Answer: D
Step-by-step explanation:
A constant rate of change means that for every x, y increases the same amount.
In the situation "The amount a person pays for gas compared with the number of gallons purchased ," the independent quantity, or x-value, is the number of gallons, and the dependent quantity, or y-value, is the amount paid. Since gas has a set price per gallon, for every gallon purchased, the total amount paid goes up the same amount. This means it has a constant rate of change.
Fifteen divided by three is 18
Answer:
New Cost of item is 
Step-by-step explanation:
Given:
Original Cost of item = 
Item was marked down by 
To Calculate Cost after marked down we need to multiply original cost by 50 and then divide by hundred.
Cost after marked down = 
Now New cost of the item is calculated by Subtracting Cost after marked down by Original Cost of item.
New cost of the item = 
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;
}
