- Jul 1, 2006
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Originally Posted by DeadsetAce
48÷2(9+3) = ???
- Thread in 'General' Thread starter Started by n1k3b4ller,
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- Apr 1, 2009
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- Apr 1, 2009
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- May 18, 2009
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You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated.
as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.
and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))
Hopefully that clears it up once and for all.
- May 18, 2009
- 5,215
- 118
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated.
as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.
and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))
Hopefully that clears it up once and for all.
balloonoboy
Banned
- Nov 13, 2009
- 9,784
- 481
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.Originally Posted by do work son
Originally Posted by balloonoboy
Word?Originally Posted by DeadsetAce
i dont know what else to tell you. you're reading the chart too literally. the only reason it doesnt go further is because there is nothing left to do in the example they provided. if there was another part of the equation in the example, you would combine before moving forward.
Taken from Wikipedia:
Given a set S and two binary operations · and + on S, we say that the operation ·
is left-distributive over + if, given any elements x, y, and z of S,
x · (y + z) = (x · y) + (x · z);
is right-distributive over + if, given any elements x, y, and z of S:
(y + z) · x = (y · x) + (z · x);
is distributive over + if it is both left- and right-distributive.
Notice that when · is commutative, then the three above conditions are logically equivalent.
They didn't combine anything or leave them in parenthesis.
wait wait wait, you didn't just quote wikipedia did you?
Spoiler [+]
balloonoboy
Banned
- Nov 13, 2009
- 9,784
- 481
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.Originally Posted by do work son
Originally Posted by balloonoboy
Word?Originally Posted by DeadsetAce
i dont know what else to tell you. you're reading the chart too literally. the only reason it doesnt go further is because there is nothing left to do in the example they provided. if there was another part of the equation in the example, you would combine before moving forward.
Taken from Wikipedia:
Given a set S and two binary operations · and + on S, we say that the operation ·
is left-distributive over + if, given any elements x, y, and z of S,
x · (y + z) = (x · y) + (x · z);
is right-distributive over + if, given any elements x, y, and z of S:
(y + z) · x = (y · x) + (z · x);
is distributive over + if it is both left- and right-distributive.
Notice that when · is commutative, then the three above conditions are logically equivalent.
They didn't combine anything or leave them in parenthesis.
wait wait wait, you didn't just quote wikipedia did you?
Spoiler [+]
- Feb 9, 2003
- 10,791
- 932
Man all hope is lost for the future lol
- Feb 9, 2003
- 10,791
- 932
Man all hope is lost for the future lol
- Apr 13, 2007
- 5,288
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you are assuming the problem is 488 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
No man. I'm not assuming anything. I got the problem from the same source you did, and applied the above priciple like I was taught.
2(9+3) is an entire statement that can't be broken up. after this is calculated, the PEDMAS applies.
- Apr 13, 2007
- 5,288
- 467
you are assuming the problem is 488 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
No man. I'm not assuming anything. I got the problem from the same source you did, and applied the above priciple like I was taught.
2(9+3) is an entire statement that can't be broken up. after this is calculated, the PEDMAS applies.
- May 17, 2007
- 25,106
- 23,891
How the hell is this 18 pages LMAOOOO x 1000
- Apr 13, 2007
- 5,288
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No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.
- May 17, 2007
- 25,106
- 23,891
How the hell is this 18 pages LMAOOOO x 1000
- Aug 7, 2006
- 942
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48 / 2(9 + 3)
48÷2(9+3) IS the original equation.
-waystinthyme
- Aug 7, 2006
- 942
- 14
48 / 2(9 + 3)
48÷2(9+3) IS the original equation.
-waystinthyme
- Apr 13, 2007
- 5,288
- 467
No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.
do work son
Banned
- May 22, 2010
- 17,118
- 5,186
after you distribute, you combine the like terms derived from the distribution. then you continue onwards.Originally Posted by balloonoboy
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.Originally Posted by do work son
Originally Posted by balloonoboy
Word?
Taken from Wikipedia:
Given a set S and two binary operations · and + on S, we say that the operation ·
is left-distributive over + if, given any elements x, y, and z of S,
x · (y + z) = (x · y) + (x · z);
is right-distributive over + if, given any elements x, y, and z of S:
(y + z) · x = (y · x) + (z · x);
is distributive over + if it is both left- and right-distributive.
Notice that when · is commutative, then the three above conditions are logically equivalent.
They didn't combine anything or leave them in parenthesis.
wait wait wait, you didn't just quote wikipedia did you?
Spoiler [+]
do work son
Banned
- May 22, 2010
- 17,118
- 5,186
after you distribute, you combine the like terms derived from the distribution. then you continue onwards.Originally Posted by balloonoboy
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.Originally Posted by do work son
Originally Posted by balloonoboy
Word?
Taken from Wikipedia:
Given a set S and two binary operations · and + on S, we say that the operation ·
is left-distributive over + if, given any elements x, y, and z of S,
x · (y + z) = (x · y) + (x · z);
is right-distributive over + if, given any elements x, y, and z of S:
(y + z) · x = (y · x) + (z · x);
is distributive over + if it is both left- and right-distributive.
Notice that when · is commutative, then the three above conditions are logically equivalent.
They didn't combine anything or leave them in parenthesis.
wait wait wait, you didn't just quote wikipedia did you?
Spoiler [+]
- Aug 7, 2006
- 942
- 14
is this real life?
a fraction is just another way to write a number...
so if 2, was written as 4/2 instead...you couldn't distribute it?
please tell me you're trollin' right now...
-waystinthyme
- May 18, 2009
- 5,215
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/ and ÷ are the same thing.
- May 18, 2009
- 5,215
- 118
/ and ÷ are the same thing.
- Aug 7, 2006
- 942
- 14
is this real life?
a fraction is just another way to write a number...
so if 2, was written as 4/2 instead...you couldn't distribute it?
please tell me you're trollin' right now...
-waystinthyme
- Feb 19, 2007
- 7,190
- 162
- Feb 19, 2007
- 7,190
- 162
