A B 5 C mm, 89 mm 7/89 = 3.4. π 3 6 π 6 6 = 6 π > 6, π > 3 : π > 3
|
|
- あかり まるこ
- 7 years ago
- Views:
Transcription
1 π π π p N p N
2 A B 5 C mm, 89 mm 7/89 = 3.4. π 3 6 π 6 6 = 6 π > 6, π > 3 : π > 3
3 . r πr πr n n n n r πr r πr = πr : (4.3 B a O C A 3: N p N N s N s N p N 3
4 N a p N = (N a/ = Na 3 OA =, BC = a OAB s N = (N a = a ( a = Na = p N ( N s N N p N N s N = p N π π.4 π a r = na S S = πr = πn a M B Ma < S = πn a < (M + Ba, M n < π < M + B n a =, r = n ( n π (. n = 4 π (M, B /4 4.5 ( ( x- y- k /4 N x- P k = N,, k =,,..., N P k y- Q k Q k = k ( k N, N 4
5 4: π 5
6 Q k R k Q k O P k P k 5: (N = A Q k Q k+ O P k P k+ 6: (N = A 6
7 Q k x- P k Q k R k /4 π/4 P k P k Q k R k k =,,..., N ( 5 /4 π/4 P k P k+ Q k+ Q k k =,,..., N ( 6 N k= ( k < π N N 4 N N k= N 4 < k= N k < π < 4 N ( k. N N N k= N k. ( N π ( 4/N N = 3.4 < π < N = ( π. N N a N b N a N b 7 OBA = EBD OAB EDB OA =, ED = AE = b, AB = b OA : ED = AB : DB, : b = b : DB, DB = bb. (3 ODC = EBD OCD EDB OD =, EB = AB AE = b b, CD = a OD : EB = CD : BD, : (b b = a : BD, BD = a(b b. (4 (3 (4 bb = a(b b, b (a + b = ab, b = ab a + b. (5 OAF = DAC OF A DCA OA =, DA = F A = a, AF = a OA : DA = AF : AC, : a = a : AC, AC = a. (6 7
8 D B a F E b a b O C A 7: EOA = AOF OAE OF A OAE DCA OA =, DC = a, AE = b OA : DC = AE : CA, : a = b : CA, CA = ab. (7 (6 (7 a = ab, ab a =. (8 N p N N q N p N = Na, q N = Nb, p N = Na, q N = Nb a = p N /N, b = q N /N, a = p N /(N, b = q N /(N (5, (8. q N = p Nq N p N + q N p N = p N q N (p N q N, (p N q N. 8
9 6, 4, p 6 = 3, q 6 = 3 q = = ( 3, p = 3 ( 3 = 6 3. p N < π < q N n =, 3.58 < π < 3.53, n = 4, 3.36 < π < 3.596, n = 48, < π < 3.46, n = 96, 3.4 < π < 3.47, n = 9, 3.44 < π < π p N, q N N =, 4, 48, 96, 9. a, b A(a, b = a + b, G(a, b= ab, H(a, b = ab a + b A(a, b a b, G(a, b, H(a, b A(a, b G(a, b a = b 4(A(a, b G(a, b = (a + b 4ab = a + ab + b 4ab ( A a, = b = a ab + b = (a b. ( a + = a + b b ab = H(a, b 9
10 ( H(a, b = A a, ( G b a, = b ab, H(a, b ab = G(a, b H(a, b G(a, b A(a, b max(a, b = { a, a b, b, a < b, min(a, b = { a, a b, b, a > b, a max(a, b, b max(a, b A(a, b = (a + b (max(a, b + max(a, b = max(a, b a min(a, b, b min(a, b ( A a, = ( b a + ( b min(a, b + = min(a, b min(a, b, H(a, b = A (, min(a, b. a b min(a, b H(a, b G(a, b A(a, b max(a, b p N < q N q N = H(p N, q N, p N = G(p N, q N p N < p N < q N < q N p 3 n q 6 α q 3 n p 6 β n q 3 n+ = H(p 3 n, q 3 n β = H(α, β = αβ α + β α + αβ = αβ, α = αβ, α = β α π
11 3 3. p N p N N 6 p N p N 7 DCA DC = a, CA = a, DA = a a + (a = (a a = p N /N, a = p N /(N p N N + p4 N 4N = p N 4 N, p4 N 4N p N + 4N p N =, p N = N ± = N ( N ± p N = N 4N 4 4N p N = N ± N N p N. N p N p N < q N < q 6 = 3 ( N, p N = N N p N ( N N p N. (9 3. (64? 78 a n = p 3 n {a n } a n+ a n a n+ a n+ a n+ a n.5 = 4 t N = N N p N N p N = Nt N, N ( t N = p N, N t N = N p N N N (9 p N = N ( N (N t N = (N p N N p N = N ( t N, (N t N = (N p N = 4N N ( t N = N ( + t N.
12 n 3 n a n+ a n+ a n a n+ a n a n+ a n : t N = ( + t N, p Nt N = N ( t N ( + t N = N ( t N = p N. p N t N = p N, t N = ( + t N ( ( N = 3 n a n = p N, a n+ = p N ( t N = p N (N a n+ a n = p N p N = p N p N t N = p N ( t N = p N( t N + t N p 3 N = (N ( + t N. a n+ a n+ = p 4N t 4N = p N a n+ a n+ = p 3 4N (4N ( + t 4N. p 3 N (4N t 3 4N ( + t 4N.
13 a n+ a n+ a n+ a n = p3 N (N ( + t N p 3 N (4N t 3 4N ( + t 4N = + t N 4t 3 4N ( + t 4N n N = 3 n p N π t N = N N p N = p N N t N, t 4N ( a n+ a n+ a n+ a n = + t N 4t 3 4N ( + t 4N + 4( + = 4 (. n a n+ a n+ a n+ a n a n π b n b n = 4a n+ a n ( 3 n b n 4π π 3 = π b n+ b n = 4a n+ a n+ 4a n+ a n 3 3 = 4(a n+ a n+ a n+ a n 3 = 4(a n+ a n 3 ( 3 an+ a n+ a n+ a n 4 n b n a n π π 6 4. a n ( b, b, b 3, b 4, b 5 3
14 ([] ( ([] 4. < r < S n = + r + r + + r n + r n rs n = r + r + r r n + r n+ S n rs n = r n+, ( rs n = r n+, S n = rn+ r n r n+ + r + r + + r n + r n + = lim n S n = r (3 r = = OA =, AC = a OAB S(a < a S(a a < r < n AC C n AC n = ar n C = C OC n B n B n OC n D n OC n C n C n C n = r n a r n a = a( rr n a( rrn OB n D n OC n C n OB n =, OC n = + a r n 4
15 : = C B D D B D 3 B B 3 C C C 3 a O 9: A 5
16 OB n D n OC n C n : ( + a r n OB n D n a( rr n ( + a r n = a( rrn n =,,... n= a( rr n ( + a r n ( a r n + a 4 r 4n a 6 r 6n + a( r a( rr a( rr a( rr3 = ( + a r ( + a r 4 ( + a r 6 ( + a r 8 + a( r ( = a r + a 4 r 4 a 6 r 6 + a( rr ( + a r 4 + a 4 r 8 a 6 r + a( rr ( + a r 6 + a 4 r a 6 r 8 + a( rr3 ( + a r 8 + a 4 r 6 a 6 r a( r ( = a r + a 4 r 4 a 6 r 6 + a( r ( + r a r 5 + a 4 r 9 a 6 r 3 + a( r ( + r a r 8 + a 4 r 4 a 6 r + a( r ( + r 3 a r + a 4 r 9 a 6 r
17 a( rr n a( r = ( + r + r + r 3 + ( + a r n n= a( r (a r + a r 5 + a r 8 + a( r + (a 4 r 4 + a 4 r 9 + a 4 r 4 + a 4 r 9 + a( r (a 6 r 6 + a 6 r 3 + a 6 r + a 6 r 7 + a( r = ( + r + r + r 3 a( r + a r ( + r 3 + r 6 + a( r + a 4 r 4 ( + r 5 + r + r 5 + a( r a 6 r 6 ( + r 7 + r 4 + r + ( a( r = r a r r + a4 r 4 3 r a6 r 6 5 r + 7 = (a a3 r + r + r + a 5 r 4 + r + r + r 3 + r a 7 r r + + r +. 6 r OB n D n OAB S(a r a n+ r n an+ + r + + rn n + S(a = (a a3 3 + a5 5 a (4 a = OAB π /8 8 π = S( = ( , 3 (. π = 4 ( π π (4 π 7
18 C G b + a F E B a O D : A OA =, AB = a, BC = b OB, OAB = 9, OBC = 9, < a, b < C OA CD CD OB E B CD BF c = DC/OD OAG S(c = S(a + S(b c a, b OAB ODE, ODE BF E, BF E CF B OAB CF B OB : CB = AB : F B, OA : CF = AB : F B CB = b OB, AB = a OB : b OB = a : F B, F B = ab OA = : CF = a : ab, CF = b CD = DF + F C = AB + F C = a + b, OD = OA DA = OA F B = ab, c = CD OD = a + b ab. 8
19 S ( a + b = S(a + S(b (5 ab (5 a =, b = = π 8 = S( = S ( + S ( 3 : S(/ + S(/3 = S( 4 (. ( π = ( (6 9
20 [ ] (5 S ( = S 5 ( + S 5 ( = S = S ( 5. (5 4S ( ( ( 5 5 = S + S = S = S (. 9 S( + S ( = S 39 S( + S = S ( ( = 4S, 39 5 ( ( S( = 4S S (. 9 S( = π 4 (4 5. (6 5 6 π 5 5. a > b a = a, b = b, a n+ = a n + b n, b n+ = a n b n n =,,,... {a n }, {b n } b < b < b < < b n < b n+ < a n+ < a n < < a < a < a.
21 b n b n+ a n+ a n lim n a n = α, lim n b n = β α = lim n a n+ = lim n a n + b n = α + β α = β α a b M(a, b. M(, n a n b n a n b n (a n+ b n+ = (a n + b n a n b n = ( an b n (a n b n = ( an +. (7 b n a n > b n < a n b n < N < a n+ b n+ < N (. a =, b = c =, a n+ = a n + b n, b n+ = a n b n, c n+ = a n b n, n =,,... π = M (, n c n n=. s n = n k= k c k, p n = a n s n
22 p p p p p π p 3 9 p 4 p 5 4 π π p π ( p, p, p 3 A a a k a = k a a k k a b = k a > b b = b n + b n + + b n + b n + b n+ + b n+ 4 + b i 99 b ( n b < n+ n b < n+ i =,,... Y i = b i + b i + + b i + b i X i X i Y i X i = c i + c i + + c i + c i, c, c,..., c i 9 X i = Z + c i, Z = c i + c i + + c i
23 Z (X i + c i = X i Y i = Y i + b i < (Y i + Z < Y i +, Z Y i (Z+ > Z+c i = X i X i (Z + > Y i = Y i + b i Y i. (Z + > Y i Z Z Y i Z = X i c i Z + Zc i + c i = X i Y i = Y i + b i, Zc i + c i (Y i Z + b i (8 (Y i Z + b i Z Q ZQ (Y i Z + b i < Z(Q + Z(Q + + (Q + > Z(Q + > (Y i Z + b i c i Q c i c i = Q c i = Q c i = Q X n Y n b < (X n + X n X n b = k a < X n + k X n a < k X n + k a k
24 c 3 7 ( + c c (c = 9 c = 8 3 ( 7 + c c ( 89 = ( 34 3, c = 3 3 ( 73 + c 3 c 3 ( 9 = 7 (7 346, c 3 = 3 = =.73 3 = = =.57 3 =.57 q = ( 3 <.68 = 3.6 4
25 7 p = = 6.57 < p < 6.58 = 3.8 p < π < q 3. < π < 3.6 B ([3] k <. K(k = E(k = dx ( x ( k x = dθ k sin θ (, k x π dx = k x sin θ dθ (. x = sin θ. a > b >, k = b a I(a, b = J(a, b = cos θ = sin θ I(a, b = J(a, b = dθ a (a b sin θ = a a (a b sin θ dθ = a dθ a cos θ + b sin θ, a cos θ + b sin θ dθ dθ k sin θ = a K(k, k sin θ dθ = ae(k. I(a, b = K(k, J(a, b = ae(k. (9 a 5. I ( a + b, ab = I(a, b. 5
26 [ ] b tan θ = u du dθ = dθ cos θ du = cos θ b cos θ = cos θ b I(a, b = = I = b + u. dθ a cos θ + b sin θ = (a + u (b + u du = ( a + b, ab = b cos θ, cos θ = + tan θ = (( a+b dθ cos θ a + b tan θ b b + u, (a + u (b + u du. + u (ab + u u = ( v ab ab + u = v 4v (ab + v du, ab + u dv = v, ( a + b + u = 4 (a + v ( + b v = 4v (a + v (b + v v u ( a + b I, ab = dv = I(a, b. (a + v (b + v du. 6. k = k K(k = π M(, k. [ ] a =, b = k n =,,... a n+ = a n + b n, b n+ = an b n m = M(, k lim a n = lim b n = m n n 5 I(, k = I(a, b = = I(a n, b n n K(k = I(, k = I(m, m = m dθ = π m = π M(, k. ( a + b J, ab J(a, b. de dk = dk (E K, k dk = kk (E k K. 6
27 . (a K(k = + k K (c E(k = + k E 7. [ ] J ( k + k ( k + k J k = b a, k = ( a + b, ab = a + b E, (b K(k = + k K + k K(k, (d E(k = ( + k E ( a + b, ab J(a, b = abi(a, b. b a, k = a b + k a + b ( 4ab (a + b, + k ( k k K(k. + k ( k J(a, b = ae(k = a + b E ( a b. a + b J(a, b = ae(k I(a, b = K(k (d a ( a + b J, ( a b ab J(a, b = (a + be ae(k a + b ( k = (a + be ae(k + k ( k = (a + b E(k + + k + k K(k ae(k = bk(k = abi(a, b. 3. lim n n c n =. [ ] (7 8. [ ] ( J(a, b = a n c n I(a, b. n= A n = n (J(a n, b n a ni(a n, b n 7 5 A n+ A n = n (J(a n+, b n+ J(a n, b n n+ a n+i(a n+, b n+ + n a ni(a n, b n = n a n b n I(a n, b n n+ a n+i(a n, b n + n a ni(a n, b n = ( n a n b n (a n + b n + an I(an, b n = n (a n b ni(a n, b n = n c ni(a, b. 7
28 n A n = a ni(a n, b n J(a n, b n = = = c n a n a n cos θ + b n sin θ dθ a n a n cos θ b n sin θ a n cos θ + b n sin θ dθ = sin θ a n cos θ + b n sin θ dθ c n a n cos θ + b n sin θ dθ (a n b n sin θ a n cos θ + b n sin θ dθ dθ a n cos θ + b n sin θ = c ni(a n, b n. < A n n c ni(a n, b n = n c ni(a, b. 3 lim n n c n = A + n= (A n+ A n = lim N A N+ =. J(a, b a I(a, b = A = (A n+ A n = n c ni(a, b J(a, b = = I(a, b ( a n= n c n, n= n c n I(a, b. n= n= 9 (. E(kK(k + E(k K(k K(kK(k = π. [ ] a =, b = k = k = 9 E ( ( ( K K = π. 8 E ( ( = n c n K n= (. 8
29 ( 6 ( K n c n K n= ( = M n c n 4M n= ( = π. π (, π (, = π. C 3. a, b, c C c n Z, n a (n = a(a + (a + n (a + n, a ( = F (a, b, c; u = n= a (n b (n n!c (n un (. f(u = F (a, b, c; u u < u( u d f df + (c (a + b + u abf = ( du du ( x / K(k. K(k = π F (,, ; k. 4. K (k = K(k K(k, K (k (k 3 k d y dk + (3k dy + ky = ( dk 9
30 K K ( 5. K (k = K(k, E (k = E(k EK + E K KK [ ] 5 W = EK + E K KK lim W K( = π k E( = W = (E KK + E K sin θ dθ = cos θ dθ = [ sin θ ] π =. lim W = lim (E k k KK + E(K( = lim (E KK + π k. lim k (E KK = (K EK = = k ( ( = kk kk ( dθ k sin θ ( k sin π θ k sin θ dθ ( k sin θ dθ k dθ cos θ + k sin θ k dθ k cos θ + k sin θ = kk π k sin θ dθ K(k dθ ( k sin θ k dθ ( k sin θ < (K EK < kk π (k. [] π. [],,, 999. [3] π 7 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... (
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 informationHITACHI 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
More informationHITACHI 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
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 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 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 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 informationB. 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
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 informationBD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B
2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :
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 informationさくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a
... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c
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 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 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 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 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 informationa (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
More information5 36 5................................................... 36 5................................................... 36 5.3..............................
9 8 3............................................. 3.......................................... 4.3............................................ 4 5 3 6 3..................................................
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 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 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 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 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 information日立液晶プロジェクター CP-AW2519NJ 取扱説明書- 詳細版-
PAGE UP DOWN D- ESC ENTER 1 1 2 2 3 COMPUTER IN1 USB TYPE A DC5V 0.5A USB TYPE B HDMI COMPUTER IN2 LAN CONTROL MONITOR OUT MIC AUDIO IN1 AUDIO IN3 AUDIO OUT R R L L S-VIDEO AUDIO IN2 VIDEO PAGE UP DOWN
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 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 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 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 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 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 information1/68 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
More information, ,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
More informationA(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 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
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 information2/8 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)
More information05‚å™J“LŁñfi~P01-06_12/27
2005 164 FFFFFFFFF FFFFFFFFF 2 3 4 5 6 7 8 g a 9 f a 10 g e g 11 f g g 12 a g g 1 13 d d f f d 14 a 15 16 17 18 r r 19 20 21 ce eb c b c bd c bd c e c gf cb ed ed fe ed g b cd c b 22 bc ff bf f c f cg
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 informationdi-problem.dvi
III 005/06/6 by. : : : : : : : : : : : : : : : : : : : : :. : : : : : : : : : : : : : : : : : : : : : : : : : : 3 3. : : : : : : : : : : : : : : 4 4. : : : : : : : : : : : : : : : : : : : : : : 5 5. :
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 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 informationuntitled
10 log 10 W W 10 L W = 10 log 10 W 10 12 10 log 10 I I 0 I 0 =10 12 I = P2 ρc = ρcv2 L p = 10 log 10 p 2 p 0 2 = 20 log 10 p p = 20 log p 10 0 2 10 5 L 3 = 10 log 10 10 L 1 /10 +10 L 2 ( /10 ) L 1 =10
More information‚å™J‚å−w“LŁñfi~P01†`08
156 2003 2 3 4 5 6 7 8 9 c f c a g 10 d c d 11 e a d 12 a g e 13 d fg f 14 g e 15 16 17 18 19 20 21 db de de fg fg g gf b eb g a a e e cf b db 22 d b e ag dc dc ed gf cb f f e b d ef 23 f fb ed e g gf
More information000 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
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 informationi 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...................
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 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 information案内(最終2).indd
1 2 3 4 5 6 7 8 9 Y01a K01a Q01a T01a N01a S01a Y02b - Y04b K02a Q02a T02a N02a S02a Y05b - Y07b K03a Q03a T03a N03a S03a A01r Y10a Y11a K04a K05a Q04a Q05a T04b - T06b T08a N04a N05a S04a S05a Y12b -
More information0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ). f ( ). x i : M R.,,
2012 10 13 1,,,.,,.,.,,. 2?.,,. 1,, 1. (θ, φ), θ, φ (0, π),, (0, 2π). 1 0.,,., m Euclid m m. 2.., M., M R 2 ψ. ψ,, R 2 M.,, (x 1 (),, x m ()) R m. 2 M, R f. M (x 1,, x m ), f (x 1,, x m ) f(x 1,, x m ).
More information17 18 2
17 18 2 18 2 8 17 4 1 8 1 2 16 16 4 1 17 3 31 16 2 1 2 3 17 6 16 18 1 11 4 1 5 21 26 2 6 37 43 11 58 69 5 252 28 3 1 1 3 1 3 2 3 3 4 4 4 5 5 6 5 2 6 1 6 2 16 28 3 29 3 30 30 1 30 2 32 3 36 4 38 5 43 6
More information4 5.............................................. 5............................................ 6.............................................. 7......................................... 8.3.................................................4.........................................4..............................................4................................................4.3...............................................
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 informationiii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................
More information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
More informationuntitled
1 2 1 2 1 1 2 2 18 1 1990 2 3 4 5 6 2006 1 19981995 1999 1993 20002004 2006 2004 2006 1 2 1970 70 1980 71 86 01 71 86 01 4 4 2 5 12 8 7 1 3 10 8 9 2 3 4 11 10 10 6 5 6 14 14 10 20063 15 4 71 86 01 71 86
More information案内最終.indd
1 2 3 4 5 6 IC IC R22 IC IC http://www.gifu-u.ac.jp/view.rbz?cd=393 JR JR JR JR JR 7 / JR IC km IC km IC IC km 8 F HPhttp://www.made.gifu-u.ac.jp/~vlbi/index.html 9 Q01a N01a X01a K01a S01a T01a Q02a N02a
More information05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
More informationuntitled
( )!? 1 1. 0 1 ..1 6. 3 10 ffi 3 3 360 3.3 F E V F E + V = x x E E =5x 1 = 5 x 4 360 3 V V =5x 1 3 = 5 3 x F = x; E = 5 x; V = 5 3 x x 5 x + 5 3 x = x =1 1 30 0 1 x x E E =4x 1 =x 3 V V =4x 1 3 = 4 3 x
More information( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED)
rational number p, p, (q ) q ratio 3.14 = 3 + 1 10 + 4 100 ( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED) ( a) ( b) a > b > 0 a < nb n A A B B A A, B 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 information50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
More information‚å™J‚å−w“LŁñ›Ä
2007 172 FFFFFFFFF FFFFFFFFF 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 c d e cc bd b fb ag ag ed ed ed bd b b ef bf f df bd f bff d D f F d f 19 bd 20 21 F C e e f b b b 22 d d e f e f bf bd 23 24 222222222222222222222222222222222222222222222222222222222222222222222222
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 information1 913 10301200 A B C D E F G H J K L M 1A1030 10 : 45 1A1045 11 : 00 1A1100 11 : 15 1A1115 11 : 30 1A1130 11 : 45 1A1145 12 : 00 1B1030 1B1045 1C1030
1 913 9001030 A B C D E F G H J K L M 9:00 1A0900 9:15 1A0915 9:30 1A0930 9:45 1A0945 10 : 00 1A1000 10 : 15 1B0900 1B0915 1B0930 1B0945 1B1000 1C0900 1C0915 1D0915 1C0930 1C0945 1C1000 1D0930 1D0945 1D1000
More informationSIRIUS_CS3*.indd
SIRIUS Innovations SIRIUS SIRIUS Answers for industry. SIRIUS SIRIUS S00 S0 SIRIUS SIRIUS ZX0-ORAZ-0AB0 7.5kW 6 S00 7 8 7.5kW 9 S00 0 8.5kW S0 8.5kW S0 5 6 7 IO-Link AS-InterfaceRT 8 8US 5 6 SIRIUS SIRIUS
More information7 9 7..................................... 9 7................................ 3 7.3...................................... 3 A A. ω ν = ω/π E = hω. E
B 8.9.4, : : MIT I,II A.P. E.F.,, 993 I,,, 999, 7 I,II, 95 A A........................... A........................... 3.3 A.............................. 4.4....................................... 5 6..............................
More information12~
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
More information., a = < < < n < n = b, j = f j j =,,, n, C P,, P,,, P n n, n., P P P n = = n j= n j= j j + j j + { j j / j j } j j, j j / j j f j 3., n., Oa, b r > P
. ϵριµϵτρoζ perimetros 76 Jones, Euler. =.,.,,,, C, C n+ P, P,, P n P, P n P n, P P P P n P n n P n,, C P, P j P j j =,,, n P n P., C.,, C. f [a, b], f. C = f a b, C l l = b a + f d P j P j a b j j j j
More informationuntitled
0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.
More informationx, 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
More informationSolutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F
Quiz 1 Due at 10:00 a.m. on April 20, 2007 Division: ID#: Name: 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T F T T F T T T F F T F T T T F T F T F F T T F F F T 2. 1.1 (1) (7) p.44 (1)-(4)
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春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
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 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 informationac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y
01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y
More information1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
More information2 1 17 1.1 1.1.1 1650
1 3 5 1 1 2 0 0 1 2 I II III J. 2 1 17 1.1 1.1.1 1650 1.1 3 3 6 10 3 5 1 3/5 1 2 + 1 10 ( = 6 ) 10 1/10 2000 19 17 60 2 1 1 3 10 25 33221 73 13111 0. 31 11 11 60 11/60 2 111111 3 60 + 3 332221 27 x y xy
More informationuntitled
...1... 3 1... 3 2... 4 3... 4 4... 5...... 6 1... 6 2... 7 3... 8 4... 9 5... 10... 12 1... 12 2... 13 3... 14 4... 16...... 19 1... 19 2... 20 3... 22 4... 24...... 25... 26 1... 26 2... 26 3... 26......
More information1 180m g 10m/s 2 2 6 1 3 v 0 (t=0) z max t max t z = z max 1 2 g(t t max) 2 (6) 1.3 2 3 3 r = (x, y, z) e x, e y, e z r = xe x + ye y + ze z. (7) v =
1. 2. 3 3. 4. 5. 6. 7. 8. 9. I http://risu.lowtem.hokudai.ac.jp/ hidekazu/class.html 1 1.1 1 a = g, (1) v = g t + v 0, (2) z = 1 2 g t2 + v 0 t + z 0. (3) 1.2 v-t. z-t. z 1 z 0 = dz = v, t1 dv v(t), v
More informationZ: Q: R: C: 3. Green Cauchy
7 Z: Q: R: C: 3. Green.............................. 3.............................. 5.3................................. 6.4 Cauchy..................... 6.5 Taylor..........................6...............................
More information190 87 28 1 212 77 1777 77 219 1 171 28 201 1 1 16 102 17 10 1 16 99 1 1 1 1 960 1 1 1 1 1 2 168 1 12 2 18 100 2 1 6 1 61 7 16 18 20 2 961 2 11 6 2 6 6 0 17 86 1 2 16 1 1 9 2 1 1 1 1 1 1 0 2 17 16 6 1
More information( ) x y f(x, y) = ax
013 4 16 5 54 (03-5465-7040) nkiyono@mail.ecc.u-okyo.ac.jp hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy
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 informationPROSTAGE[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
More information取扱説明書 [F-12C]
F-12C 11.7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 a bc b c d d a 15 a b cd e a b c d e 16 17 18 de a b 19 c d 20 a b a b c a d e k l m e b c d f g h i j p q r c d e f g h i j n o s 21 k l m n o p q r s a X
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 information200608094-101
94 A O D 1 A 1 A A 1 AO 1 95 A OA 1 a r A A 1 r A R 1 A R 1 A R 1 a a A OA R 1 96 F AO 1 A O 1 A 1 A O 1 A 1 O A 1 97 b O AO 1 O AO 1 A 1 A OA 1 AO 1 AA 1 98 A AO 1 A AO 1 b b 1 b b B B A 1 Q 1 rr 1 99
More informationr III... IV.. grad, div, rot. grad, div, rot 3., B grad, div, rot I, II ɛ-δ web page (
r 8.4.8. 3-3 phone: 9-76-4774, e-mail: hara@math.kyushu-u.ac.jp http://www.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html Office hours: 4/8 I.. ɛ-n. ɛ-δ 3. 4. II... 3. 4. 5.. r III... IV.. grad, div,
More informationMS-1J/MS-1WJ(形名:MS-1/MS-1W)取扱説明書 - 詳細- 技術情報編
720 x 400 37.9 85.0 VESA TEXT 640 x 480 31.5 59.9 VESA VGA (60Hz) 640 x 480 37.9 72.8 VESA VGA (72Hz) 640 x 480 37.5 75.0 VESA VGA (75Hz) 640 x 480 43.3 85.0 VESA VGA (85Hz) 800 x 600 35.2 56.3 VESA SVGA
More informationr 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B
1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n
More information