The DW Math Thread

Arctix - Dreamweaver · April 2013

Synta - Dreamweaver wrote: »

Uh, hmm

I like this proof, but it's a bit circular /O\

shaddup u r supposed to b perplexed by that problem D<

Veneir - Dreamweaver · May 2013

Needs more maths!
I'm doing maths reviews today for my final which is tomorrow, so I'll go ahead and throw some of those problems at the thread

Of course, that means lots of them won't be in word problem forme, but ah well; I'll fix that for a few if I can think of a way to words them~
Don't mind my little [trying to be helpful notes]
Ignore the numbers in (black font) they're there for my reference ahaha 'x3
Solving for x [insert tilde here aka basic stuffs~]:
(5) 6 - (6x+4) =18-8x

(6) x/5 = x/2 - 3 [fractions! LCDs!

]

(9) f(-3) = x^2 + 1 [f(#) is f(x) and the number in the (s) is what's plugged in for x

Solving in function forms~]

Domains:
(10 exe.v.) In the function f(x)=3x - 9, x can equal all real numbers [aka x|x is a real number], and no real numbers are outside of the domain. However, when f(3), the function solves to be 0. Why is this applicable and still inside the domain?

Solving systems to get an ordered pair of (x,y): [takes two lines to type one problem!]
2x - 3y = 2
5x + 5y = 5 (16)
[No there should not be any 'gross' decimals in that problem x3 No x or y already has a 1 in front of it so tis 'addition method'.]

Supply models..... sort of. Not really. Yeah it's really not. OTL:
Decus wants to bring cookies to an event, where the amount of cookies he wants to bring is represented by C=13(b)+105, with b being the amount of burnt cookies per dozen-- for some strange reason, b=3, but we'll just go with that. But Waffles, who only cooks old and burnt cookies, won't allow for this and says the amount of cookies should be C=15(6)+78! (No the ! is not a maths symbol 3:<) How many cookies was Decus originally planning on bringing, and how many were burnt? What fraction of the total amount of cookies does Waffles say have to be burnt?

Nariin - Dreamweaver · May 2013

Since nobody wants to solve those, I shall.

---
6 - (6x+4) =18-8x
8x-6x = 18-6+4
2x = 16
x = 8

---
x/5 = x/2 - 3
x/5 - x/2 = -3
(2x-5x)/10 = -3
-3x/10 = -3
3x = 30
x = 10

---
f(-3) = x^2 + 1 ...I assume just insert the x value?
f(-3) = (-3)^2 + 1
f(-3) = 10

---
Uh.. I don't understand the question at f(x)=3x - 9

---
2x - 3y = 2
5x + 5y = 5
Well, that clearly is 1 for x and 0 for y

And the equation stuffies should be um... Inserting one into another perhaps.
2x = 3y + 2
5y = 5 - 5x

2x = 3(5-5x) + 2
2x = 15 - 15x + 2
17x = 17
x = 1

5y = 5 - 5
5y = 0
y = 0

--
dozen = 12?

C = 13(3) + 105 = 144

C = 15(6) + 78 = 168

x/12 = 6
x = 72 ... cookies need to be burned by Waffle's demand... uh... 42,9%?

Decus was planning to bring 144 cookies, perhaps, 3 were burnt?

Nop... no idea what I'm talking about or what this task is about

Decus - Dreamweaver · May 2013

Veneir - Dreamweaver wrote: »

Domains:
(10 exe.v.) In the function f(x)=3x - 9, x can equal all real numbers [aka x|x is a real number], and no real numbers are outside of the domain. However, when f(3), the function solves to be 0. Why is this applicable and still inside the domain?

Because 3 is a real number? Am I misunderstanding? Also, I think you're confusing real numbers and rational numbers. Rationals are intuitively said to be "any number that can be expressed as a ratio of two numbers" (EDIT: So long as the denominator isn't zero). Reals are somewhat simpler: "any number that can be a point on a continuous line."

f(x) is linear and thus (x,y) ∈ R

Ergo, (3,0) is within the set of all real numbers.

Veneir - Dreamweaver · May 2013

Decus - Dreamweaver wrote: »

Because 3 is a real number? Am I misunderstanding? Also, I think you're confusing real numbers and rational numbers. Rationals are intuitively said to be "any number that can be expressed as a ratio of two numbers" (EDIT: So long as the denominator isn't zero). Reals are somewhat simpler: "any number that can be a point on a continuous line."

f(x) is linear and thus (x,y) ∈ R

Ergo, (3,0) is within the set of all real numbers.

Nono you're not misunderstanding my dear; it was more a 'simple explanation' problem than anything, and I suppose it would have had more than one explanations. I sort of thought of when you made the problem with Synta being perplexed at a maths proof/rule.
What I was thinking of when typing that was that the denominator wasn't 0, so it wasn't something/0, but other logics would work

(referring to edits!!)

Nariin - Dreamweaver wrote: »

Since nobody wants to solve those, I shall.

I'll go into my maths thing to double check the answers, Nariin [I already did them twice I don't want to do them again ahhh

] but they all look correct

Nariin - Dreamweaver wrote: »

dozen = 12?

C = 13(3) + 105 = 144

C = 15(6) + 78 = 168

x/12 = 6
x = 72 ... cookies need to be burned by Waffle's demand... uh... 42,9%?

Decus was planning to bring 144 cookies, perhaps, 3 were burnt?

Nop... no idea what I'm talking about or what this task is about

And yeah the supply model attempt problem was pretty much an abomination and for the life of me I couldn't manage to turn it into something that worked how I wanted it to Dyna-style-problem-wise OTL but you did get where I was going with it, yes x3

Oliiander - Lost City · May 2013

Olii isn't a fan of maths, nor is she any good with it. So when thrown a particularly difficult question regarding functions and whatnot, it's taking everything I have to not hang myself with a shoelace. If anyone can shed some light on this, it would be much appreciated >:

f(x+1) = f(x) - f(x-1) + 1
f(2) = 8
f(1) = 2

Calculate: f(2013)

Functions make my head spin Q.Q

Also!

ab = (a/b) + (b/a)

Calculate: (a+b)(a-b)

Thank you for your time <:

Veneir - Dreamweaver · May 2013

Bah I can't even check for f(2) = 8 and get 8 right now OTL *just got done cleaning pelt so not really thinky* I tried putting in the 2 for f in three different ways[wut] and got -2, 3, and.... f(2)+1=0, not 8, so I'm really not help right now, I'm so sorry ahhhh.
Edit: Now four and I still can't get 8 hdfhjfgj. 2/3=/=8 OTL
*pesters Decusy*

Oliiander - Lost City · May 2013

Veneir - Dreamweaver wrote: »

Bah I can't even check for f(2) = 8 and get 8 right now OTL *just got done cleaning pelt so not really thinky* I tried putting in the 2 for f in three different ways[wut] and got -2, 3, and.... f(2)+1=0, not 8, so I'm really not help right now, I'm so sorry ahhhh.
Edit: Now four and I still can't get 8 hdfhjfgj. 2/3=/=8 OTL
*pesters Decusy*

You can make more sense from it than I can.
I tackled it again and once more, didn't know how to start. Q.Q

Decus - Dreamweaver · May 2013

Oliiander - Lost City wrote: »

Olii isn't a fan of maths, nor is she any good with it. So when thrown a particularly difficult question regarding functions and whatnot, it's taking everything I have to not hang myself with a shoelace. If anyone can shed some light on this, it would be much appreciated >:

f(x+1) = f(x) - f(x-1) + 1
f(2) = 8
f(1) = 2

Calculate: f(2013)

Functions make my head spin Q.Q

Also!

ab = (a/b) + (b/a)

Calculate: (a+b)(a-b)

Thank you for your time <:

(I'll use asterisks as references, just to make it easier to read.)

So for the first, it's a recursive sequence:

f(2) = 8, so...

f(2+1) = f(2) - f(2-1) + 1 --> f(3) = f(2) - f(1) + 1
**f(3) = 8 - 2 + 1 = 7

Keep going until you see some sort of pattern develop...

f(4) = f(3) - f(2) + 1 = 7 - 8 + 1 = 0
f(5) = f(4) - f(3) + 1 = 0 - 7 + 1 = -6
f(6) = f(5) - f(4) + 1 = -6 -0 + 1 = -5
f(7) = f(6) - f(5) + 1 = -5 - -6 + 1 = 2 (Looks like it may start repeating!)
f(8) = f(7) - f(6) + 1 = 2 - -5 + 1 = 8 (Hmm... let's keep going to verify)
f(9) = f(8) - f(7) + 1 = 8 - 2 + 1 (Yep, this is identical to **)

So really, we have six unique terms: {2, 8, 7, -6, -5, 0}

Thus, every 6th term is zero. Now, 2012 isn't divisible by 6, but 2010 is; so, f(2011) would be the start of a new repeated set. Ergo, f(2011) = 2, f(2012) = 8, and f(2013) = 7. Another way to do that would be to just divide 2012 by 6, and see that the remainder is one third, meaning that it's [some multiple of 6] + 1/3 (= 2/6). Therefore, 2012 would have been the second term of the new set.

[Working on the second one but I was asked to post the first.]

Decus - Dreamweaver · May 2013

I tried this problem two different ways, and I couldn't get a nice, neat solution. The one I got is pretty ugly, which tells me I've done something wrong and/or missed something fundamental. As it's now 1:30 am, I think it's best if I return to this at a better time. I can figure it out, I just need some time. I'll leave my current work up in case it helps you out.

If ab = (a/b) + (b/a), then a = a/b^2 + 1/a --> (a^2 + b^2) / ab^2 and b = b/a^2 + 1/b --> (b^2 + a^2) / ba^2 (Just combining the fractions)

Now remember that (a + b)(a - b) is just a difference of two squares:

***(a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - b^2

Based on our above definitions of a and b...

a = (a^2 + b^2) / ab^2
a^2 = (a^2 + b^2)^2 / (ab^2)^2

b = (b^2 + a^2) / ba^2
b^2 = (b^2 + a^2)^2 / (ba^2)^2 (Same numerator as before since it's the same operation.)

****For sake of ease when writing, I'll make (a^2 + b^2)^2 = C. It's getting tedious to re-type it every time ;-;

So, according to ***, we have...

a^2 - b^2
[C / (ab^2)^2] - [C / (ba^2)^2]

Putting that under one denominator gives us:

[C*(a^2) - C*(b^2)] / (ab)^4

Which factors to:

C[a^2 - b^2] / (ab)^4 --> { (a^2 + b^2)^2 * [a^2 - b^2] } / (ab)^4 (From my shortcut at ****)

Looking at the numerator from the step above:

(a^2 + b^2)^2 * [a^2 - b^2] ~~> (a^2 + b^2) * [ (a^2 + b^2)*(a^2 - b^2) ] (Just re-writing it a different way. Look! Another difference of two squares!)

(a^2 + b^2) * [ (a^2 + b^2)*(a^2 - b^2) ] --> (a^2 + b^2) * [ a^4 - b^4] --> (a^6 -a^2*b^4 + a^4*b^2 - b^6)

[Work in progress.]

Arctix - Dreamweaver · May 2013

Decus - Dreamweaver wrote: »

Answer

I like to fiddle around LHS n RHS alot,so here's my approach {also a2=a squared n b2= b squared }

ab=a/b +b/a; now (a+b)(a-b)=a2 - b2=[ (a/b)2 - 1 ]b2.......*

by the equation ab= [ a2 + b2 ]/ab
-> (ab)2= a2 + b2
-> a2 + b2 - (ab)2 = 0
->a2 + b2(1 - a2) = 0
-> a2=b2(a2 - 1)
-> (a/b)2 = a2 - 1,..........**

replace (a/b)2 value from ** in *

(a + b)(a - b) = [(a2 - 1) -1] b2 = a2b2 - 2xb2 ( a-squared b-squared - two b-squared )

its late night for me too n I am sorry if I introduced 0/0 somewhere b:cute

Oliiander - Lost City · May 2013

Decus - Dreamweaver wrote: »

(I'll use asterisks as references, just to make it easier to read.)

So for the first, it's a recursive sequence:

f(2) = 8, so...

f(2+1) = f(2) - f(2-1) + 1 --> f(3) = f(2) - f(1) + 1
**f(3) = 8 - 2 + 1 = 7

Keep going until you see some sort of pattern develop...

f(4) = f(3) - f(2) + 1 = 7 - 8 + 1 = 0
f(5) = f(4) - f(3) + 1 = 0 - 7 + 1 = -6
f(6) = f(5) - f(4) + 1 = -6 -0 + 1 = -5
f(7) = f(6) - f(5) + 1 = -5 - -6 + 1 = 2 (Looks like it may start repeating!)
f(8) = f(7) - f(6) + 1 = 2 - -5 + 1 = 8 (Hmm... let's keep going to verify)
f(9) = f(8) - f(7) + 1 = 8 - 2 + 1 (Yep, this is identical to **)

So really, we have six unique terms: {2, 8, 7, -6, -5, 0}

Thus, every 6th term is zero. Now, 2012 isn't divisible by 6, but 2010 is; so, f(2011) would be the start of a new repeated set. Ergo, f(2011) = 2, f(2012) = 8, and f(2013) = 7. Another way to do that would be to just divide 2012 by 6, and see that the remainder is one third, meaning that it's [some multiple of 6] + 1/3 (= 2/6). Therefore, 2012 would have been the second term of the new set.

[Working on the second one but I was asked to post the first.]

That makes some form of sense! \o/
Thank you so much!

Decus - Dreamweaver wrote: »

I tried this problem two different ways, and I couldn't get a nice, neat solution. The one I got is pretty ugly, which tells me I've done something wrong and/or missed something fundamental. As it's now 1:30 am, I think it's best if I return to this at a better time. I can figure it out, I just need some time. I'll leave my current work up in case it helps you out.

If ab = (a/b) + (b/a), then a = a/b^2 + 1/a --> (a^2 + b^2) / ab^2 and b = b/a^2 + 1/b --> (b^2 + a^2) / ba^2 (Just combining the fractions)

Now remember that (a + b)(a - b) is just a difference of two squares:

***(a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - b^2

Based on our above definitions of a and b...

a = (a^2 + b^2) / ab^2
a^2 = (a^2 + b^2)^2 / (ab^2)^2

b = (b^2 + a^2) / ba^2
b^2 = (b^2 + a^2)^2 / (ba^2)^2 (Same numerator as before since it's the same operation.)

****For sake of ease when writing, I'll make (a^2 + b^2)^2 = C. It's getting tedious to re-type it every time ;-;

So, according to ***, we have...

a^2 - b^2
[C / (ab^2)^2] - [C / (ba^2)^2]

Putting that under one denominator gives us:

[C*(a^2) - C*(b^2)] / (ab)^4

Which factors to:

C[a^2 - b^2] / (ab)^4 --> { (a^2 + b^2)^2 * [a^2 - b^2] } / (ab)^4 (From my shortcut at ****)

Looking at the numerator from the step above:

(a^2 + b^2)^2 * [a^2 - b^2] ~~> (a^2 + b^2) * [ (a^2 + b^2)*(a^2 - b^2) ] (Just re-writing it a different way. Look! Another difference of two squares!)

(a^2 + b^2) * [ (a^2 + b^2)*(a^2 - b^2) ] --> (a^2 + b^2) * [ a^4 - b^4] --> (a^6 -a^2*b^4 + a^4*b^2 - b^6)

[Work in progress.]

Thank you for looking at the question >.<
This one made a little sense, then the sense took a stroll off the roof after *** >.>

Decus - Dreamweaver · May 2013

Arctix - Dreamweaver wrote: »

(a + b)(a - b) = [(a2 - 1) -1] b2 = a2b2 - 2xb2 ( a-squared b-squared - two b-squared )

its late night for me too n I am sorry if I introduced 0/0 somewhere b:cute

Yea I got that the second time I took a stab at this, but I didn't post it. I still do not think that is the proper solution.

Oliiander - Lost City wrote: »

Thank you for looking at the question >.<
This one made a little sense, then the sense took a stroll off the roof after *** >.>

It's awfully convoluted, which means it's wrong. I've still not figured it out. Do you have any idea in what form the final answer would be? As I said before, given that I've not found something succinct and neat, I've either missed something fundamental or there's a problem in my work/the problem.

Oliiander - Lost City · May 2013

Decus - Dreamweaver wrote: »

It's awfully convoluted, which means it's wrong. I've still not figured it out. Do you have any idea in what form the final answer would be? As I said before, given that I've not found something succinct and neat, I've either missed something fundamental or there's a problem in my work/the problem.

It was part of a test that I didn't take, so even if there are solutions sent to us, I wouldn't have copy ;~;
I'm going to blame the question and sweep this one under that carpet. thank you for helping with the first question! f:grin

Arctix - Dreamweaver · May 2013

Decus - Dreamweaver wrote: »

Yea I got that the second time I took a stab at this, but I didn't post it. I still do not think that is the proper solution.

It's awfully convoluted, which means it's wrong. I've still not figured it out. Do you have any idea in what form the final answer would be? As I said before, given that I've not found something succinct and neat, I've either missed something fundamental or there's a problem in my work/the problem.

Alright one more try with introduction of square roots [ a2 = a squared; r= root ]

ab = a/b + b/a ;

-> (ab)2 = a2 + b2 {as before}

->a2b2 + 2ab = a2 + b2 + 2ab

->a2b2 + 2ab = (a + b)2

-> r[a2b2 + 2ab] = a+b ......*

-> r[a2b2 - 2ab] = a-b ........**do as above with - 2ab

now (a+b)(a-b) = r[ (a2b2 + 2ab)(a2b2 - 2ab) ]

I dont think so I've done a miscalculation here

Superfeng - Lost City · May 2013

You have ab = a/b + b/a

This becomes (a^2)(b^2) = a^2 + b^2 as everyone else has observed.

Now, (a^2)(b^2) - a^2 - b^2 = 0

--> (a^2)[(b^2) - 1] - b^2 + 1 = 1

--> (a^2)[(b^2) - 1] - [(b^2) - 1] = 1

--> [(a^2) - 1][(b^2) - 1] = 1

So either, (a^2) - 1 = 1, (b^2) - 1 = 1

or (a^2) - 1 = -1, (b^2) - 1 = -1

If the latter case were true, then it'd imply a^2 = 0, so a = 0, and similarly b = 0.
But in your original problem, you are dividing by a and b, which means a and b cannot be 0.

Now the first case would imply that (a^2) = 2, (b^2) = 2

You are looking for (a+b)(a-b) = a^2 - b^2 = 2 - 2 = 0.

So yeah the answer is 0.

Superfeng - Lost City · May 2013

Give me some real math problems please.

Decus - Dreamweaver wrote: »

No one wants to do math anymore. Where are Fizban, Cheze, and Yorkin?

More 5th year grade math:

Fissile - Archosaur · May 2013

Superfeng - Lost City wrote: »

You have ab = a/b + b/a

This becomes (a^2)(b^2) = a^2 + b^2 as everyone else has observed.

Now, (a^2)(b^2) - a^2 - b^2 = 0

--> (a^2)[(b^2) - 1] - b^2 + 1 = 1

--> (a^2)[(b^2) - 1] - [(b^2) - 1] = 1

--> [(a^2) - 1][(b^2) - 1] = 1

So either, (a^2) - 1 = 1, (b^2) - 1 = 1

or (a^2) - 1 = -1, (b^2) - 1 = -1

If the latter case were true, then it'd imply a^2 = 0, so a = 0, and similarly b = 0.
But in your original problem, you are dividing by a and b, which means a and b cannot be 0.

Now the first case would imply that (a^2) = 2, (b^2) = 2

You are looking for (a+b)(a-b) = a^2 - b^2 = 2 - 2 = 0.

So yeah the answer is 0.

Lol what?

let x = a*a - 1 and y = b*b - 1

You are saying if xy = 1, then x or y must be 1 or -1 and that just makes no sense.

If you factor it to (a+1)(a-1)(b+1)(b-1)=1 you can get forms for a and b like a = 1/[(a-1)(b+1)(b-1)] - 1, then plug those into (a+b)(a-b) and see what you get. I'm guessing the answer is not 0.

Superfeng - Lost City · May 2013

I might have umm, assumed a and b were integers b:surrender

HrunsPanda - Archosaur · May 2013

Is this the "help with my homework" thread ? b:laugh

darknessofmy · August 2013

6+3+6+3+6+3+6=6666666

BloodyOne - Dreamweaver · August 2013

I Cant beleive this thread is still going on lol

ColdAsLife - Dreamweaver · October 2013

BloodyOne - Dreamweaver wrote: »

I Cant beleive this thread is still going on lol

I hear you, man, but I'm happy about it. Decus deserves even better! b:thanks

Decus - Dreamweaver · November 2013

ColdAsLife - Dreamweaver wrote: »

I hear you, man, but I'm happy about it. Decus deserves even better! b:thanks

It's nice to see you still around. You're very kind. Thank you, Grim.

As for this thread, I've not actually received any questions lately. Alongside that, I've been focusing on some more technical things as of late and so math sort of falls to the wayside

opkossy · January 2014

Decus - Dreamweaver wrote: »

PS. You can un-sticky it since I don't plan to really keep track of it and no one asked anything for a long time.

DW doesn't get any more math sticky. b:cry

Necro period begins as of this post.

Veneir - Dreamweaver · January 2014

OPKossy wrote: »

DW doesn't get any more math sticky. b:cry

Necro period begins as of this post.

*sniffles* My maths class was starting up in a few days since classes next week, too >:

opkossy · January 2014

We must stalk Decus. For maths... for honor.... FOR POST COUNTS!

ovenusarmanio · January 2014

Nuuu *tries to put the duct tape back on the thread but it isn't sticky anymore* b:cry

torquoisegamer · November 2014

Q____Q - Dreamweaver wrote: »

a lot? you should've gave me a map =[
is this a map? http://www.ducksters.com/geography/south_america_map_coloring.jpg

EDIT:! oh I THINK THE ANSWER IS 4! IDK :D here's my map: http://i.imgur.com/vn0kgJ3.png
ps i didn't know that south america had like 10 countries

it's like that thread on dyna forums with the math clock... and how the 7 on clock was 6.9 and the nine had the line thingy on it.... OTL he will never understand!

Four is correct, since Decus doesn't seem to have responded. Four color theorem states that on any flat surface (and since South America doesn't have any fragmented countries such as Russia, South America can be treated as a flat surface), a maximum of four colors is needed to fill each area with a country so that no two countries are touching each other.

SweetieBot - Lothranis · November 2014

torquoisegamer wrote: »

Four is correct, since Decus doesn't seem to have responded. Four color theorem states that on any flat surface (and since South America doesn't have any fragmented countries such as Russia, South America can be treated as a flat surface), a maximum of four colors is needed to fill each area with a country so that no two countries are touching each other.

This looks like a NECRO!

torquoisegamer replied to a message that was 10 months 19 days 34 minutes old.

Any thread over one month (30 days) old is considered to be a dead thread and you're not supposed to post in them. The person you are replying to probably doesn't care any more or can no longer be found on the forums. The topic itself could be out of date. Next time just make a new thread.

Let's see how long it takes for a mod to close this

The DW Math Thread

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