Rujukan Custom-made Matematik (Dwibahasa) Tingkatan 3 ALAF SANJUNG SDN. BHD. (516756-V) Wisma Alaf Sanjung No. 23, Jalan Sungai Besi Indah 5/2, Taman Sungai Besi Indah, 43300 Seri Kembangan, Selangor Darul Ehsan. Tel : 03-8941 0411 / 03-8941 0611 Faks : 03-8941 0041 E-mel : alafsanjung@hotmail.com © Alaf Sanjung Sdn. Bhd. (516756-V) Semua hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh dipergunakan lagi, ataupun dipindahkan, dalam sebarang bentuk atau cara lain tanpa kebenaran terlebih dahulu daripada Alaf Sanjung Sdn. Bhd. ISBN 978-629-7512-40-2 Dicetak oleh : Bookmate Sdn. Bhd. Low Ming Yike Alaf Sanjung Sdn Bhd
KANDUNGAN / CONTENTS Bab 1 : Indeks Indices 1 – 6 Bab 2 : Bentuk Piawai Standard Form 7 – 13 Bab 3 : Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang Consumer Mathematics: Savings and Investments, Credit and Debt 14 – 29 Bab 4 : Lukisan Berskala Scale Drawings 30 – 35 Bab 5 : Nisbah Trigonometri Trigonometric Ratios 36 – 42 Bab 6 : Sudut dan Tangen bagi Bulatan Angles and Tangent of Circles 43 – 57 Bab 7 : Pelan dan Dongakan Plans and Elevations 58 – 66 Bab 8 : Lokus dalam Dua Dimensi Loci in Two Dimensions 67 – 72 Bab 9 : Garis Lurus Straight Lines 73 – 79 Alaf Jawapan Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 1 Matematik T3 Bab 1 Indeks Chapter 1 Indices 1.1 Tatatanda Indeks / Index Notation Nombor dalam tatatanda indeks atau dalam bentuk indeks boleh ditulis sebagai; A number in index notation or in index form can be written as; Nilai indeks dalam suatu bentuk indeks adalah sama dengan bilangan kali asas didarab secara berulang. The value of index in an index form is the same as the number of times the base is multiplied repeatedly. Contoh / Example : ( 1 3 ) × (1 3 ) × (1 3 ) × (1 3 ) = ( 1 3 ) 4 Maka, nilai indeks sama dengan bilangan kali ( 1 3 ) didarab secara berulang. Thus, the value of index is the same as the number of times ( 1 3 ) is multiplied repeatedly. __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ a n Indeks Asas Index Base a n = a × a × a × …× a ; a ≠ 0 n faktor / factors Menyatakan pendaraban berulang dalam bentuk indeks. State repeated multiplication in index form. (a) 5 × 5 × 5 = 53 (b) 1 5 × 1 5 × 1 5 × 1 5 = ( 1 5 ) 4 (c) (−0.2) × (−0.2) × (−0.2) ×(−0.2) ×(−0.2) ×(−0.2) = (−0.2) 6 Menukarkan nombor atau sebutan algebra dalam bentuk indeks kepada pendaraban berulang. Convert the numbers or algebraic terms in index form into repeated multiplications. (a) (0.4) 2 = 0.4 × 0.4 (b) ( 1 ) 5 = 1 × 1 × 1 × 1 × 1 Nilai indeks ialah 4. The value of index is 4. Berulang empat kali Repeated four times Customise your own notes Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 2 Matematik T3 Suatu nombor boleh ditulis dalam bentuk indeks jika suatu asas yang sesuai dipilih dengan menggunakan: A number can be written in index form if a suitable base is selected by using: Kaedah pembahagian berulang / Repeated division method 81 [asas 3 / base of 3] 81 dibahagi secara berulang dengan 3. 81 is divided repeatedly by 3. Maka / Thus, 81 = 34 Kaedah pendaraban berulang / Repeated multiplication method 81 [asas 3 / base of 3] Maka / Thus, 81 = 34 Menentukan nilai bagi nombor dalam bentuk indeks, a n Determine the value of the number in index form, an Nilai a n boleh ditentukan dengan kaedah pendaraban berulang atau dengan menggunakan kalkulator saintifik. The value of an can be determined by repeated multiplication method or using a scientific calculator. Kaedah pendaraban berulang / Repeated multiplication method (−1 7 ) 5 Maka / Thus, (−1 7 ) 5 = – 1 16 807 Menukar suatu nombor kepada nombor dalam bentuk indeks Convert a number into a number in index form 3 81 27 9 3 1 3 3 3 Pembahagian diteruskan sehingga mendapat nilai 1. The division is continued until 1 is obtained. 3 × 3 × 3 × 3 9 27 81 – 1 7 × – 1 7 × – 1 7 × – 1 7 × – 1 7 1 49 – 1 343 1 2 401 – 1 16 807 Menulis nombor 0.00032 dalam bentuk indeks dengan menggunakan asas 0.2. Write number of 0.00032 in index form using base of 0.2. Maka, 0.00032 = (0.2)5 Menghitung nilai nombor (–3.2)3 dalam bentuk indeks. Calculate the value of number (–3.2)3 in index form. (–3.2)3 = –32.768 Kalkulator saintifik / Scientific calculator ( (–) 1 a 7 ) ^ 5 = ___________________________ ___________________________ ___________________________ ___________________________ Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 3 Matematik T3 Bab 1 Indeks Chapter 1 Indices 1.2 Hukum Indeks / Law of Indices Pendaraban / Multiplication Contoh / Example : ( 2 5 ) × (2 5 ) 3 × (2 5 ) 2 = ( 2 5 ) 1+3+2 = ( 2 5 ) 6 – – 1 3 5 × (–4) 2 × 1 2 3 = (− 1 3 × (−4) × 1 2 ) ( 5 × 2 × 3 ) = 2 3 5+2+ 3 = 2 3 10 Mempermudahkan nombor atau sebutan algebra dalam bentuk indeks yang mempunyai asas yang berlainan Simplify a number or an algebraic term in index form with different bases Contoh / Example : (0.5) 3 × (1.4) 5 × (0.5) × (1.4) 3 × (1.4) = (0.5) 3 × (0.5) × (1.4) 5 × (1.4) 3 × (1.4) = (0.5) 3 + 1 × (1.4) 5 + 3 + 1 = (0.5) 4 × (1.4) 9 –52 × 3 × 1 2 5 × 2m = –52 × 2m × 3 × 1 2 5 = (−5 × 2 × 1 2 ) 2 × m × 3 × 5 = –52 + 1 3 + 5 = –53 8 __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ a m × a n = a m + n Menyatakan dalam bentuk indeks paling ringkas State in the simplest index form 14 5 × 3 × 1 4 p × 4 7 4 14 5 × 3 × 1 4 p × 4 7 4 = (14 × 1 4 × 4 7 ) 5 + 1 3 + 4 = 2 6 7 2 6 × 83 × 22 × 84 2 6 × 83 × 22 × 84 = 26 + 2 × 8 3 + 4 = 28 × 87 Operasi untuk pekali Operation of the coefficients Kumpulkan asas yang sama. Group the same base. Tambahkan indeks bagi asas yang sama. Add the indices for the same base. Customise your own notes Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 4 Matematik T3 Pembahagian / Division Contoh / Example : 7 10 ÷ 76 ÷ 72 = 710 – 6 – 2 = 72 –25h 4 ÷ 5h 2 ÷ h = – 25 5 (h 4 ÷ h 2 ÷ h) = –5(h 4 – 2) ÷ h = –5h 2 – 1 = –5h Kuasa / Power Contoh / Example : Menggunakan hukum indeks untuk operasi pendaraban dan pembahagian Use law of indices to perform operations of multiplication and division Contoh / Example : (22 3 ) 4 = 2 4 2(4) 3(4) = 16 8 12 3 3× (62) 3 6 4 = 3 3× 66 6 4 = 3 3 × 66 − 4 = 3 3 × 62 (3 10) 2 = 310(2) = 320 ((–s) 7 ) 3 = (–s) 7(3) = (–s) 21 (a m b n ) q = a mq b nq (a m× b n ) q = (a m ) q × (b n ) q = a mq × b nq Jika 3 × 4 3 3 × 4 6 = 12, tentukan nilai m + n. If 3 × 4 3 3 × 4 6 = 12, determine the value of m + n. 3 × 4 3 3 × 4 6 = 12 3 − 34 − 6 = 31 × 41 m – 3 = 1 n – 6 = 1 m = 4 n = 7 maka / thus, m + n = 4 + 7 = 11 Permudahkan. Simplify. (35) 2 × 2 7 4 27 52 (35) 2 × 2 7 4 27 52 = (3 2)( 2 10) × 2 7 4 27 52 = ( 9 ×2 27 ) 2+7 −5 10+4 −2 = 2 3 4 12 ( ) = (a m ÷ b n ) q = (a m ) q ÷ (b n ) q = a mq ÷ b nq ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ (a m ) n = a mn a m ÷ a n = a m – n Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 5 Matematik T3 Suatu nombor atau sebutan algebra yang mempunyai indeks sifar akan memberi nilai 1. A number or an algebraic term with a zero index will give a value of 1. Contoh / Example : – 5–6 = – 5 6 ( 2 3 ) 7 = ( 3 2 ) –7 (42 4) 2 (2−2) 5(34) 3 = (42)(4 8) (2 5)(−105)(3 3)( 123) = 16 32 ×27 4 −(−10+12) 8 −(5+3) = 1 54 2 0 = 2 54 Contoh / Example : √2 185 7 = 2 185 1 7 (−729) 1 3 = √−729 3 (−32 768) 1 5 = (−85 ) 1 5 = –8 Contoh / Example : 1 024 2 5 = (√1 024 5 ) 2 = (4) 2 = 16 (24×36) 1 2 × √8 3 × √81 16 3 4 × 27 1 3 = (24) 1 2 × (36) 1 2 × (23) 1 3 × (34) 1 2 (2 4) 3 4 × (3 3) 1 3 = 2 2 × 33 × 21 × 32 2 3 × 31 = 2 2+1 −3 × 33+2 −1 = 2 0 × 34 = 1 × 81 = 81 a 0 = 1 ; a ≠ 0 √ = 1 ; a ≠ 0 = ( ) 1 = ( 1 ) = √ = ( √ ) y 0 = 1 Diberi bahawa x = 2 dan y = –3. Hitung nilai Given that x = 2 and y = –3. Calculate the value of 64 3 × 512(− 1 ) ÷ 81 2 64 3 × 512(− 1 ) ÷ 81 2 = 64 2 3 × 512(− 1 −3 ) ÷ 81 −3 2(2) = (2 6 ) 2 3 × (2 9 ) (− 1 −3 ) ÷ (3 4 ) −3 2(2) = 2 4 × 2 3 ÷ 3 −3 = 2 4 + 3 3−3 = 2 7 3−3 = 2 7 × 3 3 = 128 × 27 = 3 456 __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ __________________________________ a –n = 1 ; a ≠ 0 Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 6 Matematik T3 Menyelesaikan masalah yang melibatkan hukum indeks Solve problems involving laws of indices Contoh / Example : Suhu sebiji kek meningkat daripada 30˚C kepada T˚C mengikut persamaan T = 25(1.07)t apabila kek tersebut dibakar selama t minit. Hitung beza suhu di antara minit ke-10 dengan minit ke-25 dalam darjah Celsius terdekat. The temperature of a cake from 30˚C to T˚C according to equation T = 25(1.07)t when the cake was baked for t minutes. Calculate the difference in temperature between the 10th minute and the 25th minute, to the nearest degree Celsius. Memahami masalah / Understanding the problem Menghitung beza suhu di antara minit ke-10 dengan minit ke-25 dalam darjah Celsius terdekat. Calculate the difference in temperature between the 10th minute and the 25th minute, to the nearest degree Celsius. Merancang strategi / Planning a strategi Cari nilai T pada minit ke-10 dan minit ke-25 dan hitung beza suhu. Find the value of T on 10th minute and 25th minute and calculate the difference in temperature. Melaksanakan strategi / Implementing the strategy Beza suhu / The difference in temperature = T25˚C – T10˚C = 25(1.07)25 – 25(1.07)10 = 86.5070 ≈ 87˚C Membuat kesimpulan / Making a conclusion Beza suhu di antara minit ke-10 dengan minit ke-25 ialah 87˚C. The difference in temperature between the 10th minute and the 25th minute is 87˚C. Memahami masalah Understanding the problem Merancang strategi Planning a strategy Melaksanakan strategi Implementing the strategy Membuat kesimpulan Making a conclusion Encik Ravi membeli sebuah rumah dengan harga RM200 000. Selepas 4 tahun Encik Ravi ingin menjual rumah tersebut. Berdasarkan penerangan penilai hartanah, harga rumah Encik Ravi akan dihitung dengan formula RM200 000( 3 2 ) . Dalam situasi ini, p ialah bilangan tahun selepas rumah dimiliki. Berapakah nilai pasaran rumah Encik Ravi? Encik Ravi bought a house for RM200 000. After 4 years, Encik Ravi wishes to sell the house. Based on the explanation from the property valuer, the price of Encik Ravi’s house will be calculated using the formula RM200 000( 3 2 ) . In this situation, p is the number of years after the house is owned. What is the market value of Encik Ravi’s house? Nilai pasaran rumah Encik Ravi / The market value of Encik Ravi’s house = RM200 000( 3 2 ) 4 = RM 1 012 500 Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 7 Matematik T3 Bab 2 Bentuk Piawai Chapter 2 Standard Form 2.1 Angka Bererti / Significant Figures Angka bererti suatu integer atau perpuluhan merujuk kepada digit-digit dalam nombor yang dinyatakan tepat kepada suatu darjah ketepatan yang dikehendaki. The significant figures of an integer or decimal refer to the digits in the number stated accurately to a certain degree of accuracy as required. Ciri-ciri Characteristics Integer Perpuluhan Decimal Semua digit bukan sifar adalah angka bererti. All non-zero digits are significant figures. 2 345 [4 a.b. / s.f.] 2.56 [3 a.b. / s.f.] Semua sifar antara dua digit bukan sifar adalah angka bererti. The digit zero between non-zero digits is a significant figure. 30 005 [5 a.b. / s.f.] 20.05 [4 a.b. / s.f.] Digit sifar di bahagian akhir suatu integer ialah angka bererti mengikut tahap kejituan yang dikehendaki. The digit zero at the end of an integer is a significant figure according to the level of accuracy required. 15 000 [2 a.b. / s.f.] Jika tahap kejituan ialah ribu terhampir. If level of accuracy is to the nearest thousand. 15 000 [5 a.b. / s.f.] Jika tahap kejituan ialah sa terhampir. If level of accuracy is to the nearest one. Digit sifar di bahagian akhir suatu perpuluhan ialah angka bererti kerana menentukan tahap kejituan perpuluhan tersebut. The digit zero at the end of a decimal is a significant figure because it determines the level of accuracy of the decimal. 1.20 [3 a.b. / s.f.] 1.200 [4 a.b. / s.f.] 1.2000 [5 a.b. / s.f.] Digit sifar sebelum digit bukan sifar yang pertama bukan angka bererti. The digit zero before the first non-zero digit is not a significant figure. 0.005000 [4 a.b. / s.f.] 0.230 [3 a.b. / s.f.] __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ Menentukan bilangan angka bererti Determining the number of significant figures Kenal pasti bilangan angka bererti bagi nombor-nombor Identify the number of significant figure for the numbers 500 [2 a.b / s.f.] Jika tahap kejituan ialah puluh terhampir. If level of accuracy is the nearest ten. 500 [3 a.b. / s.f.] Jika tahap kejituan ialah sa terhampir. If level of accuracy is the nearest one. 0.162 [3 a.b. / s.f.] 235 [3 a.b. / s.f.] Customise your own notes Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 8 Matematik T3 __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ Membundarkan suatu nombor kepada bilangan angka bererti yang tertentu Round off a number to certain numbers of significant figures 57 258 dibundarkan kepada … is rounded off to … 2 angka bererti / significant figures Digit yang ingin dibundarkan. Digit to be rounded off. 57 258 2 < 5, maka digit 7 tidak berubah. 2 < 5, thus digit 7 remains unchanged. 2, 5 dan 8 terletak sebelum titik perpuluhan. Maka, gantikan 2, 5 dan 8 dengan sifar. 2, 5 and 8 are placed before decimal point. Thus, replace 2, 5 and 8 with zero. Maka / Thus, 57 258 = 57 000 (2 a.b. / s.f.) 1 angka bererti / significant figure Digit yang ingin dibundarkan. Digit to be rounded off. 57 258 7 > 5, maka tambah 1 kepada 5. 7 > 5, thus add 1 to 5. 7, 2, 5 dan 8 terletak sebelum titik perpuluhan. Maka, gantikan 7, 2, 5 dan 8 dengan sifar. 7, 2, 5 and 8 are placed before decimal point. Thus, replace 7, 2, 5 and 8 with zero. Maka / Thus, 57 258 = 60 000 (1 a.b. / s.f.) 3 angka bererti / significant figures Digit yang ingin dibundarkan. Digit to be rounded off. 57 258 5 = 5, maka tambah 1 kepada 2. 5 = 5, thus add 1 to 2. 5 dan 8 terletak sebelum titik perpuluhan. Maka, gantikan 5 dan 8 dengan sifar. 5 and 8 are placed before decimal point. Thus, replace 5 and 8 with zero. Maka / Thus, 57 258 = 57 300 (3 a.b. / s.f.) Membundarkan suatu nombor kepada 2 angka bererti Round off a number to 2 significant figures 71.5 kg = 72 kg 603 = 600 Alaf 5.048 = 5.0 Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 9 Matematik T3 Membundarkan suatu nombor kepada bilangan angka bererti yang tertentu Round off a number to certain numbers of significant figures __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ 0.02835 dibundarkan kepada … is rounded off to … 1 angka bererti / significant figure Digit yang ingin dibundarkan. Digit to be rounded off. 0.02835 8 > 5, maka tambah 1 kepada 2. 8 > 5, thus add 1 to 2. Digit 8, 3 dan 5 digugurkan kerana digit tersebut terletak selepas titik perpuluhan. Digits 8, 3 and 5 are dropped because it is placed after the decimal point. Maka / Thus, 0.02835 = 0.03 (1 a.b. / s.f.) 2 angka bererti / significant figures Digit yang ingin dibundarkan. Digit to be rounded off. 0.02835 3 < 5, maka digit 8 tidak berubah. 3 < 5, thus digit 8 remains unchanged. Digit 3 dan 5 digugurkan kerana digit tersebut terletak selepas titik perpuluhan. Digits 3 and 5 are dropped because it is placed after the decimal point. Maka / Thus, 0.02835 = 0.028 (2 a.b. / s.f.) 3 angka bererti / significant figures Digit yang ingin dibundarkan. Digit to be rounded off. 0.02835 5 = 5, maka tambah 1 kepada 3. 3 < 5, thus add 1 to 3. Digit 5 digugurkan kerana digit tersebut terletak selepas titik perpuluhan. Digit 5 is dropped because it is placed after the decimal point. Maka / Thus, 0.02835 = 0.0284 (2 a.b. / s.f.) 1 600 20 004 1 000 800 0.6005 0.070 8.0060 0.003000 25.0006 2 angka bererti / significant figures 1 600, 0.070 4 angka bererti / significant figures 0.6005, 0.003000 5 angka bererti / significant figures 20 004, 1 000 800, 8.0060 6 angka bererti / significant figures 25.0006 Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 10 Matematik T3 Bab 2 Bentuk Piawai Chapter 2 Standard Form 2.2 Bentuk Piawai / Standard Form Bentuk piawai ialah cara menulis suatu nombor tunggal dalam bentuk; Standard form is a way to write a single number in the form; dengan keadaan 1 ≤ A ≤ 10 dan n ialah integer. where 1 ≤ A ≤ 10 and n is an integer. Menukar nombor tunggal kepada bentuk piawai Change a single number to standard form Apabila suatu nombor tunggal ditukar kepada bentuk piawai; When a single number is changed to standard form; Nombor bernilai lebih daripada 1 akan memberi indeks positif. Numbers with value more than 1 is written as a positive index. Nombor bernilai kurang daripada 1 akan memberi indeks negatif. Numbers with value less than 1 is written as a negative index. __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ 0.00546 = 5.46 × = 5.46 × = 5.46 × 10–3 Nilai tempat ialah per seribu. Place value is one thousandths. 1 1 000 1 103 Menukar nombor tunggal kepada bentuk piawai Change a single number to standard form 100 = 1.0 × 100 = 1.0 × 102 36.6 = 3. 66 × 10 0.385 = 3.85 × = 3.85 × 10–1 1 10 A × 10n 450 = 4.5 × 100 = 4.5 × 102 Nilai tempat ialah ratus. Place value is hundreds. Titik perpuluhan selepas digit bukan sifar yang pertama. Decimal point after first non-zero digit. Customise your own notes Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 11 Matematik T3 Menukar nombor dalam bentuk piawai kepada nombor tunggal Change a number in standard form to single number Apabila suatu nombor dalam bentuk piawai ditukar kepada nombor tunggal; When a number in standard form is changed to a single number; Nombor itu bernilai sama atau lebih daripada 10 jika indeksnya positif. The numbers will be equal to 10 or more if the index is positive. Nombor itu bernilai kurang daripada 1 jika indeksnya negatif. The numbers will be less than 1 if the index is negative. Operasi asas aritmetik yang melibatkan nombor dalam bentuk piawai Basic arithmetic operations involving numbers in standard form Operasi tambah dan tolak Operations of addition and subtraction __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ 5.07 × 103 = 5.07 × 1 000 = 5 070 8.504 × 10–4 = 8.504 × = 0.0008504 Nilai setiap operasi berikut. The value of each of the following operations. 7.2 × 106 – 4.25 × 106 = (7.2 – 4.25) × 106 = 2.95 × 106 4.2 × 10–4 + 8.5 × 10–3 = 4.2 × 10–1 × 10–3 + 8.5 × 10–3 = 0.42 × 10–3 + 8.5 × 10–3 = (0.42 + 8.5) × 10–3 = 8.92 × 10–3 8.5 × 10–4 – 6.1 × 10–5 = 8.5 × 10–4 – 6.1 × 10–1 × 10–4 = 8.5 × 10–4 – 0.61 × 10–4 = (8.5 – 0.61) × 10–4 = 7.89 × 10–4 6.5 ×104 –7.3 × 103 = 6.5 ×101 × 103 –7.3 × 103 = 65 × 103 – 7.3 × 103 = (65 – 7.3) × 103 = 57.7 × 103 = 5.77 × 101 × 103 = 5.77 × 104 1.2 × 105 + 3.74 × 104 = 1.2 × 101 × 104 + 3.74 × 104 = 12 × 104 + 3.74 × 104 = (12 + 3.74) × 104 = 15.74 × 104 = 1.574 × 101 × 104 = 1.574 × 105 1 10 000 104 ditukar kepada 101 × 103 untuk memudahkan pengiraan. 104 change to 101 × 103 to simplify calculation. Faktorkan 103 . Factorise 103 . Bagi operasi tambah dan tolak, tukarkan indeks bernilai kecil kepada indeks bernilai besar. For operations involving addition and subtraction, change index with small value to index with large value. 5.2 × 10–3 – 4.12 × 10–4 = 5.2 × 10–3 – 4.12 × 10–1 × 10–3 = 5.2 × 10–3 – 0.412 × 10–3 = (5.2 – 0.412) × 10–3 = 4.788 × 10–3 5.7 × 103 + 8.02 × 104 = 5.7 × 103 + 8.02 × 101 × 103 = 5.7 × 103 + 80.2 × 103 = (5.7 + 80.2) × 103 = 85.9 × 103 = 8.59 × 101 × 103 = 8.59 × 104 Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 12 Matematik T3 Operasi darab dan bahagi Operations of multiplication and division __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ Sebuah bekas berukuran 125 cm × 65 cm × 31 cm. Hitung isi padu maksimum agar-agar yang boleh diisi, dalam, liter. Nyatakan jawapan dalam bentuk piawai dan betul kepada empat angka bererti. A container measures 125 cm × 65 cm × 31 cm. Calculate the maximum volume of jelly that can be filled in litres. State your answer in standard form and correct to four significant figures. Isi padu maksimum agar-agar dalam liter The maximum volume of jelly, in litres = 125 cm × 65 cm × 31 cm = 251 875 cm3 = 2.51875 × 105 cm3 = 2.51875 × 105 cm3 ÷ 1.0 × 103 cm3 = 2.51875 × 105–3 = 2.51875 × 102 liter /litres = 2.519 × 102 liter /litres Hukum Indeks / Law of Indices Operasi darab Operation of multiplication (A × 10m ) × (B × 10n ) = (A × B) × 10m + n Operasi bahagi Operation of division (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × 10m – n (9.6 × 104 ) ÷ (1.5 × 10–3 ) = (9.6 ÷ 1.5) × 104 – (–3) = 6.4 × 104 + 3 = 6.4 × 107 (2.58 × 10–7 ) ÷ (0.3 × 10–3 ) = (2.58 ÷ 0.3) × 10–7 – (–3) = 8.6 × 10–7 + 3 = 8.6 × 10–4 (8.4 × 105 ) ÷ (2 × 102 ) = (8.4 ÷ 2) × 105 – 2 = 4.2 × 103 6.2 × 105 × 3.3 × 102 = (6.2 × 3.3) × 105 + 2 = 20.46 × 107 = 2.046 × 101 × 107 = 2.046 × 108 2.5 × 10–3 × 5 × 10–6 = (2.5 × 5) × 10–3 + (–6) = 12.5 × 10–9 = 1.25 × 101 × 10–9 = 1.25 × 10–8 Alaf Sanjung Sdn Bhd
© Alaf Sanjung Sdn. Bhd. (516756-V) 13 Matematik T3 Menyelesaikan masalah yang melibatkan nombor dalam bentuk piawai Solve problems involving numbers in standard form Purata taburan hujan sebulan di Lembah Klang ialah 200 mm. Hitung taburan hujan setahun di kawasan Lembah Klang. Nyatakan jawapan dalam bentuk piawai betul kepada tiga angka bererti. A monthly rain distribution average in Klang Valley is 200 mm. Calculate the annual rain distribution in Klang Valley area. State the answer in standard form correct to three significant figures. Memahami masalah / Understanding the problem Purata taburan hujan sebulan : 200 mm A monthly rain distribution 1 tahun / year = 12 bulan / months Merancang strategi / Planning a strategy Taburan hujan setahun = purata taburan hujan sebulan × 12 bulan The annual rain distribution a monthly rain distribution average months Melaksanakan strategi / Implementing the strategy Membuat kesimpulan / Making a conclusion Taburan hujan setahun ialah 2.40 × 103 mm. The annual rain distribution is 2.40 × 103 mm. __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ Taburan hujan setahun / The annual rain distribution = 200 mm × 12 = 2 400 mm = 2.40 × 103 mm Sekeping dinding memerlukan 400 keping bata berukuran 250 mm panjang dan 120 mm lebar. Hitung luas dinding dalam, m 2 . Nyatakan jawapan anda betul kepada tiga angka bererti. A wall requires 400 pieces of brick measuring of 250 mm length and 120 mm width. Calculate the area of the wall in, m 2 . State your answer correct to three significant figures. Luas dinding / Area of the wall = 400 × 250 cm × 120 cm = 400 × 0.250 m × 0.120 m = 12.0 m2 Memahami masalah Understanding the problem Merancang strategi Planning a strategy Melaksanakan strategi Implementing the strategy Membuat kesimpulan Making a conclusion Alaf Sanjung Sdn Bhd