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```3-5. a: D is not similar. AB = 5, BC = 4, AC =3
b: A!B!
10units,
B!C! = 8 units, and A!C! = 6 units.
=
c: A ==24 sq. units; P = 24 units
3-6. a: x = 18 b: x = 3 c: x = 6 d: x = 2 3-7. a: ≈ 30°,
≈
40°,
≈
110° b: Obtuse
scalene triangle
4
b: MU =
=x 95
41 !
+
6.40units
c: One is a ratio (slope) while the other is a length (distance).
3-8. a: 5 , y
4
5
3-9. a: triangle inequality
b: Pythagorean Theorem
c: base angles not equal
3-10. a: If a shape is an equilateral triangle, then it has 120° rotation symmetry.
b: If a shape is a rectangle, then the shape is a parallelogram.
c: If a shape is a trapezoid, then the area of the shape is half the sum of its
basesmultiplied by its height.
3-18. Result should be 12 units tall and 16 units wide. 3-19. a: The 15 corresponds to the 6, while
15
20
the 20 corresponds
the 8. Multiple
equivalent
ratios aretopossible.
One possibility:
== 2.5
68
25
b: 25 and 10;
= 2.5; yes
10
3-20. Yes they are parallel because they have the same slope: !
3-2
1.
3-2
2.
3-2
3.
a: 6x2 ! 8x
b: 2x2 + x !15
c: 4x2 ! 25
3
5
.
d: 2x3 ! 5x2 ! 3x
x = 10° ,
= 61°
y
No, this is not convincing. While the facts are each correct, the conclusion is not based on
the facts. As stated in Fact #2, a square is a rectangle because it has four right angles.
enough evidence
thatdoes
a rhombus
is to
a rectangle.
However,
a rhombus
not have
have four right angles, so therefore there is not
a: f = 9
b: g = 18
c: h = 703
3-4
Lesson
3.1.3
1.
3-4
180°factor:
! 38° !0.5;
63° =The
79°sides
and 180°
! 38°
! 79°
= 63°so
; corresponding
angles aretoequal.
3-29.
Zoom
are only
half
as long,
the side corresponding
the 16b:must
2. a: a:
all unmarkedtoangles
thebecome
same since
become Upon
8, andinspection,
the side corresponding
the 11are
must
5.5. the difference with 180° will be
the same.
b: It is 1:1 because it is congruent.
3-4
3-30.
P(original) = 18 units and P(new) = 36 units; A(original) = 18 sq. units and A(new) = 72 sq.
3.
a: Frank: 0.25x +1.95 = y; Alice: 0.40x +1.5 = y b: They will be 3 years old.
units. The enlarged perimeter is 2 times greater. The enlarged area is not 2 timesgreater. Notice
that the enlarged area is 4 times greater.
3-4 a: If a rectangle has base x and height 2x, then the area is 2x2. b: If a rectangle has base x
42
3
and
3y, then the perimeter is 2x + 6y
. c: If a rectangle has base of 2 feet and a height
4.
3-31.
a:
x =height
=
5 8.4 b: m = 22 c: t = 12.5 d: x = 2 = 1.5
of 3 feet, then the area is 864 sq. inches.
3
3-32.
a: y = 3! 5 x
3-4
5.
3-4
3-33.
6.
b: A = 7.5 sq. units; P = 8+
34 !
In theory,5 3 < x < 13 but some of these lengths are not practical.
c: y = 3+ 3 x
13.8
a: The coordinates of the image are A(–6, –4), B(10, –4), C(10, 6), D(2, 12), E(–6, 6) b:
a:
If the lines
have
Perimeters
= 28
andthe
56same
units;slope,
areasthen
= 52they
and are
208parallel.
sq. units
b: If a line is vertical, then the slope is undefined.
c: If lines have slopes
3-34.
2
3
3
and ! 2 , then they are perpendicular.
a: alt. int. angles b: vertical angles
c: corresponding angles d: supplementary and/or adjacent angles
3-2
1.
3-2
2.
3-2
3.
a: 6x2 ! 8x
3-4
3.
3-4
4.
3-4
5.
3-4
6.
c: 4x2 ! 25
d: 2x3 ! 5x2 ! 3x
x = 10° ,
= 61°
y
No, this is not convincing. While the facts are each correct, the conclusion is not based on
the facts. As stated in Fact #2, a square is a rectangle because it has four right angles.
However, a rhombus does not have to have four right angles, so therefore there is not
a: f = 9
3-4
1.
3-4
2.
b: 2x2 + x !15
b: g = 18
c: h = 703
a: 180° ! 38° ! 63° = 79° and 180° ! 38° ! 79° = 63° ; corresponding angles are equal. b:
Upon inspection, all unmarked angles are the same since the difference with 180° will be
the same.
a: Frank: 0.25x +1.95 = y; Alice: 0.40x +1.5 = y b: They will be 3 years old.
a: If a rectangle has base x and height 2x, then the area is 2x2. b: If a rectangle has base x
and height 3y, then the perimeter is 2x + 6y . c: If a rectangle has base of 2 feet and a height
of 3 feet, then the area is 864 sq. inches.
In theory, 3 < x < 13 but some of these lengths are not practical.
a: The coordinates of the image are A(–6, –4), B(10, –4), C(10, 6), D(2, 12), E(–6, 6) b:
Perimeters = 28 and 56 units; areas = 52 and 208 sq. units
3-53. a: Yes, since all trees are green and the oak is a tree.
b: No, only trees must be green according to the statement.
c: No, the second statement reverses the first.
3-54. a: Yes, AA ~. Dilate from right vertex.
b: Yes, AA ~ since all angles are 60°.
c: Yes, zoom factor of 2.5; translate so that one pair of corresponding vertices coincide,
rotate so that rays coincide, and dilate.
d: No, since corresponding angles are not equal. Note that you cannot apply zoom factor to
angles.
3-55. a: One strategy: Translate one so that the centers coincide. Then dilate so that the radius is
the same as the other circle.
b: Equilateral triangles, which from part (b) of 3-54 were similar because they have
equalangle measures. Squares or other regular polygons are also always similar.
3-56. a: There are 12 combinations. One way to systematically list them all is to list a
busnumber (such as 41) and then match it with each possible activity. This can be repeated for
each of the possible bus numbers.
9
8
1
b: i: 12 , ii: 12 , iii: 12
3-57. See graph at right. Perimeter = 44.9 units; Area = 94 sq. units
3-58. a: ABCD ~ EVOL
b: RIGHT ~ RONGW
c: One possible answer: ΔTAC ~ ΔGDO
3-65.
a: x = 20 mm b: w = 91 mm
3-66.
a: Impossible: can be rejected using Triangle Inequality or Pythagorean Theorem.
b: Possible
c: Impossible: rejected because the sum of the angles is 179°.
3-67.
8
a: 12 b:
4
8
3-68. This reasoning is incorrect. Rewrite “it is raining” in the lower left oval, and “Andrea’s
flowers must be closed up” in the right oval.
3-69.
a: Reflection, rotation, and translation
b: Rotation and translation
c: Rotation, dilated by factor of 2, and translation.
3-70.
a: Possible
b: Not possible because the sum of the measures of an obtuse and right angle is
morethan 180°.
c: Not possible because a triangle with sides of equal length obviously cannot have sidesof
different lengths.
d: Possible
3-88. a: Scalene triangle b: Isosceles triangle
c: Not possible d: Equilateral triangle
3-89. a: The two equations should have the same slope but a different y-intercept. This forces the
lines to be parallel and not intersect.
b: When solving a system of equations that has no solution, the equations combine to
create an impossible equality, such as 3 = 0. Another special case occurs when theresulting
equality is always true, such as 2 = 2. This is the result when the two lines coincide,
creating infinite points of intersection.
3-90. a: Not similar, interior angles are different.
b: Must be similar by AA ~.
c: Similar, all side lengths have the same ratio. 3-91. Perimeter = 10+10+ 4+ 3+ 4 + 3+ 4
= 38 units, height of triangle = 8 units,
area = 60 square units. 3-92. This
reasoning is correct. 3-93. a: 3(4x
!12)= 180° , x2 = 18 2 2
b: 4.9 ! 3.1 = x , x ! 3.79
!
c: x + (180° ! 51° !103°)+ 82°= 180° , x = 72
d: 3x ! 2 = 2x + 9 , x = 11
3-99.
a: SSS ~ and SAS ~ (if you show that the triangles are right triangles)
b: AA ~ and SAS ~
c: None since there is not enough information.
3-100. a:
24
= 60%
40
18
b: == 60
2
3, x
x 10
2
3-101. a: 12x ! 7x !10 b: 16x ! 8x +1
b: x = !
5
9
c: x = 3
!
!
3-102. !y = 48 because of vertical angles; !z = 48 because of reflection of !y or because of angle of
incidence = angle of reflection with !x.
3-103. a: !
5
6
b: LD = 61 ! 7.81units
c: Calculate ∆ x and ∆ y by determining the difference in the corresponding coordinates.
3-104. Original: A = 135 sq. units, P = 48 units; New: A = 15 sq. units, P = 16 units
Lesson 3.2.4
3-108. x = 137°,!y = 76° 3-109. h = 5
feet; perimeter ≈ 24.2 feet
28
3-110. a: =? 5 ; 2There are 70 animals in the bin.
b:
3
c: ? = 5%;
13+17
30 =
= 40%
75
22+8+13+15+17
You need a total of 60 animals in the bin. 3-111. a:
1
y = ! 2 x + 4 b: y = 2x !1
c: y = 2 x 7
5
5
+
d: C = 15+ 7(t !1)= 8+ 7t
3-112. ≈
13.2
miles
3-113. Possible response: Rotate WXYZ clockwise, translate it to the left, and dilate it by
afactor of 0.4. y = 7.5,!z = 9.6