Citizen Anna
By James Poniewozik
Americans love selfmade men and women. And
who was ever more literally
self-made than Anna Nicole
Smith?
Described
by
acquaintances as a flatchested teen, she shaped,
nipped and tucked herself
into a living hood ornament.
Name_
Calculus
February 15, 2013
Period B
Directions: This assignment is due on Thursday March 1st at the start of class.
1. A cylindrical can is to be manufactured to hold 50 cubic inches. Find the height
and the radius of the base so that the smallest a
The American PageantCh. 8: America Secedes from
the Empire
1. What actions did the Second Continental Congress take in 1775?
2. What traits did Washington display as commander of the Revolutionary Army?
3. What was important about the colonial invasion of
Minor Parties in the United States
NOTE: Party list is not maintained by staff. It is here solely to show you some of the ideas
citizens in the US come up with for party names. Use caution and judgment if you decide to
explore any these topics on your own
HONORS PRE-CALCULUS/TRIG
POW #5, DUE FRIDAY, SEPTEMBER 30TH
Name: _
1.) Page 161, # 18.
2.) For what value(s) k will the function f ( x) 9kx2 7 x k have imaginary roots?
3.) Find all values of k such that the graph of the function d ( x) 3x2 2kx (2k 1) wi
American PageantCh. 2The Planting of English
America-1500-1733
1. In 1600, most of the New World lay firmly within the grip of
.
2. What diverted Englands attention away from colonization in the 1 st half of the 1500s?
What changed this equation?
3. What
UNIT II Test Review Sheet
Election Unit (Political Parties & Political Spectrum)
Standards:
1. The major political ideologies present in the American political system
What does it mean to be liberal or conservative? think about their characteristics
What
Whats the word? The Things They Carried
Pick the most important word from the first paragraph of the novel.
Discuss and use evidence for support to share why it is most important.
WORD: _
Discuss the questions below.
1. What does the word you chose mean t
Your name:
Your lecture number and/or time:
Math 294 Prelim 1
Monday, June 30th, 2008. 7:00 8:30 PM
This exam should have 7 pages (not including this cover page), with 6 problems printed on pages 1
through 6. The last page is blank and can be used as scra
MATH3320 HW11 (Solution)
Outline of methods and proofs with remarks
p.153, Q1 We will repeatedly make use of Theorems (5.10) and (5.18) and then
Theorems (5.11), (5.12) and (5.13).
(70/97) = (2/97)(5/97)(7/97) = 1 1 (6/7) = (2/7)(3/7) = 1.
(263/331) = (33
MATH3320 HW10 (Solution)
Outline of methods and proofs with remarks
(1) x2 1 (mod 561) if and only if x 1 or 1 (mod 3), (mod 11)
and (mod 17) respectively. Therefore, the Chinese Remainder Theorem tells us that there are exactly 8 solutions modulo 561, wh
MATH3320 HW9 (Solution)
Outline of methods and proofs with remarks
(1) Since s18 0 (mod M19 ) and s30 0 (mod M31 ), we conclude that
M19 and M31 are primes. Next, s10 1736 0 (mod M11 ) and s22
313912 0 (mod M23 ) so we conclude that M11 and M23 are not p
MATH3320 HW8 (Solution)
Outline of methods and proofs with remarks
p.64, Q36 Chinese Remainder Theorem.
Q39 This question is tricky. We will see how to use the Chinese Remainder
Theorem to solve it. Let m = r pki . So we have
i=1 i
#cfw_0 x < m : m|x2 x =
MATH3320 HW7 (Solution)
Outline of methods and proofs with remarks
Q4 27 28 36 1 2 10 6 (mod 13).
you dont have to calculate the number 1 2 10 = 10! before doing
modulo 13 because the product is large. You can just carry out each
calculation modulo 13: 1
MATH3320 HW6 (Solution)
Q1 The fundamental solution is the smallest positive solution, which is
(10, 3).
General form is (xn , yn ) by Theorem 10.10 where xn + yn 11 =
(10 + 3 11)n .
Q3 The 4th positive solution is (32257, 12192).
Q4
(a) The least positi
MATH3320 HW3 (Solution)
Outline of methods and remarks
Q1 Since all the factors of 15 are 1, 3, 5 and 15, we have to solve for each
one for a and b where (x, y, z) = (a2 b2 , 2ab, a2 +b2 ) for possible values
of x, y and z.
For 1, there is no good values
MATH3320 HW2 (Solution)
Outline of methods and remarks
Q30 Use the identity
a3 + b3 = (a + b)(a2 ab + b2 ).
Q32 Do not use brute force. Re-write the question as:
33a + 39b + 47(100 a b) = 3998.
and solve for a, b N with 35 b 40.
Q33 First, solve 455 = 10a
MATH3320 HW1 (Solution)
Outline of methods and remarks
B solution exists if and only if the gcd divides the number on the right
hand side.
D For those who are unfamiliar with doing induction, I am going to write
out all the steps with extra details here.
MATH 3320: ASSIGNMENT #1
Due Monday September 14th. Try as many of the following questions as you can. Note that
grades will be based not only on correctness but also clarity of exposition. For the computational
questions you can use a calculator to perfo
INTRODUCTION
TO REAL ANALYSIS
William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus
Department of Mathematics
Trinity University
San Antonio, Texas, USA
wtrench@trinity.edu
This book has been judged to meet the evaluation criteria set by
the
Chapter 3 Continuous Functions
Continuity is a very important concept in analysis. The tool that we shall use to study
continuity will be sequences. There are important results concerning the subsets of
the real numbers and the continuity of the function:
4/19/2014
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Science & Mathematics > Mathematics
Math 3110 Homework 14 Solutions
Exercise 15.1.3
If f (x) changes sign from negative to positive on [a, b], then there are some h, k 2 [a, b] with
h < k, f (h) < 0, and f (k) > 0. Since f (x) is dierentiable on [a, b] and [h, k] [a, b],
f (x) is dierentiab
HW 5 Hints and Solutions
13.1.1 (a) Suppose S = [a1 , b1 ] [a2 , b2 ] [an , bn ]. If xn is an innite sequence of points in S, then one of the intervals [ai , bi ] must contain innitely
many of the xi . Let yj be an innite sub-sequence of xj such that yj i
Math 101: Selected solutions, Homeworks 6-8
Hw 6: Problem 1b
Suppose g : R ! R equals 1 on all rational points and g is continuous. Towards contradiction, suppose g(x) 6= 1 for some x 2 R. Then we may let = |g(x) 1| > 0 and nd such
that for any x0 2 (x
,