The diameter of a value for a space shuttle must be within 0.001 mm of 5 mm. Enter and solve an absolute-value equation to find the boundary value for the acceptable diameters of the valve.
An equation to represent the situation is
The acceptable boundaries for the diameters are
Inadequate data.
|d-5| <= 0.001
so
5-0.001 <= d <= 5+0.001
To find the boundary value for the acceptable diameters, we can set up an absolute value equation.
Let x be the diameter of the valve.
The absolute value equation can be written as:
| x - 5 | ≤ 0.001
To solve this equation, we can split it into two cases:
Case 1: (x - 5) ≤ 0.001
Case 2: - (x - 5) ≤ 0.001
For Case 1:
x - 5 ≤ 0.001
x ≤ 5.001
For Case 2:
-(x - 5) ≤ 0.001
-x + 5 ≤ 0.001
-x ≤ -4.999
x ≥ 4.999
Therefore, the acceptable boundaries for the diameters are:
4.999 ≤ x ≤ 5.001
To find the boundary value for the acceptable diameters of the valve, we can set up an absolute-value equation.
Let's call the diameter of the valve x.
The given condition is that the diameter must be within 0.001 mm of 5 mm. This means that the diameter can be as close as 0.001 mm smaller than 5 mm, or as close as 0.001 mm larger than 5 mm.
Therefore, the acceptable boundaries for the diameters are:
x - 5 ≤ 0.001 (for the smaller boundary)
and
5 - x ≤ 0.001 (for the larger boundary)
To solve these equations, we need to get x by itself. Let's solve each equation separately:
1. x - 5 ≤ 0.001:
Adding 5 to both sides:
x - 5 + 5 ≤ 0.001 + 5
x ≤ 5.001
So, the smaller boundary value for the acceptable diameters is 5.001 mm.
2. 5 - x ≤ 0.001:
Adding x to both sides:
5 - x + x ≤ 0.001 + x
5 ≤ 0.001 + x
Subtracting 0.001 from both sides:
5 - 0.001 ≤ 0.001 - 0.001 + x
4.999 ≤ x
or
x ≥ 4.999
So, the larger boundary value for the acceptable diameters is 4.999 mm.
Therefore, the absolute-value equation representing the acceptable boundaries for the diameters of the valve is:
|5 - x| ≤ 0.001
This means the diameter of the valve must be within 0.001 mm of 5 mm, which translates to the range [4.999 mm, 5.001 mm].