suppose that y=5x+1 and it is required that y be within 0.005 units of 7. for what values of x will this be true
1 answer:
Answer:
1.199 ≤ x ≤ 1.201
Step-by-step explanation:
You want ...
|y -7| ≤ 0.005
Substituting the expression for y, you have ...
|5x+1 -7| ≤ 0.005
|x -1.2| ≤ 0.001 . . . . . simplify, divide by 5
-0.0001 ≤ x -1.2 ≤ 0.001 . . . . . "unfold" the absolute value
1.199 ≤ x ≤ 1.201 . . . . . . . . . . . .add 1.2
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