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Mathematics 11 Online
MsBrains (ms-brains):

Help

MsBrains (ms-brains):

MsBrains (ms-brains):

@Directrix

Directrix (directrix):

@Lily2913

OpenStudy (anonymous):

What are you not understanding?

MsBrains (ms-brains):

How to solve it..?

Directrix (directrix):

How many eighths are in in one-half is what I am seeing.

OpenStudy (anonymous):

Notice how the 1/2 box is split into components so that each component is = 1/8?

OpenStudy (anonymous):

It's exactly as @Directrix says, it's asking for the equation where you would somehow get 4 1/8 components when you split the 1/2 somehow

OpenStudy (anonymous):

I believe it's asking for how many 1/8ths are in 1/2, so it's a division problem

MsBrains (ms-brains):

So what answer are you thinking? @jaeuni

OpenStudy (anonymous):

lol what answer do you have in mind and I'll let you know if I agree

MsBrains (ms-brains):

B, like @Directrix said.

OpenStudy (anonymous):

And that's because?

Directrix (directrix):

1/2 / 1/8 = 1/2 * 8/1 = 8/2 = 4. That translates to: There are four eighths in one-half.

Directrix (directrix):

@Lily2913 What are you thinking about this problem?

OpenStudy (anonymous):

lol @Directrix beat you to it I agree that it's B as well since like @Directrix states, the problem asks about how many eighths can go into 1/2 and you can find that out by division

OpenStudy (anonymous):

I agree with both of you. B

MsBrains (ms-brains):

What about this one

OpenStudy (anonymous):

If she uses 1/4th cup of sugar and she used up 8 cups of sugar...how many 1/4ths go into 8?

OpenStudy (anonymous):

It's parallel to you earlier problem when we were looking at how many 1/8ths go into 1/2

OpenStudy (anonymous):

*your

MsBrains (ms-brains):

@Directrix What do you think?

Directrix (directrix):

1/4 cup 1 batch 1 cup 4 batches 8 cups 32 batches

Directrix (directrix):

8 divided by 1/4 = 8 * 4/1 = 32

MsBrains (ms-brains):

So 32 is correct?

Directrix (directrix):

I think so. What about you @Ms-Brains

MsBrains (ms-brains):

I'm think 32 is correct. I'll post a new thread with a new problem

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