Question Number 28821 by abdo imad last updated on 30/Jan/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{4}\:{cos}\left(\mathrm{4}\theta\right)}{\mathrm{5}−\mathrm{4}{cos}\theta}\:{d}\theta\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 159881 by LEKOUMA last updated on 22/Nov/21 $${Resolve}\: \\ $$$$\mathrm{1}.\:\:{u}_{{n}} −\mathrm{3}{u}_{{n}−\mathrm{1}} =\mathrm{12}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \\ $$$$\mathrm{2}.\:{u}_{{n}} =\mathrm{2}{u}_{{n}−\mathrm{1}} +\mathrm{5cos}\:\left(\frac{{n}\Pi}{\mathrm{3}}\right),\:{u}_{{o}} =\mathrm{1}\: \\ $$$$\mathrm{3}.\:{u}_{{n}} ={u}_{{n}−\mathrm{1}} −{u}_{{n}−\mathrm{2}} +\mathrm{2sin}\:\left(\frac{{n}\Pi}{\mathrm{3}}\right) \\…
Question Number 94319 by Mozay1 last updated on 18/May/20 Commented by john santu last updated on 18/May/20 $$\left(\mathrm{1b}\right)\:\left(\mathrm{2}+\mathrm{3y}+\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{5}} =\left\{\left(\mathrm{y}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{2}\right)\right\}^{\mathrm{5}} \\ $$$$=\:\left\{\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\:\mathrm{C}\:_{\mathrm{k}} ^{\mathrm{5}}…
Question Number 159839 by abelpedro last updated on 21/Nov/21 $${q}=\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$ Answered by abdullah_ff last updated on 21/Nov/21 $${q}\:=\:\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{2}^{\mathrm{1}} \centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\…
Question Number 159737 by LEKOUMA last updated on 20/Nov/21 $${Prove}\: \\ $$$$\left.\mathrm{1}\right)\:{E}\left({x}\right)+{E}\left({y}\right)\leqslant{E}\left({x}+{y}\right)\leqslant{E}\left({x}\right)+{E}\left({y}\right)+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{E}\left({x}\right)+{E}\left({y}\right)+{E}\left({x}+\mathrm{1}\right)\leqslant{E}\left(\mathrm{2}{x}\right)+{E}\left(\mathrm{2}{y}\right) \\ $$$$\left.\mathrm{3}\right)\:{E}\left(\frac{{x}}{\mathrm{2}}\right)+{E}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)={E}\left({x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 159736 by LEKOUMA last updated on 20/Nov/21 $${Prove}\:{that} \\ $$$$\left.\mathrm{1}\right){Sup}\left({A}\cup{B}\right)={ma}\mathrm{x}\left(\mathrm{S}{up}\left({A}\right),\:{Sup}\left({B}\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{inf}\left({A}\cup{B}\right)={min}\left({inf}\left({A}\right),\:{inf}\left({B}\right)\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28626 by Cheyboy last updated on 28/Jan/18 Commented by Tinkutara last updated on 28/Jan/18 Which one? Commented by Cheyboy last updated on 28/Jan/18 $${both}\:{of}\:{them}\:{sir}…
Question Number 28600 by A1B1C1D1 last updated on 27/Jan/18 Commented by abdo imad last updated on 27/Jan/18 $${from}\:{where}\:{come}\:\frac{\mathrm{3}}{\mathrm{8}}\:{sir}\:?\:{there}\:{is}\:{no}\:{mistake}\:{in}\:{my}\:{response} \\ $$$${and}\:{if}\:{you}\:{find}\:{something}\:{inform}\:{me}\:… \\ $$ Commented by abdo…
Question Number 159675 by Tawa11 last updated on 19/Nov/21 Commented by mr W last updated on 20/Nov/21 $${i}\:{remember}\:{this}\:{question}\:{is}\:{not}\:{asked} \\ $$$${for}\:{the}\:{first}\:{time}. \\ $$ Commented by Tawa11…
Question Number 94081 by Rio Michael last updated on 16/May/20 $${P}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{representing}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\:{z}\:=\:{r}\left(\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\:\theta\right)\:\mathrm{in}\:\mathrm{an}\:\mathrm{argand}\:\mathrm{diagram}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mid{z}−{a}\mid\mid{z}\:+\:{a}\mid\:=\:{a}^{\mathrm{2}} .\:\mathrm{Show}\:\mathrm{that}\:{P}\:\mathrm{moves}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{whose}\:\mathrm{equation}\:\mathrm{is}\:{r}^{\mathrm{2}} \:=\mathrm{2}{a}^{\mathrm{2}} \:\mathrm{cos2}\theta.\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$${r}^{\mathrm{2}} \:=\:\mathrm{2}{a}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}\theta\:,\:\mathrm{showing}\:\mathrm{clearly}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the}\:\mathrm{pole}. \\…