Technical Mathematics Grade 10 P1 Eastern Cape November 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 NOVEMBER 2018 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 100 This marking guideline consists of 8 pages./ Hierdie nasienriglyn bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2018) QUESTION/VRAAG 1 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 1.1 1.1.1 3 √9 answer / antwoord (1) 1.1.2 √9 answer / antwoord (1) 1.2 √9 < √11 < √16 3 √11 lies between / lê tussen 3 and/en 4 4 (2) 1.3 7𝑥 2 − 2𝑥 − 6 − (3𝑥 2 − 5𝑥 − 7)  −3𝑥 2 + 5𝑥 + 7 = 7𝑥 2 − 2𝑥 − 6 − 3𝑥 2 + 5𝑥 + 7  answer / antwoord = 4𝑥 2 + 3𝑥 + 1 (2) 1.4 1 1 12  method / metode 1 0 0 12 1 1 1 1 1 12  1112 1 0 0 12 11012 7 1 0 0 12 1 0 0 12 1 0 0 12 0000 A n s w e r / A n tw o o r d  1 1 12  7 (3) 1.5 1.5.1 2𝑥 2 + 4𝑥 − 2𝑥 2 + 3𝑥 + 3  2𝑥 2 + 4𝑥 = 7𝑥 + 3  3𝑥 + 3  answer / antwoord (3) 1.5.2 𝑎3 + 3𝑎2 + 9𝑎 − 3𝑎2 − 9𝑎 − 27  𝑎3 + 3𝑎2 + 9𝑎 = 𝑎3 − 27  −3𝑎2 − 9𝑎 − 27  answer / antwoord (3) 1.5.3 −2 + 6𝑖 − 3𝑖 + 9𝑖 2  product / produk = −2 + 3𝑖 + 9(−1) 𝑖 2 = −1 = −11 + 3𝑖  answer / antwoord (3) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 3 QUESTION/VRAAG 2 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 2.1 2.1.1 (𝑥 2 + 9)(𝑥 2 − 9) = (𝑥 2 + 9)(𝑥 − 3)(𝑥 + 3)  (𝑥 2 + 9)(𝑥 2 − 9)  (𝑥 − 3)(𝑥 + 3) (2) 2.1.2 6𝑥 2 𝑦 − 10𝑥𝑦 + 5𝑥 − 25 grouping / groepering = (6𝑥 2 𝑦 − 10𝑥𝑦) + (15𝑥 − 25) = 2𝑥𝑦(3𝑥 − 5) + 5(3𝑥 − 5) HCF / HGF = (3𝑥 − 5)(2𝑥𝑦 + 5) factors / faktore OR / OF (6𝑥 2 𝑦 + 15𝑥) + (10𝑥𝑦 − 25) = 3𝑥(2𝑥𝑦 + 5) − 5(2𝑥𝑦 + 5) = (2𝑥𝑦 + 5)(3𝑥 − 5) (3) 2.2 2.2.1 3𝑥+2 .27𝑥−2  33 and/en 34 81𝑥 3𝑥 .32 .(33 )𝑥−2  simplification / = vereenvoudiging (34 )𝑥 𝑥+2+3𝑥−6−4𝑥 =3  3−4 = 3−4 1 1  answer / antwoord = 4= 3 81 (4) 2.2.2 𝑥 3 +𝑦 3 𝑥 3 𝑦−𝑥 2 𝑦 2 +𝑥𝑦 3  factorising sum of two 2𝑥 3 −𝑥 2 𝑦−3𝑥𝑦 2 ÷ 4𝑥 4 −9𝑥 2 𝑦 2 cubes faktorisering van som van twee derdemagte (𝑥+𝑦)(𝑥 2 −𝑥𝑦+𝑦 2 ) = ×  𝑥(2𝑥 − 3𝑦)(𝑥 + 𝑦) 𝑥(2𝑥−3𝑦)(𝑥+𝑦) 2 𝑥 (2𝑥−3𝑦)(2𝑥+3𝑦) 𝑥𝑦(𝑥 2 −𝑥𝑦+𝑦 2 )  𝑥𝑦(𝑥 2 − 𝑥𝑦 + 𝑦 2 )  𝑥 2 (2𝑥 − 3𝑦)(2𝑥 + 3𝑦) 2𝑥+3𝑦 = 𝑦  answer / antwoord (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2018) QUESTION/VRAAG 3 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 3.1 3.1.1 5𝑥 = 5−3  5−3 𝑥 = −3  answer / antwoord (2) 3.1.2 3𝑥 + 1 = 2𝑥  2𝑥 3𝑥 − 2𝑥 = −1  3𝑥 − 2𝑥 = −1 𝑥 = −1  answer / antwoord (3) 3.1.3 𝑥 = −13 𝑜𝑟/𝑜𝑓 𝑥 = 1  −13 1 (2) 𝑥 3.1.4 3𝑥 + 21 < 2 + 1 3𝑥 + 21 6𝑥 + 42 < 𝑥 + 2 6𝑥 + 42 < 𝑥 + 2 5𝑥 = −40 answer / antwoord 𝑥 < −8 OR / OF 𝑥  indicating numbers to the 3𝑥 − 2 < 1 − 21 left 5𝑥 < −20 of -8 and -8 not included / 2 𝑥 = −8 Toon aan getalle links van -8 en -8 is nie ingesluit -8 nie. (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 5 QUESTION/VRAAG 4 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 4.1 1.675 × 10−27 𝑘𝑔  answer / antwoord (2) 4.2  answer / antwoord 𝐴 𝑅 = √ + 𝑟2 𝜋 (2) 4.3 𝑆 = 𝑣𝑡  𝑠 = (𝑥 + 5)4 Car/Kar  𝑠 = 4𝑥 𝑣 =𝑥+5  4(𝑥 + 5) + 4𝑥 = 380 𝑆 = (𝑥 + 5)4  𝑥 = 45 Truck/Trok  car speed / kar spoed 𝑣=𝑥  truck speed / trok spoed 𝑠 = 4𝑥 4(𝑥 + 5) + 4𝑥 = 380 𝑥 = 45 Truck speed/Trokspoed = 45 𝑘𝑚⁄ℎ Car speed/Karspoed = 50 𝑘𝑚⁄ℎ (6) 4.4 4.4.1 Floor plan-1/Vloerplan-1 sum of the lengths= 70 𝑚 2(2𝑥 − 10) + 2(5 + 𝑦) = 70 som van die lengtes = 70 m 4𝑥 − 20 + 10 + 2𝑦 = 70 𝑦 + 2𝑥 = 40 ……….(1) 𝑦 + 2𝑥 = 40 Floor plan-2/Vloerplan-2 1 sum of the lengths = 60 𝑚 2(2𝑥 − 10) + 2[2 (5 + 𝑦)]=60 som van die lengtes = 60 m 4𝑥 − 20 + 5 + 𝑦 = 60 4𝑥 − 2𝑦 = 75 4𝑥 + 𝑦 = 75 ……….(2) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2018) 4.4.2 From equation (1) / Vanaf vergelyking (1) 𝑦 = 40 − 2𝑥 ……(3) Substitute eq (3) in eq (2) / Vervang verg.(3) in 𝑦 = 40 − 2𝑥 verg.(2) substitution / vervanging 4𝑥 + (40 − 2𝑥) = 75 2𝑥 = 35 35 ∴𝑥= = 17,5 𝑥-value / x-waarde 2 Substitute the value of 𝑥 into(3)/Vervang waarde van x in (3) 35 𝑦-value / y-waarde 𝑦 = 40 − 2 ( 2 ) 𝑦=5 OR / OF 2𝑥 + 𝑦 = 40 ……. (1) 4𝑥 + 𝑦 = 75……… (2) (2) – (1) 2𝑥 = 35 35 𝑥 = 2 = 17.5 𝑦=5 (4) [18] QUESTION/VRAAG 5 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 5.1 5.1.1 A (−2; 0)  (-2;0) B (2; 0)  (2;0) C (0; 4)  (0;4) D (0; −2)  (0;-2) (8) 5.1.2 𝑓(𝑥) = 𝑔(𝑥) equating the functions / −𝑥 2 + 4 = 𝑥 − 2 gelykstel van die funksies 𝑥2 + 𝑥 − 6 = 0 standard form / (𝑥 + 3)(𝑥 − 2) = 0 standaardvorm 𝑥 = −3 𝑜𝑟 𝑥 = 2 factors / faktore 𝑦 = −3 − 2 = −5 𝑥-values / x-waardes E(−3 − 5) 𝑦-value / y-waarde coordinates of E / koördinate van E (6) 5.1.3 CD= 𝑂𝐶 + 𝑂𝐷 4 + 2 = 4 + 2 = 6 units / eenhede answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 7 5.2 5.2.1 𝑥∈𝑅 answer / antwoord (1) 5.2.2 𝑦 ≤ 4 OR/OF 𝑦 answer / antwoord (1) 5.3 𝑗(𝑥) = −(𝑥 2 + 4) substitution / vervanging = 𝑥2 − 4 answer / antwoord (2) 5.4 shape of 𝑘 / vorm van k asymptotes of 𝑘 / asimptote van k y-intercept of ℎ y-afsnit van h shape of ℎ / vorm van h asymptote of ℎ asimptote van h (5) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2018) QUESTION/VRAAG 6 NO. SOLUTION/OPLOSSING EXPLANATION/ VERDUIDELIKING 6.1 𝐴 = 𝑃(1 + 𝑖)𝑛 formula / formule 𝑅30 000 = 𝑃(1 + 0.135)11 substitution / 𝑃 = 𝑅7450,18 vervanging answer / antwoord (3) 6.2 6.2.1 𝑅200 000 answer / antwoord (1) 6.2.2 Simple interest, because the interest is constant.  Simple interest / Enkelvoudige rente Enkelvoudige rente, omdat die rente konstant is.  reason / rede (2) 6.2.3 𝑅200 000 + 25 000 + 25 000 + 25 000 + 25 000 + method / metode 25 000 + 25 000 = 𝑅350 000 answer / antwoord It will cost 𝑅350 000 / Dit sal R350 000 kos OR/OF In 2018, it cost/kos dit 𝑅325 00 ∴ In 2019, it will cost/sal dit kos = 𝑅325 00 + 𝑅25 000 = 𝑅350 000 (2) 6.2.4 𝐴 = 𝑃(1 + 𝑖𝑛) formula / formule 325 000 = 200 000(1 + 5𝑖) substituting (any 1,625 = 1 + 5𝑖 point from the 0,625 = 5𝑖 graph) 𝑖 = 0,125 vervanging (enige Interest rate/Rentekoers = 12,5% punt vanaf die grafiek) 0,625 = 5𝑖 answer / antwoord (4) 6.3 𝐴 = 𝑃(1 − 𝑖𝑛)  substituting in a = 385000(1 − 0,06 × 5) correct formula = 𝑅269500 Vervaning in ’n korrekte formule answer / antwoord (2) [14] TOTAL/TOTAAL: 100 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief