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inessss [21]
3 years ago
11

___________is a standard for exchanging structured data over the web. It allows creating documents consisting of customized tags

, enabling the definition, transmission, validation, and interpretation of data between applications and between organizations.
Computers and Technology
1 answer:
svetlana [45]3 years ago
6 0

Answer:

The answer is "XML"

Explanation:

Extensible Markup Language is a marking language, which specifies a collection of document. It uses the coding rules, which is readable for both man and computer.  

  • It is a scalable way of creating data formats and sharing structured data through the internet or private networks digitally.
  • It permits the development of custom tags to identify, transfer, verify and view data among apps and organizations.
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When selecting font size in a document, ____ points equals one-inch-tall text?
IgorC [24]
Hello <span>Jhigg4685 
</span><span>

Answer: When selecting font size in a document, 72 points equals one-inch-tall text?


I hope this helps
-Chris</span>
3 0
4 years ago
Is a defense of a political position an argument
shutvik [7]

Answer:

yes

Explanation:

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8 0
3 years ago
The Polish mathematician Wacław Sierpiński described the pattern in 1915, but it has appeared in Italian art since the 13th cent
soldi70 [24.7K]

Answer:

/ Sierpinski.java

public class Sierpinski {

     // method to find the height of an equilateral triangle with side length =

     // length

     public static double height(double length) {

           // formula= length*sqrt(3)/2

           double h = (length * Math.sqrt(3)) / 2.0;

           return h;

     }

     // method to draw a filled triangle (pointed downwards) with bottom vertex

     // x,y

     public static void filledTriangle(double x, double y, double length) {

           // finding height

           double h = height(length);

           // filling triangle as a polygon

           // passing x, x-length/2, x+length/2 as x coordinates

           // and y, y+h, y+h as y coordinates

           StdDraw.filledPolygon(new double[] { x, x - (length / 2.0),

                       x + (length / 2.0) }, new double[] { y, y + h, y + h });

     }

     // method to draw an n level sierpinski triangle

     public static void sierpinski(int n, double x, double y, double length) {

           // checking if n is above 0 (base condition)

           if (n > 0) {

                 // drawing a filled triangle with x, y as bottom coordinate, length

                 // as side length

                 filledTriangle(x, y, length);

                 // drawing the triangle(s) on the top

                 sierpinski(n - 1, x, y + height(length), length / 2);

                 // drawing the triangle(s) on the left side

                 sierpinski(n - 1, x - (length / 2.0), y, length / 2);

                 // drawing the triangle(s) on the right side

                 sierpinski(n - 1, x + (length / 2.0), y, length / 2);

           }

     }

     public static void main(String[] args) {

           // parsing first command line argument as integer, if you dont provide

           // the value while running the program, this program will cause

           // exception.

           int n = Integer.parseInt(args[0]);

           // length of the outline triangle

           double length = 1;

           // finding height

           double h = height(length);

           // drawing a triangle (pointed upwards) to represent the outline.

           StdDraw.polygon(new double[] { 0, length / 2, length }, new double[] {

                       0, h, 0 });

           // drawing n level sierpinski triangle(s) with length / 2, 0 as x,y

           // coordinates and length / 2 as initial side length

           sierpinski(n, length / 2, 0, length / 2);

     }

}

7 0
3 years ago
Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,
Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary                                        denary or decimal

11111111      =        2⁸      =             256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s =  /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1      </em></u><u><em> 1 </em></u><u><em>      1     1     1     1</em></u>

<u><em>128  64     32    </em></u><u><em>16</em></u><u><em>    8    4     2    1</em></u>

step 3

in the ip network address: 198.16.0.0/19 <em>(subnet representation)</em> we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

          Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts<u><em>+16: 198.16.</em></u><u><em>16</em></u><u><em>.0 - 198.16.31.255</em></u>

<u><em>                   +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.47.255 et</em></u>c

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11111000.000000                /21

now we have added 5 1s in the third field to reserve 11 0s

<u><em>subnet mask: 255.255.</em></u><u><em>8.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       </em></u><u><em>1 </em></u><u><em>    1     1     1</em></u>

<u><em>128  64     32    16    </em></u><u><em>8 </em></u><u><em>   4     2    1</em></u>

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (<em>subnet representation</em>) we increment this using 8

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts <u><em>+16: 198.16.</em></u><u><em>24</em></u><u><em>.0- 198.16.31.255</em></u>

<em>                             </em><u><em> +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.112.255 et</em></u>c

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       1     1     1     1</em></u>

<u><em>128  64     32    16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts <u><em>+16: 198.16.40.0- 198.16.55.255</em></u>

<em>                          </em><u><em>    +16: 198.16.56.0- 198.16.71.255 et</em></u>c

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11100000.000000                /19

now we have added 3 1s in the 3rd field to reserve 13 0s

<u><em>subnet mask: 255.255.</em></u><u><em>32.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1      </em></u><u><em> 1 </em></u><u><em>      1       1     1     1     1</em></u>

<u><em>128  64     </em></u><u><em>32  </em></u><u><em>  16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts <u><em>+16: 198.16.72.0- 198.16.103.255</em></u>

<em>                          </em><u><em>    +16: 198.16.104.0- 198.16.136.255 et</em></u>c

5 0
3 years ago
A ____ is a key-dependent, one-way hash function that allows only specific recipients(symmetric key holders) to access the messa
Alexxx [7]

Answer: (C) MAC

Explanation:

MAC is stand for the message authentication code and it is one of the key dependent hash function. Message authentication provide the data integrity and also the origin of data authentication.

  • It basically allow the specific recipients known as the holder of the symmetric key.
  • The message authentication code uses the CBC (Cipher block chaining) mode for constructing the MAC algorithm and it also uses the block cipher for the conjunction by using the CBC operations.    

On the other hand, all the options are incorrect as they are not the key dependent hash function.

Therefore, Option (C) is correct.

7 0
3 years ago
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