Answer:
Multiple devices can be connected
Answer:
index.html:
<!DOCTYPE html>
<html>
<head>
<title>Brainly</title>
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<table id="myTable">
<caption>Duty Roster for last two days</caption>
<tr><th>Day</th><th>Morning</th><th>Afternoon</th></tr>
<tr><td>Monday</td><td colspan="2">John</td></tr>
<tr><td>Tuesday</td><td rowspan="2">Tom</td><td>Peter</td></tr>
<tr><td>Wednesday</td><td>simon</td></tr>
</table>
</body>
</html>
<u>style.css:</u>
#myTable {
background-color: Cyan;
border-collapse: collapse;
height: 200px;
width: 500px;
font-size: 20px;
}
#myTable td,th {
border: 2px black solid;
text-align: center;
}
#myTable td:first-child, th {
background-color: lightgray;
font-weight: bold;
}
JAVA programming was employed...
What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules
Solution:
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x1[i][j] = num.nextInt();
}
}// End of Array 1, x1.
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x2[i][j] = num.nextInt();
}
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:
x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
x1rules = true;
system.out.print(x1rules);
}
else
{
system.out.print(x1rules);
}//Conditional Statement
I believe the answer to this question is XML
If the only trying that appears is terms of use and change country, that means you are a free user and not a paying subscriber.
So there is no actual subscription to cancel.
If you were a paying subscriber, there would be a cancel subscription option within your account settings.
