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I really need help!!!

45 posts in this topic

Is there any mathematical book in the whole history of mathematics in which 48÷2y can be equal to [math](\frac{48}{2})y[/math]???

 

I've read enough scientific papers, and especially older ones, that I have seen it both ways. It was disconcerting, but in the end, dimensional analysis showed what the author meant. Believe me, I then scratched in my own sets of brackets into the equation to make it clear next time I read it. Back in the day, when it wasn't easy to typeset equations unless they were all in the same line, there were occasionally ambiguous terms printed.

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ermm, i bit confused here...

 

doesn't that (48/2)y = 48y / 2y

 

so, how can 48/2y = (48/2)y

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ermm, i bit confused here...

 

doesn't that (48/2)y = 48y / 2y

It indeed does not. You've probably mixed that up with the distribution law (48 + 2)y = 48y + 2y.

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It indeed does not. You've probably mixed that up with the distribution law (48 + 2)y = 48y + 2y.

 

then, what's the distribution law for '/' ?

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then, what's the distribution law for '/' ?

 

In this instance there is really only one distributive property, and that is the distribution of multiplication over addition ie: a(b+c)=ab+ac.

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ermm, i bit confused here...

 

doesn't that (48/2)y = 48y / 2y

If this were true...

 

[math]\frac{48}{2}y = \frac{48y}{2y}[/math]

 

One could divide out the y:

 

[math]\frac{48y}{2y}=\frac{48}{2}=24[/math]

 

And hence doing (48/2)y would be no different from doing (48/2), and clearly that can't be true.

 

This is what really happens:

 

[math]\frac{48}{2}y = \frac{48y}{2}=24y[/math]

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As the OP indicated by saying, "=24y or 24/y?" the he was asking how we would interoperate the expression given. As for where has 48÷2y been equal to:

 

[math](\frac{48}{2})y[/math],

 

according the standard oder of operations taught in children operations should be evaluated in the following order Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction (PEMDAS), and when operations of equivalent order are next to each other they are evaluated from left to right. Meaning that since division and multiplication of are same order we would evaluate 48÷2y as being (48÷2)y --since division furthest to the left.

 

To read about the standard order of operations see:

http://www.purplemath.com/modules/orderops.htm

http://www.mathsisfun.com/operation-order-pemdas.html

http://www.math.com/school/subject2/lessons/S2U1L2GL.html

 

How much is 48÷2 ??

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If this were true...

 

[math]\frac{48}{2}y = \frac{48y}{2y}[/math]

 

One could divide out the y:

 

[math]\frac{48y}{2y}=\frac{48}{2}=24[/math]

 

And hence doing (48/2)y would be no different from doing (48/2), and clearly that can't be true.

 

This is what really happens:

 

[math]\frac{48}{2}y = \frac{48y}{2}=24y[/math]

 

 

ahh i see, so from my understanding, (48/2)y not equal to 48/2y doesn't ??

 

 

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48/2y is kind of ambiguous, since it could mean (48/2)y or 48/(2y), depending on what the author means.

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48/2y is kind of ambiguous, since it could mean (48/2)y or 48/(2y), depending on what the author means.

 

 

How much is ,48/2 ,[math]\frac{48(x-a)^2}{2(x-a)}??[/math]

Edited by kavlas
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How much is ,48/2 ,[math]\frac{48(x-a)^2}{2(x-a)}??[/math]

 

[math]{24(x-a)}[/math]

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[math]{24(x-a)}[/math]

 

The question now is: Does Cap'n Refsmmat agree with you ??

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The question now is: Does Cap'n Refsmmat agree with you ??

 

Dollars to donuts he does. Your first question of the thread was ambiguous - your post had two phrases both clear. That's one of the benefits of using latex as soon as you have multiple levels of brackets or division

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Dollars to donuts he does. Your first question of the thread was ambiguous - your post had two phrases both clear. That's one of the benefits of using latex as soon as you have multiple levels of brackets or division

 

What was my first question of the thread .....?

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48/2y is kind of ambiguous, since it could mean (48/2)y or 48/(2y), depending on what the author means.

 

According to the above 48/2 could be 24 ,or (48/1).2 = 96 ??, depending on what the author means ??

 

Since 2 = 1.2 ??

 

That is why i asked you how much is 48/2

 

Further more [math]\frac{48(x-a)^2}{(x-a)}[/math] could be 24(x-a) ,or:

 

[math]\frac{48(x-a)^2}{2(x-a)}[/math] = [[math]\frac{48(x-a)^2}{2}[/math]].(x-a) = [math]24(x-a)^3[/math]

 

Of course it is up to you to answer or not.

 

I just wanted to show what i ment by asking those questions

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According to the above 48/2 could be 24 ,or (48/1).2 = 96 ??, depending on what the author means ??

 

Since 2 = 1.2 ??

 

No, what you have are two different expressions. Since [math]2=(1*2)[/math] you could write 48/(1*2), which would be 24.

 

Further more [math]\frac{48(x-a)^2}{(x-a)}[/math] could be 24(x-a)

 

Ahh, not it cannot.

 

[math]\frac{48(x-a)^2}{2(x-a)}[/math] = [[math]\frac{48(x-a)^2}{2}[/math]].(x-a) = [math]24(x-a)^3[/math]

 

 

No it could not unless you completely change the standard convention of what the notation of [math]/frac{a}{b}[\math] means.

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According to the above 48/2 could be 24 ,or (48/1).2 = 96 ??, depending on what the author means ??

 

Since 2 = 1.2 ??

 

That is why i asked you how much is 48/2

 

Further more [math]\frac{48(x-a)^2}{2(x-a)}[/math] could be 24(x-a) ,or:

 

[math]\frac{48(x-a)^2}{2(x-a)}[/math] = [[math]\frac{48(x-a)^2}{2}[/math]].(x-a) = [math]24(x-a)^3[/math]

 

Of course it is up to you to answer or not.

 

I just wanted to show what i ment by asking those questions

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You can simplify it easily,

 

[latex] \frac{48}{2y} [/latex]

 

Write this as,

 

[latex] \frac{48}{2}\times\frac{1}{y} [/latex]

 

Now, simply divide 48 by 2 and then multiply,

 

[latex] 24\times\frac{1}{y}=\frac{24}{y} [/latex]

 

I hope it' ll help.

Edited by deesuwalka
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