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Using the Root Test: r = lim(n→∞) |(2 - e^(1/n))^(n^2)|^(1/n) ..= lim(n→∞) (2 - e^(1/n))^n Using logarithms, ln r = lim(n→∞) n ln(2 - e^(1/n)) ......= lim(n→∞) ln(2 - e^(1/n))/(1/n) ......= lim(t→0+) ln(2 - e^t)/t, letting t = 1/n ......= l…
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Using the Root Test: r = lim(n→∞) |(2 - e^(1/n))^(n^2)|^(1/n) ..= lim(n→∞) (2 - e^(1/n))^n Using logarithms, ln r = lim(n→∞) n ln(2 - e^(1/n)) ......= lim(n→∞) ln(2 - e^(1/n))/(1/n) ......= lim(t→0+) ln(2 - e^t)/t, letting t = 1/n ......= l…
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a) The region is bounded by y = ln x to y = 1 with x in [1, e]. Rewrite this as x = 0 to x = e^y with y in [0, 1]. So, the integral equals ∫(y = 0 to 1) ∫(x = 1 to e^y) sin(y^2) * 1/x dx dy = ∫(y = 0 to 1) sin(y^2) * ln |x| {for x = 1 to e^y} …
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a) The region is bounded by y = ln x to y = 1 with x in [1, e]. Rewrite this as x = 0 to x = e^y with y in [0, 1]. So, the integral equals ∫(y = 0 to 1) ∫(x = 1 to e^y) sin(y^2) * 1/x dx dy = ∫(y = 0 to 1) sin(y^2) * ln |x| {for x = 1 to e^y} …
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Remember that φ fixes Q. If φ(4√3) = a + b * 4√3 for some a, b in Q, then φ((4√3)^2) = (a + b * 4√3)^2, since φ respects multiplication ==> 48 = (a^2 + 48b^2) + 8ab√3 ==> a^2 + 48b^2 = 48 and 8ab = 0 ==> a = 0 and b = ±1. So, φ(4√…
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Remember that φ fixes Q. If φ(4√3) = a + b * 4√3 for some a, b in Q, then φ((4√3)^2) = (a + b * 4√3)^2, since φ respects multiplication ==> 48 = (a^2 + 48b^2) + 8ab√3 ==> a^2 + 48b^2 = 48 and 8ab = 0 ==> a = 0 and b = ±1. So, φ(4√…
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Fix a positive integer n. Note that dim(Pn)= n + 1, since {1, x, x^2, ..., x^n} is a basis for Pn. ----------------------------- Tweaking notation a bit... fix a scalar a (instead of x). If we want the dimension of the subspace S = {p(x) in P3 :…
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(a) Let A = [0 1] [0 0]. Then, A^2 is the zero matrix. (Note: [0 1]^2 [1 0] is the identity matrix.) (b) This is false; let A and X be the matrix from part (a), and let Y = zero matrix. Then, AX = AY = zero matrix, but X and Y…
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We have two cases: (1) p is a positive integer. Then, p applications of L'Hopital's Rule (∞/∞ case) yields lim(n→∞) n^p/e^n = lim(n→∞) pn^(p-1)/e^n (first application) = lim(n→∞) p(p-1)n^(p-2)/e^n (second application) ... = lim(n→∞) p!/e^n…
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We have two cases: (1) p is a positive integer. Then, p applications of L'Hopital's Rule (∞/∞ case) yields lim(n→∞) n^p/e^n = lim(n→∞) pn^(p-1)/e^n (first application) = lim(n→∞) p(p-1)n^(p-2)/e^n (second application) ... = lim(n→∞) p!/e^n…
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1) By the geometric series, 1/(1-x) = 1 + x + x^2 + x^3 + ... Integrate both sides from 0 to x: - ln(1 - x) = x + x^2/2 + x^3/3 + x^4/4 + ... Replace x with -x: - ln(1 + x) = -x + x^2/2 - x^3/3 + x^4/4 + ... ==> ln(1 + x) = x - x^2/2 + x…
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Encryption: Compute r^k mod p = 6^3 = 216 mod 2551 and Pb^k mod p = 111 * 33^3 = 1794 mod 2551 So, the cyphertext is the pair (216, 1794). Decryption: 1794 / (216^13) mod 2551 = 1794 / 223 mod 2551 By Euc…
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Encryption: Compute r^k mod p = 6^3 = 216 mod 2551 and Pb^k mod p = 111 * 33^3 = 1794 mod 2551 So, the cyphertext is the pair (216, 1794). Decryption: 1794 / (216^13) mod 2551 = 1794 / 223 mod 2551 By Euc…
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1) x = (1 - t^2)/(1 + t^2), y = 2t/(1 + t^2). This is a parameterization of the unit circle, which can be recognized by noting that x^2 + y^2 = 1. Not too surprisingly, this has constant curvature 1. --------------------------- 2) r(t) = <(…
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Quem não deve não teme... Se você a pardo então é pardo. Se sua certidão diz isso então deixa quieto. Uma faculdade ou instituição não vai contestar sua certidão. Para de se preocupar se mentiu ou não agora é tarde. A situação está fora de…
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Não. Uma das provas exigido pelo governo de qualquer país é que para poder estudar em no país precisa-se passar num exame de fluência com um media ja pre-estabelecida. Em alguns países a prova pode ser em inglês invés da linguá local, pois alguns p…
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I had horrible breakouts all the time before my period would start. I tried everything you did (over the counter stuff) I started taking Yasmin Birth Control and in 3-4 months it was completely clear. It's been about 7 months and I get one zit like…
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∑(n=1)^∞ :3/(5^n) = (3/5) * ∑(n=1)^∞ : (1/5)^(n-1) This is a multiple of the geometric series with |r| = 1/5 < 1. So, the sum equals (3/5) * [1/ (1 - 1/5)] = 3/4. I hope this helps!
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∑(n=1)^∞ :3/(5^n) = (3/5) * ∑(n=1)^∞ : (1/5)^(n-1) This is a multiple of the geometric series with |r| = 1/5 < 1. So, the sum equals (3/5) * [1/ (1 - 1/5)] = 3/4. I hope this helps!
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By Partial Fractions, f(x) = 2/(x^2 - 3x + 2) = 2/(x - 2) - 2/(x - 1). Since we only need four terms, we'll do this via differentiation. I'll do this for x = -1 (the other case is similar). f(x) = 2/(x - 2) - 2/(x - …
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lo siento tienes un 6
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Using the cross product, the desired perpendicular vector is <5,-2,1> x <-3,0,-3> = <6, 12, -6>. So, the equation of the line is r(t) = <-3, 4, 2> + t <6, 12, -6>. I hope this helps!
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Using the cross product, the desired perpendicular vector is <5,-2,1> x <-3,0,-3> = <6, 12, -6>. So, the equation of the line is r(t) = <-3, 4, 2> + t <6, 12, -6>. I hope this helps!
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Pretende trabalhar como auxiliar....?? Faça o curso superior, depois de um ano cursando ciências contábeis terá mais conhecimento do que um técnico. Não tem vantagem nenhuma fazer um curso técnico que não vai te permitir a exercer as mesmas funções…
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Pretende trabalhar como auxiliar....?? Faça o curso superior, depois de um ano cursando ciências contábeis terá mais conhecimento do que um técnico. Não tem vantagem nenhuma fazer um curso técnico que não vai te permitir a exercer as mesmas funções…
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1) ∂u/∂t = -n² sin(nx) e^(-n²t) ∂u/∂x = n cos(nx) e^(-n²t) ∂²u/∂x² = -n² sin(nx) e^(-n²t) Hence, ∂u/∂t = ∂²u/∂x². ----------------- 2) We need to show that u satisfies Laplace's Equation ∂²u/∂x² + ∂²u/∂y² = 0. ∂u/∂x = (1/(1 + (y/x)²) * (…
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Assuming that the index of summation starts at n = 0, consider S = Σ(n=0 to k) 3(-1/2)^n = 3 - 3/2 + 3/4 - ... + 3(-1/2)^k. Multiply both sides by (-1/2): -S/2 = Σ(n=0 to k) 3(-1/2)^(n+1) = -3/2 + 3/4 - ... + 3(-1/2)^k + 3(-1/2)^(k+1). Subtr…
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Look at the solution to #2 in the following link: http://faculty.csuci.edu/brian.sittinger/351_hw7so... I hope this helps!
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Look at the solution to #2 in the following link: http://faculty.csuci.edu/brian.sittinger/351_hw7so... I hope this helps!
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Rewrite f as f(x,y) = ln x - ln y. So, f_x = 1/x and f_y = -1/y ==> f_xx = -1/x^2, f_yy = 1/y^2, and f_xy = 0. Hence, x f_x - y^2 f_yy = x * (1/x) - y^2 * (1/y^2) = 1 - 1 = 0. I hope this helps!