Answer:
The code will give an error that is "At least one public class is required in main file".
Explanation:
In the given code if we do not use the public access modifier to the class. It will give an error so, the correct code to this question as follows:
Program:
import java.util.HashSet; //import package
public class A //define class as public.
{
public static void main(String[ ] args) //define main method.
{
HashSet set = new HashSet(); //creating hashset object.
set.add("A"); //add alphabet in hashset
set.add("B"); //add alphabet in hashset
set.add("C"); //add alphabet in hashset
System.out.print("Size of HashSet is :"set.size()); //print the size of hashset.
}
}
Output:
Size of HashSet is : 3
Explanation of the program:
- In the above program, we define a public class that is "A" and inside the class, we define the main method.
- Inside the main method, we create a HashSet class object that is "set".
- To add elements in HashSet we use add() function that adds elements and in the last, we use the size() function that prints the size HashSet.
True I think because it helps right?
Answer:
A. Share resources and thus are not independent
Explanation:
This would be the answer. If this is wrong plz let me know
Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
