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NATIONAL SENIOR CERTIFICATE/ NASIONALE …

NATIONAL . SENIOR certificate /. NASIONALE . SENIOR SERTIFIKAAT. GRADE/GRAAD 12. JUNE/JUNIE 2016. mathematics P1/ wiskunde V1. memorandum . MARKS/PUNTE: 150. This memorandum consists of 12 Hierdie memorandum bestaan uit 12 bladsye. 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). NOTE/LET OP: If a candidate answered a question TWICE, mark the FIRST attempt ONLY. Indien kandidaat vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging. Consistent accuracy(CA) applies in ALL aspects of the memorandum . Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum . If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien kandidaat poging vir vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is.

national senior certificate/ nasionale senior sertifikaat grade/graad 12 june/junie 2016 mathematics p1/wiskunde v1 memorandum marks/punte: 150

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Transcription of NATIONAL SENIOR CERTIFICATE/ NASIONALE …

1 NATIONAL . SENIOR certificate /. NASIONALE . SENIOR SERTIFIKAAT. GRADE/GRAAD 12. JUNE/JUNIE 2016. mathematics P1/ wiskunde V1. memorandum . MARKS/PUNTE: 150. This memorandum consists of 12 Hierdie memorandum bestaan uit 12 bladsye. 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). NOTE/LET OP: If a candidate answered a question TWICE, mark the FIRST attempt ONLY. Indien kandidaat vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging. Consistent accuracy(CA) applies in ALL aspects of the memorandum . Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum . If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien kandidaat poging vir vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is.

2 The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1. 2 2 7 = 0. (2 7) = 0 factorisation / faktorisering = 0 or/of 2 7 = 0 -values / waardes 7. =2. (3). 4. 4 + + 11 = 0. standard form / standaardvorm 4 2 + 11 + 4 = 0. 11 (11)2 4(4)(4). = Penalise 1 mark for substitution / vervanging 2(4). 11 57 incorrect rounding = 8 Penaliseer 1 punt vir = 0,43 or/of = 2,32 verkeerde afronding. -values / waardes 11 57. If stopped at : max. 2 marks 8. 11 57. As stop by : maks. 2 punte 8. (4). (2 1)( 3) > 0. + +. 1. 3 critical values with method 2 kritieke waardes met metode 1 3. 2. 1. < / > 3 answer / antwoord (3). 2. 3 . 3 +1 = 27 32 +1. 32 +1 = 33 33.

3 2 + 1 = 3 equating / gelykstel =1 answer / antwoord (4). Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 3. 3 + = 2 . = 2 3 .. (1). 4 2 + 2 = 2 + 7. 4 2 + (2 3)2 = 2 (2 3) + 7 substitution / vervanging 4 2 + 4 2 12 + 9 = 4 2 6 + 7 removing brackets / verwyder 4 2 + 4 2 12 + 9 4 2 + 6 7 = 0 hakies 4 2 6 + 2 = 0 standard form / standaardvorm If formula is used, award 2 2 3 + 1 = 0. factor's mark for factors / faktore (2 1)( 1) = 0. 1 substitution. -values / waardes = 2 or/of = 1 As formule gebruik word, -values / waardes = 2 or/of = 1 ken faktore punt toe vir substitusie. (6). ( ) = ( 2)( 2 6 + 10). Consider the quadratic factor : ( 2 6 + 10). = 2 4 . = ( 6)2 4(1)(10) substitution into delta /. = 36 40 vervanging in delta = 4.

4 => < 0, therefor NO Solutions answer / antwoord = 2 is the only solution. conclusion / gevolgtrekking (4). [23]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 2/VRAAG 2. 0 ; 1 ; 1 ; 6 ; 14 ; . 1 2 5 8 - 1st differences first differences / eerste verskille 3 3 3 - 2nd differences second differences / twee verskille (2). Next term/volgende term = 25 answer/antwoord (1). 2 = 3 3 + = 1 + + = 0 3. =2. 3 3 3 11. =2 3 (2) + = 1 2. 2 + =0 = . 11. 11 2. = 2. =4 =4. 3 2 11. = +4. 2 2 answer / antwoord (4). 3 11 substitution / vervanging 30 = (30)2 (30) + 4. 2 2. = 1189 answer / antwoord (2). 3 2 = 4 3. 13 = 27 method / metode 2 = 40. = 20. values of a and b /. 13 = 7 waardes van a en b =6 (2). + ( 1) = formula / formule 6 + ( 1)(7) = 230 substitution / vervanging 7 = 231.

5 = 33 answer / antwoord (3). 1 =. 1 . = 5. 5. 1 substitution / vervanging 1 < <1. 5. 5 < 1 < 5. 6 < < 4. answer / antwoord (3). 4 < < 6. (1 ). = + . method / metode 1 . 1 20. 16 (1 (2) ) substitution / vervanging 40 = + 20(3). 1. 1 2. = 31,99 + 60 simplification / vereenvoudiging 92 answer / antwoord (4). Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 5.. 1 1. 16 ( ) . 2. =1 =1. 1 1. 16 (2) (2).. =. 1 . 16 substitution / vervanging =. 1. 1 2. answer / antwoord (2). = 32. [25]. QUESTION 3/VRAAG 3. = 2 (horizontal asymptote / horisontale asimptoot) answer / antwoord (1). = 1 (vertical asymptote / vertikale asimptoot) answer / antwoord (1). 3. 2=0 y=0. 1. 3. =2. 1. 3 = 2 2. 5 = 2 . 5 x intercept / afsnit =. 2. 3. = 2.

6 0 1. y intercept / afsnit (3). = 5. asymptotes / asimptote x-and y-intercepts / x-en y-afsnitte shape / vorm (3). (4 ; 5) (2). [10]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 4/VRAAG 4.. =. 2 . ( 2). = x-coordinate / ko rdinaat 2( 1). = 1. = ( 1)2 2( 1) + 3 y-coordinate / ko rdinaat = 1 + 2 + 3. =4 ( 1 ; 4) coordinates of C / ko rdinate van C. (3). 2 2 + 3 = 0. 2 + 2 3 = 0 standard form / standaardvorm ( + 3)( 1) = 0 factors / faktore + 3 = 0 1 = 0. = 3 = 1 both values / beide waardes ( 3; 0) (1; 0) (3). m=2 & c=6 value of m / waarde van m y = 2x + 6 value of c / waarde van c (2). 2 = (1)2 + (2)2 method / metode =5 substitution / vervanging = 5. answer / antwoord OR. C(-1 ; 4) and/en E(0 ; 6). = (0 + 1)2 + (6 4)2.

7 = 5 (3). >1 answer / antwoord (2). [13]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 7. QUESTION 5/VRAAG 5. ( ) = . substitution / vervanging 27 3. 8. = . 3 3. ( ) = 3 answer / antwoord 2. 3 (2). = 2. 3 . =( ). 2. 3 swop and / ruil en om =( ). 2. = 3 answer / antwoord 2. (2). 3 = 1 OR other method (eg. sketch). 2. equating / gelykstel 3 1. =( ). 2. 2 answer / antwoord =. 3 (2). answer / antwoord (1). y asymptote / asimptoot negative y-intercept / negatiewe y-afsnit x shape / vorm q (3). [10]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 6/VRAAG 6.. 1 + = (1 + ) formula / formule . 8,5 4. = (1 + ) 1 substitution / vervanging 400. = 0,0877. / = 8,77% . answer / antwoord (3).

8 = (1 + ) formula / formule 20. 8,5. = 12 000 (1 + ) substitusie / vervanging 400. = 18 273,54 answer / antwoord (3). = (1 ) . 12,4 substitution / vervanging 41 611,57 = 120 000 (1 ). 100. simplification / vereenvoudiging 0,3467630833 = 0,876 . correct use of logs / korrekte log(0,3467630833) gebruik van logs =. log(0,876). answer / antwoord 8 (4). 8 12 8 8 8 4. 20 000 = (1 + ) + (1 + ) + (1 + ) setting up equation / opstel 400 400 400. van vergelyking 8 12 8 8 8 4. 20 000 = [(1 + ) + (1 + ) + (1 + ) ]. 400 400 400. 20 000 x the subject of the formula /. = 12. 8 8 8 8 4 x die onderwerp van die formule [(1 + 400) + (1 + 400) + (1 + 400) ]. answer / antwoord = 5 678,05. (4). [14]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 9.

9 QUESTION 7/VRAAG 7. ( ) = 2 2 3 . ( + ) = 2( + )2 3( + ) substitute / vervang ( + ). = 2( 2 + 2 + 2 ) 3( + ). = 2 2 + 4 + 2 2 3 3 . ( + ) ( ) formula / formule ( ) = lim 0 Penalise 1 mark for incorrect use of formula. Must show ( ). 2 2 + 4 + 2 2 3 3 (2 2 3 ) Penaliseer 1 punt vir verkeerde = lim gebruik van formule. Moet ( ). 0 . toon. 2 2 + 4 + 2 2 3 3 2 2 + 3 . = lim 0 . 4 + 2 2 3 simplification / vereenvoudiging = lim 0 . (4 + 2 3). = lim common factor / gemene faktor 0 . = lim (4 + 2 3). 0. Answer ONLY: 0 marks answer / antwoord = 4 3 SLEGS antwoord: 0 punte (5). 3 Penalise 1 mark for incorrect notation. = 2 . 5 2 Penaliseer 1 punt vir verkeerde notasie. 1 3. = 2 2 1 1. 3. 5 2 2 5 1. 1 3. = 2 + 2 1. 3. 5 2 5 2. 1 3. = +. 5 2 (4). [9]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016).

10 QUESTION 8/VRAAG 8. ( ) = 3 4 2 11 + 20. ( ) = 3 2 8 11 = 0 ( ) = 0. (3 11)( + 1) = 0. 3 11 = 0 + 1 = 0 factors / faktore 11. = (3,7) = 1 x-values / waardes 3. 400 y-values / waardes = ( 14,8) = 36. 27. ( 1 ; 36) &. 11. ( ; . 400. ) / (3,7 ; -14,8) coordinates / ko rdinate 3 27 (5). ( ) = 6 8 = 0 ( ) = 0. 6 = 8. 8 1. = 6 or/of 1 3 or/of 1, 33 answer / antwoord (2). OR. 11. 1+. 3. = 2. 4. = or/of 1,33. 3. = (2) = 3(2)2 8(2) 11. = 15 = 15. (2) = (2)3 4(2)2 11(2) + 30. =0 (2) = 0. 1 = ( 1 ). 0 = 15( 2) substitution / vervanging = 15 + 30 answer / antwoord (4). 36 units downwards answer / antwoord 14,8 units upwards answer / antwoord (2). = 9 answer / antwoord (1). = 1 = 5 answers / antwoorde (2). < 1 > 5 <1 >5 (2). [18]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 11.


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