Review Module (Vectors)
All,
Operations andComponents of Vectors
#1
(2 5 -6)
-
,
H -T
:
#
R
=
A
2i 3j + 41
=
# 11
-
F
46937
=
-
24i +
·
1-2
Jail
1
-
Vector
Head
4
-
,
2
4
-
9
,
,
,
XA
-9)
DC
-9
11
*x
Si + 5j
=
-
-
costy
by
H B
=
(1)(2) + (3)(4) + ( 4)) 1)
A B
=
20
-
A
=
71
B
=
<24
3
-
= 131
17
< 10
.
A B
.
AT
Si
=
=
B
Ver VetB #9
10i
=
A (1 6
.
.
,
0
=
48
B
=
<6
A
.
B
4k
,
F AB
.
3
B
A
-
IF/cost
↑3) > VctA
-
-
5
-
47
=
Vet A
Cost
=
IF/ cost
cost
=
Abs(VctA) Abs (Vet B)
Frost
0 193
=
.
8 = 78 868
#10
.
A
=
gi
-
5j
A2 =?
A
=
=
A2
VetA
4-17
B = <5
327
#x B
=
AXB
=
BXA
=
-
.
VetA
g
=
A= <
.
11
Abs (AXB)
=
Abs
=
Abs(VetAs"
=
Fd
F (4357
=
Frostd
D
F
=
+
1-1387
D
-
.
# 13
45 Joules
or
& (1 , 4 3)
T
15
2)
,
AXB
/F1/ABI cost
d
AB X AC
/ACI/ABI sint
=
< 731 4611371 491
-
1828 6543
.
.
A23427
> o
2K
=
llB
=
10
.
-
3 37
0 707 units
=
.
1)
V
=
B441-2)
a
=
131-17
Fa + FB
b
=
< 3227
=
#14
R
(X
=
bxa
10 9527
<X
.
159 4) kN
=
,
< 4
.
y,
,
-
axb
.
< 4
-
393
< 43
-
17
Y,
-
-
397
=
297
31 92
.
d
1017
Al
d
Bi + 4j + 2k
R
IAC/sint
=
.
.
=
IAC)
=
.
,
1AC1 sint
=
Net ABI
0 571-0 7613
1 5
=&
verABX VetAC
240040 3047
,
1AI/BIsint
=
#
.
-
sint
A 10 4 2)
-
.
=
AB
⑤
,
(6
7B
d
ad
16k
7
& 108 000187 371
=
=
VetF Vet D
=
100 (VctA = Abs (VetA)) + 60 /Vet B = Abs (Vct B))
(BXA)
=
W
= 95 N-m
=
B
<5-11-297
-
7
=
60kNSGi + j
VetAxVetB
< 5
A
100kN
A2 A A
+ 2k
W
42)
FX AB
=
VetB
.
W
EFA
=
Ver
/Al /BI cost
=
4j
0)
.
.
-
,
B (2 4 1 5
20
4167
2)
AB40 8
Gi + 5j
-
10z = 16
X =
-
,
.
1
GRN
,
-
aj + zk
-
W
#12
.
(5-287
=
XB
67
0 7427
.
-
=
=
.
B
Caltech-mode-f
A B
.
123 850 82 = 42 0580
2j + 8k
-
Cost2
- 0 557
=
=
xi 4j + zk(x
=
XA
.
.
B
9k
.
.
.
%
A = 5i
# 8
15 038
=
A B = ?
.
#
.
.
97
(VetF UctOAS : ADsUctOA
=
=
0 5570 74277
.
68 22
Foo
Frosty Frost2]
<Frostx
=
-
0 371
=
=
.
.
.
costX
,
ItS
F
#5
< 0 371
=
Frost
0 5570 74277
-
.
"
Fnet
#4
70 371
=
27
-
IF/IOA) cost
=
=
Directional Cosine
28i
#3
11
14
,
,
#
(5
:
5
1-12
=
VetA
↑n =
,
5
-
Init Vector (X)
14k)
-
83
,
,
-
9j
OA
Xi
=
4/a ; y
=
-
My
9)
>
-
SITISATION :
#15
B
3i 9j + 3k
C
PQ
# 20
1 3097
=
.
U
PR
E
7
A
2i + 4j
D
+ k
(PQXPR)
AD X AB
= AB sint
(X Y 2)
2)"12)(AD) (ABsint)
=
< 3-937
=
V
13
(UctAD XVetAB)
V
=
D
(Ax) (ecost)
-x + 2y + 2
/AXB) (9)
=
·
#21
.
Volume
# 16
A
#28
/
< 241)
=
V
(AXB) (
·
·
=
225 units
=
3
=
(2
PR
=
< 4-8-27
# 22
-
m)
1
,
X + y
yy
=
0
X +
by
=
=
d
3y
-
zy
4
-
4)
X =
17
X=
25/4
A
gas
m
>
C
07
planes will be formed
,
-
-
,
9. 9 units
B
(AXB)
(AX B)
0
C
·
-
-
=
-
21m
d
/AXBles
Al +
d
(5 1 4)
21347
,
,
,
# 25
0)
A
=
=
-
h) + (y k)" + (2 +)
6 22 units
#26
.
Ax + By + Cz + D
11X + 3y
-
=
3
0
-
57
-
D
=
16
(1x +
3y 52 + 16
-
=
-
(4 10 - () ; 77 9 2)
,
,
X
it g
+
52 + D = 0
1) (1) + 3101 5 (1) + D 0
=
-
-
,
,
87
-1
: H
E (CAX(B)
CAXCB = < 11
-
16/ units
23
(CA) (CB) sint
=
=
B2 + 22
=
6
=
r3
(X 1) + (y 5) + (2 + 3) ( (4(z)3
17
,
A
-
# 29(X
,
&
a2 + 12 + 82
1)
-
+ Cz + 1
-
d
(10
0
4(1) + (5) + 81 3) 33
m + g0m
-
=
=
-
C (2 -2
A
Axi + By ,
=
do
(g)(1) + (30 m)
21m +
B
#19
0
=
1/3
A
-
:
c
9m
=
M =
# 18
·
-
-
=
g
(AXB)
< 213387
=
4x + y + 82 33
# 23
m
> 2417
* Cros90
↑
2 12
1
-
=
0
=
=
+
+ + 10
z=
St-l
+
4y
=
y
-
go
8)
#29 -
-
=
+
by + X
-u(2 + (5 + 3) + (3 2)2
=
,
30
-
(xz X ) + (Xz y ) + (22 21)"
d
mi-jumk < m
<3 &37
# 17
0
=
0
100
Infinite
57
-
-
=
=
,
y-10
=
PQ
-
C <43
=
-
0
=
PQXPR
C
·
-937V
< 3
=
B
(AxB)
=
095
62 + D =
+
Abase Height
36 74 units
=
a
3( 3) + 2(0) + 9 + D
=
AD <24
A
< 18 12
-
2y + 2 + D
3x +
/AD XAB/
=
=
A
al
=
18
-
A
AB
12
-
,
2
<
(18x + 12y
-
-
=
(PRXPQ)
Area of the base
Area
<2-1
=
(4 10 -6) + + < 3
# 27
-
=
=
+ 10
37/z 0
=
Module Exercise (Vectors)
#
A = (xX
#9
x3)
costX
(Al
5551
=
X= ?
(A)
=
5 ti
=
Jen
=
((x) + (x2)2 + (x3)2
#x
f(x)" + (x2) (x 392
&
+
0 7845
=
-
.
cos" (0 7845)
=
X-axis 8
-
38 3260 > + X-axis
=
.
-x
,
180
=
-
.
38 324
-
=
.
11 6740
.
-
5
J(X) (x)" + (x)3
A(X Y 2)
AB
,
,
=
<1
B(ey
A B
15)
-
-
#B <hy XX ya 2)
=
-
1
2y X
=
-
(y
X =
2y
-
-
1
=
2y y
y
y5
-
-
1
=
4
-
5
-
Of X
2= 1
1
#3
/X y
=
#11 A
=
B
=
R
=
1
-
X= -
y 0
=
4X
i + 3j
B(i +j + k)
9k)
-
R = (4i + 5j
8k)
-
(45
4 + 1 +X
#4 X
1)
+
5
-
#5
8 >
y
-
#6
#7
C
=
=
4
=
< 0 146
F
=
# 13
1
A
A
-
k
F
#
costy
=
FX
=
F(0 78450 5883
coste
*z
=
=
costal
.
-
-0 194
#
.
Cos
-
(-0 194)
=
.
101 303
.
,
-
,
4)
He45
Ap sint
/API /ABI sint
=
APX AB
=
-
IABI
AP sint
=
14
=
.
B
°
d
&
d
=
W
6 , 3 , -4)
5
A7678)
#X *
D310003
=
-
FXA
=
·
D
FD
15=
F
-
F D
#15 Work
F
=
3
=
Asint
=
=
3 05 N
.
I
73i + 2j
> 98
A B
.
~
2i + kj
k
zi +
#17
=
/All BI cos 90
3(2) + 2(k)
B
A
-
=
=
0
0
=
3
24 t
=
2i +kj
zk 4
=
k
4/3
=
Situation
#18
Al3 , 3 , -2)
#- 20
10
307
A(33-2)
FXA
M
# 19
M
=
150 6 16-f +
=
.
E27-10
# <z 3
M
27
-
Fz xA
=
M
#20
507
50
-
169 12 lb-ft
=
.
A
X
.
ty' + 2
=
+4x +
2y
-
. x" + y + 22 2x y +z
B
-
c
x * + y" + 22
.
-
-
4x
gx
-
2y + z
3 (1 + t)
22 + 69
-
-
=
-
3
#22
69
=
-
130
=
x
2X
-
0-
=
z + 69 =
132x
5
=
3 + 3t + 4 - +
H
X
-
=
2 ( 2 + (t) + 4t
-
St
69
2y + 22
-
D. x ty" + 2 + 2x + y
# 21
(5, ,
Trial and Error-substitute
10 Xy
-
101 3030
3
=
↑
.
=
(4 6
(2, 2 , 1)
10, 0 01
Tz
.
,
d
.
.
(4 3, +)
APXAB
0 1967
2172 8)
-B
APsing
.
.
#
d=
F = F(costX
1901 251
82
2172 8
=
↑
2 =0
j
4i + zj
=
.
d
&
=
.
(VctA = AbsUctA)
F 3814
5
S
-0 8317
.
3000
=
Fcostz
-
R
=
-1
g + 1 +z=
177
-
F F(costx costy cost2]
# 12
-1
=
Sk
0 537
.
8x3
.
=
=
-
X = - 1 4661
-
11
X
12 + 12702
=
c
#8
=
5 + 1 +y
xy2
<4
R
*
97
-
2)1
<
2(xi + yj + 2k)
,
,
-
+ ( 8)(X3)
9k
-
2i + 8j
< 3
0
=
]
9X2
-
SITUATION 2
↑ (ei + 5j
.
-
# Use the choices to
determine the value
z
6
z =
1 +
-
-
x
=
X=
y
=
1
-
1
A B
*
4(x) + ( 9)(x2)
=
0 =
-
-
.
B(XXX
6)
X ,
,
141 6740
=
x
A 44-9-87
# 10
X = 5
SIMJATION I
#2
8-
+
#14
=
-
=
5
-
5
12
=
=
9
-
=
B
22 =
2 + 2(
-
4)
2 = 16
- 10
14 at 2 = 1-3+
#
(9i +aj
-
zk)
# 23
X
3y
-
(1)
g(9) ( 5) + D
-
1)
-
3y
-
=
-
-
z
-
5
y2 plane
y
- 12 3 X 1
=
-
1
=
# 25
-
1
X =
1
,
=
-
2
2)
,
plane
X2
X
=
z
10
=
-
,
0
#Le
-
3
0
=
# 2446667
D 2
,
5
=
X
(1
+4 = 0
2
-
-
2
2
1
-
z
(-1 0
,
,
1)
3
=
=
1
-
2
-5)
Review Module (Differential
Calculus
1
substitute
#
21 + 52 + 6
Lim
6
=
12 + 1
2 -
1
substitute
#2
Lim
=
X
X + 2x
-4
+ 2x
LiM Y
=
999
Does not exist
Lim bx32xt
#5
0
-
X
y
=
-
001
#7
<xi + x 1
=
X
X 1
1
-
=
x
0
-
3X-1
3
+
=
4)(x 5) [(x 4) + 3(x-5)
y 2(3x
+
=
t
#17
xX
inx
=
*
# 18
X2 inx
=
+
x
y
**
y
d
0 125
=
-
.
·
y
2
=
-
X
=
0 25
.
.
=
-
=
2
=
r2
↑ (n2 Gh(q)
=
02
-
12t-3t2
=
V
y
2/
# 22
2x
=
y
=
y
=
-
& ( 0)
0
=
,
&2 0
y
,
=
0
1
b
-
m -
'=
10
-
#19
0 1250017
y
di + de
=
di
=
de
D
Mx + b
4
2
+ Xi
182 + (10- x)
=
(6 + x 2)0
2
=
=
y
·
5
0
=
-
b
=
(Slope
Critical Points
y
X(nx(x 12
=
o X(* ) (nx(x)
x 1 + 1nX(x 12
+
=
=
=
X
=
0
.
307
y
=
9x2 + 8x
0
:
9X2 + 8X
X
x
=
-
=
9/9jy
0jy
is
-
-
-
X
D
X +
=
(34 + x2) 12x) + (324 + (10 x() (10 x)(1)
=
to the curve
+b
.
(62 + (2 52 + /182 + 110 2 5)2
=
-
.
.
D
26 miles
=
140mi/nr
30
zero)
·
C
30 + 40t
mi
D
·
and
=
-
25 95
-
i
25 95)
-
27
Gmilhr
.
,
35
(0 -27)
,
D
.
=
-
2 5 mi
=
# 23
@f(-89
5
-
1
CR are
:
.
-
(e) (10 -X) ( 1)
by
((2)
(182 + (10 X(2)0
+
112
42(42 + xz)(tX) + (45)(18: + 110 X(2)
=
0
mx +
y
2x + 4
4
=
=
=
=
2(2) + b
-
.
D
.
X
3x3 + 4x2- 27
=
33 51
=
18 mi
35 mi
124998
unit
di
D
(1, 0)
Tangent to the Curve
2 2x
-
Vone
de
9m/s
=
6 + 0y = 0
X2
(4) +
lomi
2 m/s
=
substitute
2yy'
a(4)2]
Ucone =
3h + 12
omi
+ 10y + 5
-
-
r2
4
=
X
Vz
2x +
=
=
a
=
1
=
Uz
t
12- 6t
x + y" ex
n
=
=
a2j
=
0
= V 37z
6t2-t3
=
=
.
+3 -1
hi + 6h
Ucone ET(-h2+ an) h
Vone = h + 6h2)
- B
1 + 0 0001
issing the choices
=
-
(n-3)2
-
3678610537
-
y
32
~
.
0
(n 3)2
r2 +
=
n 3
0 0122
#13
x
= 2x (x
.
<
y
solve
-
torn
=
ng2
3
N
A
-
1
=
+
Stan" (inx)
=
y 1))()
,
Veone
3
1
Normal
ay y [x 2xinx]
= [x 2xinx]
=
+
4
=
-
= x 2(y) + 1nX(2x)
#12
-0
=
Get
Yeseexsanxseex)
Iny
#11
x
a
Iny
=ex(x
x
(2) (3x + 4) (x 5) + 6 (x-5j3x +
y
zy(z) ya)
#21
=
yzdx
,
-
use calculator
eX(2x)(x
+
-
S1
-
#10
y
+
dx
:
=
=
0
=
y-
-
=
-
=
t
=
.
Se
+
(E) + + an"(f)
an
=
0 3678794412
-
dy (3x 4) (2)(X 5) (x-5)(2)(3x 4)
=
24
-
x ex +
=
=
=
# 16
+
;X
=
y"
2X + 3
-
-
y2
rx(X
y" =
y
2X2- 2X
=
+
t
+ (1)
Da
=
X 22x
=
y
-
y (3x + 4)2(X j)2
=
#9
x2 +
+
E
t f , + te
0
6xy'
=
-
=
6X
2) + 3
I
#8
y
=
Caltech
-
-
6X
y"
2
3X
GXy' by
-
-
1(2x2 + x 1
=
=
y
=
y
,
-
No Horizontal Asymptote
X 1
1 1)
0
=
3x2 + 3y2 64
y
y
=
0
[6x(y) by]
2
&
y 2/1
=
0
=
+
-
zy Gy
+
X eX
# 15
g
=
-
+
y
=
99g9
Horizontal Asymptoti t
11
=
.001
1000
1000
3 =0
x 1
X
.
and 0
.
X = 3
#6
3X2 + 3y2
and 0 6667
-10 -
=
Vertical Asymptote
=
y
↓
bx
-
6x4
3x2 + 3y2
S
substitute
#4
x3 + y3
-
.
.
substitute
#3
# 14
and 2 001
0 6667
2/3
=
-
.
com
#20
"
value of 2
a
as
1 (3 5
=
:
.
51 35 mi
.
% 0)
60 (t -100)) + (30 + 40t)2
c
49 92mi -
.
.
.
. 50 05 mi
B
1
-
2
-
Use the choices
A
60 (t
-
D 49 98 mi
.
.
# 24
X2 + y2
↓=
13M
y
y
Z
=
132
y
2(5)*
=
A
5 m/s
+
=
0
:
0
- 12m/s
=
# 312xy + X + y
y2
,
2xy + x + y
=
exy + 2y + 1
= [b +
A = [x2) (-12) (5)(5)]
&A-59 myse
A
y
5
=
-
+
y
=
0
y)
=
0(x
,
1)
1
2[xy" y]
+
.
4
+ 1 +
+
12bh
14
or
.
4 a( 1)
=
2[Xy' y]
=
0 25
=
5
=
A
#25
122
=.
2x zy
2(12)(5) +
1
-Ginst t s
#30
+
5
X
=
132
=
+
2y + y"
+
0(x
=
,
1) : y
=
-
y" 4/3
a
=
(i + <y 12332[1 + 7132312
·
cost =
#26
13 cost
13
R
X
=
at
n
13 ↑(5)
n
=
If
=
3
-A
&
# 27
9ft3/min
=
Vone =
/
R
-
Fran
Vrone =
10ft
=
.
I
# 32@(2
# (0 2543)
.
5=
10
U
= (0 75
a
(0 75(6)2)A
.
r= 05
.
=
=
0 318f /min
+
.
.
V
=
V
=
.
n)
=
15 m/sec
(h) (3r - n)
[rn-n]
=torn-3h2]a
Y
Z
600
X
comp
22 = xi + y
2 =
&
dt
=
-
/
-
-
200t
0 5 (400 + 100t
.
**
-
=
=
2x4cS68
10 + (10t)" 2(20) (10t) Cos 60
1400 + 100t
za 12
2) 5 (a 12 (15)
=
dax
6 152 + 182
23 43 mis
200t[ (200t 200)
-
6 2 = 4
-
8 66mph
.
=
18m/
.
=
=
-
12x'
# 35
12ax
=
64 8 M/s2
.
y'
y X2
=
+
An =?
&t
,
20 mi
=
=
=
At =?
zpo
(2 1213)2
.
y
V 5 f /5
10t
5/2
or
=
+
=
.
after ehrs
# 29
2=
#36
-
0 01326 m/min
=
=
=
= 4113)(8) 3182]
&
5(2)"
Ax 0
-
.
g2 12x
=
Ay
y"
r2
=
constant
6
=
[gay y'(y))
2[(5)(9y) + 18(18)] 12(0)
0 01326
-6 m/min
-
5)
UX
2
=
-
y
2yy'
# 34
↓ dr
E
(y k)2
08 ,
25
=
4/3
12x
V
13 m
-
(1 + (4)
=
V
#33
#18
(n Y (30
413
+
1+
112
=
Vy =?
.
a
2
=
-
(x +(z)2 + (y
=
i
r
y1
-
.
k
(1 + ( 1)2)
1 +
=
It y
y (1 + y(z)
-
1-
=
(x n)
(0 5h)2 (h)
#10 25n2) (h)
Off
-
.
x
=
k
A /3
2 1213
=
-
rad
1 Jes
du
5ft
S
-
=
y
R
&X
sin)
=
=
2XXI
A
=
-
64 Sm/s"
·
Module Exercise /Differential Calculus
#
12 + X-0
Lim
#2
12
=
X
Xi +9
him
#
13 y
y
05
=
.
5X + 2XF 3
x
Y
X2
?
X
uu
(y y 1 ) m(X xi]
-
=
-
-
-
2x
-
-
y
2x 3 = 0
-
=
-
X
4
=
-
y(X 5)
Area = a sing
= sing [2a]
3
x=
X=
in
18 in
-
1/2
=
=
2
00,
68
7
=
=
3
-
S
~
&
' = 3x2 12x + 8
y' z(2) 12(2) + 8 (0 2)
X3 -X = Px +
=
=
W
# 19
=
-
-
-
#3
@y 0
X3 2x2 + 8x + 5
=
& [sinGo[2(18)(27)]
=
x3 + 3x2 2x
-
#
4y
-
=
X2
X
-
-
X (t)
# 14
2
3
t
=
(513)3 4(83)2
47"
X
St
X=
-
A 420 89 in Je
-
=
=
-X+2
X
-
X
2)
-
-
-
x3 + X2 + 2X
-
2x
2
-
2X2
y
#5
X
An
=
-
X+2
2X-2
-
-
=
4X
-
(2 x)(2x +
=
Y
x3 + 3x 2X
2x
-
-
y
-
#8
y
=
813
t
=
0
1)10 99
xz + y
# 15
0
X(x + 1)
=
A
(x
B
0
.
y
~
.
.
/25
Area
#17
x(ax
=
X(k) + (nX(x)
.
-
=
1 + (nx
=
1nX
Y
-
1
1
I
(x2 + 1)x
y
=
-
=
C
=
C
=
+
25
1
# 10
x3 + y3
zy 0
-
y
2(4)y'
y
3/4
-
y
=
y
=
ty' y
0
=
600 m2
=
bh
=
1
C
=
280h + 140b + 280b
0
=
1)3(x
=
9x2
y=
=
=
3x
y" ↓ [18 -ux)
=
(i8 6x]
-
; y
=
10
-
(d)
↓
# 21
280(7001642010s
=
3(X + 1)2
2(120) (100) + 2 (52 4) (50)
=
.
-
16 800
,
a
250 x 120
(i) 420b
+
"
180(600) (1) (b)
=
X
# 22
=
3 2t
=
-
+ 420
# 23
X= +
=
y
w
Kokm/min
E
# 24
20km
min
(20
R
+
-
x2
=
=
-
=
&
2)
2(20 - ( - ) +
=
2(20
8
-
(1 + ( 1(2)3/2
-
2
R = 1 414
3y"
=
.
2
y 2(1) 3
2(20 16t)
y
-
-
# 25
A
=
-
X-
=
) ( ) + 2(20 14t)(4)
y')1 + y'z)
=
Ru
)
-
=
4X
y"
y
R
2yy
1(I + 1)
-
+
=
3X71(1 -
-
8
t 0
,
zX
2
2t + 3
1
y"
y
) (20 16t)
+
-
(1 + (y/(2y312
=
y
-
t
-
It
+ 3t + 2
=
=
dy -5/2
=
y 5t 7
=
5
dX
-
=
+
yu
-
-
+4
00kph
130
124
=
s
0
( 16)
-
X = 3
E
600
B" 20km
-x3
+(9X2 x3 6)
[isx -]
0 =
=
=
+
=
12 y
Xi +
=
2(130)
0
280() mob
c=
+
+
120 = + 502
=
280h + 420 b
y 3)
#
=
b 20
#18
1
yx(x + 1) + (x + 1)3
=
S=
es-exey
5x + 25
as
y" 3x(2))x 1)() + (x 1) 2(3)
=
h
140 (2n + b) + 2806
=
y -
x(3)(x + 1) 2() + (x + 1534)
=
y
X(x +
-
-
N
#11
S2
-
4y + 5x 41
=
=
-
-
=
5/4
-
c
X
=
342
3x2 +
=
2(3) +
S
X
X ,)
-
E(x 5)
-
16 =
-
280n + 420b
=
=
S
120km
58km
y
(140h + 1404 + 140b + 280b)
=
&
Py
0
I won
Al
#9X* + y2
PKPMA
.
280b
-
(x4 + 3.x" + 5)
=
=
(y 4)
=
-
=
m(X
=
,
0
=
140b
=
82
as
:
1
=
0 25
.
0 25
=
o
.
(y -y )
4)
=
2(5) + 2(y) y
-
.
9 481im
-
# 20
Cyy
2x +
20002
-
y
y
t
-
19998
non
#7
37"
1
H
.
=
y
:
2
x
#
0
=
2yy' 44
=
4X2yy" + ya
=
4
2(2)(y") + 1(2) 4
=
t
C
=
1 05 min
J (20
=
2
=
2(2)y' 4(1)y"
.
4
-
=
11
05))2+ (20 1611 0557
-
.
y'
.
#26
b
=
y+
H
=
1
=
05
.
# 27
V
=
5 + 4 sint
d
#28
#29
3X
X- sing3 + Xe-cos22 = 39
-
#30
62 + 8
Veylinder
TrTh
joint
link
↓
=
-
Hin/s
h join
=
-
-
Module rise
(Integral cals a
Use the choices
I
A -0 1667
dX
B
#
2j2(
jj(4x
0 04762
.
.
0 2143
.
D
.
-
.
C
0 04762
=
.
-
0 04762/
.
.
(4x +543)dydx
y
+
Ey4)ax
(2(4x3 + Ex)dx
A
24
.
y 4x
#3
; ( y
=
-
( m
=
=
·
19 6)
,
2y2 4y + 48
=
(4 -4)
,
er
A
=
=
A
) (X2-Xc) dy
j-(du
=
125/3
sq
.
units
y 9X
#4
=
B
.
9/4
a=
T
I All
(R
I
=
y
=
9/
/
# 5 r = Isint + Ecst
j
X
Y
A
x
=
πrz
=
+ y = 2x + 24
r=
A π(v2)2
=
I
=
2
-
Icost
52
(34-
y
r
=
=
3 + 3 cost
~
r
dr
d
=
-bin
i
inta
=
Ey
-
=
ds
=
Y
-
"
I
-
-5
-
7
[x
-) x (ax
X
a
+
Ix
0 , 4)
i
Ax
=
.
dy
Jay)dy
-
dA
=
+x
=
-
·
.
) X Y dA
Ixy 2) xyydx
=
!) "x(*)
,
Ixy
# 17
21 33 units
=
.
FO
100 N /On
x
=
W
10 000 N/m
=
,
.
.
=
!
#18
.
k = 10, 000
Im
106
10
0 2
8m
90 " (10 000x)dx
0 2
,
x 3/4
A
ax
=
[xi dA
.
c
,
dA ydX
=x(4-Xd
F
12 19 units
=
=
j-(4 -XY)dx
A 14/3
(2, 0
:
=
[xn
#16
J y 14-sty) de
=
(Xc dA
=
&
J" y'dA
=
51 C units
=
O
3-
·
=
=
3
-
dA
=
f) () Jax
·
z
O
X2 44
# 15
In
-
=
Ey
X=4
2π)d-d)
4
Fy
4 ,4
xdAax
dy
-
SA
)
=
=
N
=
GOON -m
.
I
#
I
04
.
Mode 3-3
X
Y
00
00
w
=
All
v=
211 +
ast)
Prost
by observation issing the graph
W
=
2π
i
X
=
2
y
=
# 19
8
=
5
=
·I
-
?
2π
N
W = 4 Nim
y 84/55/
W
.
.
=
#
63 795
=
B
.
0
z= 0
-
=
X
-
!
X= 0 6
2
-
=
8) "((x)(8 x)dy
q 7
and choices
W
#
iT (1) 2
Im
3M
9
.
.
g zl
·
X 44
#14
af(x) -2 cost y Isint
e
=
Yet
1 sint
X=
75
y
m
-
/111/IIIIIIIIIIIIIIII dy
·
·
28- Isint
X =
2 to
2) Jetecost+ Isintat
in
a
=
S Fxdx
=
(350(4)(50 x)dx
X
W
50m
=
W=
+
315 000 N-m
,
315k)
(Differential Equations)
Module Exercise
#1
Order
#2
X
xyy
2
=
2
=
y
,
Degree
;
=
-M
1
=
=
1 +y
Xy 1 yz
yay
=
-
zydy
2[
y 5/2x2
1 +
B.
+
=
du
=
I dy =
jeay 2)
H+ y
It (3)
C
#
2
((X)
=
2y 8x2
-
X
2
=
=
&(2)
=
X
-
3y Xyz
8X
+
-
y
=
gidy
xdx
=
2
-
2
-
Jj 2 dy (xdx
=
A
.
=
+C
*
-
yX
.
+1
y(xz
=
y(x* + 2)
-
-
+ 2)
2
=
=
y x
Xy + Cy
=
yx2
0
&
D
=
=
2
.
=
YX2
y
=
=
=
(x
=
=
A
=
k
Vy(y) (Vdy
T
=
Ts-T
2x* + C
=
y(n(y)
y
.
B
=
500F
=
100 60 400F
=
-
100 T
-
79 520F
=
.
Am Mf-in-Me-out
=
=
3
=
an
.
a 500
-0 05dm
.
-
=
Qin-QCout
am
am 3(-3(d)
am-3-0
c
.
=
solve fore
+
am
m
obs In 15-003m)
·
due
&
=
fat
t + c
=
m
when
=
0
1
and =o
36 2854
-
.
solve for time when m = 2 and c -31 9878
=
.
A
#B
GP
= (P)
pindP
(p
y"( 1)
+
2x3 y + y
y (2x + y?)
=
-
,
c
=
100 50
Q=
i (2(yi) 1)(c)
C
I
.
-
+
=
.
-
U = 5 0 05 M
(2vi 1)(C)
=
.
=
du
(n(2v + 1) + c
=
.
Ts T
vQ
+
=
c
i
In 1202 + 1) + 2
=
Y
0 3466
3 17 hours
=
To
y3u 2v 1
↑
-
20 48
=
=
T
t
.
loogal xibsalt
0
(tydy =F-dV
-
%
y
.
A
kt
vy3dU
-
y3
-
=
20
2x + C
0
=
-
=
Pre- (2)
=
=
[vidy vyaV y dy
gdy(2v" + 1) + vydV 0
=
k
yz)dy 0
+
=
Po &
=
vey' dy + vyV + Vydy + yody
(n14)
Ce
-
to
ya dy (2V + 1)
20
Poe-t
=
U 3UUU(t)
.
#11
# 12
Po C)"
= Doe
0 3464
-
.
yar) + (vy +
+
=
20 48
vdy + ydv
P
3464(3)
k(2)
-
8
.
=
↓C
P Lekt
.
↳Pr Poe
A
+
Po e0
=
2xdx
H
.
=
,
=
126 23 hours
=
Po 7071 OGTS
.
#10
Vy
=
20 000
=
,
XydX + (xi +y)dy 0
X
%3
01824+
=
,
,
t
.
.
=
2 12 200
20 000
X
A
=
t 3
2(n(x)
2
20
=
.
Po
=
&
-
(x 8xdX
(1-0 90)
.
=
=
k(38)
0 01824
=
P
t 0
#9
t
=
=
Po(1-90%) kt
Poekt
:
k
Xdy
X
P4
38hr
=
& Noe
0
=
dx
.
=
=
/P(X/dX
-
Pn
(t)
1566 6542 years
=
halflife
0
) Toe"4244X10
=
-
=
4
.
58%
-
+
.
#8
+
=
A
C
4 4244x10-
=
Fo(1
xdY/dx
2(n(x/
I
= EX
=
E xya
#4
=
2) kaXX
5/2
=
=
8x =
y 3
,
E
=
-
-
MIx a
+
y
k
%
+=
(2y - 8xzdx
(25)
=
.
(Cy 8x2)dx + xdy
2(n(x) + C
empty
=
-
- 2
=
Pn Poet
40(1-1 1 % ) Poe
2(y 4x2)dx + Xdy
#
R
&
It
Exact
zex +
+
&R
#7
2
=
-
=
(n)( + y))
:
Xi + 4yX +
1 +y
=
N
2
=
84
D ly2
&
3
1 + yz
=
(X + 24)dx + (2x + dy =
y)
#5
-
2dp
pl
kdt
=
=
(kdt
t 0 338 minutes
=
.
.
di
k(P
=
k(100)"2
20
=
k
2
=
&t 0
=
,
2To
4 = 100
k(0) + C
=
C = 20
25
&t 12
=
=
2(12) + 20
A P 484 rabbits
=
.
(Analytic Geometry
Module Exercise
#1 d
(1 2)
#9
,
(5 6) and (7 , 2)
,
,
A
d
-
t
=
5
&
e
A
:
(Xz X , ) + (Yz Y )
-
,
=
5x + Ty
6
13
=
# 21
,
5x
+
·
⑨
&
5 x + Ty
y
=
y
18
=
-
A
3) + (7 y)2
-
-
5 D.
(2 5 4 5)
# 10
.
.
,
(X 6)
19
=
,
=
6 +4 5
74 +30 =0
x
.
Xc-X ,
fan 450
-
Ly
S
>
12 = 05x
T
q
B
Y2 -Y,
tant
-
12 isnits"
=
-
Fi
&
132
intersects each other
determined the
distance between them
Thus , we cannot
0
.
D
+
-
=
⑧
((4 10 + 14) (10 42 + 2)]
=
Line 1 and Line 2
12=0
-
7y + 30
-
10
.
=
B.
#22
X2 + 2 5
.
(X b)
#2
(9 4)
+
,
X2
,
Midpoint
(7 , 3)
=
x+ 9
=
2
7
;
2
y
5
=
2+ 4
2nd
#3
·
(3 5)
Xr
,
3159 +(x y)
d
,
=
2/
f(2 3) (
-
5
=
5)2yr
-
3(5) 3
-
2)
(yz y )
&
555
yu 1
=
( 1 1)
(-1 3) and (6 5)
-
7
X=
-7
52
m =
:
Xzx,
(q y)) m(X
-
-
=
-
(6 g) + (3 7)
Midpoint
5 units
u =
-
+5
=
.
C
#6
3V
(10 ,5)
r=
7y
Equation of
a
Ym
=
Xo = X , + r(Xz
(4 2)
4)
,
a
A(d b)
,
yr
2
=
b
y,
+
r(4z
b + q(5
-
y)
,
-
-
-
X
X,)
# 15
=
-
1
C
=
0
=
0
63
9X
=
-
9(0) 7y + g 0
Y 1 29 A.
=
Ty
k)"
26
0C
.
d=
between
(1 6) and (3 3)
.
,
(l-3x + 16 3)
-
02
=
13 A
=
.
0
=
Al + B2
v=
131342
r =
( 5)
·
=
-
v = 5 units
+ 9
"
Yz -41
(6 , 4)
(0 y)
y
-
y
,
d
.
=
Ax + By + c
r =
=
·
-
-
,
d
54
-
-
circle
-
(3 -2) ; 3x + 4y
# 25
y-intercept
-
=
(x 3) + (y 3)
.
m(X2-X1)
=
4 =
=
-
=
#26
70 6)
-
Downward a
y
=
-
x +x + 1
-
4
distance between (6 ,4) and
b)
(2 1)
=:
1
=
#
-
2
=
Xi)
+(10 a)
a +
=
=
-
-
+
radius ,
(X n)" + (y
-
0 = gX
.
&
12
4
=
-11-5x-by
,
3
=
.
iTz
*
2722-10x by !
radius rdistance
=
Xm
is 4
# 149x 7y + 9
X-intercept
9X T(0) + 9
9/7
=
21 =
21 : :
center
=
=
5
I
-
q(X 6)
y g
-
radius
# 24
D = 1/2
y 4X + "/2
-
18
=
+
(5 , 0)
and
,
ey=8 le
the slope of line +
,
y2 41
M=
,
=
-
(6, 9) and (13 18)
# 13
(6, 3) and (9 7)
r=
=z
50r3
.
(1 6)
4(-1) + b
=
-
-
#5
1
=
1
3-0
=
,
,
-
-
slope
A
,
=
=
=
0 3
=
⑨
7
-
-
-
32 + 5X + 3y
y mx + b
E
=
Ym
⑧
-
=
-
-1/1
midpoint
,
,
5 + 3/5(5 + 5)
=
2 [0 4y 2x) [28 3x 2y)
(6 44-TX-28-3X- 2y)
(1
:2)
6 C
.
Xm
-
12-4X
=
=
-
X=
,
7
-
8 X
-
I
y, +
=
-
-7
M=
5 1
-1
=
-
(5 0) and (7 3)
#12
-
,
2 +
-
=
-
315
d
#4
+
+
=
Xr
m
.
X + r(Xz X , )
=
=
je -5)
3
=
C
,
,
r
-
(5 0)
·
=
21
,
0
=
A
,
,
M=
X
( -4 3)
·
.
1 (X ; 7) and (8 7)
3
=
# 23
.
(-5 ,2) and (1 -4)
#11
6+
8 B
=
(0 -4)
i
↑
.
.
(6 0)" + (4 +4)2
-
·
#
(1 1)
,
↑ S
=
d
(0 8) , (4 5) and 73 ,4)
,
,
,
=
d
#17
(1 4) + (1 5)2
=
=
-
-
d= A
Y
A
=
25 units"
4x by 12 0
4X 3y + 8 0
-
d
=
d
# 18 (x + 3y + 4 0 d
3x + 4y 6 0
=
!8
y))5 + 32 3) (4 24 + 4))
A=
A
(1 1) (4 8) (7 2)
,
,
A
= 1
A
=
A
=
A
=
,
,
,
and
-
-
=
25 units
(3 5)
·:
O
+((2
#19
.
(1 3)
,
d
=
(1 1) (7 2) (4 8) (3 5)
59 5 units
?
B
,
A + B2
y" 4y
-
ex +
,
y2 4y
is the only letter that
.
=
-
8(X-1)
+4 =
=
by
-
# 29
11 0
=
Ax + By + C
8x. + 4
110ft
100
10t
Al + B2
2(1) + 5(3) 11
n
-
10
loft
-
J2 + 52
#20
=
distance from (2 , 3) to (3, 8) minus radius
n = 26 ++ c.
1 11 units B
.
=
.
x + y" + 4x 6y 12
-
X + 4x + 2= y
-
=
0
d
9
by
(X + 2) + (y 3)
-
+
-
=
12 + 4 + 9
=
j2
V= 5
S
=
(2 3) +13 8)
-
1 d 2 07 c.
=
.
-
-
5
# 30
Im
52
·
(5 8)
,
distance
=
d -r
=
32 + 22
c= 4
.
3m
"Im Ima
I'm
x=
2C
X = 2(4)
X= 8 m
B
.
o
A
By inspection
8x + S
match the condition .
-
,
-
d
.
C
t
,
(y-2)2
# 28
Fowl is & (4 , 0)
=
,
,
↑
Ax + By + c
since the two points are on the line 2x + 3y + 4 0 ,
check Whether the points are on the line given.
·
3) (7 + 8 24 + 5))
-
=
Using the choices
2
/Yi
-
a= 4
4 units C.
=
=
-
-
[11
+ 56 + 20
=
=
42 + 32
.
YXXX,
=
=
-
B
=
08
,
y2 16 x 16 49
#27
1 12 8)
-
=
-
or
#8
A
-
=- between two points
,
10 units
,
1/
-3
=
Isse (1 , 1) and (4 5)
since the figure is square
directrix is DX 3
=
,
A
.
(Analytic Geometry
Module Exercise
X2 + 4y" 2x
# 31
-
(x2
Sy + 1 0
=
-
1) 4(yz
2x +
-
+
y + 1)
-
(x 1) + 4(y 1)
=
-
-
=
1 + 1 + 4(4)
-
4
4
↑
=
↑
a= 2
b
LR
=
1
2
=
LR
# 92
2 06
.
,
=
e
(1
(
LR
=
a=
g2
# 33
a= b + c
1:2
=
.
3
=
B
-
1 unit
=
b
b + 22
=
=
=
=
Eb
4 27 units B.
.
5
.
-a
# 34 e =
9xz 4y2
-
-
36x
+
9 (x2 4x + 4)
-
-
9)(X
-
2)
Sy 4
=
4(yz 2y + 12)
4(y 1)
-
-
=
-
=
4 + 9(4) 4()
-
36
36
(X-2) L
-
b
C
2
# 35m
2
y2 +7z
=
:
C
C
=
1 80
A
.
.
T3
-
=
=
y y,
-
y
-
X,]
E(x)
3y
=
2x
# 36
m(X
=
=
-
2x
3y
=
0 C
.
(7 38%
,
Y
X
=
r cost
=
Tc0) 38
X
=
=
55
y
.
sinf
=
Tsin (38)
4%
=
.
(5 5 4 3) A
,
.
# 37
0= 4
.
.
(4 0)
,
(x 4)2 + (y)" 42
=
-
"so"
X
-
,
8x ++ + y 1
=
x2 8x + yz
-
0
=
et)" S(rcos) +Using O
↓
=
-
COS45 = sin 45
r
# 38
X , = TcoS60
Y,
=
Tsin60
=
35
.
=
=
& cost
(3
5+ 4
.
,
=
0
33)
+
A
.
16 06
.
6 06
.
Y3
2
d
-
d
=
8 60 units
.
A.
-
2
.
532
Since the person
Vo
.
a =
7
-
.
zero (0)
92m/s202
S = ?
6 72 +
=
mr
=
&1500)(2) 2(250)(4)
-
f
B.
MN
=
↓ = 490 5
point
X
200sin 30 = 390 5
-
.
.
-
=
0 30/390 5) =
0
=
[Fx
200c0s30
a
.
c
100(X- 07
A
.
ma
=
.
S = 1 83m
=
117 15 N
.
.
-
117 15
50 a
=
.
# 27 m , (vi) + Me(r)
A.
1 12 m/s"
=
.
O
=
Impixive Momentum
Area + t
.
.
Vf
# 26
V 303
2 - 7 92)S
.
200X
T
.
1200singon
14
Vf == Vo + zas
6 70 m/s
=
in the
.
-
0
=
time
and end
displacement is
the
#2
=
start
50(9 81) = 490 5
# 16
displacement
Average Velocity
#I
Use
(Physics)
Module Exercise
=
m , (V) + Mc (vi)
. 0132730) + 2(0) = 0 013 (170) + 2(X)
0
.
Vo
#3
Uf
13 4 m/s
=
.
0
"
Vo2 +eas
=
13 42 +
2(7 92)S
B
=
.
S =
# 17
11 33 m
=
Vo
=
Vf 5 m/s
jz
:
.
.
[Fx
Vf" Vo + 2as
0
=
S = 5m
02 + 29(5)
=
ma
=
X
Pcos30 -0 30(490 5
..
P
W = mg
Psin30)
-
.
.
a = 2 5 /s
.
267 857NA
=
.
=
S
:
ax = 0
-
Vo = 10 m/s
Not +
19 6
-
.
lumi, a
.
t
#
2 second
=
Vfy
.
D
#28 e
.
Vfy
19 62
-
V=
ViRhM , 194)
.
LM
.
630
-
S
Vot
=
at
+
b
Voyt
Eat
+
=
2
zW
o
migh Em[r2-voY
9 81(10) E(v = 152
-
(0 7685)
=
-
.
.
5 78mc
V
.
At
#8
-
5 3666m/s
=
4 sin30
r
.
.
IOON
-
14
f
Fnet
·
.
20a
=
.
15
V 8 m/s
=
,
25 , V +: :
=
Uf = Vo +
2 503m/5
a=
19 6 NC
.
as
Vf= 82 + 2(2 5)(10)
.
.
.
Ve
=
.
.
=
&t
.
.
=
0 25 (146 2) = 36 55
50 0525N
=
50 0525
0 80
CF
=
.
&
muz
ma
MN
=
146 IN
N =
CF Eso)
v2 gr
0 So
=
-
N
3
V
# 10
#
radius , u
you
# 20
D.
.
md
#9
.
c.
.
2019 81) = 196 2
3 5 m/s2
=
A.
.
19 32kph
=
4 COS 30
=
At
F
312 5 m
=
3 6
+
.
252
=
10
=
677m/sc
.
.
v" (9 81)(0 8)
=
.
V
# 11
t 0
c =
30rad/s2
A
,
=
.
2 80 m/s
.
24 rad/s
=
a =
A
.
#21
N=
5
=
zm
f
-
8
F
Not + Eat
=
24(2) + E(30)(2)2
=
108 rad
A+
=
W+
#It
"
=
0
t
+
↑ net
.
=
81)
W=
19 62 N
.
F d
-
-
=
-
100(x)2
Wfriction
-
19 92 (2 + x)
Wspring
19 42 (2 + x)
-
.
=
a
=
-
8 17 rad/s
B
.
X
.
#22
84 + 7 8 17)t
=
Em(Uf2 Vo2
-
.
-
Not at
=
.
842 + 2(a)(432)
02
84 rad/s
=
.
=
:
=
0 25(8)(9
We
Wfwo" + 2 at
Eky2
- (200) (x)
W=
M(mg)
=
spring
work due to
MN
Wfriction
242 + (2)(30)(108)
=
Wf
# 13
=
=
Wo' + Lat
=
Wf
Faet
=
540 rad
MN
Work due to friction
432 + 108
=
=
N
z
=
8+
MXM
<
n
t = 0 > t = Is
NIm
200
Ma
mi
Enet
432 rad
=
Fn(t)
=
0 045(45)
,
.
Fact)
# 25
m
=(8)/02 53
.
(v)
=
100(X)
0 688 m A
=
=
=
mcVs
.
10 28s
=
In (0 00 35)
2 + 10 28
Fn(t)
.
t += 12 28s
.
2 025N-S
.
# 14
Ut
=
No Area
Vt 30
-
# 24
+
4/2 5 m/s A
=
=
15
a
48
=
M(V)
Ac
=
-27250((4)
.
578 57Nc
=
.
.
.
S
So Area-moment
=
=
30
S =
(15)(15)(15 + =(15)) +
4500mC
.
(20)(15)/
&
A
-
250
.
500
=
-
=
500
# 15
t(S)
F(t)
(2(500)
=
0 045(45)
500
Area
(15)(15) + (20)(15)
=
20
Al
Fr
A
B
.
a (M/s
15
=
=
.
.
.
Total Time
1
=
-
8
=
4(1)
=
8 =
2VA + 2UB
2VA + 2VB
VA
m/r= V 2)
=
V
An =
.
e
VA-VB
.
e=
8 = 2(VA) + 2 (2 4)
=
1820s20t
.
2VB
2VA + 2VB
=
VB = 4 M/s D
# 30
-
"
va
#7
# 29
10m
0 7685s
S = 8 cos (201
S =
MA(Val + Mis(Vi)
2VA +
=
=
.
S =
=
VA 2 M/s C
VB 2 M/S
E 15m
.
t
=
=
sm/s
8sin20t + E(-g 81) Its
-
=
4
VA-VB
VB-VA
.
,
S=
8
D.
.
-
VA = 0
#19
.
solve for time t
-
.
0 84
M=
Scos20(t)
S=
2) (4) + 20)
26 39 M/s
=
below the horizontal
C
#O
0
Ye = VA-VB
MM A s
.
-
-
f
=
A
VA-VB
VAV-
=
.
( 19 42)" + (10)2
=
0
MA (V ,) + MB (V )
F
M/s
.
=
e
.
= Vot at s (2)
:
69 m/s
.
#18
2
' ( 9 81)(t)2
=
.
19 6 m
a+
.
50 (2 5)
↓
#4
0
=
max
velocity
using the
A,
=
Impulse Momentum
=
V = 5 m/s
#25
=
-
=
is
my
- 1500s (2)
"
is at
-
100
(V-No (
B
.
Velocity will decrease
Thus, max
after
velocity will b @t
=
2 seconds
2 seconds
Il
Mr
=
1 6 MIS
.
·
Module Exercises (Data Analytics 1)
SITUATION :
I
# 20
15 ! x 5 ! )(2 ! )
99 79 85 #2 C8
66 93 98 9969 54 56 68,69 6 10 ,88
,
,
,
,
,
,
,
,
,
,
* Write
in
order
54 , 56
#
,
,
2314567
I
Range
#h
#3
,
,
,
,
,
99 54
-
=
R
70 19 85 88 998 99
15
Highest-Lowest
=
R
9
I
I
61 , 66 , 48 69 69
,
,
,
B S
5
45C
=
.
Caltech
STD
Mode 3-1
-
XOn-1
Variance
Sx
0
=
Free on
-
15 718 A
=
.
Q2
=
15 7182
=
.
Sx
#4
Mean
7
#5
Median
=
#6
Mode
=
#7
Q 1 and Q3
Index
.
.
.
10
D
.
.
69 and 99
= (15+
Index
= (15+ 1)
=
4
Index = 12
Q
=
66
·:
Q3
93
=
C
.
De and Dgo
= (15+ 1)
Index
Index
:
0 32
=
Index
.
=
Index
#9
76 933 B
=
A and C
.
Index
..
#8
247 067
=
=
D2
=
54
(15 + 1)
Dgo
:
14 4
.
IQR and SIQR
99
=
47 + 47 + 48 +?
# 18
IQR = Q3-Q1
=
=
=
93-66
47 + 47 + 48 + X + 1
=
possible since the set is in order (lowest to
make X as small as
highest)
.
B
.
,
,
the value of X can be as low as 49
5
y = 74
&xx1
.
Make
X = 49 , then solve
for y
4) + 47 + 48 + 49 + y
120 ways
=
SITISATION :
B
A
=
53
.
.
0
, 1 , 2 , 3 4 , 5 , 6 ,7 , 8 9
,
,
358
#12
soT
# 13
10 x 10 x 10 x 10
# 14
7x6x5x4
# 15
10 , 000
=
Tot A
D
.
=
.
BG BG B
31X2! X2! X 1 ! XI !
24 ways A.
=
# 16
ASSASINATE
#
# 17
5! x Y!
2880 ways A.
# 18
5 ! x 41x4 ! x2 ! 4 !
# 19
53
i
27/2
SIQR = 13 5
#11
?
* 53
IQR = 27
SIQR
+
=
(4 ! x 4 ! ) 2 !
=
=
100 , 800
=
way
165 888 ways D
1152 ways
,
A
.
=
1024
:
45cc
nCr
# 21
12(4
# 23
# 35
2880 ways
=
10
#36
11 25
#25
Cace1
WA - 923
Cast2
WNA-gC5 =
462 ways
-
P(p)
70 % L
P(f)
20 % V
10 % M
=
50%
=
50%
P(p)
=
40%
P(t)
=
60%
=
50%
P(p)
g(g 10 80)" 10 20)
P
=
0 176
=
05
=
0 5
P
.
=
n2r poque
.
.
.
10 (g (0 5)"20 5)4
=
.
P
84
=
P(t)
=
P((p)
=
P((p)
=
0 205
=
.
WWWWL
# 37
P(w)
=
P (2)
=
p
=
0 57
1
=
.
.
P(4)
.
.
P(Det)
# 38
420
97 %
=
Thin
Thin / garlic and
,
oregano ,
13
13
Standard
# 30
# 31
New sauce c/mgarlic
,
13
Y3
p
=
(3 + 13
p
=
19
Thin / bits
P(s)
=
P(f)
of cheese
3r
New
,
P
(4(0)(0 03)" 10 97)"
P
=
0 0266
.
,
P
-
38
(113) (518)
+
36
5B
Pl
a
(13)(2/3) +
(113)(45)
:
t
IR-2
PLAR)
P
#4)
P(AIB)
0 82
.
0 78
.
=
0 82
P(W)
P (w)
=
P(m)
4%
=
=
=
-
60%
P(P) -1 %
-E
P(f)
96 %
-
BOXA-1 2 3 4 5 6 ,7 8 9
,
,
,
,
,
,
12
BOXB- 1 2 3, 4 5
,
,
p
=
p
=
=
00%
=
40%
,
.
.
PIN
=
.
.
P
0 022
.
=
#43p
=
Eve
+ #44
P
p
"
102 x (0 60) 10 40)10-Y
.
(X
=
7
to
10)
.
0 382
.
(4(2)(6(1)
10C3
=
0 3
.
sample space
5225
=
P
=
(26(2) (26(3)
5225
.
= (0(x) (0 60)4(0
=
=
P
nCopique
5/48
comple spaceoa
p
# 45
Solve for P
28C5
=
40)10 X
-
.
p
=
0 325/
.
P
p
13129117234
=
p
19145
P
20(5
20
.
3/11
10/19
P
.
17 NV
((4(a)
=
0 0586
00
0 01/0 6)
=
p
"2
=
3V
taller thana ft
(0 6) (0 01) + 10 4)(0 04)
.
%
40
,
P
13/52
>
-
# 42
taller than
6 ft probability of Men and women
140
.
SITISATION :
6ft
Taller than
P(M)
0 155
1/52SH3() (i) (e) (i)
15 =
1H > 13/52
ID > 13/52
2)
.
.
.
=
-
0 g5
#
0 78
=
G
.
.
=
5/10
-
26 0%
=
0 83
=
.
3/10
-
zeren
=Bs
Isrn2"
.
.
0 1852
=
2R-40
#40
PLAR)
9
0 03
-
13
Yu
p
x
4 (4(2)(0 70)(0 36)" x 0 70
wi fresh basil
sauce
UrnB"
P(AND)
# 34
.
.
.
=
-
#33
,
F,S , S, F S
10 %
30 %
=
13
JonAs-5W-5
P(A)
# 32
ND , ND , ND D
,
=
.
t Probingmencoe
P(D)
0 0195
1425
=
XV
# 29
.
.
ND
3%
:
.
# 39
# 28
0 57
x
.
.
PINDe
.
.
.
0 70 x 0 50
50%
·
.
(4(0)(0 51) (0 43)"
%
0 43
+ 10 2)(0 4) + 10 10/10 I
# 27 4(p) 10 1)(0 5)
=
3
.
.
ways B.
#26
p
9
#24
=
nCpgr
%
210 ways
=
495 ways
=
=
=
P
PE80
1725
=
2025
=
228
Module Exercises (Data Analytics 2)
#1
X=
less than 4
X
5
=
↑ (xx4)
(xe
=
=
5X2
↑ (x(4)
#2
X
5
0 2650
=
.
3 life insurance
=
-
X !
X =0
per week
X = # of
policies
P(x)1)
= 1 P(X
P(X 0)
=
>
-
-
-
0!
P(X > 1)
#3
.
Mode 3-1
=
O
0 9502
=
M 70
4
45
=
.
-
.
5
65 5
,
65 - 2
=
-
#
1 11/
2 227
.
80
↑ (65 < x80)
=
Q( 1 (11) + Q(2 222)
↑ (654x480)
=
0 8536
↑ (65 < x480)
=
1 2hrs
=
70
65
.
80 + z =
.
85 36 %
.
2- score
.
M = 40hrs
z=
.
.
Illin
513
P(X > 42)
↑ (x(42) =
R(5/3)
P(x > 42) = 4 8 %
.
#5
Trial and Error
Use the choices
H
.
001
.
B
#6
N=03
P(X 5)
=
↑ (x 5)
=
#T
C
/
↑ (x
=
.
O=4=300
=
D
.
= nx=
10 Black)
2 8%
.
=
n(xp q" Y
+
-
probability of drawing #10 black card
9
=
2/52p(x
9 = 50/52
=
P
#10 Black) =
=
3
.
846 %
(1(1)()' (
IModule Exercise (Basic Transportation Engineering
SITISATION 1 :
#1
Speed (kph)
Time (s)
i
58
3
40
Mt
Arithmetic Mean
M+
45
5
=
200
5
=
40kph
=
#2
Ms
35
=
5
Mean
harmonic
Ms
=
50
5
+
+
4 5 3
+
+
38 72 kpr
=
.
30
200
Ms
#3
9
q = Msk
20kph
=
3800
=
2hrs
ven-1900 very
1988 ven/nu
k
#4
Ms
30
=
.
Msk
=
M/n
48 28032
14 ven/km
k=
q
=
9
48 28032x14
=
.
675 92
=
.
ven/nr
Space headway
#5
#6
My
50 kph
=
qmax
qmax
=
ven
3200
=
hr
(((
3200
=
k+
=
(E) (E)
256 ven/km
# vena
3600
=
kf
Kf
240
=
A + 0
A
=
248
44
=
u
0; M
-
248
=
+= 60 whenkt
41
-
60
=
0
=240 + B (60S
B=
k = 240
wen/km
=
Bo
k = A +
240
(5)(E)
=
M+
when
240
=
>
-
-
4
equation
k
-
9 = Mk
3344
=
(40 74)k
k = 88
-
ven/km
SITISATION 2 :
& t
X
M
=
&t
Omin
St
=
X
=
=
20min
((t
-
20)
00/15 +
=
=
4t
-
8220)
8t
-
/20]
<
It
↑
21-20
⑮
6: 00
7 : 00
# 10
#S
8(20) +
21 + 20)
=
-
t = 60mins
;
=
#9
qmax
=
St
=
8120)
-
-
9 max = 80 ven
4t
4/20)
4t
t = 7 : 00 AM
# Vehicles
Affected
=
I
TIME
DEMAND
9 18
4006
cumu
CAPACITY
4000
2968
2960 -
1040
in
3500
7508
2960
5920
-
15801
2500
10
, 000
2960
8880
-
1120
12-1
2000
12 , 000
2968
11848-
-
1
2
-
14800
2 3
2960
17760
3 4
2960
20120
-
-
#11
11 AM
# 12
1580 ven
9
SITI ATION 4
R
(6000 500x)(3 50 + 0 5x)
=
-
.
-
2) , 000 + 3000x
I
2) ,000 + 1250 x
8
1250
=
-
Traffic
=
3 50 + 0 5X
=
.
3 50 + 0 5 (2 5)
.
Traffic
.
.
$4 75 ven
6000
=
6000
=
4750
.
=
=
250x2
# 15
.
Toll ,
-
250x2
500X
-
2 5
X =
#14
Toll
.
1750X
-
-
-
500 X
500 (2 5)
.
vere/day
.
# 10
(4750) (4 75)
R
=
R
=
.
$22 562 5
,
.
SITIATION 5
# 17
6 : 00
-
375 + 380 + 412 +
7 : 00
#1S
PHF
#19
DAY
=
PHF
398
=
1557 ven/ur
4
0 945
=
,
=
1557
=
0 945
.
DHV
# 28
R
1648
=
A(100 x 10%
=
ADTX Nx 365xL
0 0
A , (100 X 104
=
.
0 0
.
36000(4)(365)(10)
#,
=
Az
.
=
15 768 = 7 884(1 + 0 25X/
.
.
.
X
#21
IV
4
=
0 4x2 36x + 1000
=
.
X = 20
N = 440 accidents
#22
SR
Injury + Fatal
=
Property Damage fingury + Fatal
3
0 24863 =
+
(13 +x)
.
961 + 293 + (13+ x)
X =
fatalities
12
#13
AADT
#24
↓. 14
=
AADT
910017
=
1482
2200
# 25
Dractor
Dractor
=
=
2200 + 1300
0 63
.
* (100 X 104)
2
36000(2)(365)(10)
7 884
=
4t = 4 (0)
# Veh Affected = 240 ven
Cumu
n
20km/ no (k)
=
g5 ven/km
=
SITUATION 3
15 768
.
160
Module Exercise) Elementary Surveyinga
ElevB
#17
Elev A + harp + his
=
=
250
Temperature correction
#1
Ct
0 005
ElevB
"m/ (20-is) (30)
11 6x10-
=
.
.
T
Ct
=
.
#
.
.
#2 (
19se
48024'
-
=
8
=
50z8'
-
11 6x10 0 (5 5 63218)(472 90)
=
Ct
=
.
3 4679x18
-
·
.
3
.
472 90
=
3 47x10-
-
.
MDC
I
64 25
3
64 18
#4
SITISATION
0 0351
.
03
=.
0
:
Em
0
=
Em
Azimuth
=
Dep
⑧
+ 565 6854
+ 565 6884
800
700S40DE
6745(SE)
4 5600
N45000E
-
1 % 01
-
& 20000'W
-
.
W
# Measurements
LEC
2
Y2
4
14
71
3
13
LINE
118
6
10
A-B
1 25
B-C
870481
EA
361
87048"
=
2
=
C
=
CB =
12
2)
FS
3 87
Eler
H
192 23
196 .
.
.
.
=
h
191 81
196 84
4 25
192 59
194 07
.
1 9714
.
5 25
14
5 03
188 99 1194 24
.
.
8 19
3 57
DEP
LAT
56 56
-
111
-
.
119 43
=
LEC
=
RE
DISTANCE
.
.
196 41
2 94
200 76
.
/205 7
-
=
196 4)
-
.
188 99
.
1 42m
=
.
STA 5
190 67m
=
.
o
#15
tan"
=
-
=
0
.
(1
.
90
1
-
.
56
.
85)
206 26
=
206 26
-
DISTANCE
677 97
-
BC
616 05
CD
690 88
DE
783 82
EA
970
-
.
.
-
.
.
759 93
192 06
-
.
.
.
.
=
=
(131 02)
=
690 89
-
x
tan"
=
8
=
-
=
M
+ 1678
.
tanso
40 s
X
401 3422
.
X=
tan 180301
+ 16
+
.
=Lat = 0 68
.
=
1199 4854m
X
1 2kM
=
.
.
358)2
79 07
N19 870E
.
CLatitude
=
249 90-249 90
.
.
.
mode 3 1
-
I
Y
1868 940 68
.
X =
0 42
z Dep =
.
( ,835
=
.
B
.
ID
# 29
.
=
-0 09
.
-
-
=
.
SITUATION :
Elevation B
Elevation C = 111 356m
B C 111 356
B 111 356 + 289 9842
-
2
.
In
# 20
C
:
CT
343
.
=Dep
.
Corrected Distance
+13608
ExX3900
R = 3000 mm
= 1803,
-
.
0 1
=
180
R
206 26
961 55
48
26/127
.
.
Lat,
=
249 St
.
# 30
CDep
Bearing
.
678 94131 02678 358
191 04
.
ep
593 21
166 19
.
.
Fat
117 75
.
.
=
.
Dep,
344 90-364 905
.
.
=
Lsin 23341
*
B
2 25
.
L
-
.
.
=
228 003
.
228 001 cos 23034
=
:
.
208 99 m
=
.
y
228 003 sin 23034
=
.
=
91 16m
.
ElevA +
Im
DEP
LAT
667 70
.
=
L
)
Escost +
2m
"
X
250341
23-341
1
=
.
45-y-2 25
.
=
ElevB
63 50 m
.
# 39
1 : 23
=
AB
# 27
.
d
=
H = Los
>
49
:
E
LINE
6
BS
:
ElevB
.
Flat
.
0572963540"
.
(99 5)(3 50 1) cos (23037'
=
# 38
19 49
BS =
=
.
98 69
↳
S
.
# 14
-
.
KS + C
=
y
-
ITATION
144 g5
L
# 37
.
SITUATION
Sta 4
.
.
&
.
.
#12
.
.
GO
.
.
.
.
n = Sta 6
.
.
=
36
# 36 H
.
RE
121 5608
580 5 W
=
DPDcD
36124 9003
.
10 83)"+ (19 47)
=
x=
.
*
1758
52 46750 4438
-
.
# 25 LEC
-
.
h
1198 86
190 67
2 45
Ec
A
.
.
-
7
he is
-12.560
.
=
19
200 76 190 67
=
h
.
.
.
.
DPD
8438
.
20 4315
=
tan"
=
.
sug
.
# 11
.
Bearing
1%
H
STA 5
-
n = 10 09 m
.
.
.
.
.
Dep
.
71 962053 7170
.
y
965 0009
6902 ERETE b.34)
.
# 26
# 10 n STA T
4 29
7 29
-
.
.
Dep = 1 6902
-
3
89 8
=
y
-
.
.
SITISATION :
25
20 4315x
-
-
N36 74"E
47
.
58
=
Im
109 870
=
-
#
RE = 1 : 1758
#98
-
88 - 874811
BS
42 3935
# 33
=
.
=
2045
SITUATION
.
205 2121
SITISATION :
.
-
LB = 88
Lat
So
50
-
12 =
-
88
D
North
IN
50 8558
-
.
11 0148 + 11
=
C
B
B
.
.
=Lat = 1 0148
.
:
-
1800 + 46 % 81
606 2178
+
543 8156
.
3466 34
92
8
.
.
N580E
420 yz
.
.
350
-
-
966 34586059'W
5 1
B
=
.
1-16182 4063/
182 87 = k(1 85) + 0 30
Lat
400
A
t
11371 3116
.
8091 2032
=
LINE
#35 L
BEARING
Due North
23
VALISE
6
183 29
16182 4063
=
A-B
# 34
400
-
4608'
=
=
DISTANCE
.
ANGLES
5
-
.
y
#
SITISATION
.
29
.
304 12
-
3
-
B-C
Azimuth = 2250081
3- 4
.
331030'
MD
12
0 01368
=
#22
0 0203
.
MD
LINE
-
=
7 30
D A
I
.
SE
+
-
/I
.
SE
°
-
#21
1045
2
I
2A
C D
SITISATION :
.
SITUATION :
STA
.
.
.
TRIALS
=
36
-
%
3
472 8965 m
=
360
=
Azimuth
↳ 30000
MD I Ct
=
TD
#
33279 8516
SITUATION :
two short
3 47 mm
TD
#6
-
.
=
A
Azimuth
# 20
m
.
#5
304 12
.
.
+ 183
.
# 32
S
-
.
Ct
=
62 04
42046'
-
0
-
KDt(
=
=
#3
-
-
# 31 DMP
.
z
0 00022m
-
.
.
.
62 46
-
DMD x Lat
120 835726 1337
-
.
109 43
3 1
540 25 m
63218)(38)
0 21 mm too short
=
2000 tan 8015 :
.
=
DM
120 83
-
.
2000 +an 2015
0 0675(212 + 250 +
=
Elev B
DEP
T
47 39
-
23
here + Eleva +
.
1-2
m
.
=
5 63218
=
= A 4(5 5
-
C
Elect
# 18
SITUATION
1200 tan 180901
.
.
651 61
=
kAtL
=
0 0675(1 2)" +
+
364 12
.
Nig OE
.
H
#(
tan002230)
H
=
183 34 m
.
) =
Module Exercise (Route Surveying
T
#1
F
# II
140
-
1145 916
240 = 420
.
R
=
.
L
=
GRLS
803
6(190 986)(80)
152 7832
=
190 986
3
R
=
R
D
X sc
soon
R + an (85/2)
=
282
R=
R + an( = (2)
=
=
E
.
.
X Sc = 5 5851
.
D
1145
,
D
=
9
428
, Xx
=
.
cos() 190 986 + E
.
.
=
.
.
E = 15 08 m
.
SITISATION :
#12
#2
:
160
tan c
250
i
=
YSI
T = Rtan (#(2)
Y
8
250 = R +an (22 6199)
R
60
=
18 70
.
TR
Lc =
424 06
=
.
D
a=
z
37 3801
.
(c =
#5
=
.
37 3801
a=
#4
22 6198
(658)(37 3801)
.
=
180
.
50-1000
1145 916
Re
.
0
↳
.
R
286 479
=
.
TRf
180
Lu =
/
iT (186 479)(40)
.
=
150
Lc =
#6
300
TitTz = TP
190
=
70
180
,
T2 = R2 tan (*(z)
=
Pr
it
180 tan (182)
T2 = 28 5092
.
180
T1
:
=
41 4908
.
, = Ritan (F(2)
T
41 4908 = R , tan(1/2)
.
Ri = 394 76
.
#7
> 160
Tz = Rztan (F2/2)
T + Tz = 84 5
.
84 5
200
.
38
,
38 6 + T = = 84 5 45 9 = Ratan (202)
U
.
.
.
.
Ry = 260 3
Tz = 45 9
.
.
Pr
in
#S
SITUATION
..
.
LC
in
229 8
=
.
, 50
#9
Y
150
300 4
-
COS 15 =
300
,
y
10 2222
=
.
R 30 + 10 2222
-
COS 15 =
15/2 007 5
.
.
R
Long Chord
38
7 51
.
Ri = 580 4
.
# 10
:
sin 7 5
i
ZOOM
1
S
sint
=
8
15 4668
=
30
.
I = 15 4660 X2
,
I
°
=
30 56'
=
=
L-
80-
PORTL , "
805
40(190 986)2(80/2
.
YS
600m
=
-
# 38
.
.
1380u
-
5 5851
15
.
a)
Short tangent + Long Tangent
Xx =
a = 11 3099
a
1 3943
190 986 + 1 3963
7 50
=
P=
cos(E) =
=
79 649/
.
-
-
-
-
-
-
-
ENGINEERING
MATHEMATICS
Differential Calculus
#1
-