The sizing handles is known to be the small squares that appear in the corners and in the middle of the sides of a selected object's border.
<h3>What is sizing handles in MS Word?</h3>
The term sizing handles is known to be a tool that a person can use to be able to change a picture's size.
Note that if the mouse pointer is placed on a key that is of the sizing handles, the pointer alters to a double-headed arrow and a person can be able to then alter the size or shape of the image by dragging the sizing handle.
Note that it is often called handle, drag handle, sizing grip, resize corner, and therefore, The sizing handles is known to be the small squares that appear in the corners and in the middle of the sides of a selected object's border.
Learn more about sizing handles from
brainly.com/question/8806816
#SPJ1
Well the result shown in C19 will show the sum of all cells from C5 to C18:
ie: C5 + C6+ C7+ C8+ C9+ C10+ C11+ C12+ C13+ C14+ C15+ C16+ C17+ C18
:)
Answer:
see explaination
Explanation:
The following code is in python 3.5 and above:
# Create a main method
def main():
# Accept name from the user
name = input("Enter your name: ")
# Accept describe yourself from the user.
describe = input("Describe yourself: ")
# Create a file object
f = open('person.html','w')
# Creat a string to store the html script.
message = """<html>
<head>
</head>
<body>
<center>
<h1>"""+name+"""</h1>
</center>
<hr/>"""+describe+"""<hr/>
</body>
</html>"""
f.write(message)
f.close()
main()
The answer is C because your sideview and rearview mirrors are there specifically to check behind you, while the vanity mirror is not supposed to be used for the same purpose.
Answer:
Explanation:
The following code is written in Java and creates the recursive function to find the longest common substring as requested.
static int lengthOfLongestSubsequence(String X, String Y) {
int m = X.length();
int n = Y.length();
if (m == 0 || n == 0) {
return 0;
}
if (X.charAt(m - 1) == Y.charAt(n - 1)) {
return 1 + lengthOfLongestSubsequence(X, Y);
} else {
return Math.max(lengthOfLongestSubsequence(X, Y),
lengthOfLongestSubsequence(X, Y));
}
}