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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

User Archlight
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1 Answer

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Answer:

1.199 ≤ x ≤ 1.201

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

User Espen Riskedal
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