2.1 The Rectangular Coordinate Systems and Graphs
x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .
125 = 5 5 125 = 5 5
( − 5 , 5 2 ) ( − 5 , 5 2 )
2.2 Linear Equations in One Variable
x = −5 x = −5
x = −3 x = −3
x = 10 3 x = 10 3
x = 1 x = 1
x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .
x = 1 3 x = 1 3
m = − 2 3 m = − 2 3
y = 4 x −3 y = 4 x −3
x + 3 y = 2 x + 3 y = 2
Horizontal line: y = 2 y = 2
Parallel lines: equations are written in slope-intercept form.
y = 5 x + 3 y = 5 x + 3
2.3 Models and Applications
C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650
L = 37 L = 37 cm, W = 18 W = 18 cm
2.4 Complex Numbers
−24 = 0 + 2 i 6 −24 = 0 + 2 i 6
( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i
5 2 − i 5 2 − i
18 + i 18 + i
−3 −4 i −3 −4 i
2.5 Quadratic Equations
( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1
( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.
( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.
( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4
x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1
x = 4 ± 5 x = 4 ± 5
x = 3 ± 22 x = 3 ± 22
x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3
2.6 Other Types of Equations
{ −1 } { −1 }
0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2
1 ; 1 ; extraneous solution − 2 9 − 2 9
−2 ; −2 ; extraneous solution −1 −1
−1 , −1 , 3 2 3 2
−3 , 3 , − i , i −3 , 3 , − i , i
2 , 12 2 , 12
−1 , −1 , 0 0 is not a solution.
2.7 Linear Inequalities and Absolute Value Inequalities
[ −3 , 5 ] [ −3 , 5 ]
( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )
x < 1 x < 1
x ≥ −5 x ≥ −5
( 2 , ∞ ) ( 2 , ∞ )
[ − 3 14 , ∞ ) [ − 3 14 , ∞ )
6 < x ≤ 9 or ( 6 , 9 ] 6 < x ≤ 9 or ( 6 , 9 ]
( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )
| x −2 | ≤ 3 | x −2 | ≤ 3
k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) . ( − ∞ , 1 ] ∪ [ 7 , ∞ ) .
2.1 Section Exercises
Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.
The y -intercept is the point where the graph crosses the y -axis.
The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .
The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .
The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .
y = 4 − 2 x y = 4 − 2 x
y = 5 − 2 x 3 y = 5 − 2 x 3
y = 2 x − 4 5 y = 2 x − 4 5
d = 74 d = 74
d = 36 = 6 d = 36 = 6
d ≈ 62.97 d ≈ 62.97
( 3 , − 3 2 ) ( 3 , − 3 2 )
( 2 , −1 ) ( 2 , −1 )
( 0 , 0 ) ( 0 , 0 )
y = 0 y = 0
not collinear
A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )
| 1 | |
| 0 | 2 |
| 3 | 3 |
| 6 | 4 |
| –3 | 0 |
| 0 | 1.5 |
| 3 | 3 |
d = 8.246 d = 8.246
d = 5 d = 5
( −3 , 4 ) ( −3 , 4 )
x = 0 y = −2 x = 0 y = −2
x = 0.75 y = 0 x = 0.75 y = 0
x = − 1.667 y = 0 x = − 1.667 y = 0
15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter
6 .0 42 6 .0 42
Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.
2.2 Section Exercises
It means they have the same slope.
The exponent of the x x variable is 1. It is called a first-degree equation.
If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).
x = 2 x = 2
x = 2 7 x = 2 7
x = 6 x = 6
x = 3 x = 3
x = −14 x = −14
x ≠ −4 ; x ≠ −4 ; x = −3 x = −3
x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution
x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2
y = − 4 5 x + 14 5 y = − 4 5 x + 14 5
y = − 3 4 x + 2 y = − 3 4 x + 2
y = 1 2 x + 5 2 y = 1 2 x + 5 2
y = −3 x − 5 y = −3 x − 5
y = 7 y = 7
y = −4 y = −4
8 x + 5 y = 7 8 x + 5 y = 7
Perpendicular
m = − 9 7 m = − 9 7
m = 3 2 m = 3 2
m 1 = − 1 3 , m 2 = 3 ; Perpendicular . m 1 = − 1 3 , m 2 = 3 ; Perpendicular .
y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40
y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 , y max = 10 y min = −10 , y max = 10
y = − A B x + C B y = − A B x + C B
The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 . The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 .
Yes they are perpendicular.
2.3 Section Exercises
Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.
2 , 000 − x 2 , 000 − x
v + 10 v + 10
Ann: 23 ; 23 ; Beth: 46 46
20 + 0.05 m 20 + 0.05 m
90 + 40 P 90 + 40 P
50 , 000 − x 50 , 000 − x
She traveled for 2 h at 20 mi/h, or 40 miles.
$5,000 at 8% and $15,000 at 12%
B = 100 + .05 x B = 100 + .05 x
R = 9 R = 9
r = 4 5 r = 4 5 or 0.8
W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14
f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21
m = − 5 4 m = − 5 4
h = 2 A b 1 + b 2 h = 2 A b 1 + b 2
length = 360 ft; width = 160 ft
A = 88 in . 2 A = 88 in . 2
h = V π r 2 h = V π r 2
r = V π h r = V π h
C = 12 π C = 12 π
2.4 Section Exercises
Add the real parts together and the imaginary parts together.
Possible answer: i i times i i equals -1, which is not imaginary.
−8 + 2 i −8 + 2 i
14 + 7 i 14 + 7 i
− 23 29 + 15 29 i − 23 29 + 15 29 i
8 − i 8 − i
−11 + 4 i −11 + 4 i
2 −5 i 2 −5 i
6 + 15 i 6 + 15 i
−16 + 32 i −16 + 32 i
−4 −7 i −4 −7 i
2 − 2 3 i 2 − 2 3 i
4 − 6 i 4 − 6 i
2 5 + 11 5 i 2 5 + 11 5 i
1 + i 3 1 + i 3
( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1
5 −5 i 5 −5 i
9 2 − 9 2 i 9 2 − 9 2 i
2.5 Section Exercises
It is a second-degree equation (the highest variable exponent is 2).
We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.
One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .
x = 6 , x = 6 , x = 3 x = 3
x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3
x = 5 , x = 5 , x = −5 x = −5
x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2
x = −2 , 3 x = −2 , 3
x = 0 , x = 0 , x = − 3 7 x = − 3 7
x = −6 , x = −6 , x = 6 x = 6
x = 6 , x = 6 , x = −4 x = −4
x = 1 , x = 1 , x = −2 x = −2
x = −2 , x = −2 , x = 11 x = 11
z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2
x = 3 ± 17 4 x = 3 ± 17 4
One rational
Two real; rational
x = − 1 ± 17 2 x = − 1 ± 17 2
x = 5 ± 13 6 x = 5 ± 13 6
x = − 1 ± 17 8 x = − 1 ± 17 8
x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535
x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7
a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a
x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.
maximum at x = 70 x = 70
The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.
2.6 Section Exercises
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
x = 81 x = 81
x = 17 x = 17
x = 8 , x = 27 x = 8 , x = 27
x = −2 , 1 , −1 x = −2 , 1 , −1
y = 0 , 3 2 , − 3 2 y = 0 , 3 2 , − 3 2
m = 1 , −1 m = 1 , −1
x = 2 5 , ±3 i x = 2 5 , ±3 i
x = 32 x = 32
t = 44 3 t = 44 3
x = −2 x = −2
x = 4 , −4 3 x = 4 , −4 3
x = − 5 4 , 7 4 x = − 5 4 , 7 4
x = 3 , −2 x = 3 , −2
x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3
x = 2 , −2 x = 2 , −2
x = 1 , 5 x = 1 , 5
x ≥ 0 x ≥ 0
x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8
2.7 Section Exercises
When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
( − ∞ , ∞ ) ( − ∞ , ∞ )
We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.
( − ∞ , 3 4 ] ( − ∞ , 3 4 ]
[ − 13 2 , ∞ ) [ − 13 2 , ∞ )
( − ∞ , 3 ) ( − ∞ , 3 )
( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]
All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )
( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )
( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )
No solution
( −5 , 11 ) ( −5 , 11 )
[ 6 , 12 ] [ 6 , 12 ]
[ −10 , 12 ] [ −10 , 12 ]
x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 , ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 , ( − 2 , + ∞ )
x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ [ 1 , ∞ ) x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ [ 1 , ∞ )
( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )
[ −11 , −3 ] [ −11 , −3 ]
It is never less than zero. No solution.
Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.
( − ∞ , −3 ) ( − ∞ , −3 )
Where the blue line is above the orange line; always. All real numbers.
( − ∞ , − ∞ ) ( − ∞ , − ∞ )
( −1 , 3 ) ( −1 , 3 )
( − ∞ , 4 ) ( − ∞ , 4 )
{ x | x < 6 } { x | x < 6 }
{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }
( −2 , 1 ] ( −2 , 1 ]
( − ∞ , 4 ] ( − ∞ , 4 ]
Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .
Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .
x = 2 , − 4 5 x = 2 , − 4 5
( −7 , 5 ] ( −7 , 5 ]
80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400
[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]
Review Exercises
x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )
y = 5 3 x + 4 y = 5 3 x + 4
72 = 6 2 72 = 6 2
620.097 620.097
midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )
| 0 | −2 |
| 3 | 2 |
| 6 | 6 |
x = 4 x = 4
x = 12 7 x = 12 7
y = 1 6 x + 4 3 y = 1 6 x + 4 3
y = 2 3 x + 6 y = 2 3 x + 6
females 17, males 56
x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4
horizontal component −2 ; −2 ; vertical component −1 −1
7 + 11 i 7 + 11 i
−16 − 30 i −16 − 30 i
−4 − i 10 −4 − i 10
x = 7 − 3 i x = 7 − 3 i
x = −1 , −5 x = −1 , −5
x = 0 , 9 7 x = 0 , 9 7
x = 10 , −2 x = 10 , −2
x = − 1 ± 5 4 x = − 1 ± 5 4
x = 2 5 , − 1 3 x = 2 5 , − 1 3
x = 5 ± 2 7 x = 5 ± 2 7
x = 0 , 256 x = 0 , 256
x = 0 , ± 2 x = 0 , ± 2
x = 11 2 , −17 2 x = 11 2 , −17 2
[ − 10 3 , 2 ] [ − 10 3 , 2 ]
( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )
Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.
( 3.5 , ∞ ) ( 3.5 , ∞ )
Practice Test
y = 3 2 x + 2 y = 3 2 x + 2
| 0 | 2 |
| 2 | 5 |
| 4 | 8 |
( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )
( − ∞ , 9 ] ( − ∞ , 9 ]
x = −15 x = −15
x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1
x = 3 ± 3 2 x = 3 ± 3 2
( −4 , 1 ) ( −4 , 1 )
y = −5 9 x − 2 9 y = −5 9 x − 2 9
y = 5 2 x − 4 y = 5 2 x − 4
5 13 − 14 13 i 5 13 − 14 13 i
x = 2 , − 4 3 x = 2 , − 4 3
x = 1 2 ± 2 2 x = 1 2 ± 2 2
x = 1 2 , 2 , −2 x = 1 2 , 2 , −2
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Physics 181/182 Homework Solutions
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Chapter 2 Homework Solutions - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. The document provides answers to review questions about controls and properties in Visual Basic. Some key points covered include: - Using text boxes versus labels to display output - Independent option button groups using frames - Setting images in image controls ...
Problem 2.9: Problem 2.9: Scott Thomas A = 3 7 -4 12 -5 9 10 2 6 13 8 11 15 5 4 1 Part (a): Sort each column, store result in B
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98 Chapter 2 Differentiation 24. 4 x x x 2 lim x→0 4 x x x lim x→0 4x 4 x x x x x x x x x lim x→0 4 x 4 x x x x x x x x x x x x lim x→0 4 x x 4 x f xx lim x→0
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College Algebra Chapter 2. Close. Contents Contents. Highlights. Print. Table of contents. Preface; 1 Prerequisites. Introduction to Prerequisites; 1.1 Real Numbers: Algebra Essentials; 1.2 Exponents and Scientific Notation; ... This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the ...
Chapter 2 Homework Solution - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The power absorbed by the heater is proportional to the square of the applied voltage. Using P1/V1^2 = P2/V2^2, we can calculate: P1 = 1000 W V1 = 250 V V2 = 210 V P2 = (1000 W) × (210 V/250 V)^2 = 864 W So the power absorbed from the 210 V source is 864 W. P 2.4-7 An electric heater is ...
1.1 Solution 5. Correctness Verification We want to see whether the derivation is correct or not For this we set n = 3 and we want to see if the right part of the derivation is equal to the left part. SL = ( 3 1 ) ∑. 3 j=1 aj)( ∑ k=1 ak. SL = ( 3 j=1 aj)( 1 1. + + 1.
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Step-by-step solution. Step 1 of 3. Consider the provided statement to justify answer is True or False. Provided condition is, if then the right hand rule underestimates the integral. Step 2 of 3. It is assumed that, The value of interval is. Now by using MAPLE to find the value of integral is shown as below; Therefore, by using Right Hand rule ...
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2 Math 372: Homework #2: Solutions by Nick Arnosti and Thomas Crawford (2010) 8 3 Math 372: Homework #3: Carlos Dominguez, Carson Eisenach, David Gold 12 ... Due Friday, September 22: Chapter 1: Page 24: #16abc, #24, #25ab. Chapter 2: (#1) We proved Goursat's theorem for triangles. Assume instead we know it holds for any rectangle; prove it ...
2-3 2-5C Solution We are to define the state postulate. Analysis The state postulate is expressed as: The state of a simple compressible system is completely specified by two independent, intensive properties. Discussion An example of an intensive property is temperature. 2-6C Solution We are to discuss the applicability of the ideal gas law.
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Homework #2: singular homology Exercises from Hatcher: Chapter 2.1, Problems 11, 12, 16, 17a (S2 only, using 2.14), 18. 11. Suppose that A is a retract of X. That means that there exists a map r : X !A such that r i = id A, where i : A !X is the inclusion. Then r ( i = id H n A), so i is injective. 12.
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