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xz_007 [3.2K]
3 years ago
12

PLS HELP WITH MY PYTHON HW ILL GIVE YOU BRAINLIEST

Computers and Technology
1 answer:
Nezavi [6.7K]3 years ago
4 0

Answer:

class Cat:

   isHungry = None

play = Cat().isHungry = True

eat = Cat().isHungry = False

if play:

   print("I am hungry, I cannot play")

else:

   play = Cat().isHungry = False

if eat:

   print("I am full, I cannot eat")

else:

   eat = Cat().isHungry = True

Explanation:

Just simple if and else statements.

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True or false? To help improve SEO, your URL should match the title of your blog post, word for word.
Vsevolod [243]

Improving SEO, by ensuring the URL matches the title of your

blog post, word for word is False.

<h3>What is SEO?</h3>

This is referred to as Search engine optimization. It is used to

improve a site by ensuring that is more visible when people

search for certain things or words.

The URL should contain only key words and unnecessary ones

should be eliminated which is why it isn't compulsory for the title

to be word for word.

Read more about Search engine optimization here brainly.com/question/504518

7 0
3 years ago
John travels a lot and he needs to access his documents and services on the go. Which of these technologies allows his to access
kirza4 [7]

<u>Answer:</u>

<u>Cloud computing</u><em> allow the user to access software or any document from the remote place.</em>

<u>Explanation:</u>

Let us understand what does the word cloud actually means. In simple terms, we get rain from cloud, but we don’t know actually which cloud burst to give rain.

In a similar way, <em>the cloud computing is the concept of storing files in multiple servers and multiple location</em> and it provide access when the <em>client needs the source. </em>

Cloud computing enable user to work on <em>software online</em> or to download document or <em>edit / create documents online</em>. Certain services are <em>free and few other are paid.</em>

6 0
3 years ago
The open items on your computer are displayed here.
koban [17]
The GUI or Graphical <u /><u></u><em />User Interface.
7 0
3 years ago
Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the elemen
Verdich [7]

Answer:

Replace <STUDENT CODE> with

for (i = 0; i < SCORES_SIZE; ++i) {

       if(lowerScores.at(i)<=0){

           lowerScores.at(i) = 0;

       }

       else{

           lowerScores.at(i) = lowerScores.at(i) - 1;

       }  

   }

Explanation:

To do this, we simply iterate through the vector.

For each item in the vector, we run a check if it is less than 1 (i.e. 0 or negative).

If yes, the vector item is set to 0

If otherwise, 1 is subtracted from that vector item

This line iterates through the vector

for (i = 0; i < SCORES_SIZE; ++i) {

This checks if vector item is less than 1

       if(lowerScores.at(i)<1){

If yes, the vector item is set to 0

           lowerScores.at(i) = 0;

       }

       else{

If otherwise, 1 is subtracted from the vector item

           lowerScores.at(i) = lowerScores.at(i) - 1;

       }  

   }

Also, include the following at the beginning of the program:

<em>#include <vector></em>

8 0
3 years ago
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

7 0
3 years ago
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