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AllQuestion and Answers: Page 76

Question Number 205808 Answers: 2 Comments: 3

Question Number 205794 Answers: 1 Comments: 1

A = { (k/2^n ) ∣ 1≤ k ≤ 2^n , n∈N } find . A^( −) = ?

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{A}\:=\:\left\{\:\frac{{k}}{\mathrm{2}^{{n}} }\:\mid\:\mathrm{1}\leqslant\:{k}\:\leqslant\:\mathrm{2}^{{n}} \:,\:{n}\in\mathbb{N}\:\right\} \\ $$$$\:\:\:\:\:{find}\:.\:\:\overset{\:−} {\mathrm{A}}\:=\:? \\ $$$$ \\ $$

Question Number 205790 Answers: 0 Comments: 0

Question Number 205789 Answers: 0 Comments: 0

Question Number 205784 Answers: 0 Comments: 1

Question Number 205775 Answers: 1 Comments: 0

calcu/ limit/n→+oo ∫_0 ^(+oo) arctan((x/n))e^(−x) dx

$${calcu}/\:\:\:\:{limit}/{n}\rightarrow+{oo} \\ $$$$\:\:\int_{\mathrm{0}} ^{+{oo}} {arctan}\left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx} \\ $$

Question Number 205774 Answers: 2 Comments: 0

−−−−−−− Ω = Σ_(n=0) ^∞ (( (−1)^( n) )/((−1)^( n) −n)) = ? −−−−−−−

$$ \\ $$$$\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} }{\left(−\mathrm{1}\right)^{\:{n}} \:−{n}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:−−−−−−− \\ $$

Question Number 205772 Answers: 2 Comments: 0

Question Number 205770 Answers: 0 Comments: 0

If x,y,z>0 and xyz = 1 Prove that: ((((√2)x)^2 )/((1+xz)(1+xy))) + ((((√2)y)^2 )/((1+yz)(1+xy))) + ((((√2)z)^2 )/((1+xz)(1+yz))) ≥ (3/2)

$$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{xyz}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{yz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{yz}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 205767 Answers: 1 Comments: 0

a,b,c ∈ℜ^+ a+b+c=1 a^2 /(1+b+c) + b^2 /(1+a+c) + c^2 /(1+a+b)≥k find k max. hint : inequality cauchy schwarz

$$ \\ $$$${a},{b},{c}\:\in\Re^{+} \:\: \\ $$$${a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:{a}^{\mathrm{2}} /\left(\mathrm{1}+{b}+{c}\right)\:+\:{b}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{c}\right)\:\:+\:{c}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{b}\right)\geqslant{k} \\ $$$${find}\:\:\:{k}\:{max}. \\ $$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \\ $$$$ \\ $$

Question Number 205750 Answers: 2 Comments: 3

∫_0 ^(+∞) (1/(1+e^(2x) ))dx=?

$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}=? \\ $$

Question Number 205749 Answers: 0 Comments: 0

whats the suficient condition to became the question σ^2 (1−a_i )[λ_i (1+a_i )−(1−a_i )(1−d_i )+(λ_i +k)] < 0

$${whats}\:{the}\:{suficient}\:{condition}\:{to}\:{became}\:{the}\:{question}\: \\ $$$$ \\ $$$$\sigma^{\mathrm{2}} \left(\mathrm{1}−{a}_{{i}} \right)\left[\lambda_{{i}} \left(\mathrm{1}+{a}_{{i}} \right)−\left(\mathrm{1}−{a}_{{i}} \right)\left(\mathrm{1}−{d}_{{i}} \right)+\left(\lambda_{{i}} +{k}\right)\right]\:<\:\mathrm{0}\: \\ $$

Question Number 205746 Answers: 1 Comments: 0

Question Number 205734 Answers: 5 Comments: 0

Question Number 205733 Answers: 1 Comments: 0

If a,b,c∈R^+ and a+b+c=6 Prove that: ((a^2 −4)/(4a^2 −9a + 6)) + ((b^2 −4)/(4b^2 −9b + 6)) + ((c^2 −4)/(4c^2 −9c + 6)) ≤ 0

$$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$

Question Number 205727 Answers: 1 Comments: 4

calculate lim_(x→0) ((e^x −cosx)/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{e}^{{x}} −{cosx}}{{x}^{\mathrm{2}} } \\ $$

Question Number 205716 Answers: 2 Comments: 0

lim_(n→∞) (((2n+1)(2n+3)...(4n+1))/((2n)(2n+2)...(4n))) = ?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)...\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)...\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$

Question Number 205726 Answers: 1 Comments: 0

101 is chosen arbitrarily from the numbers 1,2,3,...,199,200. Prove that two of these selected numbers can be found such that one is divisible by the other.

$$ \\ $$101 is chosen arbitrarily from the numbers 1,2,3,...,199,200. Prove that two of these selected numbers can be found such that one is divisible by the other.

Question Number 205685 Answers: 2 Comments: 0

Question Number 205683 Answers: 1 Comments: 0

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Question Number 205682 Answers: 1 Comments: 2

Question Number 205681 Answers: 2 Comments: 0

Question Number 205680 Answers: 1 Comments: 0

solve ⌊x ⌋ + ⌊ x^2 ⌋ = ⌊ x^3 ⌋

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\lfloor{x}\:\rfloor\:+\:\lfloor\:{x}^{\mathrm{2}} \rfloor\:=\:\lfloor\:{x}^{\mathrm{3}} \:\rfloor \\ $$$$ \\ $$

Question Number 205690 Answers: 0 Comments: 3

Question Number 205673 Answers: 0 Comments: 0

Question Number 205672 Answers: 1 Comments: 0

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