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Someone remind me...

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Old Apr 11, 2005 | 10:04 PM
  #1  
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chris is the devil
 
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Default Someone remind me...

Its been too long..

12 - l x-5 l ≤ 9

those are absolute value things around the x-5.

thanks

EDIT:

Is X ≥ 8 the answer?
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Old Apr 11, 2005 | 10:20 PM
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take the 12 to the other side

- |x-5| ≤ -3

divide by -1, and when you do that you have to invert the ≤

3 ≤ |x - 5|

now find the critical values

3 = x - 5

and

3 = -(x-5)

solve equations to find critical values, then set up your number line and solve
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Old Apr 11, 2005 | 10:23 PM
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chris is the devil
 
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oh duh... jeez im blind.. thanks!
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Old Apr 11, 2005 | 10:26 PM
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ha, no prob


took me alittle while to recall too, havent had an algebra class in over 4 years!
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Old Apr 11, 2005 | 10:28 PM
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im trying to help with homework, yet no idea anymore

factor
12x ^4 – 17x ^3 + 6x ^2

i remember having to multiply the outer ones together, and taking the common multiples or something..
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Old Apr 11, 2005 | 10:32 PM
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well first of all you can take an x^2 out

so it is

x^2 (12x^2 - 17x - 6)

then its just trying to figure out the 12x^2 - 17x - 6 part.... i'll get back to you on that one h:
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Old Apr 11, 2005 | 10:43 PM
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x^2 (12x^2 - 17x - 6)


ok, it factors to

(ax + b)(cx - d)

because the first - in 12x^2 - 17x - 6 means ax*d + b*cx must be negative, so obviously there must be one -... but they cant both be - because b * d must be - from the - 6... so one has to be positive and one must be negative

and reading that over you probably arent following at all

but im going with it anyways because im on a roll

so then its gotta be either
(4x + 2)(3x - 3) or
(4x + 3)(3x - 2) or
(4x + 1)(3x - 3) or
(4x + 6)(3x - 2) or

(3x + 2)(4x - 3) or
(3x + 3)(4x - 2) or
(3x + 1)(4x - 6) or
(3x + 6)(4x - 1) or

(12x + 2)(x - 3) or
(12x + 3)(x - 2) or
(12x + 1)(x - 6) or
(12x + 6)(x - 1) or

(x + 2)(12x - 3) or
(x + 3)(12x - 2) or
(x + 1)(12x - 6) or
(x + 6)(12x - 1) or

(6x + 2)(2x - 3) or
(6x + 3)(2x - 2) or
(6x + 1)(2x - 6) or
(6x + 6)(2x - 1) or

(2x + 2)(6x - 3) or
(2x + 3)(6x - 2) or
(2x + 1)(6x - 6) or
(2x + 6)(6x - 1)

because of the 12, and the x coefficients are multiples of 12
and because of the 6, and the a and d numbers are multiples of 6

wait, none of the work... fawk!
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Old Apr 11, 2005 | 10:53 PM
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damn i did that type of shit frosh year... hard as hell (thats why i had to retake the class soph year and almost again this year)
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Old Apr 11, 2005 | 11:38 PM
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I suck at Math horribly. h:
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Old Apr 11, 2005 | 11:58 PM
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Originally Posted by AcuraFanatic
I suck at Math horribly. h:
..
fuggin werd.
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