

11062008, 10:45 PM

#85

You have gnarly titties.

I have no clue what the hell you are saying. Please speak in coherent sentences. I was already helping that other guy with calculus, and now I'm going to my gf's house.
If you're looking for help posting, pretend you're trying to communicate to somebody by speaking, and just type those words you would say.



11062008, 10:48 PM

#86

Hip 2 be Square
Join Date: Oct 2008
Location: Surrey, BC

Quote:
Originally Posted by BVCballa
that dont help me if anyone could i would be so thankful

I suggest you grow a ****ing cerebrum you pathetic dolt. Actually a better suggestion would be for you to leave this thread, subforum and forum entirely. Goodday.



11062008, 10:49 PM

#87

Give Me More...
Join Date: Mar 2005
Location: Denham Springs, LA

Quote:
Originally Posted by Strawberryblonde.
I tried it again, using u=3x and v=sec(3x), then du=3 and dv=3*sec(3x)tan(3x), then applying the product rule:
f'(x)=udv+vdu
=(3x)(3*sec(3x)tan(3x))+(3)(sec(3x))
=9x*sec(3x)tan(3x)+3sec(3x)
That's the exact same answer I got the first time, so I think I got it right.

Ah, my bad. I didn't mean that one. I mean f(x) = sqrt(sin2x)
Its first on my WS.
__________________
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LFC FTW
"Past experience tells me to hold the phone a ways from my ear when I hear the little click of the trigger"



11062008, 10:51 PM

#88

You have gnarly titties.

I'm pretty sure I did that right, but it would be really helpful if you would post what you got, so I could use it as a reference. Alas, I must be going now, I'll be back around noon or so, my calc II class is cancelled for tomorrow. (yay.)



11062008, 10:55 PM

#89

OHIO
Join Date: Jul 2008
Location: Arcadia, Oh

school help
alright if anyone could help me in a bcs problem solution paper i would be thankful. i need three solutions and a counter argument please help. thanks



11062008, 10:58 PM

#90

You have gnarly titties.

I solved the derivative and then took the antiderivative and got the original function, so I'm pretty sure I did that one right, too.
There we go.
I converted the square root to the exponent 1/2, then just applied the power rule, and used the chain rule twice.
After that, it was simple algebra.
Last edited by Strawberryblonde. : 11062008 at 11:00 PM.



11062008, 10:58 PM

#91

Give Me More...
Join Date: Mar 2005
Location: Denham Springs, LA

Quote:
Originally Posted by Strawberryblonde.
I'm pretty sure I did that right, but it would be really helpful if you would post what you got, so I could use it as a reference. Alas, I must be going now, I'll be back around noon or so, my calc II class is cancelled for tomorrow. (yay.)

f(x) = (sin2x)^(1/2)
= (1/2)(sin2x)*cos2x
= (1/2)(cos2x)(sin2x)
Thats what i got...lol.
__________________
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LFC FTW
"Past experience tells me to hold the phone a ways from my ear when I hear the little click of the trigger"



11062008, 11:02 PM

#92

You have gnarly titties.

Quote:
Originally Posted by JoeBum
f(x) = (sin2x)^(1/2)
= (1/2)(sin2x)*cos2x
= (1/2)(cos2x)(sin2x)
Thats what i got...lol.

Your problem is that you only subtracted 1/2 from the exponent, and you need to subtract 1.
So d/dx[sin(2x)^(1/2)]=(1/2)[sin(2x)^(1/2  1)]*cos(2x)*2=(1/2)[sin(2x)^(1/2)]*cos(2x)*2
=cos(2x)/[sin(2x)^(1/2)]
You have to use the chain rule twice, I believe.
Last edited by Strawberryblonde. : 11062008 at 11:05 PM.



11062008, 11:10 PM

#93

Give Me More...
Join Date: Mar 2005
Location: Denham Springs, LA

Quote:
Originally Posted by Strawberryblonde.
Your problem is that you only subtracted 1/2 from the exponent, and you need to subtract 1.
So d/dx[sin(2x)^(1/2)]=(1/2)[sin(2x)^(1/2  1)]*cos(2x)*2=(1/2)[sin(2x)^(1/2)]*cos(2x)*2
=cos(2x)/[sin(2x)^(1/2)]
You have to use the chain rule twice, I believe.

Ahh, I completely forgot that. I'll rework it and see what I get.
__________________
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LFC FTW
"Past experience tells me to hold the phone a ways from my ear when I hear the little click of the trigger"



11092008, 04:04 PM

#94

hey baby
Join Date: Mar 2005
Location: mass

does anyone know the organelles that are in a melanocyte?



11092008, 04:21 PM

#95

Double Up
Join Date: Oct 2006
Location: Massachusetts

Quote:
Originally Posted by BVCballa
hey does anyone have a good problem in sports right now? I dont want to do steroids because some one in my class is doing that already. The paper is a problemsolution paper and i need three solutions to my problem. I want to do something with sports because I usually do the best on my papers when I am actually interested in what im writing about. Any help would be greatly apprectiated.

Should college athletes be paid or not, easy plenty of things to consider. Just take a stance. Do some research on how much some schools make on college games, you'd be shocked, think millions. I do not think they should be payed though, I don't have the breath to explain.



11092008, 05:02 PM

#96

Join Date: Mar 2006
Location: Chicago, IL

I'm trying to solve this circuit for Isupply, Vbc, and I1. The circuit I was given is drawn in black and anything I solved is in red... Idk if that is even right haha.
Any and all help appreciated.



11092008, 08:12 PM

#97

It's fizzix.
Join Date: Apr 2005
Location: Illinois

A steel belted radial automobile tire is inflated to a gauge pressure of 30.0 lb/in^2 when the temperature is 61 F. Later in the day, the temperature rises to 100 F. Assuming that the volume of the tire remains constant, what is the tire's pressure at the elevated temperature? [hint: remember that the ideal gas law uses absolute pressure.]
The answer is 33.4 lb/in^2, but I haven't been able to get that.
Edit: I got it now. I wasn't subtracting 1 atm from the final pressure so it would read as the gauge pressure.
Last edited by nflvikings : 11092008 at 08:36 PM.



11102008, 09:51 PM

#98

Voice of the Soul
Join Date: Jun 2007
Location: LSU

approximate the intergral(2x^27x+1 dx) using n=6 subdivisions and midpoints for sample points. a=0(bottom),b=2(top)
The answer is 20/3, I need an explanation of how to solve it in the above method,I've used the antiderivative substitution and limit to find and confirm the answer so I just need an explanation of how to get there.
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11102008, 09:55 PM

#99

You have gnarly titties.

Looks like you're using Riemann sums. Lemme look up how to do those again.



11102008, 09:58 PM

#100

Voice of the Soul
Join Date: Jun 2007
Location: LSU

Quote:
Originally Posted by Strawberryblonde.
Looks like you're using Riemann sums. Lemme look up how to do those again.

Yeah,thanks in advance
__________________
NCAA ThreadCome feel the hate



11102008, 09:59 PM

#101

You have gnarly titties.

My damn laptop power cord is about to break.
Are you looking for the approximate integral from 0 to 2?



11102008, 10:16 PM

#102

You have gnarly titties.

If it's from 0 to 2, you want to find ∆x, the length of each rectangle under the curve. The equation for this is ∆x=(ba)/n.
ba = 20
n=6
So, (20)/6 = 1/3.
Now, you want to imagine the function divided into 6 rectangles of equal length from 0 to 2, or each 1/3 units long. At the midpoint of each of these rectangles, you want to find the value of ƒ(x), so you know the height of the rectangle and can find the area of each rectangle and eventually the approximate area underneath the curve.
So you take R1, it stretches from 0 to 1/3, the midpoint is at (1/3)/2 = 1/6.
The midpoint of R2 is at 1/6+(∆x)=1/6+1/3=1/2
I guess the general equation for this would be R(N)=(N)*∆x∆x/2=∆x(N1/2). But I just made that up, so I could be wrong.
Now you find the value at each midpoint, and all these added up and multiplied by the length of each rectangle should give you the sum of the areas of the rectangles.
ƒ(x)=2(x^2)7x+1
∆x[ƒ(1/6)+ƒ(1/2)+ƒ(5/6)+...+ƒ(11/6)]



11112008, 09:11 AM

#103

Voice of the Soul
Join Date: Jun 2007
Location: LSU

Thanks,you're a lifesaver. It was a takehome quiz. I did two other methods of solving it and got the same answer. Thanks.
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NCAA ThreadCome feel the hate



11112008, 01:38 PM

#104

5 Santa Hats

Hey Strawberry, although I don't need your help I think it's awesome you are helping people out. Great to know there are still good people out there.
My hat is off to you, sir.
__________________
Why not Zoidberg?



11112008, 02:17 PM

#105

You have gnarly titties.

Thanks a lot, I love to help when I can, and I love to help with basic calc. I would help that other guy up there, but I'm terrible at circuit analysis due to complete laziness during a full semester of electricity and magnetism.
Facepalm to me.



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