How many 3 number combinations are possible if you can't repeat the numbers? The numbers range from 0-39

49600

Step-by-step explanation: There are 40 ways to choose the first number

and there are 40 ways to choose the second number.

Now we take some examples to determine how many choices there are for

the third number

If the second number is 23, then the 3rd number cannot be any of these

9 numbers 19,20,21,22,23,24,25,26,or 27, since they are not at least 5

numbers away from 23.  Same for all other 40 second numbers.

Let's look at a second example which includes some on both sides of 0.

If second number is 2, we couldn't choose any of the 9 numbers 38,39,0,1,2,3,4,5,6.

So thats 40-9 or 31 choices for the 3rd number.

Answer 40*40*31 = 49600 ways.

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

answer: 4 (:

step-by-step explanation: