6/2(1+2)= whats this equal
What do you think?
i need help answering it I forget how to do it
Well, 6/2 = 3 when you simplify it, and 1 + 2 = 3. Of course, you need to multiply both those numbers, and the answer would give you 9.
thats what i was thinking to
@gracie101507 1 would not be the answer...
how did you get 9
but wouldnt you do 2 times 1 first
You got to put it like this, (6/2) x (1+2)
Do PEMDAS and you can get your answer.
\[6/2(1+2)= ?\] According to PEMDAS: Parentheses Exponents Multiplication Division Addition Subtraction We look to the parentheses first, simplifying to -> \[6/2(3)= ?\]| Since the only thing remaining is multiplication and division, we can do them in either order. Starting with division is 'prettier' though. \[3(3) = 9\]
Just like Shadow did.
but wouldn't you do 2 times whats in the parenthesis first
Shadow is right, just look at the steps...
PEMDAS means you start what's INSIDE the parentheses first, not outside.
ok thanks
No problem
ok im still confused
With?
that same question
with that same question
What part of the question is confusing you?
everything
i still dont really know if it is 9
the part where it says parenthesis first
if you didn't do that you would have got 1
Nevermind I see the problem
whats the problem
We were viewing it as a division function, which in reality, it's a fraction.
Actually wait, that doesn't make sense.
|dw:1569359974952:dw|
This is one of those "screenshot if from the book" questions
Lol
no i just did that
Oh I see
You did the Distributive Property
yea
I think the Distributive Property counts as multiplication, as opposed to being considered a parentheses function. Which is why you would do the addition inside of the parentheses first.
Since parentheses is ranked higher than multiplication in PEMDAS.
I don't see where it was screenshot from the original text
The way you wrote the original problem is ambiguous
Yeah if you can take a picture and upload it, or take a screenshot, that'd clear things up @katwhirles
ok
Since some will interpret it as \(\dfrac{6}{2}(1 + 2)\) and others will interpret it as \(\dfrac{6}{2(1 + 2)}\) which are two completely different expressions
Yeah but in order for it to be the second, the parentheses would have to include 2 and the 1 + 2, so that it would all be in the denominator, under 6. But it's just 6/2 multiplied by (1+2).
Of course, taking what @katwhirles wrote as what the textbook/problem depicts.
i think it is 1
do u think it is 1?????
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