Answer:
1.Lighter Weight. ...
Emission Sensors. ...
LED Lights. ...
Hybrid Engines. ...
Radar Alerts. ...
Air Filtration Systems. ...
2. Some techniques you can use are Limit AC usage, Obey the speed limit, Use cruise control, Maintain the radiator, and Change your air filters.
Answer: they hack in your device and change all your passwords so that so you can’t get in maybe i don’t know
Explanation:
Here you go,
Import java.util.scanner
public class SumOfMax {
public static double findMax(double num1, double num2) {
double maxVal = 0.0;
// Note: if-else statements need not be understood to
// complete this activity
if (num1 > num2) { // if num1 is greater than num2,
maxVal = num1; // then num1 is the maxVal.
}
else { // Otherwise,
maxVal = num2; // num2 is the maxVal.
}
return maxVal;
}
public static void main(String[] args) {
double numA = 5.0;
double numB = 10.0;
double numY = 3.0;
double numZ = 7.0;
double maxSum = 0.0;
/* Your solution goes here */
maxSum = findMax(numA, numB); // first call of findMax
maxSum = maxSum + findMax(numY, numZ); // second call
System.out.print("maxSum is: " + maxSum);
return;
}
}
/*
Output:
maxSum is: 17.0
*/
Answer:
return value =2.
Here the function f() returns the length of the substring we traversed before we find the same character at the equal index of two substrings.
Take the inputs s= “abcd” and t= “bccd”.
• Now, p1 points to s1, i.e., p1 points to the character ‘a’ of “abcd”. And similarly, p2 points to ‘b’ of “bccd”.
• Then we compare the values at p1 and p2, are not equal, so p1 and p2 both are incremented by 1.
• Now the characters ‘b’ and ‘c’ of “abcd” and “bccd” respectively are compared. They are not equal. So both p1 and p2 both are incremented by 1.
• Now, p1 points to ‘c’ of “ abcd” that is the element at index 2 of s. And p2 points to ‘c’ of “bccd” that is the element at index 2 of t. Here value at p1 and p2 becomes equal. So the break statement is executed. We stop moving forward.
• As p1 is pointing to index 2 and s is pointing to the base that is index 0, so p1-s = 2.
Explanation:
#include<stdio.h>
int f(char *s, char *t);
void main()
{
int k = f("abcd", "bccd");
printf("%d", k);
}
int f(char *s, char *t)
{
char *p1, *p2;
for(p1 = s, p2 = t; *p1 != '\0'&& *p2 != '\0'; p1++, p2++)
{
if (*p1 ==*p2)
break;
}
return (p1-s);
}
OUPUT is given as image
Answer:
yall keep giving me the wrong answer
Explanation: