O E ( ) A a A A(a) O ( ) (1) O O () 467
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1 ( 1 1 ) 1 466
2 O E ( ) A a A A(a) O ( ) (1) O O () 467
3 ( ) A(a) O A 0 a x ( ) A(3), B( ), C 1, D( 5) DB C A x A(1), B( 3) A(a) B(b) d ( ) A(a) B(b) d AB d = d(a, B) = AB ( ) ( ) A(a) B(b) d d = d(a, B) = b a 468
4 (1) d(a, B) A B A B b a d d(a, B) () A, B AB ( ) (3) d distance d (4) ( ) A(), B( 3) d(a, B) d(a, B) = ( 3) = 5 = 5 ( ) 17 (1) A( 3), B() () P( ), Q( 1) ( ) d(a, B) ( ) d(a, B) > = 0 d(a, B) = 0 A = B ( ) d(a, B) = d(b, A) ( ) d(a, B) < = d(a, C) + d(c, B) ( ) A(a), B(b) ( ) d(a, B) = b a d(a, B) > = 0 d(a, B) = 0 A = B ( ) a = a d(a, B) = 0 b a = 0 b a = 0 b = a A = B d(a, B) = b a = (a b) = a b = d(b, A) 469
5 ( ) a + b < = a + b d(a, B) = b a = (b c) + (c a) < = b c + c a = d(a, C) + d(c, B) ( ) (1) () A B B A (3) a + b < = a + b a + b < = a + b C (d(a, C) + d(c, B)) (d(a, B)) (< = ) ( ) ( ) P AB AP PB = m : n P AB m : n P AB m : n m > 0, n > 0 ( ) A m P n B A B ( ) 470
6 A B 5 x AB ( ) 4 3 AB 1 P p A P B 1 p 5 x AP PB = : 1 AP = p PB= 5 p p : 5 p = : 1 1 p = 5 p a : b = x : y b x a y bx ay bx = ay a : b = x : y a b, x y a b = x y bx = ay ( ) 471
7 < p < 5 p = p, 5 p = 5 p p = (5 p) p = ( ) A(a) B(b) AB m : n P p p = na + mb m + n na + mb a, b, m, n AB m : n na mb A a m P p n B b x ( ) (I) a < b a < p < b AP = p a, PB = b p AP PB = m : n (p a) : (b p) = m : n m(b p) = n(p a) p p = na + mb m + n 118 a > b 47
8 p = na + mb m + n ( ) AB 1 1 ( ) A(a) B(b) M m m = a + b A( 1) B(4) AB 3 P p p = 3 ( 1) = 1 P(1) M m m = = 3 ( ) M 3 ( ) 18 A( ) B( 9) (1) AB 3 4 () AB ( ) Q AB AQ QB = m : n Q AB m : n Q AB m : n m > 0, n > 0 m n ( ) (1) AB () m = n 473
9 ( ) AB A B m, n m n m n m < n m > n m > n Q AQ QB = m : n AQ QB Q AB A A B n Q m m < n B Q 119 ( ) ( ) A(a) B(b) AB m : n Q q q = na + mb m n (I) a < b m > n Q B a < b < q (q a) : (q b) = m : n m(q b) = n(q a) q q = na + mb m n 474
10 10 3 (II) a < b m < n (III) a > b m > n (IV) a > b m < n q = na + mb m n (I) (IV) q = na + mb m n ( ) A() B(8) AB 4 3 Q q q = = 6 Q(6) ( ) 19 A( ) B(4) (1) AB () AB q = na + mb m n ( n)a + mb q = m + ( n) AB m : ( n) m > 0, n > 0 n < 0 m > 0, n > 0 AB m : ( n) ( ) P AB AP PB = m : n P AB m : n P AB m : n ( ) 475
11 P AB m, n m n ( ) P(p) AB m : n p = na + mb m + n 11 m > 0, n > 0 m : ( n) ( m) : n ? A (a 1, a ) 3 ( x y ) A (a 1, a ) A A(a 1, a ) (0, 0) x y (a 1, a ) a 1 x a x y ( ) y x y 3 476
12 y a A O a 1 x ( ) ( ) ( ) x y x y 477
13 ( ) 4 x y ( ) A B d d = d(a, B) = AB AB AB ( ) ( ) A(a 1, b ) B(b 1, b ) d d = d(a, B) = (b 1 a 1 ) + (b a ) 4 478
14 A d d = a 1 + a A x B y C y b B a A C O a 1 b 1 x ABC C 90 ( ) AB = AC + BC AB = AC + BC AC = b 1 a 1, BC = b a a = a AB = (b 1 a 1 ) + (b a ) B O(0, 0) A OA = a 1 + a ( ) A B x A(a 1, 0) B(b 1, 0) AB = (b 1 a 1 ) + (0 0) = (b 1 a 1 ) 479
15 a = a AB = b 1 a 1 ( ) A(1, 4) B(4, ) d d = (4 1) + ( 4) = = 45 = 3 5 ( ) O A( 4, 3) d d = ( 4) + ( 3) = 5 = 5 ( ) 0 (1) A(, 5) B( 4, 4) () A( 1, 3) O A(a 1, a ) B(b 1, b ) AB m : n P(p 1, p ) ( ) 3 l, l, l m, n AA : A A = BB : B B 480
16 m n A B l A B l A B l y P B A O A P B a 1 p 1 b 1 x P AB m : n AP : PB = m : n AP : PB = A P : P B P A B m : n p 1 = na 1 + mb 1 m + n A P B y p = na + mb m + n 481
17 13 ( ) A(a 1, a ) B(b 1, b ) AB m : n P ( ) na1 + mb 1 m + n, na + mb m + n M ( a1 + b 1, a + b ) A(, 5) B(, 3) AB : ( ) x = y = 1 ( ) , ( ) x = 0 y = 4 (0, 4) ( ) 1 A(, 3) B(4, 1) (1) AB 1 3 () AB ( ) A(a 1, a ) B(b 1, b ) AB m : n Q ( na1 + mb 1, m n na + mb m n ) 14 A(, 3) B(4, 1) AB
18 AM 8 ABC BC M AB + AC = (AM + BM )? ABC 3 A(a 1, a ) B(b 1, b ) C(c 1, c ) M AB 15 B C BC 0? AB + AC = (AM + BM ) 483
19 ABC BC x M BC y A(a 1, a ) B( b 1, 0) C(b 1, 0) y A(a 1, a ) B( b 1, 0) O (M) C(b 1, 0) x AB + AC = {(a 1 + b 1 ) + a } + {(a 1 b 1 ) + a } = (a 1 + b 1 + a ) AM = a 1 + a, BM = b 1 (AM + BM ) = (a 1 + a + b 1 ) AB + AC = (AM + BM ) ( ) 3 ABC BC 1 D AB + AC = 3(AD + BD ) 16 D m : n 484
20 14 ( ) ( ) ( ) ( ) 3 A(a 1, a ) B(b 1, b ) C(c 1, c ) ABC G G ( a1 + b 1 + c 1 3, a + b + c 3 ) AB? 5 ( ) ABC BC M G AM 1 M ( b1 + c 1, b + c ) 1 a 1 + b 1 + c 1 G x + 1 y = a 1 + b 1 + c
21 G ( a1 + b 1 + c 1 3, a + b + c 3 ) ( ) ( ) 3 A(0, 6) B(6, ) C(9, 5) ABC G G(5, 3) x : = 5, y : = 3 ( ) 4 3 A(, 8) B( 3, ) C(7, 3) ABC G 83 ABC AB BC CA L M N ABC LMN A(a 1, a ) B(b 1, b ) C(c 1, c ) ABC G ( ) a1 + b G 1 + c 1 a, + b + c 3 3 L ( a1 + b 1, a + b ) ( b1 + c, M 1, b + c LMN G a 1 + b 1 + b 1 + c 1 x 3 + c 1 + a 1 ) ( c1 + a, N 1, = a 1 + b 1 + c 1 3 y ( ) G a1 + b 1 + c 1 a, + b + c 3 3 c + a ABC LMN ( ) ) 486
22 5 ABC AB BC CA m : n L M N ABC LMN ( ) ( ) 1 ax + by + c = 0 (a, b 0 ) A(a 1, a ), B(b 1, b ) ax+by+c = 0 A(a 1, a ), B(b 1, b ) ax + by + c = 0 a, b, c A(, 1), B(6, 7) ax + by + c = 0 a, b, c 487
23 A B 1 A(, 1) x y 1 a b + c = 0 (1) B(6, 7) 6a + 7b + c = 0 () a, b, c (1) 3 a () (1) 3 10b c = 0 c c = 5b (1) c a b + 5b = 0 a a = b a : b : c = b : b : 5b = : 1 : 5 a = b, c = 5b 1 bx + by + 5b = 0 (3) b = 0 a = b a = 0 a, b 0 b 0 (3) b x + y + 5 = 0 488
24 ( ) 0 a, b, c ( ) 6 (1) A(1, ), B( 1, 1) () A(, 0), B( 3, 5) A(, 1) B( 3, 1) x y = ( ) ( y = 1 ) ax + by + c = 0 A(, 1) a + b + c = 0 (1) B( 3, 1) 3a + b + c = 0 () (1) () b c a = 0 (1) b = c cy + c = 0 (3) c = 0 b = 0 c 0 (3) c y + 1 = 0 y = 1 ( ) 7 (1) A(3, ), B(100, ) () A( 3, 4), B( 3, ) 489
25 18 ( ) A(a 1, a ), B(b 1, b ) ax + by + c = 0 8 m 1 A(a 1, a ) 1 y n y = mx + n m n m a 1, a 1 A(a 1, a ) a = ma 1 + n n = a ma 1 y = mx + a ma 1 ( ) A(a 1, a ) m y a = m(x a 1 ) (, 1) y 1 = (x ) y = x 3 ( ) 8 (, 3) (1) 1 () 1 (3)
26 8 (1 ) 1 1 (x, y) A m = y a x a 1 (x, y) A ( ) ( ) (a 1, a ), (b 1, b ) a 1 b 1 y a = b a a 1 = b 1 x = a 1 b 1 a 1 (x a 1 ) b a b 1 a 1 (1, 3), (3, 1) y ( 3) = 1 ( 3) 3 1 y = x 5 (x 1) ( ) (a 1, a ), (b 1, b ) a 1 = 1, a = 3, b 1 = 3, b = 1 ( ) 19? 491
27 (, 1), (, 4) x x = ( ) 9 (1) (1, 4), (4, 7) () (, 1), (1, 1) (3) ( 1, 4), ( 1, 0) (4) (7, 3), ( 11, 3) ( ) y = mx + n, y = m x + n m = m ( BC B C x AB=A B ) l l C C A B A B x l l CAB= C A B BC B C x AB=A B ABC A B C BC=B C BC AB = B C A B 49
28 AB=A B BC=B C BC B C x ABC A B C CAB= C A B l l ( ) 1 y = mx + n ax + by + c = 0 ( ) ax + by + c = 0, a x + b y + c = 0 ab a b = 0 ax + by + c = 0, a x + b y + c = 0 ( ) 84 (, 1) 3x y + 1 = 0 3x y + 1 = 0 y = 3x y + 1 = 0 y = 3x (, 1) y 1 = 3(x ) y = 3x 5 ( ) ( ) 30 (1) ( 1, ) y = x 1 () (, 1) x + 3y 1 = 0 493
29 1..3 ( ) l : y = mx + n, l : y = m x + n mm = 1 ( ) 6 ( AB x OH 1 ) y l A O H x B l l AOB= 90 AB = OA + OB AOH BOH OA = OH + AH OB = OH + BH AB = OH + AH + BH (1) l 6 494
30 OH = 1, AH = m, BH = m AB = m m 7 (1) (m m ) = + m + ( m ) mm = mm = 1 mm = 1 AB = (m m ) = + m + m OA + OB = + m + ( m ) = + m + m AB = OA + OB AOB O 90 l l ( ) 1 ax + by + c = 0 ( ) ax + by + c = 0, a x + b y + c = 0 aa + bb = 0 ( ) ( ) ( ) 7 BH = m l BH 495
31 85 (, 1) x 3y + 1 = 0 y = x 3y + 1 = m m 1 3 = 1 m = 3 y + 1 = 3(x ) y = 3x + 5 ( ) ( ) 31 (1) (, 1) x y + 1 = 0 () (, 3) x + y + 1 = 0 (3) (1, 3) 3x y + 1 = A(4, 5), B(, 3) AB AB AB AB M ( 4 +, 5 3 M(3, 4) ) 496
32 AB m m = = 1 1 y + 4 = x 3 y = x 7 ( ) ( ) 3 A(6, 3), B(, 7) AB 87 x 5y 16 = 0 A(1, 3) B(x, y) AB x, y AB AB x, y B(x, y) AB x y = 0 x 5y = 45 (1) 5 AB 5 y 3 x 1 = 1 497
33 5x + y = 11 () (1) () x = 5, y = 7 ( ) ( ) 33 x y = 0 A( 1, 1) 1..5 ( ) A l d A l H d = AH ( ) (1) ( ) A l l ( ) l A H () l P AP A l
34 ( ) ( ) (x 0, y 0 ) ax + by + c = 0 d d = ax 0 + by 0 + c a + b (x 0, y 0 ) ax + by + c = 0 H(x 1, y 1 ) ( ) AH AH = (x 1 x 0 ) + (y 1 y 0 ) (I) b 0, a 0 a b AH y 1 y 0 x 1 x 0 ( y 1 y 0 a ) = 1 x 1 x 0 b y 1 y 0 x 1 x 0 x 1 x 0 a k x 1 = x 0 + ak, = b a = y 1 y 0 b x 1 x 0 = ak, y 1 y 0 = bk (1) y 1 = y 0 + bk (x 1, y 1 ) ax + by + c = 0 a(x 0 + ak) + b(y 0 + bk) + c = 0 k k = ax 0 + by 0 + c a + b () 499
35 (1) AH AH = (ak) + (bk) = (a + b )k () (ax0 + by AH = 0 + c) a + b AH> 0 AH = ax 0 + by 0 + c a + b (II) b = 0 ( a 0) ax + c = 0 ( c ) a, 0 d ( d = x 0 c ) ax = 0 + c a a ax 0 + by 0 + c a + b = ax 0 + c a (III) a = 0 ( b 0) y = ax 0 + c a 130 ( ) (1, ) 4x 3y 3 = 0 d ( ) 3 d = 4 + ( 3) = 7 5 = 7 5 ( ) 34 (1) (, ) 4x 3y 3 = 0 () ( 1, 3) x + y 3 = 0 (3) 3x y + 1 = 0 500
36 ??? ( ) ( ) C (a, b) r P(x, y) CP = r (x a) + (y b) = r (x a) + (y b) = r P(x, y) CP = r 9 ( ) C(a, b) r 9 P (x a) + (y b) = r 501
37 (x a) + (y b) = r ( 1, 3) {x ( 1)} + (y 3) = (x + 1) + (y 3) = 4 ( ) 35 (1) (, 1) 3 () ( 1, 1) 5 (3) 1 (x 1) + (y + ) = 9 (x 1) + {y ( )} = 3 (1, ) 3 ( ) 36 (1) (x ) +(y 1) = 1 () (x+) +(y 3) = (3) (x + 1) + y = 16 (x a) + (y b) = r r x + y ax by + (a + b r ) = 0 l = a, m = b, n = a + b r x + y + lx + my + n = 0 50
38 x + y + lx + my + n = 0 x + y + lx + my + n = 0? 131 x + y 6x 4y 3 = 0 x 6x y 4y 9 4 ( 3 3 ) (x 6x + 9) + (y 4y + 4) = (x 3) + (y ) = 4 (3, ) 4 ( )? x + y 6x 4y + 13 = 0 (x 3) + (y ) = 0 0? ( ) a, b a + b = 0 a = b = 0 x 3 = 0, y = 0 503
39 x = 3, y = x + y 6x 4y + 13 = 0 (3, ) ( ) x + y + lx + my + n = 0? x + y 6x 4y + 14 = 0 (x 3) + (y ) = ( ) 88 (1) x + y + x 4y 4 = 0 () x + y + 4x y + 6 = 0 (3) x + y x y + 1 = 0 0 (1) (x + 1) + (y ) = 9 ( 1, ) 3 () (x + ) + (y 1) = 1 (3) ( x 1 ) ( + y 1 ) = 0 ( ) 1, 1 ( ) 10 ( ) 504
40 37 (1) x + y x + y 1 = 0 () x + y 4y + 4 = 0 13 ( ) x + y + lx + my + n = 0 l, m, n x + y + lx + my + n = 0 1 y = mx + n m, n l, m, n l, m, n (1, 1), (, ), (4, ) x + y + lx + my + n = 0 3 l, m, n x + y + lx + my + n = 0 (1, 1) l 1 + m 1 + n = 0 l + m + n + = 0 (1) 505
41 (, ) l m + n + 8 = 0 () (4, ) 4l + m + n + 0 = 0 (3) (1), (), (3) l = 6, m = 0, n = 4 x + y 6x + 4 = 0 ( ) ( ) 38 3 (1, 1), (3, 5), (5, 1) ( ) d d < 0 d = 0 d > 0 506
42 4x 3y + 5 = 0 x + y = 3 x + y = 3 ( ) 4x 3y + 5 = 0 d d = ( 3) = = d < 3 ( ) 39 (1) x + y + 1 = 0 x + y = 1 () y = 3x (x + 1) + (y + 1) = 1 (3) x + y 5 = 0 x + y = 5 ( ) D D > 0 D = 0 D < 0 ( ) y = 3x (x + 1) + (y + 1) = 1? 507
43 (x + 1) + (3x 1) = 1 10x 4x + 1 = 0 D D/4 = ( ) 10 1 = 6 < 0 ( ) 40 (1) x + y + 1 = 0 x + y = 1 () x + y 5 = 0 x + y = 5 ( ) ) ( ) 1 ( ) ( ) x + y = r (x 0, y 0 ) x 0 x + y 0 y = r (1) x x 0 y y 0 508
44 () (3) ( ) (I) x 0 0 y 0 0 m m y 0 = 1 x 0 m = x 0 y 0 y y 0 = x 0 y 0 (x x 0 ) x 0 x + y 0 y = x 0 + y 0 (x 0, y 0 ) x + y = r x 0 + y 0 = r x 0 x + y 0 y = r (II) x 0 = 0 (0, ±r) x y = ±r x 0 x + y 0 y = r x 0 = 0, y 0 = ±r ±ry = r y = ±r ( ) (III) y 0 = 0 (II) 509
45 133 (II) ( ) 134 (I) x 0 0 y 0 0 x + y = 5 (3, 4) 3x 4y = 5 ( ) (3, 4) ( ) 41 x + y = 5 (1) ( 4, 3) () (0, 5) (3) ( 5, 0) ( ) ( ) 1 ( )
46 1.4. A B A B AB A B P P AB H APH BPH 13 AH=BH P A H B P AB P AB AB H APH BPH AP=BP 14 P A B ( ) 13! 14! 511
47 ? p : P A B q : P AB p q ( ) p p q p P q Q P Q Q P p q q p ( ) ( ) A B A(c, 0), B( c, 0) c 0 P (x, y) AP = BP AP, BP AP = BP a = b = a = b (!) 51
48 (x c) + y = (x + c) + y 4cx = 0 c 0 x = 0 y AB A(c, 0), B( c, 0) AB ( ) (1) () AP = BP AP = BP ( ) 4 (, 1), ( 1, ) 90 A( 6, 0), B(3, 0) AP : BP = : 1 P P (x, y) AP : BP = : 1 BP = AP 4BP = AP a, b a = b a = b 513
49 4{(x 3) + y } = (x + 6) + y 3x 36x + 3y = 0 3 x 1x + y = 0 (x 6) + y = 36 P (6, 0) 6 ( ) ( ) A B AP : BP = m : n (m > 0, n > 0, m n) P AB m : n (!) ( ) 43 A(, 0), B(3, 0) AP : BP = 3 : P
50 ( ) 515
76 3 B m n AB P m n AP : PB = m : n A P B P AB m : n m < n n AB Q Q m A B AQ : QB = m : n (m n) m > n m n Q AB m : n A B Q P AB Q AB 3. 3 A(1) B(3) C(
3 3.1 3.1.1 1 1 A P a 1 a P a P P(a) a P(a) a P(a) a a 0 a = a a < 0 a = a a < b a > b A a b a B b B b a b A a 3.1 A() B(5) AB = 5 = 3 A(3) B(1) AB = 3 1 = A(a) B(b) AB AB = b a 3.1 (1) A(6) B(1) () A(
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18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
０４年度LS民法Ⅰ教材改訂版.PDF
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高校生の就職への数学II
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空き容量一覧表（154kV以上）
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X線-m.dvi
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05秋案内.indd
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Contents
Contents 6-1 6-2 780 630 440 385 355 325 295 205 80 1-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9 1-10 1-11 1-12 1-13 1-14 1-15 1-16 1-17 1-18 1-19 1-20 1-21 1-22 1-23 1-24 1-25 1-26 1-27 1-28 1-29 1-30 MEMO G
koji07-01.dvi
2007 I II III 1, 2, 3, 4, 5, 6, 7 5 10 19 (!) 1938 70 21? 1 1 2 1 2 2 1! 4, 5 1? 50 1 2 1 1 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 3 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k,l m, n k,l m, n kn > ml...?
1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1
1/5 ( ) Taylor ( 7.1) (x, y) f(x, y) f(x, y) x + y, xy, e x y,... 1 R {(x, y) x, y R} f(x, y) x y,xy e y log x,... R {(x, y, z) (x, y),z f(x, y)} R 3 z 1 (x + y ) z ax + by + c x 1 z ax + by + c y x +
15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = N N 0 x, y x y N x y (mod N) x y N mod N mod N N, x, y N > 0 (1) x x (mod N) (2) x y (mod N) y x
A( ) 1 1.1 12 3 15 3 9 3 12 x (x ) x 12 0 12 1.1.1 x x = 12q + r, 0 r < 12 q r 1 N > 0 x = Nq + r, 0 r < N q r 1 q x/n r r x mod N 1 15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = 3 1.1.2 N N 0 x, y x y N x y
HITACHI 液晶プロジェクター CP-EX301NJ/CP-EW301NJ 取扱説明書 -詳細版- 【技術情報編】 日本語
A B C D E F G H I 1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 K L J Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C RS-232C RS-232C Cable (cross) LAN cable (CAT-5 or greater) LAN LAN LAN LAN RS-232C BE
1 Abstract 2 3 n a ax 2 + bx + c = 0 (a 0) (1) ( x + b ) 2 = b2 4ac 2a 4a 2 D = b 2 4ac > 0 (1) 2 D = 0 D < 0 x + b 2a = ± b2 4ac 2a b ± b 2
1 Abstract n 1 1.1 a ax + bx + c = 0 (a 0) (1) ( x + b ) = b 4ac a 4a D = b 4ac > 0 (1) D = 0 D < 0 x + b a = ± b 4ac a b ± b 4ac a b a b ± 4ac b i a D (1) ax + bx + c D 0 () () (015 8 1 ) 1. D = b 4ac
新たな基礎年金制度の構築に向けて
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[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
6-1 6-2 1-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9 1-10 1-11 1-12 1-13 1-14 1-15 1-16 1-17 1-18 1-19 1-20 1-21 1-22 1-23 1-24 1-25 1-26 1-27 1-28 1-29 1-30 2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 2-10 2-11 2-12
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2002 2 2 2 2 22 2 3 3 3 3 3 4 4 5 5 6 6 7 7 8 8 9 Cramer 9 0 0 E-mail:hsuzuki@icuacjp 0 3x + y + 2z 4 x + y
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7 27 7.1........................................ 27 7.2.......................................... 28 1 ( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a -1 1 6
26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7
12 重度訪問介護重度障害者等の場合 ( 重度訪問介護 Ⅰ)
請求サービスコードと決定サービスコードの対応表 網掛けは 平成 30 年 4 月報酬改定等により 追加及び変更になったものサービス種類決定サービスサービス種類サービス内容請求サービスコードコードコード 11 居宅介護居宅における身体介護 111111~111999 111000 112000~112058 112061~112364 11A001~11BW60 通院介助 ( 身体介護を伴う場合 )
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6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
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koji07-02.dvi
007 I II III 1,, 3, 4, 5, 6, 7 5 4 1 ε-n 1 ε-n ε-n ε-n. {a } =1 a ε N N a a N= a a
mugensho.dvi
1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim