Western International University Material s1

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: w  Cr 2 , where C is a constant, and r is the distance that the object is from the center of Earth.

a. Solve the equation w  Cr 2 for r.

b. Suppose that an object is 150 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.)

c. Use the value of C you found in the previous question to determine how much the object would weigh in

i. Dark Valley (184 feet below sea level).

ii. The top of Mount MoleHill (22,340 feet above sea level).

C C C a) w  Cr 2  w   r 2   r  r 2 w w C Answer: r  w b) w  Cr 2 , w  150, r  3,963  150  C(3,963) 2  C  150(3,963) 2  2,355,805,350

Answer: C=2,355,805,350

c) i) 1 mile = 5,280 feet so 184 feet = 184/5,280 miles then: 184 feet = 0.0348 miles

r  3,963  0.0348  3,962.9652

w  Cr 2  2,355,805,350(3,962.9652)2  150.0026

Answer: 150.0026 pounds

MAT/117

ii) 1 mile = 5,280 feet so 22,340 feet = 22,340/5,280 miles then: 22,340 feet = 4.2311 miles

r  3,963  4.2311  3,967.2311

w  Cr 2  2,355,805,350(3,967.2311) 2  149.6802

Answer: 149.6802 pounds

2. The equation D  1.2 h 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.

b. Tall´s peak is 13,225 feet in elevation. How far can you see to the horizon from the top Tall´s peak? c. From the top of Tall´s peak can you see a park that is about 104 miles away? Explain your answer.

2 2 D 2 D D a) D  1.2 h  h    h   h  1.2 (1.2) 2 1.44 D2 Answer: h  1.44 b) D  1.2 h,h  13,225  D  1.2 13,225  138

Answer: we can see 138 miles away from the top of Tall´s peak and we can see the park that is 104 miles away because 138 > 104

MAT/117