Question Number 79479 by jagoll last updated on 25/Jan/20 $$\mathrm{how}\:\mathrm{do}\:\mathrm{16}\:\mathrm{people}\:\mathrm{play}\:\mathrm{3}\:\mathrm{matches} \\ $$$$\mathrm{in}\:\mathrm{teams}\:\mathrm{of}\:\mathrm{4}\:\mathrm{but}\:\mathrm{must}\:\mathrm{only}\:\mathrm{be} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{team}\:\mathrm{once}\:?\: \\ $$ Commented by mr W last updated on 25/Jan/20 $${i}\:{don}'{t}\:{understand}\:{the}\:{question}…
Question Number 79472 by jagoll last updated on 25/Jan/20 $$\mathrm{what}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}}\:+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{14x}+\mathrm{65}} \\ $$ Commented by john santu last updated on 25/Jan/20 $${yes}\:{sir}.\:{haha}\:{this}\:{method}\:{not}\:…
Question Number 13508 by tawa tawa last updated on 20/May/17 $$\mathrm{5}!!\:=\:? \\ $$$$\mathrm{please}\:\mathrm{workings},\:\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{15} \\ $$ Answered by mrW1 last updated on 20/May/17 $${n}!!={n}×\left({n}−\mathrm{2}\right)×\left({n}−\mathrm{4}\right)×…×\mathrm{2}\:\:{or}\:\:×\mathrm{1} \\ $$$$…
Question Number 78799 by Pratah last updated on 20/Jan/20 Commented by mathmax by abdo last updated on 20/Jan/20 $${convergence}?\:\:\:{let}\:{f}\left({x}\right)=\frac{{ln}\left({x}\right)}{{x}^{\frac{\mathrm{5}}{\mathrm{4}}} }\:\:\:{with}\:{x}\geqslant\mathrm{2}\:\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)=\frac{\frac{\mathrm{1}}{{x}}{x}^{\frac{\mathrm{5}}{\mathrm{4}}} −\frac{\mathrm{5}}{\mathrm{4}}{x}^{\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}} {lnx}}{{x}^{\frac{\mathrm{5}}{\mathrm{2}}}…
Question Number 131464 by bemath last updated on 05/Feb/21 $$\:\begin{array}{|c|c|}{{old}\:{plate}\::\:{EEU}\:\mathrm{874}}\\{{new}\:{plate}\::\:\mathrm{1}{BXK}\:\mathrm{267}}\\\hline\end{array} \\ $$$${Old}\:{California}\:{license}\:{plate}\: \\ $$$${consisted}\:{of}\:{a}\:{sequence}\:{of} \\ $$$${three}\:{letters}\:{followed}\:{by}\:{three} \\ $$$${digits}\:\left({see}\:{figure}\:{above}\right). \\ $$$${Assuming}\:{that}\:{any}\:{sequence} \\ $$$${of}\:{letters}\:{and}\:{digits}\:{was}\:{allowed} \\ $$$$\left({though}\:{actually}\:{some}\:{combinations}\right. \\…
Question Number 131459 by bramlexs22 last updated on 05/Feb/21 $${Eight}\:{eligible}\:{bachelor}\:{and}\:{seven}\:{beautiful} \\ $$$${models}\:{happen}\:{randomly}\:{to}\:{have}\: \\ $$$${purchased}\:{single}\:{seats}\:{in}\:{the}\:{same}\:\mathrm{15}−{seats} \\ $$$${row}\:{of}\:{theather}.\:{On}\:{the}\:{average}\:,\:{how}\:{many} \\ $$$${pairs}\:{of}\:{adjacent}\:{seats}\:{are}\:{ticketed} \\ $$$${for}\:{marriageable}\:{couples}\:? \\ $$ Answered by bemath…
Question Number 131248 by bramlexs22 last updated on 03/Feb/21 $${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{coefficient}\: \\ $$$${of}\:{x}^{\mathrm{8}} \:{and}\:{x}^{−\mathrm{24}} \:{respectively}\: \\ $$$${in}\:{the}\:{expansion}\:{of}\:\left[\:{x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\right]^{\mathrm{10}} \\ $$$${in}\:{powers}\:{of}\:{x}\:{then}\:\frac{\alpha}{\beta}\:{is}\:{equal}\:{to}\: \\ $$ Answered by EDWIN88…
Question Number 131162 by EDWIN88 last updated on 02/Feb/21 $${Five}\:{children}\:{sitting}\:{one}\:{behind} \\ $$$${the}\:{other}\:{in}\:{a}\:{five}\:{seater}\:{merry}−{go}−{round} \\ $$$$,{decide}\:{to}\:{switch}\:{seats}\:{so}\:{that}\:{each} \\ $$$${child}\:{has}\:{new}\:{companion}\:{in}\:{front}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$ Commented by mr W last…
Question Number 66 by newuser last updated on 15/Nov/14 $$\mathrm{How}\:\mathrm{many}\:\mathrm{unique}\:\mathrm{arrangements}\:\mathrm{are}\: \\ $$$$\mathrm{possible}\:\mathrm{for}\:\mathrm{a}\:\mathrm{2}×\mathrm{2}\:\mathrm{Rubic}\:\mathrm{cube}? \\ $$ Commented by 123456 last updated on 14/Dec/14 $$\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$ Commented…
Question Number 23 by user1 last updated on 25/Jan/15 $$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\frac{{x}^{{n}} }{{n}!}\:,\:\mathrm{where}\:\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree}\:{r}\:\mathrm{in}\:{n}. \\ $$ Answered by user1 last updated on…