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Answer:
The contents in ax and bx registers are as following:-
ax=3432
bx=4670
Explanation:
mov command in 8086 is an important command in 8086 microprocessors.It moves the data from the source to destination.
mov destination,source
add it adds the data to the destination with the content of destination.
add a,b means add b to a.
in operation mov ax,3432 we are moving 3432 to ax register similarly moving 1238 to bx.
Now adding ax to bx so the content of bx =1238+3432 that is 4670.
I would say the answer is <u>Logistics Manager, Material Handlers, and Inventory Managers.</u>
<u></u>
<u><em>Logistics Manager:</em></u><em> person in charge of overseeing the purchasing </em>
<em> and distribution of products in a supply chain</em>
<em />
<em> </em><u><em>Material Handlers:</em></u><em> responsible for storing, moving, and handling </em>
<em> hazardous or non-hazardous materials</em>
<em />
<em> </em><u><em>Inventory Managers:</em></u><em> oversee the inventory levels of businesses</em>
<em> </em>
<u></u>
Hope that helps!
Answer:
The Admin should Create a rule with three conditions, each that compares the configuration attribute filed with a static value.
Answer:
#include<iostream>
using namespace std;
int main(){
int x1,x2,x3;
int y1,y2,y3;
cout<<"Enter the value of first point(x1,y1): ";
cin>>x1>>y1;
cout<<"\nEnter the value of second point(x2,y2): ";
cin>>x2>>y2;
cout<<"\nEnter the value of third point(x3,y3): ";
cin>>x3>>y3;
}
Explanation:
first include the library iostream for use the input/output commands
then, write the main function. within the main function declare the variable which store the value of points.
after that, use the 'cout' for output. It has the function which print the value or message on the screen.
and 'cin' is used to store the value in the variable.
So, in the above code cout ask for enter the value of point and after enter value by user, cin store in the variables.
Note: you have to enter two value with space or by enter.