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# Appendix E mat 117

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Axia College Material
Appendix E
Application Practice
Answer the following questions. Use Equation Editor to write mathematical expressions and equations.
First, save this file to your hard drive by selecting Save As from the File menu. Click the white space
below each question to maintain proper formatting.
Hint: Pay attention to the units of measure. You may have to convert from feet to miles several times in
this assignment. You can use 1 mile = 5,280 feet for your conversions.
1. Many people know that the weight of an object varies on different planets, but did you know that
the weight of an object on Earth also varies according to the elevation of the object? In particular,
the weight of an object follows this equation:
2
= Crw
, where C is a constant, and r is the
distance that the object is from the center of Earth.
a. Solve the equation
2
=
Crw
for r.
1
r
2
w=
C
r
2
r
2
=
C
w
r can only be a positive value because it represent the distance between object and Earth:
r=
C
w
b. Suppose that an object is 50 pounds when it is at sea level. Find the value of C that
makes the equation true. (Sea level is 3,963 miles from the center of the Earth.)
We are given r=3963 and w=50. So we solve the equation for C using this information:
r
2
=
C
w
C=wr
2
C=50×3963
2
C
¿ 785268450
c. Use the value of C you found in the previous question to determine how much the object
would weigh in
i. Death Valley (282 feet below sea level). Round to 3 decimal places (thousandths)
1 mile=5280 feet
MAT/117

This means that 282 feet =
282
5280
miles=0.053 miles
=>so the distance between the object situated in Death Valley and the center of
the Earth is:
r=3963-0.053=3962.947 miles
Now we solve the initial equation for w:
w=
C
r
2
w=
785268450
3962.947
2
w=50.001 pounds
ii. the top of Mount McKinley (20,320 feet above sea level). Round to 3 decimal
places (thousandths)
1 mile=5280 feet
This means that 20320 feet=
20320
5280
=3.848 miles
=>so the distance between the object situated on Mount McKinley and the center
of the Earth is:
r=3963+3.848=3966.848 miles
Now we solve the initial equation for w:
w=
C
r
2
w=
785268450
3966,848
2
w=49.903 pounds
2. The equation
hD 2.1=
gives the distance, D, in miles that a person can see to the horizon
from a height, h, in feet.
a. Solve this equation for h.
h=
D
1.2
h=
D
2
1.2
2
h=
D
2
1.44
b. Pikes Peak in Colorado is 14,110 feet in elevation. How far can you see to the horizon
from the top of Pikes Peak? Can you see Cheyenne, Wyoming (about 89 miles away)?
MAT/117

We need to find D knowing the height h=14,110. So the equation becomes:
D=1.2
h
D=1.2
14110
D=142.5 miles
You can see Cheyenne from the mountain peak because you can see up to 142.5 miles
from there and Cheyenne is only 89 miles away.
MAT/117

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