Moving your Sprite from right to left is considered the X coordinate?

How to get the blade out of a pencil sharpener?

agasfer [191]

Poke hole with thin but strong object
Or buy new sharpener

7 0

4 years ago

Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and

lana [24]

Answer:

Explanation:

The following code is written in Java and uses a for loop with a series of IF ELSE statements to check the next and previous characters in a string. Checking to make sure that there are no asterix. If so it adds that character to the String variable output. Which is returned to the user at the end of the method. Two test cases have been created and the output can be seen in the attached image below.

class Brainly {

public static void main(String[] args) {

System.out.println(starOut("sm*eilly"));

System.out.println(starOut("ab**cd"));

}

public static String starOut(String str) {

String output = "";

for (int i = 0; i < str.length(); i++) {

if ((i != 0) && (i != str.length()-1)) {

if ((str.charAt(i-1) != '*') && (str.charAt(i+1) != '*') && (str.charAt(i) != '*')) {

output += str.charAt(i);

}

} else {

if ((i == 0) && (str.charAt(i) != '*') && (str.charAt(i+1) != '*')) {

output += str.charAt(i);

} else if ((i == str.length()-1) && (str.charAt(i) != '*') && (str.charAt(i-1) != '*')) {

output += str.charAt(i);

}

return output;

}

4 0

3 years ago

What can the tab key do

bulgar [2K]

Answer: it indents

Explanation:

8 0

3 years ago

Your program will read a word (or a whole line) from the user. It will then count the number of vowels in the word (or line) and

Softa [21]

Answer:

Following are the program in the Python Programming Language.

def vowel(): #define function

print("Enter END to Break") #print message

while(True): #set while loop

word=input("Enter words: ") #get input from the user

count=0 #set count to 0

if(word=="END"): #set if condition

break #terminate the loop

else:

for i in range(len(word)): #set the for loop

c=word[i] #initialize in c

if(c=='a'or c=='e' or c=='i'or c=='o'or c=='u'or c=='A'or c=='E'or c=='I'or c=='O'or c=='U'):

count+=1 #increament in count by 1

print("The word " , word , " contains " , count, end=" ")#print message

if(count==1):#check condition

print("vowel.")

else:

print("vowels")

#call the function

vowel()

Output:

Enter END to Break

Enter words: Vowel

The word Vowel contains 2 vowels

Enter words: END

Explanation:

Here, we define a function "vowel()" inside it.

Print message for the user.
Set the while loop and pass condition is True then, get input from the user in the variable "word" then, set variable "count" and initialize to 0.
Set if condition to check if the user input "END" then, the loop will terminate.
Otherwise, set for loop which continues from the length of the word.
Then, we set if statement which checks the vowel in the word, then increment in count by 1.
Print the message in the following format which is mentioned in the question.
Then, set if-else statement for check count is equal to 1 then print "vowel." Otherwise, it prints "vowels."

Finally, we call the following function "vowel()".

4 0

3 years ago

In which order do the DHCP messages occur when a client and server are negotiating address configuration?

Leona [35]

One two three and four

8 0

3 years ago