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2 MATHEMATICS P2 NOVEMBER 2013,QUESTION VRAAG 1, 1 1 100 143 150 155 164 171 171 180 182 188 190 min and max min en maks. min 100 Q1 150 Q2 171 Q3 182 max maks 190 Q1,100 150 171 182 190 4. 1 2 Skewed to the left skuins na links answer antwoord. 1 3 100 answer antwoord 1,QUESTION VRAAG 2,2 1 Mean Gemiddelde answer antwoord. 2 2 equating gelykstel,x 36 answer antwoord 2,2 3 8 88 answer antwoord 2. 2 4 outside buite 36 8 88 36 8 88 method metode,27 12 44 88.
3 people persone answer antwoord 2,QUESTION VRAAG 3. Interval Frequency Cumulative frequency,Frekwensie Kumulatiewe frekwensie. 2 2 first three correct,7 9 eerste drie korrek,14 23 remaining two correct. 12 35 oorblywende twee korrek,points accurate punte. 0 10 20 30 40 50 60,shape vorm 3,NOVEMBER 2013 WISKUNDE V2 3.
3 3 35 learners leerders answer antwoord, Accept 36 or 34 learners Aanvaar 36 of 34 leerders 2. QUESTION VRAAG 4,4 1 4 1 1 A 4 1 B 1 2 F k p method metode. 4 1 2 formula formule,substitution instelling,answer antwoord. OR OF formula formule,substitution instelling,answer antwoord. 4 1 3 B 1 2 and en,formula formule,substitution instelling.
equation in any form,vergelyking in enige vorm,4 2 C 2 5 A 4 1 F 2 5. formula formule,substitution instelling, CAF is scalene ongelyksydig scalene ongelyksydig 6. 4 3 tan m tan 1,4 MATHEMATICS P2 NOVEMBER 2013, 4 4 If BC were perpendicular to AF the triangle would have to explanation. be isosceles As BC loodreg op AF was sou die driehoek verduideliking. gelykbenig moes wees, C does not satisfy the equation of the perpendicular bisector. C maak nie die vergelyking van die middelloodlyn waar. 4 5 y 0 as D must be the midpoint of FC aangesien D die y 0 2. middelpunt van FC is 24,QUESTION VRAAG 5,5 1 x 2y 6 0.
y x 3 y x 3,y 5 x 2 substitution instelling,answer antwoord 4. 5 2 K 3 5 L 2 3 N 5 9 substitution instelling,OR OF OR OF. K L M are NOT collinear is NIE saamlynig NIE,not collinear nie. saamlynig nie,QUESTION 6 VRAAG 6,6 1 6 1 1 5tan 4 0. x 5 y 4 in,y 5 diagram,2cos 180o 2cos 2cos,answer antwoord.
6 1 2 sin2 90o sin2 cos2 sin2 cos2,substitution instelling. answer antwoord,NOVEMBER 2013 WISKUNDE V2 5,6 2 4cos2x tan 45 0. 4cos2x 1 tan 45 1,any two correct angles,x 60 or of x 120 or of x 240 or of x 300. enige twee korrekte hoeke,remaining two correct,angles oorblywende twee. korrekte hoeke,QUESTION VRAAG 7,7 2 2 x 90o 90o,71 57 k 180 71 57 k 180.
6 MATHEMATICS P2 NOVEMBER 2013,QUESTION VRAAG 8,8 1 a 2 a 2. r 45 r 45 4,8 2 f x g x 0 75,f x g x 105,x 75 or of 105 2. 8 3 360 360,8 4 g x sin x 45o,answer only full marks slegs. antwoord vol punte,QUESTION VRAAG 9,Sine rule sinus re l. answer antwoord 6,9 2 9 2 1 76 72 and en x 48 metres meter.
substitution instelling,MN 27 16 metres meter answer antwoord. 9 2 2 Area of KLN KL LN sin formula formule,48 88 sin 72o substitution instelling. 2 008 63 m2 answer antwoord,NOVEMBER 2013 WISKUNDE V2 7. QUESTION VRAAG 10, 10 1 10 1 1 Equal to angle in the alternate segment answer antwoord. Gelyk aan die hoek in die teenoorstaande 1, 10 1 2 Interior opposite angle teenoorstaande answer antwoord 1.
Constr Join OQ and OS Teken OQ en OS construction konstruksie. Proof Bewys,angle at centre hoek by middelpunt reason rede. angle at centre hoek by middelpunt,Hence dus 5,10 3 10 3 1. 5 1 reason,A S T rede,tan chord raaklyn koord reason. QR RS rede 4,10 3 2 sum of angles of a triangle reason rede. 10 3 3 cyclic quad koordevierhoek reason rede,8 MATHEMATICS P2 NOVEMBER 2013.
10 3 4 angle at centre middelpunthoek reason rede,QUESTION VRAAG 11. 11 1 OD 25 cm OC 25 cm 18cm 7 cm OC 7 cm,Pythagoras. reason rede,OD AB OD AB,answer antwoord 5,11 2 11 2 1 PR QB alt angles verw hoeke reason. Subtended by onderspan deur RB rede,Subtended by onderspan deur PQ OR reason. OF alt angles verw hoeke rede,11 2 2 angle at centre middelpunthoek reason rede.
11 2 2 angles of hoeke van,b reason rede 2,11 2 2 reason rede. 11 2 2 angle in semi circle hoek in answer antwoord. d half sirkel reason rede,QUESTION VRAAG 12,12 1 12 1 1 F. tangent is perp to a diameter,NOVEMBER 2013 WISKUNDE V2 9. 12 1 2 angle at centre middelpunthoek reason,12 2 alt angles verw hoeke AC DB reason. subtended by onderspan deur CR rede,both albei reason.
Hence PDBR is a cyclic quadrilateral rede, Ext angle int opp angle buitehoek teenoor binnehoek. NATIONAL SENIOR CERTIFICATE GRADE GRAAD 11 NOVEMBER 2013 MATHEMATICS P2 WISKUNDE V2 MEMORANDUM MARKS PUNTE 150 This memorandum consists of 9 pages