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asambeis [7]
3 years ago
8

Using symbols (%, $, #, etc.) can make it easier to take notes.

Computers and Technology
2 answers:
Tju [1.3M]3 years ago
7 0

Answer: Ture

Explanation: edgenuity 2020 just took the test

frez [133]3 years ago
4 0

false because is more esay what you learn

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Answer:

C. Double-click the database icon.

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Create a new class MyArray_DE and copy and paste the previous program. Be sure to rename the class to MyArray_DE. Modify the pro
vova2212 [387]

Answer:

public class MyArray_DE

{

public static void main(String[] args) {

    int[] numbers = {28, 7, 92, 0, 100, 77};

    int total = 0;

    for (int i=0; i<numbers.length; i++){

        if(numbers[i] % 2 == 0)

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 for (int i=numbers.length-1; i>=0; i--){

        System.out.print(numbers[i] + " ");

    }

}

}

Explanation:

Since you did not provide the previous code, so I initialized an array named numbers

Initialize the total as 0

Create a for loop that iterates through the numbers

Inside the loop, if the number % 2 is equal to 0 (That means it is an even number), add the number to total (cumulative sum)

When the loop is done, print the total

Create another for loop that iterates through the numbers array and print the numbers in reverse order. Note that to print the numbers in reverse order, start the i from the last index of the array and decrease it until it reaches 0.

4 0
3 years ago
What is the initial condition in this set of code?
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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:

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Fill in the blank with the correct response.<br> A submenu in a text editor is indicated by a <br> .
igomit [66]

Answer:

triangle

Explanation:

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