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GRADE / GRAAD 11 NOVEMBER 2012 …

Province of the EASTERN CAPE. EDUCATION. NATIONAL SENIOR. CERTIFICATE. NASIONALE. SENIOR SERTIFIKAAT. GRADE / GRAAD . 11. NOVEMBER 2012 . mathematics P1 / wiskunde V1. MEMORANDUM. MARKS: 150. PUNTE: This memorandum consists of 12 pages Hierdie memorandum bestaan uit 12 bladsye. 1. 2 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). QUESTION 1 / VRAAG 1. ( )( ). ( )( ) ( ) simplify LHS. vereenvoudig LK. ( )( ) standard form standaardvorm ( )( ) factorisation faktorisering or/of for both values of x vir beide x waardes (4). factorisation faktorisering ( )( ) -4 and/en 1. notation/ notasie (3). standard form standaardvorm ( ) ( ). method/Substitution ( ) ( )( ) metode/vervanging ( ). ( ).. simplification vereenvoudiging . one mark for each answer een punt vir elke antwoord (5).

1 province of the eastern cape education national senior certificate nasionale senior sertifikaat grade / graad 11 november 2012 mathematics p1 / wiskunde v1

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Transcription of GRADE / GRAAD 11 NOVEMBER 2012 …

1 Province of the EASTERN CAPE. EDUCATION. NATIONAL SENIOR. CERTIFICATE. NASIONALE. SENIOR SERTIFIKAAT. GRADE / GRAAD . 11. NOVEMBER 2012 . mathematics P1 / wiskunde V1. MEMORANDUM. MARKS: 150. PUNTE: This memorandum consists of 12 pages Hierdie memorandum bestaan uit 12 bladsye. 1. 2 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). QUESTION 1 / VRAAG 1. ( )( ). ( )( ) ( ) simplify LHS. vereenvoudig LK. ( )( ) standard form standaardvorm ( )( ) factorisation faktorisering or/of for both values of x vir beide x waardes (4). factorisation faktorisering ( )( ) -4 and/en 1. notation/ notasie (3). standard form standaardvorm ( ) ( ). method/Substitution ( ) ( )( ) metode/vervanging ( ). ( ).. simplification vereenvoudiging . one mark for each answer een punt vir elke antwoord (5).

2 X=3 y substitution ( ) ( ) vervanging 2(9 6y + ) simplification vereenvoudiging simplification vereenvoudiging ( )( ) factorisation faktorisering both values of/. albei waardes van y each value of/ elke waarde van x (8). ( NOVEMBER 2012 ) wiskunde V1 3. ( ) ( ) ( ) substitution vervanging simplification ( ) vereenvoudiging ( ) quadratic expression kwadratiese uitdrukking completion of square /. kwadraatsvoltooiing (4). 2. = 50 divide by 2. deel deur 2. same base x = 2 selfde basis answer/. antwoord (3). [27]. 3. 4 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). QUESTION2 / VRAAG 2. (0 ; 1) y-intercept y-afsnit (1). Asymptotes of : f(x): y = 0 1 mark for each h(x): x = 0 correct answer y = 5 1 punt vir elke korrekte antwoord (3). f and h are decreasing.

3 F en h verminder (2). 1 mark for each h correct curve /.. f 5 1 punt vir elke korrekte kromme 4. 3 1 mark for asymptote 2 y=5. 1 g 1 punt vir asimptoot y=5. -2 -1 -0 1 2 3 4 5. (4).. (1). answer antwoord (1). [12]. ( NOVEMBER 2012 ) wiskunde V1 5. QUESTION 3 / VRAAG 3. Tn = 3n2 + 2 substitution in T2 = 3(2)2 + 1 general term = 13 vervanging in T2 12 algemene term False/Vals simplification vereenvoudiging T2 12. deduction / afleiding (4). Shape number, n 1 2 3 4 5. Patroon nommer Number of white triangles 1 3 6 10 15. Aantal wit driehoeke Number of black triangles 0 1 3 6 10. Aantal swart driehoeke Total number of triangles 1 4 9 16 25. Totale aantal driehoeke 1 mark per correct entries 1 punt per korrekte inskrywing (6). = 144 triangles/driehoeke . answer/ antwoord (2).

4 5. 6 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). There is a common second difference of 1, so the sequence is quadratic. d=1. Daar is tweede gemene verskil van 1, daarom is die reeks kwadraties. T = a + bn + c a = value of a waarde van a =. equation / vergelyking b = OR/OF. value of b = 1 0- waarde van b = - equation / vergelyking c =. value of c = 0- ( ) waarde van c = 0. general term ( ) algemene term (7). equating to 190. ( ). gelykstel aan 190. ( ) simplification vereenvoudiging ( )( ) standard form standaardvorm factorisation faktorisering n = 20 (5). [24]. ( NOVEMBER 2012 ) wiskunde V1 7. QUESTION 4 / VRAAG 4. ( ) substitution of 12% and 6. ( ) vervanging van 12% en 6. substitution of R15 000. vervanging van R15 000. answer antwoord (3). / per maand answer antwoord (1).

5 1+ (. (). ). substituting into correct formula 1+ ( ) vervanging in die korrekte formule simplifying ( ) vereenvoudiging 0,16075. = 0,16075. value of r r = waarde van r = (4). ( ). substitution ( ) vervanging simplifying OR/OF ( ). vereenvoudiging ( ) answer antwoord OR/OF R7 097,59 (4). substitution ( ) vervanging simplifying vereenvoudiging ( ) answer antwoord (4). ( ) substituting vervanging ( ) ( ) ( ) simplifying vereenvoudiging answer antwoord (4). 7. 8 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). ( ) (( ). ( ). ( ) ( ) . ( ). method / metode answer antwoord (5). [25]. ( NOVEMBER 2012 ) wiskunde V1 9. QUESTION 5 / VRAAG 5. ( ) simplify vereenvoudig . (0 ; 1) (3). ( ). y = 0. ( ). ( ). both answers/. albei antwoorde (3). y 7 f: x-intercepts f x-afsnitte 6.)

6 Y-intercept 5 y-afsnit 4 turning point 3 draaipunt shape / vorm 2. 1 x -3 -2 -1 1 2 3 4. g: intercepts 5 6 7. -1 afsnitte asymptotes -2 (4;-2) asimptote -3 shape / vorm -4. -5. g -6. (7). ; . (2). -2 -2 (1). x 4. (2). [18]. 9. 10 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). QUESTION 6 / VRAAG 6. f(x) = 1 + a. substitution (0 ; 0) 0 = 1 + a. vervanging 0 = 1 + simplifying a = -1 vereenvoudiging (2). f(x) = 1 . substitution f(-15) = 1 vervanging answer = 0,99997 antwoord (2). f(x) = 1 . substitution (x ; 0,5) 0,5 = 1 vervanging = 1 0,5. express with same = 0,5 = base uitdruk met dieselfde basis =. answer antwoord x = -1 (3). h(x) = f(x 2) h(x). = 1 (2). [9]. ( NOVEMBER 2012 ) wiskunde V1 11. QUESTION 7 / VRAAG 7. equating to 0. gelykstel aan 0. ( )( ) factors /faktore x = -3 x = 1.

7 AB 4 units/eenhede AB = 4 units/eenhede /eenhede OC = 3 units/eenhede (5). x = -1 x = -1. (2). f(-1) = -(-1)2 -2(-1) + 3 substituting = -1 + 2 + 3 vervanging =4 simplification vereenvoudiging answer antwoord (3). mAC = numerator teller =1. denominator noemer answer / antwoord (3). m=1 m=1. subs/verv. (1;0) in y = x + c y =x+c 0=1+c 0 =1+c c = -1 c = -1. y =x 1 (4). [17]. 11. 12 MATHEMATIC P1 / wiskunde V1 ( NOVEMBER 2012 ). QUESTION 8 / VRAAG 8.. 0 (4). y x = 35. y = 20. Number of table models / Getal tafelmodelle 50. x + y = 50. 40. 4x + 5y = 200. 30. feasible region gangbare gebied 20. 10. 10 x 20 30 40 50. Number of wall models Getal muurmodelle (5). (25 ; 20) (25 ; 20). (30 ; 20) (30 ; 20). (35 ; 15) (35 ; 15). (35 ; 12) (35 ; 12) (4). P = 20x+ 10y P = 20x + 10y (1).

8 Max. profit: R850 for 35 wall models and 15 table models for max profit Mak. wins: R850 vir 35 muurmodelle en 15 tafelmodelle vir mak. wins Min. profit : R700 for 25 wall models and 20 table models for min profit Min, wins : R700 vir 25 muurmodelle en 20 tafelmodelle vir mak. wins (4). [18]. TOTAL/TOTAAL: 150.


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