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r-ruslan [8.4K]
3 years ago
7

Please help me with coding!

Computers and Technology
2 answers:
tresset_1 [31]3 years ago
6 0

Answer:

key code space

Explanation:

the transformation

elena-14-01-66 [18.8K]3 years ago
6 0
The answer is the third question
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An asymmetric encryption system utilizes how many keys?
ivanzaharov [21]
Two. One for encryption, and one for decryption. RSA is an example of an asymmetric encryption algorithm.
7 0
3 years ago
2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of
allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0
3 years ago
How does a MIPS Assembly procedure return to the caller? (you only need to write a single .text instruction).
AnnyKZ [126]

Answer:

A MIPS Assembly procedure return to the caller by having the caller pass an output pointer (to an already-allocated array).

8 0
4 years ago
Flesh out the body of the print_seconds function so that it prints the total amount of seconds given the hours, minutes, and sec
fiasKO [112]

Answer:

Step by step explanation along with code and output is provided below

Explanation:

#include<iostream>

using namespace std;

// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three  

void print_seconds(int hours, int mints, int seconds)

{

   int total_seconds= hours*3600 + mints*60 + seconds;

   cout<<"Total seconds are: "<<total_seconds<<endl;

}

// test code

// user inputs hours, minutes and seconds and can also leave any of them by  entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.

int main()

{

   int h,m,s;

   cout<<"enter hours if any or enter 0"<<endl;

   cin>>h;

   cout<<"enter mints if any or enter 0"<<endl;

   cin>>m;

   cout<<"enter seconds if any or enter 0"<<endl;

   cin>>s;

   print_seconds(h,m,s);

  return 0;

}

Output:

enter hours if any or enter 0

2

enter mints if any or enter 0

25

enter seconds if any or enter 0

10

Total seconds are: 8710

8 0
4 years ago
To define constructors and member functions outside of a class's original scope, the operator can be used.
Artyom0805 [142]

Answer: Scope resolution operator(::)

Explanation: A member function and the constructor can be called within the function easily but for the execution of the these components outside the class , a special operator is required to call the functions. The scope resolution operator(::) preceding with the name of class is thus used for defining of the function outside class.This operator maintains the cope of the function and constructor outside the class.

8 0
3 years ago
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