04-7-2011, 10:05 PM | #1 |
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Damn Math... 25k Credits For Help?
I need to know how to get the answer... I have no idea how to get the actual answer with the inequality signs.
the question is this: 3 + X > 4/X Where X cannot = 0 This is what I get: Case 1: Where x > 0 x > -4 , x > 1 Case 2: Where x < 0 x < -4, x < 1 I have no clue how to arrange the answers and we are not allowed to use our graphing calculator. I would like help in trying to understand the process of rearranging it. I'm good with the math. I hope I made it clear enough, if not just say what other clarification you may need. Thanks
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04-7-2011, 10:15 PM | #2 |
Snake Princess
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Re: Damn Math... 25k Credits For Help?
So you want to know how to graph the inequalities without the calculator?
Haha well iironiic kinda beat me to the explanation. lol
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[SIGPIC][/SIGPIC] Last edited by Sephiroth28; 04-7-2011 at 10:22 PM.. |
04-7-2011, 10:17 PM | #3 |
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Re: Damn Math... 25k Credits For Help?
no, I want to know how to rearrange the answers that I got to the cases to find the right answer. :P
All I know is that you have to eliminate some of them with the x > or < 0 but Im not sure which ones.
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04-7-2011, 10:17 PM | #4 |
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Re: Damn Math... 25k Credits For Help?
3 + X > 4/X
Let X < 0 Therefore, 3X + X^2 < 4 Or, X^2+3X - 4 < 0 Or, (X+4)(X-1) < 0 So we know that one (X-1) or (X+4) is negative and the other is positive a) Suppose (X-1)>0, or X>1 This implies that (X+4)<0 but for all X>1, that's not true. b) Suppose (X+4) > 0, or X> -4 This implies that (X-1) < 0, which is true for -4<X<1 But remember X < 0, so -4<X<0 for the first case. Now let X>0 Therefore, 3X +X^2 >4 Or, (X+4)(X-1) > 0 Which implies that they both have to share the same sign. Let (X+4) > 0 and (X-1) > 0. Or, X > -4 and X > 1. This is only true when X>1. Now let (X+4) < 0 and (X-1) < 0. Or, X < -4 and X < 1. This is only true when X<-4. This implies that -4 < X < 1 but note that X > 0. So 0 < X < 1. So the final solution is the union of the two, or -4 < X < 1. I hope this helps. (more of a spoiler oops) EDIT: Exclude 0 in the set of solutions. I forgot we're dividing by X lol. Last edited by iironiic; 04-7-2011 at 10:48 PM.. |
04-7-2011, 10:30 PM | #5 | |
I like max
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Re: Damn Math... 25k Credits For Help?
Quote:
-4 < x < 0, X > 1. :S
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04-7-2011, 10:36 PM | #6 |
Hunger Games Hunty
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Re: Damn Math... 25k Credits For Help?
You have to use discontinuities in your interval test as well.
So test, - infinity to -4, -4 to 0, 0 to 1, and 1 to infinity. If you test those you get your answer. |
04-7-2011, 10:37 PM | #7 |
Expect delays.
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Re: Damn Math... 25k Credits For Help?
That.. doesn't make sense!?
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04-7-2011, 10:37 PM | #8 | |
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Re: Damn Math... 25k Credits For Help?
Oh found my mistake rofl
Quote:
EDIT: Nice trollpost >=( |
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04-7-2011, 10:38 PM | #9 |
I am leonid
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Re: Damn Math... 25k Credits For Help?
3 + X > 4/X
If X > 0, X^2 + 3X - 4 > 0 => (X+4)(X-1) > 0 => X+4 and X-1 are both positive or both negative. Since X>0, X+4 can't be negative, so they are both positive, which means X>1 If X<0, X^2 + 3X - 4 < 0 => (X+4)(X-1) < 0 => X+4 and X-1 have different signs. Since X<0, X-1 cannot be positive, so X+4 is positive and X-1 is negative Therefore, X<1 and X>-4. Since X<0, -4<X<0 Thus, the final answer is -4<X<0 and X>1 |
04-7-2011, 11:51 PM | #10 |
Nothing.
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Re: Damn Math... 25k Credits For Help?
You know what **** you people that are good at math. Seriously I look at equations like they're written in Chinese or something. I hope I'm not alone in this haha.
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04-7-2011, 11:52 PM | #11 |
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Re: Damn Math... 25k Credits For Help?
Same for me.
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04-7-2011, 11:59 PM | #12 |
FFR Player
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Re: Damn Math... 25k Credits For Help?
I hate math but with all my GRE prep I actually get this question.
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04-8-2011, 12:03 AM | #13 |
I am leonid
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Re: Damn Math... 25k Credits For Help?
come on now are you calling this hard or something
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04-8-2011, 12:04 AM | #14 |
It's Saint Pepsi bitch
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Re: Damn Math... 25k Credits For Help?
Last time I made a thread about getting help for math homework my thread was immediately deleted and I was given a warning. How this place has changed.
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04-8-2011, 12:05 AM | #15 |
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Re: Damn Math... 25k Credits For Help?
The homework forum is dead now
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04-8-2011, 01:21 AM | #16 |
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Re: Damn Math... 25k Credits For Help?
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04-8-2011, 01:34 AM | #17 |
Guuuuuuuuuuuuuuuuuuuuuurl
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Re: Damn Math... 25k Credits For Help?
I am taking another Algrebra course right now because I failed my first one. Math is my worst subject because I just don't understand it at all so I know how you feel. (other than basic math, that shit's easy >_>)
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04-8-2011, 08:05 AM | #18 |
x'); DROP TABLE FFR;--
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Re: Damn Math... 25k Credits For Help?
I am going to take a risk and answer based on just LOOKING at each problem for sub-10 seconds each because I'm a cocky douche who gets off on it:
Case 1: X has to be greater than 1 Case 2: X has to be bound between 0 and -4 EDIT: Shit I just realized a mistake I made, fixing it for Case 2. I failed ;-; That was like 20s. Last edited by Reincarnate; 04-8-2011 at 08:08 AM.. |
04-8-2011, 08:11 AM | #19 |
I am leonid
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Re: Damn Math... 25k Credits For Help?
nub marcus no 25k for you
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04-8-2011, 05:44 PM | #20 | |
Dark Chancellor
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Re: Damn Math... 25k Credits For Help?
Quote:
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