Answer:
The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)
Then satisfying this theorem the system is consistent and has one single solution.
Explanation:
1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
Then the system is consistent and has a unique solution.
<em>E.g.</em>
2) Writing it as Linear system
3) The Rank (A) is 3 found through Gauss elimination
4) The rank of (A|B) is also equal to 3, found through Gauss elimination:
So this linear system is consistent and has a unique solution.
Code:
def myAppend( str, ch ):
# Return a new string that is like str but with
# character ch added at the end
return str + ch
def myCount( str, ch ):
# Return the number of times character ch appears
# in str.
# initiaalizing count with 0
count = 0
# iterating over every characters present in str
for character in str:
# incrementing count by 1 if character == ch
if character == ch:
count += 1
# returning count
return count
def myExtend( str1, str2 ):
# Return a new string that contains the elements of
# str1 followed by the elements of str2, in the same
# order they appear in str2.
# concatenating both strings and returning its result
return str1 + str2
def myMin( str ):
# Return the character in str with the lowest ASCII code.
# If str is empty, print "Empty string: no min value"
# and return None.
if str == "":
print("Empty string: no min value")
return None
# storing first character from str in char
char = str[0]
# iterating over every characters present in str
for character in str:
# if current character is lower than char then
# assigning char with current character
if character < char:
char = character
# returning char
return char
def myInsert( str, i, ch ):
# Return a new string like str except that ch has been
# inserted at the ith position. I.e., the string is now
# one character longer than before.
# Print "Invalid index" if
# i is greater than the length of str and return None.
if i > len(str):
print("Invalid index")
return None
# str[:i] gives substring starting from 0 and upto ith position
# str[i:] gives substring starting from i and till last position
# returning the concatenated result of all three
return str[:i]+ch+str[i:]
def myPop( str, i ):
# Return two results:
# 1. a new string that is like str but with the ith
# element removed;
# 2. the value that was removed.
# Print "Invalid index" if i is greater than or
# equal to len(str), and return str unchanged and None
if i >= len(str):
print("Invalid index")
return str, None
# finding new string without ith character
new_str = str[:i] + str[i+1:]
# returning new_str and popped character
return new_str, str[i]
def myFind( str, ch ):
# Return the index of the first (leftmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str
for i in range(length):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRFind( str, ch ):
# Return the index of the last (rightmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str from right side
for i in range(length-1, 0, -1):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRemove( str, ch ):
# Return a new string with the first occurrence of ch
# removed. If there is none, return str.
# returning str if ch is not present in str
if ch not in str:
return str
# finding position of first occurence of ch in str
pos = 0
for char in str:
# stopping loop if both character matches
if char == ch:
break
# incrementing pos by 1
pos += 1
# returning strig excluding first occurence of ch
return str[:pos] + str[pos+1:]
def myRemoveAll( str, ch ):
# Return a new string with all occurrences of ch.
# removed. If there are none, return str.
# creating an empty string
string = ""
# iterating over each and every character of str
for char in str:
# if char is not matching with ch then adding it to string
if char!=ch:
string += char
# returning string
return string
def myReverse( str ):
# Return a new string like str but with the characters
# in the reverse order.
return str[::-1]
Answer:
Intelligent Automation (IA) is a combination of Robotic Process Automation (RPA) and Artificial Intelligence (AI) technologies which together empower rapid end-to-end business process automation and accelerate digital transformation..
Explanation:
Hope it helps you..
Your welcome in advance..
(ㆁωㆁ)
Answer:
The answer is "Option A".
Explanation:
Add extra functionality, otherwise, it increases the R-square value, which is defined in the following points:
- To incorporate essential elements, R-square is explicitly promoted.
- It Increases the R-square value, which is an additional feature.
- It removes the features, which provide the value of the reduce R-square.
- After incorporating the additional features is used as the model, which is R-square, which is never reduced.
Answer: C
Explanation:
I believe this is because you cannot represent the number sixteen with only four binary bits. The highest you can go is 15.
