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This one is a doozy.. X/x^24 (greater than or equal to sign) 0. How would I solve ?
 one year ago
 one year ago
This one is a doozy.. X/x^24 (greater than or equal to sign) 0. How would I solve ?
 one year ago
 one year ago

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lgbasalloteBest ResponseYou've already chosen the best response.0
\[\frac x{x^2  4} \ge 0\] ???
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
yes that is the problem
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
cross multiply..what do you get?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
where did you get =?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
should be \(\ge\) not =
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
your initial sign was \(\ge\) so you can't change it
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
so my answer is correct ? or is there more to do ?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or 2
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Uh... @lgbasallote , that answer isn't right.
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
Oh ok, just making sure because I am supposed to solve this as an inequality and show a sign chart as well expressing in interval notation
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
then why don't you show how it's done, oh great one?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
It's \(x\in (2, 0]\cup (2,\infty)\) Well, that's what I'm getting... let me check.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i did say x cannot be 2 or 2
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
According to Mathematica, that seems to be the case. And, I'm not trying to make you look bad or anything, jeez, it's a mistake, and it happens, don't take it personally. Check the boundaries yourself.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you did read what i was saying right?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Yes, but, it's supposed to be: \[ x>2\\ 2<x\le 0 \]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
I'd look at where x^34x is positive and negative...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
simplest way I think...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i didn't say any signs did i? i said x cannot be 2 or 2
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
So how does one get to this conclusion? If you dont mind me asking.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and i was also insulting the mathematicians
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
multiply both sides by (x^24 )^2
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
I'm actually not sure what the 'standard' way is to look at these sorts of inequalities... I've forgotten maybe. Someone else might have a more kosher method.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\[\frac x{x^2  4} \ge 0\] \[\frac x{(x2)(x+2)} \ge 0\] changes sign at \(2,0,2\) so only need to check on interval. say \(x<2\) i pick \(x=3\) we get \[\frac{3}{(3)^24}=\frac{3}{5}\] which is negative so it is negative, positive, negative, positive on the respective intervals \[(\infty,2), (2,0), (0,2), (2,\infty)\]
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Here's a counter example to your solution: \[ x=1\ge0\\ \frac{1}{14}=\frac{1}{3}\not \ge\; 0 \]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
counter example to what?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you're a mathematician @LolWolf ...you know that greater than and less than are under "inequalities" => not equal to and again i repeat...i was insulting the meticulousness of mathemeticians
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
(@lgbasallote 's not yours, @satellite73 , haha)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
if it's a computer you're trying to please
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
Thank you. The other question I have goes like this : a^5/3 * b^2/3 * c^3 / C^2 * a^2/3 * b^4/3 .
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
for example on (a) would leave denominator 3 alone and just add the numnerators?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
Hope you're not asking about adding fractions :P
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
My answer is A^1 * b^6/3 * c^5
 one year ago

TombrokawBest ResponseYou've already chosen the best response.0
would this be correct?
 one year ago
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