# Math Question



## darkzelkova (May 28, 2009)

So I got a math question the other day, and I haven't been able to figure it out. Anyone got any idea how I could do it?

Here it is: Point A and point B are the origins of their respective circles. CAE and CBF are line segments. Angle F is 78 degrees (how do I do that little symbol?), What is the measure of angle C?







Any help would be greatly appreciated!

EDIT: Sorry about that, I did mean origin of the circle


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## Jokerman5656 (May 28, 2009)

do you mean that points A and B are origins of their respective circles, not triangles?


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## spdcbr (May 28, 2009)

I got it, judging by the circle, it's equal to angle F. So...

180-78+78=24 degrees.
.......(156).................

Mwaahahaa! I solved it! Point A and B is there to confuse you.


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## Jokerman5656 (May 28, 2009)

i dont think the answer is a fart piece of cow **** that sucks ****. so your wrong


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## spdcbr (May 28, 2009)

Logan said:


> I say they answer is pi........ MeMyselfandPi to be precise



What th- 
I ALRREADYY SOOLVEDD IT!!!! I DID ITTT!NOT PIIIII! NOT PIIIII! NOT MEMYSELFANDPI!!!!!!NO!!!!!!IT 24!!!


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## Lord Voldemort (May 28, 2009)

"180-78+78=24 degrees"
Bad math... that equals precisely 180 degrees.


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## shelley (May 28, 2009)

spdcbr said:


> I got it, *judging by the circle, it's equal to angle F*. So...
> 
> 180-78+78=24 degrees.
> .......(156).................
> ...



Your faulty math aside, please explain this statement.


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## blah (May 28, 2009)

x + 2x + 78 + 78 = 180
x = 8


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## chencube (May 28, 2009)

blah said:


> x + 2x + 78 + 78 = 180
> x = 8



This is right.


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## Lord Voldemort (May 28, 2009)

Which angles do x and 2x represent?


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## blah (May 28, 2009)

Lord Voldemort said:


> Let the point on the line segment AE where the circle B intercepts the triangle be point Z, and the point on line segment BC on circle A be point Y. First, draw line segment AB. AB is a radius of circle A. It is also a radius of circle B. If the same line segment is the radius of two circles, then the radius of the circles must be equal.



Your point being?


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## spdcbr (May 28, 2009)

Lord Voldemort said:


> "180-78+78=24 degrees"
> Bad math... that equals precisely 180 degrees.



Uhhh...you take away from 180 and you get precisely 180?


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## spdcbr (May 28, 2009)

blah said:


> x + 2x + 78 + 78 = 180
> x = 8



Sigh, this is wrong, 2x is just a distraction.


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## Ellis (May 28, 2009)

I got 8 also, my way was much more complex and involved drawing two radii from B to make two more triangles and using the arcs of the circle on the right. If anyone wants me to explain the way I did it, I can. But that really shouldn't be necessary.

spdcbr, shut up.


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## Siraj A. (May 28, 2009)

spdcbr said:


> Lord Voldemort said:
> 
> 
> > "180-78+78=24 degrees"
> ...



You subtracted from 180, and then added back what you subtracted.

Learn order of operations, please.


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## Poke (May 28, 2009)

spdcbr said:


> Lord Voldemort said:
> 
> 
> > "180-78+78=24 degrees"
> ...



Let's slow this down for you...

180-78=102
102+78=180
NOT 24.


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## Lord Voldemort (May 28, 2009)

blah said:


> Lord Voldemort said:
> 
> 
> > Let the point on the line segment AE where the circle B intercepts the triangle be point Z, and the point on line segment BC on circle A be point Y. First, draw line segment AB. AB is a radius of circle A. It is also a radius of circle B. If the same line segment is the radius of two circles, then the radius of the circles must be equal.
> ...



I needed to get another look at the diagram, so I posted it without finishing. Then I realized that where I was going was wrong. 



spdcbr said:


> Lord Voldemort said:
> 
> 
> > "180-78+78=24 degrees"
> ...



Uhh... You forgot parenthesis...



spdcbr said:


> blah said:
> 
> 
> > x + 2x + 78 + 78 = 180
> ...



I thought so...


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## spdcbr (May 28, 2009)

I meant this:

180-(78+78)=24
Okay? get it?


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## chencube (May 28, 2009)

Lord Voldemort said:


> Which angles do x and 2x represent?



x = Angle C.
2x = Angle CEB

Just draw a line from A to B you will see how easy it is.


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## Ellis (May 28, 2009)

spdcbr said:


> I meant this:
> 
> 180-(78+78)=24
> Okay? get it?



wrong


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## shelley (May 28, 2009)

spdcbr said:


> blah said:
> 
> 
> > x + 2x + 78 + 78 = 180
> ...



Nope, it's right. There are no "distractions" in this problem.


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## Lord Voldemort (May 28, 2009)

I realized that.
Now prove that CE = CF\

Not to you shelley


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## Ellis (May 28, 2009)

Lord Voldemort said:


> I realized that.
> Now prove that CE = CF



?

oh that was to spdcbr.


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## spdcbr (May 28, 2009)

If you look at it visually it doesn't look like 8 degrees to me.


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## shelley (May 28, 2009)

spdcbr said:


> If you look at it visually it doesn't look like 8 degrees to me.



It's not drawn to scale. You obviously haven't taken geometry so stop trying to answer questions you don't know how to do.


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## Ellis (May 28, 2009)

spdcbr said:


> If you look at it visually it doesn't look like 8 degrees to me.



If you look at it mathematically, it doesn't equal 24.


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## spdcbr (May 28, 2009)

Ellis said:


> spdcbr said:
> 
> 
> > If you look at it visually it doesn't look like 8 degrees to me.
> ...



 you're right. how can we tell who's right then? einne mienne moeny mo?


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## pcharles93 (May 28, 2009)

We can tell who's right by whoever used actual logic. That would mean blah is right.


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## Kian (May 28, 2009)

spdcbr said:


> Ellis said:
> 
> 
> > spdcbr said:
> ...



sigh.


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## Ellis (May 28, 2009)

spdcbr said:


> Ellis said:
> 
> 
> > spdcbr said:
> ...



It's called math. So far we have at least 2 ways of actually getting the correct answer, which definitely beats eyeballing it.


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## spdcbr (May 28, 2009)

Ellis said:


> spdcbr said:
> 
> 
> > Ellis said:
> ...



Hahah, you're right too. So, then what degrees are the 3 angles?


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## pcharles93 (May 28, 2009)

spdcbr said:


> Ellis said:
> 
> 
> > spdcbr said:
> ...



We just figured out 2 of them...


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## Ellis (May 28, 2009)

8, 78, and 94

learn geometry


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## Lord Voldemort (May 28, 2009)

Hmm... I still don't get it...
*facepalms self in shame*
How do you know angle E equals angle F?
That is where the two 78s are coming from, right?
*facepalms self in shame again*


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## spdcbr (May 28, 2009)

I seirously doubt that one of them is 94.


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## Ellis (May 28, 2009)

spdcbr said:


> I seirously doubt that one of them is 94.



That's because you eyeballed it.


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## shelley (May 28, 2009)

Lord Voldemort said:


> Hmm... I still don't get it...
> *facepalms self in shame*
> How do you know angle E equals angle F?
> That is where the two 78s are coming from, right?
> *facepalms self in shame again*



It doesn't.

Hint: draw segments from B to A and from B to E.


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## spdcbr (May 28, 2009)

shelley said:


> Lord Voldemort said:
> 
> 
> > Hmm... I still don't get it...
> ...



That's not much of a hint.


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## KubeKid73 (May 28, 2009)

Ellis, you're right about the angles. But if he did eyeball it, he would see that the one angle is clearly larger than 90 degrees.


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## Kian (May 28, 2009)

spdcbr said:


> shelley said:
> 
> 
> > Lord Voldemort said:
> ...



it's a very helpful hint if you didn't do that to begin with. however, you've made it cleared that you haven't learned geometry, so it's really not going to make a difference what hints you get.


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## spdcbr (May 28, 2009)

KubeKid73 said:


> Ellis, you're right about the angles. But if he did eyeball it, he would see that the one angle is clearly larger than 90 degrees.



It depends on the way you look at it.


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## blah (May 28, 2009)

spdcbr said:


> shelley said:
> 
> 
> > Lord Voldemort said:
> ...



That's because you don't have much of a brain.


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## KubeKid73 (May 28, 2009)

spdcbr said:


> KubeKid73 said:
> 
> 
> > Ellis, you're right about the angles. But if he did eyeball it, he would see that the one angle is clearly larger than 90 degrees.
> ...



E is definitely larger than 90 degrees.


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## spdcbr (May 28, 2009)

blah said:


> spdcbr said:
> 
> 
> > shelley said:
> ...



Maybe you said that because all you can say to people is "blah" your opinion doesn't matter.


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## Siraj A. (May 28, 2009)

Spdcbr, just stop.


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## Gparker (May 28, 2009)

spdcbr said:


> blah said:
> 
> 
> > spdcbr said:
> ...




It's a fact, not an opinion. Go back to English. Facts can be proven and the proof is almost every post you made in this thread.

On topic, I also came up with 4  and I'm not even in geometry


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## ThatGuy (May 28, 2009)

12 degrees. maybe. let me spend time and check 
HAHA BSing. My geometry intuition is really bad. don't listen to me.


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## gpt_kibutz (May 28, 2009)

Ok. The first step to begin to solve a geometry problem is NEVER state something jut because "it looks like it is more than 8 degrees"
If you draw AB and BE, then the triangle is split into three isoceles triangles (Because AB, AC, BE, and BF are radii).
Then assuming that ACB is x, then CBA is x, and therefore EAB is 2x. But, as trangle ABE is isoceles, BEA is also 2x. The last isoceles triangle is BEF, and as angle F is 78 degrees so is angle BEF. 
We have then that angles ECF+CFE+FEC=180
So:
x+78+2x+78=180
and:
x=8=ECF


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## spdcbr (May 28, 2009)

Siraj A. said:


> Spdcbr, just stop.



Could you tell HIM to stop?
http://www.speedsolving.com/forum/showpost.php?p=184267&postcount=42


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## Ellis (May 28, 2009)

I just looked back at the silly way I did it, and it was wrong. I was trying to incorrectly find ABE using arcs, and at the same time I made a stupid subtraction mistake. Somehow I got the correct measure of ABE though. I don't know how I managed that. And I hadn't even looked at the answer yet. But I see how it's done now. That was a facepalm moment for me.



luisgepeto said:


> But, as trangle ABE is equilateral, BEA is also 2x. The last equilateral triangle is BEF, and as angle F is 78 degrees so is angle BEF.



You mean isosceles


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## spdcbr (May 28, 2009)

http://www.speedsolving.com/forum/showpost.php?p=184251&postcount=33
This?


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## Ellis (May 28, 2009)

spdcbr said:


> http://www.speedsolving.com/forum/showpost.php?p=184251&postcount=33
> This?



And... so...?

I meant the three angles of the single triangle in the drawing.

Oh, do you mean that was my mistake? No, that's correct.

The mistake I made was that I couldn't actually do it the way I was trying to, but I made two mistakes that somehow ended up getting the right answer. Those are still the measures of the angles in the triangle.


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## blah (May 28, 2009)

spdcbr said:


> I got it, judging by the circle, it's equal to angle F. So...
> 
> 180-78+78=24 degrees.
> .......(156).................
> ...





spdcbr said:


> What th-
> I ALRREADYY SOOLVEDD IT!!!! I DID ITTT!NOT PIIIII! NOT PIIIII! NOT MEMYSELFANDPI!!!!!!NO!!!!!!IT 24!!!





spdcbr said:


> Lord Voldemort said:
> 
> 
> > "180-78+78=24 degrees"
> ...





spdcbr said:


> blah said:
> 
> 
> > x + 2x + 78 + 78 = 180
> ...





spdcbr said:


> I meant this:
> 
> 180-(78+78)=24
> Okay? get it?





spdcbr said:


> If you look at it visually it doesn't look like 8 degrees to me.





spdcbr said:


> shelley said:
> 
> 
> > It doesn't.
> ...



Okay I apologize for saying that you don't have much of a brain. No one deserves such an insult. But I just want you to know that you pissed me off when you claimed that my correct solution was wrong and continued to make a stupid statement about certain points being there to distract the rest of us, acting as if you were the only genius around here who didn't fall for that distraction and found the correct solution; as if the rest of us here trying to help you find your own mistake are intellectually inferior to you. And all this happened _with the correct solution shown in your face_.

Although I take back what I said about you not having much of a brain, I still quoted these posts to express my _opinion_ that you refuse to use your brain to think logically.


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## gpt_kibutz (May 28, 2009)

Ellis said:


> luisgepeto said:
> 
> 
> > But, as trangle ABE is equilateral, BEA is also 2x. The last equilateral triangle is BEF, and as angle F is 78 degrees so is angle BEF.
> ...



Oops yea sorry


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## ThatGuy (May 28, 2009)

Circles A and B are congruent? Yes? No?


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## daeyoungyoon (May 28, 2009)

I got 8.......


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## Ellis (May 28, 2009)

ThatGuy said:


> Circles A and B are congruent? Yes? No?



Yea, they have they same radius.


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## Lord Voldemort (May 28, 2009)

shelley said:


> Lord Voldemort said:
> 
> 
> > Hmm... I still don't get it...
> ...



That's what I did, so you get three isosceles triangles, and then you can get the values of the angles in terms of x. It didn't work for me before. That was where I was going before... was I foiled by an arithmetic error? I think so...


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## shelley (May 28, 2009)

spdcbr said:


> shelley said:
> 
> 
> > Lord Voldemort said:
> ...



Only if you don't know what you're doing. This is clearly the case with you. Go back to school and leave the math to the people who can actually do it.


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## ThatGuy (May 28, 2009)

It's 24. (short)


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## Daniel Wu (May 28, 2009)

ugh. I tried using arcs instead of the radii.


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## shelley (May 28, 2009)

ThatGuy said:


> It's 24. (short)



Not quite.


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## Ellis (May 28, 2009)

ThatGuy said:


> It's 24. (short)



spdcbr logic?


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## daeyoungyoon (May 28, 2009)

How is it 24?


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## ThatGuy (May 28, 2009)

F is 78, so angle BEF is also 78. That makes EBF 24. EBF is haha i'm a retard never mind


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## pcharles93 (May 28, 2009)

I don't know why, but I expected that to happen.


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## ThatGuy (May 28, 2009)

yeah, its 3. ME IS RETARD.


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## fanwuq (May 28, 2009)

The answer and explanation were given.
http://www.speedsolving.com/forum/showpost.php?p=184278&postcount=48

This is a nice fun geometry problem. Right from the beginning, simply by looking at it, angle is obviously less than 12. Without even solving it, you should be able to estimate that it should be around 8 or 9.


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## Dene (May 28, 2009)

spdcbr definitely doesn't have a brain. It surely cannot be possible. I mean, one cannot be so inane. I refuse to believe anything with a brain could be that inane.


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## Ellis (May 28, 2009)

spdcbr said:


> http://www.speedsolving.com/forum/showpost.php?p=184251&postcount=33
> This?



Oh wait, I think I finally understood what you were getting at... silly me. 

There's a difference between making a simple subtraction mistake in your head and pulling answers out of your ass.


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## darkzelkova (May 28, 2009)

Thanks guys! I'm pretty sure I've got it now! I had gotten to the isoceles triangles, and I was going to try subbing in 1 for the radius, I didn't really think of using just x. Thanks again for the help!


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## Ellis (May 28, 2009)

darkzelkova said:


> Thanks guys! I'm pretty sure I've got it now! I had gotten to the isoceles triangles, and I was going to try subbing in 1 for the radius, I didn't really think of using just x. Thanks again for the help!



You don't even need to put the radius as anything. x doesn't represent the radius. You just need to know that the radii are equal to each other.


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## qqwref (May 28, 2009)

Dene said:


> spdcbr definitely doesn't have a brain. It surely cannot be possible. I mean, one cannot be so inane. I refuse to believe anything with a brain could be that inane.



I agree with you. He's practically bragging about how stupid he is by insisting he's right, even though people who both understand the problem and are actually good at math have showed that he was wrong from the start.


By the way, this is a very nice problem. The isosceles triangles are not immediately obvious but they make it work out very nicely with very little algebra.


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## Bob (May 28, 2009)

If spdcbr was in my class, I would cry.


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## blah (May 28, 2009)

So I'm supposed to take my apology back now? (Since Dene the philosopher and Michael have both agreed upon my original claim?)

By the way I didn't want to reveal all of my solution in the beginning because I wanted to give the rest of you here a shot at the question, but oh well


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## Daniel Wu (May 28, 2009)

Bob said:


> If spdcbr was in my class, I would cry.



haha. that made me laugh.


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## ManasijV (May 28, 2009)

For those who haven't already found the solution here it is.
In triangle CAB, CA = CB. Theredore ACB=ABC= x This implies CAB= 180-2x.BAE=2x
In Triangle BAE BA =BE. BAE=BEA =2x.
In triangle BEF BEF=BFE=78. EBF=24.
ABE+BEA+EAB=180=EBC+ABE+EBF
2x+2x=x+24
x=8.


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## Gparker (May 29, 2009)

Hey, i dont want to make a thread for this but my math teacher gave us some really hard problems yesterday and I need help. There not for a grade but there for extra credit.

1. Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month?

2. If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y,
then x = ? 

3. One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there? 

4. If ax*x + bx +c = 0,
then what is x? 

5. What number shows up most often when you roll 10 dice? 


I seem really confused about this so any help is appreciated.


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## Lord Voldemort (May 29, 2009)

5. Weird. It seems almost like a trick question.
Theoretically, there's a 1/6 chance of any number occurring on all 10 rolls, so...

4. x = (-b + sqr(b^2-4ac))/2a or x = (-b - sqr(b^2-4ac))/2a
It's the quadratic formula.

3. x is the number of humans.
2x + (74-x)x4 = 196
-2x = -100, x = 50
50 humans, 24 horses.

1. First, she spends 1/2 her money so she has 1/2 left. Next, she spends 1/3, so 2/3 left. Then, she spends 1/4, so she had 3/4 left.
(1/2)(1/3)(1/4)x = 7777
She gets $31,108 a month. I wish I had that kind of money...

2. Not sure about this one.


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## fanwuq (May 29, 2009)

Gparker said:


> There not for a grade but there for extra credit.



No. That's even worse. Math extra credits are meant to be way too hard for the average student to solve. I've done plenty of them at my school. It actually requires you to sit there yourself and think for a few hours.

Your problems actually look too easy to be extra credit. I would have given that as classwork if I'm the teacher.

1. Very very easy. Definitely regular classwork for pre-Algebra.
2. Looks interesting. Just try it!
3. This is a 1st grade problem, literally.
4. That's extra credit? Look up the formula. Google is your friend.
5. See peterbat's comment below.


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## peterbat (May 29, 2009)

Lord Voldemort said:


> 5. Weird. It seems almost like a trick question.
> Theoretically, there's a 1/6 chance of any number occurring on all 10 rolls, so...




I think he meant if you add the results of all 10 dice each time you throw them on the table, what sum do you get the most often.

Hint: 10 and 60 occur least often.


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## shelley (May 29, 2009)

fanwuq said:


> Gparker said:
> 
> 
> > There not for a grade but there for extra credit.
> ...



How do you know what level math he's familiar with? If he were in 1st grade, these problems would be very challenging indeed.

And yeah, question 1? I make a little more than that but I have to work for it. And I have to pay rent. What kind of allowance is this?


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## fanwuq (May 29, 2009)

shelley said:


> fanwuq said:
> 
> 
> > Gparker said:
> ...



I think he's in 7th grade. Even if he is in first grade, #3 is still easy.


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## Gparker (May 29, 2009)

Yes i am in 7th grade, but weve messed around with geometry and stuff, i guess yall think its easy because yall are higher up in math


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## Lord Voldemort (May 29, 2009)

Well, 1 can be done with only knowledge of arithmetic and reasoning.
2 is the hardest for me, 3 you can do guess and check but it's easier if you can set up an equation, 4 might be hard to derive yourself, but doable, 5 is just badly worded.


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## shelley (May 29, 2009)

Lord Voldemort said:


> Do any of you have #2?
> It's an interesting problem.
> It seems like it's x=y at first glance.



Hint for problem 2: it's quite repetitive. Try making some substitutions. (That's one way to go about it anyway)

You do have the right answer. Now just figure out how to get there.


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## Gparker (May 29, 2009)

I feel so stupid 

An 8th grader had to do this.

Let f be the function given by f(x)= e^(-2x^2) (read as e raised to the negative 2 x squared)

(a) Find the first four nonzero terms and the general term of the power seriews for f(x) about x=0.

(b) Find the interval of convergence of the power seriews for f(x) about x = 0. Show the analysis that leads to your conclusion.

(c) Let g be the function given by the sum of the first four nonzero terms of the power seriews for f(x) about x=0. Show that l f(x) - g(x) l < 0.02 for -0.6 less than or equal to x less than or equal to 0.6.


And thank you to everyone that helped me. I think I figured #2 on my own before i saw voldemorts ansewer. But like he said im not that sure, ill make substitutions as shelly said.


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## Lord Voldemort (May 29, 2009)

Oh duh. It's just an infinite geometric series with starting term 1/2 and common ration 1/2. Just carry it to a few repetitions, ignore the last 1/2 and it's clear.

EDIT: Are you sure that's not your homework?
Seems a bit suspicious to me. 
Isn't power series usually with Taylor's series? I don't know what a power series is, and I don't really care to find out. How do you go from basic algebra to something like this in one year?


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## Gparker (May 29, 2009)

Lord Voldemort said:


> Oh duh. It's just an infinite geometric series with starting term 1/2 and common ration 1/2. Just carry it to a few repetitions, ignore the last 1/2 and it's clear.
> 
> EDIT: Are you sure that's not your homework?
> Seems a bit suspicious to me.
> Isn't power series usually with Taylor's series? I don't know what a power series is, and I don't really care to find out. How do you go from basic algebra to something like this in one year?



Wait what? The problems i gave were extra credit, we havnt learned much about that kind of stuff yet and she just wanted for us to try it, and plus you dont have to even try it. And the one that the 8th grader sent me was also for his extra credit.


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## Lord Voldemort (May 29, 2009)

Well, posting two sets of problems, it does seem like we might be doing your home work for you. This is more complex than the first set though, so I doubt this is from the same extra credit set. Taylor's Series are usually introduced in first year calculus as far as I know.


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## Gparker (May 29, 2009)

Lord Voldemort said:


> Well, posting two sets of problems, it does seem like we might be doing your home work for you. This is more complex than the first set though, so I doubt this is from the same extra credit set. Taylor's Series are usually introduced in first year calculus as far as I know.



Wow, i can do my own math homework thank you. The first set was my set, i didnt ask for ansewers, i asked for help. But if you like just gave me the ansewer ii would have to figure out how you got it. The second one was from a friend of mine. And plus, the last 2 weeks of school with finals over and i cant do my own math homework? Its just extra credit, thats all. She would never give it to us for a grade since it wouldent be fair. Honestly, do you think any of that is pre-algabra stuff?


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## Bryan (May 29, 2009)

Isn't extra credit something that boosts your grades? It doesn't matter if it's base homework or extra homework. The fact is if you're not figuring it out on your own and submitting our answers, you're unethically inflating your grades.


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## fanwuq (May 29, 2009)

Is the 8th grader doing typical 8th grade math? I've not encountered any of the stuff in the 2nd set and I just finished AP Calculus AB. Series does sound like a topic from BC.
The first set is pre-algebra other than 2nd and 4th problem. 4th problem would have been better if it asks you to prove it. I would say there is about a 6 year difference between the level of the 1st and 2nd set of problems.

Bryan: Agreed.


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## Gparker (May 29, 2009)

Bryan said:


> Isn't extra credit something that boosts your grades? It doesn't matter if it's base homework or extra homework. The fact is if you're not figuring it out on your own and submitting our answers, you're unethically inflating your grades.



I'm really sorry then, I know no one will believe me but I promise i wont send them in. I dont even need the extra credit anyway. I put it on here for help and to see if it was a challenge to anybody else. I'm very sorry and i aplogize deeply.


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## Lord Voldemort (May 29, 2009)

I never outright accused you, I just suggested it may have been a possibility. As for the answers/help thing, they can be rather ambiguous without anything else to go by. School isn't over for everyone, only in America. And no, as I said, it seems like calculus with isn't Pre-Algebra. 

I had a similar response to fanwuq, and it was somewhat suspicious. If someone else asked questions and then more questions, then what would you think?


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## Gparker (May 29, 2009)

Lord Voldemort said:


> I never outright accused you, I just suggested it may have been a possibility. As for the answers/help thing, they can be rather ambiguous without anything else to go by. School isn't over for everyone, only in America. And no, as I said, it seems like calculus with isn't Pre-Algebra.
> 
> I had a similar response to fanwuq, and it was somewhat suspicious. If someone else asked questions and then more questions, then what would you think?



Well, the second set was just for people to try. Have some fun you know?


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## moogra (May 30, 2009)

Let f be the function given by f(x)= e^(-2x^2) (read as e raised to the negative 2 x squared)

(a) Find the first four nonzero terms and the general term of the power series for f(x) about x=0.

e^x = 1+x+x^2/2+x^3/3!+x^4/4!+...
e^-x = 1-x+x^2/2-x^3/3!-x^4/4!+x^5/5!...
e^(-2x^2)=1-2x^2+2x^4/2-2x^6/3!-2x^8/4!+2x^10/5!-...
general term:
2x^(2n)/n!*(-1)^n

(b) Find the interval of convergence of the power series for f(x) about x = 0. Show the analysis that leads to your conclusion.

Hint: -1< e^(-2(x+1)^2) /e^(-2x^2) < 1
Solve and then look at the series when it's at 1 and -1 and if it does not diverge, then that is also included in the answer.
(c) Let g be the function given by the sum of the first four nonzero terms of the power seriews for f(x) about x=0. Show that l f(x) - g(x) l < 0.02 for -0.6 less than or equal to x less than or equal to 0.6.

Lagrange Error Form is annoying, but that's a hint.

The following problems seem very unrelated and much much easier.

2. If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y,
then x = ? 
It's just 1/2x+1/4x+1/8x+...+1/(x*2^n) so x = y
(easily shown with geometric series and it's well known as well)

5. What number shows up most often when you roll 10 dice? 
35

Also:
LOL @ SPDCBR


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## Gparker (May 30, 2009)

moogra said:


> Let f be the function given by f(x)= e^(-2x^2) (read as e raised to the negative 2 x squared)
> 
> (a) Find the first four nonzero terms and the general term of the power series for f(x) about x=0.
> 
> ...



I have no idea if thats right. But it looks good, i wonder if anyone else could do it. Was it hard?

Also, for #5, could you say how you got 35?


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## qqwref (May 30, 2009)

Here's an interesting math problem to think about. It is not as easy as you think!

Suppose you have _N_ boxes and _K_ objects. The boxes are numbered from 1 to N; for each object, we randomly choose a box number, and write that down. Now we place the objects in the boxes, according to what number each object was given.

Okay, so here's the question: we're going to choose a random object and look in the box corresponding to that object. What is the expected value of the number of objects that will be in that box (i.e. on average, how many objects will be there)?

EDIT: If you can solve that, try this harder version: at the last part, instead of choosing a random object, we're going to choose (at random) one of the boxes that is not empty. What's the expected value of the number of objects in that box now?


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## jacob15728 (May 30, 2009)

Gparker said:


> I feel so stupid
> 
> An 8th grader had to do this.
> 
> ...




That was completely unnecessary.


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## jacob15728 (May 30, 2009)

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.


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## spdqbr (May 30, 2009)

jacob15728 said:


> When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.



I have a truly remarkable proof of this fact, which this post is too short to contain. Also, dinner.


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## qqwref (May 30, 2009)

jacob15728 said:


> When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.



hahaha, nice one


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## Bryan (May 30, 2009)

jacob15728 said:


> When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.



I just state that it's true for value of n=1 or value of p=0. Then you just point back to the basic axioms.....QED.


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## Johannes91 (May 30, 2009)

jacob15728 said:


> When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it.


Actually, Martin Musatov published recently an informal and highly experimental, unorthadox proof P=NP: http://groups.google.com/group/sci.math/browse_thread/thread/16d3720a3cffc651/37a6a16911e183e9. Several other proofs of it are have also been mentioned in sci.math, so if you had known how to use the search function, you wouldn't have needed to waste those few minutes reinventing the wheel.


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## ManasijV (May 30, 2009)

1)
x[1-(1/2)-(1/2)1/3-1/2 1/3 1/4]=7777 
x=26664


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## ManasijV (May 30, 2009)

2)1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y
1/2x +1/2(y)=y
x=y
Is it difficult to visualize this?
3)Did you really have to post the number of horses question? When I was in 7th grade I would still solve it mentally i think. (But then I'm from India you are not  )
4)quadratic
5)(31+39))/2=35. This is kind of a guess. Still haven't learnt permutations and combinations


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## qqwref (May 30, 2009)

Johannes91 said:


> jacob15728 said:
> 
> 
> > When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it.
> ...


I'm interested in looking at this but it seems that (a) the entire thread is just a flamewar and (b) every link to his paper has been removed for some reason. Do you have a working copy?


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## Johannes91 (May 30, 2009)

qqwref said:


> I'm interested in looking at this but it seems that (a) the entire thread is just a flamewar and (b) every link to his paper has been removed for some reason. Do you have a working copy?


I didn't notice (b) until after I posted. I don't have a copy; I might've checked it when the thread had just started but don't really remember anything about it.

There are a lot of such threads, where someone (possibly a very bored troll) seems to honestly believe that he's found something remarkable, like a proof of P=NP or P≠NP or that reals are countable. They are sometimes quite interesting to follow.


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## trying-to-speedcube... (May 30, 2009)

> I have no idea if thats right. But it looks good, i wonder if anyone else could do it. Was it hard?
> 
> Also, for #5, could you say how you got 35?



I think that the average number that's thrown is (1+2+3+4+5+6)/6 = 21/6=3.5
Because you throw 10 dice the most appearing sum is 3.5*10 = 35


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## blah (May 30, 2009)

qqwref said:


> Here's an interesting math problem to think about. It is not as easy as you think!
> 
> Suppose you have _N_ boxes and _K_ objects. The boxes are numbered from 1 to N; for each object, we randomly choose a box number, and write that down. Now we place the objects in the boxes, according to what number each object was given.
> 
> ...



I'm not sure if I understand the question correctly, so can I just clarify something, Michael? (Highlight below to read my question.)

We pick a random box, if there's nothing in there, we pick another random box, and we just keep doing this till we hit a box with at least 1 object. Question: What is the expected number of objects in that box? - Is this question equivalent to yours? (This refers to the first "easier" question.)

Also, I'm currently seeing this as NK slots and K objects to be "slotted in" (pardon the expression, side effect of too much Fridriching), am I heading towards the right direction?

Edit: Oops, I didn't notice your edit just now, which (I think) is the exact same thing I'm asking. Coincidence? So I guess your answer to my question is no? Or is it a yes, i.e. the "harder" version in fact the exact same question, and you're just trying to trick us?


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## qqwref (May 30, 2009)

What you're talking about, blah, is the harder question. Each non-empty box is equally likely to be chosen. In the easier one, a non-empty box with M objects in it is M times as likely to be chosen as a box with 1 object.

Here's a hint to the first question, to show why it's relatively easy to figure out:


Spoiler



Since we're randomly choosing an object and looking in its box, but the objects are all treated identically, we can just label the chosen object. Then the expected value of the number of objects in that box is one (for the labeled object) plus the expected value of the number of the OTHER K-1 objects that will end up in the given box, which is pretty easy to calculate.


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## blah (May 30, 2009)

I find it amusing that I predicted the harder question by misunderstanding the simpler one 

And err... Yeah, whatever you said in the spoiler was what first came to my mind. But something told me there was a flaw in that logic but I didn't know what, so I headed in a different direction. Guess I gotta trust my first instincts huh?


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## Sin-H (May 30, 2009)

I'm not sure if it's still important, but the taylor series stuff works like this (just one way)
you have to equate the function term with a general polynom a(1)+a(2)x+a(3)x^2+a(4)x^3+...

then insert x=0, and you get a(1).
derive, reducing the polynom's grade by 1. same procedure: x=0, now you get a(2). and so on, until you have your demanded number of terms.

but wait. Since when does a 8th grader know how to find derivatives?


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## blah (May 30, 2009)

Sin-H said:


> I'm not sure if it's still important, but the taylor series stuff works like this (just one way)
> you have to equate the function term with a general polynom a(1)+a(2)x+a(3)x^2+a(4)x^3+...
> 
> then insert x=0, and you get a(1).
> ...



What's eighth grade equivalent to? 

I realize that was a stupid question. How old would the typical eighth grader be then?


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## Sin-H (May 30, 2009)

blah said:


> I realize that was a stupid question. How old would the typical eighth grader be then?


hm. As far as I know, an eighth grader in the US is 13-14 years old...


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## Gparker (May 30, 2009)

Sin-H said:


> blah said:
> 
> 
> > I realize that was a stupid question. How old would the typical eighth grader be then?
> ...



Hes actually 15 because he started school late. And he got this as an extra credit problem as well. I'm not going to show him any of this work or ansewers. I just thought it would be fun if someone else tried.


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## Lord Voldemort (May 30, 2009)

Strange teachers...
I mean, do they expect him to be able to work that out himself?
Most people (before calculus) don't even know what a power series is.


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## jacob15728 (May 30, 2009)

Like I said, it was completely unnecessary. In fact, the sole reason I claimed to solve p=np in fifth grade was to make fun of Gparker, in case nobody noticed. It's generally not cool to lie.


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## Lord Voldemort (May 30, 2009)

When did Sin-H lie?
All he did was answer a math question and say that 8th graders are 13-14 years old.


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## jacob15728 (May 30, 2009)

Oops, I meant to say Gparker. My bad.


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## Sin-H (May 30, 2009)

true. because actually, there is no way that someone exspects a 13-14-15-year old to know something about power series...


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## Lord Voldemort (May 30, 2009)

Well, I'm doing Taylor's series as part of calculus next year and I'll be 15.
Maybe he's in a similar program?


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## jacob15728 (May 30, 2009)

Lord Voldemort said:


> Well, I'm doing Taylor's series as part of calculus next year and I'll be 15.
> Maybe he's in a similar program?



That problem is most likely more advanced than one would expect in AP calculus. I'm 15 and I know what a Taylor series is, but I don't know what a power series is nor do I have any idea what that problem is even talking about. This would have to be an 8th grader in Calculus 3, which would mean he's 6 years ahead of his age peers. This is so unlikely that we might as well assume he is lying.


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## Lord Voldemort (May 30, 2009)

But what would be his motive?
I mean, all would mean is that he's trying to make his friend sound smarter than he is. Unless it's his own stuff, which I doubt. He posted a first set of Pre-Algebra questions that would be quite simple is he's learning about Power Series and stuff like that.

@ qqwref -
I'm sure I'm missing something, or else you wouldn't have posted the problem. Is it something to do with the fact that you know that there will be at least one item in each box, since you're randomly selecting an item's box number, not just any box number? 

If you're computing the average items per box, isn't it just k/n?


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## Sin-H (May 30, 2009)

hm oh well. I also acquired my knowledge of these series when I was 15, but that also was part of a special programme. 
but my doubts are rather settled at the infinitesimal stuff. I think you need it to calculate taylor polynoms... and I have no idea how early you could learn this. It's not hard, but you still normally do it pretty late...


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## Gparker (May 30, 2009)

Well, whatever. I know im not lieing. He just gave me that problem since he said it would be fun to try. None of us are his math teacher. It would make more sense as to why it was extra credit. They probably have talked about it for a couple of days since finals were over. I'm not sure if he knows anything about it or not. Maybe the math teacher never gave extra credit and decided to give it out now and no one be able to do it.

@voldemort. How am i making him sound smarter? He doesnt even know anyone on these forums as far as my knowledge. He doesnt need to prove that hes smart on here, or tell me or someone else to say hes smart. He just gave me the problem and said he got it for extra credit, so i decided to post it here. Since i didnt know it was hard or anything, since i barely know abot this stuff. I guess I need to think more about what im typing since yall are bashing on me.


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## Sin-H (May 30, 2009)

Gparker said:


> I guess I need to think more about what im typing since yall are bashing on me.


don't take it too badly, my friend 
you have to understand it really sounds strange...


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## ThatGuy (May 30, 2009)

@jacob15728 It's not entirely impossible. Not to brag but I was and 8th grader in Calc 3. But that's only because my dad teaches math at the college and wanted to teach me seperately for math. But yes, it is quite unlikely. But depending on where he lives in the US and where he goes to school, it could be possible. Some private school are crazy.


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## jacob15728 (May 30, 2009)

ThatGuy said:


> @jacob15728 It's not entirely impossible. Not to brag but I was and 8th grader in Calc 3. But that's only because my dad teaches math at the college and wanted to teach me seperately for math. But yes, it is quite unlikely. But depending on where he lives in the US and where he goes to school, it could be possible. Some private school are crazy.



Now, I'm not saying that's impossible, but I must say I have trouble believing that. For an 8th grader to even be able to comprehend mathematics at that level, they would certainly need to have an IQ of above 175. I don't really like or agree with IQ testing, but I think that is a good estimation of what one would have to score to have that level of mathematical aptitude. If your IQ is this high, you wouldn't be in 8th grade at the standard age for 8th grade unless you were a mathematical savant. Were you in many other accelerated programs as well?


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## ThatGuy (May 30, 2009)

My IQ is certainly not >175. I think i'm ~145. (OK pulling a stereotype: I'm Chinese so that may have something to do with it.) I don't agree with IQ either. I enjoy math and I enjoy challenging myself. There is another person at my school who enjoys math and could've been the same level as me. 
The only thing was an elementary school that let you test out of 4th, 5th, 6th, and 7th grade. I skipped to 6th in 4th grade, but then when I was in 6th grade my parents didn't want to drive me to the middle school for 8th grade math so my dad kept me home and taught me every day for ~1 hour. I'm sure that if some people in my class had put in the same amount of time and effort as me, they would be the same level.


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## Sin-H (May 30, 2009)

Well, for example, I am 16 and I am just about to finish school, so I have already learned the maths for 18-year-olds and I have also read quite a bit about higher maths. But that is sort of my hobby. I don't really doubt the level someone can have at maths, but I doubt the fact that a teacher would teach something like that to such a pupil, even if it's especially for him...

oh I am a bit jealous about you Americans being able to take these and that courses and so on. But I am also not sure if that system is that good either :/


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## jacob15728 (May 30, 2009)

ThatGuy said:


> My IQ is certainly not >175. I think i'm ~145. (OK pulling a stereotype: I'm Chinese so that may have something to do with it.) I don't agree with IQ either. I enjoy math and I enjoy challenging myself. There is another person at my school who enjoys math and could've been the same level as me.
> The only thing was an elementary school that let you test out of 4th, 5th, 6th, and 7th grade. I skipped to 6th in 4th grade, but then when I was in 6th grade my parents didn't want to drive me to the middle school for 8th grade math so my dad kept me home and taught me every day for ~1 hour. I'm sure that if some people in my class had put in the same amount of time and effort as me, they would be the same level.



Alright, that's certainly impressive but I see your point. I personally despise mathematical education so I don't have much experience to judge what one can do with a teacher putting more time and effort in. I managed to teach myself basic calculus with no help, maybe if I had help and put more time into it I could have learned advanced calculus as well. I suppose anyone could learn more than they would normally be taught with the right effort and motivation.


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## qqwref (May 30, 2009)

Lord Voldemort said:


> @ qqwref -
> I'm sure I'm missing something, or else you wouldn't have posted the problem. Is it something to do with the fact that you know that there will be at least one item in each box, since you're randomly selecting an item's box number, not just any box number?
> 
> If you're computing the average items per box, isn't it just k/n?



Ah, but it's a bit more complicated than that. In the 'easy' problem, a box is more likely to be chosen when it has more objects in it (so that it is not k/n but actually a little more than that), and in the 'hard' one, you are only considering the non-empty boxes (but choosing the non-empty ones with equal probability), so the expected value is again greater than k/n.

For example, in the first problem, suppose you have 4 objects and 2 boxes, and 3 objects go in the first box. You'd be 3 times as likely to choose the first box, so the expected value of the number of objects in the box you found would be 3/4 (3) + 1/4 (1) = 2.5.

If you have 4 objects and 3 boxes, furthermore, and the distribution is 3 1 and 0 objects per box, for the 'easy' problem the expected value would again be 2.5, and for the hard problem it would be 1/2 (3) + 1/2 (1) = 2. You can see that both of these are higher than 4/3. Note that the problem is asking for the expected value over all possible circumstances, though, so this isn't the answer (although it might help).



jacob15728 said:


> For an 8th grader to even be able to comprehend mathematics at that level, they would certainly need to have an IQ of above 175.



Counterexample, in 8th grade I was doing Precalc in school (it had elements of calculus in it, though, and I remember we DID cover power series and convergence). My IQ is not as high as 175 of course 

Honestly I don't think it's all that difficult for an 8th grader to be able to understand that stuff, but they do have to be interested in math and willing to learn stuff on their own (unless they have teachers who are willing to teach them at any speed the student can handle). If it's an extra credit problem it might very well require some independent research, and even if an 8th grader doesn't know what a power series is, he probably will after Googling a tutorial on it  A motivated student who is talented in math should be able to understand this stuff with a bit of work.


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## peterbat (May 30, 2009)

ThatGuy said:


> Not to brag but I was and 8th grader in Calc 3.



Aren't you the guy who said angle C was 12, 24, and 3 degrees?


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## Sin-H (May 30, 2009)

qqwref said:


> Counterexample, in 8th grade I was doing Precalc in school (it had elements of calculus in it, though, and I remember we DID cover power series and convergence). My IQ is not as high as 175 of course


Then I would honestly like to see a way to calculate the first 5 terms of the power series of e^don't know what it was without using derivatives... that's the bigger problem, the way to calculate them using means of 8th graders... I only know that way...


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## jacob15728 (May 30, 2009)

Sin-H said:


> qqwref said:
> 
> 
> > Counterexample, in 8th grade I was doing Precalc in school (it had elements of calculus in it, though, and I remember we DID cover power series and convergence). My IQ is not as high as 175 of course
> ...



I think most 8th graders could easily learn how to calculate derivatives. But applying them to advanced problems like that is definately more difficult.


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## Gparker (May 30, 2009)

Sin-H said:


> Gparker said:
> 
> 
> > I guess I need to think more about what im typing since yall are bashing on me.
> ...



I know it probably does. But i guess thats how virginia goes 

When he gave me the problem, i had almost no idea what it was. I tried some stuff but failed.


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## ThatGuy (May 30, 2009)

peterbat said:


> ThatGuy said:
> 
> 
> > Not to brag but I was and 8th grader in Calc 3.
> ...



I have no clue why i typed 3, it was supposed to be 8. But yeah, 12 was eyeballing, 24 was being lazy, and 8 was actually doing it.


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## qqwref (May 31, 2009)

Sin-H said:


> Then I would honestly like to see a way to calculate the first 5 terms of the power series of e^don't know what it was without using derivatives... that's the bigger problem, the way to calculate them using means of 8th graders... I only know that way...



The power series of e^x is 1 + x + x^2/2! + x^3/3! + ..., right? And you can plug anything in for x. So e^(2x^2) is 1 + (2x^2) + (2x^2)^2/2 + (2x^2)^3/3! + .... If you know the power series of f(x), you can computer a lot of related ones, and it's not hard to look up the power series for e^x, sin, cos, ln(x+1), and so on.


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## fanwuq (May 31, 2009)

ThatGuy said:


> peterbat said:
> 
> 
> > ThatGuy said:
> ...



Lolwut?
I laugh at people who take more advanced math than me and only get 500 on SAT math and have trouble with simple calculations. They think they are math geniuses, but they don't have common sense. I never skipped ahead in any of my math course (algebra 1 in 8th grade; I actually learned all of that by myself in 6th grade, but after that, I stopped reading ahead for math. I might do it again this summer for Calc BC if I have the time.); however, I do very well in math contests with simple and clever problems. Intuitive problems are so much better than learning formulas and theorems that you don't actually understand. I can proof every formula I know.
I suppose this also correlates to one's cubing style. I like simple elegant math contest problems, and I try to come up with such FMC solutions intuitively. Others may like very advanced topics that they don't fully understand and prefer memorizing large amounts of algorithms.
I never took Precalc, I went from Algebra 1 to geometry to algebra 2 + trigonometry + statistics to calculus AB.


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## Lord Voldemort (May 31, 2009)

Gparker said:


> @voldemort. How am i making him sound smarter?



I never mentioned that was a bad thing... I think I need to be more clear about these things. You seem to think I hate you or something :confused:



Sin-H said:


> oh I am a bit jealous about you Americans being able to take these and that courses and so on. But I am also not sure if that system is that good either :/



Well, American schools on average are apparently well behind European schools, so much that if a foreign exchange student comes here for a year, they have to take the year again once they go back. For German and French students at least, so I was assuming most of Europe. 



qqwref said:


> If you have 4 objects and 3 boxes, furthermore, and the distribution is 3 1 and 0 objects per box, for the 'easy' problem the expected value would again be 2.5, and for the hard problem it would be 1/2 (3) + 1/2 (1) = 2. You can see that both of these are higher than 4/3. Note that the problem is asking for the expected value over all possible circumstances, though, so this isn't the answer (although it might help).



What about this? I actually found this before I read your post. If there's k/n "base" expected value, and you know there must be an item in the box, that adds 1/n, right? So (k+1)/n? 

@fanwuq - How old are you?
Your school just let you skip Precalculus?

One more thing for anyone that's done with school/college - do you ever actually use Calculus in a job? We grow up and the math teachers make up random excuses as to why certain math is necessary, and yet a lot of it doesn't make sense. What, for example, is the practical purpose of finding the non real roots of the cube root of two? Is the useful stuff in calculus?


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## fanwuq (May 31, 2009)

Lord Voldemort said:


> @fanwuq - How old are you?
> Your school just let you skip Precalculus?





fanwuq said:


> I never took Precalc, I went from Algebra 1 to geometry to algebra 2 + trigonometry + statistics to calculus AB.



17. Just finished 11th grade. 
8th grade: algebra 1
9th grade: geometry 
10th grade: algebra 2 + trigonometry + statistics
11th grade: calculus AB

I've basically relaxed through all of my math classes until last year. Trig and calculus AB are the only math classes that I actually had to learn from the teacher. On everything else, I got over 85% on the diagnostic test before I started the course. I guess I never bother to get out of them because there was always something contained in the course that I didn't know.


edit: 
Voldemort: I'm not going to be a math or physics major, so I would not use calculus after college. For engineering, you might be able to use some of it, but for most people, calculus is practically useless after you leave school.


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## Gparker (May 31, 2009)

Lord Voldemort said:


> Gparker said:
> 
> 
> > @voldemort. How am i making him sound smarter?
> ...



I hate no one  . It did kind of sound like you thought that i was "advertising" for him being smart.


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## Poke (May 31, 2009)

I am trapped in Algerbra 1 until next year, but I can do arithmetic and many calculations like mad. I made the only 100% on my Standards-Of-Learning(SOL) exams without a calculator.(Have been the only 100 several other times... getting a 99.4% and needed a challenge.)


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## ManasijV (May 31, 2009)

Sin-H said:


> true. because actually, there is no way that someone exspects a 13-14-15-year old to know something about power series...



well i knew 2 months before my 16th birthday


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## blah (May 31, 2009)

ManasijV said:


> well i knew 2 months before my 16th birthday



Dude you're in India. Everyone from this part of the world knows this sort of stuff.


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## Sin-H (May 31, 2009)

blah said:


> ManasijV said:
> 
> 
> > well i knew 2 months before my 16th birthday
> ...


as I said, I also knew it when I was 15, but as I said - extracurricular...

@qqref: of course that's possible. but looking this up is quite of a cheap way to do it. still legitimate, of course


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## ManasijV (May 31, 2009)

Sin-H said:


> blah said:
> 
> 
> > ManasijV said:
> ...



This isn't topic related. But Simon did you know you are my favourite cuber?  And I requested more walkthrough or slow motion solves from you  Anytime soon?


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## Sin-H (May 31, 2009)

ManasijV said:


> Sin-H said:
> 
> 
> > blah said:
> ...



who the hell is Simon?


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## ManasijV (May 31, 2009)

Sin-H said:


> ManasijV said:
> 
> 
> > Sin-H said:
> ...


holy ****! cant believe i made that mistake  sorry the Sin H somehow led me in the wrong way. I meant Stefan ofcourse. I'm really sorry  some fan eh?


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## jacob15728 (May 31, 2009)

Poke said:


> I am trapped in Algerbra 1 until next year, but I can do arithmetic and many calculations like mad. I made the only 100% on my Standards-Of-Learning(SOL) exams without a calculator.(Have been the only 100 several other times... getting a 99.4% and needed a challenge.)



Nice job. I am also stuck in algebra 1... pretty depressing.

anyway, I did some research last night, and I now have a vague understanding of the problem that Gparker posted. It's not quite as complicated as I thought it would be, but I still remain adamant that expecting an 8th grader to solve it is absurd. 

Here's what I found out. A power series is a form of infinite series in X. For any value of X, the series can converge to a limit or diverge (the value of the limit is infinite or undefined). So for that particular series, the X values converge to a limit for an interval, and the problem is asking you to find it. This requires the use of relatively advanced calculus that I have a very vague understanding of.


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## blah (May 31, 2009)

Sin-H said:


> who the hell is Simon?



Ouch. (too short)


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## mazei (May 31, 2009)

i feel so dumb...16 n im blur about half of the stuff talked about in the last few pages..some math enthusiast i am...


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## Sin-H (May 31, 2009)

ManasijV said:


> Sin-H said:
> 
> 
> > ManasijV said:
> ...


 I kinda knew that. but I just had to ask that question  some more walkthroughs are definitely on my list of to-do-videos. but I don't cube much these days... I hope I can do some more  and thanks, of course =)


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## ManasijV (Jun 1, 2009)

You don't have your holidays now? And one thing I don't like is your finger trick for R(a)  Must be the only thing I can do faster than you  The double U turn with the left hand is much faster I think!


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## moogra (Jun 1, 2009)

jacob15728 said:


> Poke said:
> 
> 
> > I am trapped in Algerbra 1 until next year, but I can do arithmetic and many calculations like mad. I made the only 100% on my Standards-Of-Learning(SOL) exams without a calculator.(Have been the only 100 several other times... getting a 99.4% and needed a challenge.)
> ...



Since when were power series advanced calculus? They're not much different from Taylor series since most are derived from taylor series: i.e. sin x, cos x, e^x, ln(x), arctan(x)

@ThatGuy
This is all covered in BC Calculus, which is definitely not Calculus 3.

I also find that school math is a bad representation on how good one is in math. Everyone can get an A if they try in school. In Contest math, not everyone can succeed.


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## ThatGuy (Jun 1, 2009)

moogra said:


> Since when were power series advanced calculus? They're not much different from Taylor series since most are derived from taylor series: i.e. sin x, cos x, e^x, ln(x), arctan(x)
> 
> @ThatGuy
> This is all covered in BC Calculus, which is definitely not Calculus 3.
> ...



(I think my quoting failed. IT WORKED!)
BC only goes up the infinite series, but three includes polar stuff (ARGGGG I HATE IT) But yeah, everyone can get an A in school. I completely agree with you in contest math, which is why I started this year, which is also why I stopped pursuing Calc stuff and started Number Theory (sentence fail).


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## JBCM627 (Jun 1, 2009)

ThatGuy said:


> BC only goes up the infinite series, but three includes polar stuff (ARGGGG I HATE IT) But yeah, everyone can get an A in school. I completely agree with you in contest math, which is why I started this year, which is also why I stopped pursuing Calc stuff and started Number Theory (sentence fail).



BC definitely covers 'polar stuff', and even touches on some multivariable subjects. I personally found these topics a bit less dry, and enjoyed them more. And unfortunately I was pretty bored when I took a class that covered some number theory... hopefully you find it interesting, though.

And LOL @ LOLCODE


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## jacob15728 (Jun 1, 2009)

JBCM627 said:


> ThatGuy said:
> 
> 
> > BC only goes up the infinite series, but three includes polar stuff (ARGGGG I HATE IT) But yeah, everyone can get an A in school. I completely agree with you in contest math, which is why I started this year, which is also why I stopped pursuing Calc stuff and started Number Theory (sentence fail).
> ...



Alright, I guess I wouldn't really know as I haven't taken that class. I just assumed BC would not go up to series as I personally find them more difficult and confusing than other aspects of calculus.


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## Lord Voldemort (Jun 2, 2009)

Yay, my calculus book came today


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## moogra (Jun 3, 2009)

BC definitely has series as #6 this year on the AP test was about series.


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## jacob15728 (Jun 3, 2009)

Lord Voldemort said:


> Yay, my calculus book came today



Cool, what book did you get?


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## Lord Voldemort (Jun 3, 2009)

It the same one I need for my course next year, but I decided to buy it early. 
http://www.amazon.com/gp/product/0534409865/ref=ox_ya_oh_product

I'm on the 2nd chapter so far, just learning about limits and derivatives.


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## Bob (Jun 3, 2009)

I like james stewart. I used 4th edition Stewart in the calc course i taught this year

http://www.amazon.com/Calculus-James-Stewart/dp/0534359493


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