What is necessary for your camera to connect to your computer using a usb or firewire cable

Which routine is configured to execute first when the program runs?

tester [92]

It's D because of ok

4 0

3 years ago

When you register to take classes, the registrar office pulls up a file full of

anzhelika [568]

Answer:

Database

Explanation:

Database software is intended to develop databases and to maintain, alter, access, and obtain useful data stored within them. It is often used to store, manage, and organize information such as field names, documents, and file structures in a database. it helps users to create their own databases to meet their business specifications and requirements. It provides data flexibility, as the storage process and formats can be modified without changing the entire program inside the database.

6 0

3 years ago

19. in the array implementation of a queue, the pop operation is most efficient if the front of the queue is fixed at index posi

ICE Princess25 [194]

It is true that in the array implementation of a queue, the pop operation is most efficient if the front of the queue is fixed at index position. The correct option is a.

<h3>What is pop operation? </h3>

The removal of an element is referred to as a pop operation. Again, because we only have access to the element at the top of the stack, we can only remove one element. We simply take the top of the stack off.

A push operation decrements the pointer before copying data to the stack; a pop operation copies data from the stack before incrementing the pointer.

The pop operation in an array implementation of a queue is most efficient if the queue's front is fixed at index position.

Thus, the correct option is a.

For more details regarding pop operation, visit:

brainly.com/question/15172555

#SPJ1

6 0

1 year ago

Create a program to compute the fee for parking in a garage for a number of hours. The program should: 1. Prompt the user for ho

Harman [31]

Answer:

The ans will be given in the python script below. A picture of the answer is also attached

Explanation:

print("Welcome To Garage Parking Fee Calculator")

hours = float(input("Type the number of hours parked : "))

#fee per hour

rate = 2.40

#multiply rate per hour by the number of hours inputted

price = rate * hours

if price < 6:

price = 6

if price > 20:

price = 20

print("Parking fee is: $", +price)

7 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