I am still grossly upset on how much you hate me, and all of the other people who don't know enough about maths or have Google fail them.

My name is Tony Keen. You fail to give me an invitation. Now you will die!

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[ Massassiism ] [ Cynicism ] [ Fanaticism ]
Sell your soul for a cookie?

2.

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If you can read this, you need better glasses.

[EDIT: Oops, didn't see Kyle's post]

[This message has been edited by Connection Problem (edited June 20, 2004).]

Kyle is correct with 4.

How exactly did you manage to quote me after I changed that?

Ππ

[Pfft. So now the greek letters work.]

[This message has been edited by DogSRoOL (edited June 20, 2004).]

Oh wait, I remember now. Most peoples browsers are only set to view western european encoding. Needs to be viewing Greek for those to work.


ð

[This message has been edited by Darth (edited June 20, 2004).]

Most excellent. (w00t!).

paulsona AT colorado DOT edu



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"What sane person could live in this world, and *not* be crazy?"

Council of 14

"I am the signature virus! Copy me into your signature so that I can take over the world! Moohahahee!"

[whoops.. missed that it was already answered]

[This message has been edited by Vornskr (edited June 20, 2004).]

Gmail rocks.

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BV's rendition of Titanic: Let's have sex. I won't let go. I don't need this stone.
The End.
~ Wolfy

6. Suppose f(x)=e^(ax)+e^(bx). For how many different pairs (a,b) is f"-2f'-15f=0?

linear 2nd order homogeneous differential equation with constant coefficients

1 pair

Actually, if you did the math, you'd know it's four. f''-2f'-15f = 0 becomes e^(ax)(a^2-2a-14) + e^(bx)(b^2-2b-14) = 0, and since e^(ax) and e^(bx) can never be zero, the equation is only true when (a^2-2a-14) = 0 and (b^2-2b-14) = 0. Since there are two places in each case where that's true (a and b can either be -2.87 or 4.87) there are 4 possible pairs of a and b satisfying that equation. (-2.87,-2.87) (-2.87,4.87) (4.87,-2.87) and last but not least (4.87,4.87).

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"What sane person could live in this world, and *not* be crazy?"

Council of 14

"I am the signature virus! Copy me into your signature so that I can take over the world! Moohahahee!"

Darth, no sane person wants Gmail that much.

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Have a good one,
Freelancer

Haha, excellent idea to give math tasks for gmail accounts

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No signature for you.

The questions weren't THAT hard.

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WOOSH|-----@%

Some of that stuff is beyond first semester calculus, which weeds out 80% of the population right there.

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Have a good one,
Freelancer

There were a lot harder ones on there, but I just picked what was easiest to write and what didn't rely on the multiple choices that were given.

Since some of the questions were of the "Which of these fits the following conditions?" type.

It was a test designed for just high school Calculus AB classes I believe. I found it quite easy overall though, seeing as I placed second in it.

Try doing it again with -15 instead of -14 Kyle. But the equations are the same so there's only one solution, not 4.

I got 5, -3

[Edit]Now if I only had a calculator. Oh well, damn you. Put up easier equations for the people that cant do that!

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Jedi rocks rule the house...
YEAH YEAH YEAH

[This message has been edited by Jedi_Goon (edited June 22, 2004).]