First you have all ten books, but then every time after, you loose one because you can't put the same book in the group twice.
10 * 9 * 8 * 7 = 5040 groups
Answer:
The equation of the line in point slope form is equal to
Step-by-step explanation:
The picture of the question in the attached figure
step 1
Find the slope
The formula to calculate the slope between two points is equal to
we have
(-2,-2) and (2,2)
substitute the values
step 2
Find the equation of the line in point slope form
we have
substitute the values in the formula
Answer:
Step-by-step explanation:
(96 minutes)/(32 sculptures) = (96/32 minutes)/sculpture = (3 minutes)/sculpture
It takes 3 minutes to make one sculpture.
(160 balloons)/(32 sculptures) = (160/32 balloons)/sculpture = (5 balloons)/sculpture
It takes 5 balloons to make one sculpture.
(160 balloons)/(96 minutes) ≈ 1.7 balloons used per minute.
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:
desmos
Step-by-step explanation:
look up desmos. its a grphing calculator and it gives u all the awnsers. welcome bro
