# Mathematics Grade 12 Paper 2 Limpopo Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 2 question paper and memo from the Limpopo Mock Exam 2023, 34 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-limpopo-mock-exam-question-paper-and-memorandum-paper-2/09610f2c-d82a-4272-8d8d-e38a9a5f957c
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: Limpopo Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 34

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## About this document

Mathematics · Grade 12 · Limpopo Mock Exam · 2023. Question paper and memorandum, 34 pages. Read online or download the PDF.

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## Document text

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### Page 1

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LIMPOPO

PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

Tee
DEPARTMENT OF

EDUCATION

NATIONAL

SENIOR CERTIFICATE

MARKS: 150

TIME: 3 HOURS

A

This question paper consists of 14 pages and an information sheet.

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INSTRUCTIONS AND INFORMATION

Read fhe-following instructions carefully before answering the questions.

L This question paper consists of 10 questions.
2. Answer ALL the questions in the SPECIAL ANSWER BOOK.
3. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in

determining your answers.

4. ANSWERS ONLY will not necessarily be awarded full marks.

S. You may use an approved scientific calculator (non-programmable and non-

graphical), unless stated otherwise.

6. If necessary, round answers off to TWO decimal places, unless stated otherwise.
q Diagrams are NOT necessarily drawn to scale.
8. Write legibly and present your work neatly.

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QUESTION 1

A mathematics teacher wants to make an unbiased prediction of her Grade 12 learners’ final

marks. She uses their SBA mark and notes the final mark. The results are as follows:

SBA MA FINAL MARK (%) SBA MARK (%) | FINAL MARK (%)
aps 31 B 59
35 43 72 85
69 76 57 63
62 73 25 35
83 85 65 59
75 72 68 75

1.1 Draw the scatter plot for the data on the grid provided in the ANSWER BOOK. (4)
1.2 Calculate the correlation coefficient for the data. (2)

1.3. Is the SBA mark a reliable predictor of the final mark? Provide a reason for your

answer. (2)
1.4 Determine the equation of the least squares regression line. (3)
1.5 Predict Toby’s final mark if his SBA mark was 66%. (2)

[13]

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QUESTION 2

The following set of data: 3;4;4;4;6;10; 12; 123 y has

2.1

22

a mean of 7.

etermine:
1.1 the value of y
.1.2 the median of this set of data points
Two additional numbers, 7—n and 7 +77, are added to the data set.

2.2.1 Calculate the mean of these eleven numbers.

2.2.2 Determine the standard deviation if the data points, that are within ONE

standard deviation of the mean, lie in the interval 3<.x <1 1.

(2)

(2)
7]

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QUESTION 3

In the diagram below A, B (—2 ; —7), C (4; 0) and D are the vertices of a kite. E is the

midpoint of the diagonal BD and AC | BD at E. The equation of AC is y= -3H+ 2.

y+
D
A
oO £14; 0) 2
B(-2;;=-7)
Determine:
3.1 The equation of BD. (4)
3.2 The coordinates of E. @)
3.3. Ifthe ratio CE: EA= 1: 3, determine the coordinates of A. (2)
3.4 Kite PQRS is obtained after the measurements of kite ABCD is d by a scale
factor 2, Calculate the area of kite PQRS. (5)

[14]

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QUESTION 4

4.1 In the diagram, the centre of the circle is N (2 ; m) where m <0. The radius of the

circle is 17 units. R (—13;5) and S(—13 ; —11) are two points on the circle,

Ya
R
\ mx
P 1 4 Oo
*N(2; m)
s
Ne
v
4.1.1. (a) Determine the numerical value of m. (4)

(b) Determine the equation of the circle in the form

(x-a) +(y-b) =7? (1)

4.1.2 Determine the gradients of:

(a) NR (2)

(b) NS ()

4.1.3 The tangents at S and R intersect at P. Calculate the size of B. (6)
Ha

4.1.4 Circle N is reflected about the x—axis and the slated 2 units upwards

to obtain circle M. Determine the equation le M in the form

(x-c) +(y-d) =r a

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4.2 An infinite number of circles, each touching the next, are drawn between C and O.
The centres of all the circles lie on the negative x-axis. The radius of the largest

circle, centred at A, is 4 units and the radius of each circle thereafter is halved. B is a

point e largest circle.
rea
B
_— ee —>.
[al 1 2 A 4 @) *
v
4.2.1 Show that OC = 16 units. (2)

4.2.2 If BC is a tangent to circle A at B, write down the size of ABC, providing

a reason for your answer. (2)

4.2.3 Hence, determine tan c. (4)
4.2.4 Determine the equation of BC. (3)
[27]

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Mathemati

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QUESTION 5
5.1 In the diagram below P (3 ; 4) and R (m ; —12) are two points as indicated.
POW =a and ROW=f.
Ss
f P(3 ; 4)
* (o| a ———> x
[e) WwW
R(m ; - 12) |
Answer the following questions without using a calculator.
5.1.1 Write down the value of tana. ()
5.1.2 Determine the value of sin(90°+@). (3)
5.1.3. Determine the value of m if it is given that 12+13sin 8 =0. (4)
(3)

5.1.4 Determine the value of cos(a +f).

5.2 Simplify the following:

5.2.1 y4'"2°*"°* without using a calculator. (5)

tan(180°+x)cosx , taba as ;
sin(180° + x) cos. x—cos(540°+ x) cos(90° + x) © 2 single trigonometric

expression. 0)

I—cos2x-sinx
5.3. Prove that: ———————-= tanx (4)
sin 2x—cos.x

5.4 It is given that P and Q are both acute angles, solve for P and Qi

7)

NRi-

I r
sinPsinQ —cosPcosQ = a and sin(P-Q)=

[33]

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h ics B2 . . 4 202
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QUESTION 6

In the diagram, the graphs of J (x) =asinx and g(x) =coshx are drawn for x [0° ; 180°].

ya
° L
1
&
* 0 90° 180° ™
6.1 Determine the values of a and b. Q)

6.2 Consider the interval x €[0° ; 180°]:

6.2.1 Calculate the value(s) of x where asinx — cosbx =0. (2)
6.2.2 For which value(s) of x will g(x)./"(x) 20. (2)
6.2.3 Determine the values(s) of y for which p=27/°". (2)

[8]

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QUESTION 7

The diagram shows AACE with A=0 and E=a. Points B, D and F lie on AC, CE and

AE respectively so that BC =3x , CD=4x, DE=* and AF=y. BD 1 AC and

D=90°. ABF=30° and DFE=30°.

7.1 Write BF in terms of @ and y. (3)

72 Write DF in terms of @ and x. (2)

7.3 Hence, prove that BD* =4x" sin? @+4y" sin? @ ()
_ | 4y’.sin? 0

7.4 Hence, prove that x = Pay rae (3)

19]

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Mathematics P2
Downloaded from Stanmorephysics.com LimpopoDob/September 2023
QUESTION 8

In the diagram, ABC is the tangent to the circle cent
ntre

circle. EF=BF. ABF=37°.

A c
Determine, with reasons, the sizes of the following:
81 EB, (2)
_ 40s

82 F i (2)

83 (2)

84 6, (2)

asf, ®)
[10]

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Mathematics P2 . 2 LimpopoDoE/September 293
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QUESTION 9

9.1 Oiis the centre of the circle. Points K, L, Mand N are on the circle. Use the diagram

to prove the theorem that states that the opposite angles of a cyclic quadrilateral are

supplementary, ie. K-+M =180° (5)

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9 © diag: i i
9.2 Inthe diagram, RS is a tangent to the circle at R. SAB is a line that passes through

the circle and RT || SAB, MT = AB.

Ss

9.2.1 If ABR=.x, write down THREE other angles in the diagram which are also

equal to x. Provide reasons. (6)

9.2.2 If ARB= y, provide a reason why MRT = y? (1)
9.2.3 (a) Write A, in terms of x and y. (1)
(b) Write N, in terms of x and y. (1)

9.2.4 Prove that ASAR II! AKNR. (GB)
9.2.5. Prove that SAKR is a cyclic quadrilateral. (2)
[19]

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Mathematics P2 14
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QUESTION 10

In the diagram, P is a point on side AB of AABC. The circle through P, B and C cuts AC at

Q. QP prddaged cuts the circle passing through A, B and C at R.

Prove that:

10.1 2 =A,+B, (3)
(5)
[10]

RAND TOTAL: 150

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LimpopoDoE/September 2023

INFORMATION SHEET: MATHEMATICS

=b+Vb? ~ dae

2a
A=P(1+ni) A= P(1=ni)
T, =a+(n-
tea s _alr"-1

parley -1

I

(x) = tim LE+M-f@)
Fe) ect h

d=¥(x,-%,)? +(y,-y,)?
y=me+e y-y =m(x-x;)
(x-a)’ +(y-b) =r?

In AABC: se ycih 2c

sind sinB sinC
a* =b +c —2be.cos A

areaAABC = 5 absin Cc

sin(a + f)=sina.cos#+cosa.sin 8

cos(a + B)= cosa.cos #—sin a.sin 8
cos’ a@—sin’ @

cos2a =41—2sin’?a@

2cos*a@-1

Copyright reserved

r-1

A=P(l-i)" A=P(1+i)"
By = 5 Ga+(n-Nd)
y rel § 2-2 peret
l-r
paclstt) "J
T
M x +X Yy+V2
2 " 2
Pe as m=tanO
amas |

sin(a — 8) = sin a.cos 8 — cosa.sin B

cos(a— £)=cosa.cos B +sin a.sin B

sin 2a@ = 2sina.cosa@

Ye-x

oa

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LIMPOPO

PROVINCIAL GOVERNMENT
REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

NATIONAL

SENIOR CERTIFICATE

MARKS/PUNTE: 150

This memorandum consists of 19 pages/Hierdie nasienriglyne bestaan uit 19 bladsye.

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QUESTION/VRAAG 1

Ll
PREDICTION OF FINAL MARKS

=

FINAL MARK
i
HIE
TW

0 10 20 30 40 50 60 70 80 90 100

v¥vvVall dots
correct/alle punte
korrek
¥v¥v9-—11 dots
correet/ punte
korrek

¥V6-8 dots
correct/ punte
korrek

¥ 1 —6 dots correct/
punte korrek

SBA MARK
(4)
12 | r=0,94 ¥ ¥ answer/antw
(2)
1.3. | The strong correlation between the SBA mark and the final v reason/rede
mark implies that the points lie close to the least squares ¥ conclusion/konklu
regression line. Hence the prediction is reliable. sie
(2)
14 | a=14,49 ¥ value/waarde a
5=0,86 ¥ value/waarde b
y =14,49+0,86x Y equation/vgl a)
1.5 | y=14,49+0,86(66) Y subst/vervang
y=71,25% ¥ answer/antw
Q)
OR/OF
y=7L18% ¥ ¥ answer/antw
(2)
[13]

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QUESTION/VRAAG 2

Copyright reserved/Kopiereg voorbehou

2.1.1 34+44+4464+10+12412+44y _ 4
9
554 v>t¥_4
9 9
55 3
Vz ¥ value of/waarde y
(2)
2.1.2 | Median = 6 v6
Q)
2.2.1 padre 44+ 6484104124124 7-n4 74H
11
_ 77 v71
y=
11
z-7 “1
2
OR/OF @)
x Oot FontTen v77
11
__77
y=
11
=7 v7
(2)
2.2.2 | ¥- 6, =3 ¥ equation/vg/
7-0, =3
o,=4 Y answer/antw
(2)
OR/OF
¥+o,=11 ¥ cquation/vg/
Tto, =11
o,=4 Y answer/antw
(2)
7]

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QUESTION/VRAAG 3

ya
D
A
« 63,43)
o C4; 0) x
B(- 2; - 7)

3.1 I 1
YY, =m(x—-x,) m(AC)=—— ¥ m(AC) =-5
y-(-7) = 2(x-(-2)) m(BD) = 2 ¥ m(BD) =2
yt7=2(x+2) ¥ subt m and point B

/verv m en punt B
BD: y=2x-3
vanswer/antw
(4)
OF/OR
1
y=mx+e m(AC) =~
-7 =2(-2)+e m(BD) = 2 bt m and point B
—7=-4+¢
c=-3
BD: y=2x-3

(4)

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3.2 1
~3tt2 =2x-3 ¥ equating/vg/
-x+4=4x-6
—S5x=-10
¥ value of/waarde x
BD: y = 2(2)—3
y¥ value of/waarde y
(3)
3.3. | A(-4; 4) through translation Y value of/waarde x
¥ value of/waarde y
(2)
OR/OF
Find a point P so that CE = EP ¥P(0; 2)
PO; 2)
Then CP = PA VA(—4;4)
A(-4; 4) (2)
34 | aca (-4-4y +(4-0) =4V5 V length/lengte AC
BE= (2-2) +(-7-1) =45 Y length/lengte BE
Area of kite = 2x area of AABC
_ 1
7 2«( F455 | v method/metode
=80 v80
Enlarged by scale factor 2: 80 X 4 = 320 sq unit Y answer/antw
(5)
OR/OF
2 v
AC= (-4-4y +(4-0) =4/5 length//engte AC
PR =8V5 ¥ length/lengte PR
BE =,/(-2—2)' +(-7-1) =4V5
“.QS=8V5 ¥ length/lengte OS
Area of kite = 2xarea of APQR
=2x (3 8V5 x85 } thod/metode
=320 wer/antw
aA (5)
[14]

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QUESTION/VRAAG 4

4.1

R(- 13;5)

Ya

\

S(- 13; - 11)

and

*N(2: m)

4.1.1 | (13-2) +(5-m) =177
@) 225+25-10m+m°? = 289
m -10m-39 =0
(m-13)(m+3)=0
m=13 or m=-3

NA

OR/OF

(-13-2)' +(-ll-m)" =177
225+(-11-m)’ =289
(-ll-my =64
-ll-m=48

m=-19 or m=-3

NA

¥ subst R and N/verv
RenN

¥ simplify/vereenv

Y std form/std vorm

¥ value/value m

(4)

¥ subst S and N/verv
SenN

plify/vereenv

form/std vorm

lue/waarde m

(4)

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OR/OF
(-13-2)' +(-I1-m)’ = (-13-2)' +(5—m) v¥ equating
2254? +22m+121=2254+m? -10m4 25 ¥ simplify/vereenv
32m
m= ¥ value/waarde m
(4)
OR/OF
RSI y-axis“. x =-13 same x — values Vuse line x =—13
Draw line NK | RS v¥ Use NK
.. NK bisect RS at (-13 ; -3) line from centre of circle 1 | “midpt of NK
to chord
“N(2; 3) vy=m=-3
(4)
4.11} (x2) +(y +3) = 289 Y answer/antw
(b) (1)
4.1.2 3-5 ¥ subst in gradient
(a) m(NR)= 2—(-13) formula/verv in
8 gradiént formule
m(NR)=— ¥ gradient of NR
15 (2)
4.12 4 —3-CI)
| ™ONS)= 555)
m(NS) = & ¥ gradient of NS/
15 gradiént van NS
i)
4.1.3 | NRLPR NS LPS
15 15 vtana def
“.m(PR)= e .m(PS) = | ¥ value/waarde a
tana = 25 tanp=—-> Vtan PB def
8 8 Yvalue/waarde B
a =61,93° B =180°-61,93°
B=118,07°
P, =118,07°—61,93° = 56,14° ¥ method/metode
«. P, =180°—56,14° = 123,86° swer/antw
(6)

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OF/OR

Inclination of NS = 28,07°
Inclination of NR = 151,93°

Opp 2's suppl

¥ 28,07°

V151,93°

Vv RNS = 2x 28,07°
v56,14°

¥ Opp 2's cyclic
quad/teenoorst 4'e
kvh

Y answer/antw
(6)
4.1.4 | Reflection about x — axis: (2; —3)—>(2; 3)
Shift 2 units up: (2 ; 3) (2 ; 5)
Circle M: (x-2)" +(y-5) = 289 ¥ ¥ equation/vgl
(2)
4.2 ,
”
B
«

4.2.1 | Diameter =8+4+24+1 =

“OC =16

¥ sum of diameters/
som middellyne

¥ subst in sum
formula/verv in som
formule

laa? (2)

4.2.2 | 90°, radius 1 tangent

Vianswer/antw
Ia Cs ason/rede
lanay (2)

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4.2.3 | AC=16-4=12 vAC
AB=4 vAB
BC? = AC? — AB? (Pyth)
BC? 2_ 4?
BC?
BC “BC
4 2
tanC =e ¥ tan ratio
(4)
4.2.4 | y-y,=m(x-%)
y-0= 22 C16) ¥ subst/verv m
4 Y subst point/verv
B punt
y= 7 x 44/2 ¥ equation/vgl
(3)
OR/OF
a2 vee
y 4 ¥ subst/verv m
2
0= we (-16)+c¢ ¥ subst point/verv
punt
c=4V2
Lye Bey 42 ¥ equation/vg/
(3)
[27]

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QUESTION/VRAAG 5

5.1
My
P35 4)
MA
+ 0 i
R(m; - 12)
A
5.1.1 4
tana = 3 ¥ answer/antw
()
5.1.2 | sin(90°+@)
= cosa ¥ reduction/reduksie
3 vr=5
“5 ¥ answer/antw
(3)
5.13 | 12+13smB=0
. 12
sin B= “B ¥ std form/std vorm
m? =13? -(-12") (Pyth) ¥ subst into Pyth
2
=25 Y simp]/vereenv
m=t5
m=—5 ¥ answer/antw
(4)
5.1.4 | cos(a+f)
=cosacos #-sinasin B ¥ expansion/witbrei
3)\(-5 4\(-12
_{3)\(—5)_(4)\(/-2 v
(2)( 3] (2)( B } subst/vervang
3 48
=—-—+—
13. 65
_ 33
65 swer/antw
(3)

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5.2.1 | [asint50° 93tan225°
2 8in30° Shana Vj °
= (2 ) Bian sin 30
¥tan 45°
¥ special values/
spesiale waardes
¥ simpl/vereenv
¥ answer/antw
(5)
5.2.2 tan(180° + x) cos.x
sin(180° + x) cos x —cos(540° + x) cos(90° + x)
-— (tan x)(cos x) Vtan x
(sin x)(cos x) —(—cos x)(—sin x) v —sinx
sin x ¥ —cos x
COS X P
_ cos x vy —sinx
—sin xcos x —cos xsin x y Sin
sinx cos x
~ —2sin xcosx
--_! ¥ answer/antw
2cosx (6)
5.3

1—cos2x—-sinx
sin2x—cosx

_ 1-(1=2sin? x)—sinx
~ 2sinxcosx—cos x
_ L-142sin’ x-sinx
~ 2sinxxcosx—cosx
__2sin’? x—sinx

~ 2sinxcosx—cosx
_ sin x(2sin x—1)

7 cos x(2sin x—1)

sinx
COS NX
=tanx

Ycos 2x expansion/
uitbrei
vsin 2x expansion/
uitbrei

¥ simpl/vereenv

Y factors/faktore

(4)

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5.4

sin Psin Q cos PeosQ =>

-.cosPcosQ sin PsinO=

Y rearrange terms/
herrangskik terme

¥ cos identity/

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identiteit
¥ 2" quadrant/
kwadrant
¥ equation/vgl
v‘1* quadrant/
7. 2P =150° kwadrant
P=75°
° Vv 75°
Q=45 v45°
(2)
[33]
QUESTIONVRAAG 6
6.1 a=2 and b=1 ¥ value/waarde a
Y value/waarde b
(2)
6.2.1 | 2sinx—cosx=0
2sin x =cosx
sinx 1
cosx 2
1 v 1
tan x =— tanx = 3
x = 26,57° ¥ answer/antw
(2)
6.2.2 | xe [oc : 180°] Vv interval
¥ notation/notasie
(2)
6.2.3 | paper
€ _l 57
y 2° vv answer/antw
(2)
[8]

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QUESTION/VRAAG 7

A
7 In AABF: BF = AF
snA sinB
BF y .
iar ¥ correct subst in
sin@  sin30° .
. sine rule/korrekte
pr-232 a verv in sinreel
1 ¥ subst special
2 value/ verv
BF=2ysin0 spasiale waarde
¥ answer/antw
QB)
re In AEDF: DF = DE
snE sinE
DF =- * = ¥ correct subst in
sin@ sin 30 sine rule/korrekte
DF= ssn a verv in sinreel
2
DF=2xsing i inswer/antw
(2)
73 BD? = BF’ + FD?

(Pyth)
BD? =(2ysin 0) +(2xsina)"
BD? = 4y’ sin? 0+4x° sin’ a

FZ substiver in

Pyth

(1)

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74 In ABDC: BD? = (4x)? — (3x)?
BD? =16x° —9x?
BD? =7x? ¥ BD’ =7x°
7x * sin? 6+4x° sin? @ ¥ equating
7. 2 otal = 4 2 oin2 (]
on aan sm ¥ simpl/vereenv
x sin’ @) =4y" sin’ 6
ve 4y’ sin’? 6
(7—4sin’ a)
~ (3)
OF/OR
CD? = BD? + BC?
16x? = 4y’sin’@ + 4x°sin?a, + 9x" ¥ subst/verv in
16x* —9x* —4x°sin’a = 4y’sin’0 Pyth
Tx’ —4x’sin°a =4y’sin’®
2(4_ Acin2y) = Awrejn2
x (7 4sin a.) 4y°sin“0 ¥ simplivereenv
xe 4y’sin’0
7—4sin’a Y ito x
ve 4y’sin?0
7—4sin’a (3)

19]

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### Page 30

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QUESTION/VRAAG 8

A
A Cc
8.1 ABF =E, =32° tan-chord theorem YSVYR
— (2)
8.2 E, =B, =32° ZS opp equal sides vS/R
F=116° sum 2’s ofA vS/R
(2)
8.3 D+F=180° opp Z’s of cyclic quad YSYR
D=64°
- - (2)
8.4 O, =2xD midpt Z=2 X circumf Z YSYR
, = 128° : (2)
8.5 E,+0,+B, =180° sum Z’s ofA YSIR
E, =B, Z’s opp equal sides VS/R
2B, +128° =180°
2E, =52°
B, =26°
(2)
OR/OF
ABF+ B, + B, =90° rad tangent S/R
32°+32°+B, =90° Loy
B, =26° A
B, = E, = 26° Z’s opp equal sides — YSIR
(2)
[10]

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### Page 31

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QUESTION/VRAAG 9

9.1

Construction: Join LO and ON ¥ constr/konstr
O, =2xK Zatcentre=2 X Z at circumf YSVYR
0, =2xM Zatcentre=2  Z at circumf VYS/R
0, + 0, = 360° revolution YR
2M+2K =360° equating
M+K =180°
G)

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### Page 32

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9.2
9.2.1 B=M,=x Z’s in the same segment YSVYR
B=R,=x alt Z’s, TRU BS vVSVYR
B=R,=x tan-chord theorem “SYR
(6)
9.2.2 Equal chords equal circumf 72’s YR
()
9.2.3(a) A =x+y ext Z of A vSIR )
a
9.2.3(b) | Nv=x+y ext Zof A YSIR 1)
(
9.2.4 In ASAR and AKNR
(i) R,=R, both= x vs
(ii) A,=N, both= x+y vs
(iii) K, =8 sum Z’s ofA vs
SA ASARINAKNR ZZ 2 (3)
OR/OF
In ASAR and AKNR
(i) R,=R, both = x vS
(ii) A, =N, both = x+y vS
ASARIILAKNR ZZ 2 fire
: (3)
9.2.5 | R,=8 ASARIIIAKNR pa s
SAKR is acyclic quad ext Z of quadrilateral OR YR
converse ext Z of cyclic quad Q)
[19]

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### Page 33

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NS

QUESTION/VRAAG 10

10.1 P ext Z of cyclic quad

opp Z’s of cyclic quad

YSYR
Y¥SYR

¥ method/metode

(5)

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### Page 34

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OR/OF vSVYR
a a - VSVYR
ARB =180°-(A, +B, )
¥ method/metode
,)=180°-C
(5)
10.2 AABR v identify/
common 2 identifiseer A’s
=A, +B, proven vs
a i v
180°-(A, + B,) method/metode
R,
sum Z’s ofA
AABR LLZ vR
Ill A’s
vs
OR/OF (5)
In AARP and AABR
(i) A=A common Z ¥ Identify/
Gi) B =A,+B, proven identifiseer A’s
180°—P, =180°-(A, +B vs
oe -( + :) ¥ method/metode
B,=R, +R,
-. AARP Ill AABR ZLiz
AR _ AP Ill Ars YR
AB AR v8
AR? = AB.AP
(5)
[10]

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