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Question Number 182716 Answers: 1 Comments: 0

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Question Number 182705 Answers: 1 Comments: 0

A father has $320 to go a sporting event with his children. If he takes tickets of $50 he lacks money, if he takes tickets of $40 he has money left over. Find the number of children.

$${A}\:{father}\:{has}\:\$\mathrm{320}\:{to}\:{go}\:{a}\:{sporting}\:{event}\:{with} \\ $$$$\:{his}\:{children}.\:{If}\:{he}\:{takes}\:{tickets}\:{of}\:\$\mathrm{50}\:{he} \\ $$$$\:{lacks}\:{money},\:{if}\:{he}\:{takes}\:{tickets}\:{of}\:\$\mathrm{40}\:{he} \\ $$$$\:{has}\:{money}\:{left}\:{over}.\:{Find}\:{the}\:{number}\:{of}\: \\ $$$$\:{children}. \\ $$

Question Number 182704 Answers: 2 Comments: 0

On a path there was an odd number of rocks with a distance of 10m between them, we want to put them all where the middle one is, if we start to collect from one of the ends and when we finish we have walked 3km in total, how many rocks are there?

$${On}\:{a}\:{path}\:{there}\:{was}\:{an}\:{odd}\:{number}\:{of}\:{rocks} \\ $$$$\:{with}\:{a}\:{distance}\:{of}\:\mathrm{10}{m}\:{between}\:{them},\:{we}\: \\ $$$$\:{want}\:{to}\:{put}\:{them}\:{all}\:{where}\:{the}\:{middle}\:{one}\:{is}, \\ $$$$\:{if}\:{we}\:{start}\:{to}\:{collect}\:{from}\:{one}\:{of}\:{the}\:{ends}\:{and} \\ $$$$\:{when}\:{we}\:{finish}\:{we}\:{have}\:{walked}\:\mathrm{3}{km}\:{in}\:{total}, \\ $$$$\:{how}\:{many}\:{rocks}\:{are}\:{there}? \\ $$

Question Number 182702 Answers: 0 Comments: 1

log_2 (x+1)−x>0

$$\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)−{x}>\mathrm{0} \\ $$

Question Number 182695 Answers: 1 Comments: 2

If two events A and B are such that P(A^c )=0.3 ; P(B)=0.4 and P(A∩B^c )= 0.5 then P((B/(A∪B^c )))=? (A) 0.20 (B) 0.25 (C) 0.30 (D) 0.35

$$\:\:\mathrm{If}\:\mathrm{two}\:\mathrm{events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{such}\: \\ $$$$\:\:\mathrm{that}\:\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \right)=\mathrm{0}.\mathrm{3}\:;\:\mathrm{P}\left(\mathrm{B}\right)=\mathrm{0}.\mathrm{4}\:\mathrm{and} \\ $$$$\:\:\mathrm{P}\left(\mathrm{A}\cap\mathrm{B}^{\mathrm{c}} \right)=\:\mathrm{0}.\mathrm{5}\:\mathrm{then}\:\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}\cup\mathrm{B}^{\mathrm{c}} }\right)=? \\ $$$$\:\left(\mathrm{A}\right)\:\mathrm{0}.\mathrm{20}\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{0}.\mathrm{25}\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{0}.\mathrm{30}\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{0}.\mathrm{35} \\ $$$$\:\: \\ $$

Question Number 182694 Answers: 1 Comments: 0

Question Number 182681 Answers: 5 Comments: 6

Solve (√(39− 10 a −a^2 )) + (√(9−a^2 )) = 2 (√(14)) ; a∈ R^(+★) @Mr. Rasheed, yesterday i solved it on a draft paper When i came today to write it down on the notebook i got into a maze, do you have an easy way to deal with? “if you have time” i lost that paper, and this formula makes me laugh at myself, it′s for 13 y/o “confused face” Sure Every friend can share!

$$\:{Solve}\:\sqrt{\mathrm{39}−\:\mathrm{10}\:{a}\:−{a}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{9}−{a}^{\mathrm{2}} }\:=\:\mathrm{2}\:\sqrt{\mathrm{14}}\:\:\:;\:{a}\in\:\mathbb{R}^{+\bigstar} \\ $$$$\:@{Mr}.\:{Rasheed},\:{yesterday}\:{i}\:{solved}\:{it}\:{on}\:{a}\:{draft}\:{paper} \\ $$$$\:{When}\:{i}\:{came}\:{today}\:{to}\:{write}\:{it}\:{down}\:{on} \\ $$$$\:{the}\:{notebook}\:{i}\:{got}\:{into}\:{a}\:{maze},\:{do}\:{you}\:{have} \\ $$$$\:{an}\:{easy}\:{way}\:{to}\:{deal}\:{with}?\:``{if}\:{you}\:{have}\:{time}'' \\ $$$$\:{i}\:{lost}\:{that}\:{paper},\:{and}\:{this}\:{formula}\:{makes}\:{me} \\ $$$$\:{laugh}\:{at}\:{myself},\:{it}'{s}\:{for}\:\mathrm{13}\:{y}/{o}\:``{confused}\:{face}''\: \\ $$$$\:{Sure}\:{Every}\:{friend}\:{can}\:{share}! \\ $$

Question Number 182676 Answers: 2 Comments: 0

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Question Number 182659 Answers: 1 Comments: 0

lim_(n→∞) (((n!))^(1/n) /n)=?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{{n}!}}{{n}}=? \\ $$

Question Number 182657 Answers: 0 Comments: 0

as we can know for Q_1 , set the funvction in question as f(x) and make the first step as: a=1,f(x)=x^2 −7x+3ln x df(x)=2x−7+(3/x)=h(x) set h(x)=0⇒x=(1/2)&x=3 so,the monotonicity of f(x) is f(n)<f((1/2))_(∣n<(1/2)) ,f((1/2))>f(2),f(2)<f(p)_(∣p>2.) Q_1 has been proved finished Q_(2 ) ,set ax^2 −(a+6)x+3ln x>−6 when x∈[2,3e], make the f(x) dive two part as g(x)=ax^2 −(6+a)x&k(x)=−3(ln x+2), the middle value of g(x) is x=((a+6)/(2a)) when x=3e, k_(min) (3e)=−15, x=2, k_(max) (2)=−3(ln 2+2) g(2)=2a−12>k_(max) (x)⇒a>3−(3/2)ln 2 when x=(1/2)+(3/a), g(x)=−(((a+6)^2 )/(4a))>−3(ln (((a+6)/(2a)))+2)&a>0&2<(1/2)+(3/a)<3e

$${as}\:{we}\:{can}\:{know}\:{for}\:{Q}_{\mathrm{1}} ,\:{set}\:{the}\:{funvction}\:{in}\:{question}\:{as}\:{f}\left({x}\right) \\ $$$${and}\:{make}\:{the}\:{first}\:{step}\:{as}: \\ $$$${a}=\mathrm{1},{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{3ln}\:{x} \\ $$$${df}\left({x}\right)=\mathrm{2}{x}−\mathrm{7}+\frac{\mathrm{3}}{{x}}={h}\left({x}\right) \\ $$$${set}\:{h}\left({x}\right)=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\&{x}=\mathrm{3} \\ $$$${so},{the}\:{monotonicity}\:{of}\:{f}\left({x}\right)\:{is}\:{f}\left({n}\right)<{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mid{n}<\frac{\mathrm{1}}{\mathrm{2}}} ,{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)>{f}\left(\mathrm{2}\right),{f}\left(\mathrm{2}\right)<{f}\left({p}\underset{\mid{p}>\mathrm{2}.} {\right)} \\ $$$${Q}_{\mathrm{1}} \:{has}\:{been}\:{proved}\:{finished} \\ $$$${Q}_{\mathrm{2}\:} \:,{set}\:{ax}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){x}+\mathrm{3ln}\:{x}>−\mathrm{6}\:{when}\:{x}\in\left[\mathrm{2},\mathrm{3}{e}\right], \\ $$$${make}\:{the}\:{f}\left({x}\right)\:{dive}\:{two}\:{part}\:{as} \\ $$$$\:{g}\left({x}\right)={ax}^{\mathrm{2}} −\left(\mathrm{6}+{a}\right){x\&k}\left({x}\right)=−\mathrm{3}\left(\mathrm{ln}\:{x}+\mathrm{2}\right), \\ $$$$\:{the}\:{middle}\:{value}\:{of}\:{g}\left({x}\right)\:{is}\:{x}=\frac{{a}+\mathrm{6}}{\mathrm{2}{a}} \\ $$$${when}\:{x}=\mathrm{3}{e},\:{k}_{{min}} \left(\mathrm{3}{e}\right)=−\mathrm{15},\:\:{x}=\mathrm{2},\:{k}_{{max}} \left(\mathrm{2}\right)=−\mathrm{3}\left(\mathrm{ln}\:\mathrm{2}+\mathrm{2}\right) \\ $$$${g}\left(\mathrm{2}\right)=\mathrm{2}{a}−\mathrm{12}>{k}_{{max}} \left({x}\right)\Rightarrow{a}>\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}},\:{g}\left({x}\right)=−\frac{\left({a}+\mathrm{6}\right)^{\mathrm{2}} }{\mathrm{4}{a}}>−\mathrm{3}\left(\mathrm{ln}\:\left(\frac{{a}+\mathrm{6}}{\mathrm{2}{a}}\right)+\mathrm{2}\right)\&{a}>\mathrm{0\&2}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}}<\mathrm{3}{e} \\ $$$$ \\ $$

Question Number 182650 Answers: 0 Comments: 0