Download Chapter 5 - BISD Moodle

Chapter Lesson Chapter Lesson Set I (pages –)
A chapter of Lawrence Wright’s Perspective in
Perspective (Routledge & Kegan Paul ) is
titled “Grand Illusions” and concerns various tricks
played by architects with perspective Plato
mentions the use of such tricks in the design of
Greek temples such as the Parthenon Giotto’s
campanile in Florence is unusual in its use of
“counter
perspective” The tower widens
measurably upward and the upper levels
become successively taller toward the top
From the ground they can look almost equal
Hat Illusion.
•1. Either h > w, h = w, or h < w.
19. Because one has two equal sides and the
other does not. (One is isosceles, the other
scalene.)
Perspective Inequalities.
•20. ∠APB > ∠HPI.
21. ∠BPC > ∠CPD.
•22. ∠PIH < ∠PBA.
23. ∠PHG < ∠PGF.
24. ∠PHI > ∠PGH and ∠PGH > ∠PFG; so
∠PHI > ∠PFG.
25. So that the upper levels would appear to be
the same height when viewed from the
ground.
2. The “three possibilities” property.
Set II (pages –)
3. (Student answer.) (For most people, the hat
appears to be taller than it is wide, and so
h > w.)
SAT Problem.
4. The hat is 1.7 cm high and its brim is 1.8 cm
wide.
Pecking Order.
5. Hen A will always peck hen C.
•6. The transitive property.
Equalities and Inequalities.
7. The “three possibilities” property.
•8. Multiplication.
9. Subtraction.
•26. Transitive.
27. Addition.
28. Substitution. (Substituting AC for AB + BC
and CE for CD + DE in exercise 27.)
29. Could be false. (To see why, imagine that
DE is much longer than it is.)
•30. The “whole greater than part” theorem.
31. Transitive. (AB < BC is given and BC < BD
in exercise 30.)
Scalene Triangle.
32.
10. Substitution.
•11. Addition.
33.
12. Division.
Deceiving Appearances.
13. Isosceles. (Also acute.)
•14. AB > BC.
•15. Substitution.
16. Scalene. (Also acute.)
17. DF < DE.
18. The transitive property.
•34. ∠A ≈ 83°, ∠B ≈ 56°, ∠C ≈ 41°.
35. ∠A > ∠B > ∠C.
36. Yes. The order corresponds to the order of
the lengths of the opposite sides.
Chapter Lesson Rectangle Inequalities.
•37. 9 units. (2 ⋅ 4 + 2 ⋅ 0.5 = 9.)
(5) AB = DB. (If two angles of a triangle are
equal, the sides opposite them are equal.)
(6) AC > DB. (Substitution.)
38. Yes. The “whole greater than part” theorem.
39. 2 square units. (4 ⋅ 0.5 = 2.)
40. No. This area problem shows why: 4 > 0 and
0.5 > 0, but 4 × 0.5 = 2 < 4.
Vertical Lines and Angles.
•41. That the line is “upright” (contains the
center of the earth).
42. The sides of one angle must be opposite
rays to the sides of the other angle. (It must
also be true that two angles are equal if they
are vertical angles.)
43. Yes: ∠1 and ∠2 are not vertical angles. If
they were vertical angles, they would be
equal. Because AB = AC, ∠2 = ∠3 (if two
sides of a triangle are equal, the angles
opposite them are equal). But ∠1 < ∠3;
so ∠1 < ∠2 (substitution).
44. No. Because ∠2 = ∠3 (AB = AC) and
∠2 = ∠4, ∠3 = ∠4. The fact that two angles
are equal does not mean, however, that they
are necessarily vertical angles.
45. (1)
(2)
(3)
(4)
(5)
Given.
Addition.
Given.
Addition.
Transitive (steps 2 and 4).
46. (1)
(2)
(3)
(4)
(5)
Given.
Subtraction.
Given.
Substitution.
Addition.
Optics Figure.
47. (1)
(2)
(3)
(4)
Given.
Betweenness of Rays Theorem.
Protractor Postulate.
The “whole greater than part” theorem.
48. Proof.
(1) A-B-C. (Given.)
(2) AB + BC = AC. (Betweenness of Points
Theorem.)
(3) AC > AB. (“Whole greater than part.”)
(4) ∠ADB = ∠DAB. (Given.)
Set III (page )
In Human Information Processing Peter H
Lindsay and Donald A Norman deal with the
logic of choice Concerning the three statements
of the Set III exercises they write:
“Almost without exception people agree that
their own decision processes ought to be logical
Moreover formal theories of decision making
assume logical consistency: Preferences among
objects ought to be consistent with one another
If A is preferred to B and B to C then logically A
should be preferred to C Transitivity of this sort
is a basic property that just ought to hold
anytime different objects are compared to one
another These three rules constitute a sensible set of
postulates about decision behavior Indeed if the
rules are described properly so that the math
ematical framework is removed they simply
sound like common sense Those who study
decision making however have discovered that
there is a difference between the rules that
people believe they follow and the rules that they
actually do follow Thus although these three
assumptions form the core of most decision
theories it is also realized that they do not
always apply There are some clear examples of
instances in which each rule is violated”
(The authors then give anecdotal examples of
violations for each of the three rules)
Martin Gardner treated the subject of non
transitive paradoxes in one of his “Mathematical
Games” columns It is included in his Time Travel
and Other Mathematical Bewilderments (W H
Freeman and Company )
1. If you prefer an apple to a banana and a
banana to a cookie, then you prefer an apple
to a cookie.
2. The transitive property.
3. If you have no preference between an apple
and a banana or between a banana and a
cookie, then you have no preference between an apple and a cookie.
4. Substitution.
Chapter Lesson 5. If you have no preference between an apple
and a banana and you prefer a cookie to
nothing at all, then you would prefer having
an apple and a cookie to having just a
banana.
13. No. The exterior angles at each vertex are
vertical angles and vertical angles are equal.
14. Example figure:
6. Yes.
If C > 0, then A + C > A. (Addition.)
But A = B, so A + C > B. (Substitution.)
Chapter Lesson Set I (pages –)
Aristotle included a section on the causes of the
rainbow in De Meteorologia According to Carl
Boyer Aristotle’s work is “the first truly systematic
theory of the rainbow that has come down to us”
Boyer comments on the soundness of some of
Aristotle’s geometrical arguments and says: “Had
his successors continued his work at the same
high level the story of the rainbow might not
have been such a tale of frustration as it was
destined to be”
Garage Door.
•1. ∠BCY, ∠BAX, ∠CAD(∠YAD).
2. It gets larger.
•3. It gets smaller.
4. It gets smaller.
5. If two sides of a triangle are equal, the
angles opposite them are equal.
•6. An exterior angle of a triangle is greater
than either remote interior angle.
7. Vertical angles are equal.
8. They are always equal. Because ∠B = ∠BAY
and ∠BAY = ∠DAX, ∠B = ∠DAX by
substitution.
Exterior Angles.
•9. ∠2, ∠5, ∠8.
•10. Two.
11. Six.
12. Yes. If the triangle is equiangular, its exterior
angles must all be equal because they are
supplements of equal angles.
15. Example figure:
•16. Two.
17. Two.
18. Six.
Rainbow.
•19. ∆ROS.
•20. ∠ROA > ∠SRO and ∠ROA > ∠S.
21. ∆ROC.
22. ∠ROS > ∠ORC and ∠ROS > ∠RCO(∠RCS).
23. ∆RCS and ∆RCO.
24. ∠RCA > ∠SRC, ∠RCA > ∠S, ∠RCA > ∠ORC,
∠RCA > ∠ROC.
•25. No. It does not form a linear pair with an angle
of the triangle.
Lines and Angles.
•26. 360°.
27. 1,080°. (3 ⋅ 360 = 1,080.)
•28. 180°.
29. 180°.
30. 720°. (1,080 – 180 – 180 = 720.)
31. It indicates that the sum is 720°.
Chapter Lesson Set II (pages –)
We are indebted to Proclus for his commentary
on the first book of Euclid’s Elements It and the
work of Pappus are our two main sources of
information on the history of Greek geometry
Proclus defended Euclid from the charge that he
proved things that had no need of proof (More
on this is included in Lesson of this chapter)
47. ∠APC > ∠AXC and ∠AXC > ∠B because an
exterior angle of a triangle is greater than
either remote interior angle. So ∠APC > ∠B
by the transitive property.
Proclus’s Proof.
48. If two sides of a triangle are equal, the
angles opposite them are equal.
Exterior Angle Theorem.
49. ∠1 = ∠C and ∠2 = ∠A.
•32. A line segment has exactly one midpoint.
50. The Exterior Angle Theorem (∠1 is an
exterior angle of ∆PBC and ∠2 is an exterior
angle of ∆PBA.)
33. The Ruler Postulate (or by Construction 1,
to copy a line segment).
51. Indirect.
34. SAS.
35. Corresponding parts of congruent triangles
are equal.
36. Vertical angles are equal.
•37. The “whole greater than part” theorem.
38. Substitution.
Angle Sum.
•39. Two points determine a line.
40. An angle is an exterior angle of a triangle if
it forms a linear pair with an angle of the
triangle.
•41. The angles in a linear pair are
supplementary.
42. If two angles are supplementary, their sum
is 180°.
Set III (page )
In his History of Mathematics David Eugene
Smith includes a paragraph on “drumhead
trigonometry” He wrote: “The continual warfare
of the Renaissance period shows itself in many
ways in the history of mathematics One of
them is related to the subject now under
consideration Several writers of the th century
give illustrations of the use of the drumhead as a
simple means of measuring angles of elevation in
computing distances to a castle or in finding the
height of a tower” The illustration is from
S Belli’s Libro del Misvrar con la Vista (Book of
Measuring with Eyesight) published in Venice in
Drumhead Geometry.
1.
•43. An exterior angle of a triangle is greater
than either remote interior angle.
44. Addition.
45. Substitution.
Angle in a Triangle.
46.
2. (About 4 in; more precisely, about 4
in.)
3. About 80 ft.
Chapter Lesson Set I (pages –)
There are now many more “anamorphic artists”
painting streets than pictures The idea in using
anamorphic figures as traffic markers is clearly
not to make them difficult to recognize but
Chapter Lesson rather just the opposite The driver of a moving
car who is looking far ahead sees the markers
from a sharp angle and hence sees them regain
their “normal” shape An entire “Mathematical
Games” column focuses on various types of
anamorphic art (Scientific American January
) It is included in Martin Gardner’s Time
Travel and Other Mathematical Entertainments
(W H Freeman and Company )
“Triangles in perspective” is an interesting
topic Suppose we have some wire
frame models
of triangles of various shapes What kinds of
shadows can they cast? Can the shadow of an
equilateral triangle look scalene? Can the shadow
of an isosceles triangle look equilateral? Can the
shadow of an obtuse triangle look acute? Does
the shadow of a triangle always look like a
triangle? What determines the smallest shadow
that a triangle can cast?
16.
17. Obtuse. (Also, scalene.)
•18. ∠A ≈ 29°, ∠B ≈ 104°, ∠C ≈ 47°.
Triangle Drawing 2.
•19. DE.
20. DF.
21.
1. Bicycle lane.
2. So that it can be seen more easily to a driver
viewing it “on edge.”
•3. Each is the converse of the other.
•22. DF ≈ 4.5 cm, EF ≈ 5.4 cm.
•4. BC < AC.
Triangle in Perspective.
5. The “three possibilities” property.
6. ∠A = ∠B.
7. If two sides of a triangle are equal, the
angles opposite them are equal.
•8. ∠A > ∠B.
9. ∠A < ∠B.
•10. If two sides of a triangle are unequal, the
angles opposite them are unequal in the
same order.
11. ∠A > ∠B.
12. BC > AC.
13. Indirect.
Triangle Drawing 1.
•14. ∠B.
15. ∠A.
•23. ∠X > ∠Y.
24. If two sides of a triangle are unequal, the
angles opposite them are unequal in the
same order.
25. ∠Z > ∠Y.
26. ∠Z > ∠X (given) and ∠X > ∠Y (exercise 23);
so ∠Z > ∠Y by the transitive property.
27. XY > XZ.
28. If two angles of a triangle are unequal, the
sides opposite them are unequal in the same
order.
Chapter Lesson Set II (pages –)
41. GH.
•42. FG < GH because both are sides of ∆FGH in
which GH is the longest side.
43. No. They can’t be congruent because FG
and GH are corresponding parts of the
triangles but they are not equal.
44. Yes. They are not congruent, because the
shortest side of ∆IJK is IJ and the shortest
side of ∆JKL is JK. IJ < JK because they are
both sides of ∆IJK in which IJ is the shortest
side. As with the preceding pair of triangles,
these triangles cannot be congruent, because
IJ and JK are corresponding parts of the
triangles but they are not equal.
•45. An equilateral triangle is equiangular.
The exercises on the pairs of “not quite equilateral”
triangles are more challenging than they first
appear Some comparable drawings (above) in
which the angles are ° ° and ° instead are
revealing (Although the triangles in the second
and third pairs are not congruent they are
obviously similar We will return to these figures
in the chapter on similarity)
Folding Experiment.
29. (Triangle cut out and folded).
•30. ∠BDE.
31. ∠BDE is an exterior angle of ∆DEC.
32. ∠BDE > ∠C.
33. If two sides of a triangle are unequal, the
angles opposite them are unequal in the
same order.
•34. BE bisects ∠ABC.
35. ∆ABE ≅ ∆DBE (SAS); so ∠BDE = ∠A
(corresponding parts of congruent triangles
are equal).
Not Quite Equilateral.
•36. Yes. They are congruent by ASA.
46. Betweenness of Rays Theorem.
•47. The “whole greater than part” theorem.
48. Substitution.
49. If two angles of a triangle are unequal, the
sides opposite them are unequal in the same
order.
50. Because ∠PXA is an exterior angle of ∆PXY,
∠PXA > ∠PYX. Because PX ⊥ AB,
∠PXA and ∠PXY are right angles; so
∠PXA = ∠PXY = 90°. Therefore, in ∆PXY,
90° > ∠PYX (substitution) and ∠PXY = 90°;
so ∠PXY > ∠PYX (substitution again). It
follows that PY > PX because, if two angles
of a triangle are unequal, the sides opposite
them are unequal in the same order.
Set III (page )
The message says “HELLO.” (As Martin Gardner
explains, “hold the page horizontally [with the
bottom of the page] near the tip of your nose,
close one eye, and read the message on a sharp
slant.”)
Chapter Lesson 37. AC.
Set I (pages –)
38. BD.
In Euclid—The Creation of Mathematics
(Springer ) Benno Artmann quotes Proclus
on the Triangle Inequality Theorem:
39. They are equal because corresponding parts
of congruent triangles are equal.
40. FG.
“The Epicureans are want to ridicule this theorem
say it is evident even to an ass and needs no proof
Chapter Lesson . . .they make [this] out from the observation that
if hay is placed at one extremity of the sides an
ass in quest of provender will make his way along
the one side and not by way of the other two
sides”
9. Not possible.
Spotter Problem.
•10. PA + PB > 12, PA + PC > 12, and
PB + PC > 12.
Artmann adds some pertinent remarks of his
own:
11. (PA + PB) + (PA + PC) + (PB + PC) > 36.
“(The Epicureans of today might as well add
that one could see the proof on every campus
where people completely ignorant of mathematics
traverse the lawn in the manner of the ass)
Proclus replies rightly that a mere perception of
the truth of a theorem is different from a
scientific proof of it which moreover gives reason
why it is true In the case of Euclid’s geometry
the triangle inequality can indeed be derived
from the other (equally plausible) axioms On the
other hand the Epicureans win in the modern
theory of metric spaces where the triangle
inequality is the fundamental axiom of the whole
edifice”
12. 2PA + 2PB + 2PC > 36.
The transits of Venus (the occasions when
Venus is between the sun and Earth) are few and
far between In fact there have been only six of
them since the invention of the telescope (
) The next transits will
be on June and June In the past
these alignments were used to determine the
distance between Earth and the sun by timing the
beginning and ending of a transit from widely
separated geographical locations but such
methods are now obsolete More information
on the transits of Venus can be found in June —Venus in Transit by Eli Maor (Princeton
University Press )
Donkey Sense.
1. The sum of any two sides of a triangle is
greater than the third side.
•2. DP + PH > DH.
3. DH + HP > DP; PD + DH > PH.
13. PA + PB + PC > 18.
14. It is more than 18 km.
SAT Problem.
15. No. The sides of the triangle cannot be 2, 7,
and 3, because 2 + 3 < 7.
16. One triangle. The only integer that will
work for x is 6.
Distance and Collinearity.
•17. A, B, and C are not collinear.
18. In ∆ABC, AB + BC > AC. The sum of any
two sides of a triangle is greater than the
third side.
•19. AB + BC = AC.
20. A, B, and C must be collinear.
The Third Side.
21. Yes. Because the triangle is isosceles, the
length of the third side must be either 4 or 9.
It can’t be 4, because 4 + 4 < 9; so it must be
9.
22. Yes. From the Triangle Inequality Theorem
we know that, if x is the length of the third
side, then 5 + 7 > x, and 5 + x > 7.
So 2 < x < 12.
23. If the length of the third side is x, then either
x2 = 62 + 82 or 62 + x2 = 82. So either
x2 = 36 + 64 = 100 and x = 10 or 36 + x2 = 64
so that x2 = 28 and x =
≈ 5.3.
4. A postulate.
5. No.
Earth, Sun, and Venus.
•6. They are collinear.
•7. Yes; 160 million miles. (93 + 67 = 160.)
8. Yes; 26 million miles. (93 – 67 = 26.)
Set II (pages –)
Heron’s Proof.
•24. An angle has exactly one line that bisects it.
25. If an angle is bisected, it is divided into two
equal angles.
Chapter Lesson 26. An exterior angle of a triangle is greater
than either remote interior angle.
•27. Substitution.
•28. If two angles of a triangle are unequal, the
sides opposite them are unequal in the same
order.
Work Triangle.
•44. 5.5 m. (7 – 1.5 = 5.5.)
45. (5.5 – x) m.
46.
29. Addition.
30. Betweenness of Points Theorem.
31. Substitution.
Quadrilateral Inequality.
32.
47. (1) x + 1.5 > 5.5 – x; so 2x > 4, and so x > 2.
(2) x + (5.5 – x) > 1.5; so 5.5 > 1.5 (which
doesn’t tell us anything about x).
(3) 1.5 + (5.5 – x) > x; so 7 – x > x; so 7 > 2x,
and so x < 3.5.
48. It should be more than 2 m but less than
3.5 m.
Proof.
(1) ABCD is a quadrilateral. (Given.)
(2) Draw AC. (Two points determine a line.)
(3) AB + BC > AC and AC + CD > AD.
(The sum of any two sides of a triangle
is greater than the third side.)
(4) AB + BC + CD > AC + CD. (Addition.)
(5) AB + BC + CD > AD. (Transitive.)
Light Path.
•33. The sum of any two sides of a triangle is
greater than the third side.
Set III (page )
We are of course assuming that our plane
geometry is a reasonably good approximation for
what is actually a problem in spherical geometry
The distances in this problem are “air distances”
as reported in the World Almanac The incorrect
number is the distance between Paris and Rome;
it is actually miles
1. (London, Paris, and Cairo)
34. Betweenness of Points Theorem.
•35. Substitution.
36. SAS.
37. Corresponding parts of congruent triangles
are equal.
38. Substitution.
39. Vertical angles are equal.
40. Corresponding parts of congruent triangles
are equal.
41. Substitution.
42. AY + YB > AX + XB (the statement in
exercise 38).
43. ∠1 = ∠3 (the statement in exercise 41).
2. (London, Rome, and Cairo)
Chapter Lesson 3. (Paris, Rome, and Cairo)
10. An exterior angle of a triangle is greater
than either remote interior angle.
11. An exterior angle of a triangle is an angle
that forms a linear pair with an angle of the
triangle.
•12. One angle has a measure of 179° and the
other two angles are each less than 1°.
4. The London-Paris-Rome triangle, because
214 + 590 = 804 < 895.
The Paris-Rome-Cairo triangle, because
590 + 1,326 = 1,916 < 1,998.
5. The length that the two impossible triangles
have in common is 590, which suggests that
the distance between Paris and Rome is
wrong.
13. One angle is a right angle and the other two
angles are acute.
14. Every triangle has several exterior angles
that are obtuse; so this doesn’t tell us
anything.
Soccer Angle.
15.
Chapter Review
Set I (pages –)
The Gateway Arch in St Louis at ft is
approximately twice as tall as the Statue of
Liberty ( ft) and half as tall as the Empire State
Building ( ft) Designed by Eero Saarinen in
the shape of an inverted catenary the arch was
completed in The cross sections of its legs
are equilateral triangles with sides ft long at
ground level tapering to ft at the top Like the
top
hat illusion in Lesson the arch gives the
impression of being taller than it is wide; the two
dimensions are actually the same
1. In the tree trunk.
2. From the right edge at a sharp angle.
3. Diamond will scratch glass.
4. The transitive property.
•16. The midpoint of a line segment divides it
into two equal segments.
•17. Reflexive.
18. SSS.
19. Corresponding parts of congruent triangles
are equal.
•20. If the angles in a linear pair are equal, their
sides are perpendicular.
21.
Gateway Arch.
•5. The “three possibilities” property.
6. (Student answer.) (Most people see the arch
as looking taller than it is wide.)
•7. Both dimensions are 2.5 in.
8. Each is 630 ft. (2.5 × 252 = 630.)
Portuguese Theorem.
9. The Exterior Angle Theorem.
•22. If two sides of a triangle are unequal, the
angles opposite them are unequal in the
same order.
23. An exterior angle of a triangle is greater
than either remote interior angle.
Chapter Review
•24. Transitive.
25. If two angles of a triangle are unequal, the
sides opposite them are unequal in the
same order.
Roman Column.
•26. The small angles each appear to be equal
to 10°.
27. They increase as you look upward.
28. So that the sections would look equal to
someone standing at the base of the column.
•29. An exterior angle of a triangle is greater
than either remote interior angle.
30. Transitive.
40. (1) If the triangle is equilateral, all three
sides can be 10.
(2) If the triangle has exactly two equal
sides and the side of length 10 is one of
them, then the second side is 10 and the
third side is less than 20.
(3) If the triangle has exactly two equal
sides and the side of length 10 is not one
of them, then each of the other sides
must be more than 5.
Screen Display.
•41. SSS.
42. Corresponding parts of congruent triangles
are equal.
•43. Betweenness of Rays Theorem.
Set II (pages –)
44. The “whole greater than part” theorem.
Integers and Triangles.
45. Substitution.
31. 1 + 3 < 5 and 2 + 4 = 6. For a triangle to be
possible, both of these sums would have to
be greater than the third length.
32. They are collinear.
33. Example figure:
46. If two angles of a triangle are unequal, the
sides opposite them are unequal in the same
order.
SAT Problem.
47. x + (x – 2) > 7 – x.
x + (7 – x) > x – 2.
(x – 2) + (7 – x) > x.
48. 2x – 2 > 7 – x, so 3x > 9; so x > 3.
7 > x – 2; so 9 > x, and so x < 9.
5 > x; so x < 5.
Therefore, 3 < x < 5.
•34. n > 2. (2n + 2 > n + 4, n > 2.)
35. Yes. We’ve just shown that a triangle is
possible as long as n > 2.
36. An equilateral triangle.
Different Definition.
•37. One.
38. Our definition says that a triangle is isosceles
if it has at least two equal sides; so the third
side could also be equal.
•39. Yes.
49. Proof.
(1) AB > AC. (Given.)
(2) ∠C > ∠B. (If two sides of a triangle are
unequal, the angles opposite them are
unequal in the same order.)
(3) ∠A and ∠B are complementary. (Given.)
(4) ∠A + ∠B = 90°. (The sum of two
complementary angles is 90°.)
(5) ∠A + ∠C > ∠A + ∠B. (Addition.)
(6) ∠A + ∠C > 90°. (Substitution.)
Chapter Review
50. Proof.
(1) XB = XC. (Given.)
(2) ∠C = ∠XBC. (If two sides of a triangle
are equal, the angles opposite them are
equal.)
(3) ∠ABC = ∠ABX + ∠XBC. (Betweenness
of Rays Theorem.)
(4) ∠ABC > ∠XBC. (The “whole greater
than part” theorem.)
(5) ∠ABC > ∠C. (Substitution.)
(6) AC > AB. (If two angles of a triangle are
unequal, the sides opposite them are
unequal in the same order.)
Algebra Review (page )
•1.
=
•2.
=
•3.
=
=
=
=
•6.
=
•7.
=
•8.
=
.
=
=
=
.
.
.
=
=
.
.
=
.
•10.
=
=
.
•12.
=
•15.
and
•16.
and
•17.
•19.
=
=
and
=
.
=
and
,
and
,
,
and
.
and
.
.
and
.
,
and
or
and
or
•9.
•11.
•14.
.
=
•18.
.
=
•5.
=
.
=
•4.
•13.
.
,
and
.
.