PENERBIT ILMU BAKTI SDN. BHD.
N1 Bab 1 Indeks 1.1 Tatatanda Indeks 1 Tatatanda indeks merujuk kepada suatu nombor yang didarab dengan dirinya sendiri sebanyak beberapa kali. Index notation refers to a number which is multiplied by its own number for several times. 2 Suatu nombor boleh ditulis dalam tatatanda indeks sebagai an , dengan keadaan a ialah asas dan n ialah indeks. A number can be written in index notation as an , where a is the base and n is the index. an Indeks/Index Asas/Base Contoh/Example: (a) 34 = 3 × 3 × 3 × 3 (b) 1 1 22 5 = 1 2 × 1 2 × 1 2 × 1 2 × 1 2 (c) 13 1 52 6 = 3 1 5 × 3 1 5 × 3 1 5 × 3 1 5 × 3 1 5 × 3 1 5 (d) (–7)4 = (–7) × (–7) × (–7) × (–7) 3 Suatu nombor boleh ditulis dalam bentuk indeks menggunakan A number can be written in index form using (a) kaedah pembahagian berulang repeated division method (b) kaedah pendaraban berulang repeated multiplication method Contoh/Example: Tulis 32 dalam bentuk indeks menggunakan asas 2. Write 32 in index form using base 2. (a) Kaedah pembahagian berulang Repeated division method 2 32 2 16 2 8 2 4 2 2 1 n = 5 Maka/Hence, 32 = 25 . (b) Kaedah pendaraban berulang Repeated multiplication method 2 × 2 × 2 × 2 × 2 32 16 8 4 Maka/Hence, 32 = 25 . 1.2 Hukum Indeks 1 Berikut merupakan hukum indeks: Below are the law of indices: (a) am × an = am + n Contoh/Example: 53 × 54 = 53 + 4 = 57 (b) am ÷ an = am – n Contoh/Example: 76 ÷ 72 = 76 – 2 = 74 (c) (am)n = amn Contoh/Example: (k2 )6 = k2(6) = k12 (d) (am × an )p = amp × anp Contoh/Example: (x3 y2 z3 )2 = x3(2)y2(2)z3(2) = x6 y4 z6 (e) a0 = 1, dengan keadaan/where a ≠ 0 Contoh/Example: 80 = 1 (f) a–n = 1 an Contoh/Example: 4–3 = 1 43 NOTA EKSPRES Masteri Maths TG 3.indb 1 11/2/2023 11:50:12 AM PENERBIT ILMU BAKTI SDN. BHD.
Nota Ekspres N1 – N16 Aktiviti Pembelajaran Abad ke-21 P1 – P2 Rekod Prestasi Murid R1 – R4 Bab 1 Indeks 1 – 8 Bab 2 Bentuk Piawai 9 – 17 Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang 18 – 34 Bab 4 Lukisan Berskala 35 – 42 Pentaksiran Pertengahan Tahun 43 – 55 Bab 5 Nisbah Trigonometri 56 – 63 Bab 6 Sudut dan Tangen bagi Bulatan 64 – 74 Bab 7 Pelan dan Dongakan 75 – 84 Bab 8 Lokus dalam Dua Dimensi 85 – 93 Bab 9 Garis Lurus 94 – 101 Ujian Akhir Sesi Akademik 102 – 114 Jawapan 115 – 120 Kandungan Masteri Maths TG 3.indb 1 11/2/2023 11:50:10 AM PENERBIT ILMU BAKTI SDN. BHD.
1 Buku Teks: Halaman 2 – 3 Tatatanda Indeks 1.1 Padankan pendaraban berulang berikut dengan bentuk indeks an yang betul. SP 1.1.1 TP 1 Match the following repeated multiplications with the correct index form an . 5 × 5 × 5 × 5 4 × 4 × 4 × 4 × 4 (–4) × (–4) × (–4) × (–4) (–4)4 54 45 1 2 3 Lengkapkan jadual. SP 1.1.1 TP 1 Complete the table. Pendaraban berulang Repeated multiplication Bentuk indeks, an Index form, an 4 0.4 × 0.4 × 0.4 × 0.4 × 0.4 0.45 5 m × m × m × m × m × m × m × m × m m9 6 7 × 7 × 7 × 7 × 7 × 7 76 7 (–6) × (–6) × (–6) × (–6) (–6)4 8 1 s × 1 s × 1 s × 1 s × 1 s × 1 s × 1 s × 1 s 1 1 s 2 8 Tukarkan nombor atau sebutan algebra dalam bentuk indeks berikut kepada pendaraban berulang. SP 1.1.1 TP 1 Convert the following numbers or algrebraic terms in index form into repeated multiplications. 9 32 = 3 × 3 10 (–5.2)3 = (–5.2) × (–5.2) × (–5.2) 11 (–x)6 = (–x) × (–x) × (–x) × (–x) × (–x) × (–x) 12 b7 = b × b × b × b × b × b × b 13 1 1 d 2 4 1 1 d 2 × 1 1 d 2 × 1 1 d 2 × 1 1 d 2 14 (1.9z)5 = (1.9z) × (1.9z) × (1.9z) × (1.9z) × (1.9z) TP 1 Bab 1 Indeks Indices Bidang Pembelajaran: Nombor dan Operasi Buku Teks: Halaman 1 – 29 Masteri Maths TG 3.indb 1 11/2/2023 11:50:27 AM PENERBIT ILMU BAKTI SDN. BHD.
2 Buku Teks: Halaman 4 – 6 Tatatanda Indeks 1.1 Nyatakan setiap nombor berikut dalam bentuk indeks menggunakan asas yang diberikan. SP 1.1.2 TP 2 State each of the following numbers in index form using the given base. 1 49 (asas 7/base 7) 7 49 7 7 1 Maka/Hence, 49 = 72 2 1 331 (asas 11/base 11) 11 1 331 11 121 11 11 1 Maka/Hence, 1 331=113 3 256 625 1asas 4 5 /base 4 5 2 4 256 5 625 4 64 5 125 4 16 5 25 4 4 5 5 1 1 Maka/Hence, 256 625 = 1 4 5 2 4 4 59 049 (asas 9/base 9) 9 59 049 9 6 561 9 729 9 81 9 9 1 Maka/Hence, 59 049 = 95 5 0.0081 (asas 0.3/base 0.3) 0.3 0.0081 0.3 0.027 0.3 0.09 0.3 0.3 1 Maka/Hence, 0.0081 = 0.34 6 10 000 (asas 10/base 10) 10 10 000 10 1 000 10 100 10 10 1 Maka/Hence, 10 000 = 104 Cari nilai bagi setiap yang berikut. SP 1.1.2 TP 2 HEBAT MATEMATIK MODUL 12 Gangsa Find the value of each of the following. 7 65 = = 6 × 6 × 6 × 6 × 6 = 7 776 8 (32.9)2 = = 32.9 × 32.9 = 1 082.41 9 1 3 4 2 4 = = 3 4 × 3 4 × 3 4 × 3 4 = 81 256 Masteri Maths TG 3.indb 2 11/2/2023 11:50:27 AM PENERBIT ILMU BAKTI SDN. BHD.
3 Bulatkan jawapan yang betul. SP 1.2.1 TP 1 Circle the correct answer. 1 52 × 53 = 54 56 55 2 4u4 × 2u4 = 6u2 8u6 8u8 3 m14 × m12 = m2 m26 m13 4 2 3h3 × 6h × 2.5h4 = 12h4 10h8 10h7 5 (0.8)3 × (0.8)2 × (0.8)4 = (0.8)9 (0.8)10 (0.8)24 6 k25 × k10 × k15 = k25 k15 k50 Permudahkan setiap yang berikut. SP 1.2.1 TP 2 HEBAT MATEMATIK MODUL 31 Gangsa Simplify each of the following. 7 23 × 24 × 2 = = 23+4+1 = 28 8 (–1.5)4 × (–1.5)3 × (–1.5)3 = = (–1.5)4+3+3 = (–1.5)10 9 7a4 × a2 × 2a = = (7 × 2)(a4 × a2 × a1 ) = 14a4+2+1 = 14a7 10 8u2 × 5 6 u2 × (–12u5 ) × 1 4 u = = 1 8 × 5 6 × (–12) × 1 4 2 (u2 × u2 × u5 × u1 ) = – 20u2 + 2 + 5 + 1 = – 20u10 Permudahkan setiap yang berikut. SP 1.2.1 TP 2 HEBAT MATEMATIK MODUL 31 Gangsa Simplify each of the following. 11 r 2 × s5 × r3 × s 2 = = r 2 × r 3 × s5 × s2 = r 2+3 × s 5+2 = r 5 × s7 = r 5 s7 12 0.3 × (0.5)4 × (0.5)2 × (0.3)6 = = (0.3)1 × (0.3)6 × (0.5)4 × (0.5)2 = (0.3)1+6 × (0.5)4+2 = (0.3)7 × (0.5)6 13 –3c 3 × 4d2 ×10c × 2d4 = = –3c3 × 10c × 4d2 × 2d4 = (–3 × 10 × 4 × 2)c3 + 1d2 + 4 = –240c4 d6 14 2a3 × 0.4b4 × 10a2 × (– 8b) = = (2 × 0.4 × 10 × –8)a3 + 2b4 + 1 = –64a5 b5 Buku Teks: Halaman 6 – 8 Hukum Indeks 1.2 Masteri Maths TG 3.indb 3 11/2/2023 11:50:28 AM PENERBIT ILMU BAKTI SDN. BHD.
4 Padankan jawapan yang betul. SP 1.2.2 TP 1 Match the correct answer. 1 f 9 ÷ f 3 p2 • • 2 p6 ÷ p4 f 6 • • 3 1210 ÷ 124 820 • • 4 827 ÷ 87 126 • • Permudahkan setiap yang berikut. SP 1.2.2 TP 2 HEBAT MATEMATIK MODUL 31 Gangsa Simplify each of the following. 5 55 ÷ 52 = = 55 – 2 = 53 6 817 ÷ 86 = = 817 – 6 = 811 7 k18 ÷ k11 = = k18 – 11 = k7 8 (– 20)9 ÷ (– 20)4 = = (– 20)9 – 4 = (– 20)5 9 2335 ÷ 2325 ÷ 235 = = 2335 – 25 – 5 = 235 10 m8 ÷ m3 ÷ m = = m8 – 3 – 1 = m4 Permudahkan setiap yang berikut. SP 1.2.2 TP 3 HEBAT MATEMATIK MODUL 31 Gangsa Simplify each of the following. 11 18s6 ÷ 2s4 = = 18 2 s6 – 4 = 9s2 12 4 7 y5 ÷ 4y 2 = = 1 4 7 ÷ 42 y5 – 2 = 1 7 y3 13 60r12 ÷ 12r3 ÷ 5 6 r2 = = 1 60 ÷ 12 ÷ 5 6 2 r 12 – 3 – 2 = 6r 7 14 6m5 n3 ÷ 3m3 n2 = = 1 6 3 2m5 – 3n3 – 2 = 2m2 n 15 – 98g10h5 ÷ 24.5g5 h÷ 1 2 g3 h = = 1 – 98 ÷ 24.5 ÷ 1 2 2 g10 – 5 – 3h5 – 1 – 1 = –8g2 h3 16 15a6 b5 ÷ 5a2 b3 ÷ 3 4 ab = = 1 15 ÷ 5 ÷ 3 4 2 a6 – 2 – 1b5 – 3 – 1 = 4a3 b Buku Teks: Halaman 8 – 10 Hukum Indeks 1.2 Untuk tujuan pembelajaran Imbas kod QR atau layari https://youtu. be/r57nGvFoEas untuk menonton video tutorial melaksanakan operasi yang melibatkan hukum indeks. Video Tutorial Untuk tujuan pembelajaran Imbas kod QR atau layari https://youtu. be/tHCM6qNdVSc untuk maklumat tambahan tentang hukum indeks. Video Video Video Tutorial Masteri Maths TG 3.indb 4 11/2/2023 11:50:28 AM PENERBIT ILMU BAKTI SDN. BHD.
5 Nyatakan “Benar” atau “Palsu” bagi setiap persamaan berikut. SP 1.2.3 TP 3 State “True” or “False” for each of the following equations. 1 (33 )2 = (32 )3 Kiri/Left: (33 ) 2 = 36 Kanan/Right: (32 ) 3 = 32(3) = 36 \ Benar/True 2 (42 )6 = (43 )4 Kiri/Left: (42 ) 6 = 42(6) = 412 Kanan/Right: (43 ) 4 = 43(4) = 412 \ Benar/True 3 (23 )5 = (163 )2 Kiri/Left: (23 ) 5 = 23(5) = 215 Kanan/Right: (163 ) 2 = (24(3)) 2 = 224 \ Palsu/False Permudahkan setiap yang berikut. SP 1.2.3 TP 3 Simplify each of the following. 4 (33 )5 = = 33(5) = 315 5 (54 )8 = = 54(8) = 532 6 ((– y)3 )3 = = (–y) 3(3) = (– y) 9 Permudahkan setiap yang berikut. SP 1.2.3 TP 3 HEBAT MATEMATIK MODUL 31 Perak Simplify each of the following. 7 (23 × 52 )4 = = 23(4) × 52(4) = 212 × 58 8 (p3 q2 r 4 )3 = = p3(3)q2(3)r 4(3) = p9 q6 r 12 9 (94 × 52 × 69 )3 = = 94(3) × 52(3) × 69(3) = 912 × 56 × 627 10 1 23 52 2 4 = = 23(4) 52(4) = 212 58 11 (2c 4 d5 )5 8c12d = = 25 c4(5)d5(5) 8c12d = 32c20d25 8c12d = 32 8 c20 – 12d25 – 1 = 4c8 d24 12 (3rs3 )3 × (5r6 s5 )2 9r10s8 = = 33 r 3 s3(3) × 52 r 6(2)s5(2) 9r 10s 8 = 27r 3 s9 × 25r 12s10 9r 10s 8 = 1 27× 25 9 2r 3 + 12 – 10s9 + 10 – 8 = 75r 5 s11 Buku Teks: Halaman 10 – 13 Hukum Indeks 1.2 Masteri Maths TG 3.indb 5 11/2/2023 11:50:28 AM PENERBIT ILMU BAKTI SDN. BHD.
6 Tandakan () bagi jawapan yang betul dan () bagi jawapan yang salah pada petak yang disediakan. SP 1.2.4 TP 2 Mark () for the correct answer and () for the incorrect answer in the box provided. 1 a5 = 1 a–5 2 83 = 1 83 3 1 42 = 4–2 4 2b2 = 2 b–2 5 4c –2 7 = 7 4c2 6 1 x y 2 10 = 1 y x 2 –10 Tukarkan setiap yang berikut kepada bentuk (am) , a m , !an m dan ( !an ) m. SP 1.2.5 TP 3 Convert each of the following into the form (am) 1 – n , (a 1 – n ) m, !amn and ( !a n ) m. a (am) (a )m !a n m ( ! a n ) m 7 2 3 – 4 (23 ) 1 – 4 (21 – 4) 3 !234 1 !2 2 4 3 8 125 2 – 5 (1252 ) 1 – 5 (1251 – 5) 2 !125 5 2 1 ! 125 2 5 2 9 k 4 – 9 (k 4 ) 1 – 9 (k 1 – 9) 4 !k 9 4 1 !k 2 9 4 10 1 5 6 2 3 – 5 11 5 6 2 3 2 1 – 5 11 5 6 2 1 – 5 2 3 !1 5 6 2 3 5 1! 5 6 5 2 Cari nilai setiap yang berikut. SP 1.2.5 TP 3 Find the value of each of the following. 11 !81 4 = = 81 1 4 = (34 ) 1 4 = (3)1 = 3 12 !–243 5 = = (–243) 1 5 = (–3)5( 1 5 ) = (–3)1 = –3 13 (–1 024) 1 – 5 = = (–4)5( 1 5 ) = (–4)1 = –4 14 64 4 – 6 = = (!64 6 ) 4 = 24 = 16 15 1 296 3 – 4 = = (!1 296 4 )3 = 63 = 216 16 3 125 4 – 5 = = (!3 125 5 ) 4 = 54 = 625 1 n 1 n m n 1 n 1 n 3 Buku Teks: Halaman 14 – 19 Hukum Indeks 1.2 Masteri Maths TG 3.indb 6 11/2/2023 11:50:28 AM PENERBIT ILMU BAKTI SDN. BHD.
7 Permudahkan setiap yang berikut. SP 1.2.6 TP 3 HEBAT MATEMATIK MODUL 31 Perak Simplify each of the following. 1 3a2 b7 × (2a3 b)4 12a6 b8 = 3a2 b7 × 24 a3(4)b4 12a6 b8 = 3a2 b7 × 16a12b4 12a6 b8 = 1 3 × 16 12 2a2 + 12 – 6b7 + 4 – 8 = 4a8 b3 2 e 3 ! f × e5 f 1 – 4 1e 1 – 5 f 3 – 42 10 = e3 f 1 – 2 × e5 f 1 – 4 e 1 – 5 (10) f 3 – 4 (10) = e3 f 1 – 2 × e5 f 1 – 4 e2 f 15 —2 = e3 + 5 – 2f 1 – 2 + 1 – 4 – 15 — 2 = e6 f – 27 — 4 = e6 f 27 —4 3 ! 49xyz2 × 2x 1 – 2 y 2 z 3 ! 216xyz 3 = (49xyz2 ) 1 – 2 × 2x 1 – 2 y2 z (216xyz3 ) 1 – 3 = 49 1 – 2 x 1 – 2 y 1 – 2 z 21 1 – 22 × 2x 1 – 2 y2 z 216 1 – 3 x 1 – 3 y 1 – 3 z 31 1 – 32 = 7x 1 – 2 y 1 – 2 z× 2x 1 – 2 y2 z 6x 1 – 3 y 1 – 3 z = 1 7 × 2 6 2x 1 – 2 + 1 – 2 – 1 – 3 y 1 – 2 + 2 – 1 – 3 z 1 + 1 – 1 = 7 3 x 2 – 3 y 13 — 6 z 4 81 1 – 2 × 64 1 – 3 (39 × 26 ) 1 – 3 = 3 41 1 – 22 × 2 61 1 – 32 3 91 1 – 32 × 2 61 1 – 32 = 32 × 22 33 × 22 = 32 – 3 ×22 – 2 = 3–1 × 20 = 1 3 × 1 = 1 3 5 (53 × 26 ) 1 – 3 × ! 125 3 × ! 256 4 25 × 64 = 5 31 1 – 32 × 2 61 1 – 32 × 5 31 1 – 32 × 4 41 1 – 42 52 × 26 = 51 × 22 × 51 × 22 52 × 26 = 51 + 1 – 2 ×22 + 2 – 6 = 50 × 2–2 = 1 × 1 22 = 1 4 6 256 3 – 4 × ! 343 3 × (4–6 × 74 ) 1 – 2 64 1 – 3 × 2 401 3 – 4 = 4 41 3 – 42 × 7 31 1 – 32 × 4 –61 1 – 22 × 7 41 1 – 22 4 31 1 – 32 × 7 41 3 – 42 = 43 × 71 × 4–3 × 72 41 × 73 = 43 + (–3) – 1 × 71 + 2 – 3 = 4–1 × 70 = 1 4 × 1 = 1 4 Buku Teks: Halaman 20 – 21 Hukum Indeks 1.2 Masteri Maths TG 3.indb 7 11/2/2023 11:50:29 AM PENERBIT ILMU BAKTI SDN. BHD.
8 Selesaikan masalah yang berikut. SP 1.2.7 TP 4 TP 5 Solve the following problems. 1 Selesaikan persamaan serentak berikut./ Solve the following simultaneous equations. 1 5p = 25 × 25q 64p 4q = 162 2 Hitung nilai x jika 4x + 3 × 45 = 16. Find the value of x if 4x + 3 × 45 = 16. 4x + 3 × 45 = 42 4x + 3 + 5 = 42 4x + 8 = 42 x + 8 = 2 x = –6 3 Diberi bahawa 32m = 5 dan 3n = 7. Cari nilai 9 m + 1 2 n – 2. It is given that 32m = 5 and 3n = 7 . Find the value of 9m + 1 2 n – 2. 3 21m + 1 – 2 n – 22 = 32m + n – 4 = 32m × 3n 34 = 5 × 7 81 = 35 81 1 5p = 25 × 25q 5–p = 52 × 52(q) 5–p = 52 + 2q –p = 2 + 2q.... 1 64p 4q = 162 43(p) 4q = 42(2) 43p – q = 44 3p– q = 4.... 2 1 × 3 : –3p = 6 + 6q.... 3 3 + 2 : –7q = 10 q = – 10 7 Gantikan q = – 10 7 ke dalam 1 , Substitute q = – 10 7 into 1 , –p = 2 + 21– 10 7 2 p = 6 7 Maka/Hence, p = 6 7 , q = – 10 7 . Buku Teks: Halaman 22 – 24 Hukum Indeks 1.2 Masteri Maths TG 3.indb 8 11/2/2023 11:50:29 AM PENERBIT ILMU BAKTI SDN. BHD.
9 Buku Teks: Halaman 32 – 34 Angka Bererti 2.1 Nyatakan bilangan angka bererti bagi nombor-nombor berikut. SP 2.1.1 TP 2 State the number of significant figures for the following numbers. 1 0.0034 2 angka bererti 2 significant figures 2 0.0704 3 angka bererti 3 significant figures 3 1.73 3 angka bererti 3 significant figures 4 20.84 4 angka bererti 4 significant figures 5 50 000 1 angka bererti 1 significant figures 6 123 457 6 angka bererti 6 significant figures 7 0.6500 4 angka bererti 4 significant figures 8 35.90 4 angka bererti 4 significant figures 9 6 007 4 angka bererti 4 significant figures 10 7 890 007 7 angka bererti 7 significant figures 11 65.003 5 angka bererti 5 significant figures 12 5.4200 5 angka bererti 5 significant figures Kelaskan nombor-nombor berikut mengikut bilangan angka bererti yang betul. SP 2.1.1 TP 2 Classify the following numbers according to the correct number of significant figures. 457 1.77 49.6 10 031 0.069 6.031 142.3 0.0321 500.71 92 0.0005 800 0.035 9.306 200.0 29.300 300 70 003 0.08 0.0072 Bab 2 Bentuk Piawai Standard Form Bidang Pembelajaran: Nombor dan Operasi Buku Teks: Halaman 30 – 49 Angka bererti (a.b.) Significant figures (s.f.) i-THINK Peta Pokok 13 1 a.b. 1 s.f 14 2 a.b. 2 s.f 15 3 a.b. 3 s.f 16 4 a.b. 4 s.f 17 5 a.b. 5 s.f 0.0005 800 0.08 300 0.069 92 0.035 0.0072 457 1.77 49.6 0.0321 6.031 142.3 9.306 200.0 10 031 500.71 29.300 70 003 Masteri Maths TG 3.indb 9 11/2/2023 11:50:29 AM PENERBIT ILMU BAKTI SDN. BHD.
10 Buku Teks: Halaman 35 – 36 Angka Bererti 2.1 Bundarkan setiap nombor berikut betul kepada angka bererti yang diberi. SP 2.1.2 TP 2 Round off each of the following numbers to the given significant figures. Nombor Number 1 angka bererti 1 significant figure 2 angka bererti 2 significant figures 3 angka bererti 3 significant figures 4 angka bererti 4 significant figures 1 29 700 30 000 30 000 29 700 29 700 2 340 076 300 000 340 000 340 000 340 100 3 1 476 1 000 1 500 1 480 1 476 4 677 341 700 000 680 000 677 000 677 300 5 1 000 000 1 000 000 1 000 000 1 000 000 1 000 000 6 0.9671 1 0.97 0.967 0.9671 7 0.006653 0.007 0.0067 0.00665 0.006653 8 8.9213 9 8.9 8.92 8.921 9 68.795 70 69 68.8 68.80 10 90.730 90 91 90.7 90.73 11 89.06004 90 89 89.1 89.06 Hitung setiap yang berikut. Nyatakan jawapan betul kepada angka bererti yang dinyatakan dalam kurungan. Calculate each of the following. State the answer to the significant figures stated in the brackets. SP 2.1.2 TP 2 12 45 – 1.34 + 2.098 = [2] = 43.66 + 2.098 = 45.758 = 46 13 36.99 ÷ 0.3 + 21.54 = [3] = 123.3 + 21.54 = 144.84 = 145 14 1.92 – 4.6 ÷ 2 + 0.321 = [1] = 1.92 – 2.3 + 0.321 = –0.38 + 0.321 = –0.059 = –0.06 15 40 – 21 × 3.7 – 11.7 = [3] = 40 – 77.7 – 11.7 = –37.7 – 11.7 = –49.4 16 (332 + 675) ÷ 0.8 = [5] = 1 007 ÷ 0.8 = 1 258.75 = 1 258.8 17 0.0074 + 0.0897 × 1.3 = [2] = 0.0074 + 0.11661 = 0.12401 = 0.12 Masteri Maths TG 3.indb 10 11/2/2023 11:50:29 AM PENERBIT ILMU BAKTI SDN. BHD.
11 Buku Teks: Halaman 37 – 39 Bentuk Piawai 2.2 Tulis setiap nombor tunggal berikut dalam bentuk piawai. SP 2.2.1 TP 2 Write each of the following single numbers in standard form. Nombor tunggal Single number Bentuk piawai Standard form i-THINK Peta Titi 5.47 × 102 as as as 1.006 × 103 9.8053 × 103 1 547 2 1 006 3 9 805.30 7.4 × 10–2 as as 4.07 × 10–3 6.8 × 10–4 4 0.074 5 0.00407 6 0.00068 (as = sama seperti) Tulis setiap nombor berikut sebagai nombor tunggal. SP 2.2.1 TP 2 Write each of the following numbers as a single number. Bentuk piawai Standard form Nombor tunggal Single number 447 000 as as as 86 700 15 800 000 7 4.47 × 105 8 8.67 × 104 9 1.58 × 107 0.0821 as as 0.000034 0.000706 10 8.21 × 10–2 11 3.4 × 10–5 12 7.06 × 10–4 (as = sama seperti) Tukarkan ukuran dalam sistem metrik berikut kepada unit yang dinyatakan. Nyatakan jawapan dalam bentuk piawai. SP 2.2.1 TP 2 Convert the following metric measurements to the units stated. State the answers in standard form. 13 3 125 kilometer/kilometres = meter/metres = 3.125 × 103 × 103 = 3.125 × 103 + 3 = 3.125 × 106 meter/ metres 14 0.00046 terabait/terabyte = bait/bytes = 4.6 × 10–4 × 1012 = 4.6 × 10–4 + 12 = 4.6 × 108 bait/ bytes 15 304 mililiter/millilitres = liter/litre = 3.04 × 102 × 10–3 = 3.04 × 102 + (–3) = 3.04 × 10–1 liter/ litre 16 264 gigabait/gigabytes = bait/bytes = 2.64 × 102 × 109 = 2.64 × 102 + 9 = 2.64 × 1011 bait/ bytes 17 0.000056 nanoliter/nanolitre = liter/litre = 5.6 × 10–5 × 10–9 = 5.6 × 10–5 + (–9) = 5.6 × 10–14 liter/ litre 18 0.87 attometer/attometre = meter/metre = 8.7 × 10–1 × 10–18 = 8.7 × 10–1 + (–18) = 8.7 × 10–19 meter/ metre i-THINK Peta Titi Untuk tujuan pembelajaran Imbas kod QR atau layari https://youtu.be/ bcPz3FJCSAw untuk menonton video tutorial menukarkan nombor tunggal kepada bentuk piawai dan sebaliknya. Video Tutorial Video Tutorial Masteri Maths TG 3.indb 11 11/2/2023 11:50:30 AM PENERBIT ILMU BAKTI SDN. BHD.
115 Bab 1 Indeks Halaman 1 1 54 2 45 3 (–4)4 4 0.45 5 m × m × m × m × m × m × m × m × m 6 76 7 (–6) × (–6) × (–6) × (–6) 8 1 1 s 2 8 9 3 × 3 10 (–5.2) × (–5.2) × (–5.2) 11 (–x) × (–x) × (–x) × (–x) × (–x) × (–x) 12 b × b × b × b × b × b × b 13 1 1 d2 × 1 1 d2 × 1 1 d2 × 1 1 d2 14 (1.9z) × (1.9z) × (1.9z) × (1.9z) × (1.9z) Halaman 2 1 72 2 113 3 1 4 5 2 4 4 95 5 0.34 6 104 7 7 776 8 1 082.41 9 81 256 Halaman 3 1 55 2 8u8 3 m26 4 10h8 5 (0.8)9 6 k50 7 28 8 (–1.5)10 9 14a7 10 –20u10 11 r5 s7 12 (0.3)7 × (0.5)6 13 –240c4 d6 14 –64a5 b5 Halaman 4 1 f 6 2 p2 3 126 4 820 5 53 6 811 7 k7 8 (–20)5 9 235 10 m4 11 9s2 12 1 7 y3 13 6r7 14 2m2 n 15 –8g2 h3 16 4a3 b Halaman 5 1 Benar/True 2 Benar/True 3 Palsu/False 4 315 5 532 6 (–y)9 7 212 × 58 8 p9 q6 r12 9 912 × 56 × 627 10 212 58 11 4c8 d24 12 75r5 s11 Halaman 6 1 2 3 4 5 6 7 (23 ) 1 – 4 (21 – 4)3 ! 234 1 ! 2 2 4 3 8 (1252 ) 1 – 5 (1251 – 5)2 !125 5 2 1 ! 125 2 5 2 9 (k4 ) 1 – 9 (k 1 – 9)4 ! k 9 4 1 ! k 2 9 4 10 11 5 6 2 3 2 1 – 5 11 5 6 2 1 – 5 2 3 !1 5 6 2 3 5 1 2 3 5 ! 5 6 11 3 12 –3 13 –4 14 16 15 216 16 625 Halaman 7 1 4a8 b3 2 e6 f 27 — 4 3 7 3 x 2 – 3 y 13 — 6 z 4 1 3 5 1 4 6 1 4 Halaman 8 1 p = 6 7 , q = – 10 7 2 x = –6 3 35 81 Bab 2 Bentuk Piawai Halaman 9 1 2 angka bererti/ 2 significant figures 2 3 angka bererti/ 3 significant figures 3 3 angka bererti/ 3 significant figures 4 4 angka bererti/ 4 significant figures 5 1 angka bererti/ 1 significant figure 6 6 angka bererti/ 6 significant figures 7 4 angka bererti/ 4 significant figures 8 4 angka bererti/ 4 significant figures 9 4 angka bererti/ 4 significant figures 10 7 angka bererti/ 7 significant figures 11 5 angka bererti/ 5 significant figures 12 5 angka bererti/ 5 significant figures 13 0.0005, 800, 0.08, 300 14 0.069, 92, 0.035, 0.0072 15 457, 1.77, 49.6, 0.0321 16 6.031, 142.3, 9.306, 200.0 17 10 031, 500.71, 29.300, 70 003 Halaman 10 1 30 000; 30 000; 29 700; 29 700 2 300 000; 340 000; 340 000; 340 100 3 1 000; 1 500; 1 480; 1 476 4 700 000; 680 000; 677 000; 677 300 5 1 000 000; 1 000 000; 1 000 000; 1 000 000 6 1; 0.97; 0.967; 0.9671 7 0.007; 0.0067; 0.00665; 0.006653 8 9; 8.9; 8.92; 8.921 9 70; 69; 68.8; 68.80 10 90; 91; 90.7; 90.73 11 90; 89; 89.1; 89.06 12 46 13 145 14 –0.06 15 –49.4 16 1 258.8 17 0.12 Halaman 11 1 5.47 × 102 2 1.006 × 103 3 9.8053 × 103 4 7.4 × 10–2 5 4.07 × 10–3 6 6.8 × 10–4 7 447 000 8 86 700 9 15 800 000 10 0.0821 11 0.000034 12 0.000706 13 3.125 × 106 meter/ metres 14 4.6 × 108 bait/ bytes 15 3.04 × 10–1 liter/ litre 16 2.64 × 1011 bait/ bytes 17 5.6 × 10–14 liter/ litre 18 8.7 × 10–19 meter/ metre Halaman 12 1 1.48 × 105 2 2.58 × 109 3 1.164 × 10–7 4 4.98 × 10–8 5 9.9 × 102 6 1.001 × 10–2 7 7.74 × 104 8 7.116 × 10–2 Halaman 13 1 2.8 × 107 2 4.96 × 10–9 3 5.97 × 105 4 3.93 × 1010 5 2.97 × 10–4 6 4.963 × 10–6 Halaman 14 1 1.0738 × 1011 2 3.036 × 104 3 2.697 × 101 4 1.118 × 10–10 5 6.4 × 104 6 1.3552 × 10–4 7 4.48 × 106 8 9.3 × 10–10 Halaman 15 1 2 × 105 2 1.3 × 1018 3 2.5 × 104 4 2.4 × 10–11 5 1.85 × 103 Halaman 16 1 7.67 × 104 2 1.00 × 106 3 2 × 103 4 8.010 × 102 5 1.40 × 106 6 1.6 × 1010 Halaman 17 1 5.42 × 102 cm2 2 1.27 g/cm3 3 t = 580 cm 4 RM6.96 × 106 Jawapan Masteri Maths TG 3.indb 115 11/2/2023 11:50:59 AM PENERBIT ILMU BAKTI SDN. BHD.