Years of Crisis,
19191939
Previewing Main Ideas
SCIENCE AND TECHNOLOGY In the 1920s, new scientific ideas changed
the way people looked at the world. New inventions improved transportation
and communication.
Geography Innovations in transportation allowed
Circles
11A Lines and Arcs in Circles
11-1
Lines That Intersect Circles
11-2
Arcs and Chords
11-3
Sector Area and Arc Length
11B Angles and Segments
in Circles
11-4
Inscribed Angles
Lab
Explore Angle Relationships in
Circles
11-5
Angle Relationships in Ci
CHAPTER 30 TEXTBOOK QUESTIONS
CH 30 SEC 1,2
1- Explain and give examples on how Alexander III and Nicholas II continued with the
principles of autocracy. Were these actions successful? Why or Why not?
2- What steps did Lenin take to industrialize Russia?
Description de la chane logistique de carrefour
Les fournisseurs :
Carrefour joue un rle proactif auprs de ses fournisseurs, en les accompagnant dans lamlioration
de leur dmarche socitale. Depuis la cration des produits libres en 1976, Carrefour travaille
CALCULUS III - FINAL EXAM
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Name
Given P(3, —4, l) , Q(3,2,1), R(——l, 1, ~2), and S(7,0, 1), ﬁnd
(3 pts) a set of scalar parametric equations for the line that passes through
the points P and Q .
Calculus III, Exam 1: Review Solutions
1
1
Calculus III Exam 1 Practice Problems - Solutions
1. Compute the following:
h
i
(a) limt1 (1 t2 )~ + (tet )~ + (sin t)~k
= limt1 h1 t2 , tet , sin ti = hlimt1 1 t2 , limt1 tet , limt1 sin ti = h0, e, 0i
i
R1h
(b)
Keith Evans
Contemporary math
Chapter 18 Written Assignment
Pg. 673
4a) The linear scaling factor for the dollhouse is 12
4b) If both the houses were made of the same material, their weights
would also be the same ratio of 12 to 1.
Pg. 674
10) The wording
4. Rotation of the sphere by 180 about the X -axis has the eect : (X, Y, Z ) (X, Y, Z ). Recalling that the projection from the sphere to the plane is given by : (X, Y, Z ) Thus Y X +i . 1Z 1Z X Y +i , 1+Z 1+Z
(X, Y, Z ) = (X, Y, Z ) = and 1 = (X, Y, Z )
roots, so it must be the constant zero polynomial, i.e. p(z ) = 0 for all z . Thus (z 0 ) (z n1 ) = z n 1, for all z . (b) This follows immediately be equating the degree n 1 coecients in the equation in part (a). (c) This follows immediately by equating
we have
n
cos(j ) =
j =0
n j =0
= = =
1 ei/2 ei/2 ei(n+ 2 ) ei/2 ei/2 ei/2
1 ei(n+1) 1 ei
eij
=
ei/2 ei(n+ 2 ) ei/2 ei/2
1
i ei/2 ei(n+ 2 )
1
i ei/2 ei/2
Now we know
ei ei = sin(), 2i
so i ei ei = 2 sin() is purely real. Thus i ei/2 ei(n+ 2 )
1
i
2
I.2
1. Note that the problem asks for all solutions. (c) If z = 4 1, then z 4 = 1, thus z 8 = 1, so we only need consider 8th roots of unity. Since z 4 = 1, this rules out 1, 1, i, i. A simple calculation reveals that the other four are solutions, thus
Thus the equation |z |2 2 (az ) + |a|2 ) + |a|2 = 2 , becomes (x u)2 + (y v )2 = 2 , which is the equation for a circle of radius centered at (u, v ), which in complex notation is the point a = u + iv . 5. We have and taking square roots gives |z |2 = ( z
Homework 1 From Gamelin : Complex Analysis
January 21, 2007
Throughout this assignment we make use of the notation z = x + yi where x, y R, i.e. z = x, z = y .
1
1.
I.1
a. |z 1 i| = 1, so |(x 1) + i(y 1)| = 1. Squaring both sides gives (x 1)2 + (y 1)2 = 1
5
Let us conclude these notes with a discussion of the most disconnected set possible the Cantor set. Example 20. Let C denote the Cantor set (cf. .244 in [1]). Recall that C may be described as the set of all even ternary numbers; that is, C [0, 1] consi
4
hand, E = A B, and so A B B; in particular, A B. But A B = , and this is only possible if A = , contra our assumptions. Whence, A1 = , and indeed, neither A1 nor B1 are empty. Therefore, E = A1 B1 is a separation of E, which is therefore disconnected as
3
Example 12. In Example 8, we saw that C1 = [0, 1] in the metric space [0, 1] (2, 3). Following Lemma 9, this means that for any point x [0, 1], Cx = [0, 1] (indeed, this is easy to show directly). On the other hand, if y (2, 3), the reader can readily v
2
Hence, what we have shown is: if G = A B is a separation of G, then each set U G lies entirely inside either A or B. But, by assumption, there is a non-empty common intersection for all sets U G say x U for all U G. Then either x A or x B; again, wlog,
CONNECTED COMPONENTS
Recall the definition of connectedness (2.45 in [1]). Definition 1. Let (X, d) be a metric space. A subset E X is called disconnected (or separated) if there exists nonempty A, B E such that E = A B, and A B = A B = . A subset is call
CONNECTED COMPONENTS
Recall the definition of connectedness (2.45 in [1]). Definition 1. Let (X, d) be a metric space. A subset E X is called disconnected (or separated) if there exists nonempty A, B E such that E = A B, and A B = A B = . A subset is call
p(b) = some a X such that f (a) = b. It does not matter which a we choose, but there will be such an a by denition of f [X ]. We are placing the members of f [X ] in the pigeonholes X . By the pigeonhole principle, some pigeonhole has at least two occupan
well-founded. Even though a member such as (4, 3) has infinitely many other elements below it (for example, (3, y) for every y), any decreasing chain must be finite. Hence, the Ack program always gives an answer.
48
This program is a well-known example in computer science, since the function computed by this program grows extremely rapidly. We wish to prove that this program always terminates, and therefore defines a total function. Counting down from x is not good e
5.3
From Partial to Total Orders
Given a nite partial order, we can extend it to a total order. For example, suppose we have a set of tasks T to perform. We wish to decide in what order to perform them. We are not totally free to choose, because some task
5.2
Analysing Partial Orders
The shape of the partial orders (N , ) and (Z , ) are different from each other. The number 0 is the smallest element with respect to the natural numbers, but not with respect to the integers. D E FI N I T I O N 5 . 6 ( A N A
5.1
Hasse Diagrams
Since partial orders are special binary relations on a set A, we can represent them by directed graphs. However, these graphs get rather cluttered if every arrow is drawn. We therefore introduce Hasse diagrams, which provide a compact w