Carlos, Sonya, Tanner, and Wen are sharing a cake. The cake had previously been divided into four slices ( s 1 , s 2 , s 3 , and s 4 ) . T a b l e 3 - 1 7 shows the values of the slices in eyes of each player. T a b l e 3 − 1 7 S 1 S 2 S 3 S 4 Carlos $3.00 $5.00 $5.00 $3.00 Sonya $4.50 $3.50 $4.50 $5.50 Tanner $4.25 $4.50 $3.50 $3.75 Wen $5.50 $4.00 $4.50 $6.00 a. Which of the slices are fair shares to Carlos? b. Which of the slices are fair shares to Sonya? c. Which of the slices are fair shares to Tanner? d. Which of the slices are fair shares to Wen? e. Find all possible fair divisions of the assets using s 1 , s 2 , s 3 , and s 4 as shares.
Carlos, Sonya, Tanner, and Wen are sharing a cake. The cake had previously been divided into four slices ( s 1 , s 2 , s 3 , and s 4 ) . T a b l e 3 - 1 7 shows the values of the slices in eyes of each player. T a b l e 3 − 1 7 S 1 S 2 S 3 S 4 Carlos $3.00 $5.00 $5.00 $3.00 Sonya $4.50 $3.50 $4.50 $5.50 Tanner $4.25 $4.50 $3.50 $3.75 Wen $5.50 $4.00 $4.50 $6.00 a. Which of the slices are fair shares to Carlos? b. Which of the slices are fair shares to Sonya? c. Which of the slices are fair shares to Tanner? d. Which of the slices are fair shares to Wen? e. Find all possible fair divisions of the assets using s 1 , s 2 , s 3 , and s 4 as shares.
Solution Summary: The author explains that the table for value of slices to each player is shown in table 1.
Carlos, Sonya, Tanner, and Wen are sharing a cake. The cake had previously been divided into four slices
(
s
1
,
s
2
,
s
3
, and
s
4
)
.
T
a
b
l
e
3
-
1
7
shows the values of the slices in eyes of each player.
T
a
b
l
e
3
−
1
7
S1
S2
S3
S4
Carlos
$3.00
$5.00
$5.00
$3.00
Sonya
$4.50
$3.50
$4.50
$5.50
Tanner
$4.25
$4.50
$3.50
$3.75
Wen
$5.50
$4.00
$4.50
$6.00
a. Which of the slices are fair shares to Carlos?
b. Which of the slices are fair shares to Sonya?
c. Which of the slices are fair shares to Tanner?
d. Which of the slices are fair shares to Wen?
e. Find all possible fair divisions of the assets using
s
1
,
s
2
,
s
3
, and
s
4
as shares.
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
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