B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:
|
|
- きみえ いいはた
- 5 years ago
- Views:
Transcription
1 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:
2 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 = m d l (.10) x = m + 1 d l (.11).15 x P O m x <0 x m (.11) x m = m + 1 d l.15:
3 B. 43 m +1 m x x = x m+1 ; x m = m m d l ; + 1 d l x = d l (.1) 3 d =1:0 10 ;4 ml =1:0 m x =6:3 10 ;3 m (d : P 5 A P a.16 A C B AB BC a B C.17: AB A BC B a (d 3 6:3 10 ;7 m.
4 44 1 A C 0 (.5) d = a m =1 sin =1 a = a AB A BC B AB BC =0 j sin j < a (.13) jj j sin j > a (.1).18 a 4 3 P sin = 3 a =0 1 (e.18: 10 ;7 m 1m(.1 a 1m ; rad<<rad sin jj < 10 ;7 rad (e 4 1 3
5 B ABC.19 AC' A' C' A B' 0 BC A C C' C A B C.19: t C A t AC B t c A ct B 1 ct.19 A'B'C 4 ACC' 4 CAA' AC'C= CA'A= C0 C=AA 0 = ctac 4ACC 0 4CAA 0 AC' AC A'C AC = (.14) AC' A'C = 0 =0 1 (t x) =A cos ft; x
6 46 1 x =0 (t x) =A cos ft+ x (f x k (k =1) 1 O (t x) =; 1 (t ;x) (t x) = ;A cos ft; ;x = A cos ft+ x.0 P!P' x>0 1 P'.0:.0 x =0 1 1 x =0 (t x) = 1 (t ;x) (t x) = A cos ft; ;x = A cos ft+ x.1 P!P'.1: x =0 1 6 x =0 P' O O P P x x (f
7 B. 47 x = l = 4 5 l ;y +y 1-1 +x x x =0.: a Y dy =0 (.15) dx x=0 x f x =0 Y +x Y + ;x Y ;..3.4 a,b,c x =0 Y + Y ; A a ; 5A b l x + 5A ; 5A c + 5A ; 5A l l l l b,c 0 a x =0.3: b x = l Y + Y ; ab ; 5A c + 5A ; 5A l l l
8 48 [Y ] x=l =[Y + + Y ; ] x=l =0 (.16) 1 a,b,c x = 1 5 l 3 5 ll = x x =0 5 l4 5 l =01 5 x =0x = l (g.4: c 7.6 OA BC y C O.5: 3 B D A x x (g 5 n = 4l n ; 1.6:
9 B. 49 O x y d D (l0) 6 AOD = tan = d l (x0) (xx tan ) x tan x tan (1.8) I II 1n I <n II n I >n II OB OA 1 m 8 >< x tan ; = m >: x tan ; = (m ; 1) x 8 >< x = >: m + 1 x = m tan = m d l (.1) x = d l tan = m + 1 d l.15 l x d = x l d 6.7 n =1:141 d =4:00 10 ;7 m (1) () 4:00 10 ;7 m<<7:00 10 ;7 m (h.7: (h 6 (1) n,() =4:51 10;7 m
10 50 D I II ABC 1.8 AC' A A B C BC A C B' C' C t A' C A t AC B t.8: III III c 1 c C 0 C=c 1 t AA 0 = c t A c t B 1 c t.8 A'B'C 4AC 0 C 4AA 0 C C 0 C 6 AC 0 C= 6 AA 0 C= rad 6 C 0 AC= 1 6 A 0 CA = 1 sin 1 = C0 C AC = c 1t AC sin = AA0 AC = c t AC sin 1 = c 1t sin c t = c 1 c (1.8) c 1 c = c 0 n I c 0 n II = sin 1 sin = n II n I n II n I
11 B. 51 n II sin = n I sin 1 (.17) 7 n I <n II 1 (i 8.9 n 1 d A' C' 1 A B (1) B' B" D 1 () 1.9: D A C DB 1 C 1 sin = n sin 1 (1) = 1 = c sin =1 1=n sin c sin c = 1 n (.18) 1 > = c (.) c (1) sin 1 > = 1 n 1 () sin 1 < 1 n (i 7 1 >
12 5 1 B AA' C'C D AA' B'B B"B C'C 7! A 0! B! C 0!! A! B 0! D! B"! C! B 0 B+BB" AA' B'B.8 B' B BB" C C B 0 B=BB" 4B 0 DB BD = d 6 B 0 DB = B 0 B+BB" = d cos d cos B n 1 D 1 n 8 >< d cos ; = m >: d cos ; = (m ; 1) 8 >< >: cos = cos = m d cos = 1 8 >< >: q 1 ; sin = q 1 ; n sin 1 m + 1 d s sin 1 = 1 1+ n s sin 1 = 1 n 1+ m + 1 m d d 8.9 n 1 =0 d (j (j 8 nd
13 B. 53 C L N P 1 Q N' L' A.30: B A n 3 PQ B.31 (1) AP QB 1.31: + (n ; 1) (.19) () R 1 a + 1 b = 1 f (.0) a A b B f f = R (n ; 1) (.1) (1) 6 A= 6 B= 4ABC' 6 C 0 + AC'k APC'Bk QB 4CPQ 6 C=
14 54 6 P= pi ; 1 6 Q= pi ; 1 P Q 6 C+ 6 P+ 6 Q= pi pi + ; 1 + ; = 1 + = (.) AP PC N 1 = + QB QC N' = + PQ AB LL' PC 6 C L PC N L AP AB AP 1 8 >< 1 sin + = n sin 1 >: 1 sin + = n sin 1 3 sin ( 1) + n 1 (.3) + n (.4) (.7)(.8) (.6) n + (n ; 1) () AB y PQ PQ y.33 P y R sin = y R.3:
15 B y R (.5) AB P Q y tan y a tan y b 3 tan ( 1) y a y b (.6) (1) (.3)(.9)(.30)) y a + y b (n ; 1) y R y 1 a + 1 b (n ; 1) R a!1 1 a! b (n ; 1) R b R (n ; 1) f.34 (a) F (.4) a = f.33: 1 f + 1 b 1 f 1 b 0 b!1.34 (b) F 1 R R.34 (c) F.34 (d)
16 56 F 1 R <0 (.5), f<0 (a),(b) F 1 F (c),(d) F 1 F (.4) O +x A x = ;a B x = b ab 9 I. 15cm 10cm (1) () II 15cm 30cm (3) (4) (k II II-3 (l c =3: m=s (k 9 (1) b = ;30 cm(3) b = ;10 cm (l.
17 B. 57 (m) (Hz) ; ;4 7:7 10 ; : :7 10 ;7 6:4 10 ;7 3: : :4 10 ;7 5:9 10 ;7 4: : :9 10 ;7 5:5 10 ;7 5: : :5 10 ;7 4:9 10 ;7 5: : :9 10 ;7 4:3 10 ;7 6: : :3 10 ;7 3:8 10 ;7 7: : :8 10 ; ;9 7: ; ; X > 1 10 ;1 > , d =10 ;5 10 ;6 m d.4. N N 5 1 (.13) 5.34: a = Nd (.13) j sin j < Nd N!1! 0 10 (1) d () 1cm 1000 m sin m
18 ;7 m 810 ;7 m (1).35 d sin m d sin =m sin = m d () d =10 ;3 cm ;1 =10 ;5 m V R sin V = V d =4m 10; sin R = R =8m10; d m =0 =0 1 m = cm rad 1 6 (m 3 1 m d d sin.35: : sin (10 ; ) -1 c.7 c, 1- I c I () n I () c c I () = c n I ().37 (m ;7 m
19 B = rad 9 =1:79 10 ;1 rad 1:80 10 ;1 rad1:84 10 ;1 rad.(.19),
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
More information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...............................
More information1 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
More information4 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
More information1 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
More information76 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(
More information1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ
1 (1) ( i ) 60 (ii) 75 (iii) 15 () ( i ) (ii) 4 (iii) 7 1 ( () r, AOB = θ 0 < θ < ) OAB A OB P ( AB ) < ( AP ) (4) 0 < θ < sin θ < θ < tan θ 0 x, 0 y (1) sin x = sin y (x, y) () cos x cos y (x, y) 1 c
More informationO E ( ) A a A A(a) O ( ) (1) O O () 467
1 1.0 16 1 ( 1 1 ) 1 466 1.1 1.1.1 4 O E ( ) A a A A(a) O ( ) (1) O O () 467 ( ) A(a) O A 0 a x ( ) A(3), B( ), C 1, D( 5) DB C A x 5 4 3 1 0 1 3 4 5 16 A(1), B( 3) A(a) B(b) d ( ) A(a) B(b) d AB d = d(a,
More information18 ( ) ( ) [ ] [ ) 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
More informationOABC 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
More information70 : 20 : A B (20 ) (30 ) 50 1
70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................
More information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... (
More information48 * *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
More information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
More information1 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
More information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
More information( )
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...............................
More information2 (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
More informationA (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
More information18 ( ) 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(,,
More information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
More information5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4
... A F F l F l F(p, 0) = p p > 0 l p 0 P(, ) H P(, ) P l PH F PF = PH PF = PH p O p ( p) + = { ( p)} = 4p l = 4p (p 0) F(p, 0) = p O 3 5 5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 =
More information1 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 +
More informationさくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1
... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =
More information85 4
85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V
More information0.6 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 =
More information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
More informationuntitled
2010824 1 2 1031 5251020 101 3 0.04 % 2010.8.18 0.05 % 1 0.06 % 5 0.12 % 3 0.14 % 2010.8.16 5 0.42 % 2010.7.15 25 5 0.42 % 0.426% 2010.6.29 5 0.496 % 2010.8.12 4 0.85 % 2010.8.6 10 0.900 % 2010.8.18 2.340
More information17 ( ) 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
More information[ ] 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 =
More information6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P
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
More informationIMO 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 =
More informationA 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
More information(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
More information) 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
More information29
9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n
More informationx () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
More information04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
More information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
More informationC:/KENAR/0p1.dvi
2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x
More information(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37
4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin
More information数学Ⅲ立体アプローチ.pdf
Ⅲ Ⅲ DOLOR SET AMET . cos x cosx = cos x cosx = (cosx + )(cosx ) = cosx = cosx = 4. x cos x cosx =. x y = cosx y = cosx. x =,x = ( y = cosx y = cosx. x V y = cosx y = sinx 6 5 6 - ( cosx cosx ) d x = [
More information1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x
. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +
More information13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
More information[] Tle () i ( ) e . () [].....[], ..i.et.. N Z, I Q R C N Z Q R C R i R {,,} N A B X N Z Q R,, c,,, c, A, B, C, L, y, z,, X, L pq p q def f () lim f ( ) f ( ) ( ), p p q r q r p q r p q r c c,, f ( )
More information[ ] 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(
More information1. (1) 1/
2005 11 30 2006 03 31 1-1-2 [ ] 7-12 SMBC 4 1 27 1 18 1. (1) 1/5 1 2 32 1/5 1 2006 3 11 200 2006 1 1/5 20 20 30 CM 10 TVCM15 BB 2006 3 31 26 3 5 2 1 4 3 2 3 (2) (1) 2. (1) 1 2006/03/31 1,680,877,606 1
More information1990 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,
More information0226_ぱどMD表1-ol前
No. MEDIA DATA 0 B O O K 00-090-0 0 000900 000 00 00 00 0000 0900 000900 AREA MAP 0,000 0,000 0,000 0,000 00,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000 00,000 0,000
More information1 B64653 1 1 3.1....................................... 3.......................... 3..1.............................. 4................................ 4..3.............................. 5..4..............................
More informationt θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
More information0 (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)
More information.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,
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
電気電子数学入門 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/073471 このサンプルページの内容は, 初版 1 刷発行当時のものです. i 14 (tool) [ ] IT ( ) PC (EXCEL) HP() 1 1 4 15 3 010 9 ii 1... 1 1.1 1 1.
More informationzz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
More information140 120 100 80 60 40 20 0 115 107 102 99 95 97 95 97 98 100 64 72 37 60 50 53 50 36 32 18 H18 H19 H20 H21 H22 H23 H24 H25 H26 H27 1 100 () 80 60 40 20 0 1 19 16 10 11 6 8 9 5 10 35 76 83 73 68 46 44 H11
More information31 33
17 3 31 33 36 38 42 45 47 50 52 54 57 60 74 80 82 88 89 92 98 101 104 106 94 1 252 37 1 2 2 1 252 38 1 15 3 16 6 24 17 2 10 252 29 15 21 20 15 4 15 467,555 14 11 25 15 1 6 15 5 ( ) 41 2 634 640 1 5 252
More information) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
More informationkoji07-01.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...?
More information4STEP 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,
More informationORIGINAL 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
More information(, Goo Ishikawa, Go-o Ishikawa) ( ) 1
(, Goo Ishikawa, Go-o Ishikawa) ( ) 1 ( ) ( ) ( ) G7( ) ( ) ( ) () ( ) BD = 1 DC CE EA AF FB 0 0 BD DC CE EA AF FB =1 ( ) 2 (geometry) ( ) ( ) 3 (?) (Topology) ( ) DNA ( ) 4 ( ) ( ) 5 ( ) H. 1 : 1+ 5 2
More informationT 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
More information() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
More informationR 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
More informationall.dvi
38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
More information(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
More information1 P2 P P3P4 P5P8 P9P10 P11 P12
1 P2 P14 2 3 4 5 1 P3P4 P5P8 P9P10 P11 P12 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 & 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1! 3 2 3! 4 4 3 5 6 I 7 8 P7 P7I P5 9 P5! 10 4!! 11 5 03-5220-8520
More information[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )
1 1.1 [] f(x) f(x + T ) = f(x) (1.1), f(x), T f(x) x T 1 ) f(x) = sin x, T = 2 sin (x + 2) = sin x, sin x 2 [] n f(x + nt ) = f(x) (1.2) T [] 2 f(x) g(x) T, h 1 (x) = af(x)+ bg(x) 2 h 2 (x) = f(x)g(x)
More informationdynamics-solution2.dvi
1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj
More informationPart 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
More information122 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
More informationiii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................
More information( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
More informationPSCHG000.PS
a b c a ac bc ab bc a b c a c a b bc a b c a ac bc ab bc a b c a ac bc ab bc a b c a ac bc ab bc de df d d d d df d d d d d d d a a b c a b b a b c a b c b a a a a b a b a
More information, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
More information技能継承に関するアンケートの結果概要
I 1 1 1 1 1 1 2 1 3 1 II 2 1 2 2 2 3 2007 2 4 3 III 4 1 4 4 5 6 2 7 7 8 9 3 10 _10 11 _12 _13 _14 15 4 2007 16 2007 16 17 2007 18 5 19 19 I 2007 1 2005 6 21 8 3 3000 2 292 292 9.7 3 100 1 II 1 86 2 OJT
More information福岡大学人文論叢47-3
679 pp. 1 680 2 681 pp. 3 682 4 683 5 684 pp. 6 685 7 686 8 687 9 688 pp. b 10 689 11 690 12 691 13 692 pp. 14 693 15 694 a b 16 695 a b 17 696 a 18 697 B 19 698 A B B B A B B A A 20 699 pp. 21 700 pp.
More informationi
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,
More information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
More informationf (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f
208 3 28. f fd f Df 関数 接線 D f f 0 f f f 2 f f f f f 3 f lim f f df 0 d 4 f df d 3 f d f df d 5 d c 208 2 f f t t f df d 6 d t dt 7 f df df d d df dt lim f 0 t df d d dt d t 8 dt 9.2 f,, f 0 f 0 lim 0 lim
More informationfunction2.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)
More information1 (1) (2)
1 2 (1) (2) (3) 3-78 - 1 (1) (2) - 79 - i) ii) iii) (3) (4) (5) (6) - 80 - (7) (8) (9) (10) 2 (1) (2) (3) (4) i) - 81 - ii) (a) (b) 3 (1) (2) - 82 - - 83 - - 84 - - 85 - - 86 - (1) (2) (3) (4) (5) (6)
More information- 2 -
- 2 - - 3 - (1) (2) (3) (1) - 4 - ~ - 5 - (2) - 6 - (1) (1) - 7 - - 8 - (i) (ii) (iii) (ii) (iii) (ii) 10 - 9 - (3) - 10 - (3) - 11 - - 12 - (1) - 13 - - 14 - (2) - 15 - - 16 - (3) - 17 - - 18 - (4) -
More information2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4 4 4 2 5 5 2 4 4 4 0 3 3 0 9 10 10 9 1 1
1 1979 6 24 3 4 4 4 4 3 4 4 2 3 4 4 6 0 0 6 2 4 4 4 3 0 0 3 3 3 4 3 2 4 3? 4 3 4 3 4 4 4 4 3 3 4 4 4 4 2 1 1 2 15 4 4 15 0 1 2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4
More information20 15 14.6 15.3 14.9 15.7 16.0 15.7 13.4 14.5 13.7 14.2 10 10 13 16 19 22 1 70,000 60,000 50,000 40,000 30,000 20,000 10,000 0 2,500 59,862 56,384 2,000 42,662 44,211 40,639 37,323 1,500 33,408 34,472
More informationI? 3 1 3 1.1?................................. 3 1.2?............................... 3 1.3!................................... 3 2 4 2.1........................................ 4 2.2.......................................
More information振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
More informationPart () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
More informationI II
I II I I 8 I I 5 I 5 9 I 6 6 I 7 7 I 8 87 I 9 96 I 7 I 8 I 9 I 7 I 95 I 5 I 6 II 7 6 II 8 II 9 59 II 67 II 76 II II 9 II 8 II 5 8 II 6 58 II 7 6 II 8 8 I.., < b, b, c, k, m. k + m + c + c b + k + m log
More information(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
More informationvol.31_H1-H4.ai
http://www.jmdp.or.jp/ http://www.donorsnet.jp/ CONTENTS 29 8,715 Vol. 31 2 3 ac ad bc bd ab cd 4 Point! Point! Point! 5 Point! Point! 6 7 314 611 122 4 125 2 72 2 102 3 2 260 312 0 3 14 3 14 18 14 60
More information