Math 115: midterm 1
May 8, 2013
Notes:
- You may use any result which was stated in class, unless you are asked to prove
it, or explicitly instructed otherwise.
- Give a full justication for everything you are asked to prove.
(10 points)
1. Let F be an or
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OxTail with Creamy Polenta
Coat oxtails with canola oil and salt
Put oxtails a sheet pan and in oven for 20 minutes
on 500 degrees until brown
In a pan add:
Canola oil
garlic cloves
whole cut carrots
whole cut onion
whole cut celery
salt
(cook til color
Henna for Hair
1. Mix henna powder with hot water in iron wok. let sit overnight
2. Add egg white, 1 tbsp indigo, 2 tbsp of curd, 2 drops of lemon juice,
2 tbsp of mustard oil and half of coconut milk to the mix
before applying
3. Keep this henna mas
Math 115: homework 1
Due by 2 p.m. on April 18, 2013
Notes:
- You may use any result which was stated in class, unless you are asked to prove
it, or explicitly instructed otherwise.
- Give a full justication for everything you are asked to prove.
1. Let (
Math 115: solutions for homework 3
1. a) Rewrite:
n2 + n
1
=
3
3n
3
1
1
+ 2
n n
Example 9.7-(a) says that
lim
n
1
=0
n
and
lim
n
1
=0
n2
Therefore, by theorem 9.3
lim
n
1
1
+ 2
n n
1
1
+ lim 2 = 0 + 0 = 0
n n
n n
= lim
b) Rewrite:
1
1
1
4 + n2 + 2 n + 3
Math 115: solutions for homework 2
1. The hypothesis Ik+1 Ik implies that xk+1 xk , and that yk+1 yk for every
k N. It follows by induction that for k, l N with l k we have xl xk and
yl yk . The non-empty set A = cfw_xl : x N is bounded above by y1 : for
Math 115: homework 2
Due by 2 p.m. on May 2, 2013
Notes:
- You may use any result which was stated in class, unless you are asked to prove
it, or explicitly instructed otherwise.
- Give a full justication for everything you are asked to prove.
1. For each
Math 115: solutions for homework 1
1. a) Repeatedly using the associativity axiom for addition, we get:
A1
(a + (b + c) + (d + e) = (a + b) + c) + (d + e)
A1
= (a + b) + (c + (d + e)
A1
= (a + b) + (c + d) + e)
A3
DL
b) We have a (0 + 0) = a 0 (since 0 +
Math 115: solutions for homework 3
1. a) Fix x0 R. Let = 1. First assume x0 Q, so f (x0 ) = 0. Given any > 0,
/
by density of Q in R we can nd a rational y (x0 , x0 + ) Q. Then
f (y) = 1, so |f (y) f (x0 )| = |1 0| = 1 . We have thus proved f is not
conti
Math 115: homework 4
Due by 2 p.m. on May 30, 2012
Notes:
- You may use any result which was stated in class, unless you are asked to prove
it, or explicitly instructed otherwise.
- Give a full justication for everything you are asked to prove.
1. This co
Math 115: solutions for exam 1
1. Let x F . Then x2 + 2x + 2 = (x + 1)2 + 1. Since y 2 0 for any y F , then
(x + 1)2 0. Therefore,
x2 + 2x + 2 = (x + 1)2 + 1 0 + 1 = 1 > 0
since 1 > 0 (due to 1 = 12 0 and 1 = 0). Therefore, x2 + 2x + 2 = 0 for any
x F.
2.
Math 115: supplemental exercises
In class, we identied the rational numbers with a subset of the real numbers. The
next exercise is meant to show that the rational numbers embed into any ordered
eld.
S1. Let (F, +, 0, , 1, ) be an ordered eld. You are tas
Math 115: homework 3
Due by 2 p.m. on May 16, 2012
Notes:
- You may use any result which was stated in class, unless you are asked to prove
it, or explicitly instructed otherwise.
- Give a full justication for everything you are asked to prove.
1. Find th
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