Menu Close

Category: Operation Research

A-binary-operation-has-the-property-a-b-c-a-b-c-and-that-a-a-1-for-all-non-zero-real-numbers-a-b-and-c-here-represent-multiplication-The-solution-of-the-equation-2016-6-x-100-can-be-wr

Question Number 112813 by Aina Samuel Temidayo last updated on 09/Sep/20 $$\mathrm{A}\:\mathrm{binary}\:\mathrm{operation}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{a}\ast\left(\mathrm{b}\ast\mathrm{c}\right)\:=\:\left(\mathrm{a}\ast\mathrm{b}\right)\bullet\mathrm{c}\:\mathrm{and}\:\mathrm{that}\:\mathrm{a}\ast\mathrm{a}=\mathrm{1}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\left('\bullet'\:\mathrm{here}\:\mathrm{represent}\:\mathrm{multiplication}\right). \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2016}\ast\left(\mathrm{6}\ast\mathrm{x}\right)=\mathrm{100}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\frac{\mathrm{p}}{\mathrm{q}} \\ $$$$\mathrm{where}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\…

the-force-acting-on-a-particle-P-of-mass-2kg-is-2ti-4j-N-P-is-initially-at-rest-at-point-with-position-vector-i-2j-Find-the-velocity-of-P-when-t-2-and-the-position-vector-when-t-2-

Question Number 47235 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…

the-force-acting-on-a-particle-P-of-mass-2kg-is-2ti-4j-N-P-is-initially-at-rest-at-point-with-position-vector-i-2j-Find-the-velocity-of-P-when-t-2-and-the-position-vector-when-t-2-

Question Number 47234 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…

the-force-acting-on-a-particle-P-of-mass-2kg-is-2ti-4j-N-P-is-initially-at-rest-at-point-with-position-vector-i-2j-Find-the-velocity-of-P-when-t-2-and-the-position-vector-when-t-2-

Question Number 47233 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…

y-2y-y-e-2x-2x-3-

Question Number 111105 by bobhans last updated on 02/Sep/20 $$\mathrm{y}''+\mathrm{2y}'+\mathrm{y}=\mathrm{e}^{−\mathrm{2x}} +\mathrm{2x}+\mathrm{3} \\ $$ Commented by mohammad17 last updated on 02/Sep/20 $${r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}=\mathrm{0}\Rightarrow\left({r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{r}_{\mathrm{1}} ={r}_{\mathrm{2}} =−\mathrm{1}…

donner-la-forme-trigonometrique-de-1-4-cos-9-isin-9-

Question Number 176549 by doline last updated on 21/Sep/22 $${donner}\:{la}\:{forme}\:{trigonometrique}\:{de}\:\mathrm{1}/\mathrm{4}\left({cos}\Pi/\mathrm{9}+{isin}\Pi/\mathrm{9}\right) \\ $$ Answered by Ar Brandon last updated on 21/Sep/22 $$\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{cos}\frac{\pi}{\mathrm{9}}+{i}\mathrm{sin}\frac{\pi}{\mathrm{9}}\right)}=\frac{\mathrm{1}}{\mathrm{4}{e}^{\frac{\pi}{\mathrm{9}}{i}} }=\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\frac{\pi}{\mathrm{9}}{i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\frac{\pi}{\mathrm{9}}−{i}\mathrm{sin}\frac{\pi}{\mathrm{9}}\right) \\…