Answer:
1. G=D+(A+C^2)*E/(D+B)^3
cobegin:
p1: (D+B)
p2: p1^3
p3: C^2
p4: A+ p3
p5: E/p2
p6: p4 * p5
p7: D + p6
:G
coend
2. Now The value A=2, B=4, C=5, D=6, and E=8
p1: 6+4 =10
p2: p1 ^3= 10^3= 1000
p3: c^2= 5^2 =25
p4: A + p3= 2 +25 =27
p5: 8/1000
p6: 27 *8/1000
p7: D+ P6= 6+ 216/1000
= 6216/1000
=6.216
Explanation:
The above, first bracket with power is processed, and then power inside and outside bracket. And rest is according to BODMAS, and one process is solved at a time.
What happened if the offshore team members are not able to participate in the iterations demo due to timezone/infrastructure issues-(c) No Major issue. Since offshore lead and onsite members participate in the demo with the product owner, they can cascade the feedback back to the offshore members.
Explanation:
<u>No Major issue. Since offshore lead and onsite members participate in the demo with the product owner, they can cascade the feedback back to the offshore members.</u>
From the above statement it is clear that in case the offshore team members are not able to participate in the demo with the product owner due to timezone/infrastructure issues it want be a big issue since the onsite members of the team will participate in the demo and they can give all the valuable knowledge and feedback to the offshore members.As they all are part of the very same team
<u>Hence the option(3) is the correct option</u>
Answer:
See explaination
Explanation:
public class PalindromeChange {
public static void main(String[] args) {
System.out.println("mom to non palindrom word is: "+changTONonPalindrom("mom"));
System.out.println("mom to non palindrom word is: "+changTONonPalindrom("aaabbaaa"));
}
private static String changTONonPalindrom(String str)
{
int mid=str.length()/2;
boolean found=false;
char character=' ';
int i;
for(i=mid-1;i>=0;i--)
{
character = str.charAt(i);
if(character!='a')
{
found=true;
break;
}
}
if(!found)
{
for(i=mid+1;i<str.length();i++)
{
character = str.charAt(i);
if(character!='a')
{
found=true;
break;
}
}
}
// This gives the character 'a'
int ascii = (int) character;
ascii-=1;
str = str.substring(0, i) + (char)ascii+ str.substring(i + 1);
return str;
}
}
