Answer: System 1 has an infinite number of solutions and System II has no solutions.
Step-by-step explanation:
If we look at system 1, we see that if we multiply the first equation by 2 and subtract the y value so that its on the same side as x, the equations are the same. That means that no matter the x or y value, it will be a solution.
If we look at system 2, however, we see that if we multiply the top equation by 3, the equations are the exact same except for the fact that the top equation has a -21. This means that no matter the x or y, there will never be a solution.
1)To find a Scale factor of Dilation, about the origin
We have
Pre-mage Image
(x, y) k(x,y)
For example
Pre-image Image
(2,4) 2(2,4) = (4,8)
2)When it's not about the origin then we have to count from the Projection Point
Having said this, ex
Which is not a step
a)
We can divide the x value of the image over the pre-image, not the way around.
In the example, I've given if we divide the pre-image over the image value we would have found a scale factor of 1/2. In the example, The scale factor was the inverse: 2
Answer:
Step-by-step explanation:
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: 283
Step-by-step explanation:
To do this, it is helpful to get an equation you can use to solve any term.
This equation is:
So simply plug in 31 for n to get
