Answer:
II and III only
Explanation:
In Code segment II, the output of the array will be started form arr[0] and ends at the arr[length]. Because loop starts from 0 and ends at length of array. This will print the full array data.
In code segment III, the output will be all values of array as loop starts form first index and ends at last index.
On the other hand I code segment prints all array values except last value of the array. As the loop shows that, it will terminate at second last value of the array.
Answer:
Replace /* Your solution goes here */ with:
cin>>matchValue;
numMatches = 0;
for (i = 0; i < userValues.size(); ++i) {
if(matchValue == userValues.at(i))
{
numMatches++;
}
}
Explanation:
This line gets input for matchValue
<em>cin>>matchValue;
</em>
This line initializes numMatches to 0
<em>numMatches = 0;
</em>
The following iteration checks for the number of matches (numMatches) of the matchValue
<em>for (i = 0; i < userValues.size(); ++i) {
</em>
<em>if(matchValue == userValues.at(i))
</em>
<em>{
</em>
<em> numMatches++;
</em>
<em>}
</em>
<em>}
</em>
<em>See Attachment for full source code</em>
If you mean sonar it uses echolocation but GPS is signaled from satellite to the internal antenna
Answer:
Since the language isn’t stated, I’ll give answers in the two most-used (?) languages: Java and Python.
a) To print a’s value 3 times in the same line, in Java we would do:
System.out.print(a+a+a);
In Python, we would write:
print(a*3)
b) 2 times in different lines using one print statement
In Java, we would write
System.out.println(a+”\n”+a+”\n”+a);
In Python we would write:
print(a,a,a,sep=’/n’)
Hope this helps!
