Steven uses a
caliper that has an absolute error of 0.02 millimeters. An object's width is measured
using the caliper, which records a width of 14.5 millimeters. What equation would represent the situation?

A. |x-0.02|=14.5
B. |x-14.5|=0.02
C. |0.02-x|=14.5
D. |14.5-0.02|=x

Answers

Answer 1

The relationship between the measured value, error and actual width of the object is |x - 14.5| = 0.02

The measurement error = 0.02 millimeters The measured width of caliper = 14.5 millimeters

If the actual width of the caliper is represented as x

The measurement error is an absolute value which can be calculated thus :

|Actual value - measured value| = absolute error

|x - 14.5| = 0.02

Therefore, the equation which represents the scenario described is |x - 14.5| = 0.02

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Answer:

I don't know the answer you know

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Answers

9514 1404 393

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_____

Additional comment

Square roots are simplified by factoring out the factors that are perfect squares.

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Answers

Solving for the Diameter (“D”) of a Circle:
——————————————————————

To solve for the diameter of a circle, we start off with the circumference formula because it contains the information we need to Algebraically solve for the Diameter.

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