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Index/Indeks
Paper 1 / Vraestel 1 .................................................................... 1
Products & Factorisation / Produkte & Faktorisering................................................2
Indices & surds / Eksponente & wortelvorme ...........................................................4
Quadratic equations / Kwadratiese vergelykings.......................................................6
Simultaneous equations / Gelyktydige vergelykings.................................................8
Inequalities / Ongelykhede ........................................................................................9
Patterns & sequences Pattern / Getalpatrone & rye .................................................11
Annuities / Annuïteite ..............................................................................................22
Straight line & parabola / Reguitlyn & parabool .....................................................24
Exponential graph / Eksponensiële grafiek..............................................................26
Properties of graphs / Eienskappe van grafieke.......................................................27
Differential Calculus / Differensiaalrekeninge ........................................................31
Linear Programming / Lineêre Programmering.......................................................34
Paper 2/ Vraestel 2 ................................................................... 36
Volume and surface area / Volume en buiteoppervlakte .........................................37
Analytical geometry / Analitiese meetkunde...........................................................39
Trigonometry / Trigonometrie .................................................................................42
Compound & double angles / Saamgestelde- en dubbelhoeke................................43
Solving triangles / Oplos van driehoeke ..................................................................46
Transformations / Transformasies ...........................................................................48
Data handling / Data Hantering ...............................................................................52
Compiled by / Saamgestel deur: Dr Gerrit Stols
Telephone & Fax (012) 653 5758 • Cell 082 415 7583 • [email protected]
Pap
er 1
/ V
raes
tel 1
P
atte
rns
and
sequ
ence
s / G
etal
patro
ne
Ann
uitie
s an
d fin
ance
/ A
nnuï
teite
en
finan
sies
Func
tions
and
gra
phs
/ Fun
ksie
s en
gra
fieke
Alg
ebra
and
equ
atio
ns /
Alg
ebra
en
verg
elyk
ings
Cal
culu
s / D
iffer
ensi
aalre
kene
Line
ar p
rogr
amm
ing
/ Lin
eêre
pro
gram
mer
ing
1
Pro
duct
s &
Fac
tori
sati
on /
Pro
dukt
e &
Fak
tori
seri
ng
Pro
duct
s / P
rodu
kte
Gen
eral
cas
es /
Sp
esia
le g
eval
le
a(
b +
c) =
ab
+ a
c
(a
+ b
)(c
+ d
) = a
c +
ad
+ b
c +
bd
(a
+ b
)(d
+ e
+ f)
= a
d +
ae
+ a
f + b
d +
be
+ b
f
Sp
ecia
l cas
es /
Sp
esia
le g
eval
le
D
iffe
renc
e of
squ
ares
/ V
ersk
il v
an v
ierk
ante
: (2a
2 +
3b)
(2a2
– 3b
) =
4a4 –
9b2
S
quar
e / V
ierk
ant:
(a
+ 3
b)2 =
a2 +
6ab
+ 9
b2
Fact
oris
atio
n / F
akto
rise
ring
Ste
p 1
: F
ind
th
e co
mm
on f
acto
r / H
aal g
emee
nsk
apli
ke
fak
tor
uit:
Ste
p 2
: C
oun
t th
e n
um
ber
of
term
s / T
el d
ie a
anta
l ter
me
2 T
erm
s / T
erm
e
D
iffe
rent
of
squa
res:
x2 –
y2 = (
x –
y)(x
+ y
)
S
um o
f cu
bes:
x3 +
y3 =
(x
+ y
)(x2
– xy
+ y
2 )
D
iffe
renc
e of
cub
es:
x3
y3 =
(x
y)(x
2 +
xy
+ y
2 )
3 T
erm
s / T
erm
e: 6
x2 –
19x
+ 1
0 =
(2x
– 5
)(3x
– 2
)
4 T
erm
s / T
erm
e: G
roup
ing
/ Gro
epee
r 2-
2 te
rms:
x2 –
xp –
xq
+ p
q =
x(x
– p
) –
q(x
– p)
= (
x
p)(x
q
)
Use
fac
tor
theo
rem
/ G
ebru
ik d
ie f
akto
rste
llin
g
2
Wor
kshe
et 1
1.
Sim
plif
y th
e ex
pres
sion
s/V
eree
nvou
dig
die
uitd
rukk
ings
:
a)
2
x
(x
2x2 )
b)
(x
2x2 )
2x
c)
(x
2x2 ) (
2x
)
d)
2x
(x
2x2 )
e)
2x
+ (
x
2x2 )
f)
(
2x)
(x
2x2 )
g)
(2x
– y
)3
h)
(x
+ 2
)(x2
2x
+ 4
)
i)
(4
x2 + 6
x +
9)(
2x
3)
2.
Fac
tori
se/F
akto
rise
er:
a)
x2 +
x
12
b)
x2
x
12
c)
x2 +
x +
12
d)
x2
x +
12
e)
x2
7x
+ 1
2
f)
x2 +
7x
+ 1
2
g)
(x
3)(x
2
) 1
2
h)
7x3 –
56
i)
18
– 8
(x –
y)2
j)
(2
x –
y)2 +
4(y
– 2
x)
k)
2x2 +
4y
– 8y
2 – 2
x
l)
x3
3x2
16x
+ 4
8
m
) x3
4x2
11x
+ 3
0
n)
x3
3x2 +
4
3.
Sim
plif
y th
e ex
pres
sion
s/V
eree
nvou
dig
die
uitd
rukk
ings
:
a)
2
225
2
255
2
xx
xx
x
b)
3
3
22
22
ba
bab
ab
ab
a
c)
x
xx
xx
212
42
2
d)
11
1
23
2
2
xx
xx
3
Indi
ces
& s
urds
/ E
kspo
nent
e &
wor
telv
orm
e
Def
init
ion
:
ba
b
2 = a
and
a ≥
0 &
b ≥
0
21
aa
;
nn
aa
1
;
0
n
umbe
rev
ena
x
Ex
pone
ntia
l law
s Ek
spon
entw
ette
D
efin
ition
s D
efin
isie
s Su
rds
Wor
telv
orm
e
a; b
R
and
n; m
N
o
sk
sk
aa
a
.
sk
sk
aaa
mn
nm
aa
)
(
ak .bk =
(a.
b)k
10
a
11
n
nn
aa
1
nn
nab
ba
.
nnn
baba
mn
mn
aa
4
Wor
kshe
et 2
1.
A
re th
e fo
llow
ing
true
or
fals
e? /
Is d
ie v
olge
nde
waa
r of
val
s?
a)
9
54
54
44
2.2
b)
22 +
23 =
25
c)
(
2)2 =
22
2.
Sim
plif
y w
itho
ut u
sing
a c
alcu
lato
r /
Ver
eenv
oudi
g so
nder
om
’n
sak
reke
naar
te g
ebru
ik:
a)
32
278
b)
1
3 2
22
x
xx
c)
32
893
d)
x
x
xx
5.35.4
5.35.2
1
21
e)
n
nn
n
9
6.32.
82
13
f)
)
2(
2
11
4
8.2
x
xx
3.
Sim
plif
y w
itho
ut u
sing
a c
alcu
lato
r /
Ver
eenv
oudi
g so
nder
om
’n
sak
reke
naar
te g
ebru
ik:
a)
8
.2
88
b)
80
028
832
c)
8
.2
d)
8
3218
e)
3
316
.4
f)
8
.12
32.
18.
27
g)
10
82
753
483
4.
Rat
iona
lise
the
deno
min
ator
/ R
asio
nali
seer
die
noe
mer
:
a)
22
b)
3
25
5
Qua
drat
ic e
quat
ions
/ K
wad
rati
ese
verg
elyk
ings
Met
hods
to
find
the
roo
ts /
Met
odes
om
die
wor
tels
te
bepa
al:
WR
ITE
IN
ST
AN
DA
RD
FO
RM
& G
et r
id o
f fr
acti
ons:
L
CM
(de
nom
inat
ors
0)
F
acto
risi
ng
/ Fak
tori
seri
ng
i. O
ne s
ide
= 0
/ E
en k
ant =
0
ii.
Fac
tori
se /
Fakt
oris
eer
iii.
Let
(…
) =
0 o
r (
…)
= 0
/ S
tel (
…)
= 0
of
(…
) =
0
Fo
rmul
a / F
orm
ule:
a
acb
bx
2
42
Co
mpl
etin
g th
e squ
are /
Vier
kant
svol
tooi
ing
i. D
ivid
e by
the
coef
fici
ent o
f x2
/ Dee
l deu
r ko
ëffi
siën
t van
x2
ii.
Tak
e co
nsta
nt te
rm to
rig
ht-h
and
side
/ K
onst
ante
term
na
regt
erka
nt
iii.
Add
2
21
1
of
Coef.
x to
bot
h si
des
/ Tel
alb
ei k
ante
by
2
21
1
va
n
Koëf.
x
iv.
LH
S: F
acto
rise
& R
HS
: LC
M /
LK
: Fak
tori
seer
& R
K: K
GV
v.
Ext
ract
/
Tre
k
vi.
Get
x a
lone
on
a si
de a
nd L
CM
/ L
K: K
ry x
all
een
& R
K: K
GV
6
Wor
kshe
et 3
1.
S
olve
for
x /
Los
op
vir
x:
a)
2x2 +
x
6 =
0
b)
(2 x
3
)(x
1) =
15
c)
4
3
3
4
xx
d)
(x
+ 3
)(x
5) =
9
e)
5x2
80
= 0
f)
x
18
6x(
x
5) =
0
g)
42
34
21
xx
x
h)
4
52
32
23
xx
xx
i)
1
1111
xx
2.
Sol
ve f
or x
by
com
plet
ing
the
squa
re /
Los
op
vir
x d.
m.v
. vi
erka
ntsv
olto
oiin
g:
a)
2x2 +
14x
+ 2
4 =
0
b)
2x2 +
x
6 =
0
c)
ax2 +
bx
+ c
= 0
d)
mx2 +
x2
x +
m =
1
3.
Sol
ve f
or x
, usi
ng t
he f
orm
ula
/ L
os o
p vi
r x
deur
die
for
mul
e te
ge
brui
k.
a)
3
x2 =
2x2 +
1
b)
x
x4+
2 =
0
4.
Fac
tori
se /
Fakt
oris
eer:
a)
x3
3x2 +
4x
12
b)
2x3 +
x2
13x
+ 6
c)
x3 +
6x2
9x
14
d)
x3 +
3x2
3x
9
5.
Sol
ve f
or x
/ L
os o
p vi
r x:
a)
4x5 –
3x4 –
4x
+ 3
= 0
b)
x3 +
3x
= 2
c)
x
3 + 5
x2 + 8
x =
12
6.
Sol
ve f
or x
if/ L
os o
p vi
r x
as: (
2x +
3)(
x2 5
) =
0 a
nd/e
n
a)
x
Z
b)
x
Q
7.
Giv
en/G
egee
: i
1
W
rite
the
root
s of
x2 +
2x
+ 5
= 0
in te
rms
of i.
S
kryf
die
wor
tels
van
x2 +
2x
+ 5
= 0
in te
rme
van
i.
7
Sim
ulta
neou
s eq
uati
ons
/ Gel
ykty
dige
ver
gely
king
s
Meth
ods /
Met
odes
:
El
imin
atio
n / E
limin
asie
Use
this
met
hod
if th
ere
is n
o te
rm w
ith
xy. /
Geb
ruik
hie
rdie
met
ode
indi
en d
aar
geen
term
met
xy
is n
ie.
i. W
rite
sim
ilar
term
s un
dern
eath
eac
h ot
her
/ Skr
yf g
elyk
soor
tige
term
e on
der
mek
aar
neer
ii
. G
et c
oeff
icie
nts
of x
or
y to
be
the
sam
e by
/
Kry
koë
ffis
iënt
e va
n x
of y
die
self
de
iii.
Eli
min
ate
one
vari
able
/ E
lim
inee
r ee
n on
beke
nde
iv.
Sol
ve e
quat
ion
/ Los
op
v.
Sub
stit
ute
back
and
det
erm
ine
othe
r va
riab
le /
Ver
vang
teru
g en
bep
aal a
nder
onb
eken
de
Equa
te eq
uatio
ns / S
tel v
erge
lykin
gs g
elyk
Use
this
met
hod
if y
ou c
an g
et y
alo
ne o
n a
side
of
both
equ
atio
ns.
Geb
ruik
hie
rdie
met
ode
indi
en y
all
een
aan
‘n k
ant v
an a
lbei
ver
gely
king
s ge
kry
kan
wor
d.
Su
bstit
utio
n / S
ubst
itusie
If
you
can
’t u
se m
etho
d 1
or 2
, thi
s m
etho
d al
way
s w
orks
. / H
ierd
ie m
etod
e sa
l alt
yd w
erk,
indi
en m
etod
e 1
of 2
nie
geb
ruik
kan
wor
d ni
e.
i. L
ook
for
sim
ples
t equ
atio
n / S
oek
eenv
oudi
gste
ver
gely
king
ii
. G
et x
or
y al
one
(pre
fera
bly
not f
ract
ion)
/ K
ry x
of
y al
leen
(ve
rkie
slik
nie
bre
uk n
ie)
iii.
Sub
stit
ute
in o
ther
equ
atio
n an
d de
term
ine
x or
y /
Ver
vang
in a
nder
vgl
. en
bepa
al x
of
y.
iv.
Sub
stit
ute
answ
er in
to a
ny o
f th
e or
igin
al e
quat
ions
to d
eter
min
e th
e ot
her
vari
able
/ V
erva
ng
antw
oord
om
and
er o
nbek
ende
te b
epaa
l.
8
Ineq
ualit
ies
/ Ong
elyk
hede
Ste
p 1:
One
sid
e =
0 /
Een
kan
t = 0
S
tep
2: F
acto
rise
/ F
akto
rise
er
Ste
p 3:
Cal
cula
te r
oots
/ B
erek
en w
orte
ls: L
et (
…)
= 0
or
(…
) =
0 /
Ste
l (…
) =
0 o
f (
…)
= 0
S
tep
4: M
ake
a ro
ugh
sket
ch o
f pa
rabo
la /
Maa
k ro
ww
e sk
ets
van
die
para
bool
S
tep
5: I
f (
***)
(***
) <
0 r
ead
off
the
x va
lues
of
the
grap
h un
der
the
x-ax
is
As
(***
)(**
*) <
0 le
es d
an d
ie w
aard
es v
an d
ie g
rafi
ek a
f on
der
die
x-as
.
If (
***)
(***
) >
0 r
ead
off
the
x va
lues
of
the
grap
h ab
ove
the
x-ax
is
As
(***
)(**
*) >
0 le
es d
an d
ie w
aard
es v
an d
ie g
rafi
ek a
f bo
die
x-a
s.
Exa
mp
le:
Solv
e fo
r x
/ Los
op
vir
x:
x2 x
6
Ste
p 1:
x2
x
6
0
Ste
p 2:
(x
3)(x
+ 2
)
0
Ste
p 3:
x-i
nter
scep
ts /
x-af
snit
te: 3
and
2
Ste
p 4:
See
ske
tch
/ Sie
n sk
ets
Ste
p 5:
Sol
utio
n / O
plos
sing
: x
2
or /
of x
3
2
3
9
Wor
kshe
et 4
1.
Sol
ve f
or x
/ L
os o
p vi
r x:
a)
2x
= 3
2
b
2x =
5
c)
5x
100
= 0
d)
0
182.3
1
x
e)
0
188
4.3
x
f)
12
31
2
x
g)
4
21
x
h)
5.2
x+1 –
8 =
32
i)
2.
2x 6
4 =
0
2.
If
ni
PA
1,
mak
e n
the
subj
ect
of t
he f
orm
ula
/ M
aak
n di
e on
derw
erp
van
die
form
ule.
3.
Ifii
xF
n]1
)1
[(
, m
ake
n th
e su
bjec
t of
the
for
mul
a /
Maa
k
n di
e on
derw
erp
van
die
form
ule.
4.
Det
erm
ine
the
poin
ts o
f in
ters
ecti
on o
f th
e cu
rves
of:
/
B
epaa
l die
sny
punt
e va
n di
e kr
omm
es v
an:
y
+ 2
= 2
x an
d/en
y =
x2
4x
+ 3
5.
Sol
ve s
imul
tane
ousl
y fo
r x
and
y / L
os g
elyk
tydi
g vi
r x
en y
op:
a)
2x
+ 3
y =
2 a
nd/e
n x
4y =
12
b)
3y
x
4 =
0 a
nd/e
n x2
xy
+ y
2 = 3
c)
y +
2x
= 2
and
/en
y2 + 2
x2 = 3
xy
d)
y2 +
2x2
3xy
= 0
and
/en
y +
2x
= 2
6.
Sol
ve f
or x
/ L
os o
p vi
r x:
a)
1
2x
> 0
b)
(x
+ 1
)(x
2) >
0
c)
(2x
3
)2 < 4
d)
x2
x
6
0
e)
x2 <
x +
12
f)
x2
2x
8
g)
x2 <
16
h)
x2
9
10
Pat
tern
s &
seq
uenc
es P
atte
rn /
Get
alpa
tron
e &
rye
Abb
revia
tions
/ Afk
ortin
gs
a
first te
rm of
the s
eque
nce
/ eer
ste te
rm va
n die
ry (a
= T
1) d
co
nstan
t diffe
renc
e / ko
nstan
te ve
rskil (
RR):
d = T
2 – T
1 or
d =
T 30 –
T29
r
cons
tant r
atio (
GS) /
kons
tante
verh
oudin
g (MR
): r =
12 TT or
r =
4445 TT
n t
he nu
mber
of te
rms /
die a
antal
term
e T n
the
value
of th
e nth
term
of the
sequ
ence
/ die
waar
de va
n die
n de t
erm
van d
ie ry
S n
the
sum
of the
first
n ter
ms / d
ie so
m va
n die
eerst
e n te
rme
Pat
tern
s
Ex
pone
ntial
: 1;
2; 4;
8; …
= 2
0 ; 21 ;
22 ; 23 ;
… or
30 ;
31 ; 32 ;
33 ; …
Arith
met
ic: If
1st diffe
renc
e is c
onsta
nt
T2 –
T1
= T 3
– T
2
Ge
omet
ric: I
f the r
atio i
s con
stant
= 23
12
TTTT
Re
curs
ive se
quen
ces
o 1
; 1; 2
; 3; 5
; 8; .
Fibo
nacc
i seq
uenc
e: T n
+1 =
Tn +
Tn-
1 o
1; 9
; 33;
105;
321;
…
Tn =
3(T n
1 +
2)
Qu
adra
tic: I
f 2nd
diffe
renc
e is c
onsta
nt: T
n = p.
n2 + q
.n +
r
Co
mbi
natio
n: 13
; 25
; 37
; 49
; …
Tn =
nn
12
Arith
met
ic se
quen
ces /
Rek
enku
ndig
e rye
:
2; 5
; 8; …
of
–
3; –
7; –
11; …
2) T
ES
T /
TO
ET
S:
T2
– T
1 =
T3
– T
2
3)
dn
aT n
)1(
4)
d
na
nS n
)1(
22
5) S
n =
la
n
2
Geom
etric
sequ
ence
s / M
eetk
undi
ge ry
e:
2; 1
; 21
; … o
f
2; –
4; 8
; …
1) T
ES
T /
TO
ET
S:
12 TT =
23 TT
1)
1
nn
arT
2)
1
)1(
rr
aS
n
n; r
1
3)
ra
S
1 a
s
r <
1
11
Wor
kshe
et 5
1.
Det
erm
ine
the
unit
’s d
igit
(i.e
. las
t dig
it o
f th
e nu
mbe
r) o
f th
e pr
oduc
t of
the
firs
t 200
pri
me
num
bers
. 2.
A
rub
ber
ball
is b
ounc
ed f
rom
a h
eigh
t of
90 m
and
aft
er e
ach
succ
essi
ve b
ounc
e lo
osed
30%
of
the
prev
ious
hei
ght.
a) W
hat m
axim
um h
eigh
t wil
l the
bal
l rea
ch d
urin
g th
e 5t
h bo
unce
?
b)
Aft
er h
ow m
any
rebo
unce
s w
ill t
he b
all r
each
a h
eigh
t of
less
th
an 3
met
res?
c) C
alcu
late
the
tota
l dis
tanc
e tr
avel
led
by th
e ba
ll a
fter
it c
ame
to r
est.
3.
The
fol
low
ing
num
ber
patt
ern
was
giv
en 1
, 5, 1
1, 1
9, …
a) D
eter
min
e th
e 5t
h nu
mbe
r in
the
patt
ern.
b) D
eriv
e a
form
ula
for
the
nth
num
ber
in th
e pa
tter
n.
c)
Wha
t is
the
50th
num
ber
in th
e pa
tter
n?
4.
A s
wim
min
g po
ol c
ompa
ny b
uild
s re
ctan
gula
r po
ols
and
uses
sq
uare
pav
ing
bloc
ks a
roun
d it
.
a)
How
man
y sq
uare
pav
ing
bloc
ks a
re n
eede
d fo
r an
8
5 p
ool?
b)
How
man
y sq
uare
pav
ing
bloc
ks a
re n
eede
d fo
r a
m ×
n p
ool?
c)
How
man
y sq
uare
pav
ing
bloc
ks a
re n
eede
d fo
r a
m ×
m p
ool?
d)
Wha
t is
the
larg
est p
ool t
hat c
an b
e bu
ilt w
ith
200
squa
re p
avin
g bl
ocks
?
1.
Bep
aal d
ie e
ne-s
yfer
van
die
pro
duk
van
die
eers
te 2
00
prie
mge
tall
e.
2.
‘n R
ubbe
rbal
wat
van
’n
hoog
te v
an 9
0 m
laat
val
wor
d, v
erlo
or
met
elk
e te
rugs
pron
g 30
% v
an s
y vo
rige
hoo
gte.
a) W
at is
die
mak
sim
um h
oogt
e w
at d
ie b
al ty
dens
die
5de
te
rugs
pron
g be
reik
?
b)
Na
hoev
eel s
pron
ge b
erei
k di
e ba
l ‘n
hoog
te v
an m
inde
r as
3
met
er?
c)
Ber
eken
die
tota
le a
fsta
nd w
at d
ie b
al a
flê
het n
adat
dit
tot
rus
geko
m h
et.
3.
D
ie v
olge
nde
geta
l pat
roon
wor
d ge
gee:
1, 5
, 11,
19,
…
a)
Bep
aal d
ie 5
de g
etal
in d
ie p
atro
on.
b)
Lei
‘n
form
ule
af v
ir d
ie n
-de
geta
l in
die
patr
oon.
c) W
at is
die
50s
te g
etal
in d
ie p
atro
on?
4. ’
n S
wem
bad
maa
tska
ppy
bou
regh
oeki
ge s
wem
badd
ens
en
gebr
uik
vier
kant
ige
blok
ke o
m d
ie s
wem
bad
as p
lave
isel
.
a)
Hoe
veel
vie
rkan
tige
pla
veis
el b
lokk
e w
ord
beno
dig
vir
‘n 8
× 5
sw
emba
d?
b)
Hoe
veel
vie
rkan
tige
pla
veis
el b
lokk
e w
ord
beno
dig
vir
‘n m
× n
sw
emba
d?
c)
Hoe
veel
vie
rkan
tige
pla
veis
el b
lokk
e w
ord
beno
dig
vir
‘n m
× m
sw
emba
d?
d) W
at is
die
gro
otst
e sw
emba
d w
at m
et 2
00 v
ierk
anti
ge
plav
eise
l blo
kke
gebo
u ka
n w
ord?
6
by
3 po
ol
12
This
bo
ok
will
pr
ovid
e yo
u w
ith
exce
llent
su
mm
arie
s (m
indm
aps)
, ex
ampl
es a
nd e
xerc
ises
. Ava
ilabl
e in
Eng
lish
and
Afr
ikaa
ns.
Mat
hem
atic
s E
xam
Foc
us: G
rade
12
mnd
map
s, e
xam
ples
and
ex
erci
ses.
Pap
er 1
and
2 (I
SB
N: 9
78-0
-620
-411
11-0
). 15
6 pa
ges.
W
isku
nde
Eks
amen
foku
s G
raad
12
kopk
aarte
, voo
rbee
lde
en o
efen
inge
. V
raes
tel 1
en
2. (I
SBN
978
-1-7
7021
-729
-4).
156
blad
sye.
Fo
r fur
ther
que
ries
plea
se c
onta
ct th
e C
leve
r Boo
ks o
ffice
at
Ph
one:
(0
12) 3
42 3
263
Fax:
(0
12) 4
30 2
376
e-m
ail:
helo
ise@
clev
erbo
oks.
co.z
a
web
page
: w
ww
.cle
verb
ooks
.co.
za
For
mor
e de
taile
d su
mm
arie
s an
d ex
ampl
es
orde
r M
athe
mat
ics
Exa
m F
ocus
!
Vir
bet
er o
psom
min
gs e
n vo
orbe
elde
bes
tel
Wis
kund
e E
ksam
en F
okus
!
The
next
6 p
ages
wer
e co
pied
from
this
boo
k.
Die
vol
gend
e 6
blad
sye
kom
uit
hier
die
boek
. 13
e.g
.
40 5
)3
7(r
= (
–8)
+ (
–11)
+ (
–14)
+ …
+ (
–11
3)
r =
5
r =
6
r =
7
r =
40
Numb
er of
term
s = (A
bove
si
gn)
(Belo
w
sign
) + 1
=
(40
5) +
1 =
36 te
rms
D
iffe
rent
pat
tern
s
Ex
pone
ntial
o
1;
2; 4
; 8; …
= 2
0 ; 21 ; 2
2 ; 23 ; …
T
n =
2n
1
o
30 ; 31 ; 3
2 ; 33 ; …
T
n = 3
n1
Ar
ithm
etic:
If 1st d
iffere
nce i
s con
stant
T2 –
T1
= T
3 – T
2
Ge
omet
ric: I
f the r
atio i
s con
stant
= 23
12
TTTT
Qu
adra
tic: I
f 2nd
diffe
renc
e is c
onsta
nt: T
n = a
.n2 +
b.n
+ c
Co
mbi
natio
n: 13
; 25
; 37
; 49
; …
Tn =
nn
12
Al
tern
ate s
igns
ad
d (1
)n or
(1)
n1
e.g.
1;
2;
4; 8
; 16
; …
Tn
= (1
)n .2n
1
Geom
etric
: 2;
1; 21
; … o
r 2
; –4;
8; …
r co
nstan
t rati
o: r
=12 TT
=
4445 TT
1
nn
arT
1;
1
)1(
1
1
r
rra
arn i
ni
1;
11
1
r
raar
i
i1
a =
½ (c
onsta
nt dif
feren
ce)
Sub
stitut
e 1st &
2nd
term
and s
olve s
imult
aneo
usly
for q
and r
Form
ulae
n i
n1
1
n i
nn
i1
2
)1(
Pat
tern
rec
ogni
tion
gui
delin
es
1) L
ook a
t par
ticula
r cas
es (d
raw
pictur
e if
poss
ible)
2)
Dra
w an
d com
plete
a tab
le, e.
g. Nu
mber
of sq
uare
s 1
2 3
4 Nu
mber
of m
atche
s 4
7 10
13
3) I
denti
fy a p
atter
n in t
he ta
ble (w
hat m
ust y
ou
add)
, e.g.
matc
hes:
4; 4 +
1(3)
; 4 +
2(3)
; ….
4) T
ry to
find t
he 10
0th ter
m
E.
g. T 1
00 =
4 +
99(3
) 5)
Us t
he sa
me m
ethod
to w
rite th
e nth t
erm
(gen
erali
se),
e.g. T
n = 4
+ (n
-1)(3
) 6)
Tha
t is yo
ur co
njectu
re (o
r stat
emen
t abo
ut yo
ur fin
ding)
7)
Tes
t you
r con
jectur
e with
value
s in t
able
8)
Jus
tify (o
r pro
ve) y
our c
onjec
ture (
when
aske
d)
- foc
us on
the s
tructu
re of
the s
ketch
es.
- n
otat
ion:
Indic
ates
the su
m of
the te
rms
Arith
met
ic: 2
; 5; 8
; … or
–3; –
7; –1
1; …
d
cons
tant d
iffere
nce:
d =
T2
– T 1
dn
aT n
)1(
n id
na
nd
ia
1
)1(
22
)1(
Pat
tern
s &
Seq
uenc
es
Abbr
eviat
ions
a
= T
1: F
irst te
rm of
the s
eque
nce
n: t
he nu
mber
of te
rms
T n: th
e valu
e of th
e nth t
erm
C
opie
d f
rom
M
ath
emat
ics
Exa
m F
ocu
s
14
Ver
skill
ende
pat
rone
Ek
spon
ensie
el
o
1; 2
; 4; 8
; … =
20 ; 2
1 ; 22 ; 2
3 ; …
Tn
= 2
n1
o
30 ; 31 ; 3
2 ; 33 ; …
T
n = 3
n1
Re
kenk
undi
g: A
s 1ste
versk
il kon
stant
is
T2 –
T1
= T
3 – T
2
Me
etku
ndig
: As d
ie ve
rhou
ding k
onsta
nt is
23
12
TTTT
Kw
adra
ties:
As 2
de ve
rskil k
onsta
nt is:
Tn =
a.n
2 + b
.n +
c
Ko
mbi
nasie
: 13
; 25
; 37
; 49
; …
Tn =
nn
12
Te
kens
wat
afwi
ssel
voeg
by (
1)n o
f (1
)n1
Bv.
1; 2
; 4;
8;
16; …
T
n =
(1
)n .2n
1
Vb.
40 5
)3
7(r
= (
–8)
+ (
–11)
+ (
–14)
+ …
+ (
–11
3)
r =
5
r =
6 r
=7
r =
40
Aanta
l term
e = (B
oons
te – O
nder
ste) +
1 =
(40
5
) +
1 =
36
Meet
kund
ig:
2; 1;
21; .
of
2; –4
; 8; …
r ko
nstan
te ve
rhou
ding:
r =12 TT
=
4445 TT
1
nn
arT
1;
1
)1(
1
1
r
rra
arn i
ni
1;
11
1
r
ra
ari
i1
a =
½ (k
onsta
nte ve
rskil)
Ver
vang
1ste; 2
de te
rme i
n en l
os ge
lyktyd
ig op
.
Reke
nkun
dig:
2; 5;
8; .
of –3
; –7;
–11;
. d
Kon
stante
versk
il: d
= T
2 –
T 1
dn
aT n
)1(
n i
dn
an
di
a1
)1(
22
)1(
Form
ules
n i
n1
1
n i
nn
i1
2
)1(
Pat
roon
herk
enni
ngsr
igly
ne
1) K
yk na
spes
ifieke
geva
lle (t
eken
pren
t indie
n mo
ontlik
)
2) T
eken
en vo
ltooi ʼn
tabel,
byvo
orbe
eld:
Getal
vier
kante
1
2 3
4 Ge
tal vu
urho
utjies
4
7 10
13
3) I
denti
fisee
r die
patro
on in
die t
abel
:
Bv. 4
; 4 +
1(3)
; 4 +
2(3)
; ….
4) B
epaa
l die
100ste
term
:
Bv. T
100 =
4 +
99(3
) 5)
Ver
algem
een:
Gebr
uik di
eself
de m
etode
om
die nde
term
te be
paal:
Tn =
4 +
(n-1
)(3)
6) M
aak ʼn
aflei
ding o
f bew
ering
oor jo
u bev
inding
7)
Toe
ts jou
bewe
ring m
et wa
arde
s in t
abel
8)
Reg
verd
ig of
bewy
s jou
bewe
ring (
indien
ge
vra) –
foku
s op d
ie str
uktuu
r van
die s
ketse
.
Get
alpa
tron
e en
rye
-no
tasie
:
Dui d
ie so
m va
n di
e ter
me a
an.
Cop
ied
fro
m
Mat
hem
atic
s E
xam
Foc
us
Afko
rting
s a
= T
1: E
erste
term
van d
ie ry
n: A
antal
term
e T n
die w
aard
e van
die n
de te
rm
15
Th
emb
ani s
alar
y fo
r 20
08 is
R 1
00 0
00 a
nd
his
exp
ense
s ar
e R
80
000.
H
is s
alar
y in
crea
ses
by
R 1
0 00
0 p
er a
nn
um
wh
ile
his
exp
ense
s in
crea
se b
y R
12
500
per
an
nu
m. E
ach
yea
r h
e m
anag
es t
o sa
ve t
he
exce
ss o
f in
com
e ov
er e
xpen
dit
ure
. a)
Rep
rese
nt
his
acc
um
ula
ted
sav
ings
in a
n a
rith
met
ic s
erie
s.
b)
Aft
er h
ow m
any
year
s w
ill h
is s
avin
gs b
e to
tall
y d
eple
ted
? c)
To
wh
at a
mou
nt
mu
st T
hem
ban
i red
uce
th
e ra
te a
t w
hic
h h
is
ann
ual
exp
ense
s in
crea
se s
o as
to
let
his
sav
ings
last
for
21
year
s?
Arith
met
ic se
quen
ce: a
= 1st
term
; d =
cons
tant
diff
eren
ce
a) 1
st y
ear
savi
ng =
R 2
0 00
0
Sav
ings =
Sala
ry
Expe
nditu
re
G
row
th =
R
2 5
00
R
10 00
0
R 12
500
2
0 00
0 +
175
00 +
15
000
+ …
Arith
metic
serie
s b)
20
000+
17 5
00+
15
000+
… =
0
Sav
ings t
otally
deple
ted if
sum
= 0
a =
20
000
& d
=
2 50
0
1st ter
m =
a; co
nstan
t diffe
renc
e = d
)50
0 2
)(1(
000
20(2
20
nn
U
se
n i
dn
an
di
a1
)1(
22
)1(
0
= n
(40
000
2 50
0n +
2 5
00)
Sim
plify
2
500
n2 4
2 50
0n =
0
2
500
n(n
17)
= 0
Fac
torise
n
= 0
(n/
a) o
r n
= 1
7
S
avin
gs d
eple
ted
in 1
7 ye
ars
c) M
ake
expe
nses
incr
ease
: x
a
= 2
0 00
0 &
d =
10
000
x
F
ind a
& d:
n = 17
and S
um =
0
)
000
10(20
000
20(2221
0x
Use
n i
dn
an
di
a1
)1(
22
)1(
0
= 1
0,5(
40 0
00 +
200
000
2
0x)
Sim
plify
x
= R
12
000
per
annu
m in
crea
se
Ans
wer q
uesti
on
Exam
ple
3 F
ind
th
e n
ext
thre
e te
rms
and
gen
eral
exp
ress
ion
of
each
seq
uen
ce:
a) 1
; 2;
4;
8; ..
. b
) –3
; –7
; –1
1; …
c)
2;
–4;
8; …
d
) 1;
4;
9; 1
6; …
e)
0;
3; 8
; 15
; …
Find
ing
patte
rns
a) …
; 16;
32;
64
T
n= 2
n1
G
eome
tric: m
ultipl
y by 2
b)
…;
15;
19;
23
Tn=
1
4n
Arith
metic
: add
4
c)
…;
16; 3
2;
64
Tn=
(
2)n
Geo
metric
: mult
iply b
y 2
d) …
; 25;
36;
49
T
n= n
2
1
2 ; 22 ; 3
2 ; 42 ; 5
2 ;
e) …
; 24;
35;
48;
63;
T
n= n
2 1
Qua
drati
c: T n
= a
.n2 +
b.n
+ c
9
11
13
2nd di
ffere
nce i
s con
stant:
2
2
a =
½
(2) =
1
Exam
ple
1
Cal
cula
te
30
3
)43(
kk
.
Num
ber o
f ter
ms =
(Abo
ve
sign
) (B
elow
sign
) + 1
= 28
30
3
)43(
kk
= 1
3 +
16
+ 1
9 +
…
S
ubsti
tute k
= 3;
4; 5;
… in
to 3k
+ 4
Ari
thm
etic
ser
ies
T
2 T 1
= T
3 T
2 (=
d)
a =
13
& d
= 3
1st ter
m =
a; co
nstan
t diffe
renc
e = d
n =
30
3 +
1 =
28
T
erms
= (A
bove
– Be
low) +
1
28
1
)3)(
27()
13(2228
)1(
id
ia
U
se
n i
dn
an
di
a1
)1(
22
)1(
= 1
498
Exam
ple
2
16
A r
ub
ber
bal
l is
dro
pp
ed v
erti
call
y fr
om a
bu
ild
ing
20 m
hig
h. T
he
bal
l bou
nce
s to
a h
eigh
t 4/
5 of
th
e h
eigh
t of
th
e p
revi
ous
bou
nce
.
a) F
ind
th
e m
axim
um
hei
ght
of t
he
bal
l on
ce it
has
hit
th
e gr
oun
d t
hre
e ti
mes
.
b)
Det
erm
ine
an a
lgeb
raic
exp
ress
ion
for
th
e m
axim
um
hei
ght
reac
hed
on
ce it
has
hit
th
e gr
oun
d n
tim
es
c) U
se t
he
exp
ress
ion
der
ived
in b
) to
fin
d t
he
max
imu
m h
eigh
t re
ach
ed o
nce
it h
as h
it t
he
grou
nd
10
tim
es
d)
Wh
at d
ista
nce
did
it t
rave
l th
e m
omen
t th
e b
all h
it t
he
grou
nd
th
e 10
th t
ime.
e) W
hat
is t
he
tota
l dis
tan
ce t
rave
lled
by
the
bal
l?
Geom
etric
sequ
ence
: a =
1st te
rm; r
= co
nsta
nt ra
tio
Lo
ok at
partic
ular c
ases
(dra
w pic
ture i
f pos
sible)
D
raw
and c
omple
te a t
able
P
atter
n eme
rge:
54 ev
ery t
ime:
Geom
etric
sequ
ence
a)
Max
imum
hei
ght (
afte
r hi
ttin
g th
e gr
ound
3 ti
mes
) =
20(
54)3 =
10,
24 m
F
ind ge
nera
l term
by us
ing th
e patt
ern
b) P
atte
rn e
mer
ge: T
n =
20(
54)n o
r H
(n)
= 2
0(54
)n
S
ubsti
tute n
= 10
into
gene
ral te
rm
c) T
10 =
20(
54)10
= 2
,15
m
M
ultipl
ied di
stanc
e by 2
: Up &
down
d)
Dis
tanc
e tr
avel
led
afte
r nth
hit
: 20
+ 2
[20(
54)]
+ 2
[20(
54)2 ]
+ 2
[20(
54)3 ]
+ …
B
ounc
e:
1st
2
nd
3rd
4th
G
eome
tric se
quen
ce (ig
nore
the 2
0 fro
m the
serie
s)
a =
2[2
0(54
)] =
32
& c
onst
ant r
atio
: r =
54
C
alcula
te
9 1
1
i
iar
: Sum
of th
e las
t 9 te
rms &
add 2
0 m
9 1
54
954
1
1
]1)
[(32
i
iar
= 1
38,5
3 m
T
otal
dis
tanc
e tr
avel
led
= 1
38,5
3 +
20
= 1
58,5
3 m
In
finite
conv
erge
nt ge
ometr
ic se
quen
ce
1
1
1i
i
ra
ar
e)
154
1
132
i
iar
= 1
60 m
Tot
al d
ista
nce
trav
elle
d =
160
+ 2
0 =
180
m
Exam
ple
4
20 m
h =
20
54
1st
2nd
3
rd
4th
5th
h =
20 (
54)2
Numb
er of
times
the b
all hi
t the g
roun
d 1
2 3
4
Maxim
um he
ight (
after
hittin
g the
grou
nd)
20(
54)1
20
(54
)220
(54
)3
20(
54)4
Altho
ugh t
he ba
ll mov
es on
ly in
a ver
tical
direc
tion t
he sk
etch
adds
a ho
rizon
tal di
recti
on fo
r illus
tratio
n pur
pose
s!
Cop
ied
fro
m
Mat
hem
atic
s E
xam
Foc
us
17
GR
OE
I En
kelvo
udig
: F =
P(1
+ in
) Sa
amge
steld
: F =
P(1
+ i)
n
D
EP
RE
SIA
SIE
N
egat
iew
e re
ntek
oers
(i
)
Regu
itlyn
-dep
resia
sie (H
uurk
oop)
: Geb
ruik
enke
lvoud
ige fo
rmule
: F =
P(1
in)
Depr
esias
ie op
verm
inde
rde b
alans
: Geb
ruik
saam
geste
lde fo
rmule
: F =
P(1
i)n
Indi
en d
ie re
nte m
eer a
s een
keer
ʼn ja
ar b
ygev
oeg
word
:
Aa
ntal s
aamg
estel
de ty
dper
ke: n
= (a
antal
jaar
)
m (a
antal
kere
rente
per ja
ar ge
hef)
Rente
koer
s per
tydp
erk (
desim
aal):
mii
m)
(
jaa
r
per
re
nte
kere
aa
ntal
jaar)
(in
rente
no
mina
le
Voor
beeld
e: B
elê R
100 v
ir 3 j
aar t
een
12%
per
jaar
m
aand
eliks
saam
geste
l: P
= 10
0; i
=
1212,0=
0,0
1; n
= 3
1
2 =
36
F =
100
(1 +
0,0
1)36
= R
143,
08
kwa
rtaall
iks sa
amge
stel: P
= 1
00;
i =
412,0
= 0
,03;
n =
3
4 =
12
F =
100
(1 +
0,0
3)12
= R
142,
58
half
jaarlik
s saa
mges
tel: P
= 1
00;
i =
212,0
= 0
,06;
n =
3
2 =
6
F =
100
(1 +
0,0
6)6 =
R14
1,85
Nom
inale
koer
s
i(m) :
Die n
omina
le re
nteko
ers i
s die
perio
dieke
re
nteko
ers (
uitge
druk
in ja
ar).
Bv: i
(4) =
12%
= 0,1
2 Ef
fekt
iewe k
oers
i:
Di
e effe
ktiew
e ren
tekoe
rs is
dit w
at jy
werkl
ik in
jou
sak g
aan k
ry of
betaa
l na 1
jaar
. Bv:
i = 12
,55%
Fina
nsie
s
Omse
tting:
(1+
mim)
(
)m =
1+
i
F: T
oeko
mstig
e waa
rde (
eindw
aard
e)
P: H
uidige
waa
rde (
oorsp
ronk
like w
aard
e)
n: A
antal
saam
geste
lde ty
dper
ke
i: R
entek
oers
per t
ydpe
rk (d
esim
aal)
Gebr
uik l
ogs o
m n
in ek
spon
ent o
p te
los:
ax =
b, d
an x
= lo
g a b
F =
P(1
+ i)
n
n
PFi)
1(
PF
in
)1(
log
)1
log(lo
g
in
PF
As jo
u sak
reke
naar
met
enige
basis
kan w
erk –
igno
reer
laas
te sta
p!
Uit
geh
aal u
it W
isk
un
de
Ek
sam
enfo
ku
s G
raad
12
Kop
kaa
rte,
voo
rbee
lde
en o
efen
inge
: S
ien
bla
dsy
13
vir
bes
tell
ings
18
A n
ew c
ar c
osts
R 1
50 0
00. I
f in
flat
ion
is c
alcu
late
d a
t 16
% p
er
ann
um
an
d c
omp
oun
ded
mon
thly
, wh
at w
ill b
e th
e co
st o
f th
e ca
r in
4
year
s ti
me?
Com
poun
d gr
owth
: Use
F =
P(1
+ i)
n & C
ompo
unde
d MO
NTHL
Y F
= ?
Futu
re or
final
value
of ca
r P
= R
150
000
Prin
cipal
or cu
rrent
price
of ca
r n
= 4
1
2 =
48
mon
ths
N
umbe
r of c
ompo
undin
g MON
THS
i =
1216,0=
0,1
33…
mon
thly
Inter
est r
ate pe
r mon
th (d
ecim
al for
m)
F =
P(1
+ i)
n
U
se co
mpou
nded
grow
th for
mula
=
150
000
(1 +
1216,0
)48
Sub
stitut
e valu
es
=
R 2
83 2
71,6
1
Use
calcu
lator
Th
e va
lue
of a
com
pu
ter
dec
reas
es f
rom
R 5
000
to
R 2
800
in t
hre
e ye
ars.
Cal
cula
te t
he
rate
of
dep
reci
atio
n p
er a
nn
um
if t
he
calc
ula
tion
is
don
e an
nua
lly.
Com
poun
d de
prec
iatio
n: U
se F
= P
(1 +
i)n
BUT
rate
will
be n
egat
ive
F =
R 2
800
Futu
re va
lue of
comp
uter
P =
R 5
000
Prin
cipal
or cu
rrent
price
of co
mpute
r n
= 3
Num
ber o
f com
poun
ding y
ears
i =
?
In
teres
t rate
per y
ear
F =
P(1
+ i)
n
U
se co
mpou
nded
form
ula
2 80
0 =
5 0
00 (
1 +
i)3
Sub
stitut
e valu
es
3 )1(
5000
2800
i
Divi
de by
5 00
0 both
side
s
i
1
5000
2800
3
C
ube r
oot b
oth si
des
i
1
5000
2800
3
i =
0, 1
757
U
se ca
lculat
or
The
rat
e of
dep
reci
atio
n is
17,
57%
Exam
ple
1 Ex
ampl
e 3
Mok
gad
i buy
s a
new
fri
dge
at
a ca
sh p
rice
of
R 1
500
on
a h
ire
pu
rch
ase
agre
emen
t. S
he
pay
s a
dep
osit
of
20%
an
d p
ays
the
rem
ain
ing
bal
ance
off
ove
r 4
year
s w
ith
an
inte
rest
rat
e of
18%
per
an
nu
m c
omp
oun
ded
mon
thly
. Cal
cula
te t
he
mon
thly
pay
men
ts.
Hire
pur
chas
e S
impl
e gro
wth
form
ula:
F =
P(1
+ in
) F
= ?
Futu
re va
lue of
loan
P
= 1
500
(150
0
20%
) =
120
0
Loa
n (P)
= to
tal pr
ice
depo
sit
n =
4 y
ears
(an
d 48
pay
men
ts)
N
umbe
r of y
ears
i = 0
,18
p.a.
Inter
est r
ate pe
r ann
um
F =
P(1
+ in
)
H
ire pu
rchas
e S
imple
form
ula
=
120
0[1
+ (
0,18
)(4)
] =
206
4
Sub
stitut
e valu
es
Inst
alm
ents
= R
206
4
48 =
R43
Insta
lmen
ts =
F
Numb
er pa
ymen
ts
Exam
ple
2
Anot
her w
ay to
ask t
he q
uest
ion
is to
give
the f
ract
ion
of d
epre
ciatio
n.
For e
xam
ple i
f the
car l
oses
32of
its o
rigin
al va
lue t
hen
31
PF=
(1 +
i)n
Cop
ied
fro
m M
ath
emat
ics
Exa
m F
ocu
s G
rad
e 12
: M
ind
map
s, e
xam
ple
s an
d e
xerc
ises
19
Wor
kshe
et 8
Dav
id b
elê
R1
000
vir
8 ja
ar. H
y on
tvan
g re
nte
van
12%
per
jaa
r, m
aan
del
iks
saam
gest
el, v
ir d
ie e
erst
e 2
jaar
. Die
ren
tek
oers
ver
and
er
na
24%
per
jaa
r, k
war
taal
lik
s sa
amge
stel
, vir
die
vol
gen
de
3 ja
ar, e
n v
eran
der
wee
r n
a 16
% p
er j
aar,
hal
fjaa
rlik
s sa
amge
stel
, vir
die
oo
rbly
wen
de
tyd
per
k. B
erek
en h
oeve
el g
eld
Dav
id n
a 8
jaar
sal
kry
.
Gebr
uik ʼn
tydl
yn o
m m
eer i
ngew
ikkeld
e fin
ansië
le pr
oblem
e op
te lo
s.
FIN
AL
E A
NT
WO
OR
D:
Dav
id s
al R
4 05
4,38
na
8 ja
ar o
ntva
ng
J
are:
0
12
34
56
78
Maa
ndel
iks
i =1212,0
= 0
,01
Kw
arta
allik
s i =
424,0
= 0
,06
Hal
fjaar
liks
i =216,
0=
0,0
8
2
12 =
24
maa
nde
3
4 =
12
kwar
tale
3
2 =
6 h
alfj
are
R 1
000
F 1
F 2
F 3
Om F
1 te b
erek
en: D
ie to
ekom
stig
e waa
rde v
an R
1 000
na 2
jaar
F 1
= ?
Toe
koms
tige w
aard
e
P =
1 0
00
H
uidige
waa
rde
n =
24
maa
nde
A
antal
tydp
erke
(maa
nde)
i =
1212,0=
0,0
1 pe
r m
aand
Maa
ndeli
kse r
entek
oers
F 1
= P
(1 +
i)n
Saa
mges
telde
groe
iform
ule
=
1 0
00(1
+ 0
,01)
24
Ver
vang
waa
rdes
= R
1 26
9,73
Geb
ruik
sakre
kena
ar
Om F
2 te b
erek
en: G
ebru
ik F 1
= R
1 26
9,73 a
s hui
dige
waa
rde
F 2 =
?
T
oeko
mstig
e waa
rde
P
= F
1 =
R1
269,
73
H
uidige
waa
rde
n =
12
kwar
tale
Aan
tal ty
dper
ke (k
warta
le)
i =424,
0=
0,0
6 pe
r kw
arta
al
R
entek
oers
per k
warta
al
F 2 =
P(1
+ i)
n
S
aamg
estel
de gr
oeifo
rmule
= 1
269
,73(
1 +
0,0
6)12
Ver
vang
waa
rdes
= R
2 55
4,95
Geb
ruik
sakre
kena
ar
Om F
3 te b
erek
en: G
ebru
ik F 2
= R
2 55
4,95 a
s hui
dige
waa
rde
F 3 =
?
T
oeko
mstig
e waa
rde
P
= F
2 =
R2
554,
95
H
uidige
waa
rde
n =
6 h
alfj
are
A
antal
tydp
erke
i =
216,0
= 0
,08
R
entek
oers
per t
ydpe
rk F 3
= P
(1 +
i)n
Saa
mges
telde
groe
iform
ule
=
2 5
54,9
5(1+
0,0
8)6
V
erva
ng w
aard
es
=
R4
054,
38
G
ebru
ik sa
kreke
naar
Voor
beel
d 6
Uit
geh
aal u
it W
isk
un
de
Ek
sam
enfo
ku
s G
raad
12
Kop
kaa
rte,
voo
rbee
lde
en o
efen
inge
20
Wor
kshe
et 6
1.
A n
ew h
ouse
cos
ts R
240
000
. If
infl
atio
n is
cal
cula
ted
at 1
8%
per
annu
m a
nd c
ompo
unde
d m
onth
ly, w
hat w
ill b
e th
e co
st o
f th
e ho
use
in 1
2 ye
ars
tim
e?
2.
Cal
cula
te th
e ti
me
in w
hich
an
inve
stm
ent o
f R
450
wil
l in
crea
se to
R 1
200
if it
is in
vest
ed a
t a c
ompo
und
incr
ease
of
17%
per
ann
um if
the
inte
rest
is a
dded
mon
thly
.
3.
The
val
ue o
f a
vehi
cle
decr
ease
s fr
om R
18
700
to R
12
475
in
four
yea
rs. C
alcu
late
the
rate
of
depr
ecia
tion
per
ann
um if
the
calc
ulat
ion
is d
one
annu
ally
.
4.
Mak
e P,
i an
d th
en n
the
subj
ect o
f th
e fo
llow
ing
form
ula:
F
= P
(1 +
i)n
5.
Est
er in
vest
s a
cert
ain
sum
of
mon
ey f
or 6
yea
rs. S
he r
ecei
ves
inte
rest
of
12%
per
ann
um c
ompo
unde
d m
onth
ly f
or th
e fi
rst
two
year
s. T
he in
tere
st r
ate
chan
ges
to 1
6% p
er a
nnum
co
mpo
unde
d se
mi-
annu
ally
for
the
rem
aini
ng te
rm. T
he m
oney
ap
prec
iate
s to
R80
000
at t
he e
nd o
f th
e 6-
year
per
iod.
a)
Cal
cula
te th
e ef
fect
ive
inte
rest
rat
e pe
r an
num
dur
ing
th
e fi
rst y
ear.
b)
Cal
cula
te h
ow m
uch
mon
ey E
ster
inve
sted
init
iall
y.
6.
Aft
er 4
yea
rs o
f re
duci
ng b
alan
ce d
epre
ciat
ion,
a c
ar h
as a
¼ o
f it
s or
igin
al v
alue
. The
ori
gina
l val
ue w
as R
120
000.
Cal
cula
te
the
depr
ecia
tion
inte
rest
rat
e, a
s a
perc
enta
ge. (
Cor
rect
you
r an
swer
to 1
dec
imal
pla
ce.)
1.
’n N
uwe
huis
kos
R 2
40 0
00. A
s in
flas
ie b
erek
en w
ord
teen
18
% p
er ja
ar e
n m
aand
elik
s sa
am w
ord,
wat
sal
die
hui
s oo
r 12
ja
ar k
os?
2.
Ber
eken
die
tydp
erk
waa
rin
’n b
eleg
ging
van
R 4
50 s
al o
ploo
p to
t R 1
200
as
dit b
elê
wor
d te
en ’
n sa
amge
stel
de k
oers
van
17%
pe
r ja
ar a
s di
e re
nte
maa
ndel
iks
byge
voeg
wor
d.
3.
Die
waa
rde
van
’n v
oert
uig
verm
inde
r va
n R
18
700
na R
12
475
in v
ier
jaar
. Ber
eken
die
waa
rdev
erm
inde
ring
skoe
rs p
er ja
ar
as d
ie b
erek
enin
g ja
arli
ks g
edoe
n w
ord.
4.
Maa
k P
, i e
n da
n n
die
onde
rwer
p va
n di
e vo
lgen
de f
orm
ule:
F
= P
(1 +
i)n
5.
Est
er b
elê
'n s
eker
e be
drag
gel
d vi
r 6
jaar
. Vir
die
eer
ste
twee
ja
ar o
ntva
ng s
y re
nte
van
12%
per
jaar
, wat
maa
ndel
iks
saam
gest
el w
ord.
Die
ren
teko
ers
vera
nder
na
16%
per
jaar
, wat
ha
lfja
arli
ks v
ir d
ie o
orbl
ywen
de te
rmyn
vas
gest
el w
ord.
Die
be
drag
het
tot R
80 0
00 a
an d
ie e
inde
van
die
6 ja
ar-p
erio
de
gegr
oei.
a)
Ber
eken
die
eff
ekti
ewe
rent
ekoe
rs p
er ja
ar g
edur
ende
di
e ee
rste
jaar
.
b) B
erek
en h
oeve
el g
eld
Est
er a
anva
nkli
k be
lê h
et.
6.
Na
4 ja
ar v
an a
fnem
ende
bal
ansv
erm
inde
ring
, is
'n m
otor
‘n
¼
van
die
oors
pron
klik
e w
aard
e w
erd.
Die
oor
spro
nkli
ke w
aard
e w
as R
120
000.
Ber
eken
die
waa
rdev
erm
inde
ring
sren
teko
ers,
as
'n p
erse
ntas
ie. (
Ron
d jo
u an
twoo
rd k
orre
k af
tot e
en d
esim
ale
plek
.)
21
Ann
uiti
es /
Ann
uïte
ite
Hom
e lo
ans
(bon
d re
paym
ent)
/ H
uisl
enin
gs (
Ver
band
paai
emen
te)
Inte
rest
on
a ho
me
loan
is c
alcu
late
d at
the
begi
nnin
g of
eac
h m
onth
, and
pay
men
ts a
re c
alcu
late
d on
a m
onth
ly b
asis
. / R
ente
op
‘n h
uisl
enin
g w
ord
bere
ken
aan
die
begi
n va
n el
ke m
aand
, en
beta
ling
s w
ord
op ‘
n m
aand
elik
se b
asis
ber
eken
.
Use P
rese
nt va
lue f
orm
ula t
o ca
lculat
e x / G
ebru
ik di
e Hui
dige
waa
rde f
orm
ule o
m x
te b
erek
en:
ii
xP
n]
)1(
1[
P =
Loa
n (B
ond
amou
nt)
/ Len
ing
(Ver
band
bed
rag)
n =
num
ber
paym
ents
/ aa
ntal
teru
gbet
alin
gs (
aant
al p
aaie
men
te)
i = in
tere
st r
ate
per
peri
od /
rent
ekoe
rs p
er s
aam
gest
elde
per
iode
x
= r
egul
ar p
aym
ent /
ger
eeld
e be
tali
ng
Tota
l am
ount
to b
e rep
aid / T
otale
bed
rag
wat t
erug
bet
aal m
oet w
ord:
R
egul
ar p
aym
ent ×
Num
ber
of r
epay
men
ts /
Paa
iem
ent ×
Aan
tal t
erug
beta
ling
s
Outs
tand
ing
balan
ce / U
itsta
ande
bala
ns:
Wor
king
fro
m t
he p
rese
nt t
ime
to t
he p
oint
in
tim
e w
hen
the
bala
nce
mus
t be
cal
cula
ted
/ W
erk
vana
f be
gint
yd t
ot d
ie t
ydst
ip w
aaro
p di
e ba
lans
ber
eken
moe
t w
ord:
B
alan
ce =
(lo
an +
int
eres
t) –
(su
m o
f pa
ymen
ts +
int
eres
t) B
alan
s =
(le
ning
+ r
ente
) –
(som
van
paa
iem
ente
+
rent
e)
Sin
king
fun
ds /
Sin
kend
e fo
ndse
De
finiti
on: S
tart
sav
ing
a fi
xed
mon
thly
inst
alm
ent t
o pr
ovid
e fo
r th
e pu
rcha
se o
f ne
w e
quip
men
t in
futu
re. U
se p
rese
nt e
quip
men
t as
a de
posi
t.
Defin
isie:
Beg
in v
roeg
tydi
g ‘n
ger
eeld
e pa
aiem
ent t
e sp
aar
om v
oors
ieni
ng te
maa
k vi
r di
e aa
nkoo
p va
n nu
we
toer
usti
ng in
die
toek
oms.
V
erko
op h
uidi
ge to
erus
ting
vir
‘n
depo
sito
Month
ly ins
talme
nt: C
alcula
te x i
n Futu
re V
alue F
ormu
la / M
aand
eliks
e paa
iemen
te: B
erek
en x
in die
Toe
koms
tige w
aard
e for
mule
ii
xF
n]1
)1
[(
22
Wor
kshe
et 7
1.
Tok
yo b
orro
wed
R50
000
and
agr
ees
to r
epay
the
loan
by
mea
ns
of 5
0 eq
ual
mon
thly
ins
talm
ents
. In
tere
st i
s ca
lcul
ated
at
16%
pe
r an
num
co
mpo
unde
d m
onth
ly.
Cal
cula
te
the
mon
thly
in
stal
men
t.
2.
An
amou
nt o
f x
rand
is
inve
sted
k t
imes
per
ann
um i
n an
an
nuit
y. T
he in
tere
st r
ate
is r
% p
er a
nnum
and
is c
ompo
unde
d k
tim
es p
er a
nnum
. Der
ive
a fo
rmul
a fr
om f
irst
pri
ncip
les
that
you
ca
n us
e to
cal
cula
te t
he a
mou
nt a
vail
able
in
the
annu
ity
afte
r n
year
s.
3.
Tab
u de
cide
s to
inv
est
R5
000
quar
terl
y in
an
annu
ity.
His
fir
st
paym
ent
is o
nly
at t
he e
nd o
f th
e fi
rst
quat
er.
If t
he c
ompo
und
inte
rest
he
earn
s is
cal
cula
ted
quar
terl
y at
16%
per
ann
um,
calc
ulat
e th
e va
lue
of th
is a
nnui
ty in
fiv
e ye
ars
tim
e fr
om n
ow.
4.
The
cos
t of
a bu
s is
R2
400
000.
The
cos
t is
expe
cted
to r
ise
at a
ra
te o
f 16
% p
er a
nnum
com
poun
d in
tere
st,
whi
le t
he v
alue
of
the
bus
depr
ecia
tes
at a
rat
e of
22%
com
poun
ded
depr
ecia
tion
pe
r an
num
. The
life
exp
ecta
ncy
of th
e bu
s is
5 y
ears
.
a)
Fin
d th
e sc
rap
valu
e of
the
bus.
b)
Fin
d th
e co
st o
f a
new
bus
in f
ive
year
s’ ti
me.
c)
A s
inki
ng f
und
is e
stab
lish
ed t
o pa
y fo
r th
e ne
w b
us. F
or t
he
sink
ing
fund
, pay
men
ts a
re m
ade
into
an
acco
unt p
ayin
g 16
%
per
annu
m,
com
poun
ded
mon
thly
. F
ind
the
valu
e of
the
si
nkin
g fu
nd
and
the
size
of
th
e m
onth
ly
paym
ents
, if
pa
ymen
ts s
tart
in a
yea
r’s
tim
e an
d st
op th
e m
onth
bef
ore
the
purc
hase
of
the
new
bus
.
1.
Tok
yo l
een
R50
000
en
kom
oor
een
om d
ie l
enin
g in
50
gely
ke
maa
ndel
ikse
paa
iem
ente
ter
ug t
e be
taal
. R
ente
wor
d be
reke
n te
en 1
6% p
er ja
ar m
aand
elik
s be
reke
n. B
erek
en d
ie m
aand
elik
se
paai
emen
t.
2.
'n B
edra
g va
n x
rand
wor
d k
keer
per
jaar
in 'n
ann
uïte
it b
elê
Die
re
ntek
oers
is r%
per
jaar
en
wor
d k
keer
per
jaar
saa
mge
stel
. Lei
'n
for
mul
e va
nuit
eer
ste
begi
nsel
s af
waa
rmee
u d
ie b
edra
g be
skik
baar
in d
ie a
nnuï
teit
na
n ja
ar s
al k
an b
erek
en.
3.
Tab
u be
slui
t om
kw
arta
alli
ks R
5 00
0 in
'n a
nnuï
teit
te
belê
. S
y ee
rste
bet
alin
g is
eer
s aa
n di
e ei
nde
van
die
eers
te k
war
taal
. A
s di
e sa
amge
stel
de
rent
e w
at
hy
verd
ien
teen
16
%
per
jaar
kw
arta
alli
ks
geka
pita
lise
er
wor
d,
bere
ken
die
waa
rde
van
hier
die
annu
ïtei
t oor
vyf
jaar
van
nou
af.
4.
’n B
us k
os R
2 40
0 00
0. D
ie v
erw
agte
sty
ging
in
kost
e is
16%
sa
amge
stel
de r
ente
per
jaa
r. D
ie w
aard
e va
n di
e bu
s ve
rmin
der
teen
’n
saam
gest
elde
koe
rs v
an 2
2% p
er j
aar.
Die
ver
wag
te
gebr
uiks
leef
tyd
van
die
bus
is 5
jaar
.
a) B
epaa
l die
skr
ootw
aard
e va
n di
e bu
s.
b)
Vin
d di
e ko
ste
van
die
nuw
e bu
s oo
r vy
f ja
ar.
c)
’n
Del
ging
sfon
ds w
ord
gest
ig o
m v
ir d
ie n
uwe
bus
te b
etaa
l. V
ir h
ierd
ie f
onds
gel
d di
e vo
lgen
de:
Die
paa
iem
ente
wor
d ge
maa
k in
‘n
reke
ning
wat
16%
per
jaa
r be
taal
, m
aand
elik
s sa
amge
stel
. B
epaa
l di
e w
aard
e va
n hi
erdi
e de
lgin
gsfo
nds
en
die
groo
tte
van
die
maa
ndel
ikse
pa
aiem
ent,
indi
en
paai
emen
te o
or e
en j
aar
begi
n en
ein
dig
’n m
aand
voo
rdat
di
e bu
s aa
ngek
oop
wor
d.
23
Str
aigh
t lin
e &
par
abol
a / R
egui
tlyn
& p
arab
ool
Stan
dard
form
/ Sta
ndaa
rdvo
rm: y
= m
x +
c
m
(slop
e / gr
adiën
t) =
12
12
x
xy
yxy
o
|| line
s / ly
ne: m
1 =
m2
o
lin
es / l
yne:
m1.m
2 =
–1
c i
s the
y-int
erce
pt / c
is di
e y-a
fsnit
Sket
ch (3
met
hods
) / S
kets
(3 m
etod
es)
Ta
ble: C
hoos
e x an
d dete
rmine
y / T
abel:
Kies
x en
be
paal
y
Doub
le int
erce
pt / D
ubbe
l-afsn
it meto
de:
o
y-i
nterce
pt: le
t x =
0 /
y-afs
nit: s
tel x
= 0
o
x-int
erce
pt: le
t y =
0 /
x-afs
nit: s
tel y
= 0
Gr
adien
t inter
cept
/ Gra
diënt
afsnit
:
o
Get y
-inter
cept
/ Kry
y-afs
nit
o
Use m
: Wor
k in y
then
in x
direc
tion /
Geb
ruik
m: W
erk e
ers i
n y- e
n dan
in x-
rigtin
g De
term
inin
g th
e equ
atio
n / B
epali
ng va
n di
e ver
gelyk
ing
De
termi
ne gr
adien
t of li
ne / G
ebru
ik die
grad
iënt v
an
die ly
n
Find t
he co
ordin
ates o
f a po
int on
the l
ine / B
epaa
l ko
ördin
ate va
n ‘n p
unt o
p die
lyn
Su
bstitu
te m
and t
he po
int in
/ Ver
vang
m en
die p
unt
in: (
y –
y A)
= m
(x –
xA)
Gene
ral f
orm
/ Sta
ndaa
rdvo
rm:
y =
a(x
+ p
)2 + q
Sk
etch
/ Ske
ts
Tu
rning
point
: / Dr
aaipu
nt: (p
; q)
Fo
rm / V
orm:
a va
lue / a
waa
rde
+
–
x-
inter
cepts
/ x-a
fsnitte
: Let
/ Stel
y =
0
y-
inter
cept
/ y-a
fsnit:
Let /
Stel
x =
0
Sket
ch u
sing
trans
form
atio
ns / S
kets
mbv
tra
nsfo
rmas
ies: y
= a
(x –
p)2 +
q
Dr
aw th
e gra
ph / S
kets
die gr
afiek
van:
y =
ax2
Sh
ift gr
aph q
units
in th
e ver
tical
direc
tion /
Sku
if gr
afiek
q ee
nhed
e ver
tikaa
l
Shift
grap
h (
p) un
its in
the h
orizo
ntal d
irecti
on /
Skuif
grafi
ek (
p) ee
nhed
e hor
isonta
al
Dete
rmin
ing
the e
quat
ion
/ Bep
aling
van
die v
erge
lykin
g
Turn
ing po
int +
any o
ther p
oint /
Draa
ipunt
+ en
ige
ande
r pun
t: y
= a
(x –
p)2 +
q
Tw
o roo
ts +
any p
oint /
Twee
wor
tels +
enige
punt:
y
= a
(x –
x1)
(x –
x2)
24
Hyp
erbo
la /
Hip
erbo
ol
Gene
ral f
orm
/ Alg
emen
e vor
m:
xay
; if a
> 0
grap
h is i
n the
1st an
d 3rd
quad
rants
/ as a
> 0
dan i
s gra
fiek i
n 1st
en 3rd
kwa
dran
te
If gr
aph
is no
t in
stan
dard
pos
ition
/ Ind
ien g
rafie
k nie
in st
anda
ard
posis
ie is
nie:
q
px
ay
)
(; w
ith / m
et
a
(vertic
al str
etch /
vertik
ale ui
trekk
ing)
p (h
orizo
ntal s
hift /
horis
ontal
e sku
if)
q
(vertic
al sh
ift / v
ertik
ale sk
uif)
Asym
ptot
es o
f the
gra
ph / A
simpt
ote v
an d
ie gr
afiek
: q
px
ay
)
(
Ho
rizon
tal as
ympto
te / H
oriso
ntale
asim
ptoot:
y =
q
Ve
rtical
asym
ptote
/ Ver
tikaa
l asim
ptoot:
x =
p
Sket
ch u
sing
a tab
le / S
kets
gra
fiek m
bv ta
bel
Get y
alon
e. / K
ry y
allee
n aan
‘n ka
nt.
Ta
ble: C
hoos
e x an
d calc
ulate
y. / T
abel:
Kies
x en
bere
ken y
.
Sket
ch u
sing
trans
form
atio
ns / S
kets
mbv
tran
sfor
mas
ies:
qp
xa
y
)(
Dr
aw th
e gra
ph / S
kets
grafi
ek va
n: y
= xa
Shift
grap
h q un
its in
the v
ertic
al dir
ectio
n / S
kuif g
rafie
k q ee
nhed
e in d
ie ve
rtikale
rigtin
g
Shift
grap
h (
p) un
its in
the h
orizo
ntal d
irecti
on / S
kuif g
rafie
k (p
) een
hede
in di
e hor
isonta
le rig
ting
25
Exp
onen
tial
gra
ph /
Eks
pone
nsië
le g
rafi
ek
Gene
ral f
orm
/ Alg
emen
e vor
m:
y =
bx ; b
> 0
If gr
aph
is no
t in
stan
dard
pos
ition
/ Ind
ien g
rafie
k nie
in st
anda
ard
posis
ie is
nie:
y =
a.b
(x
+ p
) +
q, w
ith / m
et
a (ve
rtical
stretc
h / ve
rtikale
uitre
kking
)
p (h
orizo
ntal s
hift /
horis
ontal
e sku
if)
q (ve
rtical
shift
/ ver
tikale
skuif
)
Asym
ptot
es o
f the
gra
ph / A
simpt
ote v
an d
ie gr
afie:
y =
a.b
(x
+ p
) +
q
Ho
rizon
tal as
ympto
te / H
oriso
ntale
asim
ptoot:
y =
q
Sket
ch u
sing
a tab
le / S
kets
gra
fiek m
bv ta
bel
Ge
t y al
one.
/ Kry
y alle
en aa
n ‘n k
ant.
Ta
ble: C
hoos
e x an
d calc
ulate
y. / T
abel:
Kies
x en
bere
ken y
.
Sket
ch u
sing
trans
form
atio
ns / S
kets
mbv
tran
sfor
mas
ies: y
= a
.b (
x +
p) +
q
Dr
aw th
e gra
ph / S
kets
grafi
ek va
n: y
= a
.bx
Sh
ift gr
aph q
units
in th
e ver
tical
direc
tion /
Sku
if gra
fiek q
eenh
ede i
n die
vertik
ale rig
ting
Sh
ift gr
aph (
p) un
its in
the h
orizo
ntal d
irecti
on / S
kuif g
rafie
k (p
) een
hede
in di
e hor
isonta
le rig
ting
Dete
rmin
ing
the e
quat
ion
/ Bep
aling
van
die v
erge
lykin
g: y
= a
.b (
x +
p) +
q
Fin
d hor
izonta
l asy
mptot
e: If e
quati
on is
y =
3 the
n q =
3 / B
epaa
l die
horis
ontal
e asim
ptoot:
As y
= 3
dan i
s q =
3
Us
e y-in
terce
pt to
find a
: If p
= 0
: a =
y-int
erce
pt / G
ebru
ik y-
afsnit
om a
te be
paal:
As p
= 0
dan i
s a =
y-afs
nit
Su
bstitu
te an
y poin
t on t
he ex
pone
ntial
grap
h tem
pora
rily in
y =
bx an
d calc
ulate
b. / V
erva
ng en
ige pu
nt tyd
elik i
n die
verg
elykin
g om
b te
bepa
al.
26
x
y
Pro
pert
ies
of g
raph
s / E
iens
kapp
e va
n gr
afie
ke
Line
ar f
unct
ion
Qua
drat
ic
Exp
onen
tial
H
yper
bola
y =
x
y
= x
2 y
= b
x y
=
x1
doma
in R
R R
(
; 0)
(0; +
)
rang
e R
[0; +
) R
(
; 0)
(0; +
)
x- int
erce
pts
0 0
- -
y- int
erce
pts
0 0
1 -
turnin
g poin
ts
- (0
; 0) is
a mi
nimum
-
-
asym
ptotes
-
- y =
0 x =
0; y
= 0
symm
etry
no sy
mmetr
y x =
0 no
symm
etry
y =
x
inter
vals
on w
hich t
he fu
nctio
n inc
reas
es
R (0
; +
) R
-
inter
vals
on w
hich t
he fu
nctio
n de
creas
es
- (
; 0)
- (
; 0)
(0; +
)
x
y
x
y
x
y
27
Wor
kshe
et 8
1.
G
iven
/Geg
ee:
f (x)
= 2
cos
(x
+ 3
0°)
F
ind
the
foll
owin
g pr
oper
ties
of
f.
a)
m
axim
um v
alue
/ m
aksi
mum
waa
rde
b)
pe
riod
/ pe
riod
e
c)
ra
nge
/ waa
rdev
ersa
mel
ing
d)
am
plit
ude
2.
Giv
en/G
egee
: f (
x) =
3ta
n x
2.
F
ind
the
foll
owin
g pr
oper
ties
of
f / B
epaa
l die
vol
gend
e ei
ensk
appe
van
f:
a)
y-
inte
rcep
t / y
-afs
nitt
e
b)
ge
nera
l equ
atio
n of
the
asym
ptot
es
alge
men
e ve
rgel
ykin
g va
n di
e as
impt
ote
c)
pe
riod
/ pe
riod
e
d)
ra
nge
/ waa
rdev
ersa
mel
ing
3.
Ske
tch
the
grap
h of
/Ske
ts d
ie g
rafi
ek v
an f
(x)
= (
x
3)2 +
2.
U
se y
our
grap
h to
fin
d th
e fo
llow
ing
prop
erti
es a
bout
f
G
ebru
ik jo
u gr
afie
k om
die
vol
gend
e ei
ensk
appe
van
f te
be
paal
:
a)
y-
inte
rcep
t / y
-afs
nitt
e
b)
x-
inte
rcep
ts /
x-af
snit
te
c)
m
inim
um v
alue
/ m
inim
um w
aard
e
d)
do
mai
n / d
efin
isie
ver
sam
elin
g
e)
ra
nge
/ waa
rdev
ersa
mel
ing
f)
li
ne(s
) of
sym
met
ry /
lyne
van
sim
met
rie
4.
Dra
w th
e gr
aph
of /
Ske
ts d
ie g
rafi
ek v
an:
f (x)
= 3
x
a)
U
se a
tran
sfor
mat
ion
to d
raw
the
grap
h of
y =
3–x
on
the
sam
e se
t of
axes
.
G
ebru
ik ’
n tr
ansf
orm
asie
om
die
gra
fiek
van
y =
3–x
op
die
self
de a
sses
tels
el te
teke
n.
b)
G
ive
the
equa
tion
s of
the
line
of
s ym
met
ry b
etw
een
th
e tw
o gr
aphs
.
G
ee d
ie v
erge
lyki
ng v
an d
ie s
imm
etri
e-as
van
die
tw
ee g
rafi
eke.
5.
G
iven
/Geg
ee: g
( x)
= x
2 and
/en
f (x)
= 3
(x –
2)²
1
a)
G
raph
the
func
tion
of
g(x)
/ S
kets
die
gra
fiek
van
g(x
).
b)
D
escr
ibe
a se
ries
of
tran
sfor
mat
ions
nee
ded
to o
btai
n th
e
grap
h of
f (x
) =
3(x
– 2
)²
1 fr
om th
e gr
aph
of g
(x)
= x
2
Bes
kryf
’n
reek
s tr
ansf
orm
asie
s w
at g
ebru
ik k
an w
ord
om d
ie g
rafi
ek v
an f
(x)
= 3
(
x +
4)2 v
anaf
die
gra
fiek
g(
x) =
x2 te
ver
kry.
c)
G
raph
the
func
tion
f (x
) by
star
ting
wit
h th
e gr
aph
of g
(x)
by u
sing
tran
sfor
mat
ions
.
Ske
ts d
ie f
unks
ie f
(x)
deur
met
die
gra
fiek
van
g(x
) te
begi
n en
deu
r tr
ansf
orm
asie
s te
geb
ruik
.
d)
G
ive
the
dom
ain
of f
(x)
Gee
die
def
inis
ie v
ersa
mel
ing
van
f (x)
.
e)
G
ive
the
rang
e of
f (x
)
Gee
die
waa
rdev
ersa
mel
ing
van
f (x)
.
28
Wor
kshe
et 9
1.
The
mon
thly
ele
ctri
city
cos
t dep
ends
on
the
num
ber
of u
nits
us
ed d
urin
g th
e m
onth
. Ann
fou
nd th
at in
Apr
il h
er e
lect
rici
ty
bill
was
R46
0 fo
r 42
0 un
its
and
in M
ay th
e bi
ll w
as R
620
for
720
unit
s.
a)
E
xpre
ss th
e m
onth
ly c
ost C
in te
rms
of th
e nu
mbe
r of
uni
ts
used
x, a
ssum
ing
that
a li
near
rel
atio
nshi
p gi
ves
a su
itab
le
mod
el.
b)
U
se a
) to
pre
dict
the
cost
of
2000
uni
ts p
er m
onth
.
c)
D
raw
the
grap
h of
the
line
ar e
quat
ion.
d)
W
hat d
oes
the
slop
e of
the
line
rep
rese
nt?
e)
W
hat d
oes
the
y-in
terc
ept o
f th
e gr
aph
repr
esen
t?
2. D
raw
the
grap
hs o
f y
= x
2 + x
6
and
y =
2x
+ 1
and
use
the
grap
hs to
ans
wer
the
foll
owin
g qu
esti
ons:
If
f (x
) =
1
2
62
xx
x, t
hen
com
plet
e:
a)
F
or w
hich
val
ue(s
) of
x is
f (x
) =
0
b)
F
or w
hich
val
ue(s
) of
x is
f (x
)
0
c)
F
or w
hich
val
ue(s
) of
x is
f (x
)
0
d)
F
or w
hich
val
ue(s
) of
x d
oes
f (x)
not
exi
st?
3.
Giv
en: f
(x)
= x
2 x
6
and
g(x
) =
x +
2
a)
S
ketc
h th
e gr
aphs
of
g(x)
and
f (x
) on
the
sam
e se
t of
axes
.
b)
Use
you
r gr
aphs
to s
olve
the
equa
tion
x2
2x
8 =
0
1. D
ie m
aand
elik
se k
oste
van
ele
ktri
site
it h
ang
af v
an d
ie a
anta
l ee
nhed
e w
at g
ebru
ik w
ord
gedu
rend
e di
e m
aand
. Ann
se
kost
e in
Apr
il m
aand
was
R46
0 vi
r 42
0 ee
nhed
e w
at s
y ge
brui
k he
t en
in M
ei w
as d
it R
620
vir
720
eenh
ede
wat
sy
gebr
uik
het.
a)
D
ruk
die
maa
ndel
ikse
kos
te C
uit
in te
rme
van
die
aant
al
eenh
ede
x w
at g
ebru
ik is
, as
aanv
aar
wor
d da
t die
mod
el
deur
’n
line
êre
verw
ants
kap
voor
gest
el k
an w
ord.
b)
Geb
ruik
die
ant
woo
rd in
a)
en v
oors
pel d
ie k
oste
as
2000
ee
nhed
e pe
r m
aand
geb
ruik
wor
d.
c)
S
tel d
ie li
neêr
e ve
rgel
ykin
g gr
afie
s vo
or.
d)
W
at s
tel d
ie g
radi
ënt v
an d
ie ly
n vo
or?
e)
W
at s
tel d
ie y
-afs
nit v
an d
ie g
rafi
ek v
oor?
2.
Tek
en d
ie g
rafi
eke
van
y =
x2 +
x
6 e
n y
= 2
x +
1 e
n ge
brui
k hu
lle
om d
ie v
olge
nde
vrae
te b
eant
woo
rd:
A
s f (
x) =
1
2
62
xx
x v
olto
oi:
a)
V
ir w
atte
r w
aard
es v
an x
is f
(x)
= 0
b)
Vir
wat
ter
waa
rdes
van
x is
f (x
)
0
c)
Vir
wat
ter
waa
rdes
van
x is
f (x
)
0
d)
Vir
wat
ter
waa
rdes
van
x s
al f
(x)
onge
defi
niee
r w
ees?
3.
Geg
ee: g
( x)
= x
+ 2
& f
(x)
= x
2 x
6
a)
S
kets
die
gra
fiek
e va
n g(
x) e
n f (
x) o
p di
esel
fde
asse
stel
sel.
b)
G
ebru
ik jo
u gr
afie
ke o
m d
ie v
erge
lyki
ng x
2 2
x
8 =
0 o
p te
los.
29
Wor
kshe
et 1
0 1.
In
the
figu
re th
e gr
aphs
of
the
foll
owin
g fu
ncti
ons
are
show
n:
In
die
fig
uur
wor
d di
e gr
afie
ke v
an d
ie v
olge
nde
funk
sies
aa
nget
oon:
E is
the
turn
ing
poin
t of
f / E
is d
ie d
raai
punt
van
f
f (
x) =
x2
10x
+ 9
and
/en
g(x)
= 2
x
2
a)
Det
erm
ine
the
coor
dina
tes
of th
e fo
llow
ing
poin
ts:
i) A
ii
) B
and
/en
C
b)
D
eter
min
e th
e le
ngth
s of
/ B
epaa
l die
leng
tes
van:
i)
O
D
ii
) D
E
iii)
AF
iv
) H
I 2.
T
he f
igur
e sh
ows
the
grap
hs o
f tw
o tr
igon
omet
ric
func
tion
s,
f(x)
and
g(x)
. Det
erm
ine
the
equa
tion
of
f(x)
and
g(x)
.
Die
fig
uur
toon
gra
fiek
e va
n tw
ee tr
igon
omet
ries
e fu
nksi
es,
f(x
) en
g(x
). B
epaa
l die
ver
gely
king
s va
n f(x
) en
g(x
).
g(x)
f (x)
A B
C
D
D
E
F H I
x
y
y
–3
60
–180
90
180
36
0
4
– 4
f
g
x
30
Dif
fere
ntia
l Cal
culu
s / D
iffe
rens
iaal
reke
ning
e
Aver
age d
eclin
e / G
emid
deld
e hell
ing:
h
xf
hx
f)
()
(
(A
vera
ge sp
eed /
Gem
iddeld
e spo
ed)
The i
nclin
e in
a poi
nt / D
ie he
lling
in ‘n
pun
t: f '
(x)
=
hx
fh
xf
h
)(
)(
lim 0
Nota
tion
/ Not
asie:
f '(x
) or
Dx (*
*) o
r dxdy
or
*
*dxd
Rate
of c
hang
e / V
eran
derin
gste
mpo
: f '(x
)
Reme
mber
s(t)
d
istan
ce &
s'(
t)
spee
d &
s''
(t)
spee
d W
ordp
robl
ems /
Woo
rdso
mm
e:
Dr
aw a
sketc
h / T
eken
‘n sk
ets
Dr
aw up
an eq
uatio
n, ba
sed o
n the
prob
lem, e
g. f(x
). / S
tel ‘n
verg
elykin
g op,
geba
seer
op di
e pro
bleem
, bv.
f(x).
De
termi
ne / B
epaa
l f '(x
)
Put f
'(x)
= 0
, to de
liver
a ma
ximum
or m
inimu
m va
lue fo
r x / S
tel f
'(x)
= 0
om ‘n
mak
simum
of m
inimu
m wa
arde
vir x
te le
wer.
Su
bstitu
te x i
n f(x)
to ge
t the m
axim
um or
mini
mum
value
. / Ve
rvang
x in
f(x) o
m die
mak
simum
of m
inimu
m wa
arde
te kr
y Sk
etch
es o
f pol
ynom
ials o
f the
third
deg
ree /
Ske
tse v
an d
erde
graa
dse p
olin
ome:
Ge
nera
l equ
ation
/ Alge
mene
verg
elykin
g: y
= a
x3 + b
x2 + c
x +
d
X-
inter
cepts
/ X-a
fsnitte
: f(x)
= 0
Fa
ctoris
e thr
ough
grou
ping /
facto
r the
orem
Fakto
risee
r deu
r gro
eper
ing / f
aktor
stellin
g
Y-int
erce
pt / Y
-afsn
it: Le
t / St
el x =
0
Stati
onar
y poin
ts / S
tation
êre p
unte:
Put
/ Stel
f '(x
) =
0
o
Solve
for x
/ Los
op vi
r x.
o
Find f
(x) fo
r abo
veme
ntion
ed x-
value
s. o
Vi
nd f(
x) vir
boge
noem
de x-
waar
des.
31
Wor
kshe
et 1
1
1. G
iven
/ G
egee
: f(x
) =
3x
– 2x
2
Det
erm
ine
the
aver
age
rate
of
chan
ge o
f th
e fu
ncti
on b
etw
een
x =
2
and
x =
3.
B
epaa
l die
gem
idde
lde
vera
nder
ings
tem
po v
an d
ie f
unks
ie tu
ssen
x
=
2 en
x =
3.
2. G
iven
/ G
egee
: f(x
) =
1
2 xx
a)
D
eter
min
e th
e av
erag
e ra
te o
f ch
ange
of
the
func
tion
bet
wee
n x
=
1 an
d x
= 2
.
Bep
aal d
ie g
emid
deld
e ve
rand
erin
gste
mpo
van
die
fun
ksie
tu
ssen
x =
1
en x
= 2
.
b)
Det
erm
ine
the
aver
age
rate
of
chan
ge o
f th
e fu
ncti
on b
etw
een
x =
a a
nd x
= a
+ h
.
B
epaa
l die
gem
idde
lde
vera
nder
ings
tem
po v
an d
ie f
unks
ie
tuss
en x
= a
en
x =
a +
h.
3. I
f f (
x) =
4x
2 , det
erm
ine
f '(x
) fr
om f
irst
pri
ncip
les.
As
f (x)
=
4x2 , b
epaa
l f '(
x) v
anaf
eer
ste
begi
nsel
s.
4. D
eter
min
e f '
(x)
/ Bep
aal f
'(x)
:
a)
f (x)
= (
x
1) (
x2 + 2
)
b)
f (
x) =
3
42
2
x
x
c)
f
(x)
= 3
x4 +
x
d)
f (x)
= 3
x3 + 2
x2 4
x +
6
e)
f (
x) =
x +
(x
+ 3
)2
5. a
) F
ind
the
equa
tion
s of
the
tang
ents
to th
e pa
rabo
la
y =
x2
2x
3 at
the
poin
ts w
here
x =
0 a
nd x
= 2
re
spec
tive
ly.
Bep
aal d
ie v
erge
lyki
ngs
van
die
raak
lyne
aan
die
par
aboo
l
y
= x
2 2
x
3 by
die
pun
te w
aar
x =
0 e
n x
= 2
res
pekt
ief
is.
b)
F
ind
the
poin
t of
inte
rsec
tion
of
thes
e tw
o ta
ngen
ts.
Bep
aal d
ie s
nypu
nt v
an h
ierd
ie tw
ee r
aakl
yne.
6.
Ske
tch
the
foll
owin
g gr
aphs
/ S
kets
die
vol
gend
e gr
afie
ke:
a)
f (
x) =
x3
4x2
11x
+ 3
0
b)
f (x)
= x
3 7
x2 + 1
5x
9
c)
f (
x) =
x3
x2
8x
+ 1
2
7. A
col
d dr
ink
com
pany
wan
ts to
opt
imis
e it
s m
anuf
actu
ring
cos
t. T
hey
wan
t to
use
a cl
osed
cyl
indr
ical
can
that
hol
ds 3
50 m
l of
cold
dri
nk to
sel
l the
ir p
rodu
cts.
Cal
cula
te th
e he
ight
and
rad
ius
that
wil
l min
imis
e th
e am
ount
of
mat
eria
l nee
ded
to m
anuf
actu
re
a ca
n.
’n K
oeld
rank
maa
tska
ppy
wil
die
ver
vaar
digi
ngsk
oste
opt
imee
r.
Hul
le w
il ’
n ge
slot
e si
lind
ries
e ho
uer
gebr
uik
wat
350
ml b
evat
om
die
pro
duk
te v
erko
op. W
at m
oet d
ie r
adiu
s en
die
hoo
gte
van
die
blik
kie
wee
s om
die
hoe
veel
heid
mat
eria
al w
at b
enod
ig w
ord
om d
ie b
likk
ie te
ver
vaar
dig
te o
ptim
eer?
32
Wor
kshe
et 1
2
1. A
bio
logi
st s
tate
s th
at w
hen
a sp
ecif
ic k
ind
of a
nti-
bact
eriu
m is
ad
ded
to a
cul
ture
of
bact
eria
, the
num
ber
of b
acte
ria
pres
ent w
ill
be g
iven
by
the
form
ula
B(t)
=
5t2 +
60
t + 1
200
whe
re B
(t)
repr
esen
ts in
mil
lion
s th
e nu
mbe
r of
bac
teri
a on
mom
ent t
.
’n
Bio
loog
bew
eer
dat w
anne
er ‘
n se
kere
soo
rt a
ntib
akte
rie
tot ’
n ku
ltuu
r va
n ba
kter
ieë
gevo
eg w
ord,
die
get
al b
akte
rieë
te
enw
oord
ig g
egee
sal
wor
d de
ur d
ie f
orm
ule
B(
t) =
5t
2 + 6
0 t +
120
0 w
aar
B(t)
in m
iljo
ene
die
geta
l bak
teri
eë
op ty
dsti
p t,
gem
eet i
n uu
r, v
oors
tel.
a) H
ow m
any
bact
eria
wer
e pr
esen
t at t
he b
egin
ning
? / H
oeve
el
bakt
erie
ë w
as a
an d
ie b
egin
teen
woo
rdig
?
b) C
alcu
late
the
tem
po o
f ch
ange
in r
elat
ion
to th
e ti
me
on m
omen
t t =
10
hour
s. /
Ber
eken
die
tem
po v
an v
eran
deri
ng m
et b
etre
kkin
g to
t die
tyd
op ty
dsti
p t =
10
uur.
c) W
as th
e ba
cter
ia p
opul
atio
n sh
rink
ing
or g
row
ing
on
mom
ent t
= 1
0 ho
urs?
/ W
as d
ie b
akte
rieë
bev
olki
ng a
an
die
afne
em o
f to
enee
m o
p ty
dsti
p t =
10
uur?
d) A
t whi
ch m
omen
t was
the
max
imum
num
ber
of b
acte
ria
pres
ent?
/ O
p w
atte
r ty
dsti
p w
as d
ie m
aksi
mum
get
al
bakt
erie
ë te
enw
oord
ig?
e) A
fter
how
man
y ho
urs
did
the
num
ber
of b
acte
ria
star
t de
crea
sing
? / N
a ho
evee
l uur
het
die
get
al b
akte
rieë
be
gin
afne
em?
2. T
he h
umid
ity
(H)
of a
ir in
rel
atio
n to
the
tem
pera
ture
(t i
n ºC
) is
gi
ven
by th
e fo
rmul
a H
=
t3 + 2
4t2
84t
+ 8
0
D
ie h
umid
itei
t (H
) va
n lu
g m
et b
etre
kkin
g to
t die
tem
pera
tuur
(t
in º
C)
wor
d ge
gee
deur
die
for
mul
e H
=
t3 + 2
4t2
84t
+ 8
0
a) D
eter
min
e th
e hu
mid
ity
at te
mpe
ratu
re 0
ºC
/ B
epaa
l die
hu
mid
itei
t by
tem
pera
tuur
0 º
C.
b) A
t whi
ch te
mpe
ratu
re is
the
hum
idit
y 0?
/ B
y w
atte
r te
mpe
ratu
ur
is d
ie h
umid
itei
t 0?
c) A
t whi
ch te
mpe
ratu
re d
oes
the
hum
idit
y re
ach
a m
axim
um in
the
inte
rval
[0º
; 20
º]?
/ By
wat
ter
tem
pera
tuur
ber
eik
die
hum
idit
eit
‘n m
aksi
mum
in d
ie in
terv
al [
0º ;
20º]
?
d) S
ketc
h th
e gr
aph
of H
wit
h re
lati
on to
t in
the
inte
rval
[0º
; 20º
]. /
Ske
ts d
ie g
rafi
ek v
an H
met
bet
rekk
ing
tot t
in d
ie in
terv
al
[0
º; 2
0º].
3.
A r
ecta
ngul
ar c
amp
PQ
RS
that
mus
t be
fenc
ed in
by
72 c
m o
f w
ire
nett
ing.
PQ
is a
n ex
isti
ng w
all.
PS, S
R a
nd Q
R m
ust b
e fe
nced
in. C
alcu
late
the
max
imum
are
a of
the
cam
p.
Die
mee
gaan
de f
iguu
r to
on ’
n re
ghoe
kige
kam
p P
QR
S w
at
omhe
in m
oet w
ord
deur
72
cm o
gies
draa
d. P
Q is
‘n
best
aand
e m
uur.
PS
, SR
en
QR
moe
t toe
geka
mp
wor
d. B
erek
en d
ie
mak
sim
um o
pper
vlak
te v
an d
ie k
amp.
QR
PS
33
Line
ar P
rogr
amm
ing
/ Lin
eêre
Pro
gram
mer
ing
Step
1 / S
tap
1: D
eterm
ine w
hat y
ou ar
e goin
g to n
ame x
and y
/ Bep
aal w
at jy
x en y
gaan
noem
St
ep 2
/ Sta
p 2
Identi
fy the
cons
traint
s and
high
light
them.
(Sea
rch fo
r wor
ds lik
e “at
most”
, “at
least”
, “no
t mor
e tha
n” or
“not
less t
han”
). Ide
ntifis
eer d
ie be
perki
ngs e
n “hig
hligh
t” dit
. (So
ek vi
r woo
rde s
oos
“hoo
gsten
s”, “m
insten
s”, “n
ie me
er as
” of “
nie m
inder
as”).
St
ep 3
/ Sta
p 3
Comp
ile th
e con
strain
ts (a
ll the
ineq
ualiti
es to
gethe
r) alg
ebra
ically
. Rem
embe
r the
impli
cit co
nstra
ints (
cond
itions
that
go w
ithou
t say
ing).
Stel
die be
perki
ngs (
al die
onge
lykhe
de sa
am) a
lgebr
aïes v
oor.
Ontho
u die
impli
siete
bepe
rking
s (vo
orwa
arde
s wat
vans
elfsp
reke
nd is
). St
ep 4
/ Sta
p 4
Repr
esen
t infor
matio
n gra
phica
lly an
d dete
rmine
the f
easib
le re
gion (
the re
gion t
hat s
atisfi
es al
l the i
nequ
alitie
s St
el inl
igting
grafi
es vo
or en
bepa
al die
gang
bare
gebie
d (die
gebie
d wat
al die
onge
lykhe
de be
vredig
). St
ep 5
/ Sta
p 5:
De
termi
ne th
e obje
ctive
func
tion /
Bep
aal d
ie do
elfun
ksie.
St
ep 6
/ Sta
p 6
Sear
ch lin
e meth
od: G
radie
nt of
objec
tive f
uncti
on
Soek
lyn-m
etode
: Hell
ing va
n doe
lfunk
sie
Name
the v
ertic
es of
the f
easib
le re
gion a
nd th
e coo
rdina
tes.
Deter
mine
the v
alue o
f the o
bjecti
ve fu
nctio
n at e
ach v
ertex
. Be
noem
die h
oekp
unte
van t
oelaa
tbare
gebie
d en d
ie ko
ördin
ate.
Bepa
al die
waa
rde v
an di
e doe
lfunk
sie by
elke
hoek
punt.
St
ep 7
/ Ste
p 7:
Ans
wer t
he qu
estio
n / B
eantw
oord
die v
raag
.
y is
not
mor
e th
an 2
0 / y
in n
ie m
eer
as 2
0 ni
e
y
20
x is
at m
ost 1
2 / x
is h
oogs
tens
12
x
12
y is
at l
east
200
/ y
is m
inst
ens
200
y
200
x is
not
less
than
12
/ x is
nie
min
der
as 1
2
x
12
twic
e x
is a
t mos
t thr
ee ti
mes
y /
twee
kee
r x
is h
oogs
tens
dri
e ke
er y
2x
3y
x
is a
t lea
st tw
ice
y / x
is te
n m
inst
e 2
keer
y
x
2y
y is
at m
ost 3
tim
es x
/ y
is h
oogs
tens
3 k
eer
x
y
3x
x , y
R
- s
hade
reg
ion
/ kle
ur g
ebie
d in
x
, y
Z -
use
dot
s / m
aak
koll
e y
> …
. S
hade
abo
ve th
e li
ne
ars
eer
bo d
ie ly
n y
< …
.. S
hade
bel
ow th
e li
ne
ars
eer
onde
r di
e ly
n
or
: sol
id li
ne/s
olie
de ly
n
or >
: dot
ted
line
/sti
ppel
lyn
34
Wor
kshe
et 1
3
1.
In th
e ac
com
pany
ing
sket
ch th
ere
is a
set
of
ineq
uali
ties
that
lead
s to
the
feas
ible
reg
ion
ABC
D
as s
how
n by
the
shad
ed a
rea.
Use
the
grap
h to
ans
wer
the
foll
owin
g qu
esti
ons:
a)
Wri
te d
own
the
set o
f in
equa
liti
es th
at d
escr
ibe
the
feas
ible
reg
ion
b)
M
axim
ise
2 x +
3y
for
the
give
n fe
asib
le
regi
on.
c)
T
he c
oord
inat
es o
f po
int C
min
imis
e th
e
func
tion
val
ue o
f c
whe
re y
= m
x +
c. W
rite
do
wn
the
poss
ible
val
ues
of m
and
c.
2.
A f
arm
er p
lans
to p
lant
mai
ze a
nd s
unfl
ower
s on
his
far
m. T
he
land
ava
ilab
le f
or th
e m
aize
is x
hec
tare
s an
d fo
r th
e su
nflo
wer
is
y h
ecta
res.
The
pla
ntin
g is
sub
ject
to th
e fo
llow
ing
rest
rict
ions
:
2
x
8
0
y
7
5x
+ 7
y
70
8x +
5y
80
a)
U
se g
raph
pap
er a
nd d
raw
on
one
set o
f ax
es th
e gr
aphs
of
all t
he r
estr
icti
ons.
Sha
de th
e fe
asib
le r
egio
n cl
earl
y,
indi
cati
ng th
e co
ordi
nate
s of
all
the
vert
ices
on
the
grap
h.
b)
If
the
prof
it P
is g
iven
by
P =
6x
+ 5
y, u
se th
e gr
aph
to
dete
rmin
e ho
w m
an h
ecta
res
of e
ach
crop
mus
t be
plan
ted
for
a m
axim
um p
rofi
t. In
dica
te o
n yo
ur g
raph
whe
re th
is
valu
e m
ay b
e.
1.
In d
ie m
eega
ande
ske
ts w
ord
die
gang
bare
geb
ied
ABC
D d
eur
’n a
anta
l ong
elyk
hede
voo
rges
tel.
Geb
ruik
die
gra
fiek
om
die
vol
gend
e vr
ae te
be
antw
oord
:
a)
Skr
yf d
ie o
ngel
ykhe
de n
eer
wat
die
gan
gbar
e
gebi
ed b
eskr
yf.
b)
Mak
sim
eer
2 x +
3y
vir
die
gege
we
gang
bare
ge
bied
.
c) B
epaa
l die
moo
ntli
ke w
aard
es v
an m
en
c
waa
rvoo
r y
= m
x +
c ’
n m
inim
umw
aard
e by
C
sal h
ê.
2.
‘n B
oer
bepl
an o
m m
ieli
es e
n so
nneb
lom
me
te p
lant
op
sy p
laas
. D
aar
is x
hek
taar
land
e be
skik
baar
vir
mie
lies
en
y he
ktaa
r vi
r so
nneb
lom
me.
Die
aan
plan
t van
die
gew
asse
is o
nder
hew
ig a
an
die
volg
ende
bep
erki
ngs:
2
x
8
0
y
7
5x
+ 7
y
70
8x +
5y
80
a)
G
ebru
ik g
rafi
ekpa
pier
en
maa
k op
een
ass
este
lsel
‘n
sket
s va
n al
hie
rdie
bep
erki
ngs.
Ars
eer
die
gang
bare
geb
ied
duid
elik
en
dui a
l die
hoe
kpun
te s
e ko
ördi
nate
aan
op
die
graf
iek.
b)
As
die
prof
yt P
geg
ee w
ord
deur
P =
6x
+ 5
y, g
ebru
ik d
ie
graf
iek
om v
as te
ste
l hoe
veel
hek
taar
van
elk
e ge
was
ge
plan
t moe
t wor
d vi
r ’n
mak
sim
um w
ins.
Dui
aan
op
die
graf
iek
waa
r di
e af
lees
punt
is.
35
Pap
er 2
/ Vra
este
l 2
Coo
rdin
ate
geom
etry
/ A
nalit
iese
mee
tkun
de
Tran
sfor
mat
ions
/ Tr
ansf
orm
asie
s
Trig
onom
etry
/ Tr
igon
omet
rie
Dat
a ha
ndlin
g / D
ata
hant
erin
g
36
Vol
ume
and
surf
ace
area
/ V
olum
e en
bui
teop
perv
lakt
e
Cyl
inde
r
Vo
lume o
f a cy
linde
r =
r2 h
Su
rface
area
(clos
ed cy
linde
r) =
2r
2 + 2r
h P
rism
Vo
lume o
f pris
m =
(are
a of b
ase)
he
ight
Su
rface
area
= A
+ p
h =
ab
+ h
(a +
b)
Con
e Vo
lume o
f con
e =
hr2
31
Su
rface
area
of cl
osed
cone
=
r2 +
rH
Pyr
amid
Vo
lume o
f pyra
mid
=
Ah31
Su
rface
area
of p
yrami
d =
A +
21pH
Sph
ere
Vo
lume o
f sph
ere =
3
34r
Su
rface
area
of sp
here
=
24
r
37
Wor
kshe
et 1
4
1. C
alcu
late
the
num
ber
of d
iago
nals
in a
pol
ygon
wit
h 12
sid
es /
Bep
aal h
oeve
el d
iago
nale
’n
veel
hoek
met
12
sye
het.
2. C
ompl
ete
the
foll
owin
g ta
ble
/ Vol
tooi
die
vol
gend
e ta
bel:
Num
ber
of s
ides
of
a po
lygo
n / A
anta
l sye
van
vee
lhoe
k 3
4 5
100
n
Sum
of
inte
rior
ang
les
/ Som
van
die
bin
neho
eke
180
36
0
In th
e ca
se o
f a
regu
lar
poly
gon
find
the
mea
sure
of
each
ang
le.
In d
ie g
eval
van
’n
reël
mat
ige
veel
hoek
bep
aal d
ie g
root
te v
an e
lke
hoek
60
90
3. D
eter
min
e th
e m
easu
re o
f ea
ch o
f th
e tw
elve
ang
les
of a
reg
ular
dod
ecag
on /
Bep
aal d
ie m
ates
van
die
bin
neho
eke
van
’n r
eëlm
atig
e tw
aalf
hoek
.
4. C
alcu
late
the
sur
face
are
a of
eac
h so
lid
corr
ect
to t
wo
deci
mal
pla
ces.
Len
gths
are
in
cm a
nd t
he h
eigh
t of
eac
h of
the
pri
sms
and
pyra
mid
s is
10
cm
.
Ber
eken
die
bui
te o
pper
vlak
te v
an e
lk v
an d
ie v
olge
nde
vast
e li
ggam
e ko
rrek
tot
tw
ee d
esim
ale
syfe
rs.
Len
gtes
is
in c
m e
n di
e ho
ogte
s va
n al
le p
rism
as e
n pi
ram
iede
s is
10c
m.
a)
b)
c)
d)
e)
3
10
10
8
6
88
6
4
84
Squa
re: 33
38
Ana
lyti
cal g
eom
etry
/ A
nalit
iese
mee
tkun
de
For
mu
lae
/ For
mu
les
Dista
nce b
etwee
n poin
ts A
and B
/ Afst
and t
usse
n twe
e pun
te A
en B
: 2
2)
()
(B
AB
Ay
yx
xAB
Mi
dpoin
t M(x M
; yM)
of lin
e seg
ment
AB / M
iddelp
unt M
(xM
; y M
) of A
B:
2B
AM
xx
x
an
d 2
BA
My
yy
Grad
ient o
f a lin
e / H
elling
(gra
diënt)
: B
A
BA
ABx
xy
ym
Angle
of in
clina
tion /
Inkli
nasie
hoek
: tan
=
m
Equa
tion o
f a st
raigh
t line
/ Ver
gelyk
ing va
n ‘n r
eguit
lyn:
y =
mx
+ c
o
f/or
y
y
A =
m(x
x
A)
Ci
rcle w
ith ce
ntre (
a; b
) and
radiu
s r /
Sirke
l met
midd
elpun
t (a;
b) &
radiu
s r:
(x
a)2 +
(y
b)2 =
r 2
39
Wor
kshe
et 1
5
1. A
(5
; 6)
, B (
t ;
2) a
nd C
(7
; 2)
are
thre
e po
ints
in a
Car
tesi
an
plan
e. C
alcu
late
a)
th
e co
-ord
inat
es o
f th
e m
idpo
int o
f A
C
b)
th
e va
lue(
s) o
f t i
f A
B =
BC
2. P
(6
; 2)
and
Q (
q ; 6
) ar
e po
ints
on
the
circ
umfe
renc
e of
a c
ircl
e w
ith
the
orig
in a
s ce
ntre
. Det
erm
ine
a)
th
e eq
uati
on o
f th
e ci
rcle
b)
th
e va
lue(
s) o
f q
c)
th
e si
ze o
f th
e an
gle
betw
een
the
line
PQ
and
the
posi
tive
x-
axis
if q
= 2
.
3. B
(3; 2
) an
d C
(3;
6) a
re p
oint
s on
the
circ
umfe
renc
e of
a c
ircl
e w
ith
cent
re Q
and
wit
h eq
uati
on x
2 + y
2 2
x +
4y
15 =
0.
a)
D
eter
min
e th
e co
ordi
nate
s of
the
cent
re o
f th
e ci
rcle
and
the
leng
th o
f th
e ra
dius
.
b)
D
eter
min
e th
e eq
uati
on o
f th
e ta
ngen
t to
the
circ
le a
t C.
4. A
(4
; 5),
B (6
; 6)
and
C (
3;
2) a
re th
e ve
rtic
es o
f A
BC
.
Cal
cula
te th
e m
easu
re o
f an
gle
A
CB
5. A
cir
cle
wit
h ce
ntre
M(4
; 2)
pas
ses
thro
ugh
the
poin
t A(8
; 2
).
Det
erm
ine
a)
th
e eq
uati
on o
f th
e ci
rcle
in th
e fo
rm x
2 + a
x +
y2 +
by
+ c
= 0
b)
th
e co
-ord
inat
es o
f th
e y-
inte
rcep
ts o
f th
e ci
rcle
6. W
hat i
s th
e ra
dius
of
the
circ
le w
ith
equa
tion
9x2 =
4
9y2 ?
1. A
(5
; 6)
, B (
t ;
2) e
n C
(7
; 2)
is d
rie
punt
e in
'n C
arte
sies
e vl
ak. B
erek
en
a)
di
e ko
ördi
nate
van
die
mid
delp
unt v
an A
C
b)
di
e w
aard
e(s)
van
t as
AB
= B
C
2. P
(6
; 2)
en
Q (
q ; 6
) is
pun
te o
p di
e om
trek
van
'n s
irke
l met
die
oo
rspr
ong
as m
idde
lpun
t. B
epaa
l
a)
di
e ve
rgel
ykin
g va
n di
e si
rkel
b)
di
e w
aard
e(s)
van
q
c)
di
e gr
oott
e va
n di
e ho
ek w
at d
ie ly
n P
Q m
et d
ie p
osit
iew
e x-
as m
aak
as q
= 2
.
3. B
(3; 2
) en
C(3
; 6)
is p
unte
op
die
omtr
ek v
an ’
n si
rkel
met
m
idde
lpun
t Q e
n m
et v
erge
lyki
ng x
2 + y
2 2
x +
4y
15 =
0.
a)
B
epaa
l die
koö
rdin
ate
van
die
mid
delp
unt v
an d
ie s
irke
l en
die
radi
us v
an d
ie s
irke
l.
b)
B
epaa
l die
ver
gely
king
van
die
raa
klyn
aan
die
sir
kel b
y C
.
4. A
(4
; 5),
B (6
; 6)
en
C (
3;
2) is
die
hoe
kpun
te v
an
AB
C
.
Ber
eken
die
gro
otte
van
hoe
k A
C
B.
5. '
n S
irke
l met
mid
delp
unt M
(4 ;
2) g
aan
deur
die
pun
t A(8
; 2
).
Bep
aal
a)
die
ver
gely
king
van
die
sir
kel i
n vo
rm x
2 + a
x +
y2 +
by
+ c
= 0
b)
die
koö
rdin
ate
van
die
y-af
snit
te v
an d
ie s
irke
l .
6. W
at is
rad
ius
van
die
sirk
el m
et v
erge
lyki
ng
9x2 =
4
9y2 ?
40
Wor
kshe
et 1
6
1. D
eter
min
e th
e co
-ord
inat
es o
f th
e ce
ntre
and
the
radi
us o
f th
e ci
rcle
wit
h eq
uati
on x
2 1
0x +
y2 +
y =
0
2. A
B a
nd D
E a
re tw
o pa
rall
el li
nes.
The
equ
atio
n of
AB
is
3 y
2x
= 1
5 an
d th
e co
-ord
inat
es D
and
E a
re D
(0 ; 1
) an
d
E( e
; 3
) re
spec
tive
ly. D
eter
min
e
a) t
he g
radi
ent o
f th
e st
raig
ht li
ne A
B
b) t
he v
alue
(s)
of e
c) t
he e
quat
ion
of th
e st
raig
ht li
ne th
roug
h th
e po
int C
(3
; 8)
whi
ch
is p
erpe
ndic
ular
to A
B, i
n th
e fo
rm a
x +
by
+ c
= 0
. 3.
K(–
4; 2
), L
(2;
2) a
nd M
(3;
3)
are
ve
rtic
es o
f K
LM
in a
Car
tesi
an p
lane
.
a) C
alcu
late
the
peri
met
er o
f K
LM
.
b) C
alcu
late
the
incl
inat
ion
angl
e of
LM
.
c) D
eter
min
e th
e co
-ord
inat
es th
e m
idpo
int o
f se
gmen
t KM
. d)
Det
erm
ine
the
equa
tion
of
the
perp
endi
cula
r bi
sect
or o
f K
M.
1. B
epaa
l die
koö
rdin
ate
van
die
mid
delp
unt e
n ra
dius
van
die
sir
kel
met
ver
gely
king
x2
10x
+ y
2 + y
= 0
2.
AB
en
DE
is tw
ee e
wew
ydig
e ly
ne. D
ie v
erge
lyki
ng v
an A
B is
3 y
2
x =
15
en d
ie k
oörd
inat
e va
n D
en
E is
ond
ersk
eide
lik
D
(0
; 1)
en
E (
e ;
3). B
epaa
l
a) d
ie g
radi
ënt v
an d
ie r
egui
tlyn
AB
b) d
ie w
aard
e(s)
van
e
c) d
ie v
erge
lyki
ng v
an d
ie r
egui
tlyn
deu
r di
e pu
nt C
(3
; 8)
wat
lo
odre
g is
op
AB
, in
die
vorm
ax
+ b
y +
c =
0.
3. K
(–4;
2),
L(2
; 2)
en
M(
3;
3) is
ho
ekpu
nte
van K
LM
op ʼn
Kar
tesi
ese
vlak
.
a) B
erek
en d
ie o
mtr
ek v
an
KL
M
b) B
erek
en d
ie in
klin
asie
hoek
van
LM
c) B
epaa
l die
koö
rdin
ate
van
die
mid
delp
unt
van
lyns
tuk
KM
.
d) B
epaa
l die
ver
gely
king
van
die
lood
regt
e ve
rdel
er v
an K
M.
41
Def
init
ions
si
n
=
ry =
st
cos
= rx
= sa
ta
n
= xy
=
at
Sim
plif
ying
wit
h re
duci
ng f
orm
ulae
: St
ep 1
Make
nega
tive
s pos
itive
the
n
cos x
is po
sitive
, the r
est is
nega
tive
/ Maa
k neg
atiew
e e p
ositie
f dan
co
s x is
posit
ief, e
n die
res n
egati
ef
e.g.
sin
(30
0) =
si
n 300
&
co
s (2
50)
= +
cos 2
50
Step
2 Ma
ke an
gle gr
eater
than
360
/ sma
ller t
han 3
60 b
y sub
tracti
ng an
y mult
iple o
f 360,
then n
othing
chan
ges.
/ Maa
k hoe
ke gr
oter a
s 360 k
leine
r as
360
deur
veelv
oude
van 3
60 a
f te tr
ek, m
aar g
een t
eken
s ver
ande
r nie.
e.g. c
os 10
00 =
cos 2
80
Step
3 Ma
ke al
l ’s
acute
angle
s (<
90).
Disc
ard m
ultipl
es of
180
or 36
0’
and d
eterm
ine th
e sign
by m
eans
of C
AST
table.
Maa
k alle
e s
kerp
hoek
e (<
90),
‘goo
i 180
of 36
0
weg’
en be
paal
die te
ken m
bv di
e CAS
T-tab
el:
e.g
. co
s 150 =
cos (
180
30
) = c
os 30
Step
4 Fo
r fur
ther s
impli
ficati
on or
if the
re is
(90 )
, then
use c
ofunc
tions
/ Vir v
erde
re ve
reen
voud
ig of
as da
ar ‘n
(90 )
, geb
ruik
ko-fu
nksie
s P
roof
of
iden
titi
es (
LHS
& R
HS
indi
vidu
ally
) / B
ewys
van
iden
tite
ite
(LK
& R
K a
part
) 1)
Cho
ose m
ost d
ifficu
lt side
/ Kies
moe
ilikste
kant
2) L
ook f
or sq
uare
iden
tities
/ Kyk
vir v
ierka
nt ide
ntitei
te (K
yk vi
r ‘1’ e
n kwa
drate
) 3)
Eve
rythin
g to s
in x a
nd co
s x /
Alles
na si
n x &
cos x
4)
Fra
ction
s: G
et LC
M / B
reuk
e: Kr
y KGV
5)
Fac
torise
or si
mplify
/ Fak
torise
er of
vere
envo
udig
6) E
merg
ency
plan
: If c
osx
1, the
n mult
iply a
bove
and b
elow
by (c
osx +
1) / N
oodp
lan: In
dien (
cosx
1)
maa
l dan
bo en
onde
r: (co
sx +
1)
7) S
ee fo
r whic
h valu
es of
x the
expr
essio
n is n
ot de
fined
/ Kyk
vir w
atter
waa
rdes
van x
is di
e uitd
rukk
ing ni
e ged
efinie
er ni
e Id
enti
ties
/ Id
enti
teit
e
tan x
=
xxco
s
sin
s
in2 x +
cos2 x
= 1
Co
func
tion
sin
(90
x
) =
cos x
cos (
90
+ x
) =
si
n x
Tri
gono
met
ry /
Tri
gono
met
rie
42
Com
poun
d &
dou
ble
angl
es /
Saa
mge
stel
de- e
n du
bbel
hoek
e
Com
poun
d an
gles
/ S
aam
gest
elde
hoe
ke
co
s (
+
) =
cos
.c
os
s
in
.sin
co
s (
–
) =
cos
.c
os
+ s
in
.sin
si
n (
+
) =
sin
.c
os
+ c
os
.sin
sin
( –
)
= s
in
.cos
–
cos
.s
in
Dou
ble
angl
es /
Dub
belh
oeke
si
n 2A
= 2
sinA
.cos
A
cos
2A =
cos
2A
– s
in 2A
cos
2A =
2co
s 2A
– 1
co
s 2A
= 1
– 2
sin 2
A
43
Wor
kshe
et 1
7
1. G
iven
/Geg
ee:
f (x
) =
sin
2x
and/
en g
(x)
= s
in (
45
+ x
)
a)
Ske
tch
the
grap
hs o
f f
and
g on
the
sam
e se
t of
axes
for
]
180
;18
0[
x
. Cle
arly
sho
w th
e co
-ord
inat
es o
f th
e in
terc
epts
wit
h th
e ax
es a
nd o
f al
l tur
ning
poi
nts.
Ske
ts d
ie g
rafi
eke
van
f en
g o
p di
esel
fde
asse
stel
sel v
ir
]18
0;
180
[
x.
Dui
die
koö
rdin
ate
van
die
afsn
itte
op
die
asse
en
die
draa
ipun
te v
an d
ie g
rafi
eke
duid
elik
aan
.
b)
Wri
te d
own
the
peri
od o
f f a
nd g
S
kryf
die
per
iode
nee
r va
n f e
n g
2.
a)
S
ketc
h th
e gr
aphs
of
f, g
end
h f
or th
e in
terv
al [1
80º;
180
º],
whe
re: /
Ske
ts d
ie g
rafi
eke
van
f, g
en h
vir
die
inte
rval
[
180º
; 18
0º ]
, waa
r: f
(x)
= 2
cos
x ;
g(x)
= 2
+ c
os x
;
h(x)
= c
os 2
x
b)
F
or w
hich
val
ue(s
) of
x w
ill c
os 2
x =
0?
Vir
wat
ter
waa
rde(
s) v
an x
sal
cos
2x
= 0
?
c)
Use
you
r gr
aphs
to d
eter
min
e th
e gr
eate
st v
alue
of
the
expr
essi
on 2
c
os x
. Ind
icat
e cl
earl
y w
here
you
hav
e re
ad
your
ans
wer
, aga
in u
sing
cap
ital
lett
ers.
Geb
ruik
u g
rafi
ek o
m d
ie g
root
ste
waa
rde
van
die
uitd
rukk
ing
2
cos
x te
bep
aal.
Too
n du
idel
ik a
an w
aar
u di
e an
twoo
rd a
fgel
ees
het.
Geb
ruik
wee
r ee
ns h
oofl
ette
rs
daar
voor
.
3.a)
S
ketc
h, o
n th
e sa
me
set o
f ax
es, t
he g
raph
s of
/ S
kets
op
dies
elfd
e as
sest
else
l, di
e gr
afie
ke v
an:
f = {
(x; y
) / y
= ½
tan
x;
180º
x
1
80º}
and
/en
g
= {
(x; y
) / y
= ta
n ½
x ; 1
80º
x
180º
}
b)
Wha
t is
the
peri
od o
f f ?
/ W
at is
die
per
iode
van
f ?
c)
F
or w
hich
val
ue(s
) of
x w
ill f
(x)
= g
(x)?
V
ir w
atte
r w
aard
e(s)
van
x s
al f
(x)
= g
(x)?
d)
For
whi
ch v
alue
(s)
of x
wil
l f (x
) >
g(x
)?
Vir
wat
ter
waa
rde(
s) v
an x
sal
f (x
) >
g(x
)?
4. I
n th
e fi
gure
, the
gra
phs
of tw
o fu
ncti
ons
/ In
die
figu
ur w
ord
die
graf
ieke
van
die
twee
fun
ksie
s:
f =
{(x
; y)
/ y =
a c
os x
+ k
; 18
0º
x
180
º} a
nd/e
n
g
= {
(x; y
) / y
= s
in b
x +
m;
180º
x
1
80º}
D
eter
min
e th
e va
lues
of
a, b
, k, a
nd m
. / B
epaa
l die
waa
rdes
van
a,
b, k
en
m.
180
180
90
90
1 y 1 2 3
x
44
Wor
kshe
et 1
8
1. S
impl
ify,
wit
hout
usi
ng a
cal
cula
tor,
the
foll
owin
g ex
pres
sion
s:
(Sho
w A
LL
the
calc
ulat
ions
.)
V
eree
nvou
dig,
son
der
die
gebr
uik
van
'n s
akre
kena
ar, d
ie
volg
ende
uit
druk
king
s (T
oon
AL
die
ber
eken
inge
.):
a)
)
90co
s(
)18
0co
s().
360
tan(
xx
x
b)
)
60co
s(.
570
sin
135
tan
.33
0co
s
c)
)
300
cos(
90si
n4
225
tan
150
cos
.24
0ta
n
d)
13
5co
s18
0co
s24
0si
n.
30ta
n2
e)
)
120
tan(
150
tan
).36
0(
cos
)90
sin(
)36
0si
n().
90co
s(3
2
x
xx
f)
xx
xco
s1
1
cos
1
1si
n2
2. D
eter
min
e, w
itho
ut u
sing
a c
alcu
lato
r, th
e va
lue
of th
e fo
llow
ing
in te
rms
of t,
if s
in 1
7° =
t / B
epaa
l, so
nder
die
geb
ruik
van
'n
sakr
eken
aar,
die
waa
rde
van
die
volg
ende
in te
rme
van
t ind
ien
sin
17°
= t:
a) s
in 5
23
b)
cos
73°
c) t
an(–
17°)
3. I
f x
[0
; 180]
sol
ve th
e fo
llow
ing
equa
tion
s / A
s x
[0
; 180]
lo
s di
e vo
lgen
de v
erge
lyki
ngs
op:
a)
cos
A =
23
b)
sin
2 x
= 1
c
os2 x
c) c
os (
2 x
72
) =
sin
(x
+ 2
3)
4. S
olve
for
x if
/ L
os o
p vi
r x,
indi
en:
a)
5co
s3x
+ 4
= 0
and
/en
x
[0°;
360
°].
b)
42
8,0
)20
sin(
2
A
; ]
270
;90
[ˆ
A
5. I
f/A
s 4.
cos
x =
3
2 ,
2si
n y
+ 1
= 0
and
/en
x,y
[18
0; 3
60]
ca
lcul
ate
with
out t
he u
se o
f a
calc
ulat
or th
e va
lue
of /
bere
ken,
so
nder
die
geb
ruik
van
‘n
sakr
eken
aar
die
waa
rde
van
a) x
+ y
b)
tan
x
6. D
eter
min
e th
e ge
nera
l sol
utio
n of
/ B
epaa
l die
alg
emen
e op
loss
ing
van:
a) 2
.cos
x.s
in x
c
os x
= 0
b)
05
sin
2co
s7
2
7. P
rove
the
iden
tity
/ B
ewys
die
iden
tite
it:
xxxx
xxco
s
tan
4
sin
1
sin
1
sin
1
sin
1
8. I
f x
and
y ar
e co
mpl
emen
tary
ang
les
and
05
cos
13
x
, ca
lcul
ate,
with
out u
sing
a c
alcu
lato
r, th
e va
lue
of t
an x
+ c
os y
A
s x
en y
kom
plem
entê
re h
oeke
is e
n 0
5co
s13
x
, ber
eken
, so
nder
om
'n s
akre
kena
ar te
geb
ruik
die
waa
rde
van
tan
x +
cos
y
45
Sol
ving
tri
angl
es /
Opl
os v
an d
rieh
oeke
Sin
rul
e / r
eël
If an
d side
oppo
site e
ach o
ther a
re kn
own +
any o
ther a
ngle
or si
de (i.
e.
s or s
s)
Indien
en
sy re
goor
mek
aar b
eken
d is +
enige
ande
r hoe
k of s
y m.a.
w.
s of s
s)
If you
wan
t to ca
lculat
e / I
ndien
ge
vra w
ord:
cC
bBa
Asi
nsi
nsi
n
If you
wan
t to ca
lculat
e a si
de / I
ndien
sy ge
vra w
ord:
CcBb
Aa
sin
sin
sin
Cos
rul
e / r
eël
If no
and s
ide op
posit
e eac
h othe
r are
know
n (i.e
. sss
or s
s)
Indien
geen
en
sy re
goor
mek
aar n
ie be
kend
is ni
e (m.
a.w. s
ss of
ss)
If s
ide / I
ndien
sy ge
vra:
a2 = b
2 + c
2 – 2
bc.c
os A
If / A
s ge
vra w
ord:
bc
ac
bA
2co
s2
22
Are
a ru
le /
Opp
reë
l If 9
0
/ Ind
ien 90
in
drieh
oek:
Area
= ½
b.h
Deter
mine
ss:
Area
A
BC
= ½
b.c
.sin
A 46
P T
Q
R
Wor
kshe
et 1
9
1. I
f in
A
BC
, c =
8 u
nits
, b =
7 u
nits
and
B=
60
, cal
cula
te th
e fo
llow
ing
with
out u
sing
a c
alcu
lato
r:
A
s in
A
BC
, c
= 8
een
hede
, b =
7 e
enhe
de e
n B
= 6
0, b
erek
en
die
volg
ende
son
der
die
gebr
uik
van
die
sakr
eken
aar:
a) t
he v
alue
of
a / d
ie w
aard
e va
n a.
b) t
he a
rea
of
AB
C f
or th
e gr
eate
st v
alue
of
a.
die
oppe
rvla
kte
van A
BC
vir
die
gro
otst
e w
aard
e va
n a.
2. I
n th
e fi
gure
Q, T
and
R a
re p
oint
s in
the
sam
e ho
rizo
ntal
pla
ne
such
that
TQ
= T
R =
y a
nd T
P re
pres
ents
a v
erti
cal p
ole
posi
tion
ed a
t T.
In
die
fig
uur
is Q
, T e
n R
pun
te in
die
self
de h
oris
onta
le v
lak
soda
t TQ
= T
R =
y e
n T
P s
tel ‘
n ve
rtik
ale
paal
voo
r.
a) P
rove
that
/ B
ewys
dat
:
PQ =
PR
. b)
If
the
angl
e of
ele
vati
on o
f
P
fro
m Q
is
and
QRP
ˆ=
/
As
die
hoog
teho
ek v
an P
van
af Q
ge
lyk
is a
an
en
QRPˆ
=
i)
expr
ess
PQ
in te
rms
of
y a
nd a
tr
igon
omet
ric
valu
e of
/
druk
PQ
uit
in te
rme
van
y en
‘n
trig
onom
etri
ese
funk
sie
van
.
ii)
Pro
ve th
at /
Bew
ys d
at:
co
s.
sin
2si
ny
QR
3. T
he d
iagr
am b
elow
is a
rep
rese
ntat
ion
of a
tree
(T
A)
with
hei
ght
of 3
0 m
and
two
obse
rver
s K
and
L o
n th
e gr
ound
. The
ang
le o
f de
pres
sion
fro
m T
to p
erso
n C
is 1
2°. T
he a
ngle
of
elev
atio
n fr
om
pers
on B
to th
e to
p of
the
tree
is 1
9°.
D
ie d
iagr
am h
iero
nder
is 'n
voo
rste
lling
van
'n 3
0 m
hoë
boo
m
(TA
) en
twee
per
sone
op
die
gron
d. D
ie d
iept
ehoe
k va
n T
na
pers
oon
C is
12°
. Die
hoo
gteh
oek
van
B n
a di
e to
p va
n di
e bo
om
19°.
a)
Cal
cula
te th
e si
ze o
f C
/ B
erek
en d
ie g
root
te v
an C
.
b) C
alcu
late
the
leng
th o
f T
B /
Ber
eken
die
leng
te v
an T
B.
c) H
ence
cal
cula
te th
e di
stan
ce b
etw
een
the
two
obse
rver
s B
and
C /
Ber
eken
ver
volg
ens
die
afst
and
tuss
en d
ie tw
ee p
erso
ne B
en
C
T AB
C
12
19
30 m
47
Tra
nsfo
rmat
ions
/ T
rans
form
asie
s
Nota
tion:
A(x
; y)
A'(x
'; y')
whe
re A
' is th
e ima
ge of
A un
der t
he tr
ansfo
rmati
on
Nota
sie: A
(x; y
)
A'(x
'; y')
waa
r A' d
ie be
eld va
n A na
die t
rans
forma
sie is
. T
rans
lati
ons
(shi
ft)
/ Tra
nsla
sie
(sku
if)
p unit
s hor
izonta
lly / p
eenh
ede h
oriso
ntaal:
: (x;
y)
(x+
p; y
)
q unit
s ver
ticall
y / q
eenh
ede v
ertik
aal: (
x; y
)
(x;
y+
q)
p unit
s hor
izonta
lly an
d q un
its ve
rticall
y / p
eenh
ede h
oriso
ntaal
& q e
enhe
de ve
rtikaa
l: (x;
y)
(x+
p; y
+q)
R
efle
ctio
n / R
efle
ksie
Ab
out th
e x-a
xis / T
en op
sigte
van x
-as:
(x; y
)
(x;
y)
Ab
out th
e y-a
xis / T
en op
sigte
van y
-as:
(x; y
)
(x
; y)
Abou
t the l
ine y
= x
/Ten
opsig
te va
n die
lyn y
= x
: (x;
y)
(y;
x)
Rot
atio
n / R
otas
ie
Arou
nd th
e orig
in thr
ough
an an
gle of
90º /
90º r
otasie
ten o
psigt
e van
die o
orsp
rong
: (x;
y)
(y
; x)
Arou
nd th
e orig
in thr
ough
an an
gle of
180º
/ 180
º rota
sie te
n ops
igte v
an di
e oor
spro
ng: (
x; y
)
(x
; y)
Ar
ound
the o
rigin
throu
gh an
angle
/ R
otasie
van ‘
n hoe
k te
n ops
igte v
an di
e oor
spro
ng:
(x; y
)
(xc
os
ysin;
yco
s +
xsi
n)
Enl
arge
men
t / V
ergr
otin
g Th
roug
h the
origi
n, by
a co
nstan
t facto
r k / V
ergr
oting
van ‘
n kon
stante
fakto
r k te
n ops
igte v
an di
e oor
spro
ng: (
x; y
)
(kx
; ky)
48
Tran
slatio
n ( x;
y)
(x+2
; y5
)Re
flecti
on
( x; y)
(x
; y)
Refle
ction
(x;
y)
(x;
y) Re
flecti
on
(x; y)
(y
; x)
Rotat
ion
(x; y
) (
x; y)
Rotat
ion
+210 a
bout
(0; 0
) Ro
tation
(x;
y)
(y;
x)Ro
tation
6
0 ab
out (
0; 0)
90
-90
+21
0
-60
Exa
mpl
es o
f T
rans
form
atio
ns
49
7 6 5 4 3 2 1
1
2
3
4
5
6
7 6
5 4
3 2
1
1
2
3
4
5
6
7
x
y
x
y
A
N B
C
D
L
K
M
P
Q
R
SE
F
G
H
Wor
kshe
et 2
0
1. B
(3; 1
), C
(1; 3
) an
d D
(1;
2)
are
the
coor
dina
tes
of th
e ve
rtic
es o
f tr
iang
le B
CD
. BC
D h
as to
be
enla
rged
wit
h th
e or
igin
as
cent
re b
y a
fact
or o
f 2.
a) G
ive
the
coor
dina
tes
of v
erti
ces
of tr
iang
le B
'C'D
' of
the
enla
rgem
ent.
b)
If
the
area
of B
CD
is y
squ
are
unit
s, d
eter
min
e th
e ar
ea o
f th
e en
larg
e
B'C
'D'.
B
(3; 2
), C
(2; –
1) e
n D
(1; 0
) is
die
koö
rdin
ate
van
die
hoek
punt
e va
n B
CD
. B
CD
moe
t ver
groo
t wor
d m
et 'n
fak
tor
2 m
et d
ie o
orsp
rong
as
mid
delp
unt.
a)
Gee
die
koö
rdin
ate
van
hoek
punt
e va
n B
'C'D
' van
die
ver
grot
ing.
b) I
ndie
n di
e ar
ea v
an
BC
D y
vie
rkan
te e
enhe
de is
, bep
aal d
ie a
rea
van
verg
roti
ng
B'C
'D'.
2. T
he d
iagr
am s
how
s qu
adri
late
ral A
BC
D a
nd it
s tr
ansf
orm
atio
ns.
a) S
tate
the
gene
ral r
ule
for
the
coor
dina
tes
of a
ny p
oint
rep
rese
ntin
g th
e tr
ansf
orm
atio
n of
qu
adri
late
ral A
BC
D to
qua
dril
ater
al K
LM
N.
b)
Sta
te th
e ge
nera
l rul
e fo
r th
e co
ordi
nate
s of
any
poi
nt r
epre
sent
ing
the
tran
sfor
mat
ion
of
quad
rila
tera
l AB
CD
to q
uadr
ilat
eral
EF
GH
.
c) D
escr
ibe
two
poss
ible
tran
sfor
mat
ions
of
quad
rila
tera
l AB
CD
to q
uadr
ilat
eral
PQ
RS
.
d) G
ive
the
coor
dina
tes
of th
e re
flec
tion
of
quad
rila
tera
l AB
CD
in th
e li
ne y
= x
.
Die
dia
gram
toon
vie
rhoe
k A
BC
D a
sook
die
vie
rhoe
k se
tran
sfor
mas
ies.
a) G
ee d
ie a
lgem
ene
reël
vir
die
koö
rdin
ate
van
enig
e pu
nt w
at d
ie tr
ansf
orm
asie
van
vi
erho
ek A
BC
D n
a vi
erho
ek K
LM
N v
oors
tel.
b)
Gee
die
alg
emen
e re
ël v
ir d
ie k
oörd
inat
e va
n en
ige
punt
wat
die
tran
sfor
mas
ie v
an
vier
hoek
AB
CD
na
vier
hoek
EF
GH
voo
rste
l.
c) B
eskr
yf T
WE
E m
oont
like
tran
sfor
mas
ies
van
vier
hoek
AB
CD
na
vier
hoek
PQ
RS
.
d) G
ee d
ie k
oörd
inat
e va
n di
e re
flek
sie
van
vier
hoek
AB
CD
in d
ie ly
n y
= x
. 3.
A(
2; 6
) is
rot
ated
120
° in
a a
nti-
cloc
kwis
e di
rect
ion
thro
ugh
the
orig
in a
s ce
ntre
of
the
rota
tion
. C
alcu
late
the
coor
dina
tes
of p
oint
A' a
fter
the
rota
tion
.
A(
2; 6
) w
ord
120°
in 'n
ant
iklo
ksge
wys
rig
ting
met
die
oor
spro
ng a
s m
idde
lpun
t van
die
rot
asie
ger
otee
r. B
erek
en d
ie k
oörd
inat
e va
n A
' na
die
rota
sie.
50
Wor
kshe
et 2
1
1. A
tran
slat
ion
map
s th
e or
igin
to th
e po
int (
1; 6
).
a)
Wha
t is
the
imag
e of
poi
nt (5
; 1)
aft
er th
e sa
me
tran
slat
ion?
b) W
hat i
s th
e pr
e-im
age
of p
oint
(6;
2)?
‘n T
rans
lasi
e be
eld
die
oors
pron
g op
die
pun
t (1;
6)
af.
a)
Wat
is d
ie b
eeld
van
die
pun
t (5
; 1)
na
die
tran
slas
ie T
?
b) B
epaa
l die
voo
rafb
eeld
van
die
pun
t (6;
2)?
2.
Giv
e th
e co
ordi
nate
s of
the
vert
ices
of
quad
rila
tera
l wit
h ve
rtic
es P
(1; 1
), Q
(2;
0),
R(3
; –3)
and
S(6
; 2)
aft
er a
rot
atio
n w
ith
the
orig
in a
s ce
ntre
:
a) 9
0° in
a c
lock
wis
e di
rect
ion
b)
180
° in
a c
lock
wis
e di
rect
ion
c) 2
70
in a
clo
ckw
ise
dire
ctio
n
G
ee d
ie k
oörd
inat
e va
n di
e ho
ekpu
nte
van
‘n v
ierh
oek
met
hoe
kpun
te P
(1; 1
), Q
(2;
0),
R(3
; –3)
en
S(6
; 2)
na
‘n r
otas
ie, m
et d
ie o
orsp
rong
as
mid
delp
unt:
a) 9
0° in
'n k
loks
gew
ys r
igtin
g
b) 1
80°
in 'n
klo
ksge
wys
rig
ting
c) 2
70
in 'n
klo
ksge
wys
rig
ting
3.
Sta
te th
e ge
nera
l rul
e fo
r th
e co
ordi
nate
s of
a p
oint
aft
er a
rot
atio
n w
ith
the
orig
in a
s ce
ntre
of:
a) 9
0° in
a c
lock
wis
e di
rect
ion
b)
180
° in
a c
lock
wis
e di
rect
ion
c) 2
70
in a
clo
ckw
ise
dire
ctio
n
G
ee d
ie a
lgem
ene
reël
wat
die
koö
rdin
ate
van
'n p
unt n
a ‘n
rot
asie
, met
die
oor
spro
ng a
s m
idde
lpun
t:
a)
90°
in 'n
klo
ksge
wys
rig
ting
b)
180
° in
'n k
loks
gew
ys r
igti
ng
c)
270
in 'n
klo
ksge
wys
rig
ting
51
Dat
a ha
ndlin
g / D
ata
Han
teri
ng
Mea
sure
s of
Cen
tral
Ten
denc
y / B
epal
ers
van
sent
rale
nei
ging
e
Me
an (A
vera
ge) /
Gem
iddeld
o
The s
um of
a se
t of d
ata di
vided
by th
e num
ber o
f data
: / Di
e som
van d
ie ge
gewe
data
gede
el de
ur di
e aan
tal da
ta:
nx
x
Me
dian (
midd
le va
lue) /
Med
iaan (
midd
elwaa
rde)
o
Th
e valu
e half
way t
hrou
gh an
orde
red d
ata se
t. / D
ie wa
arde
halfp
ad in
die g
eord
ende
data
versa
melin
g.
Mo
de (m
ost fr
eque
nt va
lue) /
Mod
us
o
Value
that
appe
ars t
he m
ost . /
Waa
rde w
at die
mee
ste vo
orko
m.
Meas
ures
of d
isper
sion
(var
iabilit
y ) / M
aat v
an d
atav
ersp
reid
ing
Ra
nge (
Larg
est v
alue –
small
est v
alue)
/ Omv
ang (
Groo
tste w
aard
e – kl
einste
waa
rde)
Perce
ntiles
(% of
total
freq
uenc
y x)
/ Per
senti
el (%
van d
ie tot
ale fr
ekwe
nsie
x)
Perce
ntiles
divid
e the
data
in 10
0 gro
ups c
ontai
ning t
he sa
me nu
mber
of ob
serva
tions
. / Pe
rsenti
ele de
el die
data
in 10
0 gro
epe w
at die
selfd
e aan
tal ob
serva
sies b
evat.
Va
rianc
e / V
arian
sie:
nx
x
2 )
(
St
anda
rd de
viatio
n / S
tanda
ard a
fwyk
ing:
nx
x
2 )
(
52
Wor
kshe
et 2
2
1. G
iven
dat
a / G
egew
e da
ta: 5
, 9, 7
, 8, 3
, 10,
8, 1
0, 9
, 17,
33
F
ind
the
foll
owin
g fo
r th
e da
ta /
Bep
aal d
ie v
olge
nde
vir
die
data
:
a) m
ean
/ gem
idde
ld
b) m
edia
n / m
edia
an
c)
mod
e / m
odus
d) r
ange
/ om
vang
e) l
ower
qua
rtile
/ on
ders
te k
war
tiel
f)
upp
er q
uart
ile /
boon
ste
kwar
tiel
g)
int
erqu
arti
le /
inte
r-kw
arti
el
h)
sta
ndar
d de
viat
ion
/ sta
ndaa
rd a
fwyk
ing
i)
vari
ance
/ va
rian
sie
2. G
iven
dat
a/G
egew
e da
ta: 5
, 9, 7
, 8, 3
, 10,
8, 1
0, 9
, 17,
33
R
epre
sent
the
data
usi
ng a
box
and
whi
sker
dia
gram
/ S
tel d
ie
data
voo
r de
ur v
an 'n
mon
d-en
-sno
rdia
gram
geb
ruik
te m
aak.
3. T
he f
requ
ency
tabl
e re
pres
ents
the
mar
ks in
term
s of
%, o
btai
ned
by a
gro
up o
f G
rade
12
lear
ners
in a
Mat
hem
atic
s ex
amin
atio
n.
D
ie f
rekw
ensi
etab
el v
erte
enw
oord
ig d
ie p
unte
in te
rme
van
%,
deur
'n g
roep
Gra
ad 1
2-le
erde
rs in
'n W
isku
nde-
eksa
men
beh
aal.
Tes
t sco
res/
T
oets
punt
e F
requ
ency
/ F
rekw
ensi
e C
umul
ativ
e fr
eque
ncy/
K
umul
atie
we
frek
wen
sie
1-20
3
21
-40
5
41-6
0 9
61
-80
6
81-1
00
2
a)
Fin
d th
e fo
llow
ing
/ Bep
aal d
ie v
olge
nde:
i)
lo
wer
qua
rtil
e / o
nder
ste
kwar
tiel
ii) u
pper
qua
rtile
/ bo
onst
e kw
artie
l
ii
i) i
nter
quar
tile
/ in
ter-
kwar
tiel
iv
) s
emi-
inte
rqua
rtil
e ra
nge
/ sem
i-in
ter-
kwar
tiel
b)
Use
the
tabl
e on
the
diag
ram
she
et to
com
plet
e th
e cu
mul
ativ
e fr
eque
ncy
colu
mn
/ Geb
ruik
die
tabe
l op
die
diag
ram
vel o
m d
ie
kum
ulat
iew
e fr
ekw
ensi
e ko
lom
te v
olto
oi.
c)
Dra
w th
e og
ive
for
the
give
n da
ta o
n a
grid
/ T
eken
die
ogi
ef
op ‘
n ro
oste
r.
d)
Use
the
ogiv
e to
det
erm
ine
the
med
ian
mar
k / G
ebru
ik d
ie
ogie
f om
die
med
iaan
punt
te b
epaa
l.
4. T
he ta
ble
repr
esen
ts th
e nu
mbe
r of
car
s so
ld b
y a
car
man
ufac
ture
r fr
om 2
003
to 2
008:
Die
ond
erst
aand
e ta
bel s
tel d
ie a
anta
l mot
ors
wat
ver
koop
is d
eur
‘n m
otor
verv
aard
iger
200
3 to
t 200
8, v
oor:
Y
ear
/ Jaa
rC
ars
sold
/ M
otor
s ve
rkoo
p20
02
1234
20
03
1432
20
04
1672
20
05
1752
20
06
2013
20
07
2193
20
08
2345
a)
Dra
w th
e sc
atte
r pl
ot to
rep
rese
nt th
e ab
ove
data
. / T
eken
'n
spre
idia
gram
om
bog
enoe
mde
inlig
ting
voor
te s
tel.
b) E
xpla
in w
heth
er a
line
ar, q
uadr
atic
or
expo
nent
ial c
urve
wou
ld
be a
line
of
best
fit
for
the
abov
e-m
entio
ned
data
. / V
erdu
idel
ik o
f 'n
line
êre,
kw
adra
tiese
of
eksp
onen
siël
e ku
rwe
die
best
e pa
slyn
vi
r di
e bo
geno
emde
inlig
ting
sal w
ees.
c)
If
the
sam
e tr
end
cont
inue
d, e
stim
ate,
by
usin
g yo
ur g
raph
, the
nu
mbe
r of
car
s th
at w
ill b
e so
ld in
201
0. /
Indi
en d
iese
lfde
ne
igin
g/pa
troo
n vo
ortg
aan,
ska
t, de
ur jo
u gr
afie
k te
geb
ruik
, die
aa
ntal
mot
ors
wat
teen
201
0 ve
rkoo
p sa
l wor
d.
53
Solutions / Oplossings
PAPER 1 Worksheet 1
1. a) 2x2 3x
b) 2x2 x
c) 2x2 + 4x3
d) 2x2 + 4x3
e) 2x2 x
f) 2x2 + 4x3
g) 8x3 12x2y +6xy2 y3
h) x3 + 8
i) 8x3 27
2. a) (x + 4)(x 3)
b) (x 4)(x + 3)
c) cannot factorise
d) cannot factorise
e) (x 4)(x 3)
f) (x + 4)(x + 3)
g) (x 6)(x + 1)
h) 7(x 2)(x2 + 2x + 4)
i) 2(3 2x + 2y)(3 + 2x 2y)
j) (2x y)(2x y 4)
k) (x 2y)(2x + 4y 1)
l) (x 3)(x 4)(x + 4)
m) (x 2)(x 5)(x + 3)
n) (x + 1)(x 2)2
3. a) )5)(5(5
3 2
xxxx
b) ba
baba 22
c) )2(
1
xxx
d) )1)(1(
12 23
xxxx
Worksheet 2
1. a) False
b) False
c) False
2. a) 4
9
b) 16
7
c) 27 ¼
d) 17
85
e) 16
1
f) 64
3. a) 2
b) 12 2
c) 4
d) ½
e) 4
f) 23
g) 3
4. a) 2
2
b) 6
35
54
Worksheet 3
1. a) 1½ or 2
b) 1½ or 4
c) 5
d) 6 or 4
e) 4
f) 4½ or 32
g) 8
h) 3 or 10 32
i) no solution
2. a) 3 or 4
b) no real solution
c) a
acbbx2
42
d) )1(2
5841 2
m
mmx
3. a) x = 23
b) 1,24 or 3,23
4. a) (x 3)(x2 + 4)
b) (x 2)(2x 1)(x + 3)
c) (x 2)(x +7)(x + 1)
d) (x + 3)(x 3 )(x + 3 )
5. a) x = 43 or x = 1 or x = 1
b) x = 1 or x = 2
c) x = 1 or x = 6 or x = 2
6. a) no solution
b) 23
7. 1 2i
Worksheet 4
1. a) 5
b) 2,32
c) 2,86
d) 1,58
e) 1,75
f) 0,63
g) 16
h) 2
i) 5
2. PA
in )1(log
3. 1log )1( xFi
in
4. (1; 0) or (5; 8)
5. a) (4; 2)
b) (1; 1) or ( 711 ; 7
13 )
c) ( 32 ; 2
3 ) or ( 21 ; 1)
d) (0,5; 1) or ( 32 ; 3
2 )
6. a) x < 21
b) x < 1 or x > 2
c) 21 < x < 2
5
d) x 2 or / of x 3
e) 3 < x < 4
f) 2 x 4
g) 4 < x < 4
h) x 3 or / of x 3
Worksheet 5
1. 2; 3; 5; 7; 11; 13; 17; …
Product 1st; 2nd; 3rd; …
= 2; 6; 30; 210; 2310; …
Units digit is 0
2. a) 90(0,7)5 = 15,126 m
b) 11 bounces
c) 510 m
3. a) 29
b) 1; 5; 11; 19; 29; …
4 ; 6 ; 8 ; 10 ; …
55
2; 2 ; 2 ; …
Tn = rqnnT
21
2
Substitute T1 = 1 & T2 = 5:
Tn = 0,5n2 + 2,5n 2
c) T50 = 1373
4. a) 6 3 pool: 2(6 + 2) + 2(3)
= 22
8 5 pool: 2(8 + 2) + 2(5) = 30
b) m n pool:
2(m + 2) + 2(n) = 2(m + n) + 4
c) m n pool:
2(m + m) + 4 = 4m + 4
d) 49 49 pool
Worksheet 6
1. R27 228,80
2. 5,9 years
3. 9,62%
4. P = F(1 + i)n
i = 1n
PF
n =
PF
i )1(log
5. a) 12,68%
b) R34039,80
6. 29,3%
Worksheet 7
1.
1216,0
50
1216,011
50000
x
x = R1376,51
2.
kr
nkkrx
F11
3. 04,0
]1)04,1[(5000 20 F
= R 148 890,39
4. a) F = 2 400 000(1 + 0,16)5
= R 5 040 819,98
b) F = 2 400 000(1 0,22)5
= R 6 486 499,59
c)
1216,0
60
1216,0 11
61,1445679
x
x = R15 880,39
Worksheet 8
1. a) 2
b) 360
c) [2; 2]
d) 2
2. a) 2
b) x = 90 + k.180
c) 180
d) R
3. a) 11
b) no x-intercepts
c) 2
d) R
e) [2; )
f) x = 3
4. a)
56
b) x = 0
5. a) & c)
b) 3 vertical stretch
Translation 2 units to the right
Translation 1 unit downwards
d) R
e) [1; )
Worksheet 9
1. a) C(x) = 158 x + 236
b) C(2000) = R1302,67
c)
d) The rate that the cost change with respect to the units used.
e) Basic cost
2.
a) x = 3 or x = 2
b) x 3 or ½ < x 2
c) 3 x ½ or x 2
d) x = ½
3. a)
b) x2 2x 8 = 0
x2 x 6 = x + 2
(2; 0) & (4; 6)
57
Worksheet 10
1. a) A(0; 9)
B(1; 0)
C(9; 0)
b) OD = 5
DE = 16
AF = 10
HI = 20
2. f (x) = 2 sin(½x) + 2
g(x) = 4tanx
Worksheet 11
1. m = 1
2. a) m = 1
b) )1)(1(
2
aha
3. 8x
4. a) f '(x) = 3x2 2x + 2
b) f '(x) = 3
82
x
c) f '(x) = xx 2
1124
d) f '(x) = 9x2 + 4x 4
e) f '(x) = 622
1 x
x
5. a) y = 2x 3 & y = 2x 7
b) (1; 5)
6. a)
b)
c)
7. Volume = r2h h = 2
350
r
A(r) = 2(r2) + 2rh
A'(r) = 4r r
700
r = 3,81 cm & h = 7,67 cm
(1,3; 18,5)
(3,7; 14,8)
(1; 36)
(1,67;
58
Worksheet 12
1. a) B(0) = 1 200
b) B'(t) = 40
c) Decrease with 40 million per hour.
d) Shrinking
e) t = 6
f) 6 hours
2. a) H(0) = 8
b) t = 2ºC or/of t = 20ºC
c) t = 14ºC
d)
3. a) SR = (72 2x) m
b) A = 72x – 2x2
c) x = 18 m
d) A = 648 m²
Worksheet 13
1. a) x 10
y 0
y 2x + 60
y x + 80
y 61 x + 10
b) y = 32 x + 3
1 P
Pmax at A(10; 70)
P = 230 units
c) 2 < m < 61
10 < c < 60
2. a)
b) At the point C(6,8; 5,2)
6,8 ha maize & 5,2 ha sunflower
C (6,8; 5,2)
59
PAPER 2
Worksheet 14
1. nn 232
21
2.
5 100 n
540 17640 180(n 2)
108 176,4 nn )2(180
3. 150
4 a) 441,1 cm2
b) 412,6 cm2
c) 414,7 cm2
d) H = 91
r2 + rH
= (3)2 + (3) 91
= 118,2 cm2
e) (8)(6) +
)109)(8(2)116)(6(2 21
21
= 196,1 cm2
Worksheet 15
1. a) (6; 4)
b) t = 6
2. a) x2 + y2 = 40
b) q = 2
c) = 26,6
3. a) (1; 2) & r = 20
b) mradius = 2
mtangent = ½
y + 6 = ½(x 3)
y = 215
21 x
4. mAC = 1 & mBC = 94
Angle = 45 + 23,9 = 68,9
5. a) x2 8x + y2 4y 12 = 0
b) (0; 6); (0; 2)
6. r = 23
Worksheet 16
1. (x 5)2 + (y + 12 )2 = 25 1
4
M( 5; 12 ) & r = 5,025
2. a) mAB = 32
b) e = 3
c) 3x + 2y 7 = 0
3. a) 26 + 26 + 52 = 17,4
b) 11,3
c) (3,5; −0.5)
d) y = 0,2x + 0,2
Worksheet 17
1. a)
b) f (x): 180 & g(x): 360
2. a)
60
b) x = 135º ; 45º ; 45º ; 135º
c) 2 cosx = 2 + cosx 2 cosx
= (2 + cosx) (2cosx)
x = 180º; 180º
3. a)
b) 180º
c) x = 0º
d) 180º < x < 90º or 0º < x < 90
4. a = 2 & k = 1
b = 1 & m = 1
Worksheet 18
1. a) 1
b) 32
c) 3
d) 2
e) 3
f) 2
2. a) t
b) t
c) 21 t
t
3. a) 150
b) x R
c) x = 46,3
4. a) 12,3 ; 107,7
b) 212,4 ; 7,6
5. a) 555
b) 3
3
6. a) x = k.180
x = 30 + k.360
x = 150 + k.360
k Z
b) = 60 + k.360
= 300 + k360
k Z
7. Prove LHS = RHS
8. 65181
61
Worksheet 19
1. a) a = 5 or a = 3
b) Area = 310
2. a) PQT ≡ PRT (SS)
PQ = PR
b) PQ = cos
y; Use sine rule
3. a) 12
b) sin 19 = TB30
TB = 9,77 m
c)
12sin
77,9
7sin
BC
BC = 5,73 m
Worksheet 20
1.
a) B'(6; 2); C'(2; 6); D'(2; 4)
b) 4 times BCD = 4y
2. a) (x; y) (x; y)
b) (x; y) (x + 7; y + 5)
c) (x; y) (x; y)
d) A'(4;6); B'(1; 2)
C'(3; 4); D'(7; 1)
3. A'(6,2; 1,27)
Worksheet 21
1. a) (4; 5)
b) (5; 4)
2. a) P'(1; 1); Q'(0; 2)
R'(3; 3); S'(2; 6)
b) P'(1; 1); Q'(2; 0)
R'(3; 3); S'(6; 2)
c) P'(1; 1); Q'(0; 2)
R'(3; 3); S'(2; 6)
3. a) (x; y) (y; x)
b) (x; y) (x; y)
c) (x; y) (y; x)
Worksheet 22
1. a) 10,82
b) 9
c) 9
d) 30
e) 7
f) 10
g) 317
h) 8,15
i) 66,4
2.
62
3. a) i) 33 ii) 67
iii) 34
iv) 17
b)
1-20 3 3 21-40 5 8 41-60 9 17 61-80 6 23 81-100 2 25
c)
d) 50
4. a)
b) Linear
c) 3000
Cars sold / Motors verkoop
0
500
1000
1500
2000
2500
2000 2002 2004 2006 2008 2010
63
