No,
A line is considered a "line segment" when a line connects to 2 points.
If a line was infinite length, it would never end up ending at one point.
Hope this was clear for you
Answer:
x/7=x+14 / 5
Solve for X
Step-by-step explanation:
x/7=x+14 / 5
Solve for X
There really is no ‘best stock’ to invest in. But some good stocks to invest in are Amazon, Disney, and Apple. All are companies that have had tremendous growth and appear to still be growing. You can also go onto Yahoo finance, they will tell you whether they think the stock is over priced and you can also see the growth pattern over time.
Hope this helps! Please make me the brainliest, it’s not necessary but appreciated, I put a lot of effort and research into my answers. Have a good day, stay safe and stay healthy.
Answer:
Algorithm
Start
Int n // To represent the number of array
Input n
Int countsearch = 0
float search
Float [] numbers // To represent an array of non decreasing number
// Input array elements but first Initialise a counter element
Int count = 0, digit
Do
// Check if element to be inserted is the first element
If(count == 0) Then
Input numbers[count]
Else
lbl: Input digit
If(digit > numbers[count-1]) then
numbers[count] = digit
Else
Output "Number must be greater than the previous number"
Goto lbl
Endif
Endif
count = count + 1
While(count<n)
count = 0
// Input element to count
input search
// Begin searching and counting
Do
if(numbers [count] == search)
countsearch = countsearch+1;
End if
While (count < n)
Output count
Program to illustrate the above
// Written in C++
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main()
{
// Variable declaration
float [] numbers;
int n, count;
float num, searchdigit;
cout<<"Number of array elements: ";
cin>> n;
// Enter array element
for(int I = 0; I<n;I++)
{
if(I == 0)
{
cin>>numbers [0]
}
else
{
lbl: cin>>num;
if(num >= numbers [I])
{
numbers [I] = num;
}
else
{
goto lbl;
}
}
// Search for a particular number
int search;
cin>>searchdigit;
for(int I = 0; I<n; I++)
{
if(numbers[I] == searchdigit
search++
}
}
// Print result
cout<<search;
return 0;
}
