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Yuri [45]
3 years ago
15

10(            )  8(532.2 )                         

Computers and Technology
1 answer:
blagie [28]3 years ago
7 0

Answer:

346.30

Explanation:

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Dear ,you should ask it on stackoverflow and geekforgeelks , not here
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How do I delete my Brainly account?
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Write a C program to prform simple C aritlimetic calculations. The user is to enter a simple expression(integer operaior integer
larisa86 [58]

Answer:

Here is the C program:

#include <stdio.h>  //to use input output functions

//functions prototype

unsigned int mod(unsigned int a, unsigned int b);  

unsigned int mul(unsigned int a, unsigned int b);  

unsigned int sub( unsigned int a,unsigned int b);

float divide(unsigned int a,unsigned int b);  

unsigned int add( unsigned int a,unsigned int b);  

int main()  {  //start of main method

unsigned int a, b;   //declare variables to store the operands

char d;  //declare variable to store the operator

printf("Enter an operator:  ");   //prompts user to enter an operator

scanf("%c",&d);  //reads the operator from use

getchar();  //gets a character

while (d!='q')   { //keeps iterating until user enters q to quit

printf("Enter 1st operand: ");   //prompts user to enter first operand

scanf("%d",&a);   //reads first operand from user

getchar();  //reads character

printf("Enter 2nd operand: ");   //prompts user to enter second operand

scanf("%d",&b);   //reads second operand from user

getchar();  

if (d=='%')  {   //if the character of operator is a mod

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

mod(a,b);  }  //displays computed modulo of two input operands

if (d=='*')   //if the input character is for multiplication operator

{printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

mul(a,b); }  //displays computed multiplication

if (d=='+')  {  //if the input character is for addition operator

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = "); // displays =

add(a,b);  }   //displays computed addition

if (d=='/')  {  //if the input character is for division operator

printf("%d",a); // prints operand 1

putchar(d);  //displays operator

printf("%d",b);  //displays operand 2

printf(" = ");  //displays =

divide(a,b);  }   //displays computed division

if (d=='-')  {  //if the input character is for subtraction operator

printf("%d",a);  //prints operand 1

putchar(d);  //displays operator

printf("%d",b); // displays operand 2

printf(" = ");  //displays =

sub(a,b);  }  //displays computed subtraction

printf("Enter an operator: ");   //asks again to enter an operator

scanf("%c",&d);  //reads operator from user

getchar();  }  }   //gets character

unsigned int mod( unsigned int a, unsigned int b){  //function to compute modulo of two integers with no sign

     int c = a%b;  //computes mod

    printf("%d",c);  }  //displays mod result

unsigned int add(unsigned int a, unsigned int b){     // function to compute addition of two integers

    int c = a+b; //computes addition

    printf("%d\n",c);  } //displays result of addition

unsigned int mul(unsigned int a, unsigned int b){       //function to compute multiplication of two integers

    int c = a*b;  //multiplies two integers

   printf("%d\n",c); }  //displays result of multiplication

float divide(unsigned int a, unsigned int b){   //function to compute division of two integers

    float c = a/b;  //divides two integers and stores result in floating point variable c

    printf("%f\n",c);  } //displays result of division

unsigned int sub(unsigned int a, unsigned int b){       //function to compute subtraction of two integers

    int c = a-b;  //subtracts two integers

    printf("%d\n",c);  }  //displays result of subtraction

Explanation:

The program is well explained in the comments mentioned with each line of the program. The program uses while loop that keeps asking user to select an operator and two operands to perform arithmetic calculations. The if conditions are used to check what arithmetic calculation is to be performed based on user choice of operand and calls the relevant function to perform calculation. There are five functions that perform addition, modulo, subtraction, division and multiplication and display the result of computation. The screenshot of output is attached.

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George has a set of software programs installed on his computer to create a digital portfolio. He is not sure what each program
d1i1m1o1n [39]

Explanation:

I thinkcreating and edditing Images matchs to adobePhotoshop

Microsoft word is for adding and Creating text document

Creating andeditingpage layout -Adobe In Design

Desktop Publication-QuarkQuark Xpress

I hope thishelp and If wrong Inform me

Good luck

8 0
2 years ago
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Suppose we compute a depth-first search tree rooted at u and obtain a tree t that includes all nodes of g.
Temka [501]

G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

<h3>What are DFS and BFS?</h3>

An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.

The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.

First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.

If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.

In such a case:

- in the DFS tree, one of u or v, should be a prototype of the other.

- in the BFS tree, u and v can differ by only one level.

Since, both DFS-tree and BFS-tree are the very tree T,

it follows that one of u and v should be a prototype of the other and they can discuss by only one party.

This means that the border joining them must be in T.

So, there can not be any limits in G which are not in T.

In the two-part of evidence:

Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

The complete question is:

We have a connected graph G = (V, E), and a specific vertex u ∈ V.

Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.

Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.

Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)

To learn more about  DFS and BFS, refer to:

brainly.com/question/13014003

#SPJ4

5 0
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