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
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1 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 = CO + OE = BA + AF = AB + AF = a + b B C a A b O D F E [1] ABCD O AB 1 : 2 E, BC F AB, AD (1) AO (2) DO (3) AF (4) EF [2] x, y a, b (1) x + 2( x b) = 4( b 2 a) a (2) 1 2 ( b + x) + 3( x 1 2 b) = 0 (3) 2 x + 3 y = a + b, 3 x + 2 y = b 1
2 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 2k = 4 l = 6, k = 8 (2) 1 a 1 1 a = (3, 5) = (3, 5) [3] a = ( 1, 4), b = (3, 2), c = (0, 5) (1) 3 a + b (2) a 2 b + 3 c (3) 6( c 2 a) 5( 3 b + a) [4] (1) a = (1, 1), b = ( 1, 3), c = ( 3, 5) c = l a + k b l, k (2) ABCD A(1, 2), B(3, 5), C( 4, 0) D [5] a = (3, 1), b = (1, 2), c = a + t b t (1) c = 15 t (2) c t [6] A, B, C AB = a, AC = b ABC a, b 2
3 3 (1) a = (x 1, y 1 ), b = (x 2, y 2 ), a b θ a b cos θ = x 1 x 2 + y 1 y 2 (2) a = (1, 1), b = (2, x) x (1) a b (2) a b 120 (3) a b (1) O(0, 0), A(x 1, y 1 ), B(x 2, y 2 ) AB 2 = OA 2 + OB 2 2OA OB cos θ (x 2 x 1) 2 + (y 2 y 1) 2 = (x y1) 2 + (x y2) 2 2 a b cos θ a b cos θ = x 1x 2 + y 1y 2 (( )) a b = a b cos θ = x 1 x 2 + y 1 y 2 (2) (1) a b = ( 1) x = 0 x = 2 (2) a b = a b cos ( 1) x = p ( 1) x 2 ( ), 2 x = 4 + x 2 2 ( 2(2 x)) 2 = ( 4 + x 2 ) 2, 2(4 4x + x 2 ) = 4 + x 2, x 2 8x + 4 = 0, x = x = (3) b = k a (2, x) = k(1, 1), 2 = k x = k x = 2 [7] a = 4, b = 5 a b a b (1) 45 (2) 120 (3) 90 (4) 180 [8] a, b (1) a = (2, 0), b = (2, 1) (2) a = (1, 1), b = (3, 2) (3) a = (k + 2, k 1), b = (2k 4, k + 1) (4) a = (p + q, q), b = (p q, p) [9] a, b (1) a = (1, 3), b = (2, 0) (2) a = (3, 7), b = (2, 5) (3) a = ( 3 1, 3 1), b = (1, 1) (4) a = ( 2, 2), b = ( 3 1, 3 + 1) [10] (1) a = 1, b = 5, 2 a + b = 3 a b (2) a = (3, 7), b = (2, 5) a + t b t 3
4 4 ABC AB) 3 : 1 D AC 4 : 3 E BE CD P AP BC Q (1) BE, CD AB, AC (2) AP : P Q [11] OAB OA 5 : 2 C OB 3 : 4 D CD M (1) OM OA, OB (2) OM AB N ON : OM, AN : NB [12] ABCDE AB = a, BC = b CD a, b [13] ABC AB = c, BC = a, CA = a AB = c, BC = a, CA = b (1) ABC G AG b c (2) AG 2 a, b, c (3) a = 12, b = 21, c = 3 AGB 4
5 5 [1] (1) ( 3, 4) v = (1, 2) (2) (1, 3), ( 2, 4) (3) (5, 4) v = (1, 2) [2] (2, 0) ( 1, 3) [14] (1) (, 4) n = (1, 2) (2) (1, 3), ( 2, 4) [15] ( 2, 3) x y 1 = 0 [16] ABC O OA = a, OB = b, OC = c (1) A BC A AA (2) ABC H OH a, b, c 5
6 6 OABC AB 1 : 2 D CD 3 : 5 E OE 1 : 3 F AF OBC G OA = a, OB = b, OC = c (1) OE a, b, c (2) AG : F G [17] ABCD a, b, c, d 3 : 1 [18] ABCD 3 A(1, 2, 3), B(3, 2, 1), C(6, 4, 4) (1) D (2) E(1, y, 15) ABC y [19] P ABC A P BC H P A = a, P B = b, P C = c (1) a b, b c, c a (2) P H b, c (3) P ABC 6
7 7 [1] (1) (1, 2, 3) v = (5, 2, 7) (2) (1, 1, 1), ( 2, 1, 3) [2] (1) (1, 2, 3) v = (5, 2, 7) (2) 3 (2, 1, 1), (3, 1, 1), (4, 1, 1) [20] (1) (1, 2, 3) v = (2, 3, 1) (2) ( 1, 5, 2), (3, 4, 1) [21] (1) (4, 2, 1) v = (1, 1, 3) (2) 3 ( 2, 3, 1 ), (2, 2, 3), ( 4, 1, 1) 7
8 * 8 [1] x 1 3 = y [2] = z + 2, 3x + 2y + z = 1 2 (1, 1, 2), 2x y + z 1 = 0 [22] (1) x = y 3 = y 1, 2x y + 3z + 2 = 0 3 (2) x = y 1 3 [23] (1) (2, 0, 1), x + y 2z + 3 = 0 = z + 2, 2x 3y + z 1 = 0 2 (2) ( 1, 1, 1), 3x y + 2z + 1 = 0 [24] (1, 2, 3) (1) 2x + 3y + z + 1 = 0 (2) x + 2y 2z 3 = 0 8
9 * 9 (1) a b = b a (2) a ( b + c) = ( a b) + ( a c) (3) (k a) b = k( a b) = a (k b) (4) ( a b) c = ( a c) b ( b c) a (5) ( a b) a ( a b) b (6) ( a b) 2 = ( a a)( b b) ( a b) 2 [25] a b a b (1) a = (1, 1, 1), b = ( 2, 3, 1) (2) a = ( 1, 1, 2), b = (1, 0, 1) [26] a = (1, 1, 3), b = ( 3, 1, 4) [27] a = (1, 0, 2), b = ( 2, k, 4) a b = 0 k [28] 3 a, b, c c ( a b) 9
10 [1] OABC OAB P, OBC Q OA = a, OB = b, OC = c PQ a, b, c PQ = x a + y b + z c z = AOC = 60, OA = 3, OC = 2 PQ = ( ) [2] OAB OA = 2, OB = 3, AOB = 120 AB M, AM N OA = a, OB = b (1) a b (2) OM, ON a, b (3) OM ON (4) MON = θ cos θ ( ) [3] 2 a = (1, x), b = (2, 1) (1) a + b 2 a 3 b x (2) a + b 2 a 3 b x ( ) [4] ABCD AB // DC, AB = 6, CD = 4 AB, AD M, N MN P MP : PN = 1 : 3 CP AB Q AB = a, AD = b (1) AC, AP a, b (2) CP : PQ (3) AD = 5, BAD = 60 CQ ( ) 10
11 [5] OABC OA = 3, OB = 4, OC = 5 AOB, AOC, BOC 60 BC O BC OP OB = b, OC = c OP = b + c OP = A OBC AQ OQ = b + c AQ = OABC ( ) [6] 3 a = (cos α, sin α, 0), b = (sin α, cos α, t), c = (sin α, cos α, 0) α, t a, b 0 v v c θ cos θ ( ) [7] (0, 0, 0) O A(1, 0, 0), B(2, 1, 0), C(3, 4, 1) OA = a, OB = b, OC = c 3 A, B, C α (a) P(x, y, z) OP = r a + s b + t c r, s, t x, y, z (b) P α x, y, z (c) (4, 5, 7) D α H DH AB, DH BC H ( ) [8] 3 O(0, 0, 0), A(1, 1, 0), B(1, 0, 1) α (1) α OA (2) α OAB ( ) 11
( )
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...............................
熊本県数学問題正解
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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
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
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
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
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
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
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
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
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 =
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
29
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
1 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 =
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
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 +
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,
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[] 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(
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
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
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
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 =
O 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,
) 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
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,
(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
1 (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
6 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
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( )!? 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
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 = [
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
(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
高校生の就職への数学II
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数論入門
数学のかたち 共線問題と共点問題 Masashi Sanae 1 テーマ メネラウスの定理 チェバの定理から 共線問題と共点問題について考える 共線 点が同一直線上に存在 共点 直線が 1 点で交わる 2 内容 I. メネラウスの定理 1. メネラウスの定理とその証明 2. メネラウスの定理の応用 II. 3. チェバの定理とその証明 メネラウスの定理 チェバの定理の逆 1. メネラウスの定理の逆
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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
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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
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学習の手順
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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】
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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
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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
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2016年度 京都大・文系数学
06 京都大学 ( 文系 ) 前期日程問題 解答解説のページへ xy 平面内の領域の面積を求めよ x + y, x で, 曲線 C : y= x + x -xの上側にある部分 -- 06 京都大学 ( 文系 ) 前期日程問題 解答解説のページへ ボタンを押すと あたり か はずれ のいずれかが表示される装置がある あたり の表示される確率は毎回同じであるとする この装置のボタンを 0 回押したとき,
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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/
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
(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
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2018年度 神戸大・理系数学
8 神戸大学 ( 理系 ) 前期日程問題 解答解説のページへ t を < t < を満たす実数とする OABC を 辺の長さが の正四面体とする 辺 OA を -t : tに内分する点を P, 辺 OB を t :-tに内分する点を Q, 辺 BC の中点を R とする また a = OA, b = OB, c = OC とする 以下の問いに答えよ () QP と QR をt, a, b, c を用いて表せ
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K E N Z OU 2008 8. 4x 2x 2 2 2 x + x 2. x 2 2x 2, 2 2 d 2 x 2 2.2 2 3x 2... d 2 x 2 5 + 6x 0 2 2 d 2 x 2 + P t + P 2tx Qx x x, x 2 2 2 x 2 P 2 tx P tx 2 + Qx x, x 2. d x 4 2 x 2 x x 2.3 x x x 2, A 4 2
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26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7