Answer:
Sir William Crookes invented the cathode ray tube in 1878, but these discoveries took many years to merge into the common ground of television. His mechanical system used a scanning disk with small holes to pick up image fragments and imprint them on a light-sensitive selenium tube. A receiver reassembled the picture.
Answer:
They write code that translates commands to binary code.
Explanation:
Coding or programming is the process of creating instructions for computers using programming languages. Computer code is used to program the websites, apps, and other technologies we interact with every day.
Answer:
The code to this question can be given as:
Code:
int callsReceived,operatorsOnCall; //define variable as integer
Scanner ob= new Scanner(System.in);
//create object of scanner class for user input
System.out.println("Insert the value of callsReceived"); //print message.
callsReceived = ob.nextInt(); //input value.
System.out.println("Insert the value of operatorsOnCall"); //print message.
operatorsOnCall = ob.nextInt(); //input value.
if (operatorsOnCall == 0) //check number
{
System.out.println("INVALID"); //print message.
}
else
{
System.out.println(callsReceived/operatorsOnCall); //print value.
}
Explanation:
In the above code firstly we define 2 integer variable that name is already given in the question. Then we create the scanner class object for taking user input. Then we print the message for input first and second value from the user. then we use conditional statement. If the second variable that is operatorsOnCall is equal to 0. So It print INVALID. else it divide the value and print it.
Read it backwords. it helps look for spelling errors.
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:
