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- れいな はにうだ
- 5 years ago
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5 ( ) p p 2, 3, 5, 7,, p 1 Q = p + 1 Q > p (i) Q p (ii) Q Q q Q q 2 p q > p p Q.E.D.
6 n 1 ( ) p 2, 3, 5, 7,, p Q = p 1 4n 1 Q > p (i) Q p 4n 1 (ii) Q 4n 1 4n + 1 (4n 1 + 1)(4n 2 + 1) = 16n 1 n 2 + 4(n 1 + n 2 ) + 1 = 4N + 1 Q 4n 1 4n 1 q Q q 2 p q > p p 4n 1 4n 1 Q.E.D n 1
7 a, b gcd(a, b) lcm(a, b) gcd(a, b) = 1 a, b 4. a, b g = gcd(a, b) l = lcm(a, b) (1) a, b g (2) a, b l (3) ab = gl ( ) (1) (2) (3) a = ga 1, b = gb 1 gcd(a 1, b 1 ) = 1 l = ga 1 b 1 ab = g 2 a 1 b 1 = g(ga 1 b 1 ) = gl Q.E.D. 5. N, a, b, c (1) a, b N gcd(a, b) = 1 ab N (2) b ac gcd(a, b) = 1 b c (3) p ac p a, c ( ) (1) N a, b 4 (2) N l = lcm(a, b) gcd(a, b) = 1 l = ab ab N (2) b ac ac = bb gcd(a, b) = 1 b c b c (3) p a gcd(a, p) = 1 (2) p c Q.E.D.
8 ( ) 3 f(n) = (n 1)n(n + 1) 5 (1) f(n) 6 = 2 3 f(n) f(n) f(n) 3 f(n) p 5 p
9 x A x B x A x B A B A B B A A B x A x B x x A x B
10 x x 2 = 2 x ( ) 1, 2, 3, 4, 1, 4, 9, 16, 2 x x x = m n (n 1) x 2 = m2 n 2 (n 2 1) m2 2 n2 x Q.E.D x x 2 = 3 x 4. x 10 x = 2 x
11 x, y x y a x y mod a x a y a mod modulo 10. (1) (5) (1) x x mod a (2) x y mod a y x mod a (3) x y mod a y z mod a x z mod a (4) x x mod a y y mod a x + y x + y mod a (5) x x mod a y y mod a xy x y mod a ( ) (1), (2), (3) (4) x x = ma y y = na (x x) + (y y) = (m + n)a x + y x + y mod a (5) x y x y = (x+ma)(y+na) xy = (my+nx)a+mna 2 = (my+nx+mna)a xy x y mod a Q.E.D x, y, z x 2 + y 2 = z 2 x, y, z 5
12 x, y, z x 2 + y 2 = z 2 xyz p a p a p 1 1 mod p ( ) a p a, 2a, 3a,, (p 1)a p ja ka mod p (1 k < j < p) (j k)a 0 mod p j k a p a, 2a, 3a,, (p 1)a p a 2a (p 1)a (p 1)! (p 1)! a p 1 1 mod p Q.E.D p a a p a mod p p 2 5 1/p
13 n n n (n 1) 2 1 n! n 0! = 1 n P r np r := n! (n r)! n r P Permutation 16. np r = n (n 1) (n r + 1) ( ) Permutation np r = n! n (n 1) (n r + 1) (n r)! = (n r)! (n r)! = n (n 1) (n r + 1) Q.E.D. np r 1 n n r n P r np r = n 1 P r + r n 1 P r 1
14 (1), (2) (1) 6 0, 1, 2, 3, 4, (2) n 2 n + 1 0, 1, 2,, n r 0 r n np r 1 r r/2 n P r 1 + r (n 1) n 1 P r P P P 4 = 84 2 A C B D
15 n r n r n C r nc r := n P r r! n r C Combination 18. nc r = n C n r ( ) Combination nc r = n P r r! = n! r!(n r)! = n! {n (n r)}!(n r)! = n P n r (n r)! = nc n r Q.E.D. nc r 1 n n r n C r nc r = n 1 C r + n 1 C r (a + b) n a n r b r n C r
16 (x + 1) (x + 1) 1 x (x + 1) 2 x 2, x ( ) n (x + 1) n x n, x n 1, x n n r r 0 r 0 (x + 1) (x + 1) (x + 1) (x + 1) (x + 1) n r 0 ( ) ( n n + 2 r r + 1 ) ( ) n + = r + 2 ( ) n + 2 r + 2
17 (a + b) n a n r b r n C r., (a + b) n = n C 0 a n + n C 1 a n 1 b + n C 2 a n 2 b n + + n C n 1 ab n 1 + n C n b n nc 0 + n C 1 + n C n C n 1 + n C n = 2 n ( ), a = b = 1, ( ) = (1 + 1) n = 2 n ( ) = n C 0 + n C n C n 1 + n C n,. Q.E.D nc 0 + n C n C n C n n + 1 = 2n+1 1 n + 1 ( ) r nc r = r n! r! (n r)! = n (n 1)! (r 1)! (n r)! = n n 1C r 1 n 1C r 1 r = n C r n, n C r 1 r = n+1 C r n + 1
18 , ( ) = n C 0 + n C n C n C n n + 1 = n+1 C 1 n n+1 C 2 n n+1 C 3 n n+1 C n+1 n + 1 = 1 n + 1 ( n+1c 1 + n+1 C 2 + n+1 C n+1 C n+1 ) = 1 n + 1 {( n+1c 0 + n+1 C 1 + n+1 C n+1 C n+1 ) n+1 C 0 } = 1 ( 2 n+1 1 ) n + 1 = ( ) Q.E.D. 15. nc 0 n C 1 + n C 2 + ( 1) n 1 nc n 1 + ( 1) n nc n = nc nc nc n nc n = n 2 n 1
19 a, d a, a + d, a + 2d, a + 3d, a, d. a, d a n a n = a + (n 1)d a, d n S n S n = n{2a + (n 1)d} 2 ( ), S n = a 1 + a a n 1 + a n, S n = a n + a n a 2 + a 1 S n = a 1 + a a n 1 + a n +) S n = a n + a n a 2 + a 1 2S n = (a 1 + a n ) + (a 2 + a n 1 ) + + (a n 1 + a 2 ) + (a n + a 1 ). a r+1 + a n r = a + rd + {a + (n r 1)d} = 2a + (n 1)d, 2S n = n{2a + (n 1)d}. 2,. Q.E.D.
20 ( ) ( ) ( ) ( ) 0 1 n n + 1 (1) = = n (2) (3) ( ) ( ) ( ) n = 1 ( ) n = 2 ( n ( n ), (n 1) ), (n 2) (k 1)k (k 1)k(k + 1) (k 2)(k 1)k (3) = ( ) n 2 = 1 ( ) n + 2 ( ) n
21 a, r a, ar, ar 2, ar 3, a, r a, r n S n a(1 r n ) S n = 1 r na (r 1 ) (r = 1 ) ( ) S n = a + ar + + ar n 2 + ar n 1, r = 1. S n = a } + a + {{ + a } = na n r 1, S n rs n = ar + ar ar n 1 + ar n S n = a + ar + ar ar n 1 ) rs n = ar + ar ar n 1 + ar n (1 r)s n = a ar n, (1 r)s n = a(1 r n ). 1 r,. Q.E.D x n 1 x 1.
22 n S n S n = 10n+1 9n , 99, 999, 9999, 21. n S n S n = 2 (10n+1 9n 10) 27 6, 66, 666, 6666, 22. x x + 3x nx n 1 = 1 (n + 1)xn + nx n+1 1 x
23 a 1 + a a n ( ) a k k 1 n a k k i j 10 k = k 1 = (2j 1) = j=1 i 3 = n 3 i=1 27. (1) (2) (3) (a n ± b n ) = a n ± b n ( ) ca n = c a n (c ) c = nc (c ) ( ) (1) (a n ± b n ) = (a 1 ± b 1 ) + (a 2 ± b 2 ) + + (a n ± b n ) = (a 1 + a a n ) ± (b 1 + b b n ) = a n ± Q.E.D. b n
24 (2) (3) 28. (1) (2) ( ) (1) 24 k = 1 n(n + 1) 2 k 2 = 1 n(n + 1)(2n + 1) 6 k 1, 1 n k = n{2 + (n 1)} 2 = 1 n(n + 1) Q.E.D. 2 (2) (k + 1) 3 (k 1) 3 = 6k { (k + 1) 3 (k 1) 3} = ( 6k ) ( ) = (k + 1) 3 (k 1) 3 = { (n 1) 3 + n 3 + (n + 1) 3} { (n 1) 3} = n 3 + (n + 1) 3 1 = n 3 + (n 3 + 3n 2 + 3n + 1) 1 = 2n 3 + 3n 2 + 3n ( ) = 6k = 6 k 2 + 2n
25 n 3 + 3n 2 + 3n = 6 k 2 + 2n k 2 = 1 6 (2n3 + 3n 2 + n) = 1 n(n + 1)(2n + 1) 6 Q.E.D (2n 1) = n n(n + 1) = 1 n(n + 1)(n + 2) (k + 1) 4 (k 1) 4 = 8k 3 + 8k, k 3 = { } 1 2 n(n + 1) 2
26 n P(n) P(n) n (i) P(1) (ii) P(k) P(k + 1) (i) (ii) n P(n) n P(n) (i) (ii) P(1) (ii) P(2) P(2) (ii) P(3) P(3) (ii) P(4) P(4) (ii) P(5) (i) (ii) n P(n) (i) (ii) (i) (ii) n P(n)
27 n n 3 + 2n 3. ( ) P(n) n 3 + 2n 3 n (i) n = 1 n 3 + 2n = = 3 3 P (1) (ii) P(k) k 3 + 2k 3 k 3 + 2k = 3m (m ) P(k + 1) (k + 1) 3 + 2(k + 1) (k + 1) 3 + 2(k + 1) = (k 3 + 3k 2 + 3k + 1) + (2k + 2) = (k 3 + 2k) + (3k 2 + 3k + 3) = 3m + 3(k 2 + k + 1) = 3(m + k 2 + k + 1) (k +1) 3 +2(k +1) 3 P(k +1) (i) (ii) n P(n) Q.E.D n 2n 3 3n 2 + n 6 n 5 n 2 n 3
28 n 2 n 2n ( ) P(n) 2 n 2n n (i) n = 1 P (1) = 2 1 = 2, = 2 1 = 2 P(1) (ii) P(k) 2 k 2k P(k + 1) 2 k+1 2(k + 1) P(k + 1) = 2 k+1 = 2 2 k 2 k 2k 2 2k = 2k + 2k 2k + 2 = 2(k + 1) = P(k + 1) (i) (ii) n P(n) Q.E.D. 29. n 5 n 2 n > n n n 2n n + 1
29 ABC BC CA AB P Q R AP BQ CR S BP PC CQ QA AR RB = 1 A R S Q B P C ( ) ABS CAS S 1 S 2 ABS CAS BP PC = S 1 S 2 BCS S 3 ABS BCS CAS BCS CQ QA = S 3 S 1 AR RB = S 2 S 3 Q.E.D. BP PC CQ QA AR RB = S 1 S3 S2 = 1 S 2 S 1 S 3
30 ABC BC CA AB P Q R BP PC CQ QA AR RB = 1 AP BQ CR ABC BC CA AB L M N AL BM CN 2 : ABC
31 ABC A, B, C l BC CA AB P Q R A BP PC CQ QA AR RB = 1 l R Q B C P ( ) C AB l S PBR PCS BP : CP = RB : SC BP CP = BR SC QAR QCS AR : QA = CS : QC AR QA = CS QC Q.E.D. BP PC CQ QA AR RB = BP PC AR QA CQ RB = RB SC CS QC CQ RB = 1
32 A l R B Q C S P ABC BC CA AB P Q R BP PC CQ QA AR RB = 1 P Q R ABC A < B A BC P B CA Q C AB R P Q R
33 ABCD AB CD + AD BC = AC BD D A C ( ) BD E B BAE = CAD D A E C ABE ACD B ABE = ACD ABE ACD AB : BE = AC : CD AB CD = AC BE (1.1) AED ABC AD : ED = AC : BC AD BC = AC ED (1.2) (1.1) (1.2) AD CD + AD BC = AC BE + AC ED = AC (BE + ED) = AC BD
34 Q.E.D a, b, c 1 a = 1 b + 1 c
35 [ ] p p 1 7 = = = = p [ ] p mod p mod 2
36 mod 3 mod 5 [ ] a 1, a 2, a 3,, a n,
37 [ ] [ ] 1, 1, 2, 3, 5, 8, 13, 21, 34, n F n F n = F n 1 + F n 2 F 1 + F 2 = = 2, F 4 1 = 3 1 = 2 F 1 + F 2 + F 3 = = 4, F 5 1 = 5 1 = 4 F 1 + F 2 + F 3 + F 4 = = 7, F 6 1 = 8 1 = 7 F 1 + F 2 + F 3 + F 4 + F 5 = = 12, F 7 1 = 13 1 = 12
38 a n! p 2 p 1 2 p 1 5. n 2 2n n n n (n 1) n 1 n n p 1 p p (p 1) p 1 p 8. n n 2 3n + 1 n 1 9. n 10. n 6 n
39 p 2, 3, 5, 7,, p Q = p 1 6n 1 p (6n 1 + 1)(6n 2 + 1) = 6N + 1 Q Q p 6n = f(p) = p 4 1 = (p 1)(p + 1)(p 2 + 1) 2 4, 3, 5 (1) f(p) 3 p 3k + 1 f(3k + 1) = 3k(3k + 2)(9k 2 + 1) f(p) 3 p 3k 1 f(p) 3 p 5 p 3k (2) f(p) 5 p 5k + 1, 5k + 2, 5k + 3, 5k (3) f(p) 2 4 p 2k + 1 f(p) = 2 3 k(k + 1)(2k 2 + 2k + 1) k(k + 1) f(p) m, n x = m n 10m/n = 2 10 m = 2 n 2 m 5 m = 2 n 5. x, y, z 5 x ±1 mod 5 y ±1 mod 5 z ±1 mod 5 x 2 + y 2 z 2 0 mod 5 6. x, y, z 4 x 2 ±1 mod 4 y 2 ±1 mod 4 z 2 ±1 mod 4 x 2 + y 2 z 2 0 mod 4 xyz 4 7. a p a p a 1 a 2 p a = a 1 a 2 11 a p 1 a 1 mod p a p 2 a 2 mod p a p = a p 1 ap 2 a 1a 2 = a mod p a p 1 1 mod p n = (10 p 1 1)/p α = 1/p n = 10 p 1 α α α 10 p α
40 α 9. n 1P r = 10. (n 1)! (n r 1)! (n 1)! n 1P r 1 = (n r)! (1) 5 5 P 3 = P 3 = 60 2, 4 2 (4 4 P 2 ) = 96 (2) n n P r 1 0 n P r 1 r (n 1) n 1 P r P P 3 4 P n 1C r = n 1 P r r! = (n 1)! r!(n r 1)! n 1C r 1 = n 1 P r 1 (r 1)! = (n 1)! (r 1)!(n r)! 13. (a + b) n = (a + b) (a + b) a n rb r n (a + b) r b ( ) n 14. n C r 19 ( n C r + n C r+1 ) + ( n C r+1 + r nc r+2 ) = n+1 C r+1 + n+1 C r+2 = n+2 C r S = 0 nc nc nc n nc n S = n nc n + 1 nc nc 0 2S = n( n C nc n nc n ) 15 2S = n 2 n ( ) k 17. (1) = 1 ( ) 0 k n(n + 1) (2) = k n = ( 1) 2 ( ) ( ) k (k 1)k (k 1)k(k + 1) (k 2)(k 1)k 2 3 (3) = = S = + + ( 2) n + = (n 2)(n 1)n (n 1)n(n + 1) ( 6 ) (n 1)n(n + 1) (n 2)(n 1)n (n 1)n(n + 1) n + 1 S = = ( ) ( ) n n = n (x 1)S n = x n S n = (10 1) + (10 2 1) + + (10 n 1) = n n
41 S n = 10(10n 1) S n = n+1 9n 10 9 n = 2 ( ) S = 1+2x+3x 2 + +nx n 1 xs = x+2x 2 +3x 3 + +nx n (1 x)s = 1+x+x 2 + +x n 1 nx n 19 (1 x)s = 1 xn 1 x nxn ( ) = (2k 1) = 2 k = n(n + 1) n = n2 = ( ) 25. ( ) = k(k + 1) = = 1 n(n + 1)(n + 2) = ( ) k 2 + { (k + 1) 4 (k 1) 4} =, 28 (2) 6n 2 + 4n + 1, k = 1 6 n(n + 1)(2n + 1) + 1 n(n + 1) 2 (8k 3 + 8k) { (k + 1) 4 (k 1) 4} = 2n 4 + 4n 3 + (8k 3 + 8k) = 8 k 3 + 4n(n + 1) 27. (i) n = 1 2n 3 3n 2 + n = = 0 6 (ii) n = k 2n 3 3n 2 + n = 2k 3 3k + k 6 2k 3 3k 2 + k = 6m (m ) n = k + 1 2n 3 3n 2 +n = 2(k+1) 3 3(k+1) 2 +(k+1) = 2(k 3 +3k 2 +3k+1) 3(k 2 +2k+1)+k+1 = 2k 3 + 3k 2 + k = 2k 3 3k 2 + k + 6k 2 = 6m + 6k 2 = 6(m + k 2 ) n = k + 1 2n 3 3n 2 + n 6 (i) (ii) n 2n 3 3n 2 + n (i) n = 1 5 n 2 n = 5 2 = 3 3 (ii) n = k 5 n 2 n = 5 k 2 k 3 5 k 2 k = 3m (m ) 5 k = 2 k +3m n = k +1 5 n 2 n = 5 k+1 2 k+1 = 5 5 k 2 2 k = 5(2 k + 3m) 2 2 k = 5 2k + 15m 2 2 k = 3 2 k + 15m = 3(2 k + 5m)
42 n = k n 2 n 3 (i) (ii) n 5 n 2 n (i) n = 5 2 n = 2 5 = 32 n 2 = 5 2 = 25 2 n > n 2 (ii) n = k 2 k > k 2 n = k k+1 (k+1) 2 = 2 2 k (k 2 +2k+1) > 2k 2 (k 2 +2k+1) = k 2 2k 1 = k(k 1) 1 > 0 2 k+1 > (k + 1) 2 n = k n > n 2 (i) (ii) n 5 n 2 n > n (i) n = 1 ( ) = = 1 ( ) = = 1 (ii) n = k k 2k k + 1 n = k k + 1 k + 1 2k k k + 1 = 2k + 1 k + 1 ( ) 2k + 1 k + 1 2k + 1 k + 1 2(k + 1) (k + 1) + 1 2(k + 1) (2k + 1)(k + 2) 2(k + 1)2 = (k + 1) + 1 (k + 1)(k + 2) = k (k + 1)(k + 2) > 0 2k + 1 k + 1 2(k + 1) > ( ) (k + 1) k + 1 k + 1 2k + 1 2(k + 1) > k + 1 (k + 1) + 1 n = k + 1 (i) (ii) n 31. BQ CQ S AS BC P AP BQ CR S 33 BP P C CQ QA AR RB = 1 BP PC = BP P C P P BC P P AQ BQ CR AL BM CN G AG BC AG = GD D ABD NG BD GC BD ADC MG CD GB CD GBDC GD BC L
43 AG : GL = 2 : 1 BG : GM = 2 : 1 CG : GN = 2 : A, B, C BC, CA, AB I, J, K ABI CBK AB : BI = CB : BK BI BK = AB BC CJ CI = BC AC AK AJ = AC BI AB BK CJ CI AK AJ = AB BC BC AC AC AB = P BC PQ AB A, B R 37 BP PC CQ QA AR R B = 1 AR RB = AR R B R R AB R R P Q R 35. RC S AS BC AC C D ACS = RCD = BCB = CSA AC = AS RAS RBC RA : RB = AS : BC = AC : BC R AB CA : CB AR : RB = AC : BC AP A BP : PC = AB : AC BQ B CQ : QA = BC : AB AR RB BP PC CQ QA = AC BC AB AC BC = 1 38 AB P Q R 36. ABCDE ABCD 40 AB CD + AD BC = AC BD x = DA = AC = BD 1 + x = x 2 x = ABCDEFG a = AB b = AC c = AD ACDE 40 AC DE + CD EA = AD CE b a + a c = c b abc 1 a = 1 b + 1 c [ ] 1 2m p
44 m p 1 p 102m 1 mod p 10 m 1 mod p 10m p + 1 p 10m p 1 1 = p [ ] mod 3 n n = 0, 1, 2, 3 k n n a 1, a 2, a 3,, a s n 3 k a 1, 0,, 0, a 2, 0,, 0, a 3, 0, 0,, 0 0, a s a i a i k 1 n 3 k n O 3 k 1 n (x + 1) n = x n + a 1 x n a s 1 x + 1 (x + 1) 3 x mod 3 a 3 i a i mod 3 (x + 1) 3n = x 3n + a 1 x 3(n 1) + + a s 1 x n n O 2 3 k n n O 3 k 1 [ ] 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 2 1, 0, 1, 0, 3 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 5 1, 4, 4, 1, 0, 1, 4, 4, 1, 0, 1, 4, 4, [ ] A,B,C,D AB = CD = 10 AD = 5 2 BC = 2 AC = BD = 2 5 AB CD + AD BC = AC BD = 20 A,B,C,D [ ] (1) F 1 + F 3 + F F 2n 1 (2) F 2 + F 4 + F F 2n
45 (3) F 1 F 2 + F ( 1) n+1 F n (4) F 1 F 2 + F ( 1) n+1 F n (5) F F F F 2 n
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,
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
( )
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...............................
(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
熊本県数学問題正解
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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)
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
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) = =
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2012 A, N, Z, Q, R, C
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プロダクトニュース生産終了予定商品のお知らせ発行日 2016 年 3 月 1 日 押ボタンスイッチ / 表示灯 No. 2016025C 押ボタンスイッチ形 A22 シリーズ一部商品 ツマミ形セレクタスイッチ形 A22S シリーズ一部商品 形 A22W シリーズ一部商品 キー形セレクタスイッチ形 A22K シリーズ一部商品 表示灯形 M22 シリーズ一部商品生産終了のお知らせ 生産終了予定商品 押ボタンスイッチ
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０４年度LS民法Ⅰ教材改訂版.PDF
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応用数学III-4.ppt
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LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2002 2 2 2 2 22 2 3 3 3 3 3 4 4 5 5 6 6 7 7 8 8 9 Cramer 9 0 0 E-mail:hsuzuki@icuacjp 0 3x + y + 2z 4 x + y
h1-4_cs5.5.indd
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1/120 別表第 1(6 8 及び10 関係 ) 放射性物質の種類が明らかで かつ 一種類である場合の放射線業務従事者の呼吸する空気中の放射性物質の濃度限度等 添付 第一欄第二欄第三欄第四欄第五欄第六欄 放射性物質の種類 吸入摂取した 経口摂取した 放射線業 周辺監視 周辺監視 場合の実効線 場合の実効線 務従事者 区域外の 区域外の 量係数 量係数 の呼吸す 空気中の 水中の濃 る空気中 濃度限度