Mathematics Grade 9 Gauteng November 2019 Memo (Afrikaans)

GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN NOVEMBER 2019 GRADE / GRAAD 9 MATHEMATICS WISKUNDE MARKING GUIDELINES / NASIENRIGLYNE 9 pages / bladsye

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 1 1.1 B 1.2 C 1.3 B 1.4 A 1.5 B 1.6 D 1.7 B 1.8 C 1.9 C 1.10 C QUESTION / VRAAG 2 2.1 Simplify / Vereenvoudig: 1 mark / 1punt 36 84 Answer / Antwoord 1 mark /1 punt ∶ 100 100 36 12 36 84 ÷ 12 or/ of 12 ∶ 12 84 3 7 3 ∶ 7A 2.2 1 Substitution / Vervanging: 1mark / 1 punt = −4 (− 2) ²  1 = −4 (4) Answer / Antwoord: 1 mark / 1 punt = −1 𝐀 2.3 2.3.1 16 − 4² Simplify / Vereenvoudig: 1 mark /1 punt = 16 − 16𝐂𝐀 Answer / Antwoord 1 mark /1 punt = 0 A 2.3.2 2𝑥² − 𝑥 − 15 − (𝒙² − 𝟐𝒙 + 𝟏) Multiply ( )s / Vermeningvuldig ( )s /1 mark each /1 punt elk = 2𝑥² − 𝑥² − 𝑥 + 2𝑥 − 15 − 1 Simplify / Vereenvoudig:1 mark /1 punt Answer / Antwoord: 1 mark /1 punt = 𝑥² + 𝑥 − 16 𝐂𝐀 [10] 2

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 3 3.1 = 5𝑥𝑦(12x − 5y + 1) 5𝑥𝑦 :1 mark / 1 punt (12𝑥 − 5𝑦 + 1) :1 mark / 1 punt 3.2 = (𝑘 − 8)(𝑘 + 8) A (𝑘 − 8): 1 mark / 1 punt (𝑘 + 8): 1 mark / 1 punt 3.3 𝑥 2 + 7𝑥 − 18 (𝑥 − 2) 1 mark / 1 punt = (𝑥 − 2)(𝑥 + 9) A (𝑥 + 9) 1 mark / 1 punt 3.4 = 𝑎(𝑥 − 𝑦) + 4(𝑥 − 𝑦)A Grouping and common factors / = (𝑥 − 𝑦)(𝑎 + 4)𝐀 Groepering en gemeenskaplike faktore: 1 mark / 1 punt (𝑥 − 𝑦) 1 mark / 1 punt (𝑎 + 4) 1 mark / 1 punt [9] QUESTION / VRAAG 4 4.1 3𝑥 = 6M 1 mark for simplification 1 mark / 1 punt 𝑥 = 2CA vir vereenvoudiging 1 mark for answer / 1 punt vir antwoord 4.2 1 3 1 1 3 𝑥 3 = (2) M 1 mark for writing 8 as (2) / 1 punt vir 1 1 3 1 skrywe 8 as (2) / ∴ 𝑥 = CA 2 1 mark for answer / 1 punt vir antwoord 4.3 (𝑥 − 5)(𝑥 − 2) = 0M 1 mark for factorising / 1 punt vir faktorisering 𝑥 = 5 or /of 𝑥 = 2CA 1 mark for answer / 1 punt vir antwoord [6] 3

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 5 5.1 The price of 1 chicken / Die prys van 1 hoender = 𝑅𝑥 1 mark for reasoning / 1 punt vir The price of 1 turkey / Die prys van 1 kalkoen redenasie = 𝑅5𝑥. M 1 mark for reasoning / 1 punt vir redenasie 50 chickens / hoenders = R50 x and / en 1 mark for statement / 1 punt vir 20 turkeys / kalkoen𝑒 stelling = 𝑅100𝑥M 50𝑥 + 100𝑥 = 1200M 1 mark for calculation / 1 punt vir 150𝑥 = 𝑅1200 bewerking 1200 𝑥 = 150 = 𝑅8 M 1 mark for answer / 1 punt vir antwoord ∴ 1 turkey / kalkoen = 5 × 𝑅8 = 𝑅40CA 5.2 𝐴 1 mark for formula /1 punt vir A = P(1 + 𝑛𝑖) or / of 𝑃 = (1+𝑛𝑖)  M formule 15000 15000 = 𝑃(1 + 0,07 × 6) or / of 𝑃 = (1+0,07×6) M 1 mark for substitution / 1 punt vir vervanging 15000 15000 = 𝑃(1,42) or / of 𝑃 = (1,42) 1 mark for answer / 1 punt vir 𝑃 = 𝑅10 536,38 CA antwoord 5.3 D = speed / spoed × time / tyd M 1 mark for formula /1 punt vir formule = 80 × 3 M 1 mark for substitution / 1 punt vir vervanging = 240 km M 1 mark for answer / 1 punt vir antwoord 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑓𝑠𝑡𝑎𝑛𝑑 1 mark for formula /1 punt vir Time= / M 𝑠𝑝𝑒𝑒𝑑 𝑠𝑝𝑜𝑒𝑑 formule 240 = 60  CA 1 mark for substitution / 1 punt vir vervanging = 4 hrs / ure A 1 mark for answer / 1 punt vir antwoord [14] 4

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 6 6.1 𝑦 = 𝑚𝑥 + 𝑐 1 mark for 𝑐 = 6 / 1 punt vir c 6 𝑐 = 6 A 𝑦2 − 𝑦1 1 mark for substitution into gradient formula 𝑚= 𝑥2 − 𝑥1 1 punt vir vervanging in gradiënt formule 6−0 𝑚 = 0−(−1)CA 1 mark for 𝑚 = 6 / 1 punt vir 𝑚 = 6 1 mark for answer / 1 punt vir antwoord 𝑚 = 6 CA or / of ∴ Eqn of/ 𝑉𝑔𝑙𝑘 𝑣𝑎𝑛 𝑓: 𝑦 = 6𝑥 + 6CA 1 mark for substitution of 𝑐 = 6 / 1 punt vir or / of vervanging van 𝑐 = 6 substitute / vervang 𝑐 = 6, 𝑥 = −1 and / en 𝑦 = 0 into / in 1 mark for substitution of (−1; 0) into 𝑦 = 𝑚𝑥 + 𝑐 𝑦 = 𝑚𝑥 + 𝑐 / 1 punt vir vervanging (−1; 0) in 𝑦 = 𝑚𝑥 + 𝑐 0 = 𝑚(−1) + 6M 1 mark for 𝑚 = 6 / 1 punt vir 𝑚 = 6 𝑚 = 6CA ∴ Eqn of / 𝑉𝑔𝑙𝑘 𝑣𝑎𝑛 𝑓: 𝑦 = 6𝑥 + 6CA 1 mark for answer / 1 punt vir antwoord 6.2 𝑦 = −2A 1 mark for answer / 1 punt vir antwoord [5] 5

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 7 7.1.1 Stem / Stam Leaves / Blare 1 mark for each correct row / 1 2 1 punt vir elke korrekte ry 2 1 3 3 1 5 5 7 4 2 36 5 4 5 7 7.1.2 𝑅𝑎𝑛𝑔𝑒 = 57 − 12 / 𝑂𝑚𝑣𝑎𝑛𝑔 = 57 − 12 1 mark for answer / 1 punt vir 𝑅 = 45A antwoord 7.1.3 30 50% = 60 1 mark for answer / 1 punt vir ∴ 10 learners / leerdersA antwoord 7.2.1 5 1 1 mark for answer / 1 punt vir 𝑃(𝑅) = 15 = 3  A antwoord 7.2.2 3 7 3 7 𝑃(𝐺 or/𝑜𝑓 𝐵 ) = 15 + 15  1 mark for 15 + 15 3 7 / 1 punt vir 15 + 15 2 𝑃(𝐺 or/𝑜𝑓 𝐵 ) = 3  1 mark for answer / 1 punt vir antwoord 7.2.3 𝑃(𝑌) = 0 A 1 mark for answer / 1 punt vir antwoord [11] 6

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 8 8.1.1 1 mark for statement / Statement / Bewering Reason / Rede 1 punt vir bewering 65° + 45° + 𝑏 = 180° M ∠𝑠 on a str line/ ∠𝑒 op reguit 1 mark for reason / lynA 1 punt vir rede 𝑏 = 70° CA 1 mark for answer / 1 punt vir antwoord 8.1.2 1 mark for statement / Statement / Bewering Reason / Rede 1 punt vir bewering 70° + 𝑎 + 40° = 180° M sum int ∠𝑠 of ∆𝐴𝐶𝐸 /som binne 1 mark for reason / ∠𝑒 van ∆𝐴𝐶𝐸A 1 punt vir rede 𝑎 = 70° CA 1 mark for answer / 1 punt vir antwoord 8.1.3 1 mark for statement / Statement / Bewering Reason / Rede punt vir bewering 1 mark for reason / punt ∆𝐵CD is an iscosceles ∠𝑠 opposite equal sides are vir rede triangle / gelykbenige equal / teenoorstaande driehoek A gelykesye is gelyk A 8.2 1 mark for statement and Statement / Bewering Reason / Rede reason / punt vir bewering en rede In ∆ABC and ∆CDA: 1 mark for statement and reason / punt vir bewering 𝐴𝐵 = DC Given / Gegee  en rede 1 mark for statement / 𝐴𝐶 = AC Common side / Gemeenskaplike sy punt vir bewering alt ∠𝑠 and ABǁDC / verw. ∠𝑒 en 1 mark for reason / punt 𝐴̂1 = 𝐶̂1 ABǁDC  vir rede 1 mark for statement and ∆ABC ≡ ∆CDA 𝑠∠𝑠 reason / punt vir bewering en rede 7

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 8.3 Statement / Bewering Reason / Rede 1 mark for statement and reason / 1 punt vir bewering en rede In ∆CFB and ∆DFE: 1 mark for statement and reason / 1 punt vir bewering en rede Ĉ = D ̂ =40° given / gegee A 1 mark for statement and reason F̂1 = F̂2 vert. opp. ∠𝑠  / 1 punt vir bewering en rede 1 mark for reason / 1 punt vir sum int ∠𝑠 of ∆ / som binne ∠𝑒 rede ̂2 = E B ̂2 van ∆ A ∴ ∆ CFB ⫴ ∆DFE ∠∠∠  8.4 8.4.1 Construction / Konstruksie 1 mark for correct labeling of triangle / 1 punt vir korrekte C  benoeming van driehoek 1 mark for accurate measurement of sides / 1 punt 7 vir akkurate meting van sye 1 mark for accurate  6  measurement of 𝐴̂ = 65° angle / A 7 B 1 punt vir akkurate meting van 5° 𝐴̂ = 65° 8.4.2 A𝐵̂ C = 50°CA 1 mark for accurate measurement of angle / 1 punt vir akkurate meting van hoek 8.5 𝐻 ′ (0 ; 2)  A 1 mark for each co-ordinate / 1 punt vir elk korekte kordinate [22] 8

MARKING GUIDELINES / MATHEMATICS / WISKUNDE NASIENRIGLYNE GRADE / GRAAD 9 QUESTION / VRAAG 9 9.1 Statement / Bewering Reason / Rede In ∆YTZ Opposite ∠ of a rectangle / 1 mark for YT= 13 /1 punt YT= 13A Teenoorgestelde van ʼn reghoek vir YT= 13 Theorem of Pythagoras / 1 mark for 𝑌𝑍 2 = 𝑌𝑇 2 − 𝑌𝑍 2 = 𝑌𝑇 2 − 𝑇𝑍 2 Stelling van Pythagoras 𝑇𝑍 2 / 1 punt vir 𝑌𝑍 2 = 𝑌𝑇 2 − 𝑇𝑍 2 = 132 − 52 M and substitution / en substitusie = 169 − 25 1 mark for 𝑌𝑍 2 = 144 /1 punt vir antwoord = 144 CA 𝑌𝑍 2 = 144 1 mark for answer / 1 punt 𝑌𝑍 = 12CA vir antwoord 9.2 9.2.1 𝐴𝑟𝑒𝑎 𝐴𝐵𝐶𝐷 = 𝐴𝐵 × 𝐵𝐶 or / of 𝐴𝐷 × 𝐷𝐶  1 mark for formula / punt vir formule 9.2.2 𝐷𝐶×𝐸𝐹 1 1 mark for formula / punt 𝐴𝑟𝑒𝑎 ∆𝐷𝐸𝐶 = or / of 2 × 𝐷𝐶 × 𝐸𝐹  2 formule 9.2.3 𝐴𝑟𝑒𝑎 ∥𝑚 𝐶𝐷𝐸𝐺 = 𝐷𝐶 × 𝐸𝐹 or / of 𝐸𝐺 × 𝐵𝐶 1 mark for formula / punt formule 9.2.4 𝐸𝐹( 𝐴𝐸+ 𝐷𝐶) 1 mark for formula / punt Area of / van AECD =  2 formule 9.3 𝑆𝐴 = 2𝜋𝑟² + 2𝜋𝑟 × 𝐻 1 mark for formula / punt formule 22 22 1 mark for correct = 2 ( 7 ) (7)² + 2 ( 7 ) (7) × 12 substitution / 1 punt vir korrekte vervanging = 308cm² + 528cm²  1 mark for calculation / 1 𝑆𝐴 = 836 𝑐𝑚²CA punt vir bewerking or / of 1 mark for answer / 1 punt 𝑆𝐴 = 2𝜋𝑟(𝑟 + ℎ) vir antwoord. 22 = 2 ( 7 ) (7)(19)𝑐𝑚² 𝑆𝐴 = 836 𝑐𝑚²CA [13] TOTAL / TOTAAL [100] 9