november 2008 maths paper 2

MARK SCHEME for the October/November 2008 question paper 0580 and 0581 MATHEMATICS 0580/21 and 0581/21 Paper 21 (Extended) maximum raw mark 70 This mark 

MARK SCHEME for the October/November 2008 question paper 4024 MATHEMATICS 4024/02 Paper 2 maximum raw mark 100 This mark scheme is published as an aid to 

MARK SCHEME for the October/November 2008 question paper 9709 MATHEMATICS 9709/02 Paper 2 maximum raw mark 50 This mark scheme is published as an aid to 

NOVEMBER 2008 MATHEMATICS – SECOND PAPER IXESHA: 3 iiyure TIME: 3 hours TYD: 3 uur AMANQAKU: 150 MARKS: 150 PUNTE: 150 Write on the cover of the answer book after the word “Subject”- MATHEMATICS – SECOND PAPER This question paper consists of 10 pages 1 diagram sheet and a formula sheet

This mark scheme is published as an aid to teachers and candidates to indicate the requirements of the examination It shows the basis on which Examiners 

CIE IGCSE Maths (0580/21) Paper 2 (Extended) Oct/Nov 2008 Model Answers Time allowed: 90 minutes Score: /70 Percentage: /100 Grade Boundaries: 80 67 54 43 33 For this diagram write down (a) the order of rotational symmetry 2 (b) the number of lines of symmetry 0 [1] [1] The answer to this matrix multiplication is of order a × b

This question paper consists of 9 pages an information sheet and 4 diagram sheets MATHEMATICS P2 NOVEMBER 2008 NATIONAL SENIOR CERTIFICATE GRADE 12 

NOVEMBER 2008 MATHEMATICS: PAPER II Time: 3 hours 150 marks PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1 This question paper consists of 16 pages a Formula Sheet and an Answer Booklet of 4 pages (i – iv) Please check that your paper is complete 2 Write your examination number in the space provided in your Answer Booklet 3

This question paper consists of 12 questions Answer ALL the questions Microsoft Word - Mathematics P2 Eng Nov 2008 doc Author: Khoele k Created Date:

Mathematics/P2 7 DoE/November 2008 NSC - Memorandum QUESTION 2 2 1 169 (5)2 (12)2 2 2 = = + r =OQ ∴x2 +y2 =169 OR x2 +y2 =(5)2 +(12)2 =169 9 substituting (5 ; 12) into x2 +y2 9 169 9x2 +y2 =169 (3) 9 x2 +y2 =r2 9 substitution coordinates 9 169 (3) Answer only: Full marks 2 2 12 5 12 0) 5 0 = − − = PQ PQ m m y x 12 5 ∴ = 9 gradient c = 0