## may

Then $F$ may or may not fix $B$.

Here $S$ may be $P$, but it may also not be $P$.

In other algorithms, this may not be true.

In addition to a contribution to $W_1$, there may also be one to $W_2$.

Note that both sides of the inequality may well be infinite.

It may well be that no optimal time exists, as the following example shows.

It may seem strange to define $0.\infty=0$.

We may (and do) assume that ......

Since the integrands vanish at 0, we may as well assume that ......

However, $F$ is only nonnegative rather than strictly positive, as one may have expected.

Since $Z$ is a finite set, we may continue subtracting suitable scalar multiples of the $x_i$ from $x$. [Or: we can continue]

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