# Mathematics Grade 9 June Common 2010 Memo (Afrikaans)

> Grade 9 Mathematics memorandum from the June Common 2010 (Afrikaans), 6 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-9-mathematics-2010-june-common-memorandum/666a37c0-87e8-44c6-ae7b-095d48146f63
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 9
- Document type: Memorandum
- Year: 2010
- Exam period: June Common
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 9 · June Common · 2010. Memorandum, 6 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Province of the
                  EASTERN CAPE
                  EDUCATION

                        SENIOR FASE

                         GRAAD 9

                        JUNIE 2010

                     WISKUNDE
                    MEMORANDUM

PUNTE: 100

TYD: 2 uur

             Hierdie memorandum bestaan uit 05 bladsye.

### Page 2

2            WISKUNDE   (JUNIE 2010)

AFDELING A

1.   D√

### Page 3

(JUNIE 2010)                                  WISKUNDE                                            3

  2.      C√
  3.      B√
  4.      C √√
  5.      A √√
  6.      A√
  7.      C√
  8.      B√
  9.      A √√
  10      A√
  11.     D√
  12.     A √√
  13.     C√
  14.     D √√
  15.     B√
  16.     C√
  17.     C√
  18.     D √√
  19      D √√                                                                                 [26]

  AFDELING B

   VRAAG         Nr      OPLOSSING                                          PUNTE
                 1.1.1                                                      2
   1                                                                        1 punt spesifiek
                             3
                         = 3k m √√   -4                                     vir negatiewe
                                                                            indeks
                 1.1.2                                                      4
                                 2        2
                         =-[(a -4a+4) +(a +a-6)] +8√
                         = -[2 a2 – 3a – 2]+8√
                         =-2 a2+3a+2+8√
                         ==-2a2 + 3a + 10√
                 1.2     4(a + b) – x2(a+ b)                                3
                         = (a+b)(4 – x2) √
                         =(a+b) (2 – x)(2+ x) √√
                 1.3     1112-992                                           2
                         = (111 – 99)(111 + 99) √
                         =12 x 210
                         = 2 520         √
                                                                            3
                 1.4.1   3x + 15 – 6 = -2x – 2√
                         5x           = -11√
                         x            = -11/5√
                 1.4.2   3x + 1 = 27                                        3
                         3x + 1 = 33√
                         x + 1 = 3√
                         x=2 √

   2             2.1.1   Patroon 5 =32 aangesien die gemene verskil 7 is√   1

### Page 4

4                                           WISKUNDE                                            (JUNIE 2010)

        2.1.2   7x1–3 =7-3=4                  of enige logiese metode               3
                7x2–3 = 14–3 =11                                                    1 punt vir
                7x3–3 = 21–3 =18√                                                   eksplorasie
                                                                                    1 punt vir 7n en
                7xn -3= 7n-3√√                                                      1 punt vir -3

        2.1.3   7x200 -3 =1397                                                      1
        2.2.1         √                                                             1
        2.2.2                      √√ omdat ewewydige lyne dieselfde gradiënt het   2
                                                                                    Gee volle punte
                                                                                    al is daar geen
                                                                                    rede nie
        2.2.3   By die x-afsnit is y=0..                                            3
                                              √
                               √
                                     √
        2.2.4             √√                                                        2

    3   3.1.1   Som van die hoeke van ŉ Seshoek =4x 180º                            1
                                                    = 720º
        3.1.2   Aantal driehoeke in enige veelhoek = n-2√                           2
                Dit wil sê die som van die binne hoeke van die veelhoek
                =n–2 x 180º√
        3.1.3   1080= 180º x 6√                                                     2
                      =180º x (8–2)
                So dit is ŉ oktagoon is √

        3.2.1                                                                       2

                                                                                    1ste Kwadrant
                                                                                    Beeld

### Page 5

(JUNIE 2010)                                      WISKUNDE                                              5

   3.           3.2.2    Sien 3.1.1 4de kwadrant √                                    3
                         A  (2;-1), B  (3;0), C  (2;0), D  (1;-1) √√          1 punt vir
                                                                                      korrekte beeld
                                                                                      2 punte vir
                                                                                      korrekte
                                                                                      koördinate

   4            4.1      S = d/ t                                                     4
                         20 km/h = 1√
                                    t
                         t = 0,05 h√
                           = 3 min√
                         Agterkant van die trein sal om 1:06 nm of 13:06 uit tonnel
                         vaar√
                4.2         8 - 5     =135ℓ√                                          4
                           11 8
                            64-55     =135ℓ
                             88
                          9 = 135ℓ√
                         88
                         So volle kapasiteit =135ℓ x88√
                                                  9
                                             =1320ℓ√
                4.3      Volume van die silinder = πr2h√                              4
                                                =22x 64cm2x17cm√√                     1 punt vir
                                                   7                                  korrekte hoogte
                                                 = 3 419,43 cm3√

   5           5.1.1    72º         =1/5 √                                            1
                        360º
               5.1.2    Aantal leerders vir Engels = 126º x 120 = 42 leerders √√      2
                                                     360º
               5.1.3    180º -72º =108º ( hoeke op reguitlyn) √                       3
                        Aantal leerders vir Wiskunde =108º x120
                                                        360º
                                                      = 36√
                           Persentasie vir Wiskunde =36 x100 = 30%√
                                                       120
               5.2.1    Rangskik data van klein tot groot :                           2
                        15; 18; 20; 20; 23; 23; 25√
                        Mediaan is 20√
               5.2.3    Omvang is 25-15=10√                                           1
               5.2.4    Gemiddelde = 15+18+20+20+23+23+25√                            2
                                              7
                                  = 20,57√
               5.2.5                                                                  2
                             1 5 8√
                             2 0 0 3 3 5√

### Page 6

6                                 WISKUNDE                                           (JUNIE 2010)

    5.3.1                                                                 4

                                                                          1 punt vir
                                                                          korrekte skaal

                                                                          1 punt byskrifte
                                                                          vir beide asse

                                                                          2 punte vir
                                                                          korrekte skets

    5.3.2   Dit toon ŉ swak korrelasie tussen die punte vir Wiskunde en   2
            Taal aangesien die punte versprei is √√                       1punt vir
                                                                          korrekte
                                                                          gevolgtrekking
                                                                          1 punt vir rede

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