Mathematics Grade 10 P2 Gauteng November 2023 Memo (Afr)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN NOVEMBER 2023 GRADE/GRAAD 10 MARKING GUIDELINES NASIENRIGLYNE MATHEMATICS/WISKUNDE (PAPER/VRAESTEL 2) 8 pages/bladsye

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2) GRADE/GRAAD 10 QUESTION/VRAAG 1 1.1 32 ✓ Answer/Antwoord (1) 1.2 1.2.1 mean temperature/gemiddelde temperatuur sum of values/som van waardes ✓ Method/Metode number of values/aantal waardes ✓ Answer/Antwoord 500 20 = 25 (2) 1.2.2 Range/omvang = Max – Min/Maks. – Min ✓ Answer/Antwoord = 40 – 22 = 18 (1) 1.2.3 IQR/IKO = Q3 – Q1/K3 – K1 Q3/K3 ✓ = 36,5 – 27 Q1/K1 ✓ = 9,5 ✓ Answer/Antwoord (3) 1.3 ✓ Shape/Vorm ✓ Quartiles/Kwartiele (2) 1.4 20 ✓ Method/Metode x 20 = 4 100 37 C ✓ Answer/Antwoord (2) [11] 2

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 QUESTION/VRAAG 2 2.1 77 ✓ Answer/Antwoord (1) 2.2 110  x < 125 ✓ Answer/Antwoord (1) 2.3 110  x < 125 ✓✓ Answer/Antwoord (2) 2.4 31 x 100% ✓ Method/Metode 77 = 41,33% ✓ Answer/Antwoord (2) 2.5 x =  fx ✓ Method/Metode f 9242,5 ✓ Substitution/Vervanging = 77 = 120,03 ✓ Answer/Antwoord (3) 2.6 31 angle/hoek = x 360 ✓ Method/Metode 77 = 144,94 ✓ Answer/Antwoord (2) [11] QUESTION/VRAAG 3 3.1 3.1.1 PT = x2 − x1) 2 + ( y2 + y1 ) 2 = (2 − 10) 2 + (9 + 3) 2  Correct substitution/ Korrekte vervanging = ( −8) 2 + (12) 2 = 64 + 144 = 208  Answer/Antwoord (2) 3.1.2 −3 −9 mPT = 10  Correct substitution/ − 12 Korrekte vervanging mPT = −12 8 mPT = − 3  Answer/Antwoord 2 (2) 3

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 3.1.3 mPT = mAR = −3  Answer/Antwoord 2 (1) 3.1.4 M = ( xP + xT ; yP + yT ) 2 2  Substitute/Vervang M = ( 2 + 10 ; 9 − 3 )  x = 6 and/en 2 2 M (6 ; 3) y=3 (3) 3.2 y = mx + c  Correct substitution/ Sub/verv.: (2;9): …9 = − 3 (2) + c Korrekte vervanging 2  Value/Waarde c = 12  Answer/Antwoord (3) 3.3 mPT x mAR = –1 −13 − ( −3)  Gradient of/Gradiënt van −2 = 3 10 + n AR = − 3 2 –2(10 + n) = –30  Correct substitution/ –20 – 2n = –30 Korrekte vervanging 2n = 10 n=5  Answer/Antwoord (3) 3.4 RT = x2 − x1) 2 + ( y2 + y1 ) 2  Correct substitution/ = (10 − 5) + ( −3 + 13) 2 2 Korrekte vervanging = (5) 2 + (10) 2 ✓ 125 = 25 + 100  Correct substitution/ = 125 Korrekte vervanging  Area of RMT = 1 bh 2 = 1 (2 13 )(5 5 ) 2 = 40,31 units2/eenhede2  Answer/Antwoord (4) [18] 4

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 QUESTION/VRAAG 4 4.1 4.1.1 4sin 65  Answer/Antwoord = 3,63 (1) 4.1.2 cos 2 (32 − 10 )  Simplification to ratios/ tan 22 2 Vereenvoudiging na verhoudings cos 2 22 tan 22 0,859669 ... 0,40402 ... = 2,13  Answer/Antwoord (2) 4.2 3sin( θ2 – 20) = 0,85 sin( θ – 20) = 0,283333 ...  0,283333 2 θ – 20 = 16, 459 ...  2  θ – 20 = 16, 459 ...  2 θ = 36, 459 ...  2  θ = 36, 459 ...  2  = 72,92  Answer/Antwoord (4) 4.3 cos 45 + 5cosec sin 45 90 x 3tan230 ✓ 22 2 2 + 5 x 3( 1 ) 2  15 2 1 3 ✓ 3( 1 )2 3 2  Answer/Antwoord =1+5x1=6 (4) [11] 5

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 QUESTION/VRAAG 5 5.1. 5cos = –3 cos = – 3 5 ✓ Correct diagram and quadrant/Korrekte diagram en kwadrant (5)2 – 32 = y2 Pythagoras y = –4 y = –4 sin  =–4 ✓ Answer/Antwoord 5 (3) 5.2 tan2 – cot 2 2 ✓ Correct substitution/ − 4  –  − 3  2  =   − 3   − 4  Korrekte vervanging = 16 – 9 ✓ Answer/Antwoord 9 16 = 175 144 (2) [5] QUESTION/VRAAG 6 6.1 a = –1 ✓ a = –1 p=2 ✓p=2 q=2 ✓q=2 (3) 6.2 90  x  270 OR/OF  R [90 ; 270] ✓ Values/Waardes ✓ Inequalities/Ongelykhede (2) 6.3 x = 0 or/of x = 45 or/of x = 360 ✓✓ each x-value/elke x-waarde (3) 6.4 1y3;yR ✓ Values/Waardes ✓ Inequalities/Ongelykhede (2) [10] 6

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 QUESTION/VRAAG 7 7.1 ✓ Tan ratio/Tan-verhouding tan 44,3 = = DC 12 m ✓ Simplification/ DC = 12 tan 44,3 Vereenvoudiging DC = 11,71 m ✓ Answer/Antwoord (3) 7.2 AC = 2 x DC = 23,42 m ✓ CA from 7.1/CA van 7.1 tan ABˆ C = 23,42 ✓ tan ABˆ C 12 ABˆ C = 62,87  = 62,87 – 44,3 = 18,57. ✓ Answer/Antwoord The spotlight will need to swing 18,57./Die kollig sal 18,57 moet swaai. (3) [6] QUESTION/VRAAG 8 8.1 8.1.1 3x + 2x + x = 180 sum of <s of a triangle/ ✓S ✓R som e van  x = 30 (2) 8.1.2 • Scalene triangle/Ongelyksydige driehoek ✓S ✓R • three different angles/drie verskillende hoeke (2) 8.2 B̂1 = B̂3 given/gegee ✓S ✓S ✓R Â = Ĉ angles opp equal sides/e teenoor gelyke sye ✓ S/R Ê1 = Â + B̂1 ext < of a /buitehoek van ʼn  ✓S D̂ 2 = Ĉ + B̂3 ext < of a /buitehoek van ʼn   Ê1 = D̂ 2 (5) [9] 7

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE (PAPER/VRAESTEL 2)GRADE/GRAAD 10 QUESTION/VRAAG 9 [19] 9.1 D̂1 = B̂2 Alt <s A D ‖ BC/verw. binne e ; AD ‖ BC ✓S ✓R Ô1 = Ô3 vertically opp <s/regoorstaande e ✓S ✓R AD = BC opp sides of a parm/teenoorst sye van ʼn parm ✓S AOD  COB AAS/s DO = OB and/en AO = OC ✓R (6) 9.2 9.2.1 SP = SR sides of square/sye van vierkant ✓S ✓R P̂ = Ŝ <s of square = 90/e van vierkant = 90 ✓S ✓R PY = SX ✓S SPY  RSX SAS/SS ✓R (6) 9.2.2 Ŝ1  x s  ✓S ✓R (2) 9.2.3 Ŝ1 + Ŝ2 = 90 <s of square = 90/e van vierkant = 90 ✓ S ✓R Ŝ2 = 90 – x ✓S ✓R Ô 2 + Ŝ2 + R̂ 2 = 180 sum of int < of /som e van  Ô 2 + 90 + –x + x = 180 Ô 2 = 90 ✓S  SY ⊥ RX (5) [19] TOTAL/TOTAAL: 100 8