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2 years ago

When do images or graphics in Microsoft Word hurt the document rather than help

Computers and Technology

1 answer:

Vinvika [58]2 years ago

3 0

Answer:

When images or graphics excessively occupy most of the space on the page.

<u>Explanation:</u>

For example, if you observed the attached image you'll see a bad example in which the image excessively occupies most of the space on the page making the written text difficult to read.

Next, a good example of proper editing is found in the next image, which shows the written text properly aligned with the text.

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B.) Define latency and jitter for service flow in a fixed wireless system.

yuradex [85]

The answer would be between the base station and the subscriber station.

7 0

2 years ago

Write a function that implements another stack function, peek. Peek returns the value of the first element on the stack without

devlian [24]

Answer:

See explaination

Explanation:

StackExample.java

public class StackExample<T> {

private final static int DEFAULT_CAPACITY = 100;

private int top;

private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);

/**

* Returns a reference to the element at the top of this stack.

* The element is not removed from the stack.

* atreturn element on top of stack

* atthrows EmptyCollectionException if stack is empty

public T peek() throws EmptyCollectionException

{

if (isEmpty())

throw new EmptyCollectionException("stack");

return stack[top-1];

}

/**

* Returns true if this stack is empty and false otherwise.

* atreturn true if this stack is empty

public boolean isEmpty()

{

return top < 0;

}

//please replace "at" with the at symbol

Note:

peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.

3 0

2 years ago

Java uses interfaces to provide the benefits of multiple inheritance. A superclass object is a subclass object. A subclass is of

WITCHER [35]

COMPLETE QUESTION

Which of the following is false?

a. A subclass is often larger than its super class.

b. A super class is object is a subclass object.

c. The class following the extends keyword in a class declaration is that direct super class of the class being declared.

d. Java uses interfaces to provide the benefits of multiple inheritance

Answer:

B) A super class is object is a subclass object.

Explanation:

In object oriented programming, the concept of inheritance is greatly utilized, this refers to deriving a class from another class. The derived class is the sub-class and it is derived from its super class, as a matter of fact in Java programming language, all classes are derived from some other class. Objects are made from classes with their state (fields) and behavior (methods) since objects are made from classes, the objects of the super class will not be the same as that of the subclass.

4 0

2 years ago

When configuring a record type, an App Builder can configure the available value of a picklist field for the page layout. Which

Alla [95]

Answer:

Option (B) and (D) i.e., Type and Lead Source is the correct answer.

Explanation:

Because when the user configures the record type, Then, they can help that user to configure the value of the picklist fields for the page layout. So, The type and the lead source opportunity is available for the standard field that configure straightly in the type of record. That's why the following option is correct.

7 0

1 year ago

Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer

NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0

2 years ago