Answered by AST last…
Question Number 196565 by mokys last updated on 27/Aug/23 $${find}\:{the}\:{power}\:{series}\:{exponition}\:{of}\: \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}\:\:{about}\:{z}_{{o}} \:=\:{i} \\ $$ Answered by aleks041103 last updated on 27/Aug/23 $$\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−\mathrm{2}\right)\left({z}−\mathrm{1}\right)}=\frac{{A}}{{z}−\mathrm{1}}+\frac{{B}}{{z}−\mathrm{2}}…
Question Number 196504 by RoseAli last updated on 26/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196498 by Pedro2002 last updated on 26/Aug/23 Answered by Frix last updated on 26/Aug/23 $$\int\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\int\frac{\mathrm{2}}{\:\sqrt{\pi}}\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\:{x}\:+{C} \\ $$ Terms of…
Question Number 196489 by tri26112004 last updated on 26/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196490 by tri26112004 last updated on 26/Aug/23 $${Know}:\:{xf}\left({x}\right)−{f}'\left({x}\right)={x}^{\mathrm{2}} \\ $$$${Find}\:{f}\left({x}\right)=¿ \\ $$ Commented by mr W last updated on 26/Aug/23 $${i}\:{have}\:{given}\:{you}\:{the}\:{answer}\:{in}\: \\ $$$${Q}\mathrm{196398}.\:{you}\:{didn}'{t}\:{show}\:{any}\:{response}…