Technical Mathematics Grade 10 P1 Eastern Cape November 2019 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 NOVEMBER 2019 TECHNICAL MATHEMATICS P1/ TEGNIESE WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 100 This marking guideline consists of 9 pages./ Hierdie nasienriglyn bestaan uit 9 bladsye.

2 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION/VRAAG 1 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 1.1 √49 < √62 < √64 method / metode √62 lies between / lê tussen 7 and/en 8 answer / antwoord (2) 1.2 method / metode answer / antwoord 2 72 2 36 r 0 2 18 r 0 No mark awarded for 2 9 r 0 answer if base NOT 2 4 r 1 indicated/Geen punt 2 2 r 0 toegeken vir 2 1 r 0 antwoord as basis 0 r 1 NIE aangedui is NIE. ∴ 7210 = 10010002 (2) 1.3 11102 method / metode × 1112 answer / antwoord 11102 111002 1110002 11000102 (2) 1.4 8,72 × 10−4 answer / antwoord (1) 1.5 5(−3)2 + 3(−3)(2) − 2(2)2 = 19 substitution / vervanging answer / antwoord (2) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2019) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 3 QUESTION/VRAAG 2 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 2.1 2.1.1 5 7x 5y 2 5 49 x 2 70 xy 25 y 2 49x 2 245 x 2 350 xy 125 y 2 70xy 25y 2 answer simplified / antwoord vereenvoudig (4) 2.1.2 (2𝑎 + 3)(4𝑎2 − 6𝑎 + 9) = 8𝑎3 − 12𝑎2 + 18𝑎 + 12𝑎2 − 18𝑎 + 27 8𝑎3 = 8𝑎3 + 27 27 (2) 2.1.3 (3 + 4𝑖)(−2 − 5𝑖) expand / uitbreiding = −6 − 15𝑖 − 8𝑖 − 20𝑖 2 substitution / vervanging -1 = −6 − 23𝑖 − 20(−1) simplification / = −6 − 23𝑖 + 20 vereenvoudiging = 14 − 23𝑖 (3) 2.2 2.2.1 3𝑥+1 .81𝑥 definition 34 and/en 32 9𝑥+2 power law / magswet 3𝑥+1 .(34 )𝑥 division law / = (32 )𝑥+2 delingswet 𝑥+1 4𝑥 3 .3 answer / antwoord = 32𝑥+4 𝑥+1+4𝑥−2𝑥−4 =3 = 33𝑥−3 (4) 2.2.2 4 power law / (9𝑥 2 ) ×3𝑥 2 magswet 27 simplification / 6561𝑥 8 ×3𝑥 2 = vereenvoudiging 27 answer / antwoord 19683𝑥 10 = 27 = 729𝑥 10 (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION/VRAAG 3 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 3.1 𝑥 2 + 5𝑥 − 6 (𝑥 + 6) = (𝑥 + 6)(𝑥 − 1) (𝑥 − 1) (2) 3.2 −4𝑎3 + 32 −4 common factor / gemene faktor = −4(𝑎3 − 8) (𝑎 − 2) = −4(𝑎 − 2)(𝑎2 + 2𝑎 + 4) (𝑎2 + 2𝑎 + 4) (3) 3.3 x2 x 7 2x x 7 x 7 2 x 7 x 1 x 7 x2 2x 1 (𝑥 + 7) common factor / gemene 2 faktor x 7 x 1 x 1 x 1 x2 2 x 1 x 1 x 1 factorise/faktoriseer: x 1 x 1 1 simplification / vereenvoudiging (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2019) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 5 QUESTION/VRAAG 4 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 4.1 4.1.1 9𝑥−1 = 81 32 and / en 34 (32 )𝑥−1 = 34 exponent law / eksponent 32𝑥−2 = 34 wet ∴ 2𝑥 − 2 = 4 exponents equal / 2𝑥 = 6 eksponente gelyk 𝑥=3 answer / antwoord (4) 4.1.2 (𝑥 + 5)(2𝑥 − 3) = 0 −5 3 3 ∴ 𝑥 = −5 or / of 𝑥 = (2) 2 2 4.1.3 2𝑥 + 16 2𝑥 + 16 = 10𝑥 = 10 𝑥 16 = 8𝑥 2𝑥 + 16 = 10𝑥 answer / antwoord 16 = 8𝑥 2=𝑥 (3) 4.2 2(𝑥 + 4) > 𝑥 + 6 expand / uitbreiding 2𝑥 + 8 > 𝑥 + 6 answer / antwoord 𝑥 > −2 diagram / diagram -2 (3) 4.3 𝑥2 𝑦2 − 2 𝑦2 + =1 𝑏 𝑎2 𝑏 2 × 𝑎2 𝑥2 𝑦2 𝑎2 (𝑏2 −𝑦 2 ) = 1 − 𝑎2 𝑏2 𝑏2 𝑦2 answer / antwoord 𝑥 2 = (1 − 2 ) × 𝑎2 𝑏 𝑏 − 𝑦2 2 𝑥2 = × 𝑎2 𝑏2 2 𝑎2 (𝑏 2 − 𝑦 2 ) 𝑥 = 𝑏2 𝑎 𝑥 = ± √𝑏 2 − 𝑦 2 (4) 𝑏 4.4 4.4.1 𝑃 = 2(𝑙 + 𝑏) answer / antwoord 30 = 2(2𝑥 + 𝑦) 𝑃 = 2(𝑙 + 𝑏) answer / antwoord 24 = 2(𝑥 + 𝑦) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2019) 4.4.2 Substitution/Vervanging 30 = 2(2𝑥 + 𝑦)… 24 = 2(𝑥 + 𝑦)… 15 = 2𝑥 + 𝑦 15 − 2𝑥 = 𝑦… making 𝑦 the subject / Sub/Vervang in : maak 𝑦 die onderwerp 24 = 2(𝑥 + (15 − 2𝑥)) substitution / vervanging 24 = 2(𝑥 + 15 − 2𝑥) 24 = 2(−𝑥 + 15) expanding / uitbreiding 12 = −𝑥 + 15 −3 = −𝑥 3=𝑥 answer for 𝑥 / antwoord Sub/Vervang 𝑥 = 3 in : vir x 15 − 2(3) = 𝑦 9=𝑦 answer for y / antwoord vir y OR/OF OR / OF Elimination/Eliminasie: 30 = 2(2𝑥 + 𝑦)… 24 = 2(𝑥 + 𝑦).. 30 = 4𝑥 + 2𝑦… 24 = 2𝑥 + 2𝑦… expanding / uitbreiding – : method / metode 30 = 4𝑥 + 2𝑦 24 = 2𝑥 + 2𝑦 6 = 2𝑥 simplification / 3=𝑥 vereenvoudiging Sub/Vervang 𝑥 = 3 in : answer for 𝑥 / antwoord vir x 30 = 4(3) + 2𝑦 30 = 12 + 2𝑦 18 = 2𝑦 answer for y / antwoord vir y 9=𝑦 (5) 𝑥 4.5 Distance/Afstand Speed/Spoed Time/Tyd x 120 km/h 𝑥 120 𝑥 120 160 x 160 km/h 𝑥 160 equation for time / 𝑥 𝑥 vergelyking vir tyd ∴ − = 2ℎ𝑟𝑠 120 160 simplification / × 480 ∶ 4𝑥 − 3𝑥 = 960 vereenvoudiging 𝑥 = 960 𝑘𝑚 simplification / vereenvoudiging ∴ Slower truck takes/Stadiger tron neem answer / antwoord 960 ÷ 120 = 8 hrs (6) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2019) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 7 QUESTION/VRAAG 5 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 5.1 5.1.1 Hairdryer cost / 𝑅17,58 method / metode Haardroër koste = £100 × £1 answer / antwoord = 𝑅1 758 (2) 5.1.2 Hairdryer cost / 𝑅 15,00 Haardroër koste = £100 × £1 answer / antwoord = 𝑅1 500 (1) 5.1.3 Savings = 𝑅1 758 − 𝑅1 500 = 𝑅258 answer / antwoord (1) 5.2 5.2.1 Deposit / deposito = 𝑅10 000 × 12% method / metode = 𝑅1 200 answer / antwoord (2) 5.2.2 𝐴 = 𝑃(1 + 𝑖𝑛) substitution into correct 𝐴 = 8 800(1 + 25%(5)) formula / vervanging in 𝐴 = 𝑅19 800 korrekte formule ∴ 𝑇𝑜𝑡𝑎𝑙 = 𝑅19 800 + 𝑅1 200 R8 800 = 𝑅21 000 answer / antwoord answer / antwoord (4) 5.2.3 Monthly installment / maandelikse betaling method / metode =𝑅19 800 ÷ (5 × 12) answer / antwoord = 𝑅330 (2) 5.3 𝐴 = 𝑃(1 + 𝑖)𝑛 correct formula / korrekte 𝐴 = 𝑅6 000(1 + 15%)5 formule 𝐴 = 𝑅12 068,14 substitution into correct formula / vervanging in korrekte formule answer / antwoord (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION/VRAAG 6 Ques./ SOLUTION/OPLOSSING EXPLANATION/ Vraag VERDUIDELIKING 6.1.1 x -4 -3 -2 -1 0 1 2 3 4 4 4 h(x) -1 -2 -4 ∞ 4 2 1 −3 3 p(x) 1 1 1 1 1 3 9 27 81 81 27 9 3 h(x): Correct values Undefined value p(x): Correct values (3) 6.1.2 h(x): shape / vorm quadrants / kwadrante asymptotes / asymptote p(x): shape / vorm y-intercept / y-afsnit (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2019) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 9 6.2 A: 𝑓(𝑥) = 𝑥 2 + 3 𝑦 = (0)2 + 3 𝑥=0 𝑦=3 ∴ 𝐴(0; 3) 𝐴(0; 3) C: 𝑔(𝑥) = 2𝑥 + 3 0 = 2𝑥 + 3 𝑔(𝑥) = 0 −3 = 2𝑥 −3 =𝑥 2 −3 𝐶( −3 ; 0) ∴ 𝐶 ( ; 0) 2 2 B: 𝑓(𝑥) = 𝑔(𝑥) 𝑓(𝑥) = 𝑔(𝑥) 𝑥 2 + 3 = 2𝑥 + 3 𝑥 2 − 2𝑥 + 3 − 3 = 0 standard form / standaardvorm 𝑥 2 − 2𝑥 = 0 𝑥=2 𝑥(𝑥 − 2) = 0 𝑥 = 0 of 𝑥 = 2 ∴𝑥=2 𝑦 = 2(2) + 3 𝐵 (2; 7) 𝑦=7 ∴ 𝐵 (2; 7) (8) 6.3.1 𝑥 ∈ ℝ answer / antwoord (1) 6.3.2 𝑦 ≥ 3 answer / antwoord (1) 6.4.1 −3 Inequality / ongelykheid 𝑥> Endpoint / Eindpunt 2 From coordinates of C in 6.2.1 Vanaf koördinate van C in 6.2.1 (2) 6.4.2 x 0 or/of x 2 From/Vanaf 6.2.1 Answer / antwoord Answer / antwoord (2) [22] TOTAL/TOTAAL: 100 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief