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3 years ago

Identify the layout in which you will be able to view and edit the header and footer

1 answer:

6 0

The answer is (c.) Print Layout

The print layout view is the default view in Microsoft Office Word upon opening the application. In a new document, to add a header and a footer, go to Insert Tab. Under the Header and Footer section, you have the option to add a header and a footer. There are built-in designs to choose from by just clicking on the drop down arrow.

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A spreadsheet has some values entered: Cell A1 contains 10, cell A2 contains 14, cell A3 contains 7. You enter in cell A4 the fo

lesya [120]

The value displayed in A4 is 24

5 0

3 years ago

The purpose of this assignment is to determine a set of test cases knowing only the functionâ€™s specifications, reported hereaf

Nesterboy [21]

Answer:

The program is written using PYTHON SCRIPT below;

N=int(input(" Enter number of Rows you want:"))

M=[] # this for storing the matrix

for i in range(N):

l=list(map(int,input("Enter the "+str(i+1)+" Row :").split()))

M.append(l)

print("The 2D Matrix is:\n")

for i in range(N):

print(end="\t")

print(M[i])

W=[] # to store the first non zero elemnt index

T=[] # to store that value is positive or negative

L=len(M[0])

for i in range(N):

for j in range(L):

if (M[i][j]==0):

continue

else:

W.append(j) # If the value is non zero append that postion to position list(W)

if(M[i][j]>0): #For checking it is positive or negative

T.append(+1)

else:

T.append(-1)

break

print()

print("The first Non Zero element List [W] : ",end="")

print(W)

print("Positive or Negative List [T] : ",end="")

print(T)

Explanation:

In order for the program to determine a set of test cases it takes in input of 2D matrix in an N numbet of rows.

It goes ahead to program and find the column index of the first non-zero value for each row in the matrix A, and also determines if that non-zero value is positive or negative. The If - Else conditions are met accordingly in running the program.

5 0

2 years ago

Class 00 rubber gloves are used when working with voltages less than _____.

Radda [10]

Answer: 500 volts AC, 750 volts DC
Class 00 rubber gloves are given the color "beige" based on the color code of rubber gloves.
The proof test voltage for this class of rubber gloves is 2500 volts of AC voltage and 10000 volts of DC voltage

4 0

3 years ago

Which transfer of learning occurs when the performance of a maneuver interferes with the learning of another maneuver?

Nutka1998 [239]

Answer:

Negative transfer of learning

Explanation:

Negative transfer of learning occur when the knowledge you've acquired in the past is/are interfering with the one(s) you are currently acquiring.

Example: learning a new language that have some differences in pronunciation from the previous one, changing from a right-sided vehicles to a left-sided vehicles etc.

3 0

3 years ago

5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S

Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0

3 years ago