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7. If h is the variation in length then 3(x+h)=30±0.06.

3x+3h=30±0.06, x+h=10±0.02. When x=10 h=±0.02, so the length of each piece must be within 0.02 in of 10 in.

8. Let x=3+h then (3+h)²=9±1. 9+6h+h²=9±1 so h²+6h±1=0 and h=(-6±√36±4)/2=-3±2√2 or -3±√10.

If we calculate all possible values for h we get -0.1716, -5.8284, 0.1623, -6.1623. Remember that x is close to 3 so we can reject the large negative values. That leaves -0.1716 and 0.1623 making x=(2.8284,3.1623).

Now we can see that x=3 is approximately in the middle of this interval. If we ignore the h² term in the quadratic because it is small compared to 3 we get 6h=±1 and h=±⅙=±0.17 making x=(2.83,3.17). So there is justification in ignoring the h² term. We can say that the values of x that guarantee f(x) to be within 1 unit of 9 are that x is in the interval (2.8,3.2).

For 0.2 units h=±0.2/6=±0.033 so x=(2.967,3.033). This is ⅕ of the value of h above and 0.2 is ⅕ of one unit.

The remaining questions have been answered elsewhere.

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