1. Solve the system of equations by Gaussian elimination or Gauss-Jordan elimination method.
3x + y 2z = 1
a.
2x + 3y + z = 3
+ 2y z = 2
x
Ans: (0, 1, 0)
4x 3y + 7z = 14
b.
3x + y + 3z = 5
x 7y 2z = 6
Ans: (1, 1, 1)
2y + z = 1
x + 4y + z =
1
x 3y
c.
2
=
A
1. Write down the rst ve terms of the given sequence:
a. cfw_an = cfw_7
b. cfw_an = n2
c. cfw_an =
1
n
d. cfw_an =
(1)n+1 (n)
n+1
e. cfw_an =
2n
3n1
f. cfw_an =
2
n2 3
g. cfw_an = cfw_n! n
h. an+1 = an 3;
a1 = 0
i. an+1 = an + n;
a1 = 2
j. an+1 = a
1. Solve the system of equations by Gaussian elimination or Gauss-Jordan elimination method.
3x + y 2z = 1
a.
2x + 3y + z = 3
x
+ 2y z = 2
4x 3y + 7z = 14
b.
3x + y + 3z = 5
x 7y 2z = 6
2y + z = 1
x + 4y + z =
1
x 3y
c.
2
=
7x y z = 2
d.
x
+ 2z = 5
5y z =
Precalculus ( Math 1 ) HW Set #9.
Due Wednesday, April 23rd.
In order to receive a , you must attempt all problems and write out all steps leading to your answers
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Precalculus (25125) ( Math 1 ) HW Set #6.
Due Wednesday, March 12th.
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Due Wednesday, March 5th.
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neatly and legibly. You cannot simply write the
Precalculus (25125) HW Set #1. In order to receive a , you must attempt all problems and write out all
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demonstrate your
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Due Wednesday, February 26th.
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Precalculus (25125) ( Math 1 ) HW Set #7.
Due Wednesday, March 19th.
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Due Wednesday, April 9th.
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Precalculus (25125) (Math 1) HW Set 3.
Due Wednesday, February 12th.
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leading to your answers neatly and legibly. You cannot simply write the correct answer to
demonst
1. Let z = 3 4i, w = 2 + 3i, nd the following:
a. z
b. z + w
c. zw
z
d.
w
2. Perform the given operation of complex numbers. Express your result in the
form of a + bi:
a. 2i (3 5i)
b. (4 + i) (2 6i)
c. (i)(12 + 2i)
d. (4 2i)(5 3i)
e. (3 i)(3 + 2i)
f. (i)(
Matrices:
Denition: An m n matrix, Amn is a rectangular array of numbers with m
rows and n columns:
a1,1 a1,2 a1,n
a
2,1 a2,2 a2,n
Am,n = .
. .
.
.
.
.
.
.
.
am,1 am,2 am,n
Each ai,j is the entry at the ith row, jth column.
E.g
3
Am,n = 1
7
4
1
5
2
2
Complex Numbers:
Denition: A complex number is a number of the form:
z = a + bi
where a, b are real numbers and i is a symbol with the property:
i2 = 1. You may treat i = 1
We can treat i as a variable in an algebraic expression and all algebraic rules ar
Sequences:
Denition: A sequence is a function whose domain is the set of natural numbers
or a subset of the natural numbers. We usually use the symbol an to represent a
sequence, where n is a natural number and an is the value of the function on n.
Intuit
Transformation of Functions
You should know the graph of the following basic functions:
f (x) = x2
f (x) = x3
f (x) =
f (x) =
1
x
x
f (x) = |x|
If we know the graph of a basic function f (x), we can draw the graph of more
complicated functions, like 2f (x
1. Draw the graph of the given function. First start with a simple function whose
graph you know.
a. f (x) = x + 1 + 1
Ans:
b. f (x) =
Ans:
x1+1
c. f (x) = x 1 + 1
Ans:
d. f (x) =
Ans:
1
x1
2
e. f (x) =
1
x+1
2
Ans:
f. f (x) = 2
Ans:
1
x1
2
1
g. f (x) =
An Exponential Function with base b is a function of the form:
f (x) = bx , where b > 0, b = 1 is a real number.
We know the meaning of br if r is a rational number. What if r is irrational?
What we do is we approximate the value of br by using rational a
1. Draw the graph of the given function. First start with a simple function whose
graph you know.
a. f (x) = x + 1 + 1
b. f (x) = x 1 + 1
c. f (x) = x 1 + 1
d. f (x) =
1
x1
2
e. f (x) =
1
x+1
2
f. f (x) = 2
1
x1
2
1
g. f (x) = 2 x + 1
2
1 1
x+1
2 2
1
i.
Functions
Informal denition of a function:
A function between two sets is a rule that assigns to each member
in the rst set (called the domain) one and only one member in the
second set (called the range).
Intuitively, a function is a machine (or an opera