Mathematics Grade 10 Paper 2 Gauteng June 2019 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN JUNE / JUNIE 2019 GRADE / GRAAD 10 MATHEMATICS / WISKUNDE PAPER / VRAESTEL 2 MEMORANDUM 6 pages / bladsye

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM GRADE 10 / GRAAD 10 GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION / VOORBEREIDENDE EKSAMEN MATHEMATICS / WISKUNDE (Paper / Vraestel 2) QUESTION 1 / VRAAG 1 1.1.1 cos  b b   c c (1) 1.1.2 sin (90   ) b b   c c (1) 1.2.1 r 5 1 2 sin  4 4 1  4  2     5 4 5  4 4   25 25 (2) 1.2.2 LHS//LK LK RHS//RK RK 5 2    1 sec  1 2 tan  2 3  4  2 2  4  2 5    1     3  3   3  16 16 16    9 9 9 (3) [7] 2

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM GRADE 10 / GRAAD 10 QUESTION 2 / VRAAG 2 2.1 2,34 answer / antwoord (1) 2.2 cos (2 10)  0, 091 2 cos (2 10)  0,184  79, 4 2 10  79, 4  44, 7   44, 7 (2) ONLY PLACE IN PAPER WHERE LEARNER IS PENALIZED FOR INCORRECT ROUNDING OFF / ENIGSTE PLEK IN VRAESTEL WAAR LEERDER GEPENALISEER WORD VIR VERKEERDE AFRONDING 2.3 1 2sin 2 60  sin 45 . sec 45  tan10 . cot10 4 3 2   3  2  2  1 2  2      .    (1)  2  2   2   2   2 4   2  .  2  3 1    2   1  1 4 4 3 3   4 4 (4) [7] 3

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM GRADE 10 / GRAAD 10 QUESTION 3 / VRAAG 3 3.1 shape ) / vorm turning points )/ draaipunte shape ) / vorm turning points )/ draaipunte (4) 3.2.1 y [2 ;  2] answer / antwoord OR/OF OR / OF 2  y  2 answer / antwoord (1) 3.2.2 90 x  270 correct critical values / korrekte kritieke waardes correct inequality / korrekte ongelykheid (2) [7] 4

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM GRADE 10 / GRAAD 10 QUESTION 4 / VRAAG 4 4.1 Midpoint RS/ Middelpunt RS = (2;1) 2 1 (2) 4.2 2  x 1 y 2  x 1 y 2 1   2 and / en 1 2 2 2 2 x6 y 1  x6  y 1 (3) T (6;1) 4.3 RU  (2  2)2  (5 1) 2  (2  2)2  (5 1)2  32  32 (2) 4.4 1 5 y 3 M RU  M SP  1 2  2 x2  M RU = 1 1 y  x 5  y  x 5 SP  ( x  2) 2  ( y  3) 2  2 32  ( x  2)2  ( y  3)2 x 2  4 x  4  y 2  6 y  9  2 32  128 substitute / vervanging x 2  4 x  13  ( x  5) 2  6( x  5) 128 2 x 2  8 x  90  0 x 2  4 x  45  0  x 2  4 x  45  0 ( x  9) ( x  5)  0  x 9 ; x   5 x 9 or / of x5 y 95 y  55 y4 y   10  y  4 ; y   10 P(9; 4) P( 5;  10) (8) [15] 5

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM GRADE 10 / GRAAD 10 QUESSTION 5 / VRAAG 5 5.1 parallelogram answer / antwoord (1) 5.2.1 E1 = x <’s opp equal sides / S/R <’e teenoor gelyke sye E3 = 2x <’s opp equal sides / S x + 90° + 2x = 180° <’e teenoor gelyke sye <’s straight line / S x = 30° <’e op reguityn S (4) 5.2.2 A = 120° sum <’s  / som vd <’e  S C1 = 60° sum <’s  / som vd <’e  alt <’s AD||BC/ C2 = 60° verwis <’e AD||BC R S A=C opp <’s equal / oorst <’e ABCD is a parm gelyk R (4) [9] QUESTION 6 VRAAG 6 AO = OC  2 xy given / gegee  AO = OC  2xy BO = OD  x 2  y 2 given / gegee BO = OD  x2  y 2 BO 2 + OC 2  ( x 2  y 2 ) 2  (2 xy ) 2  ( x2  y 2 )2  (2 xy)2  x4  2x2 y 2  y 4  4x2 y 2  x4  2x2 y 2  y 4  ( x 2  y 2 )2  ( x 2  y 2 )2 BC2  ( x 2  y 2 )2 BC2  BO 2 + OC 2 pythagoras S/R BOC =90  90° diagonals bisect at 90 / R ABCD is a rhombus / is ʼn diagonale halveer 90 (5) ruit [5] TOTAL / TOTAAL: 50 6