# Mathematics Grade 10 P2 Eastern Cape November 2019 Memo > Grade 10 Mathematics Paper 2 memorandum from the Eastern Cape November 2019 (Afrikaans), 8 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p2-gr10-memo-nov2019-afrikaans/39dbc8fd-f1d1-418a-86e7-d500f002f90f - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 10 - Language: Afrikaans - Document type: Memorandum - Year: 2019 - Exam period: Eastern Cape November - Paper: 2 - Curriculum: CAPS - Pages: 8 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 10](https://www.hlayiso.com/library/grade-10) › [Mathematics](https://www.hlayiso.com/library/grade-10/mathematics) › [Mathematics Grade 10 P2 Eastern Cape November 2019 Memo](https://www.hlayiso.com/library/maths-p2-gr10-memo-nov2019-afrikaans/39dbc8fd-f1d1-418a-86e7-d500f002f90f) ## About this document Mathematics · Grade 10 · Eastern Cape November · 2019 · Afrikaans. Memorandum, 8 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 Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT GRAAD 10 NOVEMBER 2019 WISKUNDE V2 NASIENRIGLYN PUNTE: 100 Hierdie nasienriglyn bestaan uit 8 bladsye. ### Page 2 Downloaded from hlayiso.com 2 WISKUNDE V2 (EC/NOVEMBER 2019) Volgehoue akkurate nasien (CA) word deurgaans in ALLE aspekte van die nasienriglyn toegepas. VRAAG 1 48 50 52 59 60 68 73 76 76 76 78 79 80 81 82 82 84 91 92 98 1.1.1 76 + 78 ✓ antwoord Mediaan = = 77 (1) 2 1.1.2 60 + 68 Onderste kwartiel = = 64 ✓ onderste kwartiel 2 Boonste kwartiel = 82 ✓ boonste kwartiel (2) 1.1.3 Interkwartiel-variasiewydte = Q3 – Q1 ✓ vervanging = 82 − 64 = 18 ✓ antwoord (2) 1.1.4 Min = 48 en maks = 98 ✓ min en maks (1) 1.1.5 ✓ min en maks ✓ Q1 en Q3 ✓ Q2 (3) 1.1.6 Skeef na links of negatiewe skeefheid ✓ antwoord (1) 1.2 Tydsduur (min) Aantal oproepe (f1) Middelpunt (x1) (f 1) × (x1) 2 ≤ t < 5 47 3,5 164,5 5 ≤ t < 8 139 6,5 903,5 8 ≤ t < 11 211 9,5 2004,5 11 ≤ t < 14 102 12,5 1275 14 ≤ t < 17 58 15,5 899 17 ≤ t < 20 19 A B 576 5598 1.2.1 A = 18,5 en B = 351,5 ✓ antwoord A ✓ antwoord B (2) 1.2.2 som van f 1  x1 geskatte gemiddelde = som van f 1 ✓ som van ( f1 ) (x1 ) 5598 ✓som van ( f1 ) = ✓ antwoord 576 = 9,7 minute (3) 1.2.3 75 75ste persentiel =  576 = 432 ✓ 432 100 In die interval 11 ≤ t < 14 ✓ interval (2) [17] Kopiereg voorbehou Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER 2019) WISKUNDE V2 3 VRAAG 2 2.1 A( – 2 ; 6 ) , B( 6 ; 8 ) en C( 4 ; 0 ) dAB = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 ✓ formule = (6 − ( − 2)) 2 + ( 8 − 6 ) 2 ✓ vervanging ✓ afstand AB = 2 17 dBC = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 = (4 − 6) 2 + ( 0 − 8 ) 2 ✓ vervanging = 2 17 ✓ afstand van BC  AB = BC. (5) 2.2 ABCD is ʼn vlieër ✓ vlieër Aanliggende sye is gelyk ✓ motivering (2) 2.3 A( – 2 ; 6 ) , B( 6 ; 8 ) en C( 4 ; 0 )  x + x1 y 2 + y1  Middelpunt van BC =  2 ;  ✓ formule  2 2  ✓ vervanging −2+6 8+6 ✓ koördinate van = ( ; ) = G( 2 ; 7 ) 2 2 G, mdpt van BC  x + x1 y 2 + y1  Middelpunt van AB =  2 ;   2 2  ✓ vervanging ✓ koördinate van 4+ 6 0+8 = ( ; ) = H( 5 ; 4 ) H, mdpt van AB 2 2 (5) 2.4 BAˆ D = BCˆ D (teenoorstaande ’e van ʼn vlieër =) ✓ S ✓R AEˆ H = EDˆ B (ooreenkomstige ’e , EG || DB) ✓ SR maar EDˆ B = BDˆ C (hoeklyne van ʼn vlieër) ˆC  AEˆ G = BD ✓ 3de hoek ∆AEG ||| ∆CDB. (A A A) of rede (4) [16] Kopiereg voorbehou Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 WISKUNDE V2 (EC/NOVEMBER 2019) VRAAG 3 3.1.1 x 2 = 35 2 − 28 2 ✓sub in Pythagoras ✓x = 21 x = 21 21 35  cos = 28 ✓ 21 35 35  (3) 3.1.2 2 2 2  28   21  ✓   28 sin 2  + cos 2  =   +    35   35   35  = 1 2 ✓   21 = RHS  35  ✓1 (3) 3.2 Indien 37 sin  + 35 = 0 35  sin  = – − 35 37 ✓ sin  = x = 37 − 35 2 2 2 37 x = 12 – 12  ✓ 3de Kwadrant – 35 37 ✓ x waarde = – 12 24 sec − 70 cot  37 −12 ✓✓ vervanging = 24( ) − 70( ) −12 − 35 ✓antwoord = − 74 − 24 = − 98 (6) 3.3.1 8cos( x + 10 ) = 5 5 ✓cos( x + 10 ) cos( x + 10 ) = 8 x + 10 = 51,32 ✓ x + 10 x = 41,32 ✓ antwoord (3) Kopiereg voorbehou Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER 2019) WISKUNDE V2 5 3.3.2 cosec 2x = 2 1 1 sin 2x = ✓ sin 2x = 2 2 2x = 300 ✓ 2x = 30 x = 150 ✓ antwoord (3) 3.4 1 3 1  ✓ sin 30  tan 60 2 = 2 1 tan 30  cos 60 1 1 ✓ 3  3 2 1 ✓ =3 3 = RHS 1 ✓ 2 ✓ antwoord (5) 3.5.1 x ✓ gebruik sin 55  sin 55  = 15 ✓ antwoord x = 15  sin 55 = 12,29 (2) OF x cos 35 = ✓ gebruik cos 35 15 ✓ antwoord x = 12,29 (2) 3.5.2 4,4 tan 21  = y ✓ gebruik tan 21 4,4 y = ✓ antwoord tan 21 (2) = 11,46 OF y tan 69 = 4,4 y = 11,46 OF ✓ Pythagoras y = 12,29 – 4.4 2 2 2 ✓ antwoord y = 11,48 (2) [27] Kopiereg voorbehou Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 WISKUNDE V2 (EC/NOVEMBER 2019) VRAAG 4 4.1 y g f ✓ afsnitte ✓ draaipunte x ✓ vorm (3) 4.2 Periode van g = 3600 ✓ antwoord (1) 4.3 Waardeversameling van m(x) if m(x) = – 3f(x) + 1 Waardeversameling van – 3 f(x): – 3 ≤ y ≤ 3 ✓ notasie Waardeversameling van m(x) : – 2 ≤ y ≤ 4 ✓✓ eindpunte (3) 4.4 0 g dalend: 90 < x < 270 0 ✓ notasie ✓ eindpunte (2) 4.5 f ( x)  g ( x)  0 ✓ notasie 90  x  180  of 270   x 360  ✓ eindpunte ✓ eindpunte (3) [12] Kopiereg voorbehou Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/NOVEMBER 2019) WISKUNDE V2 7 VRAAG 5 5.1 AD̂C = 53 (’e op ʼn reguit lyn) ✓ SR DĈB = 116  (supplementêr aanliggende ’e) ✓SR CB̂A = 101 (’e op ʼn reguit lyn ) ✓SR BÂD = 360 − 53 − 116 − 101  ✓ antwoord = 90 (’e van ʼn vierhoek = 360 ) (4) Slegs antwoord: volpunte, op voorwaarde dat een rede verskaf is. 5.2 Laat DEBˆ = y en FEC ˆ =k  B̂ = 180 − 2 y en Ĉ = 180 − 2k (s van ʼn ∆✓ = 180) ✓ SR In ∆ ABC: x + 180 – 2y + 180 – 2k = 1800 2y + 2k = x + 180  + 180  − 180  ✓ SR 1 y + k = x + 90 ✓S 2 1 DEˆ F = 90 – x (’e op ʼn reguit lyn) ✓ SR 2 (4) [8] VRAAG 6 6.1.1 AP = DE en AQ = DF (gegee) ✓ gegee  = D̂ (gegee) ✓∆’e gelykvormig ∆ APQ  ∆ DEF (SAS) ✓ rede (3) 6.1.2 AP̂Q = Ê (∆ APQ  ∆ DEF) ✓ Stelling Maar B̂ = Ê (gegee) ✓ Stelling  AP̂Q = B̂  PQ || BC (ʼn ✓ paar ooreenkomstige ’e =) ✓ Rede (3) 6.1.3 AB BC AC = = (∆ABC ||| ∆DEF) ✓ SR DE EF DF 7,5 8 = ✓ vervanging 3,5 DF 8  3,5 DF = 7,5 ✓ vereenvoudig = 3,7 ✓ antwoord (4) 6.2.1 Omgekeerde van middelpuntstelling ✓ antwoord (1) Kopiereg voorbehou Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 WISKUNDE V2 (EC/NOVEMBER 2019) 6.2.2 BD = 32  AD = 32 ✓ BD = AD ✓S✓R  EF = 32 (teenoorstaande sye van ‘n parallelogram) ✓ SR  CG = 2 32 (midpt stelling) ✓ antwoord = 8 2 (5) [16] VRAAG 7 Buite-oppervlakte van ʼn keël = Buite-oppervlakte van hemisfeer  r 2 +  r s = 3 r 2 ✓ stel buite-  r s = 2 r 2 oppervlakte = aan s = 2x ( r = x) mekaar maar s = h2 + x2 2  h2 + x2 = 4x2 ✓ gebruik van  h = 4x − x 2 2 Pythagoras ✓ vervanging = 3x s = 2x ✓ h die onderwerp van die formule (4) [4] TOTAAL: 100 Kopiereg voorbehou Blaai om asseblief ## Related documents - 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