§13 6 stage by stage methods for batch rectification
Frequency of Routine and Flooding-stage Observations for Precise Annual Total Pollutant Loads and their Estimating Method in the Yodo River
... 0. 967 3 0. 965 5 0.9 565 0 .67 97 2005 1.7510･Ｑ (0.9 96) 2.3140･Ｑ (0.997) 3.9714･Ｑ (0.992) 2.71 86 Ｑ (0.999) 76. 062 ･Ｑ (0.9 86) Flooding -stage( 2005) 1.5145･Ｑ1.10 76 (0.952) 2.8183･Ｑ0.9424 (0.9 56) 3 .65 54･Ｑ0.9827 ... 0. 969 5 0.9575 0 .69 15 2003 2.3184･Ｑ (0.957) 2.3545･Ｑ (0. 960 ) 5.8022･Ｑ (0. 966 ) 3.1102･Ｑ (0.974) 66 .704･Ｑ (0.941) 1.39 46 1.1014 0.9825 1. 061 3 0.5422 2004 0.2412･Ｑ (0.834) 1.9175･Ｑ (0.8 06) 4.5 064 ･Ｑ ... DOC Flow 1200 1000 800 60 0 400 200 4/21 4/27 5/3 5/9 5/15 5/21 5/27 6/ 2 6/ 8 6/ 14 6/ 20 6/ 26 7/2 7/8 7/14 7/20 7/ 26 8/1 8/7 8/13 8/19 8/25 8/31 9 /6 9/12 9/18 9/24 9/30 10 /6 10/12 10/18 10/24 10/30...
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Tutorial Final and Construction Stage Analysis for a Cable-Stayed Bridge potx
... 115, 1 46 787.47 16, 25, 1 26, 135 772.57 6, 35, 1 16, 145 718.19 15, 26, 125, 1 36 705.01 7, 34, 117, 144 67 1. 96 14, 27, 124, 137 66 7.43 8, 33, 118, 143 61 2.34 13, 28, 123, 138 63 9.52 9, 32, 119, ... 37, 114, 147 Tension 14 17, 24, 127, 134 Tension 5, 36, 115, 1 46 Tension 15 16, 25, 1 26, 135 Tension 6, 35, 1 16, 145 Tension 16 15, 26, 125, 1 36 Tension 7, 34, 117, 144 Tension 17 14, 27, 124, 137 ... factor Deformation Scale Factor (0.3) ↵ Zoom Window (A, B in Fig 36) A B Fig 36 Checking Deformed Shape 38 FINAL AND CONSTRUCTION STAGE ANALYSIS FOR CABLE-STAYED BRIDGES Construction Stage Analysis...
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Báo cáo khoa học: "An Effective Two-Stage Model for Exploiting Non-Local Dependencies in Named Entity Recognition" pdf
... Document Level Statistics PER LOC ORG MISC 3141 64 36 188 2975 2030 PER 33830 Corpus Level Statistics LOC ORG MISC 113 153 3 469 66 6749 60 43892 223 66 2 86 Table 1: Table showing the number of pairs ... Statistics PER LOC ORG MISC 1941 167 63 22 328 819 191 14 224 365 Corpus Level Statistics PER LOC ORG MISC 9111 401 261 38 68 4 560 580 1543 221 1 968 3 5131 4752 50 12713 329 8 768 Table 2: Table showing ... Sampling) Local+Viterbi 88. 16 80.83 78.51 90. 36 85.51 Non Local+Gibbs 88.51 81.72 80.43 92.29 86. 86 Our Approach with the 2 -stage CRF Baseline CRF 88.09 80.88 78. 26 89. 76 85.29 + Document token-majority...
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![picture yourself building a website with joomla! 1.6[electronic resource] step-by-step instruction for creating a high-quality, professional-looking site with ease](https://media.store123doc.com/images/document/14/y/ob/small_oby1401382455.jpg)
picture yourself building a website with joomla! 1.6[electronic resource] step-by-step instruction for creating a high-quality, professional-looking site with ease
... Optimization 263 Joomla! 1 .6 SEF Configuration 264 What Is SEO? 265 SEO Information ... 2 76 Web 2.0 Templates 2 76 Building a Joomla! 1 .6 Web 2.0–Featured Website 2 76 Facebook, Twitter, and Joomla! 1 .6 ... to use for the installation Do Not use MySQLi Use MySQL for the database type If all has gone well, after the connection to the database, the next screen asks for FTP information You can bypass...
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Báo cáo hóa học: " Research Article Strong Convergence Theorems by Shrinking Projection Methods for Class T Mappings" pptx
... convergence theorems by hybrid methods for families of nonexpansive mappings in Hilbert spaces,” Journal of Mathematical Analysis and Applications, vol 341, no 1, pp 2 76 2 86, 2008 Fixed Point ... converges strongly to PF x0 Proof We first show by induction that F ⊂ Cn for all n ∈ N.F ⊂ C1 is obvious Suppose F ⊂ Ck for some k ∈ N Note that, by the definition of Tk ∈ T, we always have F ⊂ ... one can easily obtain hybrid methods introduced by Nakajo and Takahashi for a nonexpansive mapping Recently, Takahashi et al proposed a shrinking projection method for nonexpansive mappings Tn...
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Báo cáo hóa học: " Research Article A Two-Stage Approach for Improving the Convergence of Least-Mean-Square Adaptive Decision-Feedback Equalizers in the Presence of Severe " doc
... PEF for rapid convergence and the DFE for removing postcursor ISI as a system to mitigate narrowband interference A similar approach is discussed in [25, pages 364 - 365 ] when deriving the zero-forcing ... September, 20 06 This work was supported by the Office of Naval Research, Code 313, SPAWAR Systems Center, San Diego, and the UCSD Center for Wireless Communications (UCDG Grant no Com 06- 102 16) The authors ... the IEEE, vol 63 , no 4, pp 561 –580, 1975 [5] J R Zeidler, “Performance analysis of LMS adaptive prediction filters,” Proceedings of the IEEE, vol 78, no 12, pp 1781– 18 06, 1990 [6] L.-M Li and...
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Chapter 13: Technical Aspects of Chargers and Current Transformers and Methods for Supervision pptx
... power output stage Transformer The high-frequency AC voltage is separated galvanically from the mains and adjusted to the output voltage by a ferrite transformer (T1) The high transformation frequency ... 2 .6 2.7 1.7–1.9 1 .65 –1.85 0.85–1.1 V/cell V/cell Copyright © 2003 by Expert Verlag All Rights Reserved Figure 13.4 Thyristor controlled chargers Control Electronics A control unit topped by ... voltage for the mains supply circuit consisting of a power stage and a ferrite transformer Power Output Stage The transistor (V1) is triggered by a frequency generator and ‘‘chops up’’ the rectified...
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Quantitative Methods for Business chapter 6 pps
... 7.5 62 .5 175.0 135.0 27.5 56. 25 1 56. 25 3 06. 25 5 06. 25 7 56. 25 56. 25 781.25 3 062 .50 3037.50 7 56. 25 ∑f ϭ 23 Cost of calls ∑fx ϭ 407.5 Frequency (f ) ∑fx2 ϭ 769 3.75 ⎡ (407.5)2 ⎤ ⎡ 166 0 56. 25 ⎤ 769 3.75 ... 15)/9 ϭ 7.111 Length of service (x) 4 10 11 15 64 (64 )2 (61 6 Ϫ )ϭ ϭ 20.111 ϭ 4.485 sϭ Slugar: x2 16 16 25 49 64 100 121 225 61 6 (61 6 Ϫ 455.111) ϭ ( 160 .889) Mean ϭ (0 ϩ ϩ ϩ ϩ ϩ 10 ϩ 10 ϩ 14 ϩ 15)/9 ... service (x) 0 4 10 10 14 15 64 x2 0 16 16 49 100 100 1 96 225 702 208 Quantitative methods for business (64 )2 (702 Ϫ )ϭ ϭ 30. 861 ϭ 5.555 sϭ (702 Ϫ 455.111) ϭ Chapter (2 46. 889) The means are the same,...
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Quantitative Methods for Business chapter 13 pdf
... 0. 363 2 0.3 264 0.2912 0.2578 0.2 266 0.1977 0.1711 0.1 469 0.4 761 0.4 364 0.3974 0.3594 0.3228 0.2877 0.25 46 0.22 36 0.1949 0. 168 5 0.14 46 0.4721 0.4325 0.39 36 0.3557 0.3192 0.2843 0.2514 0.22 06 0.1922 ... 0.3015 0. 267 6 0.2358 0.2 061 0.1788 0.1539 0.4880 0.4483 0.4090 0.3707 0.33 36 0.2981 0. 264 3 0.2327 0.2033 0.1 762 0.1515 0.4840 0.4443 0.4052 0. 366 9 0.3300 0.29 46 0. 261 1 0.2297 0.2005 0.17 36 0.1492 ... 0.2514 0.22 06 0.1922 0. 166 0 0.1423 0. 468 1 0.42 86 0.3897 0.3520 0.31 56 0.2810 0.2483 0.2177 0.1894 0. 163 5 0.1401 0. 464 1 0.4247 0.3859 0.3483 0.3121 0.27 76 0.2451 0.2148 0.1 867 0. 161 1 0.1379 Suppose...
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Quantitative Methods for Ecology and Evolutionary Biology (Cambridge, 2006) - Chapter 6 pot
... 19 86 1987 1.78 1.31 0.91 0. 96 0.88 0.9 0.87 0.72 0.57 0.45 0.42 0.42 0.49 0.43 0.4 0.45 0.5 0.53 0.58 0 .64 0 .66 0 .65 0 .63 94 212 195 383 320 402 366 60 6 378 319 309 389 277 254 170 97 91 177 2 16 ... 0.27 0.2 0.31 0.15 0. 36 0. 36 0 .6 0.74 0.53 1.75 0.0 1.75 1.0 2.5 1.75 3.75 3. 76 8.0 7.0 5.0 0 .62 0.44 0.24 1.03 0.53 0. 26 0. 56 0.82 0.8 0.8 11.00 7.4 5.8 10.0 6. 5 4.0 8 .6 6.5 10.0 10.0 There is ... fishing pattern by different classes or nationalities of vessels The data are as follows Year CPUE Catch (thousands of tons) 1 965 1 966 1 967 1 968 1 969 1970 1971 1972 1973 1974 1975 19 76 1977 1978...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
... 6. 23 1/2 Show that each of the numbers z = −a + (a2 − b) satisfies the equation z + 2az + b = Hint, Solution 207 6. 8 Hints Complex Numbers Hint 6. 1 Hint 6. 2 Hint 6. 3 Hint 6. 4 Hint 6. 5 Hint 6. 6 ... Hint 6. 7 The Complex Plane Hint 6. 8 Hint 6. 9 208 Hint 6. 10 Write the multivaluedness explicitly Hint 6. 11 Consider a triangle with vertices at 0, z and z + ζ Hint 6. 12 Hint 6. 13 Hint 6. 14 Hint 6. 15 ... = 4x2 − 16x + 16 + 4y 1 (x − 1)2 + y = Thus we have the standard form for an equation describing an ellipse 187 -1 -1 -2 Figure 6. 4: Solution of |z| + |z − 2| = 6. 3 Polar Form Polar form A complex...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... ) = = = for x = 2πk for x = 2πk 1−eınx 1−eıx for x = 2πk e−ıx/2 − eı(N −1/2)x e−ıx/2 − eıx/2 for x = 2πk e−ıx/2 − eı(N −1/2)x −ı2 sin(x/2) = N −1 sin(nx) = n=1 for x = 2πk for x = 2πk for x = ... integral converges only for absolutely for (α) ≤ 1 dx = |xα | ∞ 1 x (α) dx = [ln x]∞ x1− (α) 1− (α) for (α) = 1, for (α) = ∞ (α) > Thus the harmonic series converges absolutely for 589 (α) > and diverges ... series converges absolutely for |z| < 1/|β| By the ratio test formula, the radius of absolute convergence is n/2n n→∞ (n + 1)/2n+1 n = lim n→∞ n + =2 R = lim By the root test formula, the radius of...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx
... put in a simpler form by exchanging the dependent and independent variables Thus a differential equation for y(x) can be written as an equation for x(y) Solving the equation for x(y) will give ... (c + c ln x) Solution 17.15 For a = 0, two linearly independent solutions of y − a2 y = are y1 = eax , y2 = e−ax For a = 0, we have y1 = e0x = 1, y2 = x e0x = x 9 76 if a2 > b, if a2 < b, if a2 ... solution for all a For a = 0, y2 is a linear combination of eax and e−ax and is thus a solution Since the coefficient of e−ax in this linear combination is non-zero, it is linearly independent to y1 For...
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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps
... series to find the value of ∞ 16 n=1 π 1 = n6 π −π ∞ 16 n=1 ∞ n=1 x3 π − x 3 16 = n6 945 6 = n6 945 13 76 ∞ n=1 n6 dx Solution 28.2 We differentiate the partial sum of the Fourier series and ... series for f (x) = x2 x ξ2 − ∞ π2 3 dξ = n=1 ∞ (−1)n n2 x cos(nξ) dξ x π (−1)n − x=4 sin(nx) 3 n3 n=1 We apply Parseval’s theorem for this series to find the value of ∞ 16 n=1 π 1 = n6 π −π ∞ 16 n=1 ... π2 (−1)n x = +4 cos(nx) for x ∈ (−π π) n2 n=1 ∞ n=1 n4 We apply Parseval’s theorem for this series to find the value of ∞ 2π 1 + 16 = n4 π n=1 ∞ π x4 dx −π 2π 2π + 16 = n4 n=1 ∞ n=1 π4 = n4...
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Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps
... dx for d = nπξ L for d = L , 2n L 2n The solution for u(x, t) is, 8dL2 v u(x, t) = π2c u(x, t) = v π2c ∞ n=1 ∞ n=1 n2 n(L2 cos − 4d2 n2 ) 2nπd + L sin 2nπd L sin nπξ L sin nπx sin L nπct L for ... 17 86 First we have a Sturm-Liouville eigenvalue problem for X, X = µ2 X, X (0) = X (a) = 0, which has the solutions, mπ mπx , Xm = cos , a a Now we have a Sturm-Liouville eigenvalue problem for ... = L L nπc bn = L L L ut (x, 0) sin v for |x − ξ| < d for |x − ξ| > d nπx L dx ξ+d nπx v sin dx nπc ξ−d L 4Lv nπξ nπd = 2 sin sin nπ c L L bn = The solution for u(x, t) is, 4Lv u(x, t) = π c ∞...
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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps
... periodically The above formula is useful when approximating the solution for small time, t For such small t, the terms decay very quickly away from n = A small number of terms could be used for an accurate ... alternate formula is useful when approximating the solution for large time, t For such large t, the terms in the sine series decay exponentially Again, a small number of terms could be used for an ... ξ)δ(t − τ ), G(x, t|ξ, τ ) = for t < τ, G → as x → ±∞ We take the Fourier transform of the differential equation ˆ ˆ Gt + κω G = F[δ(x − ξ)]δ(t − τ ), ˆ G(ω, t|ξ, τ ) = for t < τ ˆ Now we have an...
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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx
... 15 16 ∞ n=1 ∞ n=1 cos(3nπ) cos(14nπ) n4 (−1)n n4 53 −7π = − 15 16 720 u(3/4, 7/2) = 2033 12727 3840 Chapter 46 Conformal Mapping 2034 46. 1 Exercises Exercise 46. 1 Use an appropriate conformal ... functions for the two boundary conditions are shown in Figure 45 .6 2023 1 0.8 0 .6 0.4 0.05 0.1 0.2 0.15 0.2 0.25 1 0.8 0 .6 0.4 0.05 0.1 0.2 0.15 0.2 0.25 Figure 45 .6: Green functions for the boundary ... appropriate boundary conditions in the ζ plane by inspection Write the solution u in terms of x and y 2037 46. 2 Hints Hint 46. 1 Hint 46. 2 Hint 46. 3 Hint 46. 4 Show that w = (1 + z)/(1 − z) maps the...
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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 3 ppsx
... −1 < for |β/α| < = for |β/α| = > for |β/α| > β Thus we see that α x may be a minimum for |β/α| ≥ and may be a maximum for |β/α| ≤ Jacobi Condition Jacobi’s accessory equation for this ... + 6( ˆ )2 − (w − y )(4(ˆ )3 − 12ˆ ) y y ˆ y y = w − 6w − β α +6 β = w − 6w − w α β α β α β − (w − )(4 α −3 +3 β α β α 6 β − 12 ) α β α We can find the stationary points of the excess function by ... − y − 2xy dx, y(0) = = y(1) For this problem take an approximate solution of the form y = x(1 − x) (a0 + a1 x + · · · + an xn ) , 2 066 and carry out the solutions for n = and n = • (y )2 + y...
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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 4 ppt
... admissible extremal for all x1 The value of the integral for this extremal is x1 y1 which is larger than the integral of the quadratic we analyzed before for y1 > x1 3.5 2.5 1.2 1.4 1 .6 1.8 Figure ... y1 = x1 ≈ 1. 467 89 x1 The piecewise smooth extremal has the smaller performance index for y1 smaller than this value and the quadratic extremal has the smaller performance index for y1 greater ... motion for θ √ ˙ = α αg θ (α − x)2 ˙ Now we substitute the expression for θ into the equation of motion for x 2¨ + (α − x) x α3 g −g =0 (α − x)4 21 06 α3 −1 g =0 (α − x)3 2¨ + x 2¨ + x For small...
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Tìm thêm: hệ việt nam nhật bản và sức hấp dẫn của tiếng nhật tại việt nam khảo sát chương trình đào tạo của các đơn vị đào tạo tại nhật bản khảo sát chương trình đào tạo gắn với các giáo trình cụ thể xác định thời lượng học về mặt lí thuyết và thực tế khảo sát thực tế giảng dạy tiếng nhật không chuyên ngữ tại việt nam nội dung cụ thể cho từng kĩ năng ở từng cấp độ xác định mức độ đáp ứng về văn hoá và chuyên môn trong ct phát huy những thành tựu công nghệ mới nhất được áp dụng vào công tác dạy và học ngoại ngữ mở máy động cơ lồng sóc mở máy động cơ rôto dây quấn các đặc tính của động cơ điện không đồng bộ hệ số công suất cosp fi p2 đặc tuyến hiệu suất h fi p2 đặc tuyến mômen quay m fi p2 đặc tuyến tốc độ rôto n fi p2 đặc tuyến dòng điện stato i1 fi p2 động cơ điện không đồng bộ một pha phần 3 giới thiệu nguyên liệu từ bảng 3 1 ta thấy ngoài hai thành phần chủ yếu và chiếm tỷ lệ cao nhất là tinh bột và cacbonhydrat trong hạt gạo tẻ còn chứa đường cellulose hemicellulose chỉ tiêu chất lượng 9 tr 25