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Memorandum

Gr10 Math P1 (Afrikaans) June 2017 Possible Answers_hlayiso.com_.pdf

Subject: MathematicsGrade 1020176 pages
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Downloaded from hlayiso.com GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINSIALE EKSAMEN JUNIE 2017 GRAAD 10 WISKUNDE VRAESTEL 1 MEMORANDUM 6 bladsye 1
Downloaded from hlayiso.com WISKUNDE MEMORANDUM (VRAESTEL 1) GRAAD 10 GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINSIALE EKSAMEN VRAAG 1 1.1 Rasionaal ✓ (1) 1.2 Irrasionaal ✓ (1) 1.3 Rasionaal ✓ (1) [3] VRAAG 2 2.1 5 0   ✓antwoord  12  1 (1) 2.2 1  x 1 ✓antwoord x (1) 2.3 9 x 1.5 x  2 45 x 1  3  .5 2 x 1 x2 3 . 5 2 x 1 32 x  2.5 x  2 ✓ 32 x  2  2 x  2 x 1 ✓ 32 x2 . 5 x1 3 .5  32 x2( 2 x2).5 x2( x1) ✓antwoord (3)  51 [5] VRAAG 3 3.1 2 x 2  14 x  60 ✓gemeenskaplike faktor  2 x  10  x  3 ✓ ( x 10) ✓ ( x  3) (3) 3.2 1 3 9 x b 8 ✓eerste faktor ✓tweede faktor 1  1 1    x  b 3  x 2  b 3 x  b 6  (2) 2  4 2  [5] 2
Downloaded from hlayiso.com WISKUNDE MEMORANDUM (VRAESTEL 1) GRAAD 10 VRAAG 4 4.1 2  3x  6  4 x 4 x  3x  6  2 ✓transponeer x4 ✓antwoord (2) 4.2 x x 3x  2   2 2  x 3 x x  x  6 ✓verander teken  x x x 3x  2 x 3   x  2 x  3 x  3x  2 ✓KGV  x  2  x  3 x  x  3  x  x  2   3 x  2 ✓vereenvoudig x 2  3x  x 2  2 x x 2  3x  x 2  2 x  3x  2 ✓  8x   2  8x   2 ✓antwoord 1 x 4 (5) 4.3 3 x . 92 x1  81 3 x . 34 x  2  34 35 x2  34 ✓dieselfde basis 5x  2  4 ✓eksponent wet 2 ✓antwoord x (3) 5 4.4  4  3x  1  5  4  1  3x  5  1 ✓tel 1 beide kante by  3  3x  6 ✓  3 : albei kante 3: 1  x  2 ✓antwoord (3) [13] 3
Downloaded from hlayiso.com WISKUNDE MEMORANDUM (VRAESTEL 1) GRAAD 10 VRAAG 5 Patroon 1 Patroon 2 Patroon 3 5.1 Tn  7n  3 ✓ 7n ✓3 (2) 5.2 Tn  2n ✓ 2n (1) 5.3.1 T12  2(12) ✓antwoord  24 (1) 5.3.2 Tn  7 n  3 150  7 n  3 ✓ = 150 ✓ n  21 n  21 (2) [6] 4
Downloaded from hlayiso.com WISKUNDE MEMORANDUM (VRAESTEL 1) GRAAD 10 VRAAG 6 y f D(3; 4) B C D 0 x h 6.1 y  ax 2 (3;4) 4  a.3 2 ✓vervang (3; 4) 4  9a 4 4 ✓a  a 9 9 4 y  x2 (2) 9 6.2 2 4 2 y x2 en y x 3 9 4 2 2 x  x2 ✓stel vergelykings gelyk 9 3 9: 4 x 2  6 x  18 ✓vereenvoudig 2: 2 x 2  3x  9  0 (2 x  3)( x  3)  0 3 x  0:  xC   2 ✓korrekte x koördinaat 2 3 yC      2 ✓vervang 3 2  1 2 ✓ y koördinaat 1  3  C   ; 1  2  (5) 5
Downloaded from hlayiso.com WISKUNDE MEMORANDUM (VRAESTEL 1) GRAAD 10 6.3 x -afsnit: y  0 2 0 x2 3  3 : 0  2x  6 ✓vervang y  0 2 x  6 2: x  3 ✓antwoord (+) WO = 3 eenhede (2) 6.4 T 4 ; 3 ✓antwoord (1) [10] VRAAG 7 y  3. 2 x  1 y ✓ asimptote ✓vorm ✓ y -afsnit (1; 7) 4 6 y  1 1 x ✓vorm -6 0 x ✓ x -afsnit ✓enige ander koördinaat Snyding (1; 7) ✓✓ antwoord (8) [8] TOTAAL: 50 6

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