Matematika

Akar-akar persamaan kuadrat X2 + 4X + k = 0 adalah X1 dan X2. Jika X12 – X22= -32, maka k adalah …..-2 D. 12-6 E. 246

Jika Persamaan kuadrat ( p + 1 )X2 - 2 ( p + 3 )X + 3 p = 0 mempunyai dua akar yang sama, maka konstanta p adalah …..-3 dan 3/2 D. 2 dan -3-3/2 dan 3 E. 3 dan -91 dan 3

Kedua persamaan X2 + 2X + k = 0 dan X2 + X - 2k = 0 mempunyai akar-akar real untuk …..111111111111111111111111111111111111111111111111111111111111111111111111111111111111111-111/121 1 1d" 1 1k1 1 1d" 1 1211-111/141 1 1d" 1 1k1 1 1d" 1 1111-111/181 1 1d" 1 1k1 1 1d" 1 1111-111/181 1 1d" 1 1k1 1 1d" 1 1211-111/181 1 1d" 1 1k1 1 1d" 1 11111S1e1l1i1s1i1h1 1d1u1a1 1b1i1l1a1n1g1a1n1 1a1d1a1l1a1h1 11101.1 1P1a1d1a1 1s1a1a1t1 1h1a1s1i1l1 1k1a1l1i1 1k1u1d1a1r1a1t1 1k1e1d1u1a1 1b1i1l1a1n1g1a1n1 1m1a1k1s1i1m1u1m1,1 1j1u1m1l1a1h1 1k1e1d1u1a1 1b1i1l1a1n1g1a1n1 1t1e1r1s1e1b1u1t1 1a1d1a1l1a1h1 1& .1.11-111 1 1D1.1 1011-161 1 1E1.1 1211-12111J1i1k1a1 1X111 1d1a1n1 1X121 1a1d1a1l1a1h1 1a1k1a1r1-1a1k1a1r1 1p1e1r1s1a1m1a1a1n1 1 1X121 1 1-1 1 141X1 1 1+1 1 131 1 1=1 101 1m1a1k1a1 1p1e1r1s1a1m1a1a1n1 1k1u1a1d1r1a1t1 1y1a1n1g1 1a1k1a1r1-1a1k1a1r1n1y1a1 1 1X11121 1d1a1n1 1 1X12121 1 1 1a1d1a1l1a1h1 1& .1.11X121 1 1+1 1 11101 1X1 1 1+1 1 191 1 1=1 1011X121 1 1-1 1 11101 1X1 1 1+1 1 191 1 1=1 1011X121 1 1+1 1 141 1X1 1 1+1 1 131 1 1=1 1011X121 1 1 1-1 1 141 1X1 1 1+1 1 131 1 1=1 1011X121 1 1 1-1 1 141 1X1 1 1 1-1 1 191 1 1=1 10111J1i1k1a1 1 1P1 1`" 101 1d1a1n1 1a1k1a1r1-1a1k1a1r1 1p1e1r1s1a1m1a1a1n1 1X121 1 1+1 1P1X1 1 1+1 1 1Q1 1 1=1 101 1 1a1d1a1l1a1h1 1 1P1 1 1d1a1n1 1 1Q1 1m1a1k1a1 1P121 1 1+1 1 1Q121 1a1d1a1l1a1h1 1& .1.1121 1 1D1.1 151131 1 1E1.1 161141111± 1 1d1a1n1 1 1² 1 1a1d1a1l1a1h1 1a1k1a1r1-1a1k1a1r1 1p1e1r1s1a1m1a1a1n1 1k1u1a1d1r1a1t1 1X121 1 1+1 1 1 141X1 1 1+1 1 1a1 1 1-1 1 141 1 1=1 101.1 1 1J1i1k1a1 1 1± 1

2=2 2 232 2² 2 2m2a2k2a2 2n2i2l2a2i2 2a2 2y2a2n2g2 2m2e2m2e2n2u2h2i2a2d2a2l2a2h2 2& .2.2212 2 2D2.2 272232 2 2E2.2 26224222M2 2d2a2n2 2N2 2m2e2r2u2p2a2k2a2n2 2a2k2a2r2-2a2k2a2r2 2p2e2r2s2a2m2a2a2n2 232X222 2 2-2 2 242X2 2 2-2 2 222 2 2=2 202,2 2m2a2k2a2 2M222 2 2+2 2 2N222 2a2d2a2l2a2h2 2& .2.221262/292 2 2D2.2 26242/29222282/292 2 2E2.2 23222/292242/292222N2i2l2a2i2 2P2 2u2n2t2u2k2 2g2r2a2f2i2k2 2f2u2n2g2s2i2 2 2y2 2=2 2 2-2X222 2 2 2-2 2 2p2X2 2+2 2 212 2-2 2p2 2 2p2a2d2a2 2g2a2m2b2a2r2 2d2i2 2s2a2m2p2i2n2g2 2a2d2a2l2a2h2 2& .2.22P2 2`" 2222P2 2>2 212202 2<2 2p2 2<2 212202 2<2 2p2 2<2 222212 2<2 2p2 2<22222J2i2k2a2 2f2u2n2g2s2i2 2k2u2a2d2r2at y = f (X) mencapai minimum di titik (1, -4) dan f(4) = 5 maka f (X) = …..X2 + 2X + 3 D. -X2+2X+3 X2 - 2X + 3 E. -X2 +2X - 3 X2 - 2X - 3

Jika fungsi f (X) = X(12 – 2X)2 mempunyai nilai maksimum p dan nilai minimum q, maka p + q = ….0 D. 164 E. 1288

Grafik 2X + Y = A akan memotong grafik 4X2 - Y = 0 di dua titik bila …..A > -1/2 D. A < -1/4A > -1/4 E. A < -1A > 1

Jika grafik y = X2 + aX + b mempunyai titik puncak (1,2), maka nilai a dan b adalah …..a = 1, b = 3a = -1, b = -3a = -2, b = 3a = 0,5, b = 1,5a = 0,5, b = -1,5

Parabola dibawah ini dengan puncak (3, -1) dan melalui (2,0) memotong sumbu Y di titik …..(0,5) D. (0,8)(0,6) E. (0,9)(0,7)

Jumlah absis titik potong antara grafik fungsi f(X) = X – 1 dan grafik fungsi f(X) = X2 -- 4X + 3 , adalah …..1 D. 42 E. 53

Jika grafik fungsi Y = m X2 - 2mX + m di bawah garis Y = 2X – 3, maka …..m < 0-1 < m < 00 < m < 1m > 1m tidak ada

Nilai x yang memenuhi 8x+1 = 24x-1 adalah …..1 + 6 ²log 31 + 4 ²log 31 + 6 ²log 21 + 4 ²log 21 + 6 ²log 2

Jika P dan Q adalah akar-akar persamaan 2.92x-1 - 5.32x + 18 = 0 maka P + Q adalah …..0 D. 2 - 3log 22 E. 2 + 3log 23log 2

Jumlah akar-akar persamaan 5x+1 + 51-x = 11 adalah …..6 D. -25 E. -40

Nilai maksimum mutlak dan minimum mutlak dari y= x2/33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333 3p3a3d3a3 3s3e3l3a3n3g3 3-323 3 3d" 3x3 3d" 333 3a3d3a3l3a3h3 3& .3.339313/333 3 3 3d3a3n3 30339313/323 3 3 3d3a3n3 30339323/333 3 3 3d3a3n3 30339323/333 3 3 3d3a3n3 303393 3d3a3n3 3033323.343x3 3 3+3 3 32333-323x3 3 3=3 31373 3 3n3i3l3a3i3 3d3a3r3i3 32323x3 3 3=3 3 3& .3.33½3 3 3 3a3t3a3u3 3 3833½3 3 3 3a3t3a3u3 3 343313 3 3 3 3a3t3a3u3 3 3433-3½3 3 3a3t3a3u3 3 31363

4-4½4 4 4a4t4a4u4 4 4342444J4i4k4a4 4 4a434/424 4 4=4 4 4b4-434/424 4c434/444 4,4 4m4a4k4a4 4c4 4d4i4n4y4a4t4a4k4a4n4 4d4a4l4a4m a dan b adalah ….. 4/3 a½ b3/24/3 a½ b-3/2a½ b3/2a2/3 b-2a2 b2

Nilai x yang memenuhi persamaan 104 log x - 5(10)2 log x = -4 adalah ….1 D. 1 atau 44 E. 2 atau 41 atau 2

Jika 4log 4log x - 4log 4log 4log 16 = 2, maka …..2Log x = 8 D. 4Log x = 162Log x = 4 E. 16Log x = 84Log x = 8

Jika u = x2 dan xlog 10 = ulog (5u – 40), maka nilai u adalah …..25 D. 2826 E. 3027

Semua nilai X yang memenuhi pertidaksamaan ½ log (1- 2X) < 3 adalah …..X > 7/16 D. X > 7/18X < 7/16 E.4 4X4 4d" 474/414644X4 4<4 474/4148444J4i4k4a4 4 494l4o4g4 484 4=4 434m4,4 4n4i4l4a4i4 444l4o4g4 434 4=4 4& .4.44(444 4m4)4 4-414 4 4 4D4.4 4m4/444434(444 4m4)4 4-414 4 4 4E4.4 4m4434(424 4m4)4 4-41444D4i4k4e4t4a4h4u4i4 4d4e4r4e4t4 4l4o4g4 424 4+4 4l4o4g4 444 4+4 4l4o4g4 484 4+4 4& & 44D4e4r4e4t4 4i4n4i4 4m4e4r4u4p4a4k4a4n4 4& .4.44D4e4r4e4t4 4h4i4t4u4n4g4 4d4e4n4g4a4n4 4b4e4d4a4 4b4=4 424 44D4e4r4e4t4 4h4i4t4u4n4g4 4d4e4n4g4a4n4 4b4e4d4a4 4b4=4 4l4o4g4 4244D4e4r4e4t4 4u4k4u4r4 4d4e4n4g4a4n4 4p4e4r4b4a4n4d4i4n4g4a4n4 4r4 4=4 4244D4e4r4e4t4 4u4k4u4r4 4d4e4n4g4a4n4 4p4e4r4b4a4n4d4i4n4g4a4n4 4r4 4=4 4l4o4g4 4244B4u4k4a4n4 4d4e4r4e4t4 4h4i4t4u4n4g4 4m4a4u4p4u4n4 4d4e4r4e4t4 4u4k4u4r444J4i4k4a4 4 4a4l4o4g4 4(414-4 4 434l4o4g4 4(42474)4-414)4 4=4 424,4 4m4a4k4a4 4n4i4l4a4i4 4a4 4y4a4n4g4 4m4e4m4e4n4u4h4i4 4a4d4a4l4a4h4 4& .4.4414/484 4D4.4 4344¼4 4 4E4.4 44442444J4i4k4a4 444l4o4g4 4(444x4 4.4 444)4 4=4 424 4 4x4,4 4m4a4k4a4 4x4 4=4 4.4.44-414 4 4D4.4 4144-414/424 4E4.4 424

5½555P5a5d5a5 5s5e5l5a5n5g5 5-515 5d" 5x5 5d" 525,5 5f5u5n5g5s5i5 5y5 5=5 5x535 5 535x525 5+5 535 5m5e5m5p5u5n5y5a5i5 5n5i5l5a5i5 5m5a5k5s5i5m5u5m5 5& .5.55-565 5 5D5.5 5655-515 5 5E5.5 58553555P5a5d5a5 5s5e5l5a5n5g5 505 5d" 5x5 5d" 545,5 5j5a5r5a5k5 5t5e5r5j5a5u5h5 5d5a5r5i5 5k5u5r5v5a5 5f5(5x5)5 5=5 5x535 5 565x525 5+5 595x5 5d5e5n5g5a5n5 5s5u5m5b5u5 5x5 5a5d5a5l5a5h5 5& .5.5515 5 5D5.5 585525 5 5E5.5 515655455 5F5u5n5g5s5i5 5f5(5x5)5 5=5 5x535 5 535x525 5-51555 5t5u5r5u5n5 5u5n5t5u5k5 5s5e5m5u5a5 5x5 5yang memenuhi …..X > 0X < -2-2 < x < 00 < x < 2X < 0 atau x > 2

Turunan dari (1- x)2 (2x + 3) adalah …..(1 – x) (3x + 2)(x – 1) (3x + 2)2 (1 + x) (3x + 2)2 (x – 1) (3x + 2)2 (1 – x) (3x + 2)

Rusuk suatu kubus bertambah panjang dengan kelajuan 7 cm per detik Kelajuan bertambahnya volume pada saat rusuk panjangnya 15 cm adalah …..675 cm3 / detik1.575 cm3 / detik3.375 cm3 / detik4.725 cm3 / detik23.625 cm3 / detik

Turunan pertama dari y = cos4 x adalah …..¼ cos3 x-¼ cos3 x-4 cos3 x-4 cos35 5x5 5s5i5n5 5x5545 5c5o5s535 5x5 5s5i5n5 5x5 555 5D5i5b5e5r5i5k5a5n5 5s5u5a5t5u5 5k5u5r5v5a5 5d5e5n5g5a5n5 5p5e5r5s5a5m5a5a5n5 5y5 5=5 5f5(5x5)5 5d5e5n5g5a5n5 5f5(5x5)5 5=5 545 5+5 535x5 5 5x535 5u5n5t5u5k5 5x5 5e" 505.5 5N5i5l5a5i5 5m5a5k5s5i5m5u5m5 5d5a5r5i5 5f5(5x5)5 5a5d5a5l5a5h5 5& .5.5545 5 5D5.5 575555 5 5E5.5 58556555 5J5i5k5a5 5n5i5l5a5i5 5s5t5a5s5i5o5n5e5r5 5d5a5r5i5 5f5(5x5)5 5=5 5x535-5 5p5x525 5 5p5x5

6-616 6a6d6a6l6a6h6 6x6 6=6 6p6 6,6 6m6a6k6a6 6p6 6=6 6& .6.6606 6a6t6a6u6 616 6 6D6.6 616606 6a6t6a6u6 616/656 6 6E6.6 616/656606 6a6t6a6u6 6-61666 6P6a6d6a6 6"A6B6C6 6d6e6n6g6a6n6 6s6i6s6i6 6a6,6 6b6,6 6d6a6n6 6c6 6b6e6r6l6a6k6u6 6a626 6 6b626 6=6 6c626 6 6b6c6.6 6B6e6s6a6r6n6y6a6 6s6u6d6u6t6 6A6 6a6d6a6l6a6h6 6& .6.661656°6 6D6.6 66606°663606°6 6E6.6 67656°664656°666 6P6a6n6j6a6n6g6 6b6a6y6a6n6g6a6n6 6s6e6b6u6a6h6 6m6e6n6a6r6a6 6a6d6a6l6a6h6 61626 6m6e6t6e6r6.6 6J6i6k6a6 6s6u6d6u6t6 6e6l6e6v6a6s6i6 6m6a6t6a6h6a6r6i6 6p6a6d6a6 6s6a6a6t6 6i6t6u6 66606°6,6 6m6a6k6a6 6t6i6n6g6i6 6m6e6n6a6r6a6 6a6d6a6l6a6h6 6& .6.6646"366 6D6.6 61626"36666"366 6E6.6 61666"36686"3666J6i6k6a6 6f6(6x6)6 6=6 626 6 6s6i6n626 6x6,6 6m6a6k6a6 6f6u6n6g6s6i6 6f6(6x6)6 6m6e6m6e6n6u6h6i6 6& .6.66-626 6d" 6f6(6x6)6 6d" 6-6166-626 6d" 6f6(6x6)6 6d" 6166-616 6d" 6f6(6x6)6 6d" 6066 606 6d" 6f6(6x6)6 6d" 6166 616 6d" 6f6(6x6)6 6d" 62666 6J6i6k6a6 6f6(6x6)6 6=6 6s6i6n6 6a6x6 6 6c6o6s626 6b6x6,6 6 606 6d" 6x6 6d" 6×,6 6 6 6a6,6 6b6 6`" 606,6 6 6f6 (606)6 6=6 616,6 6d6a6n6 6f6(616/626)6 6=6 606,6 6m6a6k6a6 6a6+6b6 6=6 6& .6.6606 6 6D6.6 6 6 636616 6 6E6.6 6 6 64662666 6J6i6k6a6 626 6c6o6s626 6x6 6+6 6c6o6s6 6x6 6s6i6n6 6x6 6 6s6i6n6 626 6x6 6=6 606,6 6m6a6k6a6 6t6a6n6 6x6 6=6 6& .6.6616 6d6a6n6 6-626616 6d6a6n6 6266-616 6d6a6n6 6-6266-616 6d6a6n6 6266-626 6d6a6n6 62666P6a6d6a6 6" 6A6B6C6 6d6i6k6e6t6a6h6u6i6 6c6o6s6 6(6B6+6C6)6 6=6 696/64606.6 6J6i6k6a6 6p6a6n6j6a6n6g6 6s6i6s6i6 6A6C6 6=6 61606 6c6m6,6 6A6B6 6=6 686 6c6m6.6 6m6a6k6a6 6p6a6n6j6a6n6g6 6s6i6s6i6 6B6C6 6a6d6a6l6a6h6 6& .6.6686"26 6c6m66 6D6.6 61616"26 6c6m6696"26 6c6m66 6E6.6 61626"26 6c6m661606"26 6c6m666 6S6u6k6u6 6k6e6e6m6p6a6t6 6s6u6a6t6u6 6d6e6r6e6t6 6a6r6i6t6m6e6t6i6k6a6 6a6d6a6l6a6h6 696 6d6a6n6 6j6u6m6l6a6h6 6s6u6k6u6 6k6e6e6n6a6m6 6d6a6n6 6k6e6d6e6l6a6p6a6n6 63606.6 6J6u6m6l6a6h6 62606 6s6u6k6u6 6p6e6r6t6a6m6a6 6d6e6r6e6t6 6t6e6r6s6e6b6u6t6

7a7d7a7l7a7h7 7& .7.77270707 7D7.7 76747077474707 7E7.7 787070776707077 7 7S7u7k7u7 7p7e7r7t7a7m7a7 7d7a7n7 7s7u7k7u7 7k7e7d7u7a7 7s7u7a7t7u7 7d7e7r7et geometri berturut-turut adalah p2 dan px. Jika suku ke-5 deret tersebut adalah p18. maka x = …..1 D. 62 E. 84

Suku tengah suatu deret aritmetika adalah 23. Jika suku terakhirnya 43 dan suku ketiganya 13, maka banyak suku deret adalah …..5 D. 117 E. 139

Agar deret geometri tak hingga dengan suku pertama a mempunyai jumlah 2, maka a memenuhi …..-2 < a < 0-4 < a < 0 0 < a < 2 0 < a < 4-4< a < 4

Seutas pita dibagi menjadi 10 bagian dengan panjang yang membentuk deret aritmetika. Jika pita yang terpendek 20 cm dan yang terpanjang 155 cm, maka panjang pita yang semula adalah …..800 cm D. 875 cm825 cm E. 900 cm850 cm

Jika jumlah semua suku deret geometri tak hingga adalah 96 dan jumlah semua sukunya yang berindeks ganjil adalah 64, maka suku ke- 4 deret tersebut adalah …..4 D. 106 E. 128Nilai x yang memenuhi 5x-1 = EMBED Microsoft Equation 3.0 µ § adalah…- EMBED Microsoft Equation 3.0 µ § D. 3 EMBED Microsoft Equation 3.0 µ §- EMBED Microsoft Equation 3.0 µ § E. 4 EMBED Microsoft Equation 3.0 µ § EMBED Microsoft Equation 3.0 µ §

Jika x dan y memenuhi system persamaan EMBED Microsoft Equation 3.0 µ § Maka nilai x + y adalah… A. 0D. 4B. 2E. 5C. 3Jika x1 dan x2 memenuhi (4-log x).log x = log 1000, maka x1x2 = …A. 3D. 1000B. 4E. 10000C. 40Himpunan pertidaksamaan 3log x + 3log(2x-3) < 3 adalah…A. EMBED Microsoft Equation 3.0 µ §D. EMBED Microsoft Equation 3.0 µ §B. EMBED Microsoft Equation 3.0 µ §E. EMBED Microsoft Equation 3.0 µ §C. EMBED Microsoft Equation 3.0 µ §

x1 dan x2 merupakan akar-akar persamaan 3x2-4x-2=0, maka x12+x22 = …A. EMBED Microsoft Equation 3.0 µ §D. EMBED Microsoft Equation 3.0 µ §B. EMBED Microsoft Equation 3.0 µ § E. EMBED Microsoft Equation 3.0 µ § C. EMBED Microsoft Equation 3.0 µ §Nilai a agar grafik fungsi Y = (a-1)x2 – 2ax + (a-3) selalu berada di bawah sumbu X (definit negative) adalah…. a=1 D. a>3/4a>1 E. a < 3/4a<1

Himpunan penyelesaian dari system p8e8r8s8a8m8a8a8n8:88 8 8 8 8 8 88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 838.808 8µ8 8§8 8 8 8 8 8a8d8a8l8a8h8& 8A8.8 8{8(8-838,828)8,8(8-828,838)8}8 88D8.8 8{8(8-848,818)8,8(828,838)8}888B8.8 8{8(818,8-848)8,8(848,8-818)8}88E8.8 8{8(848,818)8,8(818,848)8}888C8.8 8{8(8-848,818)8,8(8-818,848)8}8888888H8i8m8p8u8n8a8n8 8p8e8n8y8e8l8e8s8a8i8a8n8 8p8e8r8t8i8d8a8k8s8a8m8a8a8n8:88x828 8+8 848x8 8-8 81828d"08 8 8a8d8a8l8a8h8& 8A8.8 8{8x8 8|8 8-828 8d" 8x8 8d" 868;8 8x8 8" 8R8}88888B8.8 8{8x8 8|8 8-868 8d" 8x8 8d" 828;8 8x8 8" 8R8}88888C8.8 8{8x8 8|8 8-828 8d" 8x8 8d" 8-868;8 8x8 8" 8R8}88888D8.8 8{8x8 8|8 8x8 8e" 8-828 8a8t8a8u8 8x8 8e" 8-868;8x8"R8}88888E8.8 8{8x8 8|8 8x8 8e" 868 8a8t8a8u8 8x8 8e" 8-828;8 8x8 8" 8R8}8888P8e8r8t8i8d8a8k8s8a8m8a8a8n8 88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 838.808 8µ8 8§8e" 808,88m8e8m8p8u8n8y8a8i8 8p8e8n8y8e8l8e8s8a8i8a8n8 8& 8A8.8 8x8 8e" 838888B8.8 8x8 8e" 818888C8.8 8-818 8d" 8x8 8<8 818 8a8t8a8u8 8x8 8>8 838888D8.8 8-818 8d" 8x8 8<8 818 8a8t8a8u8 8x8 8e" 838888E8.8 8-818 8<8 8x8 8<8 818 8a8t8a8u8 8x8 8>8 8388J8i8k8a8 8x8 8d8a8n8 8y8 8s8u8d8u8t8 8l8a8n8c8i8p8,8 8c8o8s8(8x8-8y8)8=8 88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 838.808 8µ8 8§8 8d8a8n8 8c8o8s8x8.8c8o8s8y8=88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 838.808 8µ8 8§8,8 8m8a8k8a8 88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 838.808 8µ8 8§8 8a8d8a8l8a8h8& 8A8.8 88 8E8M8B8E8D8 8M8i8c8r8o8s8o8f8t8 8E8q8u8a8t8i8o8n8 83.0 µ §D. EMBED Microsoft Equation 3.0 µ §B. EMBED Microsoft Equation 3.0 µ §E. EMBED Microsoft Equation 3.0 µ §C. EMBED Microsoft Equation 3.0 µ §Jika 0 < x < EMBED Microsoft Equation 3.0 µ §dan x memenuhi persamaan tg2 x – tg x – 6 = 0, maka himpunan nilai sin x adalah…A . EMBED Microsoft Equation 3.0 µ § D. EMBED Microsoft Equation 3.0 µ §B. EMBED Microsoft Equation 3.0 µ §E. EMBED Microsoft Equation 3.0 µ §C. EMBED Microsoft Equation 3.0 µ §

Sebuah segitiga ABC dengan panjang AB = 6 cm, BC = 5 cm, dan AC = 4 cm. Nilai kosinus sudut B adalah…½ D. 8/9¾ E. 11/124/5

Nilai rata-rata ulangan matematika dari 30 siswa adalah 7. Kemudian 5 orang siswa mengikuti ulangan susulan sehingga niai rata-rata keseluruhan menjadi 6,8. Nilai rata-rata yang mengikuti ulangan susulan adalah…

4,2 D. 5,64,5 E. 6,85,3

Persamaan log{2log(2x+1) +2} = log(2log 32), mempunyai penyelesaian…3 ½ D. 84 ½ E. 95 EMBED Microsoft Equation 3.0 µ §> 0 mempunyai himpunan penyelesaian masalah adalah…A. EMBED Microsoft Equation 3.0 µ §B. EMBED Microsoft Equation 3.0 µ §C. EMBED Microsoft Equation 3.0 µ §D. EMBED Microsoft Equation 3.0 µ §E. EMBED Microsoft Equation 3.0 µ §

Himpunan penyelesaian (2X-1)log(2x2-x) = 2 adalah…{ } D. {2}{1,1/2} E. {2,1}{1} EMBED Microsoft Equation 3.0 µ §-1/3 D. 1-1 E. 0-5

Nilai EMBED Microsoft Equation 3.0 µ §1/8 C. 1 E. 3/89/8 D. 0

EMBED Microsoft Equation 3.0 µ §p-q D. pqp+q E. ½ (p+q)0

EMBED Microsoft Equation 3.0 µ §0 D. ~6 E. 51/6

Sebuah peternakan kecil menyediakan rumput 24 karung per hari untuk 18 kambingnya. Jika minggu depan si peternak berencana membeli kambing 9 ekor lagi, berapa banyak karung setiap harinya yang harus disediakan si peternak?16 D. 3612 E. 3230

Sebuah segitiga mempunyai panjang alas 3y dan tinggi 2y. luas segitiga tersebut adalah 27 cm2. maka y adalah ….A. 2D.3.5B. 2.5E. 43Berapakah hasil 5 + 2 x 6 - 3 = …A.39 D. 13B. 21E. 11C. 14Perbandingan jari-jari dua buah bola adalah 2:3. Berapakah perbandingan volumnya?A. 2 : 3D. 4 : 6B. 4 : 9E. 3 : 28 : 27Setelah diskon 15% harga kaos menjadi Rp 76.500,00. Berapakah harga asli kaos tersebut?

Rp80.000,00 Rp 90.000,00 Rp 95.000,00Rp 100.000,00Rp. 120.000,00

4x-5 = 3, nilai dari 3x adalah…A. 2D. 6B. 4E. 128Sebuah keluarga memiliki lima anak. Anak terxuda berusia x-2 tahun dan anak yang tertua berusia 2x tahun. Sedang tiga orang anak lainnya berusia x, x+1, dan 2x-2. Jumlah umur kelima anak tersebut adalah 39. Berapakah umur anak keempat?A. 7D. 14B. 10E. 1612Jika rata-rata nilai kimia 5 siswa adalah 79, berapa nilai yang harus diperoleh agar rata-rata menjadi 80?A. 80D. 84B. 82E. 85 C. 90Di dalam duatu kebun terdapat mawar dan flamboyan dengan perbandingan 7 : 9. Jumlah flamboyant lebih banyak 10 buah dari mawar. Berapa jumlah flamboyant?A. 35D. 50B. 40E. 55C. 45Berapakah nilai 11 - EMBED Microsoft Equation 3.0 µ §?A. 7 EMBED Microsoft Equation 3.0 µ §D. 7 EMBED Microsoft Equation 3.0 µ §B. 6 EMBED Microsoft Equation 3.0 µ §E. 6 EMBED Microsoft Equation 3.0 µ §5Apabila suatu pekerjaan dapat diselesaikan dalam waktu 4 hari oleh 8 orang, maka berapakah tambahan orang yang dibutuhkan agar pekerjaan tersebut dapat selesai dalam waktu ½ hari ?A. 56D. 35B. 43E. 27C. 28Luas sebuah persegi panjang adalah 384 cm2 dengan ukuran panjang 24 cm. Berapakah keliling persegi panjang tersebut ?A. 80D. 16B. 40E. 56C. 32Bilangan berikut ini seluruhnya habis dibagi oleh 3, kecuali …A. 177D. 128B. 156E. 129C. 147Suatu seri angka terdiri dari : 2,3,4,4,8,6,10,7,14,9,16,10,20,12. Dua angka berikutnya adalah ?A. 13,22D. 24,12B. 22,13E. 12,20C. 20,12Suatu rangkaian angka adalah seperti berikut : 11,19,27,9,17,25,7. Tentukan dua angka berikutnya ?A.13,21D. 23,15B. 21,31E. 15,23C. 23,31Tentukan empat huruf selanjutnya dari rangkaian abjad berikut : a m n b o p c ?A. d e f gD. q r d eB. d q r eE. s r t vC. q r d sUntuk memeriksa sebuah mesin sandi, Yuri memerlukan waktu 20 menit, sementara Bambang hanya memerlukan 18 menit. Jika keduanya mulai memeriksa pada pukul 08.00, kapankah pertama kali mereka akan selesai memeriksa mesin sandi pada saat yang sama ?pukul 09.30pukul 09.42pukul 10.00pukul 11.00pukul 10.30

Jika xy = 5, x2 + y2 =7, maka berapakah nilai ( x + y )2 ?A. 21D. 14B. 17E. 24C. 15Berapakah X dari deretan bilangan berikut : 25,15,95,85,71,X ?A. 61D. 70B. 75E. 72C. 81Bilangan 2009 kalau ditulis dalam lambang bilangan Romawi ?MCMII D. MMXIMCMI E. MMIXMMII

Sejenis produk dijual dengan dua kali potongan ( diskon ) berturut-turut yaitu 20 % dan setelah itu 15 %. Berapakah jumlah seluruh diskon ?A. 30 %D. 35 %B. 32 %E. 40 %C. 34 %Jika populasi suatu negara bertambah 1 orang setiap 15 detik, berapa banyak pertambahan populasinya selama 20 menit ?A. 80D. 60B. 100E. 240C. 150Suatu seri : 0,2,6,12,20,... seri selanjutnya adalah ?A. 30D. 36B. 32E. 38C. 34Suatu seri : 4,11,14,21,24,... seri selanjutnya adalah ?A. 31D. 40B. 34E. 42C. 37Jika a=1/2b maka b+2=A. a/2D. 2a+1B. aE. 2aC. 2a+2Sebuah perusahaan bus mengurangi jumlah perjalanan (rit) untuk jalur tertentu sebesar 20% menjadi 8xsehari. Berpakah jumlah rit setiap hari sebelum ada pengurangan?A. 2D. 12B. 4E. 8C. 10Jika r=3s, s=5t, t=2u , berapa nilai dari rst/u adalah … A. 60D. 600B. 150E. 240C. 300Toro termasuk 3 siswa termuda di kelasnya ada 26 siswa yang lebih tua daripada Toro dan satu yang seusia dengannya. Berapakah banyak siswa yang ada di kelas Toro?A. 28D. 31B. 33E. 32C. 30Jika rata-rata nilai fisika 4 siswa adalah 85. berapa niloai yang harus diperoleh seorang siswa lagi supaya rata-ratanya menjadi 86?A. 90D. 86B. 88E. 92C. 87Biaya untuk membuat x kotak crayon adalah 570000+500x. crayon dijual dengan hargaRp.10000 per kotak. Berapa minimum jumlah crayon yang diproduksi supaya penjualannya sama dengan biaya produksi?A. 50D. 61B. 57E. 70C. 60Bahasa Inggris

TEXT 1PREPARED FOR THE WORST

Just outside Yokohama stadium, several hundred policeman stare down the crowd. “Quiet down,” booms a voice, in English, Spanish and Javanese, over the loudspeaker. “Disperse immediately”. Some 50 “hooligans”, faces painted and armed with burning wooden stick, throw chairs back at the front of line. Firecrackers explode all around. An electric billboard flickers the same warning. The doters grow only more threatening. Suddenly, water canon appear, and within minutes the last persistent “thug” is knocked to the ground. Another Japanese dress rehearsal for World Cup 2002 has ended in success and without casualties. As Asia open’s it’s arms to the world’s best team and nearly 1 million of their supporters it is keeping another set of arms braced to combat hooliganism. Fearing a repeat of Marseilles 1998, where nearly 50 were arrested and 30 injured in clashes between England and Tunisia supporters, Japan, and South Korea have taken every precaution. But while South Korea has decades of experience in quelling aggressive demonstrations including rampant pro-democracy rallies in the 98’s Japan has virtually had to learn from scratch. To add to it’s troubles, Japan was the unlucky winner chosen to host England, the teams most infamous for it’s “hooligans”.

So far, the crash course has gone well. Since 1998, Japanese officials have toured Europe and South America, consulting top football authorities to learn how to fight the game’s main malady and to train in the art of “spotting” locating potential troublemakers in a crowd of football supporters. And the best in the world attending a hand; about 100 of the world’s top spotters from 4 different countries are expected to attend World Cup 2002 to assist local authorities. (Beith & Takayama, Newsweek, June 3rd, 2002)

The crash mentioned in paragraph 1...

is actual event in Yokohamais an outrageous moment happened in Japanis a rehersal prepared for World Cup 2002was unsuccessful precaution for World Cup 2002were unsuccessful precautionJapan …. To host England A. has deliberately chosen B. was unfortunately given a chanceC. is practically experienced D. is the winner of all E. have been the winner of all

Japan and South Korea have to take every precaution because ….They want to educate the publicThey don’t want to repeat tragedy four years ago.They want to warn all hooligansThey are not confident enough to be the host of World Cup 2002They were not confident enough

How many casualties in clashes between England and Tunisia supporters in 1998?53080 none of the above 45

Spotting means ….to become a good host for World Cup 2002to arrest football troublemakers that could potentially become hooligansto combat hooliganism during a football matchto locate potential troublemakers in a crowd of football supportersto play football troublemakersWhich term in the text refers to hooligan’s very bad reputation?thug D. maladyrampant E. famousinfamous

If everything in the text is true, then the police officer should be careful to confiscate things upon entry to a football match except ….Knives D. long umbrellasBottles E. good throuserT-shirts

This passage is used for answering the following numbersTEXT 2E-mail (electronic mail) uses computer for communication. It has several important advantages over phones and regular mail. The main advantage of e-mail is that it takes every little time to send and receive messages. From your computer, you can contact someone far away or someone in the next office. Seconds later, they have your message. If they are at their computer, you can get a response instantly, too. Another reason people like to use e-mail is that for just a few rupiahs you san send a message to someone in another part in the world. You don’t have to worry about the time difference or slow mail delivery. You message is sent immediately, and your friends or colleagues can send a

response at their convenience. Lastly, e-mails allows you to send a single message to many people at the same time.

What is the main idea of the paragraph ….e-mail, the most efficient communication e-mail has advantages over the telephone and regular mail service e-mail saves our time e-mail makes people able to send a single message to many people at the same timeE-mail, the most secure chanel for communicating

Which of following statements is not true according to the text ….the process of sending and receiving an e-mail only requires a short timean e-mail can be sent to lots of people at oncethe e-mail can be sent only in secondse-mail can only be used to contact someone far awaypeople can send their mail through e-mail at once

You might have a respond for your sent e-mail immediately in condition that ….the person who receives your e-mail is in officethe person who receives your e-mail has a computerthe person who receives your e-mail knows you very well

the person who receives your e-mail is in computerthe person who receives your e-mail is in school

Which among the following words does not have similar meaning to instantly ….Immediatelywidely without delay quicklyfast

The expression at their convenience means ….when they have timeat their computerfrom their homesin their e-mailswhen they have no time

This passage is used for answering the following numbersTEXT 3A little more than a hundred years ago, a scientist in Medford, Massachusetts was trying to help local industry. Instead of helping local industry, however, he caused a major problem with the local environment.The scientist thought that it would be a good idea to try to develop the silk-making industry in Medford. He knew that the silk industry in asia was successful because of the silkworm, a caterpillar that ate only mulberry leaves. Mulberry trees did not grow in Medford, so the scientist decided to work on developing the type of silk-making worm that would eat the type of tree leaves in Medford.

The situation in this passage took place approximately ….a decade ago

two decades agoa century ago two centuries agoone year ago

The word “major” in paragraph 1 could be best replaced by ….Militaryhuge solvableminusculeminor

TEXT 4Poor spelling often results from bad habits developed in early school years. With work, such habits can (15) …. If you can write your name (16) …. misspelling it, there is (17) …. reason why you cannot do the (18) …. with almost any word in the English language. You can improve your spelling by (19) …. the dictionary, keeping a personal spelling list, mastering commonly confused words, understanding basic spelling rules, and studying a basic word list.

correctcorrected be correctedcorrectscorrecting

With D. in which without E. who Within

Many D. any no E. less Not

Difference D. sameness different E. equal same

Use D. used to use E. has been used using

If there is no problem with the weather, the satellite Cakrawala 1 ……… soon in France.will launchwill be launchedwould launchwould have launched should be launch

The two countries ……. to war against one another if they had a better mutual understanding.

will not godoes not gowould not gowould not goinghave would not be going

Most of books …. for accounting today are supplemented with CD-ROMs.use of D. in useusing E. to be usedare used

a cloud is a dense mass of …. water vapor or ice particles.or D. neitherboth E. noreither

Mrs. Rani …. If she had been delayed.would call D. will be callingwould have called E. will have calledwill call

The light is out in her room, she ….must go to bedmust have gone to bedmust gone to bed D. must be gone to bed E. must been gone to bed

I am now …. tea without sugar.accustomed to drinkaccustomed to drinkinghave the custom to drink accustoming in drinking have to be used in drinkimg

Tika : “Have you checked where most of the tryout participants come from?”Siska : “ yes, two-thirds of them …. from various parts of Java.”are coming D. to comecomes E. come C. to come

Toro : “Let’s have lunch.”Putra : “I’ll join you later. I had better not to stop …. on this report now.Work D. workingto work E. I’m workingI’ll work

I …. when Wulan …. three days ago.have been studying English ; came to my housestudied English ; was coming to my househad studied English ; came to my housewas studying English ; came to my house

had been studying English ; came to my house

Yuri : I can’t find my wallet in my bag ! July : well, you …. in the car. Let’s take a look !may have to left ithad to left itshould have left itmay have leftmust leave it

Picture of the moon show vast …. of crater and rockExpandexpansesexpound extantexplain

Tina wouldn’t give Seno his toy back, so he took it from her by ….Hard D. slowpower E. forcestrong

Although the boys have enough money, neither of them …. buying a book.Were D. iswas E. wouldare

TEXT 5Internet is a powerful network which consist of millions of computers connected together. Information on the internet comes from around the world and is gathered from many subjects. For example, there is complete information on literature, banking, medicine, law, engineering, etc. Once one has access to the system, he/she may copy information from internet data bases for free.The underlined word has similar meaning with ….Compoundobtain contents substancecomprise

The football game will start soon and the stadium is ….. with people.Teeming D. timingteaming E. tanningtaming

“We may have to go to Bandung this holiday.” Means ….perhaps we will go to Bandung this holidayI’m sure we will go to Bandung this holidayexactly we will go to Bandung this holidayI know we will be allowed to go to Bandung this holidayI’m certain we will go to Bandung this holiday

“Ary has his car repaired” means ….Ary wants his car repairedAry can repair his car by himselfAry let somebody to repair his carAry repairs his carAry wants to repair his car

. “He said you kicked Yoyok” “That’s not true ……”Yoyok is the one that I kickedhe isn’t the one whom Yoyok kickedI’m the one whom Yoyok kickedI’m not the one whom Yoyok kickedI’m the one who kicked Yoyok

Many people consider Egypt’s pyramids are remarkable and puzzling ancient monuments. The underlined word can replaced by ….Capable D. beautifulunusual E. goodordinary

After the ….. they will go to the beach.PreparementsPreparation Preparouspreparativeprepare

Fisika listrik & magnet

Sebuah transformator menurunkan tegangan listrik bolak balik dari 220 V menjadi 10 V. Efisiensi transformator 60%. Jika kuat arus yang mengalir pada kumparan sekunder 6,6 A maka kuat arus pada kumparan primer adalah …. A. 1 A D. 0,4 AB. 0,8 AE. 0,3 A0,5 A Sepotong kawat penghantar yang panjangya l digerakkan tegak lurus suatu medan magne1717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717171717