# Mathematics Grade 11 P2 Gauteng November 2022 Memo (Afr) > Grade 11 Mathematics Paper 2 memorandum from the Gauteng November 2022 (Afrikaans), 22 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p2-afrikaans-november-2022-possible-answers/1defb1c9-b865-4fd0-aeab-4f39e35d4daf - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: Afrikaans - Document type: Memorandum - Year: 2022 - Exam period: Gauteng November - Paper: 2 - Curriculum: CAPS - Pages: 22 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P2 Gauteng November 2022 Memo (Afr)](https://www.hlayiso.com/library/gr-11-mathematics-p2-afrikaans-november-2022-possible-answers/1defb1c9-b865-4fd0-aeab-4f39e35d4daf) ## About this document Mathematics · Grade 11 · Gauteng November · 2022 · Afrikaans. Memorandum, 22 pages. Read online or download the PDF. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINSIALE EKSAMEN NOVEMBER 2022 GRAAD 11 NASIENRIGLYNE WISKUNDE (VRAESTEL 2) 22 bladsye ### Page 2 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 INSTRUKSIES EN INLIGTING ➢ A – AKKURAATHEID ➢ CA – KONTINUE AKKURAATHEID ➢ S – BEWERING ➢ R – REDE ➢ S/R – BEWERING met REDE NOTAS: • Indien ʼn kandidaat ʼn vraag TWEE keer antwoord, merk slegs die EERSTE poging. • Indien ʼn kandidaat ʼn antwoord doodgetrek het, en nie die vraag weer gedoen het nie, merk die doodgetrekte poging. • Deurlopende akkuraatheid moet op ALLE aspekte van die nasienriglyn toegepas word. • Dit is ONAANVAARBAAR vir kandidate om waardes/antwoorde aan te neem in ʼn poging om probleme op te los. 2 ### Page 3 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 1 1.1 250 266 287 306 329 ✓ Mond (Q1 en Q3) ✓ Snor (Min. en ⚫ ⚫ Maks.) ✓ Mediaan 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 (3) 1.2 Die data is simmetries versprei, ʼn normale verspreiding. ✓ Verduideliking OF Die data is effens skeef na regs. (1) 1.3  = 23,77   (2) 1.4 ( x − ; x + ) ✓ x = 287 = (287 – 23,77 ; 287 + 23,77) ✓ Interval = (263,23 ; 310,77) ✓6  6 munisipaliteite is binne een standaard afwyking. (3) 1.5 1 S-IQR = (Q3 – Q1) ✓ Q3 – Q1 2 1 1 = (306 – 266) ✓  2 2 = 20 ✓ Antwoord (3) [12] 3 ### Page 4 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 2 2.1 p=5 ✓p q = 27 ✓q r=1 ✓r (3) 2.2 KUMULATIEWE FREKWENSIEGRAFIEK (OGIEF) 34 32 30 28 26 Kumulatiewe frekwensie van leerders 24 22 20 18 16 14 12  Grond by (20 ; 0) 10  Eindpunte 8  Verbind met vryhand 6 (3) 4 2 0 0 10 20 30 40 50 60 70 80 90 100 PUNTE 2.3 M = 46  M = 46( 2) (2) 2.4 32 – 24 ✓ KF = 24( 1) =8 ✓ 32 – KF  8 leerders behaal ʼn punt van meer as 55. (2) [10] 4 ### Page 5 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 3 y A(0 ; 7) B(10 ; 2) D(x ; y) O   x C(7 ; –4) 3.1 AB = (0 − 10) 2 + (7 − 2) 2 ✓ Vervang in korrekte formule AB = 125 ✓ AB AB = 5 5 ✓ Eenvoudigste wortelvorm (3) 3.2 7 − ( −4) ✓ Gradiënt substitusie m AC = 0−7 11 m AC = − ✓ m AC 7 (2) 3.3 11 ✓ Inklinasie hoek tan  = − 7 vervang (CA 4.2) verw.hoek = 57,528 …° ✓ Verw.hoek  = 180° – 57,528 …° ✓ θ (trek verw.hoek af van 180°)  = 122,47° (3) 5 ### Page 6 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 3.4 tan  = 2   = 63,434 …° ˆB AC ˆ B = 122,47° – 63,434 …° AC ˆB CA A ˆCB = 59,04° B ˆCD 1 m AB = 2 ˆ B = 180° – tan–1  1  – 122,74° CA 2 ˆ B = 30,69° CA ˆ D = 30,69° AC [verw. 'e, AB ‖ CD] ˆ D = 30,69° + 59 04° BC ˆ D = 89,73° BC OF OF Indien  nie afgerond is nie tan  = 2 ✓  = 63,434 …° ˆB ✓ AC A ˆCB = 122,4711 …° – 63,434 …° ✓ C ˆAB ˆ B = 59,036 …° AC ˆD ✓ BC 1 m AB = 2 C ˆAB = 180° – tan–1   – 122,74 …° 1 2 ˆ B = 30,963° CA ˆ D = 30,963° AC [verw 'e, AB || CD] ˆ D = 30,963° + 59 036 …° BC ˆ D = 90° BC (4) [12] 6 ### Page 7 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 4 y K M(4 ; 3) x O L(4 ; 3) 4.1 y = 5x + 9 en 5y + x – 19 = 0 ✓ Vervang y 5(5x + 9) +x = 19 ✓ Los op vir x 26x = –26 ✓ Vervang x terug x = –1 GEEN PUNTE vir koördinate nie (gegee) y = 5(–1) + 9 y=5  K(–1 ; 4) (3) 4.2 mKL = 5 en 1 mKM = − ✓ mKL 5  mKL  mKM = –1 ✓ mKL KL ⊥ KM ✓ Definisie van loodregte gradiënte ˆ M = 90°  LK (3) 7 ### Page 8 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 4.3 KL = ((−1 − (−2))2 + (4 − (−1) )2 ✓ KL lengte KL = 26 ✓ KN lengte KN = ( −1 − 4) 2 + ( 4 − 3) 2 ✓ Vervang in Oppervlak formule KN = 26 ✓ Opp ΔKML Area KML = 1 2 ( 26 ) ( 26 ) Area KML = 13 eenhede2 (4) − 2 + 4 3 + ( −1)  ✓ Middelpunt van NL M  4.4 ;   2 2  M (1 ; 1) 3 − ( −1) mNL = 4 − ( −2) 2 mNL = ✓ Gradiënt van NL 3 2 y – y1 = – (x – x1) ✓ Loodregte gradiënt 3 2 y–1=– (x – 1) ✓ Vervang Middelpunt 3 2 5 y=– x+ ✓ Vergelyking 3 2 (5) 4.5 mNL = m NP = mLP ✓ Definisie van kollineêr 2 y−3 = ✓ Vervang P(7 ; y) 3 7−4 y=5 (2) 8 ### Page 9 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 4.6 Q(3 ; –2) ✓ xQ ✓ yQ (2) 4.7 K̂ = 90° [bewys in 4.1] ✓ Binnehoek van 90° KL = KN = 26 [bewys in 4.2] ✓ Aangrensende sye gelyk  KLQN is ʼn [ʼn parallelogram met ʼn binnehoek vierkant van 90° en aangrensende sye gelyk] (2) [21] VRAAG 5 5.1 tan (180  − x ) cos (360  + x ) ✓ – tan x 2 cos (90 + x ) − tan x.cos x ✓ cos x = − 2 sin x ✓ –2 sin x sin x − .cos x = cos x ✓ – tan x identiteit − 2 sin x − sin x = ✓ vereenvoudig − 2 sin x 1 = ✓ antwoord 2 (6) 5.2 5.2.1 sin − 2 sin cos ✓ sin (1–2 cos) LK = 2 cos2  + cos − 1 ✓ (2cos – 1)(cos + 1) sin  (1 − 2 cos ) ✓ Teken verander/negatiewe = (2 cos − 1) (cos + 1) faktor − sin  (2 cos − 1) ✓ Vereenvoudig met = (2 cos − 1) (cos + 1) gevolgtrekking − sin  = cos + 1 = RK (4) 9 ### Page 10 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 5.2.2 Ongedefinieerd wanneer ✓ 2cos – 1 = 0 2cos – 1 = 0 ✓ QI algemene oplossing 1 ✓ QIV algemene oplossing cos = 2 Verw. hoek = 60° QI :  = 60° + k.360° ; kZ ✓ kZ QIV :  = 300° + k.360° OF wanneer cos  = –1  = 180° + k.360° ; kZ NOTA: Aanvaar ander ekwivalente algemene oplossings cos – 1= 0 ✓ cos  = –1 algemene oplossing  = 180° + k.360° ; kZ (5) 10 ### Page 11 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 5.3 • P(k ; m) y A(b ; 1) • 6 4 α T O • x 5.3.1 b2 + 12 = 42 ✓ Pythagoras substitusie b =  15 b = – 15 (QII) ✓ b = – 15 (2) 5.3.2 (a) 1 y tan = – ✓ 15 x ✓ CA x (2) (b) sin(–  – 180°) + cos(– ) ✓ sin  = sin  + cos  ✓ cos   − 15  =   +  1  ✓ Vervang x ; y en r  4   4   1 − 15  =     4  (3) 11 ### Page 12 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 5.3.3 k m k cos( – 90) = sin( – 90) = ✓ cos( – 90) = 6 6 6 k m m sin  = – cos  = ✓ sin( – 90) = 6 6 6 k 1 m  = cos  = – ✓ Komplementêre herleiding 6 4 6 k= 3 m  − 15   =  ✓ Verhoudings gelyk stel 2 6  4    3 15 m= ✓ Waardes van k en m 2 OF OF P lê in QI, daarom is k en m positief ✓ P in QI ʼn 90º rotasie beteken (x ; y) → (y ; x) ✓ (x ; y) → (y ; x) idee r2 ✓ Verhouding van radii r1 6 ✓ Pas skaal toe op x en y 4 3 ✓ Waardes van k en m 2  P  .1 ; . 15  = P(k ; m) 3 3 Antwoord alleen MAG NIE 2 2  volpunte kry NIE. 3 3 15 k= en m = 2 2 (5) [27] 12 ### Page 13 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 6 y 2 1 f M(48,05° ; 0,41) x −90° −60° −30° 0° 30° 60° 90° 120° 150° 180° -1 g -2 6.1 A=2 ✓ Antwoord (1) 6.2 120º ✓ Antwoord (1) 6.3 a = –1 ✓a b=3 ✓b p = 30° ✓p (3) 6.4 6.4.1 –90°  x < 60° ✓ Interval (1) 6.4.2 x = –30° of 48,05°  x  180° ✓ Interval ✓ x = –30° (2) 6.4.3 60° < x < 150° of 150° < x  180° ✓ 60º en 180º ✓ Nie ingesl. 150º maar 180º ingesl. (2) 6.5 t = –30° + k. 360° ; k ✓✓ t = –30° ✓ k. 360° (3) [13] 13 ### Page 14 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 7 T   h 1 2 G B A 7.1 B̂1 =  =  ✓ Antwoord (1) 7.2 In TBG ✓ Korrekte verhouding in ΔTBG h sin( + ) = ✓ BT BT h BT = ✓ Sin reël substitusie sin( +  ) In ABT ✓ BT substitusie  =  ✓ vereenvoudiging x BT = sin  sin  x h = sin  sin ( +  ) sin  h sin  x= sin ( +  ) sin  (5) h sin  7.3 x= ✓ Substitusie sin ( +  ) sin  20 sin 37 x= sin (65 + 37) sin 65 x = 13,58 m ✓ Antwoord Die kat stap 13,58 m (2) [8] 14 ### Page 15 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 8 FIGUUR A FIGUUR B 18 cm r Q 28° 1 4 3 V = r3 of V = r 3 3 P 8.1 1 V= (9)3 ✓ Substitusie (r = 9) 3 V = 763,41cm3 ✓ Antwoord (2) 8.2 V= 4 r ✓ Vervang volume in die 3 korrekte formule 268.08 = 4 r ✓ Onderwerp van die 3 formule 3  286 ,08 3 =r ✓r 4π r=4 4 tan 14° = ✓ tan 14° PQ 4 PQ = ✓ PQ tan 14 PQ = 16,04 cm (5) [7] 15 ### Page 16 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 9 9.1 O • D M ✓ Konstruksie F Konstruksie: Verbind OF en OD In OFM en ODM i. FM = MD [GEGEE] ii. OM is ʼn gemeenskaplike sy iii. OF = OD [Radii]  OFM  ODM [SSS] ✓R M1 = ˆ ˆ M2 [OFM  ODM] ✓R ˆ +M Maar M ˆ = 180° ['e op ʼn reguit lyn] ✓S 1 2  ˆ M1 = ˆ M 2 = 90°  OM ⊥ FD (4) 16 ### Page 17 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 9.2 P S O • M 2 62° 1 1 R 2 1 2 Q 9.2.1 M̂ 2 = 90° [lyn van midpunt sirkel, tot midpunt koord] ✓SR (2) 9.2.2 Q̂1 + Q̂2 = 90° ['e in ʼn halwe sirkel] ✓SR (2) 9.2.3 P̂ = 28° [binne 'e van ʼn ] ✓ S/R Ŝ = P̂ = 28° ['e in dieselfde sirkelsegment] R (2) [10] 17 ### Page 18 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 10 O x E D F B 10.1 B̂1 = D̂1 = x ['e in dieselfde sirkelsegment] ✓S B̂1 = Ê1 = x ['e in dieselfde sirkelsegment] ✓S (2) 10.2 10.2.1 Ô1 = 2x [midpunts = 2  omtreks] ✓SR (2) 10.2.2 ˆ B + B̂ Ô1 = OF [buite  van ] R 1 ˆ B+ x 2x = OF ˆB =x  OF OˆFB = x (2) 10.2.3 D̂2 = 90° – x ['e in ʼn halwe sirkel] ✓S R (2) 10.2.4 Ô2 = 180° – 2x ['e op ʼn reguitlyn] ✓SR Ô2 = Ê 2 = 180° – 2x ['e in dieselfde sirkelsegment] ✓SR (4) 10.3 Ê 2 = B̂ 3 + D̂2 [buite  van ] R  B̂3 = 90° – x  B̂3 = 90° – x  B̂3 = D̂2 [albei 90° – x] ✓S  EB = ED [sye teenoor gelyke ’e] R (4) [16] 18 ### Page 19 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 VRAAG 11 L R S • O P D T 11.1 Konstuksie: Trek middellyn SP en verbind PT  Konstruksie ˆ R = 90° – RS LS ˆP [rklyn ⊥ rad] SR ˆ R = 90° – RT ST ˆP ['e in ʼn halwe sirkel] SR ˆ P = RT RS ˆP ['e in dieselfde sirkelsegment] ˆ R = SˆTR  LS SR 19 ### Page 20 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 OF L R S • O P D T Konstuksie: Trek middellyn SP en verbind PR  Konstruksie ˆ R = 90° – RS LS ˆP [rklyn ⊥ rad]  S/R ˆ P = 90° SR [ in ʼn halwe sirkel] SR ˆP ˆ R = 90° – RS SP [binne 'e van ] ˆ R = SP LS ˆR ˆ P] [albei 90° – RS S ˆ R = ST SP ˆR ['e in dieselde segment]  S/R ˆ R = ST  LS ˆR 20 ### Page 21 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 Of L R 1 1 S 2 2 3 1 • O D T Konstruksie: Verbind OS en OR  Konstruksie Ŝ1 + Ŝ 2 = 90° [rklyn ⊥ radius] SR Ŝ 2 = 90° – Ŝ1 R̂ 1 = Ŝ 2 = 90° – Ŝ1 ['e teenoor gelyke radii] ✓ S/R Ô1 = 180° – 2[90° – Ŝ1 ] = 2 Ŝ1 [binne 'e van ] S Ô1 = 2 T̂ [midpunts = 2  omtreks ]  S ✓ R ˆ R = ST  LS ˆR (6) 21 ### Page 22 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 2) GRAAD 11 11.2 D C B A x T 11.2.1  tussen rklyn en koord ✓R (1) 11.2.2 Â4 = 180° –  3 ['e op ʼn reguitlyn] ✓ SR Ĉ1 = 180° –  3 [ tussen rklyn en koord] ✓S✓R Ĉ 2 =  3 ['e op ʼn reguitlyn] ✓ SR D̂2 =  3 [ tussen rklyn en koord] ✓ SR  Ĉ 2 = D̂2 [albei gelyk aan  3 ] ✓S D̂1 = F̂ [ tussen rklyn en koord /bewys in ✓R 11.2.1] Â1 = Â2 [binne 'e van /3de  in ] (7) [14] TOTAAL: 150 22 ## Related documents - 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