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;
}
Answer:
a line that continually approaches a given curve but does not meet it at any finite distance.
Step-by-step explanation:
comment and let me know how it helps
Just like any other variable, we complete the same process
Now we can test it, let's say k=4, d=6, and c=2, let's find the value of g
OR
How many facts does it take to make triangles congruent? Only 3 if they are the right three and the parts are located in the right place.
SAS where 2 sides make up one of the three angles of a triangle. The angle must between the 2 sides.
ASA where the S (side) is common to both the two given angles.
SSS where all three sides of one triangle are the same as all three sides of a second triangle. This one is my favorite. It has no exceptions.
In one very special case, you need only 2 facts, but that case is very special and it really is one of the cases above.
If you are working with a right angle triangle, you can get away with being given the hypotenuse and one of the sides. So you only need 2 facts. It is called the HL theorem. But that is a special case of SSS. The third side can be found from a^2 + b^2 = c^2.
You can also use the two sides making up the right angle but that is a special case of SAS.
Answer
There 6 parts to every triangle: 3 sides and 3 angles. If you show congruency, using any of the 3 facts above, you can conclude that the other 3 parts of the triangle are congruent as well as the three that you have.
Geometry is built on that wonderfully simple premise and it is your introduction to what makes a proof. So it's important that you understand how proving parts of congruent triangles work.
