Lesson Warm Up 117 1. parallel 2. (3, 4, â4) 3. C
Lesson Warm Up 117
117
c. Strong positive correlation
1. parallel
Distance
2. (3, 4, −4) 3. C Lesson Practice 117
Time
a. In A, the line has a slope that does not represent the data, so it is not a good choice. In B, the line goes through the points and the slope shows the trend of the data, so it is a good choice. In C, the line has the correct slope, but does not include the data points, so it is not a good choice.
d. about 62 weeks
b. weak, negative correlation
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–1
Saxon Geometry
Lesson
1 , K(-1.5, 4.5), 10. J -1, 3_ 2
)
(
Practice 117 1. The error is in using 5 as the height. The Pythagorean Theorem must be used to find the height. The height is 3. The correct solution is 32π, or approximately 3 100.5 units .
1 ,1 and L -2 _ 2
)
(
11. The correlation is positive and strong. y
x
2. always 3. approximately 2938 ft
1 12. tan θ = _ √ csc 2θ - 1 13. 97 m
2
4. y = 1
14. A negative correlation. As the absences increase and assignments are missed, the student’s grades can be expected to decrease.
5. RT = 24, m∠R = 83°, m∠T = 70° 6. 40.5 m 7.
117
2
S
T
S
T
V
U
V
U
15. 3.25 m 16. 1 m : 1 in.
8. (-1, 2), (1, 0), (-2, -1); sample: (0, 0)
17. approximately 18.5 hours
9. 9.5
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–2
Saxon Geometry
Lesson 18. By substituting the equation of a line into the equation of the circle, the largest possible polynomial is a quadratic in one variable. A quadratic function in one variable has at most two solutions. Hence, there are at most two points of intersection between a line and a circle. 19. zero lines of symmetry, rotational symmetry of order 2 20. D 21. 78.25° 22. The triangle was reflected in the y-axis because when the triangle matrix is multiplied by the reflection matrix ⎡-1 0 ⎤ ⎢ , , the result is the ⎣ 0 1⎦ image matrix.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
117
23. four; They are the planes containing the lines of symmetry of the square base. 24. Assume a right triangle VBN has an obtuse angle. Let ∠V be right and ∠N be obtuse. Then m∠B + m∠N = 90°, so m∠N = 90° - m∠B. By the definition of obtuse angles, m∠N > 90°. Substituting 90° < 90° - m∠B, which simplifies to 0° > m∠B. By the Protractor Postulate, a triangle cannot have an angle with a measure less than zero, so the assumption is contradicted and a right triangle cannot have an obtuse angle. 25. 50 in
2
26. 464.4 m
LSN 117–3
3
Saxon Geometry
Lesson
117
27. It is known that sin θ . Since tan θ = _ 2
cos θ
2
cos θ + sin θ = 1, 2 1 - sin θ . cos θ = √ Substitute to find that sin θ tan θ = _ 2 √ 1 - sin θ 28. 4.91 m/sec 2 at 2.34° to the horizontal 29. a. 199 b. 517 30. B
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–4
Saxon Geometry
117
c. Strong positive correlation
1. parallel
Distance
2. (3, 4, −4) 3. C Lesson Practice 117
Time
a. In A, the line has a slope that does not represent the data, so it is not a good choice. In B, the line goes through the points and the slope shows the trend of the data, so it is a good choice. In C, the line has the correct slope, but does not include the data points, so it is not a good choice.
d. about 62 weeks
b. weak, negative correlation
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–1
Saxon Geometry
Lesson
1 , K(-1.5, 4.5), 10. J -1, 3_ 2
)
(
Practice 117 1. The error is in using 5 as the height. The Pythagorean Theorem must be used to find the height. The height is 3. The correct solution is 32π, or approximately 3 100.5 units .
1 ,1 and L -2 _ 2
)
(
11. The correlation is positive and strong. y
x
2. always 3. approximately 2938 ft
1 12. tan θ = _ √ csc 2θ - 1 13. 97 m
2
4. y = 1
14. A negative correlation. As the absences increase and assignments are missed, the student’s grades can be expected to decrease.
5. RT = 24, m∠R = 83°, m∠T = 70° 6. 40.5 m 7.
117
2
S
T
S
T
V
U
V
U
15. 3.25 m 16. 1 m : 1 in.
8. (-1, 2), (1, 0), (-2, -1); sample: (0, 0)
17. approximately 18.5 hours
9. 9.5
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–2
Saxon Geometry
Lesson 18. By substituting the equation of a line into the equation of the circle, the largest possible polynomial is a quadratic in one variable. A quadratic function in one variable has at most two solutions. Hence, there are at most two points of intersection between a line and a circle. 19. zero lines of symmetry, rotational symmetry of order 2 20. D 21. 78.25° 22. The triangle was reflected in the y-axis because when the triangle matrix is multiplied by the reflection matrix ⎡-1 0 ⎤ ⎢ , , the result is the ⎣ 0 1⎦ image matrix.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
117
23. four; They are the planes containing the lines of symmetry of the square base. 24. Assume a right triangle VBN has an obtuse angle. Let ∠V be right and ∠N be obtuse. Then m∠B + m∠N = 90°, so m∠N = 90° - m∠B. By the definition of obtuse angles, m∠N > 90°. Substituting 90° < 90° - m∠B, which simplifies to 0° > m∠B. By the Protractor Postulate, a triangle cannot have an angle with a measure less than zero, so the assumption is contradicted and a right triangle cannot have an obtuse angle. 25. 50 in
2
26. 464.4 m
LSN 117–3
3
Saxon Geometry
Lesson
117
27. It is known that sin θ . Since tan θ = _ 2
cos θ
2
cos θ + sin θ = 1, 2 1 - sin θ . cos θ = √ Substitute to find that sin θ tan θ = _ 2 √ 1 - sin θ 28. 4.91 m/sec 2 at 2.34° to the horizontal 29. a. 199 b. 517 30. B
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 117–4
Saxon Geometry
