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Elena L [17]
3 years ago
9

Text box linking is for two or more text boxes in a document, when the text overflows from the first box. Which scenario is the

most useful for creating linked text boxes?
laying out newsletters with limited space and overflowing text
creating visual appeal without images
formatting images and reducing the amount of text on a page
reducing the amount of text on a page and laying out newsletters
Computers and Technology
2 answers:
Julli [10]3 years ago
8 0

Answer:

Explanation:

use the next box i think

allochka39001 [22]3 years ago
6 0

Answer:

laying out newsletters with limited space and overflowing text

You might be interested in
Chemical equations of Carbon + water​
Setler79 [48]

Answer:

Aqueous carbon dioxide, CO2 (aq), reacts with water forming carbonic acid, H2CO3 (aq). Carbonic acid may loose protons to form bicarbonate, HCO3- , and carbonate, CO32-. In this case the proton is liberated to the water, decreasing pH. The complex chemical equilibria are described using two acid equilibrium equations.

PLS MARK AS BRAINLIEST

8 0
3 years ago
Hey plz help me with this<br>What is malware short for?​
soldier1979 [14.2K]

Answer:

Malware is the collective name for a number of malicious software variants, including viruses, ransomware and spyware. Shorthand for malicious software, malware typically consists of code developed by cyberattackers.

Explanation:

4 0
2 years ago
The following SQL statement uses a(n) _____. SELECT P_CODE, P_DESCRIPT, P_PRICE, V_NAME FROM PRODUCT, VENDOR WHERE PRODUCT.V_COD
Trava [24]

Answer:

The answer is c. “old-style” join.

Explanation:

SELECT P_CODE, P_DESCRIPT, P_PRICE, V_NAME

FROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE = VENDOR.V_CODE;

The SELECT clause represents all the columns to be displayed.

The FROM clause represents the tables which are used to extract the columns.

The WHERE clause shows the common column that exists in both the tables. This represents the old-style join when JOIN keyword was not used.

The tables are joined in two ways.

1. Using JOIN keyword

The syntax for this is given as

SELECT column1, column2

FROM table1 JOIN table2

ON table1.column3 = table2.column3;

This returns all rows from two tables having the same value of common column.

Two tables are taken at a time when JOIN keyword is used.

If more tables are to be used, they are mentioned as follows.

SELECT column1, column2

FROM table1 JOIN table2

ON table1.column3 = table2.column3

JOIN table3

ON table3.column4 = table1.column4

( ON table3.column5 = table2.column5 )  

The part in italics shows that the third table can either share the same column with table1 or table2.

The given query can be re-written using JOIN as shown.

SELECT P_CODE, P_DESCRIPT, P_PRICE, V_NAME

FROM PRODUCT JOIN VENDOR

ON PRODUCT.V_CODE = VENDOR.V_CODE;

2. Equating common column in WHERE clause

SELECT column1, column2

FROM table1, table2

WHERE table1.column3 = table2.column3;

This returns all rows from two tables having the same value of the common column.

Here, no keyword is used except the general syntax of the query. ON keyword as shown in the above example is replaced with WHERE.

Tables to be used are mentioned in FROM clause separated by commas.

Tables are joined based on the same column having same values.

Multiple tables can be joined using this style as follows.

SELECT column1, column2

FROM table1, table2, table3, table4

WHERE table1.column3 = table2.column3

AND table3.column5=table1.column5

AND table3.column4 = table2.column4;

5 0
3 years ago
By arranging users into security groups, it is possible to set permissions at a group-level, avoiding the unwieldy alternative o
Anni [7]

Answer:

The department by department is arranging users into security groups to set permission at a group-level.

Explanation:

Team by team is just formed group method but the team been set will work on based on the project. Once the project is getting over then the team group assigned permission has to be disabled or deleted.

 

Department by department is just good methods where each department will have a list of users and easy to assign the group.

user-by-user. is just assigning the rights of each individual user. But if the user has multiple departments login it will be good advice.

organization-by-organization. It is just assigning rights to complete users list in the organization.  

6 0
3 years ago
Program:
Mkey [24]

Answer:

See explaination

Explanation:

#method to print menu & handle user choice

def print_menu(input_str):

#printing menu

print('MENU')

print('c - Number of non-whitespace characters')

print('w - Number of words')

print('f - Fix capitalization')

print('r - Replace punctuation')

print('s - Shorten spaces')

print('q - Quit\n')

#getting choice

choice=input('Choose an option: ').lower()

#identifying choice

if choice=='c':

#displaying number of non white space chars in input_str

print('Number of non-whitespace characters:',get_num_of_non_WS_characters(input_str))

elif choice=='w':

#displaying number of words in input_str

print('Number of words:',get_num_of_words(input_str))

elif choice=='f':

#fixing capitalization and getting updated string and count of values capitalized

input_str,count=fix_capilization(input_str)

#displaying results

print('Number of letters capitalized:',count)

print('Edited text:',input_str)

elif choice=='r':

#replacing punctuation, displaying updated text

input_str=replace_punctuation(input_str)

print('Edited text:', input_str)

elif choice=='s':

#shortening spaces and displaying updated text

input_str = shorten_space(input_str)

print('Edited text:', input_str)

#returning choice and input_str

return choice,input_str

#returns the number of non white space chars in input_str

def get_num_of_non_WS_characters(input_str):

count=0

#looping through each character in input_str

for i in input_str:

if not i.isspace():

#i is not a space

count+=1

return count

#returns the number of words in input_str

def get_num_of_words(input_str):

#splitting words into list of tokens by space

words=input_str.split(' ')

count=0

#counting all non empty strings in words list

for i in words:

if len(i)>0:

count+=1

return count

#method to fix capitalization and return updated string and count of letters updated

def fix_capilization(input_str):

count=0

beginning=True #starting letter should be capitalized

result=''

for i in input_str:

if beginning and i.isalpha():

#start of a sentence and i is alphabetic

if i.islower():

#converting i to upper case, incrementing count

i=i.upper()

count+=1

result+=i

#not start of a sentence

beginning=False

elif i in '?.!':

#i is either ? or . or !, next letter should be capitalized

beginning=True

result+=i

else:

#any other character

result+=i

return result,count

#method to replace exclamation and semicolons with period and comma respectively

def replace_punctuation(input_str,exclamationCount =0,semicolonCount=0):

result=''

for i in input_str:

if i=='!':

i='.'

exclamationCount+=1

elif i==';':

i=','

semicolonCount+=1

result+=i

print('Punctuation replaced')

#displaying replaced values counts

print('exclamationCount:',exclamationCount)

print('semicolonCount:',semicolonCount)

return result

#removes all double or more spaces in input_str

def shorten_space(input_str):

input_str=input_str.strip()

result=''

prev=None

for i in input_str:

if prev==None:

result+=i

elif i==' ':

if prev != ' ':

result+=i

else:

result+=i

prev=i

return result

if __name__ == '__main__':

#getting input, printing it

input_str=input('Enter a sample text:\n')

print('\nYou entered:',input_str)

choice=' '

#looping until choice becomes q

while choice!='q':

choice,input_str=print_menu(input_str)

7 0
3 years ago
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