48÷2(9+3) = ???

[usainboltisfast]

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

 

[usainboltisfast]

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

 

[balloonoboy]

Precisely, the 2 to the 9 and 3 as one would distribute the sinθ to the cosθ and tanθ.

As much as PEMDAS is a mathematical convention, so is distribution.

It's seen in Algebra, Trig, Calculus. So to say, "just use PEMDAS straight out" isn't the absolute truth.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

 

[balloonoboy]

Precisely, the 2 to the 9 and 3 as one would distribute the sinθ to the cosθ and tanθ.

As much as PEMDAS is a mathematical convention, so is distribution.

It's seen in Algebra, Trig, Calculus. So to say, "just use PEMDAS straight out" isn't the absolute truth.

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm
http://www.algebra-class....-property-equations.html

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

Who remembers doing identities in trig? Remember when you would see something like sinθ(cosθ+tanθ)? And you had to distribute in order to solve, there was no other way. What's more was that these angles were mostly already known, so they weren't variables in the common sense.

Some of you short bus Team 288 dudes who had remedial math in high school didn't learn that.

So, even if you don't agree that 2 is THE answer, you have to accept it as a possible answer. Since a lot of those who come to the answer of 2 are programmed to see any such coefficient-parenthetical pairing as a flag to, what, DISTRIBUTE.

Morever, accept the fact that you were wrong and not only do you not know as much as you think, but that you put so much time on a poorly written math problem, you should ask god to prevent you from offing thyselves.

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm
http://www.algebra-class....-property-equations.html

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

What are you distributing too? The 2 to the 9+3? if so that is wrong.

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

 

[yungchris504]

I have come to this conclusion

It is 

Statement 1 or Statement 2

with Statement 1 =  48÷2(9+3) = 288 (if seen as  (48÷2)(9+3)

and Statement 2 = 48 / 2(9+3) = 2

well you get 288 or 2

and now using logic

or = +

you get 288 + 2 = 290

So clearly the answer is 290
[h3]
[/h3]

 

[yungchris504]

I have come to this conclusion

It is 

Statement 1 or Statement 2

with Statement 1 =  48÷2(9+3) = 288 (if seen as  (48÷2)(9+3)

and Statement 2 = 48 / 2(9+3) = 2

well you get 288 or 2

and now using logic

or = +

you get 288 + 2 = 290

So clearly the answer is 290
[h3]
[/h3]

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

Not it shouldnt check 2nd example. The distributive property dosent create one term it creates 2. Stop trying to argue against sources when you cant provide your own.

2nd example: http://www.algebra-class....-property-equations.html

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

2(9+3) = 2(12) = (2*9+2*3) = (18+6) = 24

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

Not it shouldnt check 2nd example. The distributive property dosent create one term it creates 2. Stop trying to argue against sources when you cant provide your own.

2nd example: http://www.algebra-class....-property-equations.html

 

[balloonoboy]

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

 

[kingcrux31]

Originally Posted by yungchris504


I have come to this conclusion

It is 

Statement 1 or Statement 2

with Statement 1 =  48÷2(9+3) = 288 (if seen as  (48÷2)(9+3)

and Statement 2 = 48 / 2(9+3) = 2

well you get 288 or 2

and now using logic

or = +

you get 288 + 2 = 290

So clearly the answer is 290
[h3]
[/h3]

* IGNORED *

 

[kingcrux31]

Originally Posted by yungchris504


I have come to this conclusion

It is 

Statement 1 or Statement 2

with Statement 1 =  48÷2(9+3) = 288 (if seen as  (48÷2)(9+3)

and Statement 2 = 48 / 2(9+3) = 2

well you get 288 or 2

and now using logic

or = +

you get 288 + 2 = 290

So clearly the answer is 290
[h3]
[/h3]

* IGNORED *

 

[balloonoboy]

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

 

[usainboltisfast]

Originally Posted by balloonoboy

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

If you distribute (which is wrong because you must turn 9+3 to 12) you will get an answer of 11.6667 which is not 2. The correct way is to solve 9+3. Then go in order of operation. I have provided plenty of example problems that show this.

 

[usainboltisfast]

Originally Posted by balloonoboy

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

If you distribute (which is wrong because you must turn 9+3 to 12) you will get an answer of 11.6667 which is not 2. The correct way is to solve 9+3. Then go in order of operation. I have provided plenty of example problems that show this.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

Not it shouldnt check 2nd example. The distributive property dosent create one term it creates 2. Stop trying to argue against sources when you cant provide your own.

2nd example: http://www.algebra-class....-property-equations.html

The link you provided showed 6 * 2 + 6 * 4x, which is the same as (6 * 2 + 6 * 4x), not (6 * 2) + (6 * 4x). Looks all the same until it's preceded by another sign.
So, yeah. You're definitely wrong.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

That is wrong though. When you distribute you create 2 seperate terms.
Correctly distributing would give you this
2(9+3) = (2*9) + (2*3)
Add into this problem it would become 48/(2*9) + (2 * 3)
Look at how the parenthesis change in these problems
http://www.algebrahelp.co...ing/distribution/pg2.htm

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

Not it shouldnt check 2nd example. The distributive property dosent create one term it creates 2. Stop trying to argue against sources when you cant provide your own.

2nd example: http://www.algebra-class....-property-equations.html

The link you provided showed 6 * 2 + 6 * 4x, which is the same as (6 * 2 + 6 * 4x), not (6 * 2) + (6 * 4x). Looks all the same until it's preceded by another sign.
So, yeah. You're definitely wrong.

 

[usainboltisfast]

Originally Posted by kingcrux31

Originally Posted by usainboltisfast

Originally Posted by kingcrux31

You're wrong because both 2*9 and 2*3 should be in one set of ( ) in this problem. Not two.

Not it shouldnt check 2nd example. The distributive property dosent create one term it creates 2. Stop trying to argue against sources when you cant provide your own.

2nd example: http://www.algebra-class....-property-equations.html

The link you provided showed 6 * 2 + 6 * 4x, which is the same as (6 * 2 + 6 * 4x), not (6 * 2) + (6 * 4x). Looks all the same until it's preceded by another sign.
So, yeah. You're definitely wrong.

What are you talking about?

Look how they solved the question. If it was supposed to have parenthesis they would of finished it off with the invisible parenthesis you put around it. Your wrong dude.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

If you distribute (which is wrong because you must turn 9+3 to 12) you will get an answer of 11.6667 which is not 2. The correct way is to solve 9+3. Then go in order of operation. I have provided plenty of example problems that show this.

11.6667? Oh my.

I am done.

 

[kingcrux31]

Originally Posted by usainboltisfast

Originally Posted by balloonoboy

And therein lies the problem folks.

Do we do PEMDAS straight out?

Do we distribute first? If so, is each distribution given it's own set of parenthesis or are they grouped together. If the former, do we wait to add the distributed products until we get to the addition step?

It's a cluster****.

If you distribute (which is wrong because you must turn 9+3 to 12) you will get an answer of 11.6667 which is not 2. The correct way is to solve 9+3. Then go in order of operation. I have provided plenty of example problems that show this.

11.6667? Oh my.

I am done.