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Model Answers in Ordinary National Certificate Mathematics for Engineers
Model Answers in Ordinary National Certificate Mathematics for Engineers
Model Answers in Ordinary National Certificate Mathematics for Engineers
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Model Answers in Ordinary National Certificate Mathematics for Engineers

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Model Answers in Ordinary National Certificate Mathematics for Engineers presents a series of model answers that include all the topics covered by the many different syllabuses in Mathematics for the Ordinary National Certificate in Engineering. This book is composed 16 chapters; each chapter contains Worked Examples, Hinted Examples, and Further Examples. The opening chapters tackle the topics of logarithms, transformation and evaluation formula, progressions, binomial expansions, and algebraic equations. The succeeding chapters explore the topics of determination of laws, mensuration, equations and identities in trigonometry, solutions of triangles in trigonometry, and graphical solutions. The remaining chapters provides model answers for rates of change; maxima and minima; integration; areas, volume, centroids, and second moments of area; and Simpson’s rule. This book is directed towards mathematicians, and mathematics teachers and students.
LanguageEnglish
Release dateJun 28, 2014
ISBN9781483226934
Model Answers in Ordinary National Certificate Mathematics for Engineers

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    Book preview

    Model Answers in Ordinary National Certificate Mathematics for Engineers - D. W. Hilder

    1965

    MODEL ANSWERS

    Ordinary National Certificate Mathematics for Engineers

    LOGARITHMS

    WORKED EXAMPLES

    No. 1

    (a) Evaluate (i) log4 215, (ii) loge 0·6.

    (b) Evaluate without using tables

    (a) 

    (i) 

    (ii) 

    (b) 

    Note

    When working with logarithms the fourth figure is unreliable, since the same calculation done in different ways may give slight variations in the fourth figure. Hence it is usual to express answers correct to three significant figures.

    No. 2

    (a) Evaluate (i) (0·36)−2·1, (ii) 1·76²·⁴⁶ × 0·378¹·⁵⁶.

    (b) Solve the following equation for x

    (a) 

    (i) Let x = (0·36)−2·1

    (ii) Let x = 1·76²·⁴⁶ × 0·378¹·⁵⁶

    (b) 4·15(x−3) × 3 = 3·09(x−4).

    Taking logarithms of both sides gives

    No. 3

    (a) Solve the equations (i) (2x)²·⁵ = 50, (ii) (2·5)²x = 50.

    (b) If pvn = c, p = 120 when v = 100 and p = 760 when v = 25, calculate the values of n and

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