O E ( ) A a A A(a) O ( ) (1) O O () 467
|
|
- ありさ たかにし
- 5 years ago
- Views:
Transcription
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(
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
( )
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
?? A AB A B C AB A B A B A B A A B A 98 A B A B A B A B B A A B AB AB A B A BB A B A B A B A B A B A AB A B B A B AB A A C AB A C A A B A B B A B A B B A B A B B A B A B A B A B A B A B A B
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
2 (1) a = ( 2, 2), b = (1, 2), c = (4, 4) c = l a + k b l, k (2) a = (3, 5) (1) (4, 4) = l( 2, 2) + k(1, 2), (4, 4) = ( 2l + k, 2l 2k) 2l + k = 4, 2l
ABCDEF a = AB, b = a b (1) AC (3) CD (2) AD (4) CE AF B C a A D b F E (1) AC = AB + BC = AB + AO = AB + ( AB + AF) = a + ( a + b) = 2 a + b (2) AD = 2 AO = 2( AB + AF) = 2( a + b) (3) CD = AF = b (4) CE
高等学校学習指導要領解説 数学編
5 10 15 20 25 30 35 5 1 1 10 1 1 2 4 16 15 18 18 18 19 19 20 19 19 20 1 20 2 22 25 3 23 4 24 5 26 28 28 30 28 28 1 28 2 30 3 31 35 4 33 5 34 36 36 36 40 36 1 36 2 39 3 41 4 42 45 45 45 46 5 1 46 2 48 3
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a
1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =
B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:
B. 41 II: ;; 4 B [] S 1 S S 1 S.1 O S 1 S 1.13 P 3 P 5 7 P.1:.13: 4 4.14 C d A B x l l d C B 1 l.14: AB A 1 B 0 AB 0 O OP = x P l AP BP AB AP BP 1 (.4)(.5) x l x sin = p l + x x l (.4)(.5) m d A x P O
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +
( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n
空き容量一覧表(154kV以上)
1/3 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量 覧 < 留意事項 > (1) 空容量は 安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発 する場合があります (3) 表 は 既に空容量がないため
2/8 一次二次当該 42 AX 変圧器 なし 43 AY 変圧器 なし 44 BA 変圧器 なし 45 BB 変圧器 なし 46 BC 変圧器 なし
1/8 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発生する場合があります (3)
2012 A, N, Z, Q, R, C
2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)
さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n
1 1.1 1.1.1 A 2 P Q 3 R S T R S T P 80 50 60 Q 90 40 70 80 50 60 90 40 70 8 5 6 1 1 2 9 4 7 2 1 2 3 1 2 m n m n m n n n n 1.1 8 5 6 9 4 7 2 6 0 8 2 3 2 2 2 1 2 1 1.1 2 4 7 1 1 3 7 5 2 3 5 0 3 4 1 6 9 1
) 9 81
4 4.0 2000 ) 9 81 10 4.1 natural numbers 1, 2, 3, 4, 4.2, 3, 2, 1, 0, 1, 2, 3, integral numbers integers 1, 2, 3,, 3, 2, 1 1 4.3 4.3.1 ( ) m, n m 0 n m 82 rational numbers m 1 ( ) 3 = 3 1 4.3.2 3 5 = 2
0.6 A = ( 0 ),. () A. () x n+ = x n+ + x n (n ) {x n }, x, x., (x, x ) = (0, ) e, (x, x ) = (, 0) e, {x n }, T, e, e T A. (3) A n {x n }, (x, x ) = (,
[ ], IC 0. A, B, C (, 0, 0), (0,, 0), (,, ) () CA CB ACBD D () ACB θ cos θ (3) ABC (4) ABC ( 9) ( s090304) 0. 3, O(0, 0, 0), A(,, 3), B( 3,, ),. () AOB () AOB ( 8) ( s8066) 0.3 O xyz, P x Q, OP = P Q =
ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2
直交座標系の回転
b T.Koama x l x, Lx i ij j j xi i i i, x L T L L, L ± x L T xax axx, ( a a ) i, j ij i j ij ji λ λ + λ + + λ i i i x L T T T x ( L) L T xax T ( T L T ) A( L) T ( LAL T ) T ( L AL) λ ii L AL Λ λi i axx
Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n
Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a
17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :
9 ( ) 9 5 I II III A B (0 ) 5 I II III A B (0 ), 6 8 I II A B (0 ), 6, 7 I II A B (00 ) OAB A B OA = OA OB = OB A B : P OP AB Q OA = a OB = b () OP a b () OP OQ () a = 5 b = OP AB OAB PAB a f(x) = (log
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
iii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................
2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a
> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3
13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >
T T T T A 0 1 A 1 A P (A 1 ) = C 1 6 C 8C 3 = 15 8, P (A ) = C 6 C 1 8C 3 = 3 8 T 5 B P (A 1 B) = =
4 1.. 3. 4. 1. 1 3 4 5 6 1 3 4 5 6. 1 1 1 A B P (A B) = P (A) + P (B) P (C) = P (A) P (B) 3. 1 1 P (A) = 1 P (A) A A 4. A B P A (B) = n(a B) n(a) = P (A B) P (A) 50 015 016 018 1 4 5 8 8 3 T 1 3 1 T T
繖 7 縺6ァ80キ3 ッ0キ3 ェ ュ ョ07 縺00 06 ュ0503 ュ ッ 70キ ァ805 ョ0705 ョ ッ0キ3 x 罍陦ァ ァ 0 04 縺 ァ タ0903 タ05 ァ. 7
30キ36ヲ0 7 7 ュ6 70キ3 ョ6ァ8056 50キ300 縺6 5 ッ05 7 07 ッ 7 ュ ッ04 ュ03 ー 0キ36ヲ06 7 繖 70キ306 6 5 0 タ0503070060 08 ョ0303 縺0 ァ090609 0403 閨0303 003 ァ 0060503 陦ァ 06 タ09 ァ タ04 縺06 閨06-0006003 ァ ァ 04 罍ァ006 縺03 0403
行列代数2010A
a ij i j 1) i +j i, j) ij ij 1 j a i1 a ij a i a 1 a j a ij 1) i +j 1,j 1,j +1 a i1,1 a i1,j 1 a i1,j +1 a i1, a i +1,1 a i +1.j 1 a i +1,j +1 a i +1, a 1 a,j 1 a,j +1 a, ij i j 1,j 1,j +1 ij 1) i +j a
R R 16 ( 3 )
(017 ) 9 4 7 ( ) ( 3 ) ( 010 ) 1 (P3) 1 11 (P4) 1 1 (P4) 1 (P15) 1 (P16) (P0) 3 (P18) 3 4 (P3) 4 3 4 31 1 5 3 5 4 6 5 9 51 9 5 9 6 9 61 9 6 α β 9 63 û 11 64 R 1 65 13 66 14 7 14 71 15 7 R R 16 http://wwwecoosaka-uacjp/~tazak/class/017
48 * *2
374-1- 17 2 1 1 B A C A C 48 *2 49-2- 2 176 176 *2 -3- B A A B B C A B A C 1 B C B C 2 B C 94 2 B C 3 1 6 2 8 1 177 C B C C C A D A A B A 7 B C C A 3 C A 187 187 C B 10 AC 187-4- 10 C C B B B B A B 2 BC
i I II I II II IC IIC I II ii 5 8 5 3 7 8 iii I 3........................... 5......................... 7........................... 4........................ 8.3......................... 33.4...................
OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P
4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e
漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト
名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
122 6 A 0 (p 0 q 0 ). ( p 0 = p cos ; q sin + p 0 (6.1) q 0 = p sin + q cos + q 0,, 2 Ox, O 1 x 1., q ;q ( p 0 = p cos + q sin + p 0 (6.2) q 0 = p sin
121 6,.,,,,,,. 2, 1. 6.1,.., M, A(2 R).,. 49.. Oxy ( ' ' ), f Oxy, O 1 x 1 y 1 ( ' ' ). A (p q), A 0 (p q). y q A q q 0 y 1 q A O 1 p x 1 O p p 0 p x 6.1: ( ), 6.1, 122 6 A 0 (p 0 q 0 ). ( p 0 = p cos
a (a + ), a + a > (a + ), a + 4 a < a 4 a,,, y y = + a y = + a, y = a y = ( + a) ( x) + ( a) x, x y,y a y y y ( + a : a ) ( a : a > ) y = (a + ) y = a
[] a x f(x) = ( + a)( x) + ( a)x f(x) = ( a + ) x + a + () x f(x) a a + a > a + () x f(x) a (a + ) a x 4 f (x) = ( + a) ( x) + ( a) x = ( a + a) x + a + = ( a + ) x + a +, () a + a f(x) f(x) = f() = a
1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(
I n n A AX = I, YA = I () n XY A () X = IX = (YA)X = Y(AX) = YI = Y X Y () XY A A AB AB BA (AB)(B A ) = A(BB )A = AA = I (BA)(A B ) = B(AA )B = BB = I (AB) = B A (BA) = A B A B A = B = 5 5 A B AB BA A
取扱説明書 -詳細版- 液晶プロジェクター CP-AW3019WNJ
B A C D E F K I M L J H G N O Q P Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01 00 00 60 01 00 BE EF 03 06 00 19 D3 02 00
linearal1.dvi
19 4 30 I 1 1 11 1 12 2 13 3 131 3 132 4 133 5 134 6 14 7 2 9 21 9 211 9 212 10 213 13 214 14 22 15 221 15 222 16 223 17 224 20 3 21 31 21 32 21 33 22 34 23 341 23 342 24 343 27 344 29 35 31 351 31 352
13 0 1 1 4 11 4 12 5 13 6 2 10 21 10 22 14 3 20 31 20 32 25 33 28 4 31 41 32 42 34 43 38 5 41 51 41 52 43 53 54 6 57 61 57 62 60 70 0 Gauss a, b, c x, y f(x, y) = ax 2 + bxy + cy 2 = x y a b/2 b/2 c x
4STEP 数学 B( 新課程 ) を解いてみた 平面上のベクトル 6 ベクトルと図形 59 A 2 B 2 = AB 2 - AA æ 1 2 ö = AB1 + AC1 - ç AA1 + AB1 3 3 è 3 3 ø 1
平面上のベクトル 6 ベクトルと図形 A B AB AA AB + AC AA + AB AA AB + AC AB AB + AC + AC AB これと A B ¹, AB ¹ より, A B // AB \A B //AB A C A B A B B C 6 解法 AB b, AC とすると, QR AR AQ b QP AP AQ AB + BC b b + ( b ) b b b QR よって,P,
() () () () () 175 () Tel Fax
JPCA-PE04-02-01-02-01S JPCA PE04-02-01-02-01S 2005 () () () () () 175 () 167-0042 3122 2 Tel 03-5310-2020Fax 03-5310-2021e-mailstd@jpca.org Detail Specification for PT Optical Module 1 PT PT 12 Optoelectronic
.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
, ,279 w
No.482 DEC. 200315 14 1754,406 100.0 2160,279 w 100 90 80 70 60 50 40 30 20 10 28.9 23.8 25.0 19.3 30.4 25.0 29.5 80.7 75.0 75.0 70.5 71.1 69.6 76.2 7 8 9 10 11 12 13 23.2 76.8 14 14 1751,189 100.0 2156,574
0 (18) /12/13 (19) n Z (n Z ) 5 30 (5 30 ) (mod 5) (20) ( ) (12, 8) = 4
0 http://homepage3.nifty.com/yakuikei (18) 1 99 3 2014/12/13 (19) 1 100 3 n Z (n Z ) 5 30 (5 30 ) 37 22 (mod 5) (20) 201 300 3 (37 22 5 ) (12, 8) = 4 (21) 16! 2 (12 8 4) (22) (3 n )! 3 (23) 100! 0 1 (1)
18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (
8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin
, = = 7 6 = 42, =
http://www.ss.u-tokai.ac.jp/~mahoro/2016autumn/alg_intro/ 1 1 2016.9.26, http://www.ss.u-tokai.ac.jp/~mahoro/2016autumn/alg_intro/ 1.1 1 214 132 = 28258 2 + 1 + 4 1 + 3 + 2 = 7 6 = 42, 4 + 2 = 6 2 + 8
(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
13ィェィ 0002ィェィ 00ィヲ1 702ィョ ィーィ ィイ071 7ィ 06ィヲ02, ISSN
13 13ィェィ 0002ィェィ 00ィヲ1 702ィョ050702 0709ィーィ ィイ071 7ィ 06ィヲ02, ISSN 1992-6138 1 70306070302071 70307090303 07030209020703 1 7 03000009070807 01090803010908071 7030709030503 0300060903031 709020705 ィヲ0302090803001
HITACHI 液晶プロジェクター CP-AX3505J/CP-AW3005J 取扱説明書 -詳細版- 【技術情報編】
B A C E D 1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 H G I F J M N L K Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01
000 001
all-round catalogue vol.2 000 001 002 003 AA0102 AA0201 AA0701 AA0801 artistic brushes AA0602 AB2701 AB2702 AB2703 AB2704 AA0301 AH3001 AH3011 AH3101 AH3201 AH3111 AB3201 AB3202 AB2601 AB2602 AB0701 artistic
(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y
(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b
untitled
yoshi@image.med.osaka-u.ac.jp http://www.image.med.osaka-u.ac.jp/member/yoshi/ II Excel, Mathematica Mathematica Osaka Electro-Communication University (2007 Apr) 09849-31503-64015-30704-18799-390 http://www.image.med.osaka-u.ac.jp/member/yoshi/
(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y
(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b
1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1
I, A 25 8 24 1 1.1 ( 3 ) 3 9 10 3 9 : (1,2,6), (1,3,5), (1,4,4), (2,2,5), (2,3,4), (3,3,3) 10 : (1,3,6), (1,4,5), (2,2,6), (2,3,5), (2,4,4), (3,3,4) 6 3 9 10 3 9 : 6 3 + 3 2 + 1 = 25 25 10 : 6 3 + 3 3
21 1 1 1 2 2 5 7 9 11 13 13 14 18 18 20 28 28 29 31 31 34 35 35 36 37 37 38 39 40 56 66 74 89 99 - ------ ------ -------------- ---------------- 1 10 2-2 8 5 26 ( ) 15 3 4 19 62 2,000 26 26 5 3 30 1 13
function2.pdf
2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)
[ ] Table
[] Te P AP OP [] OP c r de,,,, ' ' ' ' de,, c,, c, c ',, c mc ' ' m' c ' m m' OP OP p p p ( t p t p m ( m c e cd d e e c OP s( OP t( P s s t (, e e s t s 5 OP 5 5 s t t 5 OP ( 5 5 5 OAP ABP OBP ,, OP t(
欧州特許庁米国特許商標庁との共通特許分類 CPC (Cooperative Patent Classification) 日本パテントデータサービス ( 株 ) 国際部 2019 年 1 月 17 日 CPC 版のプレ リリースが公開されました 原文及び詳細はCPCホームページの C
欧州特許庁米国特許商標庁との共通特許分類 CPC (Cooperative Patent Classification) 日本パテントデータサービス ( 株 ) 国際部 2019 年 1 月 17 日 CPC 2019.02 版のプレ リリースが公開されました 原文及び詳細はCPCホームページの CPC Revisions(CPCの改訂 ) 内のPre-releaseをご覧ください http://www.cooperativepatentclassification.org/cpcrevisions/prereleases.html
44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
i ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +
6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +
25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3
a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552
3 3.0 a n a n ( ) () a m a n = a m+n () (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 55 3. (n ) a n n a n a n 3 4 = 8 8 3 ( 3) 4 = 8 3 8 ( ) ( ) 3 = 8 8 ( ) 3 n n 4 n n
DVIOUT-HYOU
() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.
PROSTAGE[プロステージ]
PROSTAGE & L 2 3200 650 2078 Storage system Panel system 3 esk system 2 250 22 01 125 1 2013-2014 esk System 2 L4OA V 01 2 L V L V OA 4 3240 32 2 7 4 OA P202 MG55 MG57 MG56 MJ58 MG45 MG55 MB95 Z712 MG57
Taro10-名張1審無罪判決.PDF
-------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -1- 39 12 23 36 4 11 36 47 15 5 13 14318-2-
S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.
() 1.1.. 1. 1.1. (1) L K (i) 0 K 1 K (ii) x, y K x + y K, x y K (iii) x, y K xy K (iv) x K \ {0} x 1 K K L L K ( 0 L 1 L ) L K L/K (2) K M L M K L 1.1. C C 1.2. R K = {a + b 3 i a, b Q} Q( 2, 3) = Q( 2
5 n P j j (P i,, P k, j 1) 1 n n ) φ(n) = n (1 1Pj [ ] φ φ P j j P j j = = = = = n = φ(p j j ) (P j j P j 1 j ) P j j ( 1 1 P j ) P j j ) (1 1Pj (1 1P
p P 1 n n n 1 φ(n) φ φ(1) = 1 1 n φ(n), n φ(n) = φ()φ(n) [ ] n 1 n 1 1 n 1 φ(n) φ() φ(n) 1 3 4 5 6 7 8 9 1 3 4 5 6 7 8 9 1 4 5 7 8 1 4 5 7 8 10 11 1 13 14 15 16 17 18 19 0 1 3 4 5 6 7 19 0 1 3 4 5 6 7
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
X線-m.dvi
X Λ 1 X 1 O Y Z X Z ν X O r Y ' P I('; r) =I e 4 m c 4 1 r sin ' (1.1) I X 1sec 1cm e = 4:8 1 1 e.s.u. m = :1 1 8 g c =3: 1 1 cm/sec X sin '! 1 ß Z ß Z sin 'd! = 1 ß ß 1 sin χ cos! d! = 1+cos χ (1.) e
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1
ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD
A B 5 C 9 3.4 7 mm, 89 mm 7/89 = 3.4. π 3 6 π 6 6 = 6 π > 6, π > 3 : π > 3
π 9 3 7 4. π 3................................................. 3.3........................ 3.4 π.................... 4.5..................... 4 7...................... 7..................... 9 3 3. p
学習の手順
NAVI 2 MAP 3 ABCD EFGH D F ABCD EFGH CD EH A ABC A BC AD ABC DBA BC//DE x 4 a //b // c x BC//DE EC AD//EF//BC x y AD DB AE EC DE//BC 5 D E AB AC BC 12cm DE 10 AP=PB=BR AQ=CQ BS CS 11 ABCD 1 C AB M BD P
05秋案内.indd
1 2 3 4 5 6 7 R01a U01a Q01a L01a M01b - M03b Y01a R02a U02a Q02a L02a M04b - M06b Y02a R03a U03a Q03a L03a M08a Y03a R04a U04a Q04a L04a M09a Y04a A01a L05b, L07b, R05a U05a Q05a M10a Y05b - Y07b L08b
,2,4
2005 12 2006 1,2,4 iii 1 Hilbert 14 1 1.............................................. 1 2............................................... 2 3............................................... 3 4.............................................
untitled
( ) 200133 3 3 3 3, 7 347 57 10 i ii iii -1- -2- -3- -4- 90011001700mm -5- 4.2 1991 73.5 44.4 7.4 10.5 10.5 7.4 W 3 H 2.25 H 2.25 7.4 51.8 140.6 88.8 268.8m 5,037.9m 2 2mm 16cm916cm 10.5 W 3 H 2.25 62.8
12~
R A C D B F E H I J K A A A A A A A A A A AD B C BD AD E A DB DB ADB D D DB BD A C D B F E AD B B B B BF AD B B DB B B B B DB B DB D D ADB D D D D D AB AD D DB AB B B B F D D B B D D BF DBF B B B FD
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
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 +
さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B
1 1.1 1.1.1 1 1 1 1 a a a a C a a = = CD CD a a a a a a = a = = D 1.1 CD D= C = DC C D 1.1 (1) 1 3 4 5 8 7 () 6 (3) 1.1. 3 1.1. a = C = C C C a a + a + + C = a C 1. a a + (1) () (3) b a a a b CD D = D
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
[ ] 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 =
新たな基礎年金制度の構築に向けて
[ ] 1 1 4 60 1 ( 1 ) 1 1 1 4 1 1 1 1 1 4 1 2 1 1 1 ( ) 2 1 1 1 1 1 1 1996 1 3 4.3(2) 1997 1 65 1 1 2 1/3 ( )2/3 1 1/3 ( ) 1 1 2 3 2 4 6 2.1 1 2 1 ( ) 13 1 1 1 1 2 2 ( ) ( ) 1 ( ) 60 1 1 2.2 (1) (3) ( 9
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