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MAT 117 week_6_mat_117_appendix_e

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Axia College Material
Appendix E
Radicals
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
Because r represents the distance between the object and the center of the world, this value
can only be positive (the negative value would make no sense). So:
r=
C
w
b. Suppose that an object is 100 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.)
The following information is given: r=3963 and w=100. So by applying this information in
our equation we can find C:
r
2
=
C
w
C=wr
2
C=100×3963
2
C
¿ 1570536900
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).
We will transform feet in miles as follows:
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1 mile=5280 feet
So: 282 feet =
282
5280
miles=0.053 miles
=> the distance between the object situated in Death Valley and the center of the
Earth can now be calculated as follows:
r=3963-0.053=3962.947 miles
Now we find the object’s weight:
w=
C
r
2
w=
1570536900
3962.947
2
w=100.003 pounds
ii. the top of Mount McKinley (20,320 feet above sea level).
We will transform feet in miles as follows:
1 mile=5280 feet
So: 20320 feet=
20320
5280
=3.848 miles
=> the distance between the object situated on Mount McKinley and the center of
the Earth can be calculated as follows:
r=3963+3.848=3966.848 miles
Now we find the object’s weight:
w=
C
r
2
w=
1570536900
3966,848
2
w=99.806 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. Long’s Peak in Rocky Mountain National Park, is 14,255 feet in elevation. How far can
you see to the horizon from the top of Long’s Peak? Can you see Cheyenne, Wyoming
(about 89 miles away)? Explain your answer.
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Axia College Material Appendix E Radicals 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:, where C is a constant, and r is the distance that the object is from the center of Earth. a. Solve the equation for r. = Because r represen ...
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