Menu Close

Author: Tinku Tara

1-z-2i-z-2i-2-z-2i-3-z-2i-4-0-find-z-z-C-

Question Number 70914 by 20190927 last updated on 09/Oct/19 $$\mathrm{1}+\left(\mathrm{z}+\mathrm{2i}\right)+\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{2}} +\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{3}} +\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{z}\:,\:\mathrm{z}\in\mathrm{C} \\ $$ Commented by mathmax by abdo last updated on…

Mr-A-wants-to-deliver-7-letters-to-his-7-friends-so-that-each-gets-1-letter-All-of-the-letters-are-written-of-the-addresses-of-his-7-friends-Find-the-probbility-that-3-of-his-friends-receive-the-co

Question Number 136448 by adhigenz last updated on 22/Mar/21 $$\mathrm{Mr}.\mathrm{A}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{deliver}\:\mathrm{7}\:\mathrm{letters}\:\mathrm{to}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}\:\mathrm{so}\:\mathrm{that}\:\mathrm{each}\:\mathrm{gets}\:\mathrm{1}\:\mathrm{letter}. \\ $$$$\mathrm{All}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{written}\:\mathrm{of}\:\mathrm{the}\:\mathrm{addresses}\:\mathrm{of}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{probbility}\:\mathrm{that}, \\ $$$$\mathrm{3}\:\mathrm{of}\:\mathrm{his}\:\mathrm{friends}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{letters}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{4}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{wrong}\:\mathrm{ones}. \\ $$ Answered by mr W last updated on 22/Mar/21 $${p}=\frac{{P}_{\mathrm{3}}…

Question-70915

Question Number 70915 by Mr. K last updated on 09/Oct/19 Commented by Mr. K last updated on 09/Oct/19 $${The}\:{circles}\:{have}\:{the}\:{same}\:{radius}.\: \\ $$$${The}\:{triangle}\:{is}\:{equilateral}\:{side} \\ $$$$\mathrm{28}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right).\:{Determine}\:{the}\:{radius} \\ $$$${of}\:{the}\:{circumferences}.…

Question-70913

Question Number 70913 by TawaTawa last updated on 09/Oct/19 Commented by mathmax by abdo last updated on 10/Oct/19 $${we}\:{have}\:\frac{{z}^{\mathrm{2017}} −\mathrm{1}}{{z}−\mathrm{1}}\:=\prod_{{k}=\mathrm{1}} ^{\mathrm{2016}} \left({z}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{2017}}} \right) \\ $$$${z}=−\mathrm{1}\:\Rightarrow\mathrm{1}\:=\prod_{{k}=\mathrm{1}}…

Question-136445

Question Number 136445 by aupo14 last updated on 22/Mar/21 Answered by liberty last updated on 22/Mar/21 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \:\frac{{t}−\mathrm{3}}{{t}^{\mathrm{2}} +\mathrm{7}}\:{dt}\: \\ $$$${f}\:'\left({x}\right)=\:\frac{{x}−\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{7}}\:=\:\mathrm{0}\Rightarrow{x}=\mathrm{3} \\ $$$${f}\:''\left({x}\right)\mid_{{x}=\mathrm{3}}…