# Mathematics Grade 12 Paper 2 NW June 2025 Memo (Afrikaans)

> Grade 12 Mathematics Paper 2 memorandum from the NW June Exam 2025 (Afrikaans), 18 pages. Read online or download the free PDF.

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- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2025
- Exam period: NW June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 12 · NW June Exam · 2025 · Afrikaans. Memorandum. Read online or download the PDF.

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### Page 1

PROVINCIAL ASSESSMENT/
                                 PROVINSIALE ASSESSERING

                                        GRADE 12/GRAAD 12

                             MATHEMATICS P2/WISKUNDE V2
                                          JUNE/JUNIE 2025
                        MARKING GUIDELINES/NASIENRIGLYNE

   MARKS/PUNTE: 150

                           These marking guidelines consist of 18 pages.
                            Hierdie nasienriglyne bestaan uit 18 bladsye.

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### Page 2

Mathematics P2/Wiskunde V2                     2                                           NW/June/Junie 2025
                               Grade/Graad 12 Marking Guidelines/Nasienriglyne

NOTE:
 If a candidate answer a question TWICE, only mark the FIRST attempt.
 If a candidate has crossed out an attempt of a question and not redone the question, mark
  the crossed-out version.
 Consistent accuracy applies in ALL aspects of the Marking Guidelines. Stop marking at
  the second calculation error.
 Assuming answer values in order to solve a problem is NOT acceptable.

LET WEL:
 As ‘n kandidaat ‘n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
 As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek en nie oordoen nie, sien die doodgetrekte
   poging na.
 Volgehoue akkuraatheid word in ALLE aspekte van die Nasienriglyne toegepas.
   Hou op nasien by die tweede berekeningsfout.
 Aanvaar van antwoorde/waardes om ‘n probleem op te los, word NIE toegelaat nie.

          GEOMETRY

          S       A mark for a correct statement/’n Punt vir die korrekte stelling
                  (A statement mark is independent of a reason/’n stelling is onafhanklik
                  van die rede)

          R       A mark for a correct reason/’n Punt vir die korrekte rede
                  (A reason mark may only be awarded if the statement is correct/’n rede
                  punt mag slegs toegeken word met korrekte stelling)

          S/R     Award a mark if statement AND reason are both correct/ken ‘n punt
                  toe as die stelling EN rede beide korrek is

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### Page 3

Mathematics P2/Wiskunde V2                       3                                           NW/June/Junie 2025
                                 Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

                             y
                                         B (3; 10)

                                                              D
                                                                     C (13; 8)

                                                   α                                     θ
                                                                                                 x

                                                 A (5; – 6)

                                                                             .     E

 1.1                   10  8
                m BC 
                       3  13                                           substitution/vervanging
                           1
                  m BC                                                answer/antwoord
                           5
                                                                                                      (2)
 1.2         y  y1  m( x  x1 )
             B (3, 10)            or    C (13; 8))
                        1                           1
              y  10   ( x  3)          y  8   ( x  13)          substitution/vervanging
                        5            or             5
                    1     53                     1    53
              y   x                     y   x                     answer/antwoord
                    5      5                     5     5
                                                                                                      (2)
 1.3         m AD  5
          y  y1  m( x  x1 )                                          m AD  5
          A (5; – 6)
          y  6  5( x  5)                                             substituting/vervang
          y  5 x  31                                                   A or C
                                                                        answer/antwoord
                                                                                                      (3)
 1.4                               x  x 2 y1  y 2 
           Midpoint/Middelpunt AC  1      ;         
                                   2          2 
          Midpoint/Middelpunt AC  13  5 ; 8  6                    substitution/vervanging
                                            2         2 
          Midpoint/Middelpunt AC (9; 1)                                 answer/antwoord
                                                                                                      (2)

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### Page 4

Mathematics P2/Wiskunde V2                      4                                           NW/June/Junie 2025
                                Grade/Graad 12 Marking Guidelines/Nasienriglyne

 1.5                     1
              m BC  
                         5
                        11  8             11  10
                m PC            or m BP                              substitution/vervanging
                        2  13            23
                               1                   1
                      m PC             m BP                       gradient/gradiënt
                               5                   5
           m BC  m BP or mCP
                                                                      conclusion/gevolgtrekking
          The points are collinear/Die punte is kollinieêr                                    (3)
 1.6
               E (15; – 8)                                             x  15
                                                                       y  8
                                                                                                     (2)
 1.7                        1
                   m BC  
                            5
                         1                                                         1
               tan                                                  tan   
                         5                                                         5
                 168,69                                               168 ,69 

                      10  6
              m AB 
                       35
                 m AB  8                                             m AB  8
                tan   8
                                                                       tan   8
                  97,13
                                                                         97 ,13 
               AB̂C  168 ,69   97 ,13 
               AB̂C  71,56                                           answer/antwoord

                                                                                                     (6)
                                                                                                     [20]

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### Page 5

Mathematics P2/Wiskunde V2                         5                                           NW/June/Junie 2025
                                   Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2
                               y

                        A                                 .Q
                             P   .
                                                                                         x
                                       R

 2.1           ( x  15) 2  ( y  m) 2  64
          x-int (15; 0)                                               substituting x-int/vervang
                                                                       x-afsnit
                (15  15) 2  (0  m) 2  64
                m 2  64                                             simplification/vereenvoudiging
                m8                                                   answer/antwoord
          OR/OF
               r 2  64                                               method/metode
               r 8                                                  r=8
              m  8                                                  answer/antwoord
           OR/OF
                  (m  3) 2  (15  3) 2  5  8                      method/metode
                 m  6m  9  144  169
                    2

                 m 2  6m  16  0                                    std form/standaardvorm
                 (m  8)(m  2)  0
                                                                      answer/antwoord
                  m  8 or m  2                                                                      (3)
 2.2         P (3; 3)                                                P (3; 3)
             Q (15; 8)                                                Q (15; 8)
                                                                                                       (2)
 2.3         PQ  (8  3) 2  (15  3) 2                              substitution/vervanging
             PQ  13                                                  answer/antwoord
          OR/OF
           PQ = 8 + 5
                                                                      substitution/vervanging
               = 13
                                                                      answer/antwoord
           Answer only/slegs antwoord :
                                                                                                       (2)
           full marks

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### Page 6

Mathematics P2/Wiskunde V2                       6                                           NW/June/Junie 2025
                                 Grade/Graad 12 Marking Guidelines/Nasienriglyne

 2.4         ( x  3) 2  ( y  3) 2  25
             ( x  3) 2  ( y  3) 2  25                           x0
             (0  3) 2  ( y  3) 2  25
          9  ( y  3) 2  25 y 2  6 y  9  9  25                std form/vorm
          ( y  3) 2  16      y2  6y  7  0
           y  3  4     ( y  7)( y  1)  0
                y7   or y  1
            A(0; 7)                                                 answer/antwoord
                                                                                                     (3)
 2.5                    73
                 mAP 
                        03
                           4
                   mAP                                                           4
                           3                                        mAP  
                         3                                                         3
                mtan 
                         4                                                     3
                                                                    m tan 
                    3                                                          4
            y  7  ( x  0)
                    4
                 3                                                 substitution/vervanging
            y  x7
                 4
                                                                   answer/antwoord
                                                                                                     (4)
 2.6        P (3; 3) R(2; – 3)
           radius = 5                                               radius = 5

          PR  (3  3) 2  (2  3) 2
             = 37                                                   PR = 6,08
             = 6,08                                                radius <PR
          radius < PR                                              conclusion/gevolgtrekking
           P lies outside circle P                                                                  (4)
 2.7            0  x2      7  y2
            3          3
                   2           2
            x2  6      y 2  1
                                                                   x = 6
           R (6;– 1)                                               y = – 1
               3                                                    substitution/vervanging
          y  xk
               4
                3
           1  (6)  c
                4
                     11                                                 11
              c                                                   c
                      2                                                  2
                  11                                               answer/antwoord
               k  7
                   2                                                                                 (6)
                                                                                                     [24]

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### Page 7

Mathematics P2/Wiskunde V2                          7                                            NW/June/Junie 2025
                                    Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3

 3.1        sin(90  x). tan(360  x)
                cos 240. tan 225
                      (cos x). tan x                                       cos x
          
            cos(180  60  ). tan(180   45 )
                    
                                                                           tan x
                     sin x                                                cos 60 
              cos x                                                      tan 45 
                    cos x 
                                                                            sin x
             cos 60 . tan 45                                           
                                                                            cos x
             sin x
                                                                             1
                                                                            and 1
              1
             1                                                              2
              2
           2 sin x                                                       answer/antwoord
                                                                                                         (7)
 3.2.1      sin 223  sin(180  43 )
                                          

                              sin 43                                     sin 43 
                                                                           answer/antwoord
                             a                                                                        (2)
 3.2.2   sin 68,5  cos 21,5
                                   
                                                                           cos 21,5   

         cos 43  cos 2(21,5 )
                                   
                                                                           cos 2(21,5  )
         cos 43  2 cos 2 21,5   1                                      2 cos 2 21,5   1
           1  a 2  2 cos 2 21,5   1                                    1 a2
         2 cos 2 21,5   1  a 2  1
                              1 a2 1
         cos 2 21,5 
                                 2
                                1 a2 1                                   answer/antwoord
         cos 21,5  
                    
                                                                                                         (5)
                                   2

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### Page 8

Mathematics P2/Wiskunde V2                      8                                           NW/June/Junie 2025
                                Grade/Graad 12 Marking Guidelines/Nasienriglyne

 3.3                                                y

                                5
               3                                                       correct sketch/korrekte
                                                    A                 diagram
                               –4               O              x

           opp = 3                                                    opp = 3/teenoorst = 3

         sin( A  30  )  sinA.cos30   cosA.sin30 
                                                                       sinA.cos30   cosA.sin30 
                            3   3   4  1 
                           .          
                            5   2   5  2                      substitution/vervanging
                             43 3
                         
                               10                                      answer/antwoord
                                                                                                    (5)
                                                                                                    [19]

QUESTION/VRAAG 4

 4.1.1       sin 2 x  tan x
                               tan x
                 cos 2 x
                  sin 2 x  tan x
           LHS 
                       cos 2 x
                                    sin x                           2 sin x cos x
                   2 sin x cos x 
                                   cos x                             sin x
                        2 cos 2 x  1                              
                                                                      cos x
                   2 sin x cos 2 x  sin x                          2 cos 2 x  1
                                           (2 cos 2 x  1)
                            cos x
                   sin x(2 cos 2 x  1)         1                  LCM/KGV
                                       
                          cos x            2 cos 2 x  1
                                                                   common factor/gem faktor
                 tan x
            LHS = RHS
                                                                                                     (5)
 4.1.2      sin 120  tan 60
                               
                                                                    tan 60   
                              tan 60 
                 cos120 
                                    =   3                          answer/antwoord
                                                                                                     (2)

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### Page 9

Mathematics P2/Wiskunde V2                          9                                           NW/June/Junie 2025
                                    Grade/Graad 12 Marking Guidelines/Nasienriglyne

 4.2           cos 24  . cos x  sin 24  . sin x  sin 2 x            cos( 24  x)
                                                                                

               cos(24   x)  sin 2 x
               cos(24   x)  cos(90   2 x)                          changing to cos/verander na
                24   x  (90   2 x)  k .360                     cos
           24   x  90   2 x  k .360 
                                                                        first quad/1ste kwadrant
           x  66   k .360  ; k  Z
          or                                                            Gen solution 1/alg opl 1
          24   x  (90   2 x)  k .360 
                                                                       4th quad/4de kwadrant
           24   x  90   2 x  k .360 
            3x  114   k .360 
                                                                        Gen solution 2/alg opl 1
          x  38   k .120  ; k  Z                                   – 1 if/as k  Z is not
          OR/OF                                                        included/nie deel van die
            cos 24  . cos x  sin 24  . sin x  sin 2 x              antwoord is nie
             cos(24   x)  sin 2 x
               cos(24   x)  cos(90   2 x)
                                                                        cos( 24  x)
                                                                                

                90  2 x  (24  x)  k .360
                                                    
                                                                        changing to cos/verander na
           90  2 x  24   x  k .360 
               
                                                                       cos
            x  66   k .360 
          x  66   k .360  ; k  Z                                   first quad/1ste kwadrant
          or/OF
          90   2 x  (24   x)  k .360 
                                                                        Gen solution 1/alg opl 1
           90   2 x  24   x  k .360 
            3x  114   k .360                                     4th quad/4de kwadrant

          x  38   k .120  ; k  Z
          OR/OF
            cos 24  . cos x  sin 24  . sin x  sin 2 x               Gen solution 2/alg opl 2
            cos(24   x)  sin 2 x                                     – 1 if/as k  Z is not
                                                                       included/nie deel van die
                sin[90   (24   x)]  sin 2 x                       antwoord is nie
               sin( 66   x)  sin 2 x
                                                                        cos( 24  x)
                                                                                
           2 x  66   x  k .360 
           x  66   k.360  ; k  Z                                   changing to/verander na sin
          or
                                                                        first quad/1ste kwadrant
           2 x  180   (66   x)  k .360 
                                                                        Gen solution 1/alg opl 1
           2 x  114   x  k .360 
                                                                       2th quad/4de kwadrant
          x  38  k .120 ; k  Z
                               
                                                                        Gen Solution 2/alg opl 2
          OR/OF

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### Page 10

Mathematics P2/Wiskunde V2                       10                                          NW/June/Junie 2025
                                 Grade/Graad 12 Marking Guidelines/Nasienriglyne

         cos 24  . cos x  sin 24  . sin x  sin 2 x                 cos( 24   x)
              cos(24   x)  sin 2 x                                   changing to cos/verander na cos
              cos(24  x)]  cos(90  2 x)
                                       

                                                                        first quad/1ste kwadrant
         24   x  90   2 x  k .360
         x  66   k .360 ; k  Z
                                                                        Gen solution 1/alg opl 1
         or
         24   x  360   (90   2 x)  k .360                     4th quad/4de kwadrant
          3x  246   k .360 
                                                                        Gen Solution 2/alg opl 2
         x  82   k .120  ; k  Z                                   – 1 if/as k  Z is not included/nie
                                                                       deel van die antwoord is nie

                                                                                                               (6)
                                                                                                              [13]

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### Page 11

Mathematics P2/Wiskunde V2                        11                                          NW/June/Junie 2025
                                  Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5

    5.1     a=5                                                    a=5
            b=1                                                    b=1
                                                                                                      (2)
    5.2     Amplitude = 4                                           answer/antwoord
                                                                                                      (1)
    5.3     360   
                                                                    answer/antwoord
                                                                                                      (1)
    5.4.1   x  90   
                                                                    answer/antwoord
                                                                                                      (1)
    5.4.2   x  ( 90 ;90 ) or  90  x  90
                                                
                                                                    answer/antwoord
                                                                                                      (2)
    5.4.3   x  [180  ;180  ] or  180   x  180              answer/antwoord
                                                                                                      (2)
    5.5       8 minimum when x  0 
            g ( x)
                   8
            min                                                   4
                   4
            min  2                                                 answer/antwoord
                                                                                                      (2)
    5.6     k  5 or         k 5                                  k  5         k  5
                                                                                                      (2)
                                                                                                      [13]

\

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### Page 12

Mathematics P2/Wiskunde V2                      12                                          NW/June/Junie 2025
                                Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6

                         T

                          h

                                                                       β
                        G⅂                                                         A
                              120o

             θ
              α
       B

 6.1          Â 1    120   180                                  Â 1    120   180 
             Â 1  60                                              answer/antwoord
            Answer/antwoord only: full marks
                                                                                                      (2)
 6.2                h                                                           h
            tan                                                      tan  
                   BG                                                          BG
                   h                                                           h
            BG                                                        BG 
                 tan                                                        tan 
                                                                                                      (2)
 6.3                  h                                                        h
              tan                                                    tan  
                     AG                                                       AG
                     h                                                         h
              AG                                                      AG 
                   tan                                                      tan 
             sin  sin( 60    )                                      sin  sin( 60    )
                                                                           
              AG           BG                                            AG         BG
                       AG sin( 60    )
              sin  
                              BG
                         h
                              sin( 60    )
                       tan                                           substitution/vervanging
              sin  
                                 h
                               tan
                      tan sin( 60    )
             sin  
                             tan
                                                                                                      (4)

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### Page 13

Mathematics P2/Wiskunde V2                         13                                          NW/June/Junie 2025
                                   Grade/Graad 12 Marking Guidelines/Nasienriglyne

 6.4                 tan60  sin( 60   15  )
           sin 15                                                       substitution/vervanging
                              tan
                    tan60 sin( 60   15  )
                          
            tan  
                            sin 15 
            tan   4,732...                                              tan   4,732 ...
              78,07                                                    answer/antwoord
                                                                                                         (3)
                                                                                                       [11]

QUESTION/VRAAG 7

                                             O
                                             1 2

                K             1
                          2                                 1
                                                        2        N
                          L                       640
                                         1       2
                                          M
 7.1.1        Ô 1  M̂ 2  64  [alt  s /verw e ; KO//MN]               S/R
              N̂ 2  32  [  at centre = 2   at circumf]/              S R
                                                                                                           (3)
                          midp   2  omtreks 
 7.1.2       L̂  148        [opp  s of a cyclic quad]/                 S         R
                              teenoorst  van kvh                                                        (2)
 7.1.3      K̂ 1  N̂ 2  32  [alt  s /verw e ; KO//MN]
                                                                          S R
             N̂ 1  32  [  s opp = sides/ℎ𝑜𝑒𝑘𝑒 𝑡. 𝑜. = 𝑠𝑦𝑒]              answer/antwoord
          OR
          M̂ 2  N̂ 1  N̂ 2  64  [  s opp = sides/ ℎ𝑜𝑒𝑘𝑒 𝑡. 𝑜. =      OR
          𝑠𝑦𝑒 ]                                                           S      R
          N̂ 2  32  [proved/bewys in 7.1.1]
             N̂ 1  32                                                   answer/antwoord
                                                                                                           (3)

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### Page 14

Mathematics P2/Wiskunde V2                     14                                          NW/June/Junie 2025
                               Grade/Graad 12 Marking Guidelines/Nasienriglyne

 7.2

                                        C                       F

                                                 O .                         B

                 P

                                        A

                                                                Q

 7.2.     Draw diameter AOF and join FB/ Trek middellyn               construction/konstruksie
          AOF en verbind met FB
          Proof:/Bewys                                                S R
            FÂP  90  [rad  tan/rklyn]                             S/R
             AB̂F  90  [  sub by dia/  in a semi-
                                                                      S      R
          circle/halfsirkel]
          But/maar CÂF  CB̂F [  s in same seg/ e in
          dieselfde sirkel segment]
           CÂP  AB̂C                                                                                (6)

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### Page 15

Mathematics P2/Wiskunde V2                          15                                          NW/June/Junie 2025
                                    Grade/Graad 12 Marking Guidelines/Nasienriglyne

 7.3

                                                                                Q
                    P           1                                2
                            2                                        1

                                    W       2
                       1                1           2 1
                     T 2                             3 R

                    S

 7.3.1            sub by dia/onderspan deur middellyn OR/OF                        R
                   [  in semi-circle/  in halfsirkel]                                                     (1)
 7.3.2       QP̂S  90  [rad  tan/rad  rklyn]                                     S/R
                                                                                    S
             P̂2  90  x
                        

              R̂ 1  90  [  in semi-circle/  in halfsirkel]
              PR̂S  90  [  s on a str line/  op “n reguitlyn]                    S/R
                                                      e

            Ŝ  (90 - x)  90  180 [sum of  s of a Δ/som van 
                                                                         e       S/R
          van Δ]
              Ŝ  x
                                                                                                            (4)
 7.3.3      Ŝ  x [ proved in 7.3.2]
            Ŵ2  QP̂R  x [  s in same seg/  in dies seg]
                                                           e                        S R
           SRWT is a cyclic quad [ext  of a quad = int opp                       R
          OR converse of ext  of cyclic quad] /
           SWRT is ‘n kvh
          buite  van vierhoek  teenoorst binne van vierhoek 
                                                                                                            (3)
          OF omgekeerde buite  van kvh
                                                                                                            [22]

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### Page 16

Mathematics P2/Wiskunde V2                     16                                          NW/June/Junie 2025
                               Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 8
                                           A

                                                  S

                                                      P

                                  R

                                           T

                              B                                  C
 8.1      AS AR
                 [line || to one side of Δ/lyn ||een sy v ∆         S R
          SP RB
                  Prop theorem/ eweredigheid stelling
                  RS || BP],
             AS 3                                                    substitution/vervanging
         OR     
             SP 2                                                     answer/antwoord
              AS 3
                                                                                                  (4)
              SC 7
 8.2      RT SP                                                       S/R
                 [ line || to one side of Δ/lyn ||een sy v ∆
          TC PC
                  Prop theorem/ eweredigheid stelling
                  RS || TP],                                         substitution/vervanging

            RT 2                                                      answer/antwoord
               
            20 5
            RT  8 units
                                                                                                   (3)
 8.3                   1
                         TC.PCsinT ĈP
         Area of TPC 2                                               1
                                                                      TC.PC.sinTĈP
         Area of RSC 1                                               2
                         RC.SCsinR ĈS
                       2                                              1
                                                                      RC.SC.sinR ĈS
                         TC PC                                        2
                           .
                         RC SC
                         5 5                                              5 5
                        .                                               .
                         7 7                                              7 7
                         25                                           answer/antwoord
                       
                         49                                                                       (4)
                                                                                                  [11]

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### Page 17

Mathematics P2/Wiskunde V2                        17                                                NW/June/Junie 2025
                                  Grade/Graad 12 Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 9

                                         A
                                             2                   >>                             C
                                                                                    1
                                         1                                              2

                              2          2       3
                      D                              4
                          1          1                         >>
                                             T                                              B

              P

 9.1      In ΔPAT and ΔPCA
             P̂  P̂ [common/gemeenskaplik]                              S/R
            Â 1  Ĉ1 [tan chord theorem/rklyn-koord stelling]
                                                                         S /R
             AT̂P  PÂC [sum of  s of a Δ/som e v Δ]
              ΔPAT ||| ΔPCA [AAA/    ]                               R or/ 3rd angle/ of
                                                                        3de  v 
                                                                                                           (3)
 9.2          PA PT                                                     S       R
                   [ ΔPAT ||| ΔPCA]
              PC PA
             PA 2  PC.PT                                               S
                                                                        substitution/vervanging
              6 2  ( x  5 )x
              x 2  5 x  36  0                                                                           (3)
 9.3         x  5 x  36  0
               2

             ( x  9)( x  4)  0                                       factors/faktore
              x  9 or x  4                                           x-values/x-waardes
             PT = 4 m                                                  answer/antwoord
                                                                                                           (3)

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### Page 18

Mathematics P2/Wiskunde V2                     18                                    NW/ Junie 2025
                          Grade/Grade 12 – Marking Guidelines/Nasienriglyne

 9.4              In ΔPDT and ΔPTA
                    P̂  P̂ [common/gemeenskaplik]                   S
                  T̂1  Ĉ 1 [corresp  s /ooreenk e ; BD// AC]      S/R

              Â 1  Ĉ 1 [tan chord theorem rklyn-koord stelling]    S/R
                T̂1  Â 1                                          S
                  D̂1  PT̂A [sum of  s of a Δ/som e v Δ]
                  ΔPDT ||| ΔPTA [    ]                            S/R
                PD PT
                          [Δs ||| ]
                PT PA                                                S
                PD 4
                     
                 4     6                                              substitution/vervanging
                        8
                 PD = units or 2,67 units                            answer/antwoord
                        3
          OR
                  In ΔPDT and ΔPAC
                                                                     S
                    P̂  P̂ [common/gemeenskaplik]                    S/R
              T̂1  T̂4 [vert opp  s/regoorst e ]
          but T̂4  Ĉ1 [alt  s /verw e ; AC//TB]                   S/R
                                                                     S
                T̂1  Ĉ1
                          
                  D̂1  PAC [sum of  s of a Δ/som v e v Δ]
                  ΔPDT ||| ΔPAC [    ]                            S/R
                PD PT
                           [Δs ||| ]                                S
                PA PC
                PD 4
                                                                     substitution/vervanging
                 6     9
                        8                                            answer/antwoord
                 PD = units or 2,67 units
                        3                                                                        (8)

                                                                                              [17]

                                                                              TOTAL/TOTAAL: 150

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