Answer:
vocational school
Explanation:
These vocational schools are institution postsecondary and job training, these schools offer quickly programs, but with the knowledge to work in a company, if we're talking about technology programs we can consider, programming, network administrator, technician.
These institutes are for people like:
- People without work experience
- People want to start a new career
- People want to reenter to the working market
Answer:
Explanation:
Farmers are always both directly and indirectly connected to each other
Their network is mostly strong
Networks become weak only on the edges (ends) of the river but doesn't completely dimnish
With the available network length, the center of river bank forms the strongest network of all and becomes a key player in defining the balance property of overall network
The network is very well structurally balanced and we can see that through the below image
20 miles 10 20 30 40 50
See attachment file for diagram
Considering the total length of river as 50miles and and the center of the whole length will be at 25th mile. From that point, if we consider a farmer will be be having friends for a length of 20miles both along upstream and downstream.
By this he'll be in friend with people who are around 80% of the total population. As me move from this point the integrity increases and this results in a highly balanced structural network.
Answer:
Tags can be identified as they are written as <tagname> Something </tagname>
<tagname> this tag is called start tag.
</tagname> this tag is called as end tag.
for ex:<p> This is a paragraph </p>
There are some elements with no end tag.for ex:- <br> tag.
Attributes are used to provide additional information to an HTML element.
Every element can have attributes.
for Ex:- <h1 style="background-color:red";>MY WEBSITE </h1>
So the background of h1 will become red.
Answer:
See explaination
Explanation:
#include <fstream>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
// Fill in the code to define payfile as an input file
ifstream payfile;
float gross;
float net;
float hours;
float payRate;
float stateTax;
float fedTax;
cout << fixed << setprecision(2) << showpoint;
// Fill in the code to open payfile and attach it to the physical file
// named payroll.dat
payfile.open("payroll.dat");
// Fill in code to write a conditional statement to check if payfile
// does not exist.
if(!payfile)
{
cout << "Error opening file. \n";
cout << "It may not exist where indicated" << endl;
return 1;
}
ofstream outfile("pay.out");
cout << "Payrate Hours Gross Pay Net Pay"
<< endl << endl;
outfile << "Payrate Hours Gross Pay Net Pay"
<< endl << endl;
// Fill in code to prime the read for the payfile file.
payfile >> hours;
// Fill in code to write a loop condition to run while payfile has more
// data to process.
while(!payfile.eof())
{
payfile >> payRate >> stateTax >> fedTax;
gross = payRate * hours;
net = gross - (gross * stateTax) - (gross * fedTax);
cout << payRate << setw(15) << hours << setw(12) << gross
<< setw(12) << net << endl;
outfile << payRate << setw(15) << hours << setw(12) << gross
<< setw(12) << net << endl;
payfile >> hours ;// Fill in the code to finish this with the appropriate
// variable to be input
}
payfile.close();
outfile.close();
return 0;
}
Program p1;
var a,b,c,d : integer; {i presume you give integer numbers for the values of a, b, c }
x1, x2 : real;
begin
write('a='); readln(a);
write('b='); readln(b);
write('c=');readln(c);
d:=b*b - 4*a*c
if a=0 then x1=x2= - c/b
else
if d>0 then begin
x1:=(-b+sqrt(d)) / (2*a);
x2:=(-b - sqrt(d))/(2*a);
end;
else if d=0 then x1=x2= - b /(2*a)
else write ("no specific solution because d<0");
writeln('x1=', x1);
writeln('x2=',x2);
readln;
end.
