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Troyanec [42]
4 years ago
15

Write a loop that asks the user to enter a number. the loop should iterate 10 times and keep a running total of the numbers ente

red.
Computers and Technology
1 answer:
Blababa [14]4 years ago
6 0
int sum = 0;
int number;
for (int i=0; i<10; i++) { 
   printf("Enter an integer number: ");
   scanf_s("%d", &number);
   sum += number;
}

printf("The sum is %d", sum);

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with the aid of your own example explain how memory,registers and secondary storage all work together​
AnnyKZ [126]

Memory, registers and secondary storage all work together​ so as to create or produce an ideal storage conditions.

<h3>How does memories works together?</h3>

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6 0
2 years ago
Question is attached as image, please help :&gt;
harkovskaia [24]

Answer:

Proved

Explanation:

Given

X = (a.\bar b)+(\bar a.b)

(a + b)\ .  \frac{}{a.b}

Required

Find out why they represent the same

To do this, we simplify either (a + b)\ .  \frac{}{a.b} or X = (a.\bar b)+(\bar a.b)

In this question, I will simplify (a + b)\ .  \frac{}{a.b}

Apply de morgan's law

(a + b)\ .  \frac{}{a.b} = (a + b) . (\bar a + \bar b)

Apply distribution property

(a + b)\ .  \frac{}{a.b} = a.\bar a + a.\bar b + \bar a . b + b . \bar b

To simplify, we apply the following rules:

a.\bar a = 0 --- Inverse law

b.\bar b = 0 --- Inverse law

So, the expression becomes

(a + b)\ .  \frac{}{a.b} = 0 + a.\bar b + \bar a.b + 0

(a + b)\ .  \frac{}{a.b}  = a.\bar b + \bar a.b

Rewrite as:

(a + b)\ .  \frac{}{a.b}  = (a.\bar b)+(\bar a.b)

From the given parameters:

X = (a.\bar b)+(\bar a.b)

This implies that:

(a + b)\ .  \frac{}{a.b} when simplified is X or (a.\bar b)+(\bar a.b)

3 0
3 years ago
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