Math Question

Atrocities · #1 April 2nd, 2004, 06:29 AM

You guys in five short Posts or so just totally ruined any desire I might have had to return to school.

Baal · #2 April 2nd, 2004, 08:33 AM

Quote:

Originally posted by Atrocities:
You guys in five short Posts or so just totally ruined any desire I might have had to return to school.

Atrocities, you are killing me old boy. Wait for me to get my hands on a camera and I'll show you the most painful math you will ever see.

(I will be back with a picture in a few days; I promise you)

Dragonswrd · #3 April 2nd, 2004, 04:31 PM

I might have to take another look at my math books to be able to do trig again, but from what I remember about trig and algebra they were easier when I went back to school to get my degree.

Phoenix-D · #4 April 2nd, 2004, 04:33 PM

Quote:

Originally posted by Atrocities:
You guys in five short Posts or so just totally ruined any desire I might have had to return to school.

Should I post some of my old calc homework and really kill any desire?

Or maybe some statatics..

SD = sqrt((sum of y deviations - y average)^2)/n)

Baal · #5 April 2nd, 2004, 05:04 PM

Quote:

Should I post some of my old calc homework and really kill any desire?

You mean kind like...

[(d^2)z/(dt)^2] + [k/m*z] - [kL/m] + [C/m(dz/dt)] - [g] = [kA/m*sin(wt)]

I have a test involving this problem in about an hour and a half.

Phoenix-D · #6 April 2nd, 2004, 05:47 PM

Yep. Fortunately I didn't have to get too deep into calc for my major..even the course I took was a pain.

Cipher7071 · #7 April 2nd, 2004, 06:07 PM

Quote:

But isn't there always a positive AND negative number for any square root? If x is 0 for example you get f(0)=sqrt(4) Isn't that 2 or -2? Can someone explain why you don't count the -2?

Geo, this is probably because you are working with functions. A function may only have one value in the range for each value in the domain.
So, in this case, the function sqrt() is defined to be positve .