The student council was preparing for the school bake sale. the members divided each type of the donated items equally onto plates. each plate contained only one type of item and every plate had the exactly the same number of items. there were no leftovers. what is the greatest number of items that could have been placed on each plate?
muffins: 96
breadsticks: 48
rolls: 84

To find the answer, we have to divide each item in a way that gives us equal groups on each plate, that is, we have to find a divisor for each item that give us the same result. So, we are gonna test several division until find the

96/2=48

96/3=32

96/4=24

96/8=12

That is, dividing all muffins in 8 plates we would have 12 muffins per plate.

48/2=24

48/4=12

That is, dividing all breadsticks in 4 plates we would have 12 breadsticks per plate.

84/2=42

84/3=28

84/4=21

84/6=16

84/7=12

By dividing all rolls in 7 plates, we would have 12 rolls per plate.

Therefore, we need 8 plates for muffins, 4 plates for breadsticks and 7 plates for rolls, with this arrangement, all plates will have the same amount of items, 12 per plate.

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

21

step-by-step explanation: