Is 4/6 = 9/12 not proportionate
1 answer:
You might be interested in
I started with c, but you could’ve started with any box.
So I took the numbers in the c box (the black box in my picture) and divided to see which were multiples of two. They all were.
Then I went to a. All of a’s numbers were divisible by 7.
Then b. Two of them are divisible by 2, so that’s not the answer. None of them were divisible by 7, so there’s your answer!
Then d. Two are divisible by 7, so your rules is divisible by 2.
I hope this helps!
So I took the numbers in the c box (the black box in my picture) and divided to see which were multiples of two. They all were.
Then I went to a. All of a’s numbers were divisible by 7.
Then b. Two of them are divisible by 2, so that’s not the answer. None of them were divisible by 7, so there’s your answer!
Then d. Two are divisible by 7, so your rules is divisible by 2.
I hope this helps!