Idk but I use to know because I took computer classes
Answer:
It throws an error.
the public class needs a name.
like this:
public class G{ public static void main(String[] args) {
int x=5 , y = 10;
if (x>5 && y>=2) System.out.println("Class 1");
else if (x<14 || y>5) System.out.println(" Class 2");
else System.out.println(" Class 3"); }// end of main
}
if you give the class a name and format it, you get:
Class 2
Explanation:
Answer:
Given address = 94EA6
tag = 0 * 94 ( 10010100 )
line = 0 * 1 D 4 ( 111010100 )
word position = 0*6 ( 110 )
Explanation:
using the direct mapping method
Number of lines = 512
block size = 8 words
word offset = 
= 3 bit
index bit = 
= 9 bit
Tag = 20 - ( index bit + word offset ) = 20 - ( 3+9) = 8 bit
Given address = 94EA6
tag = 0 * 94 ( 10010100 )
line = 0 * 1 D 4 ( 111010100 )
word position = 0*6 ( 110 )
Answer:
Radio Corporation of America (RCA)
Explanation:
A corporation is not a video connector.
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:
