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Math Game!, whoot!
xoxoxx
post Apr 23 2006, 11:46 PM
Post #1


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For all the math nerds:

I'll give a number and I'll tell you what to do with it. Add, subtract, etc.

Then you solve it and add another thing to do with it.

And for all you super smart calculus AP people, don't mess this up using super hard equations.

#: 1
add 5 to the #.
 
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*mipadi*
post Apr 23 2006, 11:48 PM
Post #2





Guest
# = 6

Subtract 15 from the number and take the square root.
 
*salcha*
post Apr 23 2006, 11:50 PM
Post #3





Guest
LOL what the hale sean cheng

salcha4u: i dont get this?
salcha4u: is it -4 or something?
freshnx addict: ...
freshnx addict: 1 + 5
salcha4u: ...
salcha4u: shut up

6-15 is -9 so its 3i (imaginary)

Multiply by -2i + 8
 
*mipadi*
post Apr 23 2006, 11:52 PM
Post #4





Guest
# = 6 + 24i

Subtract 6 and multiply by i. (What have I started?!)
 
*salcha*
post Apr 23 2006, 11:52 PM
Post #5





Guest
Dunno but this is fun.

salcha4u: darn you when i thought of math SATs
salcha4u: ):
freshnx addict: LOL

#= -24
Aww, it's not fun anymore.

Square root that.
 
xoxoxx
post Apr 24 2006, 12:02 AM
Post #6


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I DONT GET THIS!

Rawr.

new number/eqn.

#: 5

Multiply that number by 10!
 
*Girthy*
post Apr 24 2006, 12:04 AM
Post #7





Guest
# = 50

Tan of Cos of Sin of #
 
*salcha*
post Apr 24 2006, 12:05 AM
Post #8





Guest
# = 1.45

Inverse tan that
 
*mipadi*
post Apr 24 2006, 12:07 AM
Post #9





Guest
QUOTE(Girthy @ Apr 24 2006, 1:04 AM) *
# = Square Root of -24

Cube root it.

Square root of 24 = 2i times the square root of 6.

# = 2i * the square root of six

New number = (3 to the 2/3rds times the square root of 2 over 2) plus (3 to the 1/6th times the square root of 2 over 2) times i

Multiply by i!



Nix that. tan-1 of 1.45 is approximately .968.

Add .032 to that.
 
*salcha*
post Apr 24 2006, 12:10 AM
Post #10





Guest
QUOTE(mipadi @ Apr 23 2006, 10:07 PM) *
Square root of 24 = 2i times the square root of 6.

# = 2i * the square root of six

New number = (3 to the 2/3rds times the square root of 2 over 2) plus (3 to the 1/6th times the square root of 2 over 2) times i

Multply by i!

HAHA he actually solved it.

3 to the 2/3rd times square root two over two i minus three to the 1/6th times root two over two.

Times i cubed!
 
*Girthy*
post Apr 24 2006, 12:10 AM
Post #11





Guest
3^(2/3)*2^(1/2)i / 2 - 3^(1/6) * 2^(1/2) / 2
Omg. no!
 
*mipadi*
post Apr 24 2006, 12:12 AM
Post #12





Guest
QUOTE(salcha @ Apr 24 2006, 1:10 AM) *
HAHA he actually solved it.

3 to the 2/3rd times square root two over two i minus three to the 1/6th times root two over two.

Times i cubed!

# = ((3 ^ 2/3 * 2^1/2) / 2) + ((3^1/6 * 2 ^ 1/2) / 2)

Multiply by 2!
 
xoxoxx
post Apr 24 2006, 12:12 AM
Post #13


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screw this!

i'll go play with my calculator.
 
*Girthy*
post Apr 24 2006, 12:13 AM
Post #14





Guest
QUOTE(mipadi @ Apr 23 2006, 10:12 PM) *
# = ((3 ^ 2/3 * 2^1/2) / 2) + ((3^1/6 * 2 ^ 1/2) / 2)

Multiply by 2!

(3 ^ 2/3 * 2^(1/2)) + (3^1/6 * 2 ^ (1/2))
 
*salcha*
post Apr 24 2006, 12:14 AM
Post #15





Guest
(3 ^ 2/3 * 2^1/2) + (3^1/6 * 2 ^ 1/2)

Multiply by root 2

Edit//
ahh kevin beat me.
 
*Girthy*
post Apr 24 2006, 12:16 AM
Post #16





Guest
Y = Kevin
X = Is Sexy

Y + X = ?
 
*salcha*
post Apr 24 2006, 12:23 AM
Post #17





Guest
Sally is sexier.

Solve equation two posts above.
 
*Girthy*
post Apr 24 2006, 12:24 AM
Post #18





Guest
Never!

Care to "Differentiate" my "natural log?"
 
*salcha*
post Apr 24 2006, 12:26 AM
Post #19





Guest
My love for you is like the slope of a concave up function because it is always increasing. mellow.gif
 
*Girthy*
post Apr 24 2006, 12:27 AM
Post #20





Guest
Are you a differentiable function? Because I'd like to be tangent to your curves!
 
*salcha*
post Apr 24 2006, 12:28 AM
Post #21





Guest
I am equivalent to the Empty Set when you are not with me.
 
*Girthy*
post Apr 24 2006, 12:29 AM
Post #22





Guest



































If you do the math, it turns out to be 69.
 
*salcha*
post Apr 24 2006, 12:32 AM
Post #23





Guest
HEY I'm not a junior yet. I learn that next year sad.gif
 
*Girthy*
post Apr 24 2006, 12:32 AM
Post #24





Guest
How about me and you go back to my place and form a covalent bond?
 
*salcha*
post Apr 24 2006, 12:33 AM
Post #25





Guest
Sure if I get to see what you look like. laugh.gif
I don't know anymore math pickup lines that I understand.
 
*Girthy*
post Apr 24 2006, 12:34 AM
Post #26





Guest
Two hydrogen atoms bumped into each other recently.
One said: "Why do you look so sad?"
The other responded: "I lost an electron."
Concerned, One asked "Are you sure?"
The other replied "I'm positive."
 
*salcha*
post Apr 24 2006, 12:35 AM
Post #27





Guest
laugh.gif laugh.gif laugh.gif

Only two people have A's in my chem class though.
 
*Girthy*
post Apr 24 2006, 12:41 AM
Post #28





Guest
Q. If a bear in Yosemite, and one in Alaska fall into water, which one would dissolve faster?
A. The bear in Alaska because it's polar.

A neutron walks into a bar, sits down and asks for a drink. Finishing, the neutron asks "How much?"
The bartender says, "For you, no charge."
 
*mipadi*
post Apr 24 2006, 12:50 AM
Post #29





Guest
Hey, let's keep it clean. Stop using Hoare semantics in the math thread.
 
*wind&fire*
post Apr 24 2006, 01:27 AM
Post #30





Guest
for trig lovers...

Consider the geometris series 1 - tan^2Θ - tan^4Θ - ...

When the limiting sum exists find its value in simplest form.

for what values of Θ in the intervel -(π/2) < Θ < π/2 does the limiting sum of a series exists


YOU CAN DO EEEEEETTTTTT!!!!
 
*mipadi*
post Apr 24 2006, 02:19 PM
Post #31





Guest
That sounds suspiciously like your math homework!
 
GREASEbaby
post Apr 24 2006, 05:23 PM
Post #32


What's my name? Janette. and ily. <3
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^x2, er, 5.

multiply by 25a(to the third power)b(to the fourth power)-83a(to the third power)
 
*Girthy*
post Apr 24 2006, 05:34 PM
Post #33





Guest
Programming is like sex:
One mistake and you have to support for a lifetime.
 
*wind&fire*
post Apr 24 2006, 08:56 PM
Post #34





Guest
QUOTE(mipadi @ Apr 25 2006, 5:19 AM) *
That sounds suspiciously like your math homework!

it sounds suspiciously like a question from my HSC maths exam.... spent a good ten mintues on that question... then gave up and went onto another question...

say when did you become head staffer?
 
*salcha*
post Apr 24 2006, 11:31 PM
Post #35





Guest
1 - tan^2Θ - tan^4Θ - ...

tan2Θ = (2tanΘ)/1-tan^2Θ
tan4Θ = (tan2Θ+tan2Θ) = ((2tanΘ)/(1-tan^2Θ))+((2tanΘ)/(1-ta^62Θ))/1-(2tanΘ)/(1-tan^2Θ)(2tanΘ)/(1-tan^2Θ)

....

If anyone is willing to solve.
 
*mipadi*
post Apr 24 2006, 11:58 PM
Post #36





Guest
QUOTE(wind&fire @ Apr 24 2006, 9:56 PM) *
say when did you become head staffer?

Someone forgot to lock the back door and I snuck in.






Seriously, though, I think it was around the end of March.
 
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