48÷2(9+3) = ???

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

You have to make the 9+3 because you have to do whats in the parenthesis first before moving on. You are taking care of the parenthesis by simplifying the problem.

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add.

So you are saying to distribute then?
Which like i said leads to
48/(2*9) + (2*3)
that does not give you 2.

It really doesn't give you 2 because AGAIN, you did it wrong. It's 48÷(2*9 + 2*3).

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

You have to make the 9+3 because you have to do whats in the parenthesis first before moving on. You are taking care of the parenthesis by simplifying the problem.

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add.

So you are saying to distribute then?
Which like i said leads to
48/(2*9) + (2*3)
that does not give you 2.

It really doesn't give you 2 because AGAIN, you did it wrong. It's 48÷(2*9 + 2*3).

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add.

So you are saying to distribute then?
Which like i said leads to
48/(2*9) + (2*3)
that does not give you 2.

It really doesn't give you 2 because AGAIN, you did it wrong. It's 48÷(2*9 + 2*3).

NO IT DOESNT. Why are you just blantatly acting like the distributive property doesnt give you 2 separate terms. Why are you just blindly not even looking at examples. WHY DONT YOU PROVE THE DISTRIBUTIVE PROPERTY GIVES 1 TERM OR @#$#. seriously you are becoming lamer and lamer. You have no proof no sources nothing. Either find some or just leave.

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add.

So you are saying to distribute then?
Which like i said leads to
48/(2*9) + (2*3)
that does not give you 2.

It really doesn't give you 2 because AGAIN, you did it wrong. It's 48÷(2*9 + 2*3).

NO IT DOESNT. Why are you just blantatly acting like the distributive property doesnt give you 2 separate terms. Why are you just blindly not even looking at examples. WHY DONT YOU PROVE THE DISTRIBUTIVE PROPERTY GIVES 1 TERM OR @#$#. seriously you are becoming lamer and lamer. You have no proof no sources nothing. Either find some or just leave.

 

[msaba07]

What ... in ... the ... world ... 111 pages?

 

[msaba07]

What ... in ... the ... world ... 111 pages?

 

[usainboltisfast]

Even ehow agrees with my distributive reasoning

http://www.ehow.com/how_8...operty-math-problem.html

THEY BECOME 2 TERMS if you are going to believe distribution is what you do first. And it still doenst give you 2.

 

[balloonoboy]

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

And if you distribute, whether you add first then go on or divide then add, you'll never get 288.

That was a fairly good $!* example, Team 288ers. Rebut.

 

[usainboltisfast]

Even ehow agrees with my distributive reasoning

http://www.ehow.com/how_8...operty-math-problem.html

THEY BECOME 2 TERMS if you are going to believe distribution is what you do first. And it still doenst give you 2.

 

[balloonoboy]

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

And if you distribute, whether you add first then go on or divide then add, you'll never get 288.

That was a fairly good $!* example, Team 288ers. Rebut.

 

[usainboltisfast]

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

 

[usainboltisfast]

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

50/2(1-1)
50/2(0)
25(0)

0

Its so disgusting that your a mod. You wont even admit you dont even know how to use the distributive property. Admit you were wrong trying to distribute in this problem.

The answer is UNDEFINED. 

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

50/2(1-1)
50/2(0)
25(0)

0

Its so disgusting that your a mod. You wont even admit you dont even know how to use the distributive property. Admit you were wrong trying to distribute in this problem.

The answer is UNDEFINED. 

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

The answer is UNDEFINED. 

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

lol now where are back to multiplying before divsion again.

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

The answer is UNDEFINED. 

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

lol now where are back to multiplying before divsion again.

 

[unkletomcruze]

Originally Posted by balloonoboy

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Why would you make 9+3=12 when distribution is just unfactoring the coefficient? And that is how to take care of the parenthesis.

Let me repeat, the 2 was originally part of the 9 and 3 before being factored out. Take care of the parenthesis by unfactoring or distribution.

Quick google searches won't work here, b. Either you know it or you don't.

You have to make the 9+3 because you have to do whats in the parenthesis first before moving on. You are taking care of the parenthesis by simplifying the problem.

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis" means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add them; you HAVE to distribute.

So now I'm certain it's not 288. Only 2 or 8.6.

The equation originally read 48÷(18+6) or 48÷18+6.

Edited

Yo you are seriously special.

P.E.M.D.A.S fam...what the first letter in P.E.M.D.A.S?

A friggin' P, which stands for Parenthesis...which means that you tackle whatever is in the parenthesis FIRST.

How is this basic fundamental rule false?

The two is an afterthought next to what's in the coefficient as far as P.E.M.D.A.S is concerned.

You honestly can't be this obtuse...you just can't be...i refuse to believe so.

I know you've realized that you're championing a false approach; you're just afraid and subsequently unwilling to admit your wrongs/mistakes.

It's ok son, it's ok.


...

 

[balloonoboy]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

I been said 70 pages ago -

- that I don't know whether you add the two distributed products together or leave them be and divide then add in the OOO.

Hence, the OG equation read 48÷(18+6) or 48÷18+6

Buyahhhhhhhhhhh!

 

[balloonoboy]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

I been said 70 pages ago -

- that I don't know whether you add the two distributed products together or leave them be and divide then add in the OOO.

Hence, the OG equation read 48÷(18+6) or 48÷18+6

Buyahhhhhhhhhhh!

 

[unkletomcruze]

Originally Posted by balloonoboy

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Why would you make 9+3=12 when distribution is just unfactoring the coefficient? And that is how to take care of the parenthesis.

Let me repeat, the 2 was originally part of the 9 and 3 before being factored out. Take care of the parenthesis by unfactoring or distribution.

Quick google searches won't work here, b. Either you know it or you don't.

You have to make the 9+3 because you have to do whats in the parenthesis first before moving on. You are taking care of the parenthesis by simplifying the problem.

You're not following. I know you stubborn AF but listen for once.

You may think "taking care of the parenthesis" means to add the two terms inside but that's false. Mainly because you have to account for the coefficient 2 somehow. And since it was the 2 that brought the two terms together in the first place. You just can't add them; you HAVE to distribute.

So now I'm certain it's not 288. Only 2 or 8.6.

The equation originally read 48÷(18+6) or 48÷18+6.

Edited

Yo you are seriously special.

P.E.M.D.A.S fam...what the first letter in P.E.M.D.A.S?

A friggin' P, which stands for Parenthesis...which means that you tackle whatever is in the parenthesis FIRST.

How is this basic fundamental rule false?

The two is an afterthought next to what's in the coefficient as far as P.E.M.D.A.S is concerned.

You honestly can't be this obtuse...you just can't be...i refuse to believe so.

I know you've realized that you're championing a false approach; you're just afraid and subsequently unwilling to admit your wrongs/mistakes.

It's ok son, it's ok.


...

 

[usainboltisfast]

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

I been said 70 pages ago -

- that I don't know whether you add the two distributed products together or leave them be and divide then add in the OOO.

Hence, the OG equation read 48÷(18+6) or 48÷18+6

Buyahhhhhhhhhhh!

So instead of trying to prove a point why dont you just sit this one out since you are unsure?

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

lol now where are back to multiplying before divsion again.

Funny how I already explained it in detail and you still ignored it. 
I fear for the future of NT.

 

[usainboltisfast]

Originally Posted by balloonoboy

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

The point isn't whether the DP gives you one or two terms, but that you must distribute.

Imagine the 2 like the sickle cell gene. And the 9 and 3 as the parents of an offspring. Whatever the offspring gets, is factored out. So as you see the 2 - sickle cell gene - will always be a part of the 9 and 3, no matter what you do. And the only way to express this gene is to distribute.

That was a fairly good $+# example, Team 288ers. Rebut.

Let me ask you one last time does distribution give you one or two terms in this situation?

I been said 70 pages ago -

- that I don't know whether you add the two distributed products together or leave them be and divide then add in the OOO.

Hence, the OG equation read 48÷(18+6) or 48÷18+6

Buyahhhhhhhhhhh!

So instead of trying to prove a point why dont you just sit this one out since you are unsure?

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

explain. Look at you though still wont say anything about the distributive comment though.

50÷2(1-1) = 50÷2(0) = 50÷0 = UNDEFINED
[font=arial, sans-serif]
[/font]

[font=arial, sans-serif]I already did all the explanation but you refuse to learn and accept the CORRECT answer. What you fail to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved first before you proceed with the rest of the problem. [/font]

lol now where are back to multiplying before divsion again.

Funny how I already explained it in detail and you still ignored it. 
I fear for the future of NT.

 

[balloonoboy]

How is 2 an afterthought when the only way there is a 2 outside the parenthesis is because there was originally a 2 inside.

Don't ever call me obtuse in your life, lil $$%@!. Tryna use big words and !$!!. ###% outta here.