MAT/116 Week 3 dq 1

timer Asked: Oct 2nd, 2013

Question Description

I don’t know how to handle this question and need guidance.

In week 3, Section 2.4 of the text
discussed some mathematical formulas that are used in various fields to solve
problems in geometry.

Here’s a problem involving the
circumference and radius of a circle.

The equation for
circumference C in terms of radius
r is: C = 2pr.

Let’s assume the earth is a perfect
sphere and we tie a rope around the earth. The rope sits tightly on the surface
(circumference) of the earth.

Now let’s cut the rope and add 1
foot to it. (We’ve added 1 foot to the circumference.) The rope no longer sits
tightly on the earth, but is now some distance away from the surface.

Question: How far away from the
surface is the rope after we’ve added the foot to the circumference?  Or asking
another way-- when I add 1 foot to the circumference, what happens to the
radius? Show the equations and how they were used to solve this

( Note: No need to bring in the
distance around the earth which is 25000 miles. Just work with the equation for

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