Q: Can the lower bound of an option be negative?

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The lower bound of 3.0 is 3.0

If I understand your question correctly, such a sequence is an = x cos(πx). It has neither an upper nor lower bound. It's divergent, but its limit is neither infinity nor negative infinity.

If the range is the real numbers, it has a lower bound (zero) but no upper bound.

The answer depends on the level of accuracy of the value 0.

no won noes * * * * * It means that there is an upper and lower bound or limit. There is the lower bound such that you exclude any smaller numbers, and an upper bound such that you exclude bigger numbers. What you do wit hnumbers that are equal to the bounds depends on the nature of the bounds.

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Option Explicit makes the declaration of Variables Mandatory while Option Base used at module level to declare the default lower bound for array subscripts. For eg. Option Base 1 will make the array lower bound as 1 instead of 0. by Munendra

Lower bound is 17.6 and upper bound is 17.8

The lower bound of 3.0 is 3.0

The answer is B.

If I understand your question correctly, such a sequence is an = x cos(πx). It has neither an upper nor lower bound. It's divergent, but its limit is neither infinity nor negative infinity.

You cannot list them: unless the inequality is trivial, since there are infinitely many real numbers in any range. You need toidentify the lower bound;determine whether or not the lower bound is included (

What is the lower and upper bound of 9.3 in 1 s.f.?

The lower bound is 16.35

-5 is lower

j

bcoz the energy of bound system is always negative

The lower bound is 0.5 less and the upper bound is 0.5 more.