Exam (elaborations)
Solution Manual For Algebra And Trigonometry, 9th Edition Michael Sullivan
- Course
- - NURSING NR326
- Institution
- Chamberlain College Of Nursing
Solution Manual For Algebra And Trigonometry, 9th Edition Michael Sullivan
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Chapter 2Graphs
Section 2.1 (f) Quadrant IV
1. 0
2. 5 − ( −3 ) = 8 = 8
3. 32 + 42 = 25 = 5
4. 112 + 602 = 121 + 3600 = 3721 = 612
Since the sum of the squares of two of the sides
of the triangle equals the square of the third side,
the triangle is a right triangle.
1 14. (a) Quadrant I
5. bh
2 (b) Quadrant III
(c) Quadrant II
6. true (d) Quadrant I
(e) y-axis
7. x-coordinate; y-coordinate
(f) x-axis
8. quadrants
9. midpoint
10. False; the distance between two points is never
negative.
11. False; points that lie in Quadrant IV will have a
positive x-coordinate and a negative y-coordinate.
The point ( −1, 4 ) lies in Quadrant II.
x + x y + y2
12. True; M = 1 2 , 1 15. The points will be on a vertical line that is two
2 2 units to the right of the y-axis.
13. (a) Quadrant II
(b) x-axis
(c) Quadrant III
(d) Quadrant I
(e) y-axis
152
, Section 2.1: The Distance and Midpoint Formulas
16. The points will be on a horizontal line that is
three units above the x-axis. 26. d ( P1 , P2 ) = ( 6 − (− 4) )2 + ( 2 − (−3) )2
= 102 + 52 = 100 + 25
= 125 = 5 5
27. d ( P1 , P2 ) = (0 − a) 2 + (0 − b) 2
= ( − a ) 2 + ( −b ) 2 = a 2 + b 2
28. d ( P1 , P2 ) = (0 − a) 2 + (0 − a)2
= (−a) 2 + (−a )2
17. d ( P1 , P2 ) = (2 − 0)2 + (1 − 0) 2 = a 2 + a 2 = 2a 2 = a 2
= 22 + 12 = 4 + 1 = 5 29. A = (−2,5), B = (1,3), C = (−1, 0)
18. d ( P1 , P2 ) = (−2 − 0)2 + (1 − 0) 2 d ( A, B ) = (1 − (−2) )2 + (3 − 5)2
= (−2) 2 + 12 = 4 + 1 = 5 = 32 + (−2) 2 = 9 + 4 = 13
d ( B, C ) = ( −1 − 1)2 + (0 − 3)2
19. d ( P1 , P2 ) = (−2 − 1) 2 + (2 − 1)2
= (−2)2 + (−3)2 = 4 + 9 = 13
= (−3) 2 + 12 = 9 + 1 = 10
d ( A, C ) = ( −1 − (−2) )2 + (0 − 5)2
20. d ( P1 , P2 ) = ( 2 − (−1) )2 + (2 − 1)2 = 12 + (−5)2 = 1 + 25 = 26
= 32 + 12 = 9 + 1 = 10
21. d ( P1 , P2 ) = (5 − 3)2 + ( 4 − ( −4 ) )
2
= 22 + ( 8 ) = 4 + 64 = 68 = 2 17
2
( 2 − ( −1) ) + ( 4 − 0 )2
2
22. d ( P1 , P2 ) =
= ( 3 )2 + 4 2 = 9 + 16 = 25 = 5
23. d ( P1 , P2 ) = ( 6 − (−3) )2 + (0 − 2) 2 Verifying that ∆ ABC is a right triangle by the
= 9 + (− 2) = 81 + 4 = 85
2 2 Pythagorean Theorem:
[ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2
( 13 ) + ( 13 ) = ( )
2 2 2
24. d ( P1 , P2 ) = ( 4 − 2 )2 + ( 2 − (−3) )2 26
= 22 + 52 = 4 + 25 = 29 13 + 13 = 26
26 = 26
25. d ( P1 , P2 ) = (6 − 4)2 + ( 4 − (−3) )
2 1
The area of a triangle is A = ⋅ bh . In this
2
= 22 + 7 2 = 4 + 49 = 53 problem,
153
,Chapter 2: Graphs
A = 1 ⋅ [ d ( A, B) ] ⋅ [ d ( B, C )]
problem,
1
2 A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C )]
2
= ⋅ 13 ⋅ 13 = 1 ⋅13
1
2 2 1
= ⋅10 2 ⋅10 2
13 2
= 2 square units
1
= ⋅100 ⋅ 2
2
30. A = (−2, 5), B = (12, 3), C = (10, − 11) = 100 square units
d ( A, B ) = (12 − (−2) )2 + (3 − 5)2
31. A = (− 5,3), B = (6, 0), C = (5,5)
= 142 + (−2) 2
d ( A, B ) = ( 6 − (− 5) )2 + (0 − 3)2
= 196 + 4 = 200
= 112 + (− 3) 2 = 121 + 9
= 10 2
= 130
d ( B, C ) = (10 − 12 )2 + (−11 − 3)2
d ( B, C ) = ( 5 − 6 )2 + (5 − 0)2
= (−2) + (−14)
2 2
= (−1) 2 + 52 = 1 + 25
= 4 + 196 = 200
= 26
= 10 2
d ( A, C ) = (10 − (−2) )2 + (−11 − 5)2 d ( A, C ) = ( 5 − (− 5) )2 + (5 − 3)2
= 102 + 22 = 100 + 4
= 122 + (−16)2
= 104
= 144 + 256 = 400
= 20 = 2 26
Verifying that ∆ ABC is a right triangle by the
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2
[ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2
( ) +( ) =( )
2 2 2
104 26 130
(10 2 ) + (10 2 )
2 2
= ( 20 )
2
104 + 26 = 130
200 + 200 = 400 130 = 130
400 = 400 1
The area of a triangle is A = bh . In this
1 2
The area of a triangle is A = bh . In this
2
154
, Section 2.1: The Distance and Midpoint Formulas
problem, problem,
1 1
A = ⋅ [ d ( A, C )] ⋅ [ d ( B, C ) ] A = ⋅ [ d ( A, C )] ⋅ [ d ( B, C ) ]
2 2
1 1
= ⋅ 104 ⋅ 26 = ⋅ 29 ⋅ 2 29
2 2
1 1
= ⋅ 2 26 ⋅ 26 = ⋅ 2 ⋅ 29
2 2
1 = 29 square units
= ⋅ 2 ⋅ 26
2
= 26 square units 33. A = (4, −3), B = (0, −3), C = (4, 2)
d ( A, B) = (0 − 4)2 + ( −3 − (−3) )
2
32. A = (−6, 3), B = (3, −5), C = (−1, 5)
d ( A, B ) = ( 3 − (−6) )2 + (−5 − 3)2 = (− 4)2 + 02 = 16 + 0
= 16
= 92 + (−8)2 = 81 + 64
=4
= 145
d ( B, C ) = ( 4 − 0 )2 + ( 2 − (−3) )2
d ( B, C ) = ( −1 − 3 ) 2
+ (5 − (−5)) 2
= 42 + 52 = 16 + 25
= (−4) + 10 = 16 + 100
2 2
= 41
= 116 = 2 29
d ( A, C ) = (4 − 4)2 + ( 2 − (−3) )
2
d ( A, C ) = ( −1 − (− 6) ) 2
+ (5 − 3) 2
= 02 + 52 = 0 + 25
= 5 + 2 = 25 + 4
2 2
= 25
= 29 =5
Verifying that ∆ ABC is a right triangle by the Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem: Pythagorean Theorem:
[ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 [ d ( A, B)]2 + [ d ( A, C )]2 = [ d ( B, C )]2
( ) + (2 ) =( )
2 2 2
( )
2
29 29 145 4 2 + 52 = 41
29 + 4 ⋅ 29 = 145 16 + 25 = 41
29 + 116 = 145 41 = 41
145 = 145 1
The area of a triangle is A = bh . In this
1 2
The area of a triangle is A = bh . In this
2
155
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