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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac x{x^2  4} \ge 0\] ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes that is the problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cross multiply..what do you get?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should be \(\ge\) not =

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your initial sign was \(\ge\) so you can't change it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so my answer is correct ? or is there more to do ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Uh... @lgbasallote , that answer isn't right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i did say x cannot be 2 or 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0According 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you did read what i was saying right?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'd look at where x^34x is positive and negative...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0simplest way I think...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i didn't say any signs did i? i said x cannot be 2 or 2

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i was also insulting the mathematicians

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0multiply both sides by (x^24 )^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac x{x^2  4} \ge 0\] \[\frac x{(x2)(x+2)} \ge 0\] 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 \[(\infty,2), (2,0), (0,2), (2,\infty)\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0counter example to what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you'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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(@lgbasallote 's not yours, @satellite73 , haha)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it's a computer you're trying to please

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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 .

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hope you're not asking about adding fractions :P

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would this be correct?
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