Answer:
A: Backstage view > Account Settings
Explanation:
Make another account. But u cant get into brainly unless u r over 26 or something.
1.)
<span>((i <= n) && (a[i] == 0)) || (((i >= n) && (a[i-1] == 0))) </span>
<span>The expression will be true IF the first part is true, or if the first part is false and the second part is true. This is because || uses "short circuit" evaluation. If the first term is true, then the second term is *never even evaluated*. </span>
<span>For || the expression is true if *either* part is true, and for && the expression is true only if *both* parts are true. </span>
<span>a.) (i <= n) || (i >= n) </span>
<span>This means that either, or both, of these terms is true. This isn't sufficient to make the original term true. </span>
<span>b.) (a[i] == 0) && (a[i-1] == 0) </span>
<span>This means that both of these terms are true. We substitute. </span>
<span>((i <= n) && true) || (((i >= n) && true)) </span>
<span>Remember that && is true only if both parts are true. So if you have x && true, then the truth depends entirely on x. Thus x && true is the same as just x. The above predicate reduces to: </span>
<span>(i <= n) || (i >= n) </span>
<span>This is clearly always true. </span>
Answer:
All flags are On ( c, z , N )
Explanation:
Given data:
4-bit operation
Assuming 2's complement representation
<u>Determine status flags that are on after performing </u> 1010+0110
1 1
1 0 1 0
0 1 1 0
1 0 0 0 0
we will carry bit = 1 over
hence C = 1
given that: carry in = carry out there will be zero ( 0 ) overflow
hence V = 0
also Z = 1
But the most significant bit is N = 1
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