Mathematics P1 Grade 10 Nov 2015 Memo Eng & Afr_hlayiso.com_.pdf

Downloaded from hlayiso.com basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA ey, NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 eet RSS RRR RRR RRR RRR RRR RR RRR RR RRR RRR RRR RR RRR REE EE By i MATHEMATICS P1/WISKUNDE V1 Hi rT] u " Ll ; NOVEMBER 2015 4 1! Ll i MEMORANDUM ’ "ee ee ee ee ee eee eee eee ” MARKS/PUNTE: 100 This memorandum consists of 9 pages. Hierdie memorandum bestaan uit 9 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 2 CAPS/KABV -— Grade/Graad 10 —- Memorandum NOTE: DBE/November 2015 If a candidate answered a question TWICE, mark only the FIRST attempt. e Ifacandidate crossed out an answer and did not redo it, mark the crossed-out answer. ¢ Consistent accuracy applies to ALL aspects of the marking memorandum. e Assuming values/answers in order to solve a problem is unacceptable. LET WEL: e As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. e As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het antwoord na. nie, sien die deurgehaalde Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Dit is onaanvaarbaar om waardes/antwoorde aan te neem om 'n probleem op te los. QUESTION/VRAAG 1 1.1.1 x* 81 = (x? -9\x? +9) = (x =3\x+3)(x? +9) v (x? ox? +9) ¥ (x-3\x+ 3\(x? + 9) (2) 1.1.2 6x? y—10xy + 15x —25 = 2xy(3x —5)+5(3x—5) , ey = (2xy +5)3x-5 a Gay + 5)Gx—5) Y (2xy+5)(3x—5) OR/OF (3) 6x? y—10xy +15x—25 Y 3x(2xy +5) = 3x(2xy + 5)—5(2xy +5) v—5(2xy +5) = (2xy + 5)(3x—5) Y (2xy +5)(3x-5) (G3) 1.2.1 3 2 21 + - a-4 a+3 a’-a-12 3 2 21 = - Vv (a- a4 a43 (a-4)(a +3) (a 4Ya+3) _ 3(a+3)+2(a-4)-21 yy 3a43)+2(a-4)-21 (a—4\Ya+3) (a—4\a+3) _3a+9+2a-8-21 7 (a—4)(a +3) 5a-20 = Ga\ar3) ¥ simplification, i-e./ a ahat vereenvoudiging, d.i. ___d(a-4) 5a-20 (a—-4Ya+3) (a—4\a+3) 5 443 Y answer/antwoord (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 3 DBE/November 2015 CAPS/KABV -— Grade/Graad 10 —- Memorandum 1.2.2 102°3,.4' 25°** Ae 2x43 2x43 92-2 92443 52x13 22x ~ gtx =2 2xt342-2x 5 2x+3-4-2x v writing bases in terms of prime factors/ skryf basisse in terme van priemfaktore Vv simplification/ vereenvoudiging v adding and subtracting =25.57 ~o indices/optel en afirek _ 32 van eksponente 5 - 62. v2.51 orlof 2 orlof 62 5 5 5 (4) 1.3.1 [97 v answer/antwoord qd) 1.3.2 [37 v answer/antwoord qd) [16] QUESTION/VRAAG 2 2.1.1 15x? -14x-8=-0 Y standard form/standaardvorm (5x+2)3x-4)=0 Y factorisation/faktorisering 5x+2=0 or 3x-4=0 9 4 vv answers/antwoorde x=- = or x= > (4) 5 3 2.1.2 ga 125 gat 5 v5 5* =5° ¥ answer/antwoord x=-3 (2) 2.2.1 3(x+7)<=+41 : Y 3x421 3x+21<—+4+1 2 Y 6x+42<x42 6x+42<x+2 5x <—40 x <8 v answer/antwoord 6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 4 DBE/November 2015 CAPS/KABV -— Grade/Graad 10 —- Memorandum 2.2.2 Y indicating numbers to the <—O : left of -8 and —8 not included/ dui getalle links van -8 aan met ¢ y -8 nie ingesluit Q) 2.3 Let the amount of money Mary had be Rx/Laat die bedrag V 1 28 geld wat Mary gehad het x wees. 3 ~ 1. = 1 —28 v lL 5 5 3x +420 = 5x v equation/vergelyking 2x = 420 x=210 ¥ 210 Mary had R210/Mary het R210 gehad. (4) [14] QUESTION/VRAAG 3 3.1.1 |-7;-12 Yo v —-12 (2) 3.12 | T=-sn+13 Sn v 13 (2) 3.1.3 T, =—-5n+13 Ty = -5(30) +13 . Substitution of/substitusie van =-137 ¥ answer/antwoord (2) 3.1.4 | —5n+13 = -492 v¥ —5n+13=-492 —5n = -505 n=101 Y answer/antwoord _ (2) 3.2.1 | 7, =2n-1 ¥ 2n v-l (2) 3.2.2 | T =(2n-1) Y (2n-1) =4n? —4n4+1 (1) 3.230 | 7 =(2n-1)-(2n-1) Y (2n-1)-(2n-1) = 2n—-1-(4n? —4n+1) Y 2n-1-(4n? -4n +1) =2n-1-4n? +4n-1 ¥ 2n-1-4n? +4n-1 =n’ +6n-2 ¥ answer/antwoord (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 5 CAPS/KABV -— Grade/Graad 10 —- Memorandum DBE/November 2015 QUESTION/VRAAG 4 41 |y=1 ¥ answer/antwoord () 42 fi v shape of f/vorm van f ¥ x-intercepts of f/ x-afsnitte van f v y-intercept (TP) of f/y-afsnit (DP) (0; 2) van f g: ¥ shape of g/vorm van g x v asymptote of g/ (-1; 0) 0 (1; 0) asimptoot van g f v y-intercept of g/ y-afsnit van g (6) 4.3 | Range of f/Waardeversameling van f: (-<0 ; 2] v (-0; 2] qd) OR/OF Range of f/Waardeversameling van f- ys2 Vys2 e) 4.4 | Maximum of 3/) will be obtained when f(x) is at maximum. Max of f(x) is 2 Y Max of f(x) is 2/ Max of h will be 3°=9 Maks van f(x) is 2 Maksimum van 3!) sal verkry word wanneer f(x) by maksimum is. ¥ Max of h=9/ Maks van f(x) is 2 Maks van h = 9 Maks van h sal 3? =9 wees. Q) 4.5 | f would have been reflected in the x-axis Vv reflected/gereflekteer Vin the x-axis/ f sou indie x-as gereflekteer gewees het in die x-as Q) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 DBE/November 2015 CAPS/KABV — Grade/Graad 10 — Memorandum QUESTION/VRAAG 5 5.1 a = gradient of g Van _-4-4 -1-2 ~ =-1-3 a =2 ¥ substituting/substitusie BG ; 4) 4 =2(3)+q q=-2 g(x) =2x-2 (2) OR/OF a=gradient of g y 4- (-4) 4—(-4) a= 3—(c1) 3-(-1) =2 v substituting/substitusie -4=2(-1)+¢ ACI; —-4) q=-2 g(x)=2x-2 (2) OR/OF g(x) =ax+q ¥ substituting both points/ 4=3a+q 1 substitusie van beide punte 4H Qeeeecceies 2 1-2: 8=4a a=2 . . Substitute in 1/Substitusie in 1: solving simultaneously/ 4- 3(2) +q los gelyktydig op a=? (2) g(x) =2x-2 5.2 1 z ~1=2x-2 Y equating/gelykstelling L =2x-1 x =? _ ! =x * Y standard form/ 2x° —x-1=0 standaardvorm (2x +1Xx-1)=0 1 x=- > or x=l1 2 Vv factors/faktore v x-values/-waardes (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 DBE/November 2015 CAPS/KABV — Grade/Graad 10 —- Memorandum 53 te peo or/of x>1 Vxe—t 2 2 vx<0 OR/OF ¥x2>1 y GQ) 1. . [-0,5 Smuts) Y0) ¥ [1 50) QB) 5.4 1 1 2 3)=—-1 Y=-lor -4 1@)=5 ; 5 2 3 Length of BE =4~- f(3) Y4—£() - +-(-3) 3 =442 3 =42 Y answer/antwoord 3 GQ) OR/OF 1 BE =2x-2-1 41 ¥ 2x-2--—+4+1 x x _ 2x? —x-1 x 2 2 (x =3) pe 28) -@)-1 y 28) -@)-1 3 3 _ 18-4 3 2 =4— ¥ answer/antwoord 3 @) 5.5 | h{x)= f(x)+3 1 ¥ answer/antwoord A(x)=—+2 (1) * [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 8 CAPS/KABV -— Grade/Graad 10 —- Memorandum DBE/November 2015 QUESTION/VRAAG 6 6.1 | d-5+d-1=0 Vd-5+d-1=0 2d =6 Vd=3 2 d=3 (2) 6.2 y =a(x—2)(x+2) ~9 =a(l—2\(1+2) Y y=al(x—2)(x+2) ~9=a(-1)3) Y subs (1; -9) —3a=-9 a=3 Vva=3 f(x) =3(x? —4) =3x? -12 c=-12 “ec =-12 (4) [6] QUESTION/VRAAG 7 7A R5000 Vv selects/kies sr = $525,29 9,518569 rands per dollar 9,515869 v answet/antwoord OR/OF (2) Y selects/kies R5000 x 0,105058 dollars per rand = $525,29 0,105058 v answet/antwoord Q) 7.2.1 | A= P(i+i)" Y formula/formule v 3 = 5000(1 +0,061)° 5000(1 + 0,061) v R5 971,95 =RS5 971,95 (3) 7.2.2 | Let the amount that Zach invests each year be x/Laat die bedrag wat Zach elke jaar belé, x wees. x(1+ 0,09) + x(1+0,09)' = 5980 ¥ x(1+0,09) x{L,09? +1,09]= 5980 v x(1+0,09) 5980 x =—.—_ 1,09? + 1,09 =R2 624,99 OR/OF Let the amount that Zach invests each year be x/Laat die bedrag wat Zach elke jaar belé, x wees. [x(1 +0,09)' +x|1+0,09)' = 5980 x(2,09\(1,09) = 5980 5980 (2,09)(1,09) =R2 624,99 vx as common factor/ as gemeenskaplike faktor ¥ answer/antwoord (4) ¥ x(1+0,09)' v [x(1+0,09)' +2] v x as common factor/ as gemeenskaplike faktor v answet/antwoord (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P 1/Wiskunde V1 9 CAPS/KABV -— Grade/Graad 10 —- Memorandum DBE/November 2015 QUESTION/VRAAG 8 8.1.1 Sample space/Steekproefruimte (64) ¥ diagram shape/ diagramvorm Soccer/sokker Rugby (28) v‘14 in correct position/ (24 in korrekte posisie v10 in correct position/ in korrekte posisie v 18 in correct position/ in korrekte posisie ¥ 22 in correct position/ in korrekte posisie 22 8.1.2 answer (in any form (a) P(Soccer and Rugby) = a = = = 0,15625 = 15,63% vapor (i onige vo) (1) 8.1.2 (by | P(Soccer or Rugby) = ee eee = = = = = 0,65625 = 65,63% | OR/OF v answer Gin any form)/ antwoord (in enige vorm) P(Soccer or Rugby) =1 2 = 21 Q) 64 32 8.1.3 | No/Nee. Y No/Nee Some boys play both soccer and rugby/Party seuns speel sokker en | ¥ Reason/Rede rugby. (2) OR/OF No/Nee v No/Nee P(S and R) #0/P(S enR) #0 v Reason/Rede (2) 8.2 P(more than 2 passengers per car) / P(meer as 2 passasiers per kar) 541 v numerator/teller 6 => v denominator/ 74+114+6+5+1 noemer 30 _ 6 v answer/antwoord 30 (accept/aanvaar = L = 0,2 = 20% Sot or/of 5 30 5 0,2 or/of 20%) @) 8.3 P(not getting a six)/P(nie 'n ses kry nie) y{1o_ 1 (z 1 } (3 * 5s] =1]-|—4+— 36 36 ; at 10 1 } + 25 36 36 36 v2 36 (3) [15] Copyright reserved/Kopiereg voorbehou TOTAL/TOTAAL: 100