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Document 1803838
Chapter 2,
continued
3. A;
6. If-then: If an angle measures 348, then it is an
acute angle.
You know that Ž1 and Ž2 are a pair of vertical
angles because mŽ1 5 mŽ2, while mŽ3 5 3mŽ1.
mŽ3 5 mŽ4 because Ž3 and Ž4 are the other pair
of vertical angles. Ž1 and Ž3 are linear pair, so
mŽ1 1 mŽ3 5 1808. Let mŽ1 5 x, then mŽ3 5 3x
and x 1 3x 5 180, or x 5 45. So, mŽ1 5 458,
mŽ2 5 458, mŽ3 5 3(458) 5 1358, and mŽ4 5 1358.
4. J;
Converse: If an angle is an acute angle then it
measures 348.
Inverse: If an angle does not measure 348, then it is not
an acute angle.
Contrapositive: If an angle is not an acute angle, then it
does not measure 348.
7. This is a valid definition because it can be written as a
T 5 c(1 1 s)
true biconditional statement.
T 5 c 1 cs
8. All the interior angles of a polygon are congruent if and
T 2 c 5 cs
only if the polygon is an equiangular polygon.
T2c
}5s
c
9. Because Ž B is a right angle it satisfies the hypothesis, so
c
T
}2}5s
c
c
10. The conclusion of the second statement is the hypothesis
the conclusion is also true. So, Ž B measures 908.
of the first statement, so you can write the following new
statement if 4x 5 12, then 2x 5 6.
T
}215s
c
11. Look for a pattern:
5. C;
1 1 3 5 4, 5 1 7 5 12, 9 1 3 5 12
Conjecture: Odd integer 1 odd integer 5 even integer
Let 2n 1 1 and 2m 1 1 be any two odd integers
(2n 1 1) 1 (2m 1 1) 5 2n 1 2m 1 2 5 2(n 1 m 1 1)
2(n 1 m 1 1) is the product of 2 and an integer
(n 1 m 1 1). So, 2(m 1 n 1 1) is an even integer.
The sum of any two odd integers is an even integer.
12.
K
6. mŽ1 1 mŽ2 1 mŽ3 1 mŽ4 5 3608
mŽ1 1 mŽ1 1 808 1 808 5 3608
2 + mŽ1 5 2008
mŽ1 5 1008
Chapter 2 Review (pp. 134–137)
1. A statement that can be proven is called a theorem.
2. The inverse negates the hypothesis and conclusion of
a conditional statement. The converse exchanges the
hypothesis and conclusion of a conditional statement.
3. When mŽ A 5 mŽ B and mŽ B 5 mŽC,
then mŽ A 5 mŽB.
4. 220,480, 25120, 21280, 2320,
44
44
44
...
44
Each number in the pattern is the previous number
divided by 4. The next three numbers are 280, 220, 25.
224
5. Counterexample: } 5 3
28
Because a counterexample exists, the conjecture is false.
C
D
E
13. B; With no right angle marked, you cannot assume
}
CD > plane P.
14. 29x 2 21 5 220x 2 87
Addition Property of
Equality
11x 5 266
Addition Property of
Equality
x 5 26
Division Property of
Equality
15. 15x 1 22 5 7x 1 62
Geometry
Worked-Out Solution Key
Given
8x 1 22 5 62
Subtraction Property of Equality
8x 5 40
Subtraction Property of Equality
x55
16. 3(2x 1 9) 5 30
6x 1 27 5 30
6x 5 3
1
x 5 }2
50
Given
11x 2 21 5 287
Division Property of Equality
Given
Distributive Property
Subtraction Property of Equality
Division Property of Equality
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
AB and ###$
AF
ŽBAC and ŽCAF are a linear pair because ###$
are opposite rays. mŽBAC 1 mŽCAF 5 1808 by the
Linear Pair Postulate. mŽCAF 5 mŽCAE 1 mŽEAF
and mŽCAE 5 mŽCAD 1 mŽDAE by the Angle
Addition Postulate. mŽCAD 1 mŽDAE 5 908
because ŽCAD and ŽDAE are complements. So,
mŽCAE 5 908, and mŽBAC 1 mŽCAF 5 mŽBAC 1
(mŽCAE 1 mŽEAF) 5 mŽBAC 1 (908 1 mŽEAF)
5 (mŽBAC 1 mŽEAF) 1 9085 1808. So, mŽBAC 1
mŽEAF 5 908.
Chapter 2,
continued
17. 5x 1 2(2x 2 23) 5 2154
3. 26, 21,
Given
5x 1 4x 2 46 5 2154
Distributive Property
9x 2 46 5 2154
Simplify.
9x 5 2108
Addition Property of
Equality
x 5 212
Division Property of
Equality
...
15 15 15 15
Each number in the pattern is five more than the previous
number. The next number is 14.
4. 100, 250,
1
25,
1
212.5,
1
...
1
3 2}2 3 2}2 3 2}2 3 2}2
1
19. Reflexive Property of Congruence
number. The next number is 6.25.
20. Transitive Property of Equality
5. If-then form: If two angles are right angles, then they
21. Given: Ž1 > Ž2 and Ž2 > Ž3
are congruent.
Prove: Ž1 > Ž3
Converse: If two angles are congruent, then they are right
angles.
Statements
Reasons
1. Given
2. mŽ1 5 mŽ2 2. Definition of congruent angles
mŽ2 5 mŽ3
Inverse: If two angles are not right angles, then they are
not congruent.
Contrapositive: If two angles are not congruent, then
they are not right angles.
6. If-then form: If an animal is a frog, then it is
3. mŽ1 5 mŽ3 3. Transitive Property of Equality
an amphibian.
4. Ž1 > Ž3
Converse: If an animal is an amphibian, then it is a frog.
4. Definition of congruent angles
22. Given: mŽ1 5 1148
mŽ3 5 mŽ1 5 1148
23. Given: mŽ4 5 578
mŽ2 5 mŽ4 5 578
mŽ1 1 mŽ2 5 1808
mŽ4 1 mŽ3 5 1808
1148 1 mŽ2 5 1808
578 1 mŽ3 5 1808
mŽ2 5 668
mŽ4 5 mŽ2 5 668
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
9,
Each number in the pattern is 2}2 times the previous
18. Symmetric Property of Congruence
1. Ž1 > Ž2
Ž2 > Ž3
4,
mŽ3 5 1238
mŽ1 5 mŽ3 5 1238
Inverse: If an animal is not a frog, then it is not an
amphibian.
Contrapositive: If an animal is not an amphibian, then it
is not a frog.
7. If-then form: If x 5 22, then 5x 1 4 5 26.
Converse: If 5x 1 4 5 26, then x 5 22.
Inverse: If x Þ 22, then 5x 1 4 Þ26.
Contrapositive: If 5x 1 4 Þ 26, then x Þ 22.
9. Because you are going to the football game satisfies the
24.
Statements
Reasons
1. Ž12 and Ž11 are
complementary.
mŽ10 1 mŽ11 5 908
1. Given
2. Ž10 and Ž11
are complementary.
2. Definition of
complementary angles
3. Ž12 > Ž10
3. Congruent Complements
Theorem
hypothesis, the conclusion is true. So, you will miss
band practice.
10. The conclusion of the first statement is the hypothesis of
the second statement, so you write the following
new statement.
Chapter 2 Test (p. 128)
1. The figure is rotated counterclockwise. The next
figure is:
If Margot goes to college, then she will need to buy a
lab manual.
11. Sample answers: Line @##$
TQ contains points N, Q, and T.
12. Sample answer: Plane Y contains points Q, R, and S.
13. Planes X and Y intersect at @##$
TQ.
14. 9x 1 31 5 223
9x 5 254
x 5 26
15. 27(2x 1 2) 5 42
2. Two pieces are added to the figure alternating between
unshaded and shaded pattern. The next figure is:
7x 2 14 5 42
Given
Subtraction Property of Equality
Division Property of Equality
Given
Distributive Property
7x 5 56
Addition Property of Equality
x58
Division Property of Equality
Geometry
Worked-Out Solution Key
51
Chapter 2,
continued
16. 26 1 2(3x 1 11) 5 218x
26 1 6x 1 22 5 218x
Given
Distributive Property
6x 1 48 5 218x
Simplify.
48 5 224x
22 5 x
Subtraction Property of
Equality
Division Property of
Equality
17. B; Symmetric Property of Congruence
18. A; Reflexive Property of Congruence
19. C; Transitive Property of Congruence
20. 7y 5 5y 1 36
}
}
}
}
}
}
11. 2Ï 180 5 2Ï 36 + Ï 5 5 26Ï 5
12. 6Ï 128 5 6Ï 64 + Ï 2 5 68Ï 2
}
}
}
}
}
}
}
}
13. Ï 2 2 Ï 18 1 Ï 6 5 Ï 2 2 3Ï 2 1 Ï 6 5 Ï 6 2 2Ï 2
}
}
}
}
}
}
14. Ï 28 2 Ï 63 2 Ï 35 5 2Ï 7 2 3Ï 7 2 Ï 35
}
}
}
5 2Ï7 2 Ï 35
}
}
}
}
15. 4Ï 8 1 3Ï 32 5 8Ï 2 1 12Ï 2 5 20Ï 2
}
}
}
16. (6Ï 5 )(2Ï 2 ) 5 12Ï 10
}
}
}
}
}
}
The measures of the angles are:
7(18) 5 1268
5(18) 1 36 5 1268
3(25) 2 21 5 548
Statements
}
Reasons
1. AX > DX,
} }
XB > XC
1. Given
2. AX 5 DX,
XB 5 XC
2. Definition of congruent
segments
3. AX 1 XC 5 AC 3. Segment Addition Postulate
DX 1 XB 5 BD
4. DX 1 XC 5 AC 4. Substitution Property of
DX 1 XC 5 BD
Equality
5. AC 5 BD
}
6. AC > BD
5. Transitive Property of
Equality
6. Definition of congruent
segments
Chapter 2 Algebra Review (p. 139)
2
5+x+x+x+x
x
5x4
1. }2 5 }} 5 }
5+4+x+x
4
20x
24 + 3 + a + b + b + b
24b2
212ab3
}
2. }
5 }}
5
2
3a
3+3+a+a+b
9a b
5(m 1 7)
5m 1 35
3. } 5 } 5 m 1 7
5
5
212m
36m 2 48m
4. } 5 } 5 22
6m
6m
k13
5. } ; cannot be simplified
22k 1 3
m14
1
m14
6. }
5}
5}
m
m(m 1 4)
m2 1 4m
2(6x 1 8)
12x 1 16
7. } 5 } 5 2
8 1 6x
8 1 6x
3+x+x+x
3x2
3x3
}
8. }
5}
5
2
5 1 8x
x(5 1 8x)
5x 1 8x
Geometry
Worked-Out Solution Key
19.
Ï(25)2 5 25
20.
Ïx2 5 x
21.
Ï(2a) 2 5 Ïa2 5 a
}
}
Ï(3y) 2 5 Ï9y 2 5 3y
22.
2(25) 1 4 5 548
}
}
}
}
}
}
}
23.
Ï32 1 22 5 Ï9 1 4 5 Ï13
24.
Ïh2 1 k 2 ; cannot be simplified
}
TAKS Practice (pp. 142–143)
1. C;
The student section has 40 rows, with 36 seats in each
row, and student admission is $4 per person. Because
student tickets sell out, to calculate the total amount of
student ticket sales, multiply 40 by 36 and then multiply
that product by $4.
2. J;
n2 1 n represents the pattern shown in the table.
3. C;
The other information that is needed to determine the
amount of discount per person is the number of students
on the trip.
4. J;
The charge for labor was $38 per hour, so the charge for
h hours of labor was 38h. This number plus the charge
for parts, $126, gives the total charge to repair Andy’s
car, $278. So, the equation 38h 1 126 5 278 can be used
to find the number of hours it took to repair his car.
5. B;
A scatter plot would be most helpful in determining
whether there is a correlation between the amount of TV
watched and the number of extracurricular activities.
6. F;
d 5 rt
r 5 3 mi/h
2
t 5 40 min 5 }3 h
d 5 31 }3 2 5 2
2
Jane can walk 2 miles in 40 minutes.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 25
52
}
}
18. (2Ï 6 ) 2 5 (2) 2 + (Ï 6 ) 2 5 4 + 6 5 24
3x 2 21 5 2x 1 4
}
}
}
}
y 5 18
}
}
}
10. Ï 75 5 Ï 25 + Ï 3 5 5Ï 3
17. (24Ï 10 )(25Ï 5 ) 5 20Ï 50 5 20 + 5Ï 2 5 100Ï 2
2y 5 36
21.
3x(x 2 2)
3x 2 2 6x
x22
9. }
5}
5}
3x(2x 2 1)
2x 2 1
6x 2 2 3x
Chapter 2,
continued
7. C;
The maximum value of the parabola shown is at (1, 3).
8. G;
Number
Number
Total
Price
Price
of desk
of wall
dollar
per wall 5
calendars + per desk 1 calendars +
amount
calendar
calendar
sold
sold
sold
9
+
d
1
3
+
w
5 70.5
4
+
d
1
6
+
w
5 71
9. C;
Cost per
Number of
Cost per
+
1
notebook
notebooks
pencil
+
2.5
1
n
0.5
Number
Maximum
+ of
a total cost
pencils
+
p
a
10
10. F;
(23, 2), (0, 8)
822
6
5 }3 5 2
m5}
0 2 (23)
Because the line passes through the point (0, 8), the
y-intercept, b, is 8. So, the equation y 5 2x 1 8
represents the line that passes through the points (23, 2)
and (0, 8).
11. A;
24(x 1 5) 1 2(2x 1 1) 5 24x 2 20 1 4x 1 2 5 218
12. J;
(3, 0), (0, 1)
120
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5 2}3
m5}
023
1
The y-intercept, b, is 1. So, the equation y 5 2}3 x 1 1
best represents the graph.
13. Of the 750 students in Hannah’s school, 72%, or
0.72(750) 5 540 students have at least one pet. Of these
540 students with pets, 30%, or 0.3(540) 5 162 students
own dogs and 5%, or 0.05(540) 5 27 students own two
or more dogs. So, 162 2 27 5 135 students own exactly
one dog.
Geometry
Worked-Out Solution Key
53
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