Question Number 57409 by Abdo msup. last updated on 03/Apr/19 $${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:+{k}}} \\ $$$${calculste}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$ Commented by maxmathsup by imad…
Question Number 57410 by Abdo msup. last updated on 03/Apr/19 $${find}\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{sin}\left({x}^{\mathrm{5}} \:+{x}−\mathrm{2}\right)}{{x}−\mathrm{1}} \\ $$ Commented by maxmathsup by imad last updated on 04/Apr/19 $${in}\:{this}\:{case}\:{its}\:{beter}\:{to}\:{use}\:{hospital}\:{theorem}…
Question Number 57408 by Abdo msup. last updated on 03/Apr/19 $${let}\:{the}\:{sequence}\:\left({a}_{{n}} \right)\:{wich}\:{verify}\:\:\:{a}_{\mathrm{1}} =\mathrm{2}\:\:{and} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} \:+\sqrt{\mathrm{1}+\frac{{a}_{{n}} }{{n}}} \\ $$$${prove}\:{that}\:\left(\frac{{a}_{{n}} }{{n}}\right)_{{n}\geqslant\mathrm{1}} \:\:\:{is}\:{convergente}. \\ $$ Terms…
Question Number 57406 by Abdo msup. last updated on 03/Apr/19 $${let}\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{cos}\left(\pi{x}\right) \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:{f}\left({x}\right)=\mathrm{0}\:{have}\:{a}\:{solurion}\:\alpha\:{inside}\:\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\left.\mathrm{2}\right)\:{use}\:{newton}\:{method}\:{to}\:{find}\:{a}\:{approximate}\: \\ $$$${value}\:{of}\:\alpha\:. \\ $$ Commented by kaivan.ahmadi last updated…
Question Number 57407 by Abdo msup. last updated on 03/Apr/19 $${let}\:{U}_{\mathrm{0}} ={cos}\left(\frac{\pi}{\mathrm{3}}\right)\:{and}\:{U}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{U}_{{n}} }{\mathrm{2}}} \\ $$$${find}\:{U}_{{n}} \:{interms}\:{of}\:{n}\:. \\ $$ Commented by Abdo msup. last updated…
Question Number 57389 by rahul 19 last updated on 03/Apr/19 $${If}\:{a}\epsilon{R}\:{and}\:{the}\:{equation}\:: \\ $$$$−\mathrm{3}\left\{{x}\right\}^{\mathrm{2}} +\mathrm{2}\left\{{x}\right\}+{a}^{\mathrm{2}} =\mathrm{0}\:{has}\:{no}\:{integral} \\ $$$${solution},\:{then}\:{all}\:{possible}\:{value}\:{of}\:{a} \\ $$$${lie}\:{in}\:{the}\:{interval}\:: \\ $$$$\left({a}\right)\left(−\mathrm{1},\mathrm{0}\right)\mathrm{U}\left(\mathrm{0},\mathrm{1}\right)\:\:\:\left({b}\right)\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\left({c}\right)\:\left(−\mathrm{2},−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\left(−\infty,−\mathrm{2}\right)\mathrm{U}\left(\mathrm{2},\infty\right) \\ $$…
Question Number 122788 by 676597498 last updated on 19/Nov/20 Answered by mr W last updated on 20/Nov/20 $$\mathrm{0}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1000}}<\mathrm{1}\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1000}}\right]=\mathrm{0} \\ $$$$… \\ $$$$\mathrm{0}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{499}}{\mathrm{1000}}<\mathrm{1}\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{499}}{\mathrm{1000}}\right]=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{500}}{\mathrm{1000}}=\mathrm{1}\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{500}}{\mathrm{1000}}\right]=\mathrm{1} \\…
Question Number 57232 by maxmathsup by imad last updated on 31/Mar/19 $${decompose}\:{tbe}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{{n}} \left({x}+\mathrm{1}\right)}\:\:{with}\:{n}\:{integr}\:{natural}. \\ $$ Commented by maxmathsup by imad last updated on 14/Apr/19 $$\:{the}\:{dcomposition}\:{of}\:{F}\:{is}\:{F}\left({x}\right)\:=\frac{{a}}{{x}+\mathrm{1}}\:+\sum_{{i}=\mathrm{1}}…
Question Number 122615 by john santu last updated on 18/Nov/20 Commented by liberty last updated on 18/Nov/20 $$\:\frac{\mathrm{1}}{{t}}\:=\:\frac{\mathrm{1}}{\mathrm{12}}\:+\:\frac{\mathrm{1}}{\mathrm{15}}\:−\:\frac{\mathrm{1}}{\mathrm{10}}\:=\:\frac{\mathrm{5}+\mathrm{4}−\mathrm{6}}{\mathrm{60}}\: \\ $$$$\:\frac{\mathrm{1}}{{t}}\:=\:\frac{\mathrm{1}}{\mathrm{20}}\:\Rightarrow\:{t}\:=\:\mathrm{20}\:{hours} \\ $$ Terms of Service…
Question Number 57020 by 121194 last updated on 28/Mar/19 $$\left[{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\right]^{\mathrm{2}} =\mathrm{4}\left[{f}\left({x}\right)−\mathrm{1}\right] \\ $$$${f}\left({x}\right)=? \\ $$$$−−−−−−−−−−−−− \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Commented by kaivan.ahmadi last updated on…