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Do this easy math problem for me real quick.

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alpinecowboy84
Posts: 3120 - Karma: 25
i'm taking quantitative anthropology (stats essentially) which is required for my major. in any case its been forever since i've done anything more complicated than division. i can't remember any of the rules according to square roots. so if you could please tell me what the answer to this question is and then if possible explain how you got there it would be appreciated

we'll just say that / means square root sign.

okay, the problem:

(solve for k) k = /y2(squared) - 4

because i know k does not equal (/y2) - 2, correct?







indie lovers unite! the independent music cult
Sep 26 2005 12:14AM
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alpinecowboy84
Posts: 3120 - Karma: 25
the problem should read k = /(y2 - 4)







indie lovers unite! the independent music cult
Sep 26 2005 12:15AM
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mnjib
Posts: 5592 - Karma: 362
so wait are you saying that its :
(solve for k) k = /y2(squared) - 4
as in
k=/Y(times)2(squared)-4

or

k=/(y2)^2-4



snap crackle pop
Sep 26 2005 12:18AM
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mnjib
Posts: 5592 - Karma: 362
so is it y^2?


snap crackle pop
Sep 26 2005 12:20AM
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alpinecowboy84
Posts: 3120 - Karma: 25
k=/[(y^2) - 4]







indie lovers unite! the independent music cult
Sep 26 2005 12:44AM
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alpinecowboy84
Posts: 3120 - Karma: 25
COME ON YOU DUMB SLUTS. YOU DO IT.







indie lovers unite! the independent music cult
Sep 26 2005 12:50AM
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alpinecowboy84
Posts: 3120 - Karma: 25
/ is supposed to mean 'square root of'







indie lovers unite! the independent music cult
Sep 26 2005 1:20AM
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squeakywaffle
Posts: 1800 - Karma: 13
uhh, isn't it already solved for k?

I can't see any way of simplifying that. If it was a perfect square in the root, then you could get rid of the root. But it isn't.

I might be missing something really obvious. I'm not really that great at math.

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Sep 26 2005 3:21AM
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frankenstien
Posts: 143 - Karma: 10
If you want to solve for y:
k=√(y2-4)
k2=y2-4
k2-4=y2
so y=
y= √(k2-4)

Thats if you wanted to solve for y was it?
i wasnt sure.....

Sep 26 2005 6:45AM
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SUpilot
Posts: 5267 - Karma: 15
well you have K already solved...so whats the question?

Sep 26 2005 7:21AM
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KEVTRON
Posts: 3545 - Karma: 1,222
lol ur funny

Sep 26 2005 7:25AM
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alpinecowboy84
Posts: 3120 - Karma: 25
hey, i don't write the problems, i just try to do them.







indie lovers unite! the independent music cult
Sep 26 2005 10:50AM
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jsg
Posts: 7701 - Karma: 535
fuck i dont kno this is way to confusinggg

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Sep 26 2005 11:08AM
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CooksCorner
Posts: 3939 - Karma: 22
I love lamp.




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Sep 26 2005 1:33PM
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Aseets
Posts: 2171 - Karma: 17
k=/(y^2-4)
k=/[(y-2)(y+2)] (difference of two squares)
k^2=+or- [(y-2)(y+2)]

k^2=(y-2)(y+2)
or
k^2=-(y-2)(y+2) i don't know what you would do with this but thats your job, and i thinkthis is right

Let us have a revolution
Sep 26 2005 2:04PM
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Aseets
Posts: 2171 - Karma: 17
you forgot that when you square both sides you have to do the + or - thing

Let us have a revolution
Sep 26 2005 2:06PM
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SteezePatrol
Posts: 10281 - Karma: 36
loud noises?

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Sep 26 2005 2:11PM
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assassin167
Posts: 908 - Karma: 10
i love couch

french is wierd too. the movie "mouse hunt" in french was called " nous reveillons pas un souris qui dors" meaning we wont wake up a mouse thats sleeping.
-Lat
Sep 26 2005 2:22PM
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clrumivier4life
Posts: 2429 - Karma: 9
I don't think you can do anything past what he said. Factoring is the only obvious thing to do, but that doesn't give you two equivalent factors. You want to solve for k, but if you look, you already have what k is equivalent to and factoring would only make the problem more complex...so...I think you're stuck.

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Sep 26 2005 2:58PM
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alpinecowboy84
Posts: 3120 - Karma: 25
not a problem. thanks for the help. i turned it in earlier this morning.







indie lovers unite! the independent music cult
Sep 26 2005 3:14PM
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sim
Posts: 1399 - Karma: -6
In your question up above did you mean the whole thing was square rooted? Because the way you wrote it, without brackets, makes it look like just the y^2 was square rooted. If that was the case, the square root and the square cancel out. However, I have a feeling you know this, but you have to write out the question right or else it gets confusing.

signatures are for pussies
Sep 26 2005 4:16PM
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sniper_6
Posts: 1919 - Karma: 1,573
do it urself u jew

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Sep 26 2005 4:39PM
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PimpDaddySkier
Posts: 4221 - Karma: 20
hes GOT IT

Sep 26 2005 4:42PM
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Norred
Posts: 3520 - Karma: 13
8


=============================

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Sep 27 2005 5:19PM
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skiierman
Posts: 28699 - Karma: -5,563
Apple.

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Sep 27 2005 5:24PM
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396222687862
Posts: 0 - Karma: 350
i am soo confused....

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Sep 27 2005 6:39PM
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eheath
Posts: 61603 - Karma: 123,089
the answer is 1

what's up now bitch



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Sep 27 2005 6:47PM
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Jake...
Posts: 11976 - Karma: 296
K= y2 -4






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Sep 27 2005 7:58PM
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oakjewlery
Posts: 4905 - Karma: 32
dude! i'm still in intro to high school math! don't ask mei i'm a dumbass

no

I am the sex


ABSTRACT 4 LIFE
Sep 27 2005 8:03PM
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k.strobel
Posts: 32 - Karma: 11
k=/((y^2)-(4))
k=/(y^2)-/(4)
k=y-2

Sep 27 2005 9:02PM
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Leckett
Posts: 0 - Karma: 13
^impossible. you cant work on one side of the equation without doing the same on the other side. if you want to solve for k you already have your answer. y on the other hand i forget who but someone solved it and it looked right to me. it doesnt need a + and - because the equation is a square root. basicly solving for y is finding the inverse of the equation. well i guess technicly u need a + and - but you have to add that {k|k>(or equal to) i think it was -4?} i dont remember. but u have to include an inequality because that equation turned out as a full parabola. square root equations are only half the parabola

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Sep 27 2005 9:09PM
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canadianskier
Posts: 601 - Karma: 184
k=y-2

Sep 27 2005 9:12PM
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skiierman
Posts: 28699 - Karma: -5,563
Parabola this! *Whips out dick and cock slaps the computer screen*

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Sep 27 2005 9:23PM
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sim
Posts: 1399 - Karma: -6
Just so you know this is not hard math. Not saying your dumb, just saying in general and it is quite easily solveable. I actually just think you wrote it down wrong up there.

signatures are for pussies
Sep 28 2005 12:38PM
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skierchild
Posts: 1743 - Karma: 35
k=/(y^2-4)

---------------
26 Productions
Sep 28 2005 12:56PM
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jtski
Posts: 805 - Karma: 21
this one

G R I P S E T
Sep 28 2005 12:59PM
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Hctiws_iks
Posts: 2077 - Karma: 25
k= square root of y+2 y -2 so it is not a perfect square so you cannot fact out the 4,
so y^2 -4 cannot be simplified.

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www.ihateny.net
Sep 28 2005 1:15PM
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Hctiws_iks
Posts: 2077 - Karma: 25
if you are trying to solve for y then the one guy already said it,
or it could be written as y^2-k^2=4
wichic is the only other way you really would want the equation for, but that way wouldnt really be used for stat i dont think anyway

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Sep 28 2005 1:18PM
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daniel_insam
Posts: 1494 - Karma: 33
fuck maths

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Sep 28 2005 2:04PM
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Astomp
Posts: 10090 - Karma: 344
julian your just retarded as it is. you wont ever figure that out. stupid cannuck.

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Sep 28 2005 3:06PM
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bermbuster49
Posts: 7445 - Karma: 90
haah if its so easy why cant you do it?


-kulpy-

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Sep 28 2005 3:21PM
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Anathema
Posts: 21048 - Karma: 5,061
and this is why i dropped out of university. it's not that hard to do, but i hate thinking.

stick that in your pipe and smoke it!

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Sep 28 2005 3:34PM
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murphski
Posts: 1259 - Karma: 13
wait so, KY jelly + 2 people = fun!?

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Sep 28 2005 3:43PM
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assassin167
Posts: 908 - Karma: 10
thats very hard

french is wierd too. the movie "mouse hunt" in french was called " nous reveillons pas un souris qui dors" meaning we wont wake up a mouse thats sleeping.
-Lat
Sep 28 2005 3:58PM
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CanadianSkiGod
Posts: 2281 - Karma: 10
Well yeah... (y^2-4) can be simplified... it's a difference of squares. Although if you want to solve for K... it's already solved. the only thing you can do is this:

k =

/ (y^2 - 4) =

/ (y-2) (y+2) = k

You can't get much simpler than that. Unless of cours you have k's value, in which case you know what to do.

What I did is a simple rule you are supposed to have learned in 10th grade: in a expression of second degree with a differnce of a squared variable (y, x, whatever), and of a natural number, you can simplify by assuming that the two factors are that variable (y or x) at the first degree less the natural number's square root for one and added to the square root for the other.

Difficult to explain on NS, but these examples should make it simple:

(x^2 - 4) = (x-2) * (x+2)

(y^2 - 49) = (y+7) * (y - 7)

(k^2 - 100) = (k+10) * (k- 10) = Easy to understand.

That problem seems a little simple. Are you sure it wasn't solve for y or for 4???

~~Phunkin Phatt Phreerider~~

Erm... yeah.
Sep 28 2005 4:23PM
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1440ski
Posts: 2049 - Karma: 1,110
Wow way to confusing for me


Sep 28 2005 4:56PM
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back_woods
Posts: 1579 - Karma: 10
www.quickmath.com

It Aint Me Babe
Sep 28 2005 5:19PM
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Hctiws_iks
Posts: 2077 - Karma: 25
^^ yea i guess i miss spoke, what i meant to say
was having y+2 and y-2 under the radical really does nothing for the problem

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Sep 28 2005 5:19PM
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GotamaBCshredder
Posts: 5062 - Karma: 424
DUDE, YOU CAN'T DO THAT, ITS ALREADY simplified. That is the answer for k.

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Sep 28 2005 5:49PM
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