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faltersainse [42]
3 years ago
5

When resizing images or objects in a presentation, why should a user not utilize the sizing handles in the middle of the sides

Computers and Technology
2 answers:
serg [7]3 years ago
8 0

Answer:

B. the one that talks about ratio

Explanation:

Shalnov [3]3 years ago
5 0

When resizing an image or an object in a presentation, a user should not utilize the sizing handles in the middle of the sides or the bottom of the image. There's nothing wrong with resizing this way.

There's nothing wrong with resizing this way

<u>Explanation:</u>

When the sizing handles in the middle of the sides are used, the image stretches out reflecting an increase in the width maintaining There's nothing wrong with resizing this way.

On the other hand, while using the sizing handles at the bottom, the image increases in height whilst maintaining the same width again disturbing the aspect ratio.

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Types of computer viruses<br>​
e-lub [12.9K]

Answer:anti various

Explanation:hope fully this will help u

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3 years ago
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Define input hardware​
luda_lava [24]
It’s a piece of hardware (you can touch it) that changes something on the screen. Examples are a keyboard putting letters on the screen or a mouse moving the cursor
6 0
3 years ago
Two routers, R1 and R2, connect using an Ethernet over MPLS service The service provides point-to-point service between these tw
liberstina [14]

Answer: b) R1 will connect to a physical Ethernet link, with the other end of the cable connected to a device at the WAN service provider point of presence.

.d) R1 will forward data link frames to R2 using an Ethernet header/trailer.

Explanation:

Here we have to understand what is MLPS. MLPS is a protocol which identifies the shortest route for the transfer of messages between routers instead of the longest route.

Here we are given a WAN which is a wide area network. Using layer 2 Ethernet service the frames are transmitted across the routers within the WAN.

Option A is incorrect as as connecting the other end of the cable to the R2 would reduce the functionality of the WAN.  Option C is incorrect as HDLC header trailer has no effect.

Option B and D are correct as Ethernet header/ trailer has the same size every frame and to maintain the functionality of the WAN one end must be connected to point of presence(PoP).  

5 0
4 years ago
Which of the following commands can be used to display socket information out to the terminal screen
Slav-nsk [51]
Linus ss

Explanation:

The ss (socket statistics) command provides a lot of information by displaying details on socket activity. One way to get started, although this may be a bit overwhelming, is to use the ss -h (help) command to get a listing of the command's numerous options. Another is to try some of the more useful commands and get an idea what each of them can tell you.

One very useful command is the ss -s command. This command will show you some overall stats by transport type. In this output, we see stats for RAW, UDP, TCP, INET and FRAG sockets.
6 0
2 years ago
Assume that a gallon of paint covers about 350 square feet of wall space. Create anapplication with a main() method that prompts
Elena-2011 [213]

Answer:

import java.util.Scanner;

public class PaintClculator {

   public static void main(String[] args) {

Scanner in = new Scanner(System.in);

       System.out.println("Enter Values for length, width, and height of the room");

       double length = in.nextDouble();

       double width = in.nextDouble();

       double height = in.nextDouble();

       double wallArea = calculateWallArea(length,width,height);

       double noGallons = calculatePaint(wallArea);

       double price = price(noGallons);

       System.out.println("Price is: "+ price);

   }

   // Creating Method to Calculate Wall Area

   public static double calculateWallArea( double length, double width, double height){

       // formular for the surface area of a room Area=2*length*heigth+2*width*height+lenth*width

       double Area = 2*length*height+2*width*height+length*width;

       System.out.println("Wall Area: "+Area);

       return Area;

   }

   //Creating method to calculate amount of paint

   public static double calculatePaint(double Area){

       double noGallons = Area/350;

       System.out.println("Number of gallons: "+noGallons);

       return noGallons;

   }

   // Creating Method Calculate Price

   public static double price(double noGallons){

       return noGallons*32;

   }

}

Explanation

  1. Created Three Methods in all; calculateWallArea(), calculatePaint(), and price()
  2. Method calculateWallArea() calculates area based on this formula Area=2*length*heigth+2*width*height+lenth*width
  3. The value of Area is passed to method calculatePaint() Which calculates the number of gallons required to cover the area given that one gallon covers 350 wall space
  4. Method price() calculates and dis[plays the price by calling method calculatePaint() that returns number of gallons. since one gallon is 32$
3 0
3 years ago
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