8005 ENGLISH MEDIEVAL MUSIC - LEMS

8005 ENGLISH MEDIEVAL MUSIC - LEMS

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  • expression écrite - matière potentielle : musicians
  • expression écrite - matière potentielle : scale movement of song
1 (LEMS 8005) ENGLISH MEDIEVAL SONGS Music of the Middle Ages Vol. 5 The 12th & 13th Centuries Russell Oberlin, Countertenor Seymour Barab, Viol Very few early English songs have been preserved; the eight on this record are most of what has been left to us. Various historical reasons for this ar possible, such as the destruction of the property of the Roman Catholic Church in England, including of course, the monastic libraries.
  • helpe at thine nede
  • part of the lyric poetry movement
  • liest ant
  • life thu shalt
  • sainte marie virgine
  • songs
  • construction of a system of music for the expression
  • song
  • music

Sujets

Informations

Publié par
Nombre de lectures 42
Langue English
Poids de l'ouvrage 1 Mo
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Answers to Exercises
CHAPTER 4 • CHAPTER CHAPTER 4 • CHAPTER angles P and D can be drawn at each endpoint4
using the protractor.
LESSON 4.1 Q
551. The angle measures change, but the sum
remains 180°.
2. 73°
40 85
P D7 cm3. 60°
4. 110° 17. The third angles of the triangles also have the
same measures; are equal in measure 5. 24°
18. You know from the Triangle Sum Conjecture6. 3 360° 180° 900°
that mA mB mC 180°, and mD 7. 3 180° 180° 360°
mE mF 180°. By the transitive property,
8. 69°; 47°; 116°; 93°; 86°
mA mB mC mD mE mF.
9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50° You also know that mA mD, and mB
10. mE. You can substitute for mD and mE in the
R longer equation to get mA mB mC
mA mB mF. Subtracting equal terms
from both sides, you are left with mC mF.
M A
19. For any triangle, the sum of the angle measures
11. is 180°, by the Triangle Sum Conjecture. Since the
G
triangle is equiangular, each angle has the same
L
measure, say x. So x x x 180°, and x 60°.
20. false12. First construct E, using the method used in
Exercise 10.
E
A R
Fold13. 21. false
ER
M A G RAL
22. false
14. From the Triangle Sum Conjecture
mA mS mM 180°. Because M is a
right angle, mM 90°. By substitution,
mA mS 90° 180°. By subtraction,
mA mS 90°. So two wrongs make a right!
23. false
15. Answers will vary. See the proof on page 202.
To prove the Triangle Sum Conjecture, the Linear
Pair Conjecture and the Alternate Interior Angles
Conjecture must be accepted as true.
24. true
16. It is easier to draw PDQ if the Triangle Sum
25. eight; 100
Conjecture is used to find that the measure of
D is 85°. Then PD can be drawn to be 7 cm, and
48 ANSWERS TO EXERCISES
Answers to ExercisesAnswers to Exercises
F15.LESSON 4.2
1. 79° D E
2. 54° C
3. 107.5°
A B4. 44°; 35 cm
P5. 76°; 3.5 cm 16.
6. 72°; 36°; 8.6 cm
M N7. 78°; 93 cm
K
8. 75 m; 81°
G H9. 160 in.; 126°
10. a 124°, b 56°, c 56°, d 38°, e 38°, 17. possible answer:
f 76°, g 66°, h 104°, k 76°, n 86°, Fold 1 Fold 3
p 38°; Possible explanation: The angles with
Fold 2
Fold 4measures 66° and d form a triangle with the angle
60105with measure e and its adjacent angle. Because d,
45
e, and the adjacent angle are all congruent,
3d 66° 180°. Solve to get d 38°. This is
18. perpendicular
also the measure of one of the base angles of the
19. parallelisosceles triangle with vertex angle measure h.
20. parallelUsing the Isosceles Triangle Conjecture, the other
21. neitherbase angle measures d, so 2d h 180°, or
76° h 180°. Therefore, h 104°. 22. parallelogram
11. a 36°, b 36°, c 72°, d 108°, e 36°; 23. 40
none
12a. Yes. Two sides are radii of a circle. Radii must
0 8 16 24 32 40be congruent; therefore, each triangle must be
24. New: (6, 3), (2, 5), (3, 0). Triangles areisosceles.
congruent.12b. 60°
25. New: (3, 3), (3, 1), (1, 5). Triangles13. NCA
are congruent.
14. IEC
ANSWERS TO EXERCISES 49USING YOUR ALGEBRA SKILLS 4 15. You get an equation that is always false, so
there is no solution to the equation.
1. false 2. true
16. Camella is not correct. Because the equation
3. not a solution 4. solution 0 0 is always true, the truth of the equation does
5. not a solution 6. x 7 not depend on the value of x. Therefore, x can be
7. y 4 8. x 8 any real number. Camella’s answer, x 0, is only
one of infinitely many solutions.19. x 4.2 10. n 2 17.
2x
11. x 2 12. t 18
x
2x2 9 13. n 14a. x 5 4 If x equals the measure of the vertex angle, then
9
14b. x ; The two methods produce identical the base angles each measure 2x. Applying the4
results. Multiplying by the lowest common Triangle Sum Conjecture results in the following
denominator (which is comprised of the factors of equation:
both denominators) and then reducing common x 2x 2x 180°
factors (which clears the denominators on either 5x 180°
side) is the same as simply multiplying each numerator
x 36°
by the opposite denominator (or cross multiplying).
The measure of the vertex angle is 36° and theAlgebraically you could show that the two methods
measure of each base angle is 72°.are equivalent as follows:
a c b d
a c bd bd b d
abd bcd b d
ad bc
The method of “clearing fractions” results in the
method of “cross multiplying.”
50 ANSWERS TO EXERCISES
Answers to ExercisesAnswers to Exercises
14. 135°LESSON 4.3
15. 72°
1. yes
16. 72°
2. no
17. a b c 180° and x c 180°. Subtract c
45
from both sides of both equations to get x 180 c9
and a b 180 c. Substitute a b for 180 c3. no
in the first equation to get x a b.56
12 18. 45°
4. yes 19. a 52°, b 38°, c 110°, d 35°
5. a, b, c 20. a 90°, b 68°, c 112°, d 112°, e 68°,
6. c, b, a f 56°, g 124°, h 124°
7. b, a, c 21. By the Triangle Sum Conjecture, the third
8. a, c, b angle must measure 36° in the small triangle, but it
measures 32° in the large triangle. These are the9. a, b, c
same angle, so they can’t have different measures.10. v, z, y, w, x
22. ABE11. 6 length 102
23. FNK12. By the Triangle Inequality Conjecture, the
24. cannot be determinedsum of 11 cm and 25 cm should be greater than
48 cm.
13. b 55°, but 55° 130° 180°, which is
impossible by the Triangle Sum Conjecture.
ANSWERS TO EXERCISES 51LESSON 4.4 13. Cannot be determined. SSA is not a congruence
conjecture.
1. Answers will vary. Possible answer: If three
14. AIN by SSS or SAS
sides of one triangle are congruent to three sides of
15. Cannot be determined. Parts do not correspond.another triangle, then the triangles are congruent
16. SAO by SAS(all corresponding angles are also congruent).
17. Cannot be determined. Parts do not correspond.2. Answers will vary. Possible answer: The picture
18. RAY by SASstatement means that if two sides of one triangle
are congruent to two sides of another triangle, and 19. The midpoint of SD and PR is (0, 0).
the angles between those sides are also congruent, Therefore, DRO SPO by SAS.
then the two triangles are congruent. 20. Because the LEV is marking out two triangles
If you know this: then you also know this: that are congruent by SAS, measuring the length
of the segment leading to the finish will also
approximate the distance across the crater.
21. 22.
3. Answers will vary. Possible answer:
23. a 37°, b 143°, c 37°, d 58°,4. SAS
e 37°, f 53°, g 48°, h 84°, k 96°,5. SSS
m 26°, p 69°, r 111°, s 69°; Possible 6. cannot be determined
explanation: The angle with measure h is the vertex
7. SSS
angle of an isosceles triangle with base angles
8. SAS measuring 48°, so h 2(48) 180°, and h 84°.
9. SSS (and the Converse of the Isosceles Triangle The angle with measure s and the angle with
Conjecture) measure p are corresponding angles formed by
10. yes, ABC ADE by SAS parallel lines, so s p 69°.
11. Possible answer: Boards nailed diagonally in 24. 3 cm third side 19 cm
the corners of the gate form triangles in those 25. See table below.
corners. Triangles are rigid, so the triangles in the 13 3 26a. y 6 b. y c. y x 24gate’s corners will increase the stability of those 3
corners and keep them from changing shape. 27. (5, 3)
12. FLE by SSS
25. (Lesson 4.4)
Side length 12345 … n …20
Elbows 4444444
T’s 04 812 16 76
Crosses 01 49 16 361
24n 4 (n 1)
52 ANSWERS TO EXERCISES
Answers to ExercisesAnswers to Exercises
CALIFORNIA
O REG O N
N E VA D A
I DAHO
ARIZ ONA
U TA H
21. The construction is the same as the LESSON 4.5
construction using ASA once you find the third
1. If two angles and the included side of one triangle angle, which is used here. (Finding the third angle
are congruent to the corresponding side and angles is not shown.)
of another triangle, then the triangles are congruent.
2. If two angles and a non-included side of one
triangle are congruent to the corresponding side
and angles of another triangle, then the triangles
are congruent.
22. Construction will show a similar but larger
If you know this: then you also know this: (or smaller) triangle constructed from a drawn
triangle by duplicating two angles on either end of a
new side that is not congruent to the corresponding
side.
23. Draw a line segment. Construct a perpendicular.3. Answers will vary. Possible answer:
Bisect the right angle. Construct a triangle with
two congruent sides and with a vertex that
measures 135°.
24. 125
4. ASA 25. False. One possible counterexample is a kite.
5. cannot be determined 26. None. One triangle is determined by SAS.
6. SAA L
7. cannot be determined
8. ASA
9. cannot be determined K M
10. FED by SSS
11. WTA by ASA or SAA 27.
12. SAT by SAS
13. PRN by ASA or SAS; SRE by ASA
14. Cannot be determined. Parts do not
correspond.
15. MRA by SAS
28a. about 100 km southeast of San Francisco
16. Cannot be determined.AAA does not guarantee
28b. Yes. No, two towns would narrow it downcongruence.
to two locations. The third circle narrows it down
17. WKL by ASA
to one.
18. Yes, ABC ADE by SAA or ASA.
19. Slope AB slope CD 3 and slope BC Boise
1 Eureka slope DA ,so AB BC,CD DA, and
3
BC DA. ABC CDA by SAA.
ElkoReno
Sacramento
San 20. Francisco
Las
Vegas
Los Angeles
Miles
0 50 100 200
0 100 400
Kilometers
ANSWERS TO EXERCISES 53


13. cannot be determinedLESSON 4.6
14. KEI by ASA
1. Yes. BD BD (same segment), A C
15. UTE by SAS(given), and ABD CBD (given), so DBA
16. DBC by SAA. AB CB by CPCTC.
2. Yes. CN WN and C W (given), and
RNC ONW (vertical angles), CNR
WON by ASA. RN ON by CPCTC.
3. Cannot be determined. The congruent parts
lead to the ambiguous case SSA. 17.
4. Yes. S I, G A (given), and TS IT
(definition of midpoint), so TIA TSG by
SAA. SG IA by CPCTC.
5. Yes. FO FR and UO UR (given), and UF
UF (same segment), so FOU FRU by SSS. 18. a 112°, b 68°, c 44°, d 44°, e 136°,
O R by CPCTC. f 68°, g 68°, h 56°, k 68°, l 56°, m 124°;
Possible explanation: f and g are measures of base6. Yes. MN MA and ME MR (given), and
angles of an isosceles triangle, so f g.The vertexM M (same angle), so EMA RMN by
angle measure is 44°, so subtract 44° from 180° andSAS. E R by CPCTC.
divide by 2 to get f 68°. The angle with measure 7. Yes. BT EU and BU ET (given), and
m is the exterior angle of a triangle. Add the remote UT UT (same segment), so TUB UTE by
interior angle measures 56° and 68° to get SSS. B E by CPCTC.
m 124°.
8. Cannot be determined. HLF LHA by
19. ASA. The “long segment in the sand” is aASA, but HA and HF are not corresponding sides.
shared side of both triangles
9. Cannot be determined. AAA does not guaran-
20. (4, 1)tee congruence.
21. See table below.10. Yes. The triangles are congruent by SAS.
22. Value C is always decreasing.11. Yes. The triangles are congruent by SAS, and
23. x 3, y 10the angles are congruent by CPCTC.
12. Draw AC and DF to form ABC and DEF.
AB CB DE FE because all were drawn with
the same radius. AC DF for the same reason.
ABC DEF by SSS. Therefore, B E by
CPCTC.
21. (Lesson 4.6)
Number of sides 34 5 6 7 … 12 … n
01 2 3 4 … 9 … n 3Number of struts needed
to make polygon rigid
54 ANSWERS TO EXERCISES
Answers to Exercises




Answers to Exercises
3. See flowchart below.LESSON 4.7
4. See flowchart below.
1. See flowchart below.
2. See flowchart below.
1. (Lesson 4.7)
1 SE SU
? Given
ESM USO
2 4 5? ?E U MS SO
? ?Given ASA Congruence CPCTC
Conjecture
3 1 2
? Vertical Angles Conjecture
2. (Lesson 4.7)
1 3 I is midpoint of CM CI IM

Given Definition
of midpoint
2 4 6 7 ? ? ?I is midpoint of BL IL IB CL MB
? ? ?Definition ofGiven CPCTC CIL MIB
midpoint by SAS
5 1 2
? Vertical Angles Conjecture
3. (Lesson 4.7)
1 3NS is a median NS NS
Given Same segment
EESN
2 4 6 7? ?S is a midpoint WS SE WSN W
? ?Definition Definition SSS CPCTC
of median of midpoint
5 WN NE
? Given
4. (Lesson 4.7)
1 2NS is an ?1 2
angle bisector
?Definition of Given angle bisector WNS ENS
7 NEW is3 5 6? ?W E WN NE isosceles
? ? ?Given SAA CPCTC ? Definition of
isosceles triangle
4 NS NS
? Same segment
ANSWERS TO EXERCISES 555. See flowchart below. 12. ACK by SSS
6. Given: ABC with BA BC,CD AD 13. a 72°, b 36°, c 144°, d 36°, e 144°,
f 18°, g 162°, h 144°, j 36°, k 54°,Show: BD is the angle bisector of ABC.
m 126°
14. The circumcenter is equidistant from all threeA
vertices because it is on the perpendicular bisector
1 D
B of each side. Every point on the perpendicular2
bisector of a segment is equidistant from the
C endpoints. Similarly, the incenter is equidistant
from all three sides because it is on the angle
BA BC CD AD BD BD bisector of each angle, and every point on an angle
Same segmentGiven Given bisector is equidistant from the sides of the angle.
15. ASA. The fishing pole forms the side.
“Perpendicular to the ground” forms one angle.ABD CBD
SSS “Same angle on her line of sight” forms the other
angle.
1 2 216.
CPCTC 7
17.

BD bisects ABC
Definition of angle
bisector
7. The angle bisector does not go to the midpoint
of the opposite side in every triangle, only in an
18. y
isosceles triangle.
4 Y X8. NE, because it is across from the smallest angle
BOin NAE. It is shorter than AE, which is across
x
B' Y'4from the smallest angle in LAE.
O' X'9. The triangles are congruent by SSS, so the two
central angles cannot have different measures.
10. PRN by ASA; SRE by ASA
11. Cannot be determined. Parts do not
correspond.
5. (Lesson 4.7)
1 3SA NE 3 4
? AIA ConjectureGiven
ESN ANS
42 6 71 2 ? ? ? ?SE NA SA NE
AIA Conjecture? ? ? ASA ? CPCTCGiven
This proof shows that in a parallelogram,
5 opposite sides are congruent.SN SN
Same segment
56 ANSWERS TO EXERCISES
Answers to Exercises
Answers to Exercises
1 2 37.LESSON 4.8 AC BC CD CD D is
midpointGiven Same segment
of AB1. 6
Given2. 90°; 18°
5 4ADC BDC AD BD3. 45°
SSS Def. of midpoint
4. See flowchart below.
6 7ACD DCB CD is angle5. See flowchart below.
bisector of ACBCPCTC
6. 1 2 Def. of angle bisectorIsosceles ABC ADC BDC
8with AC BC CD is altitudeConjecture A 9
CD ABand CD bisects of ABC(Exercise 4)
C Conjecture B Def. of altitude
(Exercise 5)Given
3 4 8. Yes. First show that the three exterior trianglesAD BD CD is a median
are congruent by SAS.CPCTC Def. of median
4. (Lesson 4.8)
1 2 ? 1 2CD is the bisector of C
Given Definition of
angle bisector
3 ABC is isosceles 5 ADC BDC
with AC BC
? SAS
Given
4 CD CD
Same segment
5. (Lesson 4.8)
1 ABC is isosceles with
2 4 AC BC, and CD is ADC BDC 1 2
the bisector of C ? CPCTCConjecture A
6 1 and 2 are Given
right angles
Congruent
3 5 supplementary 1 and 2 form 1 and 2 are
angles are 90 a linear pair supplementary
Definition of Linear Pair
7 linear pair Conjecture CD AB
Definition of ?
perpendicular
8 ?CD is an altitude
Definition of
altitude
ANSWERS TO EXERCISES 57