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Chapter 1 Questions
1. Draw a detailed Venn diagram representing the interrelationships of the following sets
of numbers:
Complex numbers, digits, Real numbers, imaginary numbers, rational numbers, irrational
numbers, transcendental numbers, natural numbers, whole numbers
2. The binary operation # is defined as a#b = 2a2 - b
a) Find 3#(2#1)
=3#(2(2)2 – 1) =
3#7=
2(3)2 – 7 = 18 – 7 = 11
b) Is this operation commutative? Associative? Justify your answer
4#2 = 2(4)2 – 2 = 32 – 2 = 30
2#4 = 2(2)2 – 4 = 16 – 4 = 12
Not commutative
(4#3)#2 = (2(4)2 – 3)#2 = 29#2 = 2(29)2 – 2 = 1680
4#(3#2)=4#(2(3)2 – 2) = 4#(16) = 2(4)2 – 16 = 0
3. Provide an example of each of the following:
a) A 1st degree binomial
x+y or 3x -1
b) A quartic monomial
x3y or 7xyzw or
c) A 6th degree polynomial with 3 different variables
x4yz + xy or
2x2y2z + xy4 + z3 + xy + 1
4. If x2 = 6x, solve to find the value of x
x2 – 6x = 0
x(x – 6) = 0
x = {0,6}
Do Not divide both sides by x because x could have a value of 0.
Chapter 2 Questions
1.
Plot the graph of the given equation in the indicated domain. Describe the
corresponding range and whether or not the relation is a function. (6 points each)
a)
y2 + 2x = 6,
domain = x: –3 < x Š 3, x  integers

Range   12 ,  10,  8,  6, 2,  2, 0

Not a function ( correction no +-root 12)
x  2  1  y;domain  x : 5  x  1 Range: {y: -5 < y <= 1}
b.)
( correction range from -8 to ?)
2. Plot the graph of the given equation in the given domain. Find the range and indicate
whether the relation is a function. (6 points)
y

1
2
x
0 x 4
Range: {y: y>2.25}

Chapter 3 Questions
1. Find the x and y coordinates of the intersection point or points for each of the
following pairs of functions. (5 points each)
a) y = x2 + 2.8x + 3.2 and
y = 7.8x – 2.8
Set the equations equal to each other: x2 + 2.8x + 3.2 = 7.8x – 2.8
x2 – 5x + 6 = 0
(x – 3)(x – 2) = 0
x = {2,3}
When x = 2, y = 7.8(2) – 2.8 = 12.8
When x = 3, y = 7.8(3) – 2.8 = 20.6
Check: 22 + 2.8(2) + 3.2 = 12.8
32 + 2.8(3) + 3.2 = 20.6
b) y = 3x + 5 and The line parallel to 2x + y = 5 that passes through (1,1)
2x + y = 5
y = -2x + 5
Line parallel to 2x + y = 5 has slope = -2
Point slope form: y – 1 = -2(x – 1)
Converted to slope intercept: y = -2x + 3
2. The relationship between the monthly cost of ownership of a car and the number of
miles you drive the car each month is linear. The cost is $366 per month for 300 km per
month and $510 per month for 1500 km per month
a) What linear equation, in standard form, describes the relationship the cost per month
and the number of kilometers driven per month?
Independent variable (x) = kms driven per month
Dependent variable (y) = cost per month
Slope = (510-366)/(1500-300) = 144/1200 = 3/25 = 0.12
y – 366 = 0.12(x – 300)
y – 366 = 0.12 x – 36
y = 0.12 x + 330
Standard Form: 3x – 25y = -8250
b) What would be the monthly cost if you drive the car 1,000 kilometers per month.
0.12(1000)+330 = $450
c) Suppose your cost per month is $600. How many kilometers per month do you drive.
600 = 0.12x + 330
270 = 0.12x
x = 2250
d) Explain the meaning of the slope and of the y-intercept in this particular problem.
Slope is extra cost per month for each extra kilometers driven. In this case, each extra
kilometer costs an extra $0.12
y-intercept is the cost when x = 0. In other words, the y-intercept is the cost per month of
the car, even if you do not drive it at all.
3. Modify each of the equations below, as indicated
a) 3x – 4y = 15
Transform to slope-intercept form
-4y =-3x + 15
y = ¾ x – 15/4 or y = 0.75x – 3.75
b) y – 7 = -4(x – 0.5)
y – 7 = -4x + 2
y = -4x + 9
Transform to slope-intercept form
c) y = 0.35x + 4.25
Transform to standard form
-0.35x + y = 4.25
7x – 20y = -85 (all coefficients are integers and x-coefficient is positive)
x y
 1
7 3
3x + 7y = 21
d)
Transform to standard form (from intercept form)
Intercepts: (7,0) and (0,3)
Chapter 4 Questions
1. f(x) = x2 + 2x –3
g(x) = 3+2x
h(x) = 7x+3
a) What is g(h(f(3)))?
f(3) = 32 + 23 – 3 = 12
h(12) = 7(12) + 3 = 87
g(87) = 3 + 2(87) = 177
b) Find g(g(x))
3+ 2(3 + 2x)) = 3 + 6 + 4x = 9 + 4x
c) If j(h(x)) = x, what is j(x)? (j is a linear function).
j is the inverse of h(x). To find inverse, reverse variables: x = 7(j(x)) + 3 
(x – 3)/7 = j(x)
d) g(h(x)) = -4. What is x?
3 + 2(7(x) + 3) = -4  3 + 14x + 6 = -4  9 + 14x = -4  14x = -13  x = -13/14
2. Find the values of a and b:
 2 b   b  12 
 3b a    4    32 

    
2b + 4b = 12  6b = 12  b = 2
3b2 + 4a = 32  3(2)2 + 4a = 32  12 + 4a = 32  4a = 20  a = 5
3. The matrix below has no inverse. Calculate the value of x.
x

1

x  2

x  2
A square matrix has no inverse if its determinant = 0 or x(x – 2) = 1(x – 2)
x2 – 2x = x – 2
x2 – 3x + 2 = 0
(x – 2)(x – 1) = 0; x = 1 or 2
4. Use augmented matrices to solve the following system (use graphing calculator and
explain the steps you used to find the solution). Your solutions should be in the form of
fractions.
3x + 7y + 5z = 20
15 

1
0
0
2x – 5y + 2z = 15

2 


x – 5y – z = -10
 0 1 0 5 

3 

65 
0 0 1


6 
5. Given the following shaded feasible region, find the following:
a) The point that would maximize profits given a profit equation of P = 10x + 12y. Show
your work. Is a profit of 100 feasible?
Two profit lines are shown as dashed
lines in figure to the right. Higher line
represents profit of 100. This line does
not intersect feasible region.
Lower profit line represents profit of 84
and is the line with highest profit that
still intersects the feasible region.
b) List the inequalities that represent the constraints in this system.
x<4 and y < -0.5x + 6 and x > 0 and y > 1 and y > -x + 3 and y < -2.5x + 12