MGMT 390
Strategic Management
Evaluating a Companys
External Environment
Thinking Strategically:
The Three Big Strategic Questions
1. What do we want to be?
Businesses to be in and market positions
to stake out
Buyer needs and groups to serve
Outcomes to
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Vocabulary and Dialogs in English
% 04.11.2013
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& Anastasia Koltai
' English Vocabulary
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A Typical Day
This material contains words, phrases and a sample story for your story about your typical day. (
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Time expressions
usually, often, every week
usually; regularly; often; frequently; sometimes; rarely; seldom; never
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Places in the City
Cities and towns
city; town; capital; metropolis; village; hamlet; settlement; port; resort;
health resort; seaside resort; winter resort; mountain resort; ski resort;
big city; large city; small town; densely popu
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Useful Phrases for Parent/Teacher Meetings
The following is taken from
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16. Meeting Child's Teacher at School
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A: It's nice
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Appearance and Character
Note: Words and phrases on one and the same line are NOT synonyms.
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Appearance
attractive, good-looking, beautiful, handsome, pretty, cute, nice;
plain, plain-looking, unattractive, ugly;
well-dressed, nic
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Business English Vocabulary for English for Special Purposes
Human Resources
Key Vocabulary and Phrases
By Kenneth Beare
English as 2nd Lan
2
H! WRITE YOUR NAME, STUDENT ID AND LECTURE N. BELOW !
NAME : ID :
1. (40pts) Let f(x, y) =11n(x2 y).
(a): Find the domain and range of the f(x, y).
9mm 7. )5 Cw/2).E'X59, X27g
IaavQ : ZR: ('00 , +50)
(b): Sketch the domain of f (:0, y), and the level cu
(2610 means problem 26 in 10-th edition textbook, 309 means problem 30 in 9-th edition textbook )
1
12.3
2610 = 309 : Note the acceleration is ~a = 32~j. The path of the projectile is
described by the vector-valued function:
~r(t) = h450 2 t, 3 + 450 2 t
1
15.3
30: F~ (x, y) = M (x, y)~i + N (x, y)~j =
conservative vector field:
2x
~
(x2 +y 2 )2 i
+
2y
~
(x2 +y 2 )2 j
satisfies the test for
M
N
8xy
=
= 2
.
2
3
y
(x + y )
x
To find a potential f for F~ , solve the equations:
fx =
(x2
2y
2x
, fy = 2
.
2
2
+
(2810 means problem 24 in 10-th edition textbook, 289 means problem 24 in 9-th edition textbook )
1
14.1
2810 = 289 :
Z
0
/4
Z
3 cos
/4
Z
rdrd
1 2
r
2
=
0
3
3 cos
d =
3
3
2
/4
Z
(cos2 1)d =
0
/4
3
1 cos(2)
3
d =
2 0
2
4
3 3
3 1
.
( )=
4 4
2
8
16
Z
=
1
1.5
10: Find sets of (a) parametric equations and (b) symmetric equations of the
line through the two points. (if possible). (For each line, write the direction
numbers as integers.)
(0, 4, 3), (1, 2, 5).
Solution: (a) Parametric equation:
x = t, y = 4
MAT 203 FALL 2013 MIDTERM I
NAME : ID :
RECITATION DAY (WED OR FRI):
THERE ARE SIX (6) PROBLEMS. THEY HAVE THE INDICATED VALUE.
SHOW YOUR WORK
DO NOT TEAR-OFF ANY PAGE
NO CALCULATORS NO CELLS ETC.
ON YOUR DESK: ONLY test, pen, pencil, eraser.
!
i
Opts
I!
2
2! WRITE YOUR NAME, STUDENT ID AND LECTURE N. BELOW !
NAME : ID :
1' (40Pts) Let fwd!) =.1n(y x2)
I
(a): Find the domain and range of the f (x, y).
05mm ; (MWQPKHQ /, 37202?
Ratgf jR 36:1731b00).
(b): Sketch the domain of f (x, y), and the level curve C
! WRITE YOUR NAME, STUDENT ID BELOW !
NAME : ID :
1. (40pts)
Uzi5+1; 17=f+j
(a)(20pts): Find the angle between ii and 17.
058:- TZ? : IHH) I Z
Wu Wu J; . J; 0
rs) 9:121: Cfoo
(b)(20pts): Find the area of the parallelogram spanned by 11 and 17.
glue 93% I
MAT 203
FALL 2013
NAME :
Practice Final
ID :
RECITATION NUMBER:
THERE ARE TEN (10) PROBLEMS. THEY HAVE THE INDICATED VALUE.
SHOW YOUR WORK
DO NOT TEAR-OFF ANY PAGE
NO CALCULATORS
NO CELLS ETC.
ON YOUR DESK: ONLY test, pen, pencil, eraser.
1
2
3
4
5
6
7
8
1
1.1
8. Find the vectors whose initial and terminal points are given. Show that ~u
and ~v are equivalent.
~u : (4, 1), (11, 4);
~v : (10, 13), (25, 10)
Solution: ~u = h15, 3i, ~v = h15, 3i. So ~u = ~v .
28. The vector ~v and its initial point are given.
(210 means problem 24 in 10-th edition textbook, 149 means problem 24 in 9-th edition textbook )
1
13.6
210 = 149 : f (x, y) = y/(x + y), P (3, 0), = /6.
x
1
y
= 0, fy =
= .
fx (3, 0) =
2
2
(x + y) (3,0)
(x + y) (3,0)
3
So
1
1
1
D~u f (3, 0) = fx cos
Figure 1: z =
1
2
p
x2 + y 2
3
c=0
2
1
-3
-2
c=2
c= 3
-1
1
2
c
=
1
3
-1
-2
-3
Figure 2: x2 + y 2 = 9 c2
1
13.1
27: f (x, y) = arccos(x + y). Domain=cfw_(x, y) R2 ; |x + y| 1. Range=[0, ].
38: z =
cone:
1
2
p
x2 + y 2 . The graph of this function is the up
1
1.3
14: Find the angle between the vectors (a) in radians and (b) in degrees
~u = 3~i + 2~j + ~k,
Solution: cos =
~
u~
v
k~
ukk~
vk
~v = 2~i 3~j.
= 0. So =
2
= 90 .
42: (a) Find the projection of ~u onto ~v , and (b) Find the vector component of
~u orth
(610 means problem 24 in 10-th edition textbook, 69 means problem
24 in 9-th edition textbook )
1
14.3
610 = 69 : In rectangular coordinate, x2 + (y 2)2 4 x2 + y 2 2y 0.
Substituting x = r cos , y = r sin , we get
r2 2r sin 0,
0
1810 = 189 :
Z a Z a2 x2
Z
(610 means problem 6 in 10-th edition textbook, 69 means problem
6 in 9-th edition textbook )
1
14.6
610 = 69 :
Z 4 Z e2 Z
Z
ln z dydzdx
1
1
1410 = 169 :
ZZZ
ZZ
dV =
Z
ZZ
p
dzdA =
cfw_x2 +y 2 16
2 Z 4 p
0
Z
Z 16x2 y2
16 x2 y 2 dA
x2 +y 2 16
0
16
=
4
=
0
(The numbering for most problems is according to 9-th edition. However,
2610 means the problem 26 in edition 10.)
1
12.1
Homework: 2, 21-24, 28, 56, 74, 78
2
12.2
Homework: 6
Homework: 2610 : Find (a) ~r 0 (t), (b) ~r 00 (t), (c) ~r 0 (t) ~r 00 (t), (d) ~
(2610 means problem 26 in 10-th edition textbook, 269 means problem 26 in 9-th edition textbook )
1
14.7
2610 = 269 : The mass of the cone is given:
ZZZ
ZZ
dV
h(1r/r0 )
Z
=
Q
Rxy
0
r0
Z
0
0
r3
r2
2h
2
3r0
=
2
Z
dzdA =
r0
0
r
h 1
r drd
r0
r2 h
= 0 .
3
The