Answer:
import java.util.Scanner;
public class FindMatchValue {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_VALS = 4;
int[] userValues = new int[NUM_VALS];
int i;
int matchValue;
int numMatches = -99; // Assign numMatches with 0 before your for loop
matchValue = scnr.nextInt();
for (i = 0; i < userValues.length; ++i) {
userValues[i] = scnr.nextInt();
}
/* Your solution goes here */
numMatches = 0;
for (i = 0; i < userValues.length; ++i) {
if(userValues[i] == matchValue) {
numMatches++;
}
}
System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
}
}
Answer:
prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S
Explanation:
Given that S = { p q |p, q are prime numbers greater than 0}
E = {0, −2, 2, −4, 4, −6, 6, · · · }
To prove by constructing a bijection from S to E
detailed solution attached below
After the bijection :
<em>prove that | S | = | E |</em> : every element of S there is an Image on E , while not every element on E has an image on S
∴ we can say sets E and S are infinite sets
Answer:

Explanation:
Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to
, where m is the mass and M the molar mass of the gas, and the density is
.
For air
and 
So, 

Answer:

Explanation:
The Young's module is:


Let assume that both specimens have the same geometry and load rate. Then:

The displacement rate for steel is:



Answer:
oh I am sorry I can't understand your question.