Answer:
<h2>Option B: #VALUE</h2>
is the correct answer.
Explanation:
<h3>Reasons of #VALUE error:</h3>
- When cells are not given the expected type of value.
- If cells are left blank unnoticeable or for giving a null value.
- For entering dates and other numerical data in text form.
<h3>Correcting and fixing #VALUE error:</h3>
This type of error can only be fixed by finding the cell in which there is wrong data entered and correct it.
Fixing the #VALUE error is tricky as some functions automatically ignore the data that is invalid.
<h3>EXAMPLE:</h3>
- Suppose a cell contains the value as cost of the object and in order to make it 0, NA is inserted into the cell.
- While making total of all the costs, that particular cell might cause #VALUE error.
- Following image attach will help you clear the concept.
I hope it will help you!
Answer:
Wacom Intuos S
Explanation:
If you're looking for a good tablet that isn't too expensive the Wacom Intuos S is a great starter, I personally use it and really enjoy it. It doesn't have a graphic screen. If you're looking for one with a graphic screen get the Wacom Cintiq 22, It's expensive but great quality. Most tablets with a graphic screen are this expensive though. But if you REALLY want one with a graphic screen that isn't TOO expensive I recommend the HUION Kamvas pro 12.
Hope I could help!
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:
The answer is "SUBSTITUTE(D1,"c","C",1)"
Explanation:
In the given question, some information is missing, that is an option so, the correct option can be described as follows:
- Many of us use the Bulletin feature, which is available in MS-Excel. In these function the SUBSTITUTE is a method, that is used in Builtin to aim the functions to locate a string or character and to replace with the selected module.
- In the case of the end-user, it would like to replace "c" with a capital letter "C." And displays cell shifts in D1. This method uses 4 parameters, in which first describes cell reference, second string to located, third replacing string, and fourth its length.
Answer:
c
Explanation:
I think c is correct answer if I answer wrong
