S P I B U S F e R A M M R 4 5 V x x x x シリー ズ
1.7 D D 2 100m 10 9 ev f(x) xf(x) = c(s)x (s 1) (x + 1) (s 4.5) (1) s age parameter x f(x) ev 10 9 ev 2
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古代ペルシア楔形文字フォント ( ラピュタ文字 B7uX フォント ) A b c d e f g a b c d e f g ch x h I j k l m n h i j k l m n o p q r s t u o p q r s t u θ ku v w x y z v w x y z
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1 1 Lambert Adolphe Jacques Quetelet ( ) [ ] 1 (1 ) n x 1, x 2,..., x n x a 1 a i a m f f 1 f i f m n 1.1 ( ( ))
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u τ = 2 u x 2 u(x, 0) = max[e ( 2r σ 2 1)x/2 e ( 2r σ 2 +1)x/2, 0] lim u(x, τ) = x lim u(x, τ) =0 x 1 u(x, τ) V (S, t) V = E 1 2 (1+k) S 1 2 (1 k) e 1
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z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
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A g ( v x ) i i { v ( m m) }{ v ( m m) } v i vav ( m m)( m m) i ( m m)( m m) v ( m m)( m m) SS within g ( v x v x ) i g { v ( X ) m v ( m m) } g { v (
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A F e s t i v e C h r i s t m a s
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F u k u o k a W o m e n s U n i v e r s i t y Greeting
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1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx
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Sae x Sae x 1: 1. {x (i) 0 0 }N i=1 (x (i) 0 0 p(x 0) ) 2. = 1,, T a d (a) i (i = 1,, N) I, II I. v (i) II. x (i) 1 = f (x (i) 1 1, v(i) (b) i (i = 1,
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L i q u e f i e d P e t r o l e u m G a s 33
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Margulus Margulus G E (4) )} ( { B A B A E m x x b a x x G + = (2),(3) B A B A B A E m x bx x x b a x x G + + = )} ( { (5) B A B A A B E m x bx x x b
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Java (5) 1 Lesson 3: x 2 +4x +5 f(x) =x 2 +4x +5 x f(10) x Java , 3.0,..., 10.0, 1.0, 2.0,... flow rate (m**3/s) "flow
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Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
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MPC MPC R p N p Z p p N (m, σ 2 ) m σ 2 floor( ), rem(v 1 v 2 ) v 1 v 2 r p e u[k] x[k] Σ x[k] Σ 2 L 0 Σ x[k + 1] = x[k] + u[k floor(l/h)] d[k]. Σ k x
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eq2:=m[g]*diff(x[g](t),t$2)=-s*sin(th eq3:=m[g]*diff(z[g](t),t$2)=m[g]*g-s* 負荷の座標は 以下の通りです eq4:=x[g](t)=x[k](t)+r*sin(theta(t)) eq5:=z[g](t)=r*cos(the
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x p v p (x) x p p-adic valuation of x v 2 (8) = 3, v 3 (12) = 1, v 5 (10000) = 4, x 8 = 2 3, 12 = 2 2 3, = 10 4 = n a, b a
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年代 i における炭素 14 年代キャリブレーションデータ (r i ±s i ) と試料の炭素 14 年代 (r m ±s m ) との分散 (v i ) と確率 (p i ) は, v i = (r m - r i ) 2 s 2 m + si 2... ( 3 ) p i e-vi/2 sm
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(1) x 4.0 m/s x x=0 t=0 t=8.0 s 12 m/s x t=0 t v t v v-t x x=0 t x x=0 t=8.0 s x x =0 m (2) F k2 Q1, Q2 2 r F= Vt Vq I I= (1)
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( [2], 1 p.38.) 1. [1] C R n y C u = (u 1,, u n ) α n u i y i > α i=1 n u i x i α, x C i=1 α 1 2 f(x) g(x) f(x) g(x) 1 ( 1 ) A B a b O a O b A B v a v
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