see below
be first and win , complements of yours truly,
1. Buy 0 bags of stickers. Then there won't be any left over, and everybody gets the same amount. (That's just a guess, it might be a trick question, but I don't know how many friends she has, lol)
2. 14
Common Core... Ack! We're stuck in the middle of the "curriculum"... Not only do we not know what they are asking for, the teachers themselves don't know either.
+1
My master degreed brain thinks that's correct. Then again I don't have a masters in mathematics :hick:
I suck at math with more letters then numbers. I know why 7+7=14, but I don't know how to do the work.
I've often wondered why you need to figure out a problem if you have the answer already. biggest thing I hated about algebra is that it is a study of rules and laws rather then math. I hated getting an answer wrong because I didn't show my work or did it improperly.
My sister makes fun of me for it. I just asked her what kind of accountant is told to not use a calculator and show the work for getting the answer.
How many friends does Juanita have? How many stickers are in each bag? I think due to lack of information Juanita should not buy any stickers at all.
The other night my third grader had me helping her do her math homework, it was ALL like question number 2. Im fairly certain all questions were answered right even though I showed her a different way to get the answers. She must have told me 10 times " thats not the way my teacher does it".
These poor kids stuck in a system that wasn't broke so they "fixed" it.
the apparently 'expected answer" to the first question is 6x=4y
but according to some teachers protesting this stuff,, its fairly certain that the answer does not matter in common core, as long as you can explain how you arrived at your answer.
not sure I wanna call this 4th grade math either.
the irony here is that "expected answer has nothing to do with the "final question within this question".
I don't get it. If 6 is a number of bags, and 4 is a number of bags, x and y have to be the number of stickers in each bag, right? Then if they equal each other, it won't matter how many bags she buys, she will have the same amount of stickers.
WAIT! There is an answer to the first question! I'm so stupid, I can't believe I didn't see it earlier. The only thing she's not sure of is if she needs 4 or 6 bags, right? How many stickers could she buy so that there are none left over? 12. 6 bags with 2 stickers, or 4 bags with 3 stickers. Of course 24 would work too. 6 bags with 4 stickers each, or 4 bags with 6 stickers. I suppose you could go with 48 or any other number that is divisible by both 4 and 6 evenly. But that way each friend gets the same amount if she used 4 or 6 bags, and there are none left over.
There. I showed my work too. It's way more gratifying than 6x=4y (which I don't understand). I think a more logical solution would be "12X". Any factor of 12 would give you an even number of stickers per bag whether it be 4 or 6.
Well I'm her friend and I don't want no ed stickers, so you're gonna have to figure that in as well...
I saw that to but the only way it worked and made sense is if the word "Could" were substituted with "should".
screw this bull shiznit math,,,!!!!!!!!!!!!!!!!!