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2 years ago

Var name = prompt("Enter the name to print on your tee-shirt");

1 answer:

3 0

Willy Shakespeare has 17 characters. It's higher than 12,so the output will be Too long. Enter a name with fewer than 12 characters.

Letter a.

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To adjust the height of cells, position the pointer over one of the dividing lines between cells. When the pointer changes to th

Sergio [31]

Answer:

double arrow shape

Explanation:

To adjust the height of the cells

1. We have to position the mouse pointer over one of the column line or the one of the row line.

2. As we place the pointer between the dividing lines, the cursor of the mouse pointer change from singe bold arrow to double arrow symbol.

3.Now press or click the left mouse button and drag the dividing lines of the cells to the desired position to have the required width or height of the cell.

5 0

2 years ago

A __________ error does not prevent the program from running, but causes it to produce incorrect results. a. syntax b. hardware

Lunna [17]

Answer:

1) syntax error

2)task

3)algorithm

4) pseudocode

5)flowchart

5 0

2 years ago

Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS

BlackZzzverrR [31]

Answer:

hello your question lacks the C segment so here is the C segment

while ( k<n )

{v[k] = v[k+1];

k = k+1; }

Answer : while:

bge $s0, $s2, end # while (k < n)

addi $t0, $s0, 1 # $t0 = k+1

sll $t0, $t0, 2 # making k+1 indexable

add $t0, $t0, $s1 # $t0 = &v[k+1]

lw $t0, 0($t0) # $t0 = v[k+1]

sll $t1, $s0, 2 # making k indexable

add $t1, $t1, $s1 # $t1 = &v[k]

sw $t0, 0($t1) # v[k] = v[k+1]

addi $s0, $s0, 1

j while

end:

Explanation:

The MIPS assembly code corresponding to the C segment is

while:

bge $s0, $s2, end # while (k < n)

addi $t0, $s0, 1 # $t0 = k+1

sll $t0, $t0, 2 # making k+1 indexable

add $t0, $t0, $s1 # $t0 = &v[k+1]

lw $t0, 0($t0) # $t0 = v[k+1]

sll $t1, $s0, 2 # making k indexable

add $t1, $t1, $s1 # $t1 = &v[k]

sw $t0, 0($t1) # v[k] = v[k+1]

addi $s0, $s0, 1

j while

end:

4 0

2 years ago

Write a C program that right shifts an integer variable 4 bits. The program should print the integer in bits before and after th

Sindrei [870]

Solution :

#include< $\text{stdio.h}$ >

#include<conio.h>

void dec_bin(int number) {

$\text{int x, y}$ ;

x = y = 0;

for(y = 15; y >= 0; y--) {

x = number / (1 << y);

number = number - x * (1 << y);

printf("%d", x);

}

printf("\n");

}

int main()

{

int k;

printf("Enter No u wanted to right shift by 4 : ");

scanf("%d",&k);

dec_bin(k);

k = k>>4; // right shift here.

dec_bin(k);

getch();

return 0;

}

4 0

1 year ago

Network flow issues come up in dealing with natural disasters and other crises, since major unexpected events often require the

nasty-shy [4]

Answer:

Check the explanation

Explanation:

Algorithm for solving flood condition:

We suggest an algorithm to resolve the flood condition by creating a flow network graph.

Let us assume for every patient "p" there is a node "2" and for every hospital "h" there is a node "uh" and there is an edge ()T, uh) exist between patient "p" and hospital "h" with flow capacity of 1 iff patient "p" is reachable to hospital "h" within a half-hour.

Then source node "s" is made between all the patient-nodes by an edge with flow capacity of 1 and then the sink "t" is made by linking all the hospital nodes by an edge with capacity "[n/k]".

There is an approach to send patients to hospitals: when there is a source "s" to sink "t" flow of "n". We can send 1 flow-unit from source "s" to sink "t" along the paths (s, yp, uh, t) whenever a probable approach is available to send patients.

This approach of sending patients to hospitals doesn't break the capacity limitation of edges. Hence we can send patient "p" to hospital "h" with 1 flow- unit if edge(m uh) permits at least 1 flow- unit.

The running-time of this algorithm is found by finding the time needed to solve max-flow graph with nodes O(n+k) and edges O( $n^{k}$ ) edges.

5 0

2 years ago