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Originally posted by Honor93
Originally posted by ASeeker343
Originally posted by Honor93
Originally posted by ASeeker343
Are you telling me that 6/2(3) is different than 6/2*3? If so i would disagree
ummmm, where did this come from and no i didn't say such a thing, besides ... your 6/2*3 uses symbols before my time. it isn't an equation we would have seen back then. however, if you think 6/2*3 is the same as 6 divided by 2(2+1), you are sadly mistaken.
we agree on the stuff inside the parentheses: (2+1) = 3
good start
6÷2(1+2)
6÷2(3)
What I am saying is that the 2(3) does not put the multiplication before the division. 6÷2(3) = 6÷2x3.
Left to right. Parentheses state the argument inside them must be performed first, not anything adjacent as well. If you can provide me with solid proof otherwise ill eat crow but otherwise, 6÷2x3 = 9 Left to right.
www.youtube.com...
when you are dealing with a single integer, you are correct but this integer is multiplying a collective inside the parenthesis. the equation 2(2+1) is a stand alone equation, or at least it always has been ... with the progressivisms of today, i can no longer be sure. Never have i seen this: 2*(2+1), as you are implying.
however, 2(2+1) has always been defined or as another stated "distributed" as (2+1)+(2+1), which = 6.
not arguing mind you and i know better than to argue with a machine ... just try to remember, humans program them there 'machines'. and who said the programmers were up to date when these programs y'all are using were written?
For the record, I graduated (with math honors) long before any of these machines were even conceived.
reply to post by grey580
php
$var = 6;
$var2 = 2;
$var3 = 1;
$var4 = 2;
$answer = $var / $var2 * ( $var3 + $var4 );
echo $answer ;
Originally posted by ASeeker343
Originally posted by Honor93
Originally posted by ASeeker343
Originally posted by Honor93
Originally posted by ASeeker343
Are you telling me that 6/2(3) is different than 6/2*3? If so i would disagree
ummmm, where did this come from and no i didn't say such a thing, besides ... your 6/2*3 uses symbols before my time. it isn't an equation we would have seen back then. however, if you think 6/2*3 is the same as 6 divided by 2(2+1), you are sadly mistaken.
we agree on the stuff inside the parentheses: (2+1) = 3
good start
6÷2(1+2)
6÷2(3)
What I am saying is that the 2(3) does not put the multiplication before the division. 6÷2(3) = 6÷2x3.
Left to right. Parentheses state the argument inside them must be performed first, not anything adjacent as well. If you can provide me with solid proof otherwise ill eat crow but otherwise, 6÷2x3 = 9 Left to right.
www.youtube.com...
when you are dealing with a single integer, you are correct but this integer is multiplying a collective inside the parenthesis. the equation 2(2+1) is a stand alone equation, or at least it always has been ... with the progressivisms of today, i can no longer be sure. Never have i seen this: 2*(2+1), as you are implying.
however, 2(2+1) has always been defined or as another stated "distributed" as (2+1)+(2+1), which = 6.
not arguing mind you and i know better than to argue with a machine ... just try to remember, humans program them there 'machines'. and who said the programmers were up to date when these programs y'all are using were written?
For the record, I graduated (with math honors) long before any of these machines were even conceived.
Fair enough, I understand exactly what you are saying and I can understand how the problem can be interpreted as such. However I am saying that you cant separate the 2 from the division in order to distribute it. That is going out of order according to PEMDAS, unless 6÷2x3 means something different than 6÷2(3) which I am stating to be false. Most algebraic distributions are seen with addition... for example 6+ 2(1+2). In which case what you are doing is completely correct. However in this case you must distribute the 6/2. 3(2+1) = 9. You cant just pull the two off like that.
And for the record, Im not using a calculator either. I mentioned that as a side note but im doin this all with the brain God gave me. Also I am an engineering student so I have taken all of the math I can handle. I still think this is a poorly written problem that leads to interpretation issues, but Im still goin with 9
Originally posted by A.M.L.
reply to post by Honor93
I wish I could give you the benefit of the doubt, but no, I can't.
Surely, someone who has an honors in math would understand why 1 can't be the answer.
Sets and Logic as a class should answer this. The difference is extremely important to those in the know.
Lets make 6 equal x.
x/2(2+1) is (1/2)*x*(2+1).
That equals .5*x*3 which when 6 is substituted for x is .5*6*3.
.5*6*3 = 9.
Originally posted by grey580
reply to post by Whatsreal
A string instrument.
That the tone is made from hammers striking the strings is irrelevant.
Originally posted by Honor93
solve for X
X/2 = 4[1+6(5-1)]
please?
Originally posted by Earthscum
Originally posted by Honor93
solve for X
X/2 = 4[1+6(5-1)]
please?
200
X/2=4[1+6*4]
x/2=4[1+24]
X/2=4*25
X/2=100
2*100=200edit on 2-5-2011 by Earthscum because: (no reason given)
Originally posted by prolific
Originally posted by _BoneZ_
Originally posted by prolific
it's still 6÷2(3) or otherwise know as 6÷2x3 = 9.
The only way to get one is if you do multiplication first which is an invalid move.
Because you work from left to right, just as you read from left to right.
Thus 6÷2(1+2)= 6÷2(3)= 3(3)= 9
It really doesn't get more simple than that. Well, besides:
6÷2(1+2)=
is the same as:
(6÷2)(1+2)=
Yes that is correct. He is working from right to left which violates the rules.
This thread scares me and makes me want to .