Select one:

a. If the length of the wire is within 0.003 inch distance of 10, then the area of the square formed is within 0.06 inch of 100 inches.

b. If the length of the wire is within 10 inches distance of 10, then the area of the square formed is within 0.06 inch of 100 inches.

c. If the length of the wire is within 9.997 inches distance of 10, then the area of the square formed is within 0.06 inch of 100 inches.

d. If the length of the wire is within 10.002 inches distance of 10, then the area of the square formed is within 0.06 inch of 100 inches.

If the length of wire is 4x then the square formed will have an area of x²=100 sq in and x=10 in.

If we make a small change to x and call it h then (x+h)²=x²+k. That is, the small change in the length of the wire will create a small change k in the area of the square. Expand (x+h)²=x²+2xh+h² which has to equal x²+k. So k=2xh+h². We know x=10, so k=20h+h²=0.06 sq in. So we could solve the quadratic h²+20h-0.6=0. But because k is small we can expect h to be small, too, in which case h² would be even smaller, so we could ignore it and write 20h=0.06, making h=0.06/20=0.003. This approximate answer justifies our reason for ignoring h², which is very small=0.000009 in. So the answer is answer choice a. The other choices are wrong interpretations. When we say “within” 0.003 in of 10 in, we mean it can be almost as low as 9.997 in and almost as high as 10.003 in. Using a calculator let’s prove this: 9.997²=99.940009 sq in and 10.003=100.060009 sq in. Now you can see that the area is indeed within 0.06 of 100 sq in when we ignore the tiny discrepancy 0.000009 (9 millionths of a square inch). This calculation gives us confidence in our solution.

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