Can you please explain and answer this question to me, (don't mind the blue dots, they are the answers of what I think, please l
et me know if I'm right)
1 answer:
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!
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Hmmmm to convert from improper to mixed, simply do the division of both, the quotient gets upfront, the remainder as the numerator, for example,
how many times 3 goes into 81? well, 27 times, with a remainder of 0, thus
how many times does 2 go into 29? well 14 times, with a remainder of 1, thus
how many times 3 goes into 81? well, 27 times, with a remainder of 0, thus
how many times does 2 go into 29? well 14 times, with a remainder of 1, thus