3.1.2 ( a 3 b −2 c ) 5 × ( b c 2 ) −1 = a 15 b −11 c 3 = a 15 c 3
____ b 11
3.1.3 2 2n − 1 × 2 n + 4
___________ 2 3n + 3 = 2 3n + 3
______ 2 3n + 3 = 1
3.2.1 0,000006754 = 6,754 × 1 0 −6 3.2.2 765 000 000 000 = 7,65 × 1 0 11
4.1.1 a = 99 4.1.2 b = 16 4.1.3 c = −66 4.1.4 d = 25
4.2.1 Number of hexagons (h) Number of triangles (t)
Perimeter of shape (p) (cm) 4.2.2 t = 6h 4.2.3 p = 10h + 5 4.2.4 i) p = 10 ( 250 ) + 5 = 2 505 cm
ii) 10h + 5 = 495 h = 49 t = 6h t = 6 ( 49 ) = 294 triangles
5.1.1 16x − 3xy + 5 ( 4x − 2y ) = 16x − 3xy + 20x − 10y = 36x − 3xy − 10y
5.1.2 −3k ( 2k + 2 ) + k ( k − 1 ) = −6 k 2 − 6k + k 2 − k = −5 k 2 − 7k
5.1.4 2x − 5 [ x − ( 3x + 4 ) − 3 2 ] = 2x − 5 ( − 2x − 13 ) = 2x + 10x + 65 = 12x + 65
5.1.5 ( 6x − 4 ) ( 3x + 2 ) = 18 x 2 − 12x + 12x − 8 = 18 x 2 − 8
5.1.6 (a − b ) 2 = a 2 − 2ab + b 2 5.2.3 x
5.2.1 x + 3 ( x + 1 ) = 2x + 8 x + 3x + 3 = 2x + 8 2x = 5 x = 2,5
__ 2 − 1
__ 4 = 7
__ 3
___ 12
6x − 3 = 7 6x = 10 x = 5
Chapter 17: Programme of Assessment
5.1.3 3 + 4a ( a − 2b ) − 2 + 5b ( a − 3b ) = 3 + 4 a 2 − 8ab − 2 + 5ab − 15 b 2 = 4 a 2 − 3ab − 15 b 2 + 1
1 6
15 2
12 25
3
18 35
4
24 45
5
30 55
(1) (1) (1)
(2)
(2) (1)
(1) [8] (1) (1) (1) (1) (3)
(1) [11] (2) (2) (2) (3)
(2) (1)
(2)
(2) [16] 325
CHAPTER 17
Page 1 |
Page 2 |
Page 3 |
Page 4 |
Page 5 |
Page 6 |
Page 7 |
Page 8 |
Page 9 |
Page 10 |
Page 11 |
Page 12 |
Page 13 |
Page 14 |
Page 15 |
Page 16 |
Page 17 |
Page 18 |
Page 19 |
Page 20 |
Page 21 |
Page 22 |
Page 23 |
Page 24 |
Page 25 |
Page 26 |
Page 27 |
Page 28 |
Page 29 |
Page 30 |
Page 31 |
Page 32 |
Page 33 |
Page 34 |
Page 35 |
Page 36 |
Page 37 |
Page 38 |
Page 39 |
Page 40 |
Page 41 |
Page 42 |
Page 43 |
Page 44 |
Page 45 |
Page 46 |
Page 47 |
Page 48 |
Page 49 |
Page 50 |
Page 51 |
Page 52 |
Page 53 |
Page 54 |
Page 55 |
Page 56 |
Page 57 |
Page 58 |
Page 59 |
Page 60 |
Page 61 |
Page 62 |
Page 63 |
Page 64 |
Page 65 |
Page 66 |
Page 67 |
Page 68 |
Page 69 |
Page 70 |
Page 71
