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Monday, March 4, 2013

Appendix E Math 117

Associate Level Material
Appendix E

Radicals

Application Practice

Answer the interest questions. Use Equation Editor to write mathematical expressions and equations. First, present this file to your hard drive by selecting Save As from the File menu. Click the white space below each(prenominal) question to maintain proper formatting.

Hint: Pay assistance 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.

Many hatful k immediately that the weight of an intent varies on different planets, precisely did you know that the weight of an prey on Earth overly varies according to the elevation of the object? In particular, the weight of an object follows this equation:[pic], where C is a constant, and r is the distance that the object is from the refer of Earth.

Solve the equation [pic]for r. wr^-2=c. I and so had to water parting by W on both so now it is r^-2=c/w. Now the square root of c/w=r


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.

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I plugged in the numbers where they go and then multiplied and the answer is C=1,570,536,900


Use the value of C you found in the previous question to determine how much(prenominal) the object would weigh in

Death Valley (282 feet below sea level). First multiply 3963 * 5280 which equals 20,924,640. Next, w=1570536900(20924640 282)^-2. Now subtract 20924640-282 and that is (20924358). So the equation now looks like W=1570536900(20924358)^-2. This shows that W=100.0018 approximately.

the vizor of Mount McKinley (20,320 feet above sea level).

The equation looks like r=3963+20320/5280 plug in the 1,570,536,900 into the equation so now its W=1,570,536,900/(3963+20320/5280)^2. Once I worked out this problem following the rules of trading operations now have the answer of W=99.806 pounds.



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