mat117_appendix_e

Suppose that an object is 100 pounds when it is at

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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.) = C 100 3963² = = C 100 3963 3963² = 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) = w Cr - 2 = • (( -{ / }) w 1570536900 3963 282 5280 - ) 2 = • (( - . ) w 1570536900 3963 0534 - ) 2 = • (( . ) w 1570536900 3962 9466 - ) 2 = • ( . ( w 1570536900 6 37 10 - )) 08 = . w 100 04 ii. The top of Mt McKinley (20,320 feet above sea level) W=cr -2 = • (( +{ / })- ) w 1570536900 3963 20320 5280 2 = • (( + . )- ) w 1570536900 3963 3 848 2 = • (( . ) w 1570536900 3966 848 - ) 2 = • ( . ( - )) w 1570536900 6 35 10 8 = . w 99 73 MAT 117

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2. The equation 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. = . D 1 2 h . = d1 2 h b. Long’s Peak in the 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. = . D 1 2h = . D 1 214255 Since a person can see 143.27 miles from the top of Long’s Peak, I  would have to say yes a person could definitely see Cheyenne because  89 miles is much less than 143 miles. MAT 117
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