Engineering Mathematics by John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE, FIIE, FIMA, FCollT doc

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Engineering Mathematics by John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE, FIIE, FIMA, FCollT doc

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[...]... Indices,standard form and engineering notation 17 Section 1 18 Engineering Mathematics Table 2.1 Prefix Meaning T tera multiply by 1 000 000 000 000 (i.e × 1012 ) G giga multiply by 1 000 000 000 (i.e × 109 ) M mega multiply by 1 000 000 (i.e × 106 ) k kilo multiply by 1000 (i.e × 103 ) m milli divide by 1000 (i.e × 10−3 ) μ micro divide by 1 000 000 (i.e × 10−6 ) n nano divide by 1 000 000 000 (i.e ×... introductory/access/foundation course involving Engineering Mathematics at University, Colleges of Further and Higher education and in schools Each topic considered in the text is presented in a way that assumes in the reader little previous knowledge of that topic Engineering Mathematics 5th Edition provides a follow-up to Basic Engineering Mathematics and a lead into Higher Engineering Mathematics 5th Edition This textbook... the publishers via the internet – see below A list of Essential Formulae is included in the Instructor’s Manual for convenience of reference Learning by Example is at the heart of Engineering Mathematics 5th Edition JOHN BIRD Royal Naval School of Marine Engineering, HMS Sultan, formerly University of Portsmouth and Highbury College, Portsmouth Free web downloads Additional material on statistics Chapters... value, 4375 10 000 By cancelling 4375 875 175 35 7 = = = = 10 000 2000 400 80 16 7 i.e 0.4375 = 16 285 57 (b) Similarly, 4.285 = 4 =4 1000 200 Problem 20 Express as decimal fractions: (a) 9 7 and (b) 5 16 8 (a) To convert a proper fraction to a decimal fraction, the numerator is divided by the denominator Division by 16 can be done by the long division method, or, more simply, by dividing by 2 and then... multiplication of matrices 61.3 The unit matrix 61.4 The determinant of a 2 by 2 matrix 61.5 The inverse or reciprocal of a 2 by 2 matrix 61.6 The determinant of a 3 by 3 matrix 61.7 The inverse or reciprocal of a 3 by 3 matrix 62 The solution of simultaneous equations by matrices and determinants 62.1 Solution of simultaneous equations by matrices 553 Section 11 555 515 515 Differential Equations 520 63... material will provide engineering applications and mathematical principles necessary for advancement onto a range of Incorporated Engineer degree profiles It is widely recognised that a students’ ability to use mathematics is a key element in determining subsequent success First year undergraduates who need some remedial mathematics will also find this book meets their needs In Engineering Mathematics 5th... (ii) 4 decimal places 37.81 ÷ 1.7 = 37.81 1.7 The denominator is changed into an integer by multiplying by 10 The numerator is also multiplied by 10 to keep the fraction the same Thus 37.81 ÷ 1.7 = = 37.81 × 10 1.7 × 10 378.1 17 Section 1 Revision of fractions,decimals and percentages 7 Section 1 8 Engineering Mathematics The long division is similar to the long division of integers and the first four... of × 7 15 Dividing numerator and denominator by 3 gives: 1 14 14 1 × 14 ¡ = × × = & 7 7 5 7×5 155 & Dividing numerator and denominator by 7 gives: Multiplying both numerator and denominator by the reciprocal of the denominator gives: 13 & 21 3 ¡ & 3 3 × 3 & 12 7 &4 7 = 1¡ = 4 = 1 12 1 12 1 4 & & & 21 & × 21 & & 12 21 & 1 1& This method can be remembered by the rule: invert the second fraction and change... to the nearest 1% Working in minute units, 2 hours = 120 minutes 25 Hence 25 minutes is ths of 2 hours By cancelling, 120 25 5 = 120 24 5 ˙ Expressing as a decimal fraction gives 0.2083 24 Section 1 Revision of fractions,decimals and percentages 9 Section 1 10 Engineering Mathematics Multiplying by 100 to convert the decimal fraction to a percentage gives: ˙ 0.2083 × 100 = 20.83% Thus 25 minutes is... × 54 Dividing each term by the HCF (i.e highest common factor) of the three terms, i.e 32 × 53 , gives: 33 × 5 3 32 × 55 + 2 2 × 53 32 × 55 + 33 × 53 3 × 53 = 3 4 × 54 4 × 54 3 3 32 × 53 = 3(2−2) × 5(5−3) + 3(3−2) × 50 3(4−2) × 5(4−3) 30 × 52 + 31 × 50 32 × 5 1 28 1 × 25 + 3 × 1 = = 9×5 45 = Section 1 Indices,standard form and engineering notation 13 Section 1 14 Engineering Mathematics Problem 12 Find . page iii Engineering Mathematics Fifth edition John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE, FIIE, FIMA, FCollT AMSTERDAM • BOSTON • HEIDELBERG • LONDON. of that topic. Engineering Mathematics 5th Edition provides a follow-up to Basic Engineering Mathematics and a lead into Higher Engineering Mathematics 5th Edition. This

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Mục lục

  • Engineering Mathematics, Fifth edition

  • Copyright Page

  • Contents

  • Preface

  • Section 1 Number and Algebra

    • Chapter 1 Revision of fractions, decimals and percentages

      • 1.1 Fractions

      • 1.2 Ratio and proportion

      • 1.3 Decimals

      • 1.4 Percentages

      • Chapter 2 Indices, standard form and engineering notation

        • 2.1 Indices

        • 2.2 Worked problems on indices

        • 2.3 Further worked problems on indices

        • 2.4 Standard form

        • 2.5 Worked problems on standard form

        • 2.6 Further worked problems on standard form

        • 2.7 Engineering notation and common prefixes

        • Chapter 3 Computer numbering systems

          • 3.1 Binary numbers

          • 3.2 Conversion of binary to decimal

          • 3.3 Conversion of decimal to binary

          • 3.4 Conversion of decimal to binary via octal

          • 3.5 Hexadecimal numbers

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