Menu Close

Category: Relation and Functions

let-the-sequence-a-n-wich-verify-a-1-2-and-a-n-1-a-n-1-a-n-n-prove-that-a-n-n-n-1-is-convergente-

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…

let-f-x-x-1-x-2-cos-pix-1-prove-that-f-x-0-have-a-solurion-inside-0-1-2-use-newton-method-to-find-a-approximate-value-of-

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…

If-a-R-and-the-equation-3-x-2-2-x-a-2-0-has-no-integral-solution-then-all-possible-value-of-a-lie-in-the-interval-a-1-0-U-0-1-b-1-2-c-2-1-d-2-U-2-

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-122788

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} \\…

decompose-tbe-fraction-F-x-1-x-n-x-1-with-n-integr-natural-

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-122615

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…

f-x-y-2-f-x-f-y-f-2-f-x-

Question Number 57010 by 121194 last updated on 28/Mar/19 $${f}\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\frac{{f}\left({x}\right){f}\left({y}\right)}{{f}\left(\mathrm{2}\right)} \\ $$$${f}\left({x}\right)=? \\ $$ Commented by maxmathsup by imad last updated on 29/Mar/19 $${x}={y}\:\Rightarrow{f}\left({x}\right)=\frac{{f}^{\mathrm{2}} \left({x}\right)}{{f}\left(\mathrm{2}\right)}\:\:{with}\:{f}\left(\mathrm{2}\right)\neq\mathrm{0}\:\Rightarrow{f}^{\mathrm{2}}…