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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

Question Number 182675    Answers: 2   Comments: 0

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

Question Number 182649    Answers: 1   Comments: 0

f(x;y)=((x^2 +y^2 )/(x^2 +y^4 )) lim_(x→∞) (lim_(y→0) f(x;y))=? lim_(y→∞) (lim_(x→∞) f(x;y))=?

$$\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)=\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{4}} }\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\right)=? \\ $$$$\underset{\boldsymbol{{y}}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\right)=? \\ $$

Question Number 182648    Answers: 3   Comments: 0

∫_0 ^(π/4) ((sec^2 x)/(tan x−sec x)) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 182646    Answers: 0   Comments: 2

prove that lim_(n→∞) [((((n+1)!∙(2n+1)!!))^(1/(n+1)) /(n+1))−(((n!∙(2n−1)!!))^(1/n) /n)]=(2/e^2 )

$${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!\centerdot\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}{{n}+\mathrm{1}}−\frac{\sqrt[{{n}}]{{n}!\centerdot\left(\mathrm{2}{n}−\mathrm{1}\right)!!}}{{n}}\right]=\frac{\mathrm{2}}{{e}^{\mathrm{2}} } \\ $$

Question Number 182644    Answers: 0   Comments: 5

solve for x>0 (1+(1/x))^x >(5/2)

$${solve}\:{for}\:{x}>\mathrm{0} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} >\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Question Number 182640    Answers: 0   Comments: 0

which software is the best for mathematics problems solving by video edit maker?

$${which}\:{software}\:{is}\:{the}\:{best}\:{for} \\ $$$${mathematics}\:{problems}\:{solving}\:{by}\:{video} \\ $$$${edit}\:{maker}? \\ $$

Question Number 182636    Answers: 0   Comments: 2

∫_(−1) ^3 (1/x^2 ) dx=?

$$\underset{−\mathrm{1}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{dx}=? \\ $$

Question Number 182632    Answers: 1   Comments: 0

If log _(11) (3p)=log _(13) (q+6p) = log _(143) (q^2 ) find (p/q).

$$\:\:\mathrm{If}\:\mathrm{log}\:_{\mathrm{11}} \left(\mathrm{3p}\right)=\mathrm{log}\:_{\mathrm{13}} \left(\mathrm{q}+\mathrm{6p}\right)\:=\:\mathrm{log}\:_{\mathrm{143}} \left(\mathrm{q}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{find}\:\frac{\mathrm{p}}{\mathrm{q}}. \\ $$

Question Number 182629    Answers: 0   Comments: 0

Π_(n=2) ^∞ e(1−(1/n^2 ))^n^2 = ?

$$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\boldsymbol{\mathrm{e}}\left(\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{n}}^{\mathrm{2}} } \:=\:\:? \\ $$

Question Number 182627    Answers: 1   Comments: 0

(((1+(√5))/2))^(15) =((a+b(√5))/2) find a+b

$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{15}} =\frac{{a}+{b}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${find}\:{a}+{b} \\ $$

Question Number 182623    Answers: 2   Comments: 1

Question Number 182620    Answers: 1   Comments: 0

Question Number 182613    Answers: 2   Comments: 0

Question Number 182608    Answers: 2   Comments: 1

Question Number 182600    Answers: 3   Comments: 0

Question Number 182595    Answers: 0   Comments: 1

Question Number 182593    Answers: 1   Comments: 1

Question Number 182586    Answers: 0   Comments: 0

Question Number 182584    Answers: 1   Comments: 0

Transform into factorial form the following A = 1×3×5×7×9×...×(2n−1) . Example: 4×3×2×1=4!

$$\:\:{Transform}\:{into}\:{factorial}\:{form}\:{the}\:{following} \\ $$$$\:\:{A}\:=\:\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{9}×...×\left(\mathrm{2}{n}−\mathrm{1}\right)\:.\: \\ $$$${Example}:\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{4}! \\ $$

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