# Mathematics Grade 12 P1 Eastern Cape Mock 2019 Memo (Afr) > Grade 12 Mathematics Paper 1 memorandum from the Eastern Cape Mock Exam 2019 (Afrikaans), 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2019-english-afrikaans/1ebc5faf-3cdc-48cd-bf4e-e2320655091b - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2019 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 19 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 P1 Eastern Cape Mock 2019 Memo (Afr)](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2019-english-afrikaans/1ebc5faf-3cdc-48cd-bf4e-e2320655091b) ## About this document Mathematics · Grade 12 · Eastern Cape Mock Exam · 2019 · Afrikaans. Memorandum, 19 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 SEPTEMBER 2019 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 19 pages./ Hierdie nasienriglyn bestaan uit 19 bladsye. ### Page 2 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) NOTE/LET WEL: If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging. Consistent accuracy applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn. If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is. The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word vir substitusie in die korrekte formule toegeken. QUESTION 1/VRAAG 1 1.1.1 𝑥 2 − 3𝑥 − 4 = 0 (𝑥 + 1)(𝑥 − 4) = 0 factors/faktore 𝑥 = −1 or/𝑜𝑓 4 𝑥 = −1 𝑥=4 Answers only: Antwoorde alleen (2/3) OR/OF Can use quadratic formula / Kan kwadratiese formule gebruik b b 2 4ac x 2a ( 3) ( 3) 2 4(1)( 4) correct substitution / 2(1) korrekte vervanging 3 25 2 answers / antwoorde x 4 or / of x 1 (3) 1.1.2 2𝑥 2 − 𝑥 − 7 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 ( 1) ( 1) 2 (4)(2)( 7) substitution/vervanging x 2(2) 1 57 Penalise 1 mark for incorrect rounding off 4 Penaliseer 1 punt vir verkeerde afronding 𝑥 = 2,14 𝑥 = −1,64 x 2,14 or/of 1, 64 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 3 1.1.3 5x 1 5x 2500 5 x.51 5 x 2500 5 x (5 1) 2500 factorisation/faktorisering 5 x.4 2500 x 5 625 5x 625 x 4 5 5 x 4 x 4 (3) 1.1.4 ( x 3)( x 1) 12 x 2 2 x 3 12 0 x 2 2 x 15 0 standard form/standaardvorm ( x 5)( x 3) 0 factorisation/faktorisering OR/OF + − + −3 5 −3 5 3 x 5 (accuracy) (akkuraatheid) 3 x 5 OR / OF x ( 3;5) (4) 1.2 y 2x 1 .....(1) y 2x 1 2 2 3x xy y 1 .....(2) (1) into 2 3x 2 x(2 x 1) (2 x 1) 2 1 substitution/vervanging 3x 2 2x2 x (4 x 2 4 x 1) 1 3x 2 2x2 x 4x2 4x 1 1 0 3x 2 5 x 2 0 standard form/standaardvorm 3x 2 5 x 2 0 (3 x 2)( x 1) 0 factorisation/faktorisering 2 x or/of x 1 x-values/waardes 3 2 y 2 1 or/of y 2(1) 1 3 y-values/waardes 1 (6) y or/of y 1 3 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) OR/OF y 1 x 2 y 1 x ...(1) 2 3 x xy y 2 2 1 ... 2 substitution/vervanging (1) into 2 , 2 y 1 y 1 3 y y2 1 2 2 y2 2y 1 y2 y 3 y2 1 4 2 3y2 6 y 3 2 y2 2 y 4 y2 4 0 3y2 4y 1 0 standard form/standaardvorm 3y2 4 y 1 0 (3 y 1)( y 1) 0 factorisation/faktorisering 1 y or y 1 y-values/waardes 3 /of 1 3 1 1 1 x or/of x 2 2 2 x-values/waardes x or/of x 1 3 (6) 1.3 f ( x) x 2 2 px 8 2 p For equal roots :Vir gelyke wortels : b2 4ac 0 b2 4ac 0 substitution/vervanging ( 2 p)2 4(1)(2 p 8) 0 4 p 2 8 p 32 0 2 p 2p 8 0 ( p 2)( p 4) 0 p 2 or / of p 4 p values / waardes but / maar : p 0 p 2 So, f ( x) x2 4x 4 h( x ) x2 4x 1 h( x) x2 4 x 1 x2 4x 4 4 1 ( x 2) 2 3 answer in coordinate form/ TP : ( 2; 3) antwoord in koördinaatvorm (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 5 1.3 OR/OF b 2 4ac 0 substitution/vervanging b2 4ac 0 ( 2 p) 2 4(1)(8 2 p ) 0 4 p2 8 p 32 0 p2 2p 8 0 ( p 4)( p 2) 0 p 4 or/of p 2 p values/waardes Turning point of / Draaipunt van f is 2;0 ( 2 ; 0) Turning point of / Draaipunt van h is 2; 3 ( 2 ; 3) (5) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) QUESTION 2/VRAAG 2 2.1.1 3 ; 1 ; −3 ; −9 −2 −4 −6 −2 −2 both terms/beide terme 17 ; 27 (1) 2.1.2 2a 2 3a b 2 a b c 3 a 1 3( 1) b 2 1 1 c 3 a 1 b 1 c 3 b 1 c 3 Tn n2 n 3 Tn n2 n 3 (4) 2.1.3 n 2 n 3 809 equating Tn to −809 2 stel Tn gelyk aan −809 n n 812 0 factors/faktore (n 29)(n 28) 0 choosing/kies 𝑛 = 29 n 29 (3) 2.2.1 Tn 2n 3 T53 2(53) 3 substituting into 𝑇53 / vervanging in 𝑇53 answer/antwoord 103 OR/OF T53 a 52d substituting into 𝑇53 1 52(2) vervanging in 𝑇53 103 (2) 103 2.2.2 Sn 2 [ 2a n (n 1)d ] substitution into correct formula S 29 29 2 [2( 1) 28(2)] vervanging in korrekte formule 783 783 (2) 2.2.3 29 29 (2n 3) 783 2n 3 n 1 n 1 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 7 2.3 T4 a 3d and/en T10 a 9d T10 T4 6d T10 T4 6d 6d (8x 2 y) (2 x y) 6d (8 x 2 y ) (2 x y) d x 12 y 6x 3y substitution/vervanging d x 1 2 y T4 a 3d value of a / waarde van a 2x y a 3( x 1 2 y) 2x y a 3x 3 2 y (5) a 5 2 y x [19] QUESTION 3/VRAAG 3 3.1 T1 ( x 1) T2 ( x 1) 2 r x 1 for convergence :/ vir konvergensie 1 r 1, 1 r 1 1 x 1 1 0 x 2 answer/antwoord (2) 3.2 When / Wanneer : x 2 3 , p ( 23 1) ( 23 1) 2 ( 23 1)3 ... substituting for x p ( 1 ) ( ) ( 1 1 ) ... vervanging vir x 3 9 27 values for a and r a 1 3 and / en r 1 3 waardes vir a en r a S 1 r substituting into S formula 1 3 vervanging in S formule 1 ( 1 3 ) answer/antwoord (4) 1 [6] 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) QUESTION 4/VRAAG 4 4.1 x 3 x 3 y 1 y 1 (2) 4.2 2 1 0 substitution/vervanging x 3 2 1 x 3 2 x 3 x 5 x-intercept/x-afsnit 2 y 1 0 3 5 y-intercept/y-afsnit 3 (3) 4.3 asymptotes / asimptote x-intercept / x-afsnit y-intercept / y-afsnit shape / vorm (4) 4.4 2 2 h( x ) 1 h( x ) 1 x 3 x 3 point of intersection of asymptotes/ snypunt van asimptote ( 3 ; 1) or / of y ( x p) q y x ( 3) 1 or / of y x 3 1 substitute point of intersection of asymptotes / vervang die snypunt van asimptote y x 2 answer/antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 9 4.4 OR/OF 2 h( x ) 1 x 3 2 h( x ) 1 x 3 point of intersection of asymptotes/ snypunt van asimptote substitute point of intersection of ( 3 ; 1) asymptotes / vervang die snypunt van asimptote y x k 1 3 k answer/antwoord k 2 (4) y x 2 [13] QUESTION 5/VRAAG 5 5.1 (0 ; 8) answer / antwoord (1) 5.2 y mx c c 8 y mx 8 substituting T(9;10) into 10 9m 8 equation of line / vervanging van T(9;10) in m 2 vergelyking van lyn equation / vergelyking y 2x 8 OR/OF 10 ( 8) substituting T and Q into 𝑚 𝑇𝑄 mTQ 9 0 vervanging van T en Q in 𝑚 𝑇𝑄 m 2 m=2 equation y 2x 8 (3) 5.3 y x 2 7x 8 x2 7 x ( 7 2 2) 8 ( 7 2 2) completing the square / (x 7 2 ) 81 vierkantsvoltooiing 2 4 equation / vergelyking (2) 5.4 x- coordinate/koördinaat ( 72 ; 81 4 ) y- coordinate/koördinaat (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) 5.5 Ave gradient/Gem. gradiënt y 10 1 method/metode x 9 y 10 x 9 making y the subject y x 1 maak y die onderwerp f ( x) x2 7 x 8 equating 2 equations x 1 x2 7 x 8 gelykstel van 2 vergelykings factors/faktore 0 x2 8x 9 0 ( x 9)( x 1) x 9 or / of 1 specifying coordinates for W / spesifiseer W se koördinate y 10 or / of 0 ∴ W(−1 ; 0) OR/OF x 2 7 x 8 (10) x 2 7 x 8 (10) x (9) 1 x (9) equating to 1 / gelykstel aan 1 standard form/standaardvorm x 2 7 x 18 x 9 factors/faktore x2 8x 9 0 ( x 9)( x 1) 0 specifying coordinates for W. x 9 or / of x 1 spesifiseer W se koördinate y 10 or / of y 0 ∴ W(−1 ; 0) OR/OF f ( x) 2x 7 f ( x) 2x 7 f (9) 2(9) 7 f (9) 11 11 f (9) f ( x) average gradient = 1 1 2 gemiddelde gradiënt = 1 11 2 x 7 substitution/vervanging 1 2 2x 4 1 2 x 1 1 x 1 y 0 coordinates of W / W 1; 0 koördinate van W (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 11 5.6 x2 7 x 8 0 ( x 8)( x 1) 0 P( 1; 0) and / en R (8 ; 0) x intercepts of f y 2x 8 x-afsnitte van f 0 2x 8 x intercept of g V( 4 ; 0 ) x-afsnit van g x 1 accuracy/akkuraatheid x 1 or / of 4 x 8 4 x 8 accuracy/ OR / OF akkuraatheid x ( ; 1) (4 ; 8) (4) [17] QUESTION 6/VRAAG 6 6.1 f ( x) log m x substitution/vervanging 3 log m 64 m3 64 answer/antwoord m 3 4 3 (2) m 4 6.2 f ( x) log 4 x interchanging x and y f 1 :x log 4 y omruiling van x en y y 4x answer / antwoord (2) 6.3 𝑓 −1 y-intercept/y-afsnit 1 shape and asymptote vorm en asimptoot (2) 6.4 y 2 answer/antwoord OR/OF OR/OF y ( 2; ) answer/antwoord (1) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) No penalty for rounding off in this question. QUESTION 7/VRAAG 7 Geen penalisering vir afronding in hierdie vraag nie. 7.1 A P(1 i) n substitution into correct R 26 700 R 40 000(1 i ) 5 formula vervanging in korrekte formule simplification / 26700 5 1 i vereenvoudiging 40 000 0,0777 i value for r / waarde van r r 7,77% p.a. (3) 7.2.1 x[1 (1 i ) ] n P i 0,12 115 and/en n 180 i 115 180 x[1 (1 0,12 ) ] substituting into correct R1 200 000 0 ,115 formula 12 vervanging in korrekte formule 1 200 000( 0,12 115 ) x 0 ,115 180 [1 (1 12 ) ] answer/antwoord R14 018,28 (3) 7.2.2 x 1 (1 i ) n n (a) Balance 1 i Balans 𝑛 = 105 for / vir P and/en i 𝑛 = 5 for / vir A 115 105 R14 018,28 1 (1 0,12 ) 0 ,115 5 substituting into correct P formula 0 ,115 1 12 vervanging in korrekte P formule 12 substituting into correct A formula vervanging in korrekte A formule 0 ,115 5 R925 435,98 1 12 𝑃(1 + 0,115 12 )5 R970 637,89 answer/antwoord (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 13 OR/OF 𝑛 = 75 for both formulae / Outstanding Balance after 75 months: vir albei formules substituting into correct F formula A Fv vervanging in korrekte F formule 75 substituting into correct A formula 11,5% 75 14018.28 1 1 vervanging in korrekte A formule 11,5% 12 1200 000 1 12 11.5% 12 2 453828,34 1528392, 76 R 925 435,58 Outstanding Balance after 80 months : 11,5% 5 𝑃(1 + 0,115 12 )5 925 435,58 1 12 answer/antwoord R 970 637, 48 (5) 7.2.2 x[1 (1 i ) ] n (b) P i R14 018, 28[1 (1 0,115 12 ) n] P = R970 637,89 R970 637,89 0,115 substituting into correct formula 12 vervanging in korrekte formule 970 637,89( 0,115 12 ) 1 (1 0,115 12 ) n 14 018, 28 0,3364416715 2423 ) ( 2400 n log 0, 3364416715 correct use of logs / n log 2423 korrekte gebruik van logs 2400 114, 2130673 final answer/finale antwoord (4) n 115 months/maande [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) QUESTION 8/VRAAG 8 Penalise once for notation in this question Penaliseer een keer vir notasie in hierdie vraag 8.1 f ( x) 3 2 x 2 f ( x h) 3 2( x h) 2 3 2( x 2 2hx h 2 ) 3 2 x 2 4hx 2h2 2 2 3 2x 4hx 2h 𝑓(𝑥+ℎ)−𝑓(𝑥) f ' ( x) lim ℎ→0 ℎ substitution / vervanging 3−2𝑥 2 −4ℎ𝑥−2ℎ2 −(3−2𝑥 2 ) = lim ℎ→0 ℎ simplification / vereenvoudiging 3−2𝑥 2 −4ℎ𝑥−2ℎ2 −3+2𝑥 2 = lim factorisation / faktorisering ℎ→0 ℎ ℎ(−4𝑥−2ℎ) = lim answer / antwoord ℎ→0 ℎ = lim −4𝑥 − 2ℎ ℎ→0 = −4𝑥 + 2(0) = −4𝑥 (5) 8.2.1 D x x( x 2) 2 D x x( x 2 4 x 4) x3 4 x 2 4 x 3 2 Dx x 4x 4x 3𝑥 2 −8𝑥 +4 2 (4) 3x 8x 4 8.2.2 3 2x y ax 7 3 x 3 2x y ax 7 3 1 1 x2 −2𝑥 2 3 1 ax 7 2x 2 3 3 4 1 𝑎𝑥 −7 −𝑥 −2 7 (derivative of constant must be dy 3 4 1 zero to get 3rd mark) ax 7 x 2 dx 7 (afgeleide van die konstante moet nul wees om 3de punt te kry) (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 9/VRAAG 9 9.1 x 1 3 and/en x 1 x 1 3 x 1 (2) 9.2 x (1 ( 1 )) 2 3 1 answer/antwoord 3 (2) 9.3 g ( x) is increasing when g ( x) 0 g ( x) is stygend wanneer g ( x) 0 1 1 x 1 OR / OF x ;1 answer (accuracy) 3 3 antwoord (akkuraatheid) (2) 9.4 y a( x x1 )( x x2 ) a( x 1 3 )( x 1) 1 a (0 1 )(0 1) substituting all intercepts 3 vervanging van alle afsnitte 1 1 3 a a 3 a 3 y 3( x 1 )( x 1) y 3( x 1 3 )( x 1) 3 2 g ( x) 3x 2x 1 2 2 1 3 x x 3 3 g' x 3x 2 2x 1 OR/OF y a( x x1 )( x x2 ) a (3 x 1)( x 1) 1 a (3(0) 1)(0 1) substituting all intercepts 1 a vervanging van alle afsnitte a 1 a 1 y (3x 1)( x 1) y 1(3 x 1)( x 1) g ( x) 3x 2 2x 1 2 3x 2x 1 g' x 3x 2 2x 1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) 9.5 g ( x) ax 3 bx 2 cx d g' x 3ax 2 2bx c g ' x 3ax 2 2bx c 3x 2 2x 1 3a 3 3a 3 2b 2 c 1 2b 2 a 1 ; b 1; c 1 a 1 b 1 3 substitute (0 ; 0) into g ( x) 1 y x x2 x d 1 vervanging van (0 ; 0) in 3 2 0 0 0 0 d 1 g ( x) 1 d 1 (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 17 QUESTION 10/VRAAG 10 10.1 Let the two numbers be x and y Laat die twee getalle x en y wees : x y 18 x y 18 y 18 x 2 yx Product/Produk :P( x) yx 2 substitution and simplification 18 x x2 vervanging en vereenvoudiging 18 x 2 x 3 Product is maximum when: P '( x) 0 Produk is ' n maksimum wanneer : P '( x ) 0 P ' ( x) and equating to 0 P '( x) 36 x 3 x 2 P (x) en gelykstel aan 0 36 x 3 x 2 0 3 x(12 x) 0 x 0 or x 12 x-values/waardes y-values/waardes y 18 0 18 or/of y 18 12 6 selection of the 2 numbers P is maximum when x 12 keuse van die 2 getalle P is ' n maksimum wanneer x 12 (if/as x = 0, Product/Produk = 0) the two numbers are :12 and 6 die twee getalle is :12 en 6 OR / OF Let the two numbers be x and y Laat die twee getalle x en y wees x y 18 x y 18 2 y 18 x xy substitution and simplification Product/Produk : P( x ) xy 2 vervanging en vereenvoudiging x(18 x) 2 x(324 36 x x2 ) 324 x 36 x 2 x3 Product is maximum when: P'(x) 0 Produk is ' n maksimum wanneer : P '( x ) 0 P '( x) 324 72 x 3 x 2 3x 2 72 x 324 0 P ' ( x) and equating to 0 x 2 24 x 108 0 P (x) en gelykstel aan 0 ( x 18)( x 6) 0 x 18 or x 6 x-values/waardes y 18 18 0 or/of y 18 6 12 y-values/waardes selection of the two numbers The two numbers are 12 and 6 Die twee getalle is 12 en 6 keuse van die 2 getalle (P = 0 when/wanneer x = 18) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 18 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2019) QUESTION 11 /VRAAG 11 11.1.1 a 111 answer/antwoord b 106 answer/antwoord (2) 11.1.2 P(a boy who plays cricket) / P(’n seun wat krieket speel) (a) 108 54 numerator/teller or / of 530 265 denominator/noemer (2) 11.1.2 P(A or / of B) P(A) P(B) P(A and/en B) formula/formule (b) P(girl or not a tennis player) / P(meisie of nie ' n tennisspeler nie) 288 445 231 substitution into correct formula 530 530 530 vervanging in korrekte formule 502 answer / antwoord 530 251 or / of 265 or / of 94,72% OR/OF P(Girl or not Tennis ) method/metode 1 P( Boy and Tennis ) 28 substitution/vervanging 1 530 answer/antwoord 502 530 251 or / of 265 or / of 94,72% (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 (EC/SEPTEMBER 2019) MATHEMATICS P1/WISKUNDE V1 19 11.2.1 99 or / of 387 420 489 99 (1) 11.2.2 If vowels are together/As die vokale saam is: 6! 4! 6! × 4! If vowels are not all together: subtracting from 9! As die vokale nie almal saam is nie : aftrekking vanaf 9! 9! (6! 4!) answer/antwoord (3) 345600 11.2.3 Vowels in odd spaces / Vokale in onewe spasies 4 5 3 4 2 3 1 2 (4 3 2 1) (5 4 3 2) 4! × 120 4! 120 2880 Vowelsin odd spaces Vokalein onewe spasies / 2880 (9 8 7 6 5 4 3 2) (9 8 7 6 5 4 3 2) Probability / Waarskynlikheid (9 8 7 6 5 4 3 2) 2880 answer/antwoord 362880 (4) 1 [15] 126 TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Mathematics · Addendum · 2025 · Afr - [MATHS P1 QP GR12 SEPT 2021 Afrikaans](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-afrikaans/ecc6ca73-048f-4257-8f9a-db12f35059e1) — Mathematics · QP · 2021 · Afr - [MATHS P1 QP GR12 SEPT 2018 Afrikaans](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2018-afrikaans-1/48ef1811-7641-496b-9091-7e47b50d9aae) — Mathematics · QP · 2018 · Afr - [MATHS P1 GR12 QP SEPT2017 Afrikaans](https://www.hlayiso.com/library/maths-p1-gr12-qp-sept2017-afrikaans/b7b5f8fb-f791-4893-b24f-2cabdd8f677e) — Mathematics · QP · 2017 · Afr - [Gr12 Mathematics P1 (Afrikaans) SEP 2022 Possible Answers](https://www.hlayiso.com/library/gr12-mathematics-p1-afrikaans-sep-2022-possible-answers/38fb7466-deff-4faa-9c6c-825fcf932aae) — Mathematics · Memo · 2022 · Afr - [Mathematics P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2019-memo-afr-eng/eef87169-da21-443f-af20-6207010a2376) — Mathematics · Memo · 2019 · Eng ## More from Grade 12 Mathematics - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [MATHS P2 QP GR12 SEPT 2024 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-sept-2024-afrikaans/8572c34f-22e2-48bc-9d8a-f78c7e29781d) — QP · 2024 · Afr - [FS Gr 12 Sept 2024 Afr en Eng P1](https://www.hlayiso.com/library/fs-gr-12-sept-2024-afr-en-eng-p1/4d7be3dc-5f53-4da3-a833-8fa9c4d73be1) — QP · 2024 · Eng - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — QP · 2024 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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