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Question Number 141122 Answers: 0 Comments: 4

M=<a;a+1;a+2;...;a+n> N=<a;a^2 ;a^3 ;...;a^n > be an ideals in Q[a] ; where n∈2Z M/N=?

$${M}=<{a};{a}+\mathrm{1};{a}+\mathrm{2};...;{a}+{n}> \\ $$$${N}=<{a};{a}^{\mathrm{2}} ;{a}^{\mathrm{3}} ;...;{a}^{{n}} > \\ $$$${be}\:{an}\:{ideals}\:{in}\:{Q}\left[{a}\right]\:;\:{where}\:\:{n}\in\mathrm{2}\mathbb{Z} \\ $$$${M}/{N}=? \\ $$

Question Number 141111 Answers: 1 Comments: 1

A gap must be left between steel rails to allow for thermal expansion. How large a gap is needed if the maximum temperature reached is 50° above the temperature at which the rails were laid. The length of the rail is 10m and the α_(steal) =12×10^(−6) /°c

$$\mathrm{A}\:\mathrm{gap}\:\mathrm{must}\:\mathrm{be}\:\mathrm{left}\:\mathrm{between}\:\mathrm{steel}\: \\ $$$$\mathrm{rails}\:\mathrm{to}\:\mathrm{allow}\:\mathrm{for}\:\mathrm{thermal}\:\mathrm{expansion}. \\ $$$$\:\mathrm{How}\:\mathrm{large}\:\mathrm{a}\:\mathrm{gap}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{maximum}\:\mathrm{temperature}\:\mathrm{reached} \\ $$$$\:\mathrm{is}\:\mathrm{50}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{at}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{rails}\:\mathrm{were}\:\mathrm{laid}.\:\mathrm{The}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rail}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\alpha_{\mathrm{steal}} =\mathrm{12}×\mathrm{10}^{−\mathrm{6}} /°\mathrm{c} \\ $$

Question Number 141110 Answers: 1 Comments: 0

sin^2 x+sinx=(4/3) , x∈[0,π] equation sum of roots in range?

$${sin}^{\mathrm{2}} {x}+{sinx}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:,\:\:\:{x}\in\left[\mathrm{0},\pi\right] \\ $$$${equation}\:\:{sum}\:\:{of}\:{roots}\:{in}\:{range}? \\ $$

Question Number 141107 Answers: 0 Comments: 0

Question Number 141104 Answers: 1 Comments: 0

Question Number 141098 Answers: 0 Comments: 0

Question Number 141087 Answers: 2 Comments: 0

hi, masters ! look at this thing carefully : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ................. find the line and the column where the number 795471 will appear !

Question Number 141085 Answers: 2 Comments: 0

Question Number 141062 Answers: 1 Comments: 4

prove that:: φ:=∫_0 ^( ∞) ((e^(cos(x)) sin(sin(x)))/x) dx=(π/2)(e−1)

$$\:\:\:\:\: \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{cos}\left({x}\right)} {sin}\left({sin}\left({x}\right)\right)}{{x}}\:{dx}=\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$

Question Number 141057 Answers: 2 Comments: 0

.....Nice ...... ......Calculus..... prove that: Ω(x):=Σ_(n=1) ^∞ a^n .((sin(nx))/(n!))=e^(acos(x)) sin(asin(x)) ....m.n

$$ \\ $$$$\:\:\:\:\:\:\:.....\mathscr{N}{ice}\:......\:\:......\mathscr{C}{alculus}..... \\ $$$$\:\:\:{prove}\:{that}: \\ $$$$\:\Omega\left({x}\right):=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}^{{n}} .\frac{{sin}\left({nx}\right)}{{n}!}={e}^{{acos}\left({x}\right)} {sin}\left({asin}\left({x}\right)\right) \\ $$$$\:\:\:....{m}.{n} \\ $$

Question Number 141076 Answers: 1 Comments: 0

t+x=(c/2) t^2 +x^4 =(c^2 /4) find x or t . Given 0<c<(2/(3(√3)))

$${t}+{x}=\frac{{c}}{\mathrm{2}} \\ $$$$\:{t}^{\mathrm{2}} +{x}^{\mathrm{4}} =\frac{{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${find}\:{x}\:{or}\:{t}\:.\:{Given}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\: \\ $$

Question Number 141050 Answers: 1 Comments: 0

Question Number 141077 Answers: 2 Comments: 0

Calculate the lim_(x→1) ((((√x^x ) − (√x))/(x^x − x))).

$$\:{Calculate}\:\:{the}\:\:\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\sqrt{{x}^{{x}} }\:−\:\sqrt{{x}}}{{x}^{{x}} \:−\:{x}}\right). \\ $$

Question Number 141082 Answers: 3 Comments: 0

........ nice ....... calculus ........ 𝛗:=Σ_(n=2) ^∞ ((ζ ( n ))/(n . 4^n ))=?

$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:........\:{nice}\:\:.......\:\:{calculus}\:........ \\ $$$$\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\zeta\:\left(\:{n}\:\right)}{{n}\:.\:\mathrm{4}^{{n}} }=? \\ $$$$ \\ $$

Question Number 141045 Answers: 0 Comments: 4

Question Number 141034 Answers: 0 Comments: 2

I don′t recover my old phone documents. please advise me in briefly how to restore my old phone documents in my new phone. plese help me.

$${I}\:{don}'{t}\:{recover}\:{my}\:{old}\:{phone}\:{documents}. \\ $$$${please}\:{advise}\:{me}\:{in}\:{briefly}\:{how}\:{to}\:{restore} \\ $$$$\:{my}\:{old}\:{phone}\:{documents}\:{in}\:{my}\:{new} \\ $$$$\:{phone}. \\ $$$$ \\ $$$${plese}\:{help}\:{me}. \\ $$

Question Number 141031 Answers: 0 Comments: 0

∫_(π/6) ^( π/3) (√(1+((cos^2 x)/(sin x)))) dx ?

$$\:\:\:\:\:\:\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{3}} \:\sqrt{\mathrm{1}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}}\:{dx}\:?\: \\ $$

Question Number 141030 Answers: 1 Comments: 1

→⟨Σ_(n=1) ^∞ (1/(9n^2 +3n)) =? ⟩←

$$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$

Question Number 141024 Answers: 3 Comments: 0

Given X=((√(√x))+(1/( (√(√x)))))^n . What is the coefficient of x^(5−(n/4) ?) propositions: a. 5((n!)/(10!)) b. ((n!)/(5!)) c. ((n),((10)) ) d. ((n),(5) )

$${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}−\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$

Question Number 141017 Answers: 2 Comments: 1

Question Number 141004 Answers: 0 Comments: 0

Let 0 ≤ a,b < 1. Prove that (1/4)∙(((2−a)(2−b))/((1−a)(1−b))) ≥ ((4+a+b)/(4−a−b))

$$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140998 Answers: 1 Comments: 0

Question Number 140997 Answers: 0 Comments: 1

Let a,b,c ≥ 0. Prove that (1/8)∙(((2+a)(2+b)(2+c))/((1+a)(1+b)(1+c))) ≥ ((4−a−b−c)/(4+a+b+c))

$$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140996 Answers: 1 Comments: 0

Evaluation of :: Ω :=Σ_(n=0) ^∞ (((−1)^n )/(1+n^2 )) solution:: Ω:=1+Σ_(n=1) ^∞ (((−1)^n )/(n^2 −i^2 )) =(1/(2i)){Σ_(n=1) ^∞ (((−1)^n )/(n−i))−(((−1)^n )/(n+i))} :=1+(1/(2i)) (Φ−Ψ) where Φ:=Σ_(n=1) ^∞ (((−1)^n )/(n−i)) and Ψ :=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) Φ:=Σ_(n=1) ^∞ (1/(2n−k)) −Σ_(n=1) ^∞ (1/(2n−1−i)) :=(1/2){Σ_(n=1) ^∞ (1/(n−(i/2)))−Σ_(n=1) ^∞ (1/(n−((1+i)/2)))} :=(1/2){ψ(1−((1+i)/2))−ψ(1−(i/2))} :=(1/2)(ψ(((1−i)/2))−ψ(1−(i/2))).... Ψ:=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) =Σ_(n=1) ^∞ (1/(2n+i))−Σ_(n=1) ^∞ (1/(2n−1+i)) :=(1/2){Σ_(n=1) ^∞ (1/(n+(i/2)))−Σ_(n=1) ^∞ (1/(n+((i−1)/2)))} :=(1/2)(ψ(((1+i)/2))−ψ(1+(i/2))) .... Φ−Ψ:=(1/2){ψ(((1−i)/2))−ψ(((1+i)/2))} +(1/2){ψ(1+(i/2))−ψ(1−(i/2))} :=(1/2)(−πcotπ(((1−i)/2)))+(1/2)((2/i)−πcot(π(i/2))) :=−(π/2)tan(((πi)/2))−(π/2)cot(((πi)/2))−i :=−i−(π/(sin(πi)))=−i−((2iπ)/(e^(−π) −e^π )) :=−i+πicsch(π) .... Ω :=1+(1/(2i))(−i+πicsch(π))=(1/2)+(π/2) csch(π) ... Ω:=(1/2)+(π/2) csch(π)....✓✓✓

Question Number 140988 Answers: 1 Comments: 0

∫((−csc^2 x)/((cscx+cotx)^3 ))dx

$$\int\frac{−{csc}^{\mathrm{2}} {x}}{\left({cscx}+{cotx}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 141012 Answers: 0 Comments: 0

s^2 (s+1)^2 +(c/2)s(s+1)^2 −(c^2 /2)(s+1) +(c^3 /2) = 0 solve for s in terms of 0<c<(2/(3(√3))) .

$$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

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