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6/2(2+1)=???

6/2(2+1)=???

You still haven't responded to my function where we replace (1+2) with x
This is also exactly how functions are made.
6/2x=1
6=2x
3=x

If the answer is 1, then the algebraic expression also becomes true.

2(1+2) is one value and will remain 1 value.


Math doesn't change like this.

This one statement is true and will remain true:
First X÷Y(a+b) is not equal to (X/Y)(a+b). Here, X÷Y(a+b) = X÷(Ya+Yb) If its (X÷Y)(a+b) then it's equal to (X/Y)(a+b). Those who still think the ans is 9 then first learn the difference between X÷Y(a+b) and (X÷Y)(a+b).
you go left to right, so it doesn’t matter...

Edit: ok so i just looked it up, @Cake does the “6/2” mean 6 halves or 6 divided by two???
 
you go left to right, so it doesn’t matter...
That is if division and brackets were on the same scale, which they aren't.

Brackets come before division, it doesn't matter how you eliminate the brackets, they just do.
 
OK EVERYONE SHUT UP, THE ANSWER IS THAT THERE IS NO ANSWER BECAUSE MATHEMATICIANS ARE DUMB AND CAN'T AGREE AND ALSO MATH CREATES PARADOXES
 
OK EVERYONE SHUT UP, THE ANSWER IS THAT THERE IS NO ANSWER BECAUSE MATHEMATICIANS ARE DUMB AND CAN'T AGREE AND ALSO MATH CREATES PARADOXES
We should just use a fraction symbol or multiply by a number smaller than 1 and greater than 0
 
That is if division and brackets were on the same scale, which they aren't.

Brackets come before division, it doesn't matter how you eliminate the brackets, they just do.
Yes, they arent, but you end up with 3 with the brackets regardless. 6/2 3 = 9
 
That is if division and brackets were on the same scale, which they aren't.

Brackets come before division, it doesn't matter how you eliminate the brackets, they just do.
6/2(3) IS THE SAME AS 6/2x3 U IGNORANT ****
 
6/2(3) IS THE SAME AS 6/2x3 U IGNORANT ****
No it's not, brackets go first no matter what, the brackets don't magically become a multiplication symbol, you just have to multiply to eliminate brackets. In 6/2(3) it becomes 1, if you have 6/2*3 you get a 9.
In the first one brackets come first, in the second one division comes first.
 
No it's not, brackets go first no matter what, the brackets don't magically become a multiplication symbol, you just have to multiply to eliminate brackets. In 6/2(3) it becomes 1, if you have 6/2*3 you get a 9.
In the first one brackets come first, in the second one division comes first.
are you actually ksiding me rn bro

when there is brackets but no sign, it means multiplication. You’re eliminating the brackets by adding 1 and 2.
 
Yes, they arent, but you end up with 3 with the brackets regardless. 6/2 3 = 9
6/2(3) means 6 is divided by 2(3) it ends up looking like this in a fraction

6
----
2(3)

6
---- = 1
6


are you actually ksiding me rn bro

when there is brackets but no sign, it means multiplication. You’re eliminating the brackets by adding 1 and 2.
How many times do I have to debunk your argument
First X÷Y(a+b) is not equal to (X/Y)(a+b). Here, X÷Y(a+b) = X÷(Ya+Yb) If its (X÷Y)(a+b) then it's equal to (X/Y)(a+b). Those who still think the ans is 9 then first learn the difference between X÷Y(a+b) and (X÷Y)(a+b).

You just end up saying "no" like that counters my argument.
 
6/2(3) means 6 is divided by 2(3) it ends up looking like this in a fraction

6
----
2(3)

6
---- = 1
6



How many times do I have to debunk your argument
First X÷Y(a+b) is not equal to (X/Y)(a+b). Here, X÷Y(a+b) = X÷(Ya+Yb) If its (X÷Y)(a+b) then it's equal to (X/Y)(a+b). Those who still think the ans is 9 then first learn the difference between X÷Y(a+b) and (X÷Y)(a+b).

You just end up saying "no" like that counters my argument.
u know what idc you can fail ur math test by putting 1 as ur answer
 
No it's not, brackets go first no matter what, the brackets don't magically become a multiplication symbol, you just have to multiply to eliminate brackets. In 6/2(3) it becomes 1, if you have 6/2*3 you get a 9.
In the first one brackets come first, in the second one division comes first.
The brackets don't magically become a multiplication BECAUSE THE BRACKETS IS A MULTIPLICATION IN THAT SITUATION
 
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