Download Rate of Change and Slope

Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(For help, go to Lessons 8-3 and 10-3.)
Complete each equation.
1. a 3 = a2 • a
2. b 7 = b6 • b
3. c 6 = c3 • c
4. d 8 = d4 • d
Find the value of each expression.
5.
4
8.
49
6.
169
11-1
7.
25
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Solutions
1. a3 = a(2 + 1) = a2 • a1
2. b7 = b(6 + 1) = b6 • b1
3. c6 = c(3 + 3) = c3 • c3
4. d8 = d(4 + 4) = d4 • d4
5.
4=2
6.
169 = 13
7.
25 = 5
8.
49 = 7
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
243 =
=
= 9
81 • 3
81 •
3
243.
81 is a perfect square and a factor of 243.
3
Use the Multiplication Property of Square Roots.
Simplify
11-1
81.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
28x7 =
=
4x6 • 7x
4x6 •
= 2x3
7x
28x7.
4x6 is a perfect square and a factor of 28x7.
7x
Use the Multiplication Property of Square Roots.
Simplify
11-1
4x6.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
12 •
12 •
32
32 =
12 • 32
Use the Multiplication Property of
Square Roots.
=
384
Simplify under the radical.
=
64 • 6
64 is a perfect square and a factor of 384.
=
64 •
= 8
6
6
Use the Multiplication Property of
Square Roots.
Simplify
11-1
64.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b. 7
5x • 3
8x
7
5x • 3
8x = 21
40x2
Multiply the whole numbers and
use the Multiplication Property of
Square Roots.
= 21 4x2 • 10
factor of 40x2.
4x2 is a perfect square and a
= 21 4x2 • 10
Square Roots.
Use the Multiplication Property of
= 21 • 2x
Simplify
= 42x
10
10
Simplify.
11-1
4x2.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Suppose you are looking out a fourth floor window 54 ft above
the ground. Use the formula d = 1.5h to estimate the distance you
can see to the horizon.
d =
1.5h
=
1.5 • 54
Substitute 54 for h.
=
81
Multiply.
=9
Simplify
81.
The distance you can see is 9 miles.
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
b.
13
64
13
=
64
13
64
Use the Division Property of Square Roots.
=
13
8
Simplify
49
=
x4
49
x4
Use the Division Property of Square Roots.
64.
49
x4
7
= x2
Simplify
49 and
11-1
x4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
120
10
120
=
10
12
Divide.
=
4•3
4 is a perfect square and a factor of 12.
=
4•
=2
3
3
Use the Multiplication Property of Square Roots.
Simplify
11-1
4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b.
75x5
48x
75x5
=
48x
25x4
16
=
25x4
16
=
=
25 •
16
5x2
4
Divide the numerator and denominator by 3x.
Use the Division Property of Square Roots.
x4
Use the Multiplication Property of
Square Roots.
Simplify
25,
11-1
x4, and
16.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
3
7
3
=
7
=
=
3
•
7
7
7
Multiply by
7
7
to make the denominator a
perfect square.
3
7
49
3
7
7
Use the Multiplication Property of Square Roots.
Simplify
11-1
49.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
Simplify the radical expression.
b.
11
12x3
11
=
12x3
11
•
12x3
3x
3x
=
33x
36x4
Use the Multiplication Property of Square Roots.
=
33x
6x2
Simplify
Multiply by
3x to make the denominator a
3x
perfect square.
11-1
36x4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
pages 581–583 Exercises
1. 10
2
3. 5
3
3
13. 20
2
2. 7
12. –4b2
27. 17 mi
14. 18
15. 11
2
16. 84
3
5
5. –6
30
18. –30
6. 40
5
19. 6n
2
7. 2n
7
20. 14t
2
21. 3x2
17
8.
6b2
9. 6x
10. 2n
11.
2a2
3
21
7
29. 3 3
2
30. 5
2
28.
4. –4
17. 7
26. 12 mi
3
3
31. 2 30
11
32.
5
3a
33.
7
4c
22. 80t 3
3
n
5a
23. 3a3
24. –6a2
25. 3 mi
11-1
2
2
34. 5 3a
7
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
35. 2n
2n
9
36.
5
54. Simplest form; radicand has
no perfect-square factors
other than 1.
21x
7x
47. 2 10n
5n
48. 9 2
4
46.
3
37. 3
2
38. 2
45.
3
39. –2
40. 2x
5
7
49. 2
55. Simplest form; radicand has
no perfect-square factors
other than 1.
18 • 10 = 180 =
36 • 5 = 6 5
b. Answers may vary.
Sample: a = 36,
b = 5; a = 9, b = 20
56. a.
3
41. s
50.
2b
b
2
42. a 3
51.
55y
2y
43. 3
y
52. not simplest form; radical in
the denominator of a fraction
44. 3 2
53. not simplest form; radical in
the denominator of a fraction
3
2
2
11-1
57. 30
58. 2
13
59. 3 2
4
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
69. –3 ± 3
60. –2 a
a2
61. 2
15
62. x
y
70. 1 ±
3
64. 4
5
65. 2ab
66. ab2c
3m
67.
4m
68. 8 6a
3a
78.
5
3
50 = 25 • 2 =
25 • 2 = 5 2
b. The radicand has no
perfect-square factors
other than 1.
72. a.
5b
abc
seconds
79. C
71. 2 ± 10
y2
63. –
77. 30a4
2
73. Answers may vary.
Sample: 12, 27, 48.
74. a. 2 6 in.
b. 4.90 in.
80. F
81. B
82. I
83. [2] A = 96 ft2
s = 96 =
16 • 6 = 4 6 ft
[1] correct answer,
without work shown
84. quadratic; y = 0.2x2
75. 12x
85. exponential; y = 4(2.5)x
76. 10b2
86. linear; y = –4.2x + 7
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
87.
89.
88.
90. 3n2 + 5n + 5
91. 3v2 – v – 9
92. 5t 3 + 8t 2 – 14t – 11
93. –3b2 – 23b – 21
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
1.
16 •
8 8
4.
2
a5
2
a
a3
2 2. 4
5.
144
3x
15x3
11-1
48
5
5x
3.
12
36
3
3
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
(For help, go to Lesson 10-4.)
Simplify each expression.
2. 92 – 42
3. (3t)2 + (4t)2
4. c2 = 36
5. 24 + b2 = 49
6. a2 + 16 = 65
7. 12 + b2 = 32
8. 80 = c2
9. 100 = a2 + 52
1. 52 + 62
Solve each equation.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
Solutions
1. 52 + 62 = (5 • 5) + (6 • 6) = 25 + 36 = 61
2. 92 – 42 = (9 • 9) – (4 • 4) = 81 – 16 = 65
3. (3t)2 + (4t)2 = (32 • t2) + (42 • t2) = 9t2 + 16t2 = 25t 2
4. c2 = 36
c = ± 36 = ±6
5. 24 + b2 = 49
6. a2 + 16 = 65
b2 = 49 – 24
a2 = 65 – 16
b2 = 25
a2 = 49
b = ± 25 = ±5
a = ± 49 = ±7
7. 12 + b2 = 32
8. 80 = c2
b2 = 32 – 12
c = ± 80 = ± 16 • 5 = ±
b2 = 20
b = ± 20 = ± 4 • 5 = ± 4 • 5 = ± 2 5
9.
100 = a2 + 52
100 – 52 = a2
48 = a2
a = ± 48 = ± 16 • 3 = ± 16 • 3 = ± 4 3
11-2
16 •
5 = ±4
5
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
What is the length of the hypotenuse
of this triangle?
a2 + b2 = c2
Use the Pythagorean Theorem.
82 + 152 = c2
Substitute 8 for a and 15 for b.
64 + 225 = c2
289 =
17 = c
c2
Simplify.
Find the principal square root of each side.
Simplify.
The length of the hypotenuse is 17 m.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
A toy fire truck is near a toy building on a table such that the
base of the ladder is 13 cm from the building. The ladder is extended 28
cm to the building. How high above the table is the top of the ladder?
Define: Let b = height (in cm) of the ladder from
a point 9 cm above the table.
Relate: The triangle formed is a right triangle.
Use the Pythagorean Theorem.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
(continued)
Write:
a2 + b2 = c2
132 + b2 = 282
169 + b2 = 784
Substitute.
Simplify.
b2 = 615
Subtract 169 from each side.
b2 =
Find the principal square root of each side.
b
615
24.8
Use a calculator and round to the
nearest tenth.
The height to the top of the ladder is 9 cm higher than 24.8 cm,
so it is about 33.8 cm from the table.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
Determine whether the given lengths are sides of a right
triangle.
a. 5 in., 5 in., and 7 in.
52 + 52
72
25 + 25 49
50 =/ 49
Determine whether a2 + b2 = c2,
where c is the longest side.
Simplify.
This triangle is not a right triangle.
b. 10 cm, 24 cm, and 26 cm
102 + 242 262
Determine whether a2 + b2 = c2,
where c is the longest side.
100 + 576 676
Simplify.
676 = 676
This triangle is a right triangle.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
If two forces pull at right angles to each other, the
resultant force is represented as the diagonal of a rectangle,
as shown in the diagram. The diagonal forms a right triangle
with two of the perpendicular sides of the rectangle.
For a 50–lb force and a 120–lb force, the resultant force is
130 lb. Are the forces pulling at right angles to each other?
502 + 1202
2500 + 14,400
1302
Determine whether a2 + b2 = c2
where c is the greatest force.
16,900
16,900 = 16,900
The forces of 50 lb and 120 lb are pulling at right angles to each other.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
pages 587–590 Exercises
12. 0.6
24. no
1. 10
13. 1.2 m
25. yes
2. 25
14. about 15.5 ft
26. 1.5
3. 17
15. about 5.8 km
27. 4 or 0.3
4. 26
16. yes
28. 3
5. 2.5
17. no
29. 6
6. 1
18. no
30. 2.6
7. 4
19. yes
31. 7.0
8. 5
20. no
9. 12
21. yes
32. a. 6 5 ft
b. 80.5 ft2
10. 7.1
11. 7.5
22. yes
23. no
11-2
15
33. yes
34. no
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
35. yes
36. yes
37. 4.2 cm
38. 1000 lb
39. 559.9
40. 9.0
44. a. 6.9 ft
b. 89.2 ft2
c. 981 watts
45. 12.8 ft
46. a. Answers may vary.
Sample: 5, 20, 5
b. 5 units2
47. a. 13.4 ft
b. 17.0 ft
These lengths could be
c. 10.6 ft
2 legs or one leg and
d. No; the hypotenuse d must
the hypotenuse.
be longer than each leg.
about 12.8 in. or 6 in.
48. An integer has 2 as a factor;
62 + 82 = 36 + 64 = 100 = 102
the integer is even; if an integer
is even, then it has 2 as a factor; true.
5; 12; 7; 41
Answers may vary. Sample: 10, 24, 26
41. 9.7
42. a.
b.
43. a.
b.
c.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
49. A figure is a square;
the figure is a rectangle;
if a figure is a rectangle
then the figure is a square;
false.
54. 10
50. You are in Brazil; you are
south of the equator; if you
are south of the equator you
are in Brazil; false.
58. a.
51. An angle is a right angle;
its measure is 90°; if the
measure of an angle is 90°,
then it is a right angle; true.
52. 52
55. 4
3
56. 5
57. n2 + (n + 1)2 = (n + 2)2; 3, 4, 5
b.
74
59. a. a2 + 2ab + b2
b. c2
c. ab
2
1
d. (a + b)2 = c2 + 4 ab ; a2 + 2ab + b2 =
2
2
2
2
2ab + c ; a + b = c2
e. This equation is the same as the
Pythagorean Theorem.
units2
53. 6 in.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
60. D
69. 2b2 10b
61. H
70.
63. C
2
2x2
71. 2 6v
v4
64. A
72. 3 and 4
65. [2] It is a right triangle. Substitute 17,
the longest side, for c and substitute
the other lengths for a and b in the
Pythagorean Theorem
82 + 152 172 –> 64 + 225 289 –> 289 = 289
[1] incorrect equation OR incorrect explanation
73. 8 and 9
62. B
66. 4
67.
3
75. 11 and 12
76. rational
77. irrational
78. irrational
6
3
68. 5
74. –8 and –7
79. rational
2
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
80. 8x2 – 4x
81. 12a2 + 15a
82. 18t 3 – 6t 2
83. –10p4 + 26p3
84. 15b3 + 5b2 – 45b
85. –7v4 + 42v2 – 7v
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
1. Find the missing length
to the nearest tenth.
2. Find the missing length
to the nearest tenth.
16.6
5.7
3. A triangle has sides of lengths
12 in., 14 in., and 16 in. Is the
triangle a right triangle?
no
4. A triangular flag is attached to a post. The bottom of the flag is 48 in.
above the ground. How far from the ground is the top of the flag?
57 in.
11-2
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
(For help, go to Lessons 11-2 and 2-7.)
Find the length of the hypotenuse with the given leg lengths.
If necessary, round to the nearest tenth.
1. a = 3, b = 4
2. a = 2, b = 5
3. a = 3, b = 8
4. a = 7, b = 5
For each set of values, find the mean.
5. x1 = 6, x2 = 14
6. y1 = –4, y2 = 8
7. x1 = –5, x2 = –7
8. y1 = –10, y2 = –3
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Solutions
1. a2 + b2 = c2
2.
32 + 42 = c2
9 + 16 = c2
25 = c2
c = 25 = 5
The length of the hypotenuse is 5.
3. a2 + b2 = c2
4.
32 + 82 = c2
9 + 64 = c2
73 = c2
c = 73 8.5
The length of the hypotenuse is
about 8.5.
6 + 14 20
= 2 = 10
2
6. mean = –4 + 8 = 4 = 2
2
2
5. mean =
a2 + b2 = c2
22 + 52 = c2
4 + 25 = c2
29 = c2
c = 29 5.4
The length of the hypotenuse is about 5.4.
a2 + b2 = c2
72 + 52 = c2
49 + 25 = c2
74 = c2
c = 74 8.6
The length of the hypotenuse is
about 8.6.
–12
–5 + (–7)
=
2 = –6
2
8. mean = –10 + (–3) = –13 = –6.5
2
2
7. mean =
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the distance between F(6, –9) and G(9, –4).
d=
( x2 – x1)2 + (y2 – y1)2 Use the distance formula.
d=
(9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2)
and (6, –9) for (x1, y1).
d=
32 + 52
Simplify within parentheses.
d=
34
Simplify to find the exact
distance.
d
5.8
Use a calculator. Round to
the nearest tenth.
The distance between F and G is about 5.8 units.
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the exact lengths of each side of quadrilateral EFGH.
Then find the perimeter to the nearest tenth.
The perimeter =
EF =
=
=
=
[4 – (–1)]2 + (3 + 5)2
52 + (–2)2
25 + 4
29
FG =
=
=
=
(3 – 4)2 + (–2 – 3)2
(–1)2 + (–5)2
1 + 25
26
GH =
|–2 – 3| = 5
EH =
=
=
=
[–2 – (–1)]2 + (–2 – 5)2
(–1)2 + (–7)2
1 + 49
50
29 +
26 + 5 +
50
11-3
22.6 units.
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the midpoint of CD.
x1+ x2 y1+ y2
2 ,
2
(–3) + 5 7 + 2
=
, 2
2
2 9
= 2,2
Substitute (–3, 7) for
(x1, y1) and (5, 2) for
(x2, y2).
Simplify each numerator.
1
Write 9 as a mixed
= 1, 42
2
number.
1
The midpoint of CD is M 1, 4 2 .
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
A circle is drawn on a coordinate plane. The endpoints
of the diameter are (–3, 5) and (4, –3). What are the coordinates of the
center of the circle?
x1+ x2 y1+ y2
2 ,
2
=
(–3) + 4 5 + (–3)
,
2
2
1 2
= 2,2
1
= 2, 1
1
The center of the circle is at 2 , 1 .
11-3
Substitute (–3, 5) for
(x1, y1) and (4, –3) for
(x2, y2).
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
pages 594–597 Exercises
22. MN
1. 15
12. (–2, 3)
2. 14
13. 5, –5 1
3. 10
14. – 2 1 , –1
5. 2.8
23. JK = 4; KL
24. TU
VT
2
2
4. 11.3
5.1; NP
15. (–4, 4)
2.2; LJ
7.6; UV
13.3
25. a. OR =
3.6; MP = 5
16.5;
29, ST =
b. 5 ; 5
2 2
6. 8.1
16. 8 1 , –9
7. 16
17. 10.6
8. 21.6
18. 9.4
9. (1, 6)
19. AB
4.1; BC
3.2; AC = 5
10. (–1, 12) 20. DE
6.1; EF
3.6; DF
5.7
11. (0, 0)
3.2; ST
5.7; RT
5.1
c. yes
2
21. RS
26. a. (20, 80), (–40, 30)
b. 78.1 ft
11-3
3.6
c. (–10, 55) or 55L10
29
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
27. Answers may vary. Sample:
Suppose you have (1, 1) and
(4, –3). To find the distance,
square the difference between
the x-coordinates. Square the
difference between the
y-coordinates. Find the sum
and take the square root,
so 9 + 16 = 5. To find the
midpoint, add x-coordinates
together and divide by 2.
Repeat for y-coordinates.
So, 1 + 4 1 – 3
5 , –1
,
=
2
2
2
28. Check students’ work.
29. a.
b. (0, – 1 )
2
c. one half mile south of the substation
30. about 9.5 km apart
31. a. 38.1 mi
b. 20 mi, 21.2 mi
c. 15 min, 16 min
32. Yes; all sides are congruent.
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
33. a. R(–27, –5)
b. PR = 13 3.6
RQ = 13 3.6
34. a. about 4.3 mi
b. about 17.4 mi
39. a. y = – 4 x + 1
3
b. 9 , – 7
5
5
c. 3
47. 11.7
48. 11.2
49. 10.2
50. 26
35. a. M(–0.5, 3); N(5.5, 3)
b. They are equal.
40. Yes; the distance from
each point to the
center is 5.
36. (x, y)
41. 19.1
53. 4.1
37. They are opposites.
42. 5
54. –14, 14
38. a. 5 units
b. Answers may vary. Sample:
(5, 0), (0, 5), (–5, 0), (0, –5),
(3, –4), (–3, 4), (–3, –4)
c. circle
43. 22.9
55. –10, 10
44. –5
56. no solution
45. 3
57. –5, 5
46. –8
58. no solution
11-3
51. 3.5
52. 8
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
59. – 2 , 2
3 3
60.
k2
+ 11k + 24
61. v2 + 2v – 35
62. 2p2 – 17p – 9
63. 8w4 + 19w2 + 11
64. 7t3 + 5t2 + 5t – 2
65. 6c3 – 27c2 + 33c + 24
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth.
7.2
2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth.
7.1
3. Find the midpoint of AB, A(3, 6) and B(0, 2).
(112 , 4)
4. Find the midpoint of CD, C(6, –4) and D(12, –2).
(9, – 3)
5. Find the perimeter of triangle RST to the
nearest tenth of a unit.
9.5 units
11-3
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(For help, go to Lesson 11-1.)
Simplify each radical expression.
1.
52
2.
200
3. 4
54
Rationalize each denominator.
5.
3
11
6.
5
8
7.
11-4
15
2x
4.
125x2
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Solutions
1.
52 = 4 • 13 = 4 • 13 = 2 13
2.
200 = 100 • 2 = 100 • 2 = 10 2
3. 4 54 = 4 9 • 6 = 4 • 9 • 6 = 4 • 3 • 6 = 12
4.
125x2 =
5.
3 =
11
5 =
8
15 =
2x
6.
7.
25 • 5 • x2 =
3 •
11
5 •
8
15 •
2x
11 =
11
8 =
8
2x =
2x
25 •
5 •
3 • 11 = 33
11 • 11
11
5•8 =
40 =
8
8•8
15 • 2x =
30x
2x
2x • 2x
11-4
x2 = 5x
6
5
4 • 10 =
8
4•
8
10 = 2 10 =
8
10
4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify 4
4
3+
3=4
3+1
= (4 + 1)
=5
3
3+
3
3
3.
Both terms contain
3.
Use the Distributive Property to
combine like radicals.
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
5–
Simplify 8
8
5–
45.
45 = 8
5+
9•5
9 is a perfect square and a factor of 45.
=8
5–
9•
=8
5–3
Use the Multiplication Property of
Square Roots.
Simplify 9.
= (8 – 3)
=5
5
5
5
5
Use the Distributive Property to
combine like terms.
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
5(
8 + 9) =
=
=2
5(
40 + 9
4•
8 + 9).
5
10 + 9
10 + 9
Use the Distributive Property.
5
Use the Multiplication Property
of Square Roots.
5
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify (
(
6–3
6 – 3 21)( 6 + 21)
= 36 + 126 – 3 126 – 3
21)(
441
6+
21).
Use FOIL.
=6–2
126 – 3(21)
Combine like radicals and
simplify 36 and 441.
=6–2
9 • 14 – 63
9 is a perfect square factor of 126.
=6–2
9•
Use the Multiplication Property of
Square Roots.
=6–6
14 – 63
= –57 – 6
14 – 63
Simplify
14
Simplify.
11-4
9.
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
8
7–
=
3
7+
7+
•
3
3
8
7–
3
.
Multiply the numerator and
denominator by the conjugate
of the denominator.
=
8( 7 + 3)
7–3
Multiply in the denominator.
=
8( 7 +
4
Simplify the denominator.
= 2(
7+
=2
7+2
3)
3)
3
Divide 8 and 4 by the common
factor 4.
Simplify the expression.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
A painting has a length : width ratio approximately equal to the
golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the
exact width of the painting in simplest radical form. Then find the
approximate width to the nearest inch.
Define:
51 = length of painting
x = width of painting
Relate: (1 +
Write:
5) : 2 = length : width
(1 + 5) = 51
x
2
x (1 +
5) = 102
x(1 + 5) =
102
(1 + 5)
(1 + 5)
Cross multiply.
Solve for x.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(continued)
x=
(1 –
102
•
(1 –
(1 + 5)
x=
102(1 – 5)
1–5
x=
102(1 –
–4
5)
x = – 51(1 –
5)
2
x = 31.51973343
5)
5)
Multiply the numerator and the
denominator by the conjugate
of the denominator.
Multiply in the denominator.
Simplify the denominator.
Divide 102 and –4 by the
common factor –2.
Use a calculator.
x 32
The exact width of the painting is – 51(1 –
2
5)
inches.
The approximate width of the painting is 32 inches.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
pages 603–606 Exercises
1. 5
12. 4
13. –2
6
2. 18
10
14.
3. –2
5
15. 8
4. 2
2
5
7
10
16. 4 – 4
7
5. 14
2
17. 9 +
6. –8
3
18. 6 – 2
2
3
3
7. yes
19. 3
5+2
8. yes
20. 3
2+6
21. 6 – 5
9. no
10. 4
11. –3
2
3
3
6
22. –9 – 14
23. 58 – 10
11-4
6
30
24. 11 – 4
7
25. 43 + 4
30
26. 32 + 9
11
27. 23 – 5
13
28. 2
7+2
29. –6
2
30. –4
6 – 12
31.
3(
10 +
5
3
2
5)
32. –5
11 – 5
3
33. 18
3+9
11
34. 10
2 + 10; 24.1
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
35. – 4 ; –1.3
46. 4
36. 6 – 4
47.
3
2; 0.3
3+4
48. 8
38. 5
10
49. 10 + 10
39. 6
2+6
50. 6
41. 8 + 2
42. 13 +
65 +
130 + 5
2
8
14
44. –24
45. –
2 units
10 units
10 units
52. Answers may vary. Sample:
8 2 + 4 3, 2 7 + 9 3,
6 5+3 7
53. a. The student simplified 48
as 2 24 instead of 2 12
or 4 3.
b. 2 6 + 4 3
54. a. 2 2 or 2.8 ft
b. s 2
15
43. 15 + 4
2 units
51. 4x + x
3–6
40. 22
6+6
10
5
37. 7.4 ft
3
2+3
2
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
55. 9.1%
63.
56. 12.8%
b
2
65. 23 7
21
n
59.
a+
64. 9 2
57. 15.5%
58. a. x 2
n–1
b. x 2
a + b =/
x
66. 8 15
15
ab
b
67. 2
60. about 251 years
68. 10
61. They are unlike radicals.
62. a. 1, 0, 1, 1;
4, 1, 5, 17;
5, 3, 8, 34;
8, 6, 14, 10;
10, 9, 19, 181
b. No; the only values it
worked for were 0 and 1.
2
69. 70
70. 2
2–
6–
71. a. 2 6
b. 2 13
c.
2(p + q)
72. B
11-4
3+3
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
73. I
74. [2] (3 5 – 2)( 5 + 5 2) =
3 25 + 15 10 – 10 – 5 4 =
3(5) + 15 10 – 10 – 5(2) =
15 + 15 10 – 10 – 10 = 5 + 14 10
[1] correct technique, but with a computational error
75. [4]
5
7+
21
•
7–
7–
21 Multiply the numerator and
21
denominator by the conjugate
of the denominator. Simplify the
denominator.
[3] correct steps but answer not completely simplified
[2] correct technique, but with a computational error
[1] correct answer but no work shown
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
76. 9.2 units
88. 4p2 + 28p + 49
77. 6.7 units
89. 25g2 – 49
78. 26.2 units
90. 9x2 – 1
79. (3, 5)
91. 1 k2 – 81
80. (–2, 6.5)
92. d 2 – 2.2d + 1.21
9
81. 0, 7
82. –2, 9
83. –9, –3
84. –4, 6
85. –15, –2
86. –3, – 1
87.
b2
2
+ 22b + 121
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify each expression.
1. 12 16 – 2
40
4. (
16
3–2
21)(
–123 + 3
7
2.
3+3
20 – 4
–2 5
21)
5
5.
3.
16
5–
–8
11-4
2( 2 + 3
2+3 6
7
5–8
7
3)
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(For help, go to Lesson 10-3.)
Evaluate each expression for the given value.
x – 3 for x = 16
1.
2.
x + 7 for x = 9 3. 2
x + 3 for x = 1
Simplify each expression.
4.
(
3)2
5.
(
x + 1) 2
11-5
6.
(
2x – 5) 2
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solutions
1.
x – 3 for x = 16:
2.
x + 7 for x = 9:
16 – 3 = 4 – 3 = 1
9+7=
3. 2
x + 3 for x = 1: 2
4. (
3)2 = 3
5. (
x + 1)2 = x + 1
6. (
2x – 5)2 = 2x – 5
16 = 4
1+3=2
4=2•2=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each equation. Check your answers.
a.
x–5 =4
x=9
(
x)2 = 92
Isolate the radical on the left side
of the equation.
Square each side.
x = 81
Check:
x–5
–5
9–5
= 4
4
4
Substitute 81 for x.
4=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
x–5 =4
b.
(
x – 5)2 = 42
x–5=9
Square each side.
Solve for x.
x = 21
Check:
x–5
21– 5
16
= 4
= 4
= 4
Substitute 21 for x.
4=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
On a roller coaster ride, your speed in a loop depends on the
height of the hill you have just come down and the radius of the loop in
feet. The equation v = 8 h – 2r gives the velocity v in feet per second
of a car at the top of the loop.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
The loop on a roller coaster ride has a radius of 18 ft.
Your car has a velocity of 120 ft/s at the top of the loop.
How high is the hill of the loop you have just come
down before going into the loop?
Solve v = 8
120 = 8
120
= 8
8
15 =
h – 2r for h when v = 120 and r = 18.
h – 2(18)
Substitute 120 for v and 18 for r.
h – 2(18)
8
h – 36
(15)2 = ( h – 36)2
225 = h – 36
261 = h
The hill is 261 ft high.
Divide each side by 8 to isolate the radical.
Simplify.
Square both sides.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve
(
3x – 4 =
3x – 4)2 = (
2x + 3)2
3x – 4 = 2x + 3
3x = 2x + 7
x=7
Check:
3x – 4 =
3(7) – 4
17 =
2x + 3.
Square both sides.
Simplify.
Add 4 to each side.
Subtract 2x from each side.
2x + 3
2(7) + 3
17
Substitute 7 for x.
The solution is 7.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve x =
x + 12.
(x)2 = (
x + 12)2
Square both sides.
x2 = x + 12
x2 – x – 12 = 0
Simplify.
(x – 4)(x + 3) = 0
Solve the quadratic equation by factoring.
(x – 4) = 0 or (x + 3) = 0
x = 4 or
x = –3
Use the Zero–Product Property.
Solve for x.
Check:
x =
x + 12
4
4 + 12
4 = 4
–3
–3 =/
–3 + 12
3
The solution to the original equation is 4.
The value –3 is an extraneous solution.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve
3x + 8 = 2.
3x = –6
(
3x)2 = (–6)2
Square both sides.
3x = 36
x = 12
Check:
3x
3(12)
36
6
+8=2
+8 2
+8 2
+ 8 =/ 2
Substitute 12 for x.
x = 12 does not solve the original equation.
3x + 8 = 2 has no solution.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
12. –2
24. 2
1. 4
13. 4
25. no solution
2. 49
14. 2.5
26. 4
3. 36
15. 2
27. 1.25 or 5
4. 137
16. –4
5. 15
17. none
28. 1 , 1
4
6. 16
18. – 1
7. 576 ft
19. –7
8. 602 watts
20. none
9. 4.5
21. 3
10. 3
22. 5
11. 7
23. no solution
pages 610–612 Exercises
4
2
11-5
29. a. 25
b. 11.25
30. about 2.5 in.
31. An extraneous solution is
a solution of a new equation
that does not satisfy the
original equation.
32. Answers may vary. Sample:
x – 2 = 7 – 2x , 3x = 3
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
33. 1600 ft
34. 3
35. no solution
36. no solution
37. 1, 6
38. 1.5
39. 11
44. a. 68 ft
b. 20.5 mi/h
c. As radius increases,
velocity decreases.
As height decreases,
velocity decreases.
d. Velocity depends upon
the difference of the
height and the radius.
45. a.
40. 0, 12
41. 3, 6
42. 44
43. no solution
b. approximately (6, 3.6)
c. 6; it is the x-coordinate of
the point of intersection.
46. a. V = 10x2
b. x =
c. 2, 3, 4, 5, 6, 7
47. a. – 7, 7
b. 49
c. In both cases 3 is
added to each side.
To solve the first
equation you find the
square roots of each
side, and in the second
equation you find the
square of each side.
48. –2, 8
49. 0
11-5
V
10
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
51. –1
52. Subtract 2x from each side.
Square both sides. Solve for x.
Check the solution if there is one.
60. C
15 – 5x = 4x – 3
15 – 5x = 4x – 3
–9x = –18
x =2
Check: 15 – 5(2)
4(2) – 3
5= 5
The solution is 2.
[1] correct technique with a minor error
OR correct answer, no work shown
61. [2]
53. The square of x – 1 will have
only 2 terms while x – 1 squared
will have 3 terms.
54. a. about 2.0 m
b. about 32.4 m
55. C
56. G
57. B
58. A
62. 5
5
63. 3
2+4
64.
65. 32
59. B
11-5
2
3
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
66. 4(
5–
67. 54
2
3)
78. 3(d – 1)(d + 5)
79. (4v – 5)(v – 5)
68. 2.5, –4.5
69. 8.4, –0.4
70. –0.2, –4.8
71. –10.7, 0.7
72. –11.7, 1.7
73. –1.6, 3.1
74. (x + 12)(x – 2)
75. (m – 13)(m – 1)
76. (b + 18)(b – 2)
77. (2p + 1)(p + 7)
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each radical equation.
25
1.
7x – 3 = 4
3.
2x + 7 =
5.
3x + 4 + 5 = 3
2.
7
5x – 8 5
3x – 2 =
4. x =
no solution
11-5
2x + 8
x+2
4
2
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
(For help, go to Lessons 10-1 and 10-3.)
Graph each pair of quadratic functions on the same graph.
1. y = x2, y = x2 + 3
2. y = x2, y = x2 – 4
Evaluate each expression for the given value of x.
3.
x for x = 4
4.
x + 7 – 3 for x = 2
5. 3
x + 2 for x = 9
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Solutions
1. y = x2, y = x2 + 3
2. y = x2, y = x2 – 4
3.
x for x = 4: 4 = 2
4.
x – 7 – 3 for x = 2: 2 + 7 – 3 = – 3 = 9 – 3 – 3 = 0
5. 3 x + 2 for x = 9: 3 9 + 2 = 3(3) + 2 = 9 + 2 = 11
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Find the domain of each function.
a. y =
x+5
x+5>
–0
Make the radicand >
– 0.
x>
– –5
The domain is the set of all numbers greater than or equal to –5.
b. y = 6
4x – 12
4x – 12 >
–0
Make the radicand >
– 0.
4x >
– 12
x>
– 3
The domain is the set of all numbers greater than or equal to 3.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
The size of a television screen is the length of the screen’s
diagonal d in inches.The equation d = 2A estimates the length of a
diagonal of a television with screen area A.
Graph the function.
Domain
2A >
–0
A>
–0
Screen
Area
(sq. in.)
0
50
100
200
300
400
Length of
Diagonal
(in.)
0
10
14.1
20
24.5
28.3
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph y =
y=
x + 4 by translating the graph of
x.
For the graph y =
the graph of y =
11-6
x + 4,
x is shifted 4 units up.
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph ƒ(x) =
y=
x + 3 by translating the graph of
x.
For the graph ƒ(x) =
the graph of y =
11-6
x + 3,
x is shifted to the left 3 units.
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
pages 616–619 Exercises
1.
x>
–2
2.
3
x>
–
3.
x>
–0
4.
x>
– –7
5.
x>
– –3
6.
x>
–5
7.
5
x>
––
8.
x>
– –2
9.
4
x>
–
13. x
0
3
5.3
y
0
3
4
11. x ƒ(x)
0
0
1
2
4
4
14. x
0
1
4
ƒ(x)
0
3
6
12. x
2
3
6
15. x
0
1
4
y
0
–3
–6
10. x
0
2
4.5
4
y
0
2
3
3
3
y
0
2
4
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
16.
h
0
1
4
v
0
8
16
22.
25.
23.
26.
24.
27.
17. D
18. A
19. C
20. B
21.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
28.
33. a-d.
a.
b.
c.
d.
29.
30. x >
– 4; y
31. x <
– 4; y
Answers may vary. Samples:
y= x+2
y= x+2
y=2 x
Check students’ work.
34. Translate the graph of
y = x 8 units to the left.
>
–0
>
–0
32. Form an inequality setting
the radicand >
– 0. Solve for x.
Answers may vary. Sample:
y= x–2
Domain: x – 2 >
–0
x>
–2
35. Translate the graph of
y = x 10 units down.
36. Translate the graph of
y = x 12 units up.
37. Translate the graph of
y = x 9 units right.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
38. x
2.5
3.5
6.5
y
0
1
2
39. x
0
1
2
4
ƒ(x)
0
4
5.7
8
40. x
–6
–5
–2
0
y
0
1
2
2.4
41. x
0
2
4
8
y
0
1
1.4
2
42. x
2
3
6
y
3
4
5
43. x
–2
–1
2
ƒ(x)
–4
–3
–2
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
44. x
0
1
2
3
y
3
4.4
5
5.4
47. B
48. D
49. A
50. C
45.
x
–3
–2
–1
0
y
1
2.4
3
3.4
46. x
1
2
3
4
y
–2
–0.3
0.4
1
51. a. p > 0
b.
c. about 45 lb/in.2
52. a. no
b. Answers may vary.
Sample: The graph
of y = x is the first
quadrant of the graph
of x = y2.
c. y = – x
53. y = 3 x rises more
steeply because
3 x > 3x for all
positive values of x.
54. False; x must equal 81.
55. False; only combine
like terms.
56. true
57. False; x = –1.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
58. a. about 213 cameras
b. month 4
61. a. i.
ii.
59. a.
iii.
iv.
b. y = |x| + 5
b. The greater the absolute value of n,
60. Translate the graph of
the steeper the graph. If n < 0, then the
y = x right 2 units and up 3 units.
graph lies in Quadrant II. If
n > 0, the graph lies in Quadrant I.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
62. Check students’ work.
70. 16
63. B
71. 7
64. H
72. 169
65. C
73. 14.76
66. H
74. no solution
67. B
75. 2
3
68. H
69. [2] x
y
6
0
7
1
8 1.4
9 1.7
10 2
[1] incorrect coordinates on graph
11-6
76. –2 – 3 2 , –2 + 3 2
2
2
77. 4 –
39, 4 +
39
78. no solution
11, –1 + 11
3
3
79. –1 –
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
80. no solution
421 , –13 + 421
18
18
81. –13 –
82. (2x + 1)(x – 4)
83. (3x – 5)(x + 2)
84. (2x + 1)(2x + 9)
85. 2(x – 8)(x + 3)
86. 4(x2 – x – 15)
87. x(x – 13)(x + 1)
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
1. Find the domain of the function ƒ(x) =
2. Graph y = 3
3. Graph y =
2x – 4.
x.
x – 3.
4. Describe how to translate the graph of y = x
to obtain the graph of the function y = x – 15.
Shift the graph to the right 15 units.
11-6
x>2
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(For help, go to Lessons 1-6 and 4-2.)
Let c = A , s = O, t = O . Calculate c, s, and t for the given values.
H
H
A
1. A = 3, O = 4, H = 5
2. A = 5, O = 12, H = 13
Solve each equation.
3.
15 0.75
= 1
x
5. 0.84 = 21
1
x
x
0.34
4. 20 = 1
6.
x
14
=
0.52
1
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Solutions
1. For A = 3, O = 4, H = 5:
c=
A
3
O 4
O 3
= ,s= = ,t=
=
H
5
H
5
A 4
2. For A = 5, O = 12, H = 13:
c = A = 5 , s = O = 12, t = O = 12
H 13
15 = 0.75
x
1
3.
H
13
A
5
4.
x = 0.34
1
20
0.75x = 15
x = 20(0.34)
x = 20
5. 0.84 = 21
1
x
x = 6.8
6. x
0.52
0.84 = 21
= 14
1
x = 14(0.52)
x = 25
x = 7.28
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Use the triangle. Find sin A, cos A, and tan A.
opposite leg
6
3
sin A = hypotenuse =
=
10 5
adjacent leg
8
4
cos A = hypotenuse =
=
10 5
opposite leg
tan A = adjacent leg = 6 = 3
8
4
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find sin 40° by using a calculator.
To find sin 40°, press
Use degree mode when finding
trigonometric ratios.
Rounded to the nearest ten-thousandth, the sin 40°
is 0.6428.
11-7
40
.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find the value of x in the triangle.
Step 1: Decide which trigonometric ratio to use.
You know the angle and the length of the hypotenuse.
You are trying to find the adjacent side. Use the cosine.
Step 2: Write an equation and solve.
adjacent leg
hypotenuse
x
cos 30° =
15
cos 30° =
x = 15(cos 30°)
15
30
12.99038106
x 13.0
The value of x is about 13.0.
Substitute x for adjacent leg
and 15 for hypotenuse.
Solve for x.
Use a calculator.
Round to the nearest tenth.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Suppose the angle of elevation from a rowboat to the top of a
lighthouse is 708. You know that the lighthouse is 70 ft tall. How far
from the lighthouse is the rowboat? Round your answer to the nearest
foot.
Draw a diagram.
Define: Let x = the distance from the boat to the lighthouse.
Relate: You know the angle of elevation and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write:
opposite leg
tan A = adjacent leg
70
tan 70° = X
x(tan 70°) = 70
Substitute for the angle and the sides.
Multiply each side by x.
x=
70
tan 70
Divide each side by tan 70°.
x
25.4779164
Use a calculator.
x
25
Round to the nearest unit.
The rowboat is about 25 feet from the lighthouse.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
A pilot is flying a plane 15,000 ft above the ground. The pilot
begins a 3° descent to an airport runway. How far is the airplane from
the start of the runway (in ground distance)?
Draw a diagram.
Define: Let x = the ground distance from the start of the runway.
Relate: You know the angle of depression and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write:
opposite leg
tan A = adjacent leg
tan 3° =
15,000
x
x(tan 3°) = 15,000
Substitute for the angle and the sides.
Multiply each side by x.
x=
15,000
tan 3
Divide each side by tan 3°.
x
286217.05
Use a calculator.
x
290,000
Round to the nearest 10,000 feet.
The airplane is about 290,000 feet (or about 55 miles)
from the start of the runway.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
pages 625–627 Exercises
1.
2.
3.
4.
5.
6.
3
5
4
5
3
4
4
5
3
5
4
3
7. 0.5299
8. 0.5736
9. 1.2799
10. 0.9962
11. 0.9659
23. b = 15; sin A = 8 ;
12. 5.5
17
15
cos A = ; tan A =
17
24. a = 24; sin A = 12 ;
13
5
cos A = ; tan A =
13
13. 10.4
14. 19.2
15. 38.1
16. 66.0
25. AC
6; AB
17. 21.1
26. AC
36; BC
18. about 2.0 mi
27. BC
6; AB
18
19. about 172 ft
28. AC
4; AB
50
20. about 816 ft
29. about 55 m
21. about 0.4 mi
30. about 4.4 m
22. c = 29; sin A = 21 ;
31. a. 1,720,000 ft
b. 326 mi
cos A = 20 ; tan A = 21
29
29
11-7
20
8
22
8
15
12
5
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
40. Answers may vary.
Sample:
about 8.2 cm
32. 12 m
33. about 6.8 m
34. 514.3
35. 4.5
36. 78.4
37. q
6.1; r
7.9
38. a. about 177 ft
b. about 86 ft
39. a. about 252 ft
b. about 377 ft
49. [2] sin A 0.6271,
cos A 0.7881,
tan A 0.7957
[1] at least one correct
equation
50. [4]
850
41. about 203 ft
42. about 5938 m
43. about 57 ft
tan 56° = x
x 573.3
573 ft
44. 550 ft
45. 0.25
46. a.
b.
47. C
about 229 km
about 4 km
48. G
11-7
[3] correct equation, but
minor computational error
[2] incorrect equation used
[1] no work shown
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
51.
54. 2
55. 0
56. 1
57. (n – 20)(n + 20)
52.
58. (x – 15)2
59. (10p – 7)(10p + 7)
60. 1 d – 3
4
53.
1d + 3
4
2
2
61. 2(7w – 8)(7w + 8)
62. (x + 13)2
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
1. Use the figure to find sin A, cos A, and tan A.
21 20 21
,
,
29 29 20
2. Find the value of x to the nearest tenth.
6.9
3. A group of skateboarders wants to build a ramp with an
angle of incline of 14°. What should the rise be for every
10 meters of run? about 2.5 m
4. A park ranger on a 220 ft tower spots a fire at an angle
of depression of 4°. How far is the fire from the base of
the tower? about 3146 ft
5. A wheelchair ramp is to have an angle of 3.5° with the ground.
The deck at the top of the ramp is 18 in. above the ground.
How long should the ramp be? about 294.3 in.
11-7
Radical Expressions and Equations
ALGEBRA 1 CHAPTER 11
1. yes
2. no
12. – 1 , – 2
13.
3. no
4. yes
5. 57.8 cm
6. 9.8 units
7. 4.1 units
8. 19.8 units
9. 7.6 units
10.
21, 2
2
11.
1
1
,–3
2
2
2
– 1 , 5 (– 1 , 5)
2
2
14. 29.5
15. 27.0
23. 11
3
24. 35
5 – 15
25. 8
2–8
26. 5
3
27. 2
16. 13 ft
10 –
7
3
6
17. 15 mi
28. Answers may vary. Sample:
2 5+4 5=6 5
18.
29. B
2
19. 3
30. 4
5
20. 4
3
31. 13
21. 4
6
32. 6
22. 17
2
33. 7
11-A
Radical Expressions and Equations
ALGEBRA 1 CHAPTER 11
5
35.
36. 2
41. 24 cm
38. x >
– 0;
34. 9
42. The graph of y = x
is shifted 3 units down.
10.65
5 ft
43. r =
37. x >
– 0;
V
h
44. about 32 ft
45. about 8.1
39. x >
– 4;
46. about 12.1
47. about 1.106
48. about 0.7431
40. x >
– –9;
49. BC
6.2, AC
50. about 5 ft
11-A
3.3