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3 years ago

Why is removing presentation-oriented markup from one's html document considered to be a best practice?

Computers and Technology

1 answer:

Gelneren [198K]3 years ago

3 0

Answer

Its for the purpose of maintenance and readability

Explanation

A presentation-oriented markup web application generates interactive web pages containing various types of markup language.

Markup is a language designed for the purposes of processing definition and presentation. It specifies code for formatting, both the layout and style, within a text file. Markup describes the structure while the styling is how it looks. it uses the code to specify the formatting which are called tags.

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Why are there so many unit testing tools? Are they efficient? Why or why not?

sineoko [7]

Answer:

Unit testing is the software testing method in which the individual source code are associate to control data. Unit testing basically test the code to ensure that the data or information meets its design.

The aim of unit testing that each part isolate in the program and display the correct individual parts.

Unit testing tool are efficient as they provide several benefits in the development cycle. The basic efficiency of unit testing tools depend upon its type of testing. Unit testing basically validate the units of source code in the program.

For example: when the loop and function in the program work efficiently.

5 0

2 years ago

Have a great day!

valina [46]

Answer:

thank you so much!

Explanation:

you seem like such a nice person! have a great weekend!

5 0

2 years ago

Read 2 more answers

Newt Corporation, headquartered in Los Angeles, is a nationwide provider of educational services to post-graduate students. Due

zysi [14]

Answer:

Option B i.e., Circuit level gateways only enable data to be inserted into a network which is the product of system requests within the network.

Explanation:

In the above question, some details are missing in the question that is options.

Option B is valid because Circuit level gateways are not the transmission inspection, always require information into such a server resulting through system appeal inside the server through maintaining a record for connections that are sent into the server and only enabling information in this is in answer to such queries.

Other options are incorrect because they are not true according to the following scenario.

5 0

3 years ago

Given the value x=false ,y=5 and z=1 what will be the value of F=(4%2)+2*Y +6/2 +(z&&x)?

Bezzdna [24]

<h2>Answer:</h2>

F = 13

<h2>Explanation:</h2>

Given:

x = false

y = 5

z = 1

F = (4%2)+2*Y +6/2 +(z&&x)

We solve this arithmetic using the order of precedence:

i. Solve the brackets first

=> (4 % 2)

This means 4 modulus 2. This is the result of the remainder when 4 is divided by 2. Since there is no remainder when 4 is divided by 2, then

4 % 2 = 0

=> (z && x)

This means (1 && false). This is the result of using the AND operator. Remember that && means AND operator. This will return false (or 0) if one or both operands are false. It will return true (or 1) if both operands are true.

In this case since the right operand is a false, the result will be 0. i.e

(z && x) = (1 && false) = 0

ii. Solve either the multiplication or division next whichever one comes first.

=> 2 * y

This means the product of 2 and y ( = 5). This will give;

2 * y = 2 * 5 = 10

=> 6 / 2

This means the quotient of 6 and 2. This will give;

6 / 2 = 3

iii. Now solve the addition by first substituting the values calculated earlier back into F.

F = (4%2)+2*Y +6/2 +(z&&x)

F = 0 + 10 + 3 + 0

F = 13

Therefore, the value of F is 13

5 0

2 years ago

Convert the following C program to C++.

ki77a [65]

Answer:

The program equivalent in C++ is:

#include <cstdio>

#include <cstdlib>

#define SIZE 5

using namespace std;

int main(int argc, char *argv[]) {

int numerator = 25;

int denominator = 10;

FILE * inPut = fopen(argv[1], "r");

FILE * outPut = fopen(argv[2], "w");

float result = (float)numerator/denominator;

fprintf(outPut,"Result is %.2f\n", result);

float arr[SIZE];

for(int i = 0; i < SIZE; i++) {

fscanf(inPut, "%f", &arr[i]);

fprintf(outPut, "%7.4f\n", arr[i]);

}

return 0;

}

Explanation:

See attachment for explanation.

Each line were numbered with equivalent line number in the C program

Download cpp

8 0

3 years ago