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Sloan [31]
3 years ago
14

There are two String variables, s1 and s2, that have already been declared and initialized. Write some code that exchanges their

values. Declare any other variables as necessary.
Computers and Technology
2 answers:
Ivan3 years ago
8 0

Answer:

//here is code in java.

import java.util.*;

class Solution

{

// main method of class

public static void main (String[] args) throws java.lang.Exception

{

   try{

    // declare an initialize first string variables

     String st1="hello";

     // declare an initialize first string variables

     String st2="world";

     // create another string variable

    String st3;

    // exchange the value of both string variables

    st3=st1;

    st1=st2;

    st2=st3;

    System.out.println("value of first String after exchange: "+st1);

    System.out.println("value of second String after exchange: "+st2);

   }catch(Exception ex){

       return;}

}

}

Explanation:

declare and initialize two string variables.Create another string variable "st3". first assign value of "st1" to "st3" after then value of "st2" to "st1" and then  assign value of "st3" to "st2". This will exchange the values of both the string.

Output:

value of first String after exchange: world

value of second String after exchange: hello

Lunna [17]3 years ago
5 0

Answer:

The following are the code

string s1="san";

string s2="ran";

string temp; // other variable

temp=s1;// statement 1

s1=s2;// statement 2

s2=temp;//statement 3

Explanation:

In this code we declared two string s1 , s2 and initialized them after that we declared an extra variable temp .

The statement 1  temp=s1   means that temp will store the value of s1 i.e "san". so temp="san";

The second statement 2  s1=s2   means that s1 will store the value of s2

i.e s1="ran";

The third statement 3 s2=temp means that s2 will store the value of temp.

i.e s2="san"

we see that s1 and s2 will exchange their value

Following are the code in c++

#include <iostream> // header file

#include<string>

using namespace std;

int main()// main function

{

string s1="san";

string s2="ran";

string temp;

cout<<"before exchange:"<<s1<<s2<<endl; // display before exchanging the value

temp=s1;

s1=s2;

s2=temp;

cout<<"after exchange:"<<s1<<s2<<endl;// display after exchanging the value

return 0;

}

Output:

before exchange:sanran

after exchange:ransan

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Explain the developments RAM since it was created in 1968 til today.
xenn [34]

Answer:

From DRAM to DDR4

Explanation:

RAM stands for <em>Random Access Memory.</em> In 1968, Mr. Robert Dennard at IBM's Watson Research obtained the patent for the one-transistor cell that will eventually substitute the old magnetic core memory allocated in computers of the time. By 1969 Intel released the TTL bipolar 64-bit SRAM (Static Random-Access Memory) as well as the ROM "Read Only Memory"; also in 1969 it evolved into "<em>Phase - change memory - PRAM - </em>". However this evolution was not commercialized, Samsung expressed its interest in developing it. In 1970 the first DRAM product was commercially available; it was developed by Intel. In 1971 it was patented EPROM; in 1978 George Perlegos developed EEPROM.

By 1983 a nice breakthrough happened with the invention of SIMM by Wang Labs. In 1993 Samsung came up with KM48SL2000 synchronous DRAM (SDRAM), this variation soon turned into an inductry standard.

In 1996 DDR began a revolution in the memory sector, then in 1999 RDRAM. Both DDR2 SDRAM. DDRR3 and XDR DRAM were commercialized. Finally in 2007 and 2014 the developments of DDR3 and DDR4 were available for the general public.

6 0
3 years ago
Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
Ede4ka [16]

Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

v = \frac{(xf-xo)}{(t-to)}

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

t =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

4 0
3 years ago
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<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Geo tagging

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➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0
3 years ago
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BlackZzzverrR [31]

Answer:a machine loom

Explanation:

8 0
2 years ago
ProcessName2
RoseWind [281]

Explanation:

This is easily solvable with a for loop. Something like:

(I assume c++)

#include <iostream>

#include <string>

int main() {

take_input: //tag

std::string input;

cin >> input; //take the input

int spaceCount = 0;

char checking;

for(unsigned int i = 0; i == input.length(); ++i) {

checking = spaceCount[i];

if(checking == ' ')

spaceCount++;

}

if(spaceCount >= 1 && input.length >= 5)

std::cout << "Your name is " + input;

else

goto take_input; // reasks for input if the conditions are not met

return 0;

};

**remove all spaces before using the code, the if statements are messed up

5 0
3 years ago
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