There are different software that has been developed today. The most time-saving and cost-effective way is to Design the application’s security features after the application’s initial build is complete.
Accenture is known for their work in improving business needs. They are constantly shifting to a new method of delivering information technology.
They are known to embed security into the product development life cycle helps secure the business and also keeping speed and assisting to remove friction.
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brainly.com/question/25737623
Here is some ranting.
You haven't specified programming language. And every type of loop runs only when a condition is true (that is the definition of a loop).
And here is the answer.
Well, there is [code]for[/code], [code]while[/code] and in some languages even [code]do while[/code].
Hope this helps.
Answer:
all but 3
Explanation:
3 is strange and I need points sorry
Answer:
I can give you the perimeter "algorithm" but not the flowchart.
Here you go:
p = w * 4
p = perimiter,
w = width/height
4 = the amount of sides needed.
perimeter = width * 4
to include both width and height, we would instead use:
perimeter = 2(width+height)
This also works with rectangles ^
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To find an area, it is just width * height. This is also compatible with rectangles.
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>
