I'll solve any math problem!

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deactivated-57ef6a3ad2935

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#1 deactivated-57ef6a3ad2935
Member since 2004 • 5346 Posts
The problem has to be reasonable. No non sense.
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schoeffmaster

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#2 schoeffmaster
Member since 2005 • 10674 Posts
Ok, how can you make 2+2=
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someotherguy654

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#3 someotherguy654
Member since 2004 • 15122 Posts
What is infinity divided by zero?
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cool_baller

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#4 cool_baller
Member since 2003 • 12493 Posts
If 2+2=5 then what is 5-2?
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schoeffmaster

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#5 schoeffmaster
Member since 2005 • 10674 Posts
lol didnt even show the equation...
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deactivated-57ef6a3ad2935

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#6 deactivated-57ef6a3ad2935
Member since 2004 • 5346 Posts
4,0
Cmon guys i NEED A CHALLENGE!
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cool_baller

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#7 cool_baller
Member since 2003 • 12493 Posts

or another one:

zero divided by zero=? (0/0)=? 

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Nick11478

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#8 Nick11478
Member since 2006 • 1628 Posts
3+3 PLEASE I'M IN NEED OF THE ANSWER!!! When I have a calculator in my computer along with everyone else here.
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schoeffmaster

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#9 schoeffmaster
Member since 2005 • 10674 Posts
Ok, how can you make 2+2=
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schoeffmaster

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#10 schoeffmaster
Member since 2005 • 10674 Posts
Ok, how can you make 2+2=fish schoeffmaster
damnit show UP!!!
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Sandro909

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#11 Sandro909
Member since 2004 • 15221 Posts

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

 ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

 dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

 dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

 partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

 Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0\ -betagamma c & gamma & 0 & 0\ 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1 end{pmatrix}

 beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

 eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN! 

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the_foreign_guy

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#12 the_foreign_guy
Member since 2005 • 22657 Posts
Pi to the billionth plus one digit.
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#13 thirstychainsaw
Member since 2007 • 3761 Posts

On an episode of Futurama, Fry was awared 4.3 billion dollars I beleive. I've been to lazy to go over the numbers and for some reason I really want to know if 4.3 billion is the correct figure.

Principal: .93
Interest: 2 and 1/4
Time: 1000 years

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#14 CptJSparrow
Member since 2007 • 10898 Posts
Forget that. Recommend me quality online math resources.
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#15 thirstychainsaw
Member since 2007 • 3761 Posts

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0 -betagamma c & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{pmatrix}

beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN!

Sandro909

*Brain implodes* 

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#16 Def_Jef88
Member since 2006 • 17441 Posts
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+!9+20+21+22+23+24+25+26+27+28+29+3+31+32+33+34+35+36=?
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#17 CRS98
Member since 2004 • 9036 Posts
What is 1,000,000 times 1,000,000? And don't use a calculator.
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#18 darkjedirpm
Member since 2004 • 2453 Posts

The problem has to be reasonable. No non sense.jpazhman

:lol: good luck with that here

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#19 CptJSparrow
Member since 2007 • 10898 Posts
What is 1,000,000 times 1,000,000? And don't use a calculator.CRS98
1,000,000,000,000.
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#20 BEAN_LARD_MULCH
Member since 2006 • 4720 Posts
[QUOTE="Sandro909"]

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0 -betagamma c & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{pmatrix}

beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN!

thirstychainsaw

*Brain implodes*


But those are just formulas....
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#21 CRS98
Member since 2004 • 9036 Posts

[QUOTE="CRS98"]What is 1,000,000 times 1,000,000? And don't use a calculator.CptJSparrow
1,000,000,000,000.

You win...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NOTHING!

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#22 Sandro909
Member since 2004 • 15221 Posts

[QUOTE="CRS98"]What is 1,000,000 times 1,000,000? And don't use a calculator.CptJSparrow
1,000,000,000,000.

That's like... one thousand billion. :o :lol: 

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#23 CRS98
Member since 2004 • 9036 Posts

[QUOTE="CptJSparrow"][QUOTE="CRS98"]What is 1,000,000 times 1,000,000? And don't use a calculator.Sandro909

1,000,000,000,000.

That's like... one thousand billion. :o :lol: 

That sir is One Trillion.

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#24 Sandro909
Member since 2004 • 15221 Posts
[QUOTE="thirstychainsaw"][QUOTE="Sandro909"]

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0 -betagamma c & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{pmatrix}

beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN!

BEAN_LARD_MULCH

*Brain implodes*


But those are just formulas....

To the untrained eye. >_> 

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#25 Sandro909
Member since 2004 • 15221 Posts
[QUOTE="Sandro909"]

[QUOTE="CptJSparrow"][QUOTE="CRS98"]What is 1,000,000 times 1,000,000? And don't use a calculator.CRS98

1,000,000,000,000.

That's like... one thousand billion. :o :lol:

That sir is One Trillion.

I'm aware. :P 

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#26 Aznsilvrboy
Member since 2002 • 11495 Posts

Square ABCD has sides of length 14 units. A circle is drawn through A and D, so that it is tangent to BC as shown in the figure. What is the radius of the circle?

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#27 BEAN_LARD_MULCH
Member since 2006 • 4720 Posts
[QUOTE="BEAN_LARD_MULCH"][QUOTE="thirstychainsaw"][QUOTE="Sandro909"]

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0 -betagamma c & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{pmatrix}

beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN!

Sandro909

*Brain implodes*


But those are just formulas....

To the untrained eye. >_>


Isee your point.
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#28 CptJSparrow
Member since 2007 • 10898 Posts
The OP hasn't answered a single one of these yet.
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Mumbles527

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#29 Mumbles527
Member since 2004 • 7706 Posts
Solve for me the Riemann Hypothesis. Clearly it shouldn't be a problem for such a mathematical mind who claims to be able to solve any math problem! It may have been unsolved for well over a century, but I have a feeling that that will all change tonight!
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#30 Tolwan
Member since 2003 • 2575 Posts

Two points, (2, 5) (19, 12)

Find the distance using basic formula

D=Square root of X Subroot 1 minus X subroot 2 plus Y subroot 1 minus Y subroot 2

Begin!

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schoeffmaster

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#31 schoeffmaster
Member since 2005 • 10674 Posts

Square ABCD has sides of length 14 units. A circle is drawn through A and D, so that it is tangent to BC as shown in the figure. What is the radius of the circle?

Aznsilvrboy
I would say about 8 units for the radius...give or take...
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#32 schoeffmaster
Member since 2005 • 10674 Posts

Two points, (2, 5) (19, 12)

Find the distance using basic formula

D=Square root of X Subroot 1 minus X subroot 2 plus Y subroot 1 minus Y subroot 2

Begin!

Tolwan
you can just graph those to find the difference...
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#33 Aznsilvrboy
Member since 2002 • 11495 Posts
[QUOTE="Aznsilvrboy"]

Square ABCD has sides of length 14 units. A circle is drawn through A and D, so that it is tangent to BC as shown in the figure. What is the radius of the circle?

schoeffmaster

I would say about 8 units for the radius...give or take...

That was random, but in math, there's no such thing as give or take. Also, the answer to this problem is not a number that has a lot of decimals, so you can get the exact answer.8)

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#34 buldog300
Member since 2003 • 2152 Posts
alright ill bite. what is x if 5175=x^3-2x^2+4x-500? Remember no calculator.
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#35 Mumbles527
Member since 2004 • 7706 Posts
[QUOTE="Tolwan"]

Two points, (2, 5) (19, 12)

Find the distance using basic formula

D=Square root of X Subroot 1 minus X subroot 2 plus Y subroot 1 minus Y subroot 2

Begin!

schoeffmaster
you can just graph those to find the difference...

Is the answer -4.9?
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#36 gotdangit
Member since 2005 • 8151 Posts
Ok what 1+2x3? One chance.
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cool_baller

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#37 cool_baller
Member since 2003 • 12493 Posts
Ok what 1+2x3? One chance.gotdangit
7
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#38 Zaeryn
Member since 2005 • 9070 Posts

Two points, (2, 5) (19, 12)

Find the distance using basic formula

D=Square root of X Subroot 1 minus X subroot 2 plus Y subroot 1 minus Y subroot 2

Begin!

Tolwan
No!
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schoeffmaster

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#39 schoeffmaster
Member since 2005 • 10674 Posts
[QUOTE="schoeffmaster"][QUOTE="Aznsilvrboy"]

Square ABCD has sides of length 14 units. A circle is drawn through A and D, so that it is tangent to BC as shown in the figure. What is the radius of the circle?

Aznsilvrboy

I would say about 8 units for the radius...give or take...

That was random, but in math, there's no such thing as give or take. Also, the answer to this problem is not a number that has a lot of decimals, so you can get the exact answer.8)

well i do know that circle takes the square...
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#40 gotdangit
Member since 2005 • 8151 Posts

[QUOTE="gotdangit"]Ok what 1+2x3? One chance.cool_baller
7

Whyd you answer.. you ruined it, it was for him... 

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194197844077667059316682358889

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#41 194197844077667059316682358889
Member since 2003 • 49173 Posts
[QUOTE="BEAN_LARD_MULCH"][QUOTE="thirstychainsaw"][QUOTE="Sandro909"]

ds^2 = dx_1^2 + dx_2^2 + dx_3^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2

dx_1^2 + dx_2^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 - c^2 dt^2

dx_1^2 + dx_2^2 + dx_3^2 = c^2 dt^2ds^2 = 0 = dx_1^2 + dx_2^2 + dx_3^2 - c^2 dt^2

partial_0 phi = frac{partial phi}{partial t}, quad partial_1 phi = frac{partial phi}{partial x}, quad partial_2 phi = frac{partial phi}{partial y}, quad partial_3 phi = frac{partial phi}{partial z}.

Lambda^{mu'}{}_nu = begin{pmatrix} gamma & -betagamma/c & 0 & 0 -betagamma c & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{pmatrix}

beta = frac{v}{c}, gamma = frac{1}{sqrt{1-beta^2}}.

eta_{alphabeta} = eta_{mu'nu'} Lambda^{mu'}{}_alpha Lambda^{nu'}{}_beta !

 

AND BEGIN!

Sandro909

*Brain implodes*


But those are just formulas....

To the untrained eye. >_> 

It looks like a derivation of the Lorentz transform off the top of my head, which is a transform and not really a problem to be solved. Let me know if I'm off, though. And to the TC, here's a problem for you. The royal family of Sweden will even give you a prize if you do it!