Understanding Engineering Mathematics

Thehighest common factor(HCF) of a set of integers is the largest integer which is a factor of all numbers of the set. For small ...

(v) 3, 6, 15, 27. Here 3 is the only number that divides them all and is the HCF in this case. C. (i) Since 3, 7 are both primes ...

Brackets can be used if we want to override such rules. For example: 2 ×( 3 + 5 )= 2 × 8 = 16 In general, an arithmetic expressi ...

(viii) − 2 −( 4 − 5 )=− 2 −(− 1 ) =− 2 + 1 =− 1 (ix)( 4 ÷(− 2 ))× 3 − 4 = ( 4 − 2 ) × 3 − 4 =(− 2 )× 3 − 4 =− 6 − 4 =− 10 (x)( 3 ...

this way is thepercentage. Thus we usually express 32/100 as ‘32 percent’ rather than as its equivalent, ‘8 out of 25’! Fraction ...

or, with 6 the LCM of 2 and 3 = 3 6 + 12 6 + 4 6 = 19 6 (−^1 ) and so the equivalent resistance is R= 6 19  Finally, on fracti ...

(iii) 7 3 × 4 7 = 4 3 (iv) 7 5 × 3 14 = 1 5 × 3 2 (cancelling 7 from top and bottom, as in (iii)) = 1 × 3 5 × 2 = 3 10 (v) 3 4 ÷ ...

B. Ifa:b=3:2 then a b = 3 2 sob= 2 a 3 .So,ifa=6thenb= 2 × 6 3 =4. 1.2.6 Factorial and combinatorial notation – permutations and ...

Each of these is the samecombinationof the objects A, B, C – that is a selection of three objects in which order is not importan ...

Solution to review question 1.1.6 A. (i) 3!= 3 ( 3 − 1 )( 3 − 2 )= 3 × 2 × 1 = 6 (ii) 6!= 6 × 5 × 4 × 3 × 2 × 1 = 720 (iii) 24! ...

am an =am−n (am)n=amn (ab)n=anbn Note that for any indexn,1n=1. Examples 23 × 24 = 23 +^4 = 27 35 × 30 = 35 ( 52 )^3 = 56 ( 22 ) ...

and in general √na=an^1 This fits in with the rules of indices, since ( a 1 n )n =a 1 n×n=a^1 =a Fractional powers satisfy the s ...

(vi)(− 6 )^2 ( − 3 2 ) 3 =(− 1 )^262 (− 1 )^3 33 23 =− 6233 23 on using(− 1 )^2 =1,(− 1 )^3 =− 1 =− ( 2 × 3 )^233 23 =− 223233 2 ...

Thus: (√ 3 + 2 √ 2 ) (√ 3 − √ 2 ) × (√ 3 + √ 2 √ 3 + √ 2 ) = (√ 3 + 2 √ 2 )(√ 3 + √ 2 ) (√ 3 − √ 2 )(√ 3 + √ 2 ) = (√ 3 ) 2 + 3 ...

representation of √ 2 is non-terminating: √ 2 = 1. 4142135623 ... that is, the decimal part goes on forever. All quantities meas ...

When we ‘round’ a number we change the last non-zero digit not removed according to the size of the digits dropped. Specifically ...

(d) By long division we find: 1 7 =^0.^142857142857 ...=^0. ̇ 142857 ̇ and the string of digits 142857 recur indefinitely as den ...

calculator, or we need to get a rough order of magnitude check on a messy calculation. In such situations the engineer’s most po ...

1.3 Reinforcement 1.3.1 Types of numbers ➤➤ 25 ➤ A.Say what you can about the type and nature of the following numbers: (i) 2 (i ...

1.3.3 Highest common factor and lowest common multiple ➤➤ 38 ➤ A.Express in terms of prime factors (i) 2 (ii) − 6 (iii) 21 (iv) ...