In Exercise 29 through 34 choose from the following answers and provide a short explanation for your answer using Euler’s theorems. 1. the graph has an Euler circuit. 2. the graph has Euler path. 3. the graph has neither an Euler circuit nor an Euler path. 4. the graph may or may not have an Euler circuit. 5. the graph may or may not have an Euler path. 6. there is no such graph. 5 . 1 . F i g . 5 - 4 5 ( a ) _ 2 . F i g . 5 - 4 5 ( b ) _ 3. A graph with six vertices: one vertex of degree 0 and five vertices of degree 2. F i g u r e 5 - 4 5
In Exercise 29 through 34 choose from the following answers and provide a short explanation for your answer using Euler’s theorems. 1. the graph has an Euler circuit. 2. the graph has Euler path. 3. the graph has neither an Euler circuit nor an Euler path. 4. the graph may or may not have an Euler circuit. 5. the graph may or may not have an Euler path. 6. there is no such graph. 5 . 1 . F i g . 5 - 4 5 ( a ) _ 2 . F i g . 5 - 4 5 ( b ) _ 3. A graph with six vertices: one vertex of degree 0 and five vertices of degree 2. F i g u r e 5 - 4 5
Solution Summary: The author explains that the graph has neither an Euler circuit nor an euler path.
موضوع الدرس
Prove that
Determine the following groups
Homz(QZ) Hom = (Q13,Z)
Homz(Q), Hom/z/nZ, Qt
for neN-
(2) Every factor group of
adivisible group is divisble.
• If R is a Skew ficald (aring with
identity and each non Zero element is
invertible then every R-module is free.
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
Chapter 5 Solutions
Excursions in Modern Mathematics, Books a la carte edition (9th Edition)
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