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kupik [55]
3 years ago
7

What is a best practice when approaching an automation effort?

Computers and Technology
1 answer:
NARA [144]3 years ago
3 0

Answer:

improving the current process before automating it

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Write a program to input an integer num from the keyboard and: : find the area of a square whose side length is num. : find the
NeX [460]

Answer:

# include<iostream>  

# include<math.h>  

using namespace::std;  

void calc (int num)  

{

     cout<<"Area of Square" <<num*num;   

     cout<<"Volume of a sphere" << (4/3) *3.14*num*num*num;

     cout<<"Side of a Square" <<sqrt(num);  

     }  

int main ()  

{  

     int num;

     cout<<"Enter the Number"; cin>>num;  

      calc(num);  

      return 0;  

}  

Explanation:

I have used the swtich for the selection purpose. Also, I have used sqrt() function, and for that I have included math,h library.  And the various formula comes from mensuration. Rest part is self explanatory.

3 0
4 years ago
When using a public computer or network, you should always _______.
yaroslaw [1]
When using a public computer or network, you should always browse appropriately and log out of personal accounts when you're done. 
6 0
3 years ago
Read 2 more answers
How is primary storage different from secondary storage? Select the TWO correct statements.
Zina [86]
Primary storage refers to the main storage of the computer or main memory which is the random access memory or RAM. Secondary storage, on the other hand, refers to the external storage devices used to store data on a long-term basis.
7 0
3 years ago
Im coding and need help please answer or comment to help. any skilled coders comment below
RoseWind [281]

Answer:

Whats up?

Explanation:

6 0
3 years ago
Assume a computer that has 32-bit integers. Show how each of the following values would be stored sequentially in memory, starti
finlep [7]

Answer:

Adresses                          Big endian                                    Little Endian

0x100                                  AB                                                 56

0x101                                   12                                                  34

0x102                                  34                                                  12

0x103                                  56                                                 AB

0x104                                  00                                                 76

0x105                                  2B                                                 F8

0x106                                  F8                                                 2B

0x107                                 76                                                 00

0x108                                00                                                 A1

0x109                                08                                                B0  

0x10A                                B0                                                08

0x10B                                A1                                                 00

0x10C                               00                                                  01

0x10D                               00                                                 00  

0x10E                               00                                                  00

0x10F                                01                                                   00

0x110                                FE                                                   34

0x111                                 DC                                                  12

0x112                                12                                                    DC  

0x113                                34                                                   FE

Explanation:

As, it is mentioned that all the integers have 32 bits and all the data given in the question is in Hexadecimal. In Hexadecimal each number consists of 4 bits.

Each memory location contains 1 byte = 8 bits of data. So we store 2 digits of Hexadecimal number each location from each number in given data. If any number contains less than 32 bits in binary, we will add zeros to complete the number in 32 bits.

In big endian order, data is stored from upper most bit to lower most bit in sequential order and in Little Endian order it is vise versa.

6 0
3 years ago
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