Sentence examples for Rest sufficiently from inspiring English sources

Exact(5)

"If you don't get eight and a half hours, your eye muscles just don't rest sufficiently, and if the muscles aren't rested they don't oxygenate properly and the vessels dilate and you get puffiness and discoloration".

The bigger problem, he says, is bad driving from chasers who are either distracted by the storm, do not park far enough off the road to keep it clear, or who do not rest sufficiently before chasing.

One must rest sufficiently in a good, comfortable position without disturbing their biological clock to function at 100%.

The discharge criteria were: mild pain (VAS < 30 at rest) sufficiently controlled by oral analgesics, ability to walk with elbow crutches, ability to climb 8 stairs, ability to eat and drink, and no evidence of any surgical complications.

The patient was considered ready for discharge when all of the following criteria were fulfilled: no or mild pain (VAS < 30 at rest) sufficiently controlled by oral analgesics; ability to walk with the aid of elbow crutches, to climb 8 stairs, and to eat and drink normally; and no evidence of any surgical complications.

Similar(55)

Now, for any p ≥ 2, we take r sufficiently large such that r > p in (4.7).

For any 1 < p < 2, we take r sufficiently small such that 1 < r < p in (4.7).

Then, for any given ε > 0, there exists a set E ⊂ ( 1, + ∞ ) of finite linear measure such that, for all z satisfying | z | = r ∉ [ 0, 1 ] ∪ E and r sufficiently large, exp { − r ρ + ε } ≤ | f ( z ) | ≤ exp { r ρ + ε }.

If condition (2.4) does not hold, then the numerator in (2.6) is negative for z = r sufficiently close to 1. Hence there exists z 0 = r 0 in ( 0, 1 ) for which the quotient in (2.6) is negative.

Then for any given ε > 0, there is a set E ⊂ [ 1, + ∞ ) having finite linear measure such that for all z satisfying | z | = r ∉ [ 0, 1 ] ∪ E and r sufficiently large, we have exp { − r σ + ε } ≤ | f ( z ) | ≤ exp { r σ + ε }.

By Lemma 2.3, for any given ε ( 0 < ε < n − max 1 ≤ j ≤ k σ j ), there is a set E ⊂ [ 1, + ∞ ) having finite linear measure such that for all z satisfying | z | = r ∉ [ 0, 1 ] ∪ E and r sufficiently large, we have exp { − r σ j + ε } ≤ | D j ( z ) | ≤ exp { r σ j + ε } ( 1 ≤ j ≤ k ).

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