500k/question to the person who gives the best answer (math)

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thread starter TheGreyBow Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

Haii so i have a couple questions left on my summer math packet.. wondering if you guys can help cause i'm reallyyyy rusty >.< Like the title says... 500k per question to the person who gives the best (detailed, comprehendable, explain work) answers. This is in Bera... If you're not in bera please feel free to help anyways 1. Find the zeros, both real and complex, of each polynomial function: f(x)=x^4-625. There are two 2 real and 2 complex zeros. QUESTION #1 ANSWERED BY Sleepyxdude and NejiEl 2. (same directions): f(x)= x^3 - 5x^2 + 5x - 25 (Hint: Use x-5 as the binomial and synthetic division to find the first zero. Then use the quadratic formula.) QUESTION #2 ANSWERED BY Sleepyxdude 3. (simplify) 6x/3x+1 + 9/2x - x+1/x-1 4. x^2/25 + y^2/36 = 1 Label the vertices and the foci with both coordinates 5. x^2/9 - y^2/16 = 1 Label the vertices and the foci with both coordinates. Write an equation in slope-intercept form for the asymptote with the positive slope. Uh, I think thats it for now I seem to have completed the bajillion other questions

JennaQ 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Oh! You were actually serious about the meso thing. I wonder if anyone here is actually going to care about the mesos or NX or whatever you mean. I think it's better if you ask someone else who isn't on a forum dedicated to a brainless game. Like your parents or teachers.

LightStruck 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I'm not going to do your homework for virtual money.

dattyboy4916 09/10/09 + PM | QUOTE | PERMALINK | REPORT

oof tough stuff. I'm guessing this is....parabolas? i don't remember that far off...

WolfXLord 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I would help but I don't understand how you set it up on the computer.

GEEK 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I'd do it if it was 500k irl

Xynth 09/10/09 + PM | QUOTE | PERMALINK | REPORT

...mah brain hurtses

Applesnacks 09/10/09 + PM | QUOTE | PERMALINK | REPORT

GEEK : I'd do it if it was 500k irl lol geek

TheAwesomeOne 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Whoa... o.0

gamefreakdude76 09/10/09 + PM | QUOTE | PERMALINK | REPORT

What grade are you in? That looks very complicated... (Im in grade 9)

Mist 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I've got my own physics, stat, and whatnot hw to do. You have teachers and you have the internet. Learn to use your resources.

nejiEL Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

I don't feel like doing my hw, so... 1) 5, -5, 5i, -5i 2) 5, (squareroot of 5)i, -(squareroot of 5)i 3) just multiply to get a common denominator 4)vertices (0,6) (0,-6), foci (0, sqrroot of 11) (0, -sqrroot of 11) 5) I don't wanna look up the equations for a hyperbole, but vertices might be (0, [+- 4]) and foci might be (0, +-5) slope: y=+-x(4/3)

TheGreyBow 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I'm going to be a senior in highschool but it's not like an advanced class or AP or anything lol

sleepyxdude Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

1. x^4 - 625 = (x^2+25)(x^2-25) = (x+5i)(x-5i)(x+5)(x-5) Set each term equal to 0 to find the 0's x+5i = 0 x-5i = 0 x+5 = 0 x-5 = 0 x = 5i, -5i, 5, -5 2. x^3 - 5x^2 + 5x - 25 (x-5)(x^2 + 5) = 0 x = 5, sqrt(5)i, -sqrt(5)i 3. Not sure if this is right, got confused a bit [6x/3x+1 + 9/2x - x+1/x-1] (3x+1)(2x)(x-1) = 6x(2x)(x-1) + 9(3x+1)(x-1) - (x+1)(3x+1)(2x) = 12x^2(x-1) + 9(3x^2-2x-1) - (3x^2+4x+1)(2x) = (12x^3-12x^2) + (27x^2-18x-9) - (6x^3+8x^2+2x) = (12x^3-6x^3) + (-12x^2+27x^2-8x^2) + (-18x-2x) - 9 = 6x^3 + 7x^2 -20x - 9 4. Verticies = (5,0) (-5,0) (0,6) (0,-6) Foci = (sqrt(11), 0) (-sqrt(11), 0)

NoobCakes 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Do you're own homework. When the test come, you won't be able to go on Basil and make another thread being like "oh ma gawd, help m3 an$w3r d1$ f0r ma t3$t!@@@" =/