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Dennis_Churaev [7]
3 years ago
11

What are the advantages and drawbacks of using solar energy

Computers and Technology
1 answer:
Ray Of Light [21]3 years ago
5 0

upfront price, but dependent on how long you use it you will save money, also the power it provides per square inch is low but that will be solved with time.

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Which of the following features of a network connection between a switch and server is not improved by link aggregation?
egoroff_w [7]

Answer: Speed

Explanation:

 Link aggregation is the process of a network connection between the switch and the server which are used for connections of the multiple parallel network. Line aggregations provides various advantages like increases the bandwidth, high availability and provides the fault tolerance but not speed because it increases the number of lanes not the speed as it is expensive and it does not increases the throughput of the system.

3 0
3 years ago
Groups to which we belong are defined late in our development true or false
Anna71 [15]
The answer is false......
7 0
3 years ago
Write a program in C that reads a 3-bit desired light pattern from the 3 input switches connected to pins P2.3-P2.5 and displays
laiz [17]

Answer:

Check the explanation

Explanation:

#include<stdio.h>

main()

{ int k=1,p,l,i;

int p2[8],p1[8];

for(i=0;i<8;i++){

p1[i]=0;

p2[i]=0;

}

while(k==1)

{

printf("enter the pin inputs from p2.3-p2.5:");

scanf("%d%d%d",&p2[3],&p2[4],&p2[5]);

 

printf("Press Logic 0 for on and 1 for off on switch port p2.0\n");

scanf("%d",&l);

p1[3]=p2[3];

p1[4]=p2[4];

p1[5]=p2[5];

if(l==0)

{p=1;

for(i=0;i<8;i++)

printf("p1.%d : %d\n",i,p2[i]);

printf("\n press either 1 to continue or 0 to stop :");

scanf("%d",&p);

printf("\n");

while(p==1)

{ printf("enter the pin inputs from p2.3-p2.5:");

scanf("%d%d%d",&p2[3],&p2[4],&p2[5]);

p1[3]=p2[3];

p1[4]=p2[4];

p1[5]=p2[5];

for(i=0;i<8;i++)

printf("p1.%d : %d\n",i,p2[i]);

printf("\n press either 1 to continue or 0 to stop :");

scanf("%d",&p);

printf("\n");

}

}

else

{

printf("After Rotation of pattern :\n");

for(i=0;i<8;i++)

{ if(p2[i]==0)

printf("p1.%d : 1\n",i);

else

printf("p1.%d : 0\n",i);

}

}

printf("press 0 for on and 1 for off on switch port p2.1 :\n");

scanf("%d",&l);

if(l==0)

{ printf("\nLeft Rotation of pattern from read mode :\n");

printf("p1.0 : %d\n",p1[0]);

printf("p1.1 : %d\n",p1[5]);

printf("p1.2 : %d\n",p1[6]);

printf("p1.3 : %d\n",p1[3]);

printf("p1.4 : %d\n",p1[2]);

printf("p1.5 : %d\n",p1[1]);

printf("p1.7 : %d\n",p1[6]);

printf("p1.7 : %d\n",p1[7]);

}

else

{printf("\nRight rotation of pattern from read mode: \n");

printf("p1.0 : %d\n",p1[0]);

printf("p1.1 : %d\n",p1[1]);

printf("p1.2 : %d\n",p1[2]);

printf("p1.3 : %d\n",p1[7]);

printf("p1.4 : %d\n",p1[6]);

printf("p1.5 : %d\n",p1[5]);

printf("p1.7 : %d\n",p1[4]);

printf("p1.7 : %d\n",p1[3]);

}

printf("press 0 for on and 1 for off on switch port p2.2 :\n");

scanf("%d",&l);

if(l==0)

printf("Make Rotation fast\n");

else

printf("Make Rotation slow\n");

printf("Enter 1 to continue or 0 to exit:");

scanf("%d",&k);

printf("\n");

}

}

8 0
3 years ago
X = 10<br> y =<br> 20<br> х» у<br> print("if statement")<br> print("else statement")
MrMuchimi

Answer:

"else statement"

Explanation:

Given

The above code segment

Required

The output of the program

Analyzing the program line by line, we have:

x = 10 ----> Initialize x to 10

y = 20 ----> Initialize y to 20

if x > y ----> check if x is greater than y

    print("if statement") ----> Execute this line if the condition is true

The condition is false because 10 is less than 20, so: the statement will not be executed. Automatically, the else condition will be executed

<em>else</em>

<em>    print("else statement") </em>

<em>"else statement" without the quotes will be printed because the if condition is false</em>

8 0
3 years ago
Write a function that returns a pointer to the MAXIMUM value of an array of double values. If the array is empty, return NULL. d
marusya05 [52]

Answer:

double* maximum(double *a, int size) {

if(size == 0)

return NULL;

int i;

double max = 0.0;

double *mp;

for(i = 0; i < size; i++){

if(a[i] > max){

max = a[i];

mp = &a[i];

}

}

return mp;

}

Explanation:

You iterate through the vector and find the maximum value. You go updating the pointer throughout the loop. I am going to write a C function

double* maximum(double *a, int size) {

if(size == 0)

return NULL;

int i;

double max = 0.0;

double *mp;

for(i = 0; i < size; i++){

if(a[i] > max){

max = a[i];

mp = &a[i];

}

}

return mp;

}

8 0
3 years ago
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