4 3 5 2 3 5 5 ¨ ¸ ¨ ¸ PQRS xy © ¹ © ¹ § · § · 5 4 5 4 33 ... · SOALAN 3 1. a) 2 2 2...
Transcript of 4 3 5 2 3 5 5 ¨ ¸ ¨ ¸ PQRS xy © ¹ © ¹ § · § · 5 4 5 4 33 ... · SOALAN 3 1. a) 2 2 2...
SKEMA PERCUBAAN SPM 2017
MATEMATIK TAMBAHAN KERTAS 2
SOALAN 1
1. a)
2
4 3 5
4 3( 1) 5
4 8
2
2 3 5
2(2) ( 1) 3(2)( 1) 5
3
x y
k
k
k
x py xy
p
p
b)
2
2
5 4
3
5 4 5 42 3 5 5
3 3
2 9 10 0
2 5 2 0
5, 2
2
5, 1
3
5 5,
2 3
xy
x xx x
x x
x x
x x
y
2.
2
Luas 28 224
8
1 22Perimeter 2 28 2 14 72
2 7
36 36
836 36
36 36 8 0
3 1 3 2 0
1 2,
3 3
24, 12
PQRS xy
yx
STR y x x
y x
xx
x x
x x
x x
y y
3. a)
2
2
2
(5 ) 5 1
5 5 1
4 4 0
( 4) ( 4) 4(1)(4)
2(1)
2
h h h
h h h
h h
h
h
b)
2 2
2 2
2 2
2 2
2
2
2 2 1
1 2 11
3 1
3 1 1 2(3 1) 11
4 2 2(9 6 1) 11
4 2 18 12 2 11
7 16 4 0
( 16) ( 16) 4(7)(4)
2(7)
2
7
2 13 1
7 7
x y x y
x y x y
x y
y y y y
y y y y
y y y y
y y
y
y
x
4. 2(y + 1)
2
32x – xy = 63
x + x + y + y = 14
y = 7 – x atau setara
2(7 – x + 1)
2
32x – x(7 – x) = 63
(3x – 13)(x – 3) = 0
x = 3, abaikan nilai 3
13x
y = 4, abaikan nilai 3
8y
5.
354
4
35
75.8
xy
xy
Luas
xy
yx
yx
Perimeter
6
6
1222
12
1
2
x Gantikan dalam
5.2,5.3
5.3,5.2
2
7,
2
5
07252
035244
035424
3564
2
2
yy
xx
xx
xx
xx
xx
xx
Ukuran Bilik, Panjang = 3.5 + 2 = 5.5 m
Lebar = 2.5 + 2 = 4.5 m
SOALAN 2
1. (a) a = 6000
r = 1.022
n = 2017 − 2011 + 1
= 7
T7 = 6000 × 1.0227 − 1
= RM6836.86
(b) Tn > 18000
6000 × 1.022n − 1
> 18000
1.022n − 1
> Error!
(n − 1) log10 1.022 > log10 3
n − 1 > Error!
n − 1 > 50.4842
n > 51.4842
∴ n = 52
(c) T5 = 6000 × 1.0225 − 1
= RM6545.68
Total interest earned
= 6545.68 − 6000
= RM546
2. a) 15 1 14(2) 29T
b) 20
10
1 19(2) 39
10(3 39) 210
2
T
S
c) 2(1) ( 1)(4) 435
2
15
2 15 30
nn
n
3. a) (2k + 1) – (k + 3) = (k + 3) – (k – 1)
2
1
k = 6 d = 4
b) a + (6 – 1)(4) = 5
a = –15
c) )4)(1()15(2
2 n
nSn
nnSn 172 2
4. a)
16
12
27623
30424
28
842
a
d
d
da
da
dadaa
cm
S
4000
12241622
2525
b)
292
12231624
T
cm
x
3
124
12
292
5. a) Number of sheep sold
d = 40, a = 40
for the first 21 month = 21 40 20(40)T
= 840
The number of sheep left after 21 month is 1000 – 840 = 160
b)
12
2 ( 1)2
122(1000) (12 1)( 40)
2
9360
9360 1 30
280800
n
nS a n d
S
1
2
SOALAN 3
1. a)
22
2
120
12 710
1510
x
x
x
b)
2 2
2
2
120 18 138
1510 18 1834
1834 1389.34
11 11
new
new
new
x
x
2. a)
2 2
2
2
15015
10
247215
10
22.2
xx
N
xx
N
b) i)
15016
11
26
xx
N
k
k
ii)
2 2
222472 26
1611
5.494
xx
N
3. a) 1
2
140 15
220.5 1014
24.07
m
N FL C
f
b) 22
228550 960
40 40
137.75
11.74
fxx
f
4. a) L = 19.5 atau F = 16 atau fm = 15 atau c = 5
)5(15
162
50
5.19
22.5
b)
81115106
)32(8)27(11)22(15)17(10)12(6
atau
81115106
)32(8)27(11)22(15)17(10)12(6 22222
2)5.22(50
27225
38.25
c) 41.25
5. a) i) 4
16
64
x
x
ii) 22
2
3 416
400
x
x
b)
2
22
16 12 28
64 84 148
400 480 880
1485.286
28
8805.286
28
3.487
N
x
x
x
SOALAN 4
1. a) i)
3 (1)( 6) 3 (1)(9)3,6 ,
3 1 3 1
6, 5
(6,5)
x y
x y
R
ii) 1(18 48 20 24) ( 24 36 48 10)
2
28
b) 2 2
2 2
( 2) ( 4) 2
4 8 19 0
x y
x x y y
2. a) 1(4 80 8) (20 8 16)
2
24
b) 1 21, 1
5 1( 7)
2
m m
y x
y x
c) 2 2
2 2
( 4) ( 8) 5
8 16 55 0
x y
x y x y
3. a) )0(1)5(0)4(2)2(5)1(4)0(0
2
1
3
b)
23
)1(3)2(2,
23
)5(3)0(2
5
7,3
c) AP = 2PC atau 2222 )1()5(2)2()0( yxyx
(x – 0)2 +(y – 2)
2 = 4[(x – 5)
2 + (y – 1)
2]
3x2 + 3y
2 – 40x – 4y + 100 = 0
4. a) Q(2, 0) or P(0, -6)
(2, 0) =
32
182,
32
02 yx
R = (5, 9)
b) 5)0()2( 22 yx
25)2( 22 yx
021422 xyx
5. a)
216
4082
1
0
0
2
8
5
4
0
0
2
1
unit
AOBLuas
b)
5
4,
5
16
23
5223,
23
4283
2:3:
C
CBAC
c)
0231267233
161642566442510168
28254
2
22
2222
2222
yxyx
yyxxyyxx
yxyx
PBPA
SOALAN 5
1. a)
b) 4 3cos 2
3 2
4 3cos 2
3 2
3 4 3cos 2
2 3 2
3 2 9cos 2
2 4
x x
x x
x x
x x
No of solution = 3
2. a)
b)
2cos 0
2cos
xx
xx
No. of solutions = 2
3. a) Bentuk graf –sin x
Amplitud 3
2
-2
0 2
Lakaran dalam julat 0 ≤ x ≤ 2π
b) Persamaan
xy 12
Garis lurus
xy 12 dalam julat 0 ≤ x ≤ 2π
3 penyelesaian
4.
5. a) xy 2sin2
Sine curve seen
One and a half cycle in
3π0
2x
Max value
32
, Min value
32
OR
Sketching the straight line from the
*equation involving x and y. K1
N1
2 9π 4
y x 2 9 3
sin2π 4 2
x x
or equivalent
y
xππ
2O 3π
2
3sin 2
2y x
32
32
π4
3π4
5π4
94
34
2 9π 4
y x
Number of solutions = 3
Curve and straight line sketched
correctly
b)
xy
xx
xx
1
2sin2
2sin2
Draw
xy 1 on the same axes
Number of solution = 4
SOALAN 6
1. a) 2
49
xdx
dy atau 0
49
2
x
3
2p dan
3
2q (kedua-dua jawapan betul)
b)
12,
3
2 dan
12,
3
2 (kedua-dua jawapan betul)
32
2 8
xdx
yd
27
3
2
83
dan 27
3
2
83
(kedua-duanya betul)
12,
3
2 adalah titik minimum
12,
3
2 adalah titik maksimum
2. a)
2
22
0
3
3 6 3
6 3 1
lim
6 1
x
y y x x x x
y y x x x x x x
yx x
x
dy yit
dx x
dyx
dx
b)
2
2
0
6 1 0
1 1,
6 12
1 1,
6 12
6 0
1 1,
6 12
dy
dx
x
x y
P
d y
dx
P
3. a)
2
32
312
3
16
xxdx
dy
xxy
b)
3
97,4sin
3
97
3
1464,4
3
1,0sin
3
1
3
1060,0
4,0
043
0312
0
2
2
ganPuTitik
yx
ganPuTitik
yx
xx
xx
xx
dx
dy
c)
3
97,4
0124612,4
2
1,0
0120612,0
612
2
2
2
2
2
2
dx
ydxWhen
dx
ydxWhen
xdx
yd
4. a)
2
2
2
12
12 9
1 12 9
3
dypx
dx
px
p
p
b)
2
3
3
3
3 12
12
4 1 12 1
15
12 15
dyx
dx
y x x c
c
c
y x x
5. 2
2
2 2
2
3 2
2 2
5
1
52 2
1
2 4 3
solvesimultaneous
2 4 3 3 2
2 0
0, 2
3, 3
(0,3)
2 3
(2,3)
2 7
y x x
dyx
dx
ym
x
yx
x
y x x
x x x x
x x
x x
y y
y x
y x
SOALAN 7
1. a) 0.6 0.4p q
i)
4 26
4( 4) 0.6 0.4
0.31104
P x C
ii)
4 2 5 1 6 06 6 6
4 5 6
( 4) ( 4) ( 5) ( 6)
0.6 0.4 0.6 0.4 0.6 0.4
0.54432
P x P x P x P x
C C C
b) 300 20
i)
285 320
285 300 320 300
20 20
0.75 1
1 ( 0.75) ( 1)
1 0.2266 0.1587
0.6147
P x
P z
P z
P z P z
ii) 18000.72
2500
( ) 0.72
3000.72
20
3000.583
20
288.34
P x y
yP z
y
y
2. a) i) 8 3 53( 3) (0.1) (0.9) 0.0331p x C
ii) 8 6 2 8 7 1 8 8 0
6 7 8
( 5) ( 6) ( 7) ( 8)
(0.9) (0.1) (0.9) (0.1) (0.9) (0.1)
0.9619
p y p y p y p y
C C C
b) i) 35 25( 35)
8
( 1.25)
0.1057
p x p z
p z
ii) 100( )
600
250.1667
8
250.967
8
17.26
p x t
tp z
t
t
3. a) i) 3 5
83
3 2( 3)
5 5
0.1239
P X C
ii) 0 8 1 7
8 80 1
( 2) ( 0) ( 1)
3 2 3 2
5 5 5 5
0.1064
P X P X P X
C C
b) i)
60 65( 60)
5
1
0.84134
P X P Z
P Z
ii)
75 65( 75)
5
2
0.97725
0.97725250
244.31
244
P X P Z
P Z
x
x
x
4. a) i) 15 13 2 15 14 15 15
13 14 15( 13) (0.75) (0.25) (0.75) (0.25) (0.75)
0.2361
P x C C C
ii) (0.75)(0.25) 10.2
554 / 555
n
n
b) i)
80 5880 ( )
15
0.03593
P x P z
ii) 58( ) 0.1
15
581.281
15
38.785
wP x
w
w
5. a) i) 5 1 4
1( 1) (0.6) (0.4)
0.0768
P X C
ii) 5 2 3
2( 2) (0.6) (0.4)
0.2304
P X C
b) i) 2 1.5( 2) 1
0.2
1 0.00621
0.9938
P X P
ii) ( ) 0.65
1.51 0.65
0.2
1.50.385
0.2
1.423
P x m
mP Z
m
m
SOALAN 8
1. a)
t
s 71.0 60.3 49.2 40.0 30.0 20.0
b) Skala pada kedua-dua paksi seragam
Semua titik diplot betul
Garis lurus penyuaian terbaik
c) i)
btat
s
Kelihatan garis menyentuh paksi-t
s= 80
a = 80
Keceruanan = 06
8020
b = –10
ii) 97.5
2. a)
x 1 2 3 4 5 6
1/y 3.70 5 6.25 7.69 9.09 10
Correct and uniform scale
All points correctly plotted
Line of best fit
b) i)
a
bx
ay
a
bx
y
11
1
gradienta
1
a
11.3
a = 0.7692
1.846
4.27692.0
int
b
b
erceptYa
b
ii) 0.1429
3. (a) x² 1 4 9 16 25 36
xy 89 82 72 58 38 15 2xy qx p
(b)
2.11 , 91
91
m q c p
p
(c) 2 20.25 , 9.56x y
4. (a) 10log x 0.13 0.21 0.37 0.48 0.58 0.75 0.87
10log y 0.98 0.64 1.18 1.26 1.34 1.48 1.58
10 10 10log log logy n x a
(b)
10
0.8
log 0.88
7.59
m n
c a
a
5. (a) p 1 2 3 4 5 6 7
p q 0.77 0.94 1.15 1.28 1.46 1.62 1.80
p q ap b
(b)
0.17
0.60
m a
c b
(c) 1.81 1.34 , 4.3p
SOALAN 9
1. a) yxTR 104 , yxSQ 510
b) )510( yxmSU atau ymxmSU 510
)104()10(5
2yxnxSU
ynxnxSU 1044 atau ynxnSU 10)44(
c) 4 – 4n = 10m dan 5m = 10n
6
1n ,
3
1m
d)
nTR
TR
TU
PQ 2
1
atau
yx
yx
1046
1
1042
1
3 : 2
2. a) i) 2 26 10
8
m
m
ii)
10 11
OQ OP PQ
i j
b)
2 2
3 3
2 (10 11 )
3
MN MR RN
QR RO
i j
c) i) ( )
3 4
PT PR RT PO OR RT
i j
ii) 6 8
3 4
2
OP i j
PT i j
OP PT
3. a) i)
1
2
12 ( 3 )
2
32
2
QL QR RL
QR QP
b a
b a
ii)
3 2
SN SR RN
PQ QR
a b
b) i)
32
2
32
2
QM hQL
h a b
h a hb
ii)
2
2(2 ) ( 3 2 )
3 (4 2 )
QM QN NM
QR kNS
b k a b
ka k b
Equating the coefficients of a and b
33 2 4 2
2
4 2,
3 3
hk h k
h k
c) 3 2PQ a PS b
Area of parallelogram PQRS
1
2 5.1 3 sin402
9.835
4. a) i) AC AB BC
p q
ii) 1
4
1
4
AE AD
q
iii)
1
4
BE BA AE
p q
b) i)
1
4
1
4
BF kBE
k q p
kq kp
ii)
AF hAC
h p q
hp hq
c)
1
4
11 ( )
4
AF AB BF
p kq kp
k p kq shown
5. a) i) 1
2
3
OF OQ
y
ii)
1
3
13 6
3
2 2
OE OP PE
OP PQ
OP x y
x y
b) i)
3 1 3
PG hPF
PO OG h PO OF
OG h PO OF PO
h x hy
ii)
2 2
2 2
OG kOE
k x y
kx ky
3 1 3 2 2
3 1 2 3 2
3 1 3
1
2
3
4
h x hy kx ky
h k h k
h h
h
k
SOALAN 10
1. a) )10(2
360
38 ATAU
90
19)10(
10 + 10 + )10(2360
38 ATAU 10 + 10 +
90
19)10(
atau 9
19205
45
194 atau
315
1678 atau 5.327
b) 6635.4
90
19327.5
5(4.6635)
5(4.6635) + 5 + 5
43.3125
c)
90
19)5(
2
1 2 atau
90
19)10(
2
1 2 ATAU 2)5(360
38 atau 2)10(
360
38
90
19)10(
2
1 2 -
90
19)5(
2
1 2 ATAU 2)10(360
38 - 2)5(
360
38
24.881
2. a)
b) Arc AB = K1 (arc AB or CD)
Arc CD
OE =
OD
Perimeter = 6 + 4.713 + 2 + 5.5644 + 4
c) Area of sector AOB =
Area of OCE
3. a) i) 4 1cos
12 3
70 32'
1.231
WPX
WPX
rad
ii)
180 90 70 32' 90
109 28'
1.911
WQY WQZ YQZ
rad
b) Perimeter of the shaded region
2 28(1.231) 12 4 4(1.911)
28.81
WX XY WY
cm
c) Area of the shaded region
2 2
2
1 1 18 4 128 8 1.231 4 1.911
2 2 2
13.20cm
4. a) tan 1
4
AB
OB
rad
b)
2 2
2
2 2
2.828
AB AB cm
OA
cm
c) 1
3
3 6
4
6 2.828
3.172
6 4.7134
OB
OQ
OQ OB cm
BQ cm
AP OP OA
PQ cm
Perimeter shaded region
3.172 2 4 4.713
13.89
AP AB BQ PQ
cm
d) Area of shaded region = Area sector OPQ – Area of AOB
2
2
1 16 2 2
2 4 2
12.14cm
5. a) = kos
-1 ( )
26
4
=81.15˚
= 1.417 rad
b) QP = 26.31 cm atau ∠QRY = 1.725 rad QR = 11 x 1.725 atau PR = 15 x 1.417
QR = 18.975 atau PR = 21.255
18.975 + 21.255 + 26.31
66.54
c) ½(11+15)(26.31) atau ½(112 )(1.725)
atau ½(152 )(1.417)
½(11+15)(26.31) - ½(112 )(1.725)-½(15
2 )(1.417)
181.755
SOALAN 11
1. a) 24 xxy
b) 244 xxx
1b
c) )3)(3(
2
1
4
1
2 )4( xx atau
3
)1()1(2
3
)4()4(2
32
32
3
)1()1(2
3
)4()4(2
32
32
- )3)(3(2
1
2
9
d)
1
0
22 )4( xx
5
)0()1(2
3
)0(16
5
)1()1(2
3
)1(16 54
354
3
15
53
2. a)
2
2
1
2
13 1
2
1 5
2 2
dyx
dx
m
y x
y x
b)
4 1
2
3
4
3
2
3
(4 )
4
31
2
2
3
y dy
y
c) 4
3
2 2
(4 )
4 3(4(4) ) (4(3)
2 2
1
2
y dy
3. a) 3
1
4
1
4
127.5 2
2127.5
4
1127.5
2 2
4
k
k
x dx
x
k
k
b) (i)
2
2
(0, 2)
2
2 0
2 1 0
2, 1
(1,1)
B
x x
x x
x x
x x
A
(ii)
1 2 2
0 1
1 22 3
2
0 1
2
4 22 3
1 8 18 8 4 2
2 3 3
5
6
y dy y dy
y yy y
4. a)
2
2
20 33
2
2 3 14 0
2 7 2 0
7, 2
2
2
xx
x x
x x
x x
k
b)
22
10
3
2
2
0
( 3)
26
3
1 14 497
2 3 3
26 49
3 3
25
33
L x dx
L
xx
c)
1
22
7
3
72
3
8
1(2) 3 4
3
8 4
12
( 3)
32
V
V
y dx
yy
5. a) 2
2
22
1
23
1
3 , 4
( 4)( 1) 0
4, 1
4 0
2
11 3 4
2
34
2 3
3 8 18 4
2 3 3
13
6
y x y x
x x
x x
x
x
x
xx
b) 2
2 2 2
1
23 5
1
3 5 3 3
1(3) (1) (4 )
3
83 16
3 5
8(2) 2 8(1) 13 16(2) 16(1)
3 5 3 5
86
15
x dx
x xx
SOALAN 12
1. a) 30
120
3 4 600
y
x y
x y
c) i) 150
ii) 0.5 0.6
30, 160
0.5(160) 0.6(30)
98
x y k
x y
2. a) 100
2
30
x y
y x
y x
c) i) y max = 80 y min = 50
ii)
(20,80)
40(20) 20(80)
2400
3. a)
3
50
1000
y x
y x
x y
c) i) RM720
ii) (25,75)
25 75
100
x y k
4. a) 2 5 30
3 2 24
2
x y
x y
x y
c) i) 4
ii) 200 250
200(5) 250(4)
2000
x y k
5. a)
165 150 6600
105 120 840
20
x y
x y
x y
c) i) 3 37y
ii) (0,44)
12 12
12(0) 12(44)
528
x y k
SOALAN 13
1. a) 150100
60.0
90.0 Xx
12010050.1
Xy
8010040.0
Xz
b)
100
)15(80)30(150)20(120)35(150
= 133.50
c) i) 50.133100
2000
2014 XP
26702014
RMP
ii) 50.133
100
115X
= 153.525
2. a) i) Price index I = Error!× 100
p = Error!× 100
= 126
ii) 87 = Error!× 100
q = 9.80
b) I,¯ = Error!
= Error!
= 113
c) i) I2010 = Error!× 113
= 207.92
ii) P2010 = Error!× 22.3
= RM46.37
3. a) i)
45.11
10610080.10
06
06
RMP
P
ii)
17.2
11510050.2
04
04
RMP
P
RM
b)
160
13395357795
95.13320352520
)20140()35()25115()20106(
M
M
M
I
I
I
c) i)
74.160
100
12013395
I
ii)
93.28
74.16010018
08
08
RMP
P
4. (a) 960100
120
800
m
(b) 10
09
100P
IP
x = 110 , y = 140
(c) 110(6) 125(4) 98(3) 110(2)
15
1686
15
111.6
(d) 4500111.6
100
5022
5. a) x =36
y = 54
z = 80
b) 5 8 15 4 100
5
125(5) 120(8) 105(3) 80(4)
20
111
n n n
n
c)
2000
2000
280100 111
252.25
P
P RM
d) 2008
2008
100 120280
336
336100
252.25
133.20
P
P
I
SOALAN 14
1. a) i) AC2 = (7)
2 + (6)
2 – 2(7)(6)kos(80)
8.3913
ii)
3913.8
35sin
12
sin
ACD
180 – 55.1093
124.8907
iii) 358907.124180 CAD
80sin)6)(7(2
1 atau 1093.20sin)12)(3913.8(
2
1
37.9911
b) i)
ii) 145°
2. a) i) 10sin 30
QT 20
QT
cm
ii) 3
20
81 22
cos
b) i) 12 9
sin sin50B
B = 113 11
ii) CD2 = 9
2 + 8.5
2 – 2 (9)(8.5) cos 47
0 49
’
CD = 7.107 cm.
iii) '1 1(8.5)(9)sin 46 49 (12)(9)sin16 49
2 2
42.730
3. a) 61
14
sin94 sin61
15.97
TSU
SU
SU
b) 2 2 26 12 15.97 2(12)(15.97)cos
18.70
RUS
RUS
c) 2 2 2
43.7
12 14 2(12)(14)cos43.7
9.853
RUT
RT
RT
d) 1 1(12)(15.97)sin18.7 (14)(15.97)sin25
2 2
77.97
4. a) i) 2 2 2
115
4 5 2(4)(5)cos115
7.610
QRS
SQ
SQ
ii) sin sin115
5 7.610
36.55
QSR
QSR
iii) 1 1(5)(5)sin50 (5)(4)sin115
2 2
18.64
b) i)
ii) ' ' ' 180 36.55 143.55Q S R
5. a) 15 4 sin 7
2
44.43
R
R
b) 2 2 24 5 (2 4 5cos44.43 )
3.526
QS
QS
c) 3.526 8
sin sin45
18.16
180 45 18.16
116.84
P
P
PQS
d) 13.526 8 sin116.84 7
2
12.58 7
19.58
SOALAN 15
S’
S R
Q
1. a)
max
16 8 0
2
4(2)(4 2) 16
dva t
dt
t
v
b) 2
2 3
2 3
2 3
16 4
48 0, 0
3
48
3
48(3) (3)
3
36
s t t dt
t t c t s
s t t
c) 2 3
2
48 0
3
48 0
3
0 6
s t t
t t
t t
d) 4 (4 ) 0
0 4
t t
t
2. a)
2
2
2
max
2
2 3
, 2 2 0
1
2 12 3
2
4
max
v t t c
v t t
v a t
t
v
b)
2
2
0
2 3 0
2 3 0
3 1 0
3
v
k k
k k
k k
k
c) 3
2
2
33
2
2
2 3
33
5
3
s t t dt
tt t
d) 3
2
2 3
2
0
3 03
3 9 0
3 9 0
4.854
s
tt t
t t t
t t
t
3. a) 15 3 0
0 5
t
t
b) 2
2
315
2
315(5) (5)
2
37.5
s t t
c) 2
2
315
2
315(8) (8)
2
24
2(37.5) 24 37.5 (37.5 24)
51
s t t
or
d)
4. a) 2
dva
dt
b) 0
2 8 0
4
v
t
t
c)
2
2
2
2 8
8
0, 0
8
2
2 8 2
3
3 8 3 15
15 3
12
s t dt
s t t c
t s
s t t
v
t
t
s