Here’s a simple arithmetic question:

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

A bat and ball cost a dollar and ten cents. The bat costs a dollar more than the ball.
How much does the ball cost? xD

inky · **Joined:** Nov 2006 **Posts:** 4025 **Location:**

5 cents

PureStr · **Joined:** Nov 2007 **Posts:** 1244 **Location:** GTA5

x = cost of ball
y = cost of bat

1) x+y=1.10
2) y=x+1

so: x+(x+1)=1.10 then 2x+1=1.10 so 2x = 1.10-1 then 2x = 0.10 so x = 0.05

Bring it on bro.
My 5 cents. (see what i did there?)

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

Here's One More
Square root(x)/3 = 243/x value of x ?

The Invisible · **Joined:** Jan 2011 **Posts:** 2626 **Location:** Home ._.

*BlackFox wrote:

Here's One More
Square root(x)/3 = 243/x value of x ?

That would be 81

Still don't see where this is going

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

The Invisible wrote:

That would be 81

^ FAIL!
Well... Just wait and see

Shomari · **Joined:** Jul 2007 **Posts:** 2341 **Location:** Limbo

Whats 2+2

The Invisible · **Joined:** Jan 2011 **Posts:** 2626 **Location:** Home ._.

*BlackFox wrote:

The Invisible wrote:

That would be 81

^ FAIL!
Well... Just wait and see

If the question is : (square root(x))/3 = 243 / x, then the answer is : 81

Edit: I wanted to get sure so i checked it just now

http://www.wolframalpha.com/input/?i=Sq ... 3D+243%2Fx

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

^ ROFL
But.. still Not the answer I wanted. lulz

poehalcho · **Posted:** Fri Jun 15, 2012 9:30 pm

*BlackFox wrote:

^ ROFL
But.. still Not the answer I wanted. lulz

81
It is the answer. Otherwise you've failed at phrasing it correctly.

perhaps 56.1622.... is what you're looking for, but that's not what you asked.
apparently that equals 27*3^(2/3).

Anyway, while were at it.

Prove that 0.999... = 1

CrimsonNuker · **Joined:** Aug 2006 **Posts:** 13791 **Location:**

Thank God nobody said a dollar :palm:

Hapjap · **Posted:** Sat Jun 16, 2012 1:29 am

poehalcho wrote:

*BlackFox wrote:

^ ROFL
But.. still Not the answer I wanted. lulz

81
It is the answer. Otherwise you've failed at phrasing it correctly.

perhaps 56.1622.... is what you're looking for, but that's not what you asked.
apparently that equals 27*3^(2/3).

Anyway, while were at it.

Prove that 0.999... = 1

penfold1992 · **Joined:** Apr 2007 **Posts:** 4060 **Location:**

*BlackFox wrote:

^ ROFL
But.. still Not the answer I wanted. lulz

+ or - 81?

EDIT: no that is wrong... its just 81. bad maff my bad!

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

What about this answer?
x = -9
or
x = 9

Phaedra · **Joined:** May 2009 **Posts:** 730 **Location:** Yes.

*BlackFox wrote:

What about this answer?
x = -9
or
x = 9

uhm, no

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

Damn.. It wasn't meant to be a Mathematics thread.
But I'm Just curious about this.

3 *x*x=243
x*x(squared)=243/3
x*x=81
x=9
?

poehalcho · **Posted:** Sat Jun 16, 2012 1:50 pm

*BlackFox wrote:

Damn.. It wasn't meant to be a Mathematics thread.
But I'm Just curious about this.

3 *x*x=243
x*x(squared)=243/3
x*x=81
x=9
?

Where's that square root you asked for??

Quote:

Square root(x)/3 = 243/x value of x ?

root(x)/3 = 243/x ---> (x^(1/2))/3 = 243/x

x^1.5 = 3*243 = 729
x = 729^(2/3) = 81

@hapjap that's correct btw, although that isn't the clearest description...
if x = 0.999...
then 10x = 9.999...
so 10x - x = 9.999... - 0.999...
thus 9x = 9
and finally x = 1
so x is both 0.999.. and 1 which means 0.999... = 1

That's probably a little bit clearer.

*BlackFox · **Joined:** Sep 2008 **Posts:** 7923 **Location:**

poehalcho wrote:

Where's that square root you asked for??

Ah, well... Never mind it wasn't important! xD Thanks anyway.

McLovin1t · **Joined:** Apr 2008 **Posts:** 1825 **Location:**

poehalcho wrote:

*BlackFox wrote:

Prove that 0.999... = 1

Let x = .999....

10x = 9.999....

10x - x = 9.999.... - .999...
9x = 9
x = 1
x = .999....
.999.... = 1

Squirt · **Joined:** Jan 2008 **Posts:** 8186 **Location:**

McLovin1t wrote:

poehalcho wrote:

*BlackFox wrote:

Prove that 0.999... = 1

Let x = .999....

10x = 9.999....

10x - x = 9.999.... - .999...
9x = 9
x = 1
x = .999....
.999.... = 1

That proves nothing but that you know algebra >.>

KillAndChill · **Joined:** Mar 2008 **Posts:** 770 **Location:**

McLovin1t wrote:

poehalcho wrote:

*BlackFox wrote:

Prove that 0.999... = 1

Let x = .999....

10x = 9.999....

10x - x = 9.999.... - .999...
9x = 9
x = 1
x = .999....
.999.... = 1

Not really true considering 1/3 cannot be represented 100% accurately in decimal form.

penfold1992 · **Joined:** Apr 2007 **Posts:** 4060 **Location:**

KillAndChill wrote:

Not really true considering 1/3 cannot be represented 100% accurately in decimal form.

let x = 1/9

1/9 = 0.1111111 forever,
x 9 = 0.9999999999 forever...

1/9 x 9 = 1

1=0.999 recurring

1/3 in decimal is 0.333 recurring the same way that 1/9 is 0.1111 or 3/9 is 0.3333 and 9/9 = 0.9999

you must realize the size of infinity to this problem.
if you consider... 0.99999 goes on forever... it can never be just 1 away from 1 (0.001 or 0.00001 or 0.00000001) the reason is because there should be no concept of end of infinite series of numbers.

KillAndChill · **Joined:** Mar 2008 **Posts:** 770 **Location:**

penfold1992 wrote:

KillAndChill wrote:

Not really true considering 1/3 cannot be represented 100% accurately in decimal form.

let x = 1/9

1/9 = 0.1111111 forever,
x 9 = 0.9999999999 forever...

1/9 x 9 = 1

1=0.999 recurring

1/3 in decimal is 0.333 recurring the same way that 1/9 is 0.1111 or 3/9 is 0.3333 and 9/9 = 0.9999

you must realize the size of infinity to this problem.
if you consider... 0.99999 goes on forever... it can never be just 1 away from 1 (0.001 or 0.00001 or 0.00000001) the reason is because there should be no concept of end of infinite series of numbers.

What i'm trying to say is that you cannot pinpoint .333333333 repeating on a number line because every additional 3 changes the location on the number line; therefore any number multiplied by it is not 100% accurate and cannot be used to show equivalence.

penfold1992 · **Joined:** Apr 2007 **Posts:** 4060 **Location:**

you cannot show square root of -1 on the number line yet that exists as a number...also irrational numbers are used all the time to prove equations and such

NuclearSilo · **Posted:** Wed Jun 20, 2012 9:57 am

0.99999..∞ = 2-(0.9999..∞) = 1.0000..∞..1 ?

penfold1992 · **Joined:** Apr 2007 **Posts:** 4060 **Location:**

KillAndChill wrote:

What i'm trying to say is that you cannot pinpoint .333333333 repeating on a number line because every additional 3 changes the location on the number line; therefore any number multiplied by it is not 100% accurate and cannot be used to show equivalence.

you dont need to...
let x = 0.99999
10x=9.99999 you can do this by just moving the decimal point across... the .999 on the end stays the same

9x = 10x - x
9x = (9.99999) - (0.9999) you are just subtracting one from the other, both are infinate series
9x = 9
x=1

to be honest it doesnt matter if you believe it or not, it exists... a bit like oxygen... you cant see it but i can prove that it does exist.

KillAndChill · **Joined:** Mar 2008 **Posts:** 770 **Location:**

penfold1992 wrote:

KillAndChill wrote:

What i'm trying to say is that you cannot pinpoint .333333333 repeating on a number line because every additional 3 changes the location on the number line; therefore any number multiplied by it is not 100% accurate and cannot be used to show equivalence.

you dont need to...
let x = 0.99999
10x=9.99999 you can do this by just moving the decimal point across... the .999 on the end stays the same

9x = 10x - x
9x = (9.99999) - (0.9999) you are just subtracting one from the other, both are infinate series
9x = 9
x=1

to be honest it doesnt matter if you believe it or not, it exists... a bit like oxygen... you cant see it but i can prove that it does exist.

In your example if x = 0.99999, then 10x = 9.9999, not 9.99999.

10x will always have 1 less decimal place which proves that you are completely wrong.

penfold1992 · **Joined:** Apr 2007 **Posts:** 4060 **Location:**

there is not one less decimal... the decimal goes on forever and ever. there is no one extra decimal

Goseki · **Joined:** Apr 2008 **Posts:** 3452 **Location:**

This thread makes me sad about people's lvl of basic math...

Azilius · **Joined:** Oct 2006 **Posts:** 4236 **Location:** CS:GO

KillAndChill wrote:

penfold1992 wrote:

KillAndChill wrote:

What i'm trying to say is that you cannot pinpoint .333333333 repeating on a number line because every additional 3 changes the location on the number line; therefore any number multiplied by it is not 100% accurate and cannot be used to show equivalence.

you dont need to...
let x = 0.99999
10x=9.99999 you can do this by just moving the decimal point across... the .999 on the end stays the same

9x = 10x - x
9x = (9.99999) - (0.9999) you are just subtracting one from the other, both are infinate series
9x = 9
x=1

to be honest it doesnt matter if you believe it or not, it exists... a bit like oxygen... you cant see it but i can prove that it does exist.

In your example if x = 0.99999, then 10x = 9.9999, not 9.99999.

10x will always have 1 less decimal place which proves that you are completely wrong.

infinity * 10 doesn't have one less decimal than infinity (which is in itself, not comprehensible). The proof he's talking about is true and has been around for a long time.

Anyway blackfox did you get this problem from an article based on cognitive bias? I'll post it after if I can find it..

Silkroad Online Forums

Here’s a simple arithmetic question:

Who is online