import java.util.Scanner;
public class U2_L3_Activity_Four {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a sentence.");
String sent = scan.nextLine();
int count = 0;
for (int i = 0; i < sent.length(); i++){
char c = sent.charAt(i);
if (c != ' '){
count++;
}
else{
break;
}
}
System.out.println("The first word is " + count +" letters long");
}
}
We check to see when the first space occurs in our string and we add one to our count variable for every letter before that. I hope this helps!
<span>A simplified main program used to test functions is called <u><em>formula</em></u>
</span><span />
I would probably say C, Though you might want a second, person to clarify. But My answer is C..
Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
