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-- The Math Thread (4 N3rdS!!!)
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hmmm.. Have you tried checking your TC/IP settings? delete temporary internet files? If that doesn't work then there is no answer to your problem.

quote:

Originally posted by Orbital32 hmmm.. Have you tried checking your TC/IP settings? delete temporary internet files? If that doesn't work then there is no answer to your problem.

EA games downloader support nightmares all over again

quote:

Originally posted by Orbital32 hmmm.. Have you tried checking your TC/IP settings? delete temporary internet files? If that doesn't work then there is no answer to your problem.

what the fuck??? are u talking about

good luck. I done all the calc's plus diff eq I,II

disp vel acell jerk
(X^2)'->(2x)''->(2)'''->0

you try to confuse.

math makes me

Re: Little homework help (Calculus)

quote:

Originally posted by ::TranceVanDyk:: original problem is. (y+2)/y = 1/y-5

y + 2 = 1 - 5y => 1 = -6y => y=-1/6

yes?

unless you wrote it down incorrectly (i assume you did) there is no y^2


what you have come up with, however, has 2 solutions of 2+sqrt(14), 2-sqrt(14)

Re: Little homework help (Calculus)

quote:

Originally posted by ::TranceVanDyk:: ive been working this one problem, and have worked it all the way to this point. y = y^2 -3y -10 i have no clue how to find y now, even after factoring it. y = (y-5)(y+2) original problem is. (y+2)/y = 1/y-5 WHAT THE IS Y!!!!???? edit:: ACTUALLY ITS precalculus, not all the way to the big leagues yet

wtf? we do this shit in my Algebra II class.

GIVEN: (y+2)/y = 1/y-5



SOLUTION:
(a) y = (y+2)(y-5)
(b) y = y^2 -3y -10
(1) y=y^2-3y-10
(2) y^2-3y-10-y=0
(3) y^2-4y-10=0
(4) a=1, b=-4, c=-10
(5) (-(-4)+/- sqrt((-4)^2-4(1*-10))/(2*(1))
(6) (4 +/- sqrt(16+40))/(2)
(7) (4 +/- sqrt(56))/2
(8) Y1 = 5.74165
Y2 = -1.7416

CHECK:
5.74165 = 5.74165^2 -3(5.74165)-10
-1.7416 = -1.7416^2 -3( -1.7416)-10

For me, this is MATH II in HS. Quadratic Eq ish

its always 42

Why don't you just download Matlab or other math software. it will show you exactly how to do it.

well, they meld this stuff in with pre-calc, precalc is basically algebra II review and the beginnings of basic calculus. im doing this for somebody who's paying me 100 bucks to take an online course for him. but the guy hands over to me his portfolio, which he's 1 month behind, has a 52%. i brought that shit up to a 63% and he's now only 2 weeks behind. he smokes pot like no one else, and doesnt know a damn thing, he cheats his way thru everything, but i dont care, im gettin paid.

ahhh...

brings back memories, of walking out on my maths class (teacher didn't give a shit)...

some how i Still graduated with a B+....

i think there's a special way to get derivatives of fractions, or maybe fractions of powers.

if i find my old texts, i'll PM u....

but that probably won't happen.

we got the answer on page two guys, chill!!

ill probably come back to this thread with more problems, when i continue working later.

quote:

Originally posted by Krypton we got the answer on page two guys, chill!! ill probably come back to this thread with more problems, when i continue working later.

another problem on the first page of the thread, post 1...

i know the answer to this one, but can u find it?

Find all the real RATIONAL zero's in this equation. (In other words, find how many time "x" crosses the "x" axis on a graph.)(or find all the real solutions of "x" that makes the equation = to zero. youll get a PRIZE!!!

x^3 - 4x^2 + x + 6 = 0

Damn it, a third power!!! There's a method to do this but I forgot how, lemme ask my sis, she just passed algebra so it's still fresh in her head...

quote:

Originally posted by ::TranceVanDyk:: another problem on the first page of the thread, post 1... i know the answer to this one, but can u find it? Find all the real RATIONAL zero's in this equation. (In other words, find how many time "x" crosses the "x" axis on a graph.)(or find all the real solutions of "x" that makes the equation = to zero. youll get a PRIZE!!! x^3 - 4x^2 + x + 6 = 0

x =-1; x=2; x=3

or

(x+1)(x-2)(x-3)=0

quote:

Originally posted by Jocker x =-1; x=2; x=3 or (x+1)(x-2)(x-3)=0

I'm at a loss...could you please show me how to do this or at least what method you used?

quote:

Originally posted by stevieboy32808 I'm at a loss...could you please show me how to do this or at least what method you used?

finding the first root (-1) with picking of numbers (you go 0, +-1, +-2, +-3, etc... 99.9% of cubic equations in college textbooks have at least one real root in the +-3 range). then, factoring it out, and solving the quadratic equation. there is a much more complex formula for solving cubic equations (at least for real numbers, i'm not sure about the complex ones), but i don't remember it

Why Do Our Answers Differ???

So why the difference of answers...

here's one:
disprove the pythagorean theorem.

quote:

Originally posted by Jocker x =-1; x=2; x=3 or (x+1)(x-2)(x-3)=0

niceee

quote:

I'm at a loss...could you please show me how to do this or at least what method you used?

x^3 - 4x^2 + x + 6

Possible zeros: +-6, +-3, +-2, +-1 (all factors of c/all factors of the 1st coefficent

guess and check. find out that when u divide 3 by x^3 - 4x^2 + x + 6 there is no remainder. so u know x-3 is a factor and that 3 is one solution. after dividing by 3, u get x^2 - x - 2. factor that out, and u get (x-2)(x+2) and then u include the (x-3).

use the zero theorem or whatever its called
x-2 = 0 (x = 2)
x+1 = 0 (x = -1)
x-3 = 0 (x = 3)

u have your answers.

to get the complex solutions if there are any, i believe u use the quadratic equation once u find at least one real solution. i have no idea how to find solutions to an equation with all complex zeros.
x-3 = 0 (

you have a fundamental flaw

x(x^2-4x+1)=-6 doesn't mean that x=-6 and x^2-4x+1=-6 (judge yourself, then the answer would be -6*-6); you can say that different factors on the left equal the right side only when the right side is equal to 0. i.e x(x^2-4x+1)=0 would indeed mean that x=0 and x^2-4x+1=0

edit: this is to stevieboy

quote:

Originally posted by kadomony here's one: disprove the pythagorean theorem.

only if not in euclidian geometry, because in euclidian geometry (i.e. geometry that you study in school) the pythagorean theorem holds true.

quote:

Originally posted by ::TranceVanDyk::

quote:

Originally posted by Jocker

Thanks to the both of you. I totally forgot about the setting equal to zero part and NOT -6. By setting it equal to -6 that is no longer a root...so you're right again.

anybody got anything to solve?

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