Answer:
prompt("Enter a value for one edge of a cube")
Store user's value into edgeCube
area = 6 * (edgeCube * edgeCube)
volume = edgeCube * edgeCube * edgeCube
print("One side of the cube is: " + edgecube);
print("The area is: " + area)
print("The volume is: " + volume)
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
Answer:
I want brainlyest though can I have it
<span>make periodic adjustments on an as-needed basis. </span>
If you are using CSS
:
table {
border-collapse: collapse;
border: 5px solid black;
width: 100%;
}
td {
width: 50%;
height: 2em;
border: 1px solid #ccc;
}
HTML
<table>
<tbody>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
</tbody>
</table>
for HTML:
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Sample table</title>
<style>
table {
border-collapse: collapse;
border: 5px solid black;
width: 100%;
}
td {
width: 50%;
height: 2em;
border: 1px solid #ccc;
}
</style>
</head>
<body>
<table>
<tbody>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
</tbody>
</table>
</body>
</html>
