How far below the x-axis is the beam when x=0?

farkmedead · 27 Oct 2008, 3:15 PM

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Quote:

How did you find C?

I let f(b) = f(-b) but it didn't work... Got C = C hehe ^_^

did u do it properly? it took me a while to get it right too, but you find that c=o.

YoungToby182 · 27 Oct 2008, 3:16 PM

Dont worry about it, in the past now

Theres nothing you can do now.

Lanser7 · 27 Oct 2008, 3:34 PM

i got 10kb^4/24 because c=-10kb^4/24 and because it said how far BELOW the x axis that made it positve.

ps i only just realsied that simplifies to 5kb^4/12

Wingom · 27 Oct 2008, 4:06 PM

i'm kinda pissed.. coz i got the method right.. just a silly silly mistake...
which altered my answer, but I agree with it being 5kb^4/12..

Zephyrio · 27 Oct 2008, 4:11 PM

Looks good to me.

r.uss · 27 Oct 2008, 4:36 PM

-5kb^4/12 is correct. It would have been your constant when you integrated f'(x).

I'm seriously really wondering how on earth they're going to differentiate between everyone...

danz90 · 27 Oct 2008, 5:04 PM

lol i think me the idiot wrote,

it is -k5b^4/12 units below or whatever, when x=0

i hope they don't penalise me for putting that minus.
maybe i took the minus away who knows

sparkerasp · 27 Oct 2008, 8:32 PM

I got a bit stuck halfway through this question

Oh well, no use worrying.

r.uss · 28 Oct 2008, 7:08 AM

I think they're actually going to have to start taking marks off for putting minuses in the wrong places etc. because that is the only way they'll be able to differentiate between everyone.

I've heard that occasionally they become lenient towards the end because just about nobody usually gets them fully out, so the criteria is flexible. But in this case, I dont think they will!

anonomis · 29 Oct 2008, 2:50 PM

Quote:

Originally Posted by ipod21

i left my answer as y= - k5b^4/12

im wrong casue i left it like that...fuck!

y is that wrong? i did the same thing i think

mitchellstama · 2 Nov 2008, 10:41 PM

sub in x = b and y = 0 to find c...the first 2 posts have the right answer but i gt it wrong

27 Oct 2008, 3:16 PM	#17 (permalink)
YoungToby182 New Member HSC: 2008 Gender: Male Join Date: Aug 2008 Posts: 14 Last Activity: 3 Nov 2008, 1:39 PM	Re: How far below the x-axis is the beam when x=0? Dont worry about it, in the past now Theres nothing you can do now.

27 Oct 2008, 3:34 PM	#18 (permalink)
Lanser7 New Member HSC: 2008 Gender: Male Join Date: Jun 2008 Posts: 4 Last Activity: 21 Jan 2009, 11:44 AM	Re: How far below the x-axis is the beam when x=0? i got 10kb^4/24 because c=-10kb^4/24 and because it said how far BELOW the x axis that made it positve. ps i only just realsied that simplifies to 5kb^4/12

27 Oct 2008, 4:06 PM	#19 (permalink)
Wingom New Member HSC: 2008 Gender: Male Join Date: Apr 2007 Posts: 11 Last Activity: 18 Dec 2008, 3:46 PM	Re: How far below the x-axis is the beam when x=0? i'm kinda pissed.. coz i got the method right.. just a silly silly mistake... which altered my answer, but I agree with it being 5kb^4/12.. __________________

27 Oct 2008, 4:11 PM	#20 (permalink)
Zephyrio Exalted Member HSC: 2008 Gender: Male Join Date: Oct 2006 Posts: 950 Last Activity: 12 Feb 2010, 5:18 PM	Re: How far below the x-axis is the beam when x=0? Looks good to me. __________________ MBBS/BA @ UNSW Tutoring offered by UNSW Medicine/Arts student in English (up to 4U), maths, modern history, extension history and chemistry. Drop me a word at lawsonung@gmail.com.

27 Oct 2008, 4:36 PM	#21 (permalink)
r.uss Assistant Member HSC: 2008 Gender: Male Join Date: May 2007 Posts: 55 Last Activity: Yesterday, 10:31 PM	Re: How far below the x-axis is the beam when x=0? -5kb^4/12 is correct. It would have been your constant when you integrated f'(x). I'm seriously really wondering how on earth they're going to differentiate between everyone...

27 Oct 2008, 5:04 PM	#22 (permalink)
danz90 Supreme Member HSC: 2008 Gender: Male Location: Sydney Join Date: Feb 2007 Posts: 1,464 Last Activity: Today, 1:05 AM	Re: How far below the x-axis is the beam when x=0? lol i think me the idiot wrote, it is -k5b^4/12 units below or whatever, when x=0 i hope they don't penalise me for putting that minus. maybe i took the minus away who knows __________________ *B.Pharm. II*

27 Oct 2008, 8:32 PM	#23 (permalink)
sparkerasp Member HSC: 2008 Gender: Male Join Date: Feb 2007 Posts: 196 Last Activity: 18 Oct 2009, 5:01 PM	Re: How far below the x-axis is the beam when x=0? I got a bit stuck halfway through this question Oh well, no use worrying.

28 Oct 2008, 7:08 AM	#24 (permalink)
r.uss Assistant Member HSC: 2008 Gender: Male Join Date: May 2007 Posts: 55 Last Activity: Yesterday, 10:31 PM	Re: How far below the x-axis is the beam when x=0? I think they're actually going to have to start taking marks off for putting minuses in the wrong places etc. because that is the only way they'll be able to differentiate between everyone. I've heard that occasionally they become lenient towards the end because just about nobody usually gets them fully out, so the criteria is flexible. But in this case, I dont think they will!

2 Nov 2008, 10:41 PM	#26 (permalink)
mitchellstama New Member HSC: 2008 Gender: Male Join Date: Jul 2006 Posts: 2 Last Activity: 27 Oct 2009, 1:36 PM	Re: How far below the x-axis is the beam when x=0? sub in x = b and y = 0 to find c...the first 2 posts have the right answer but i gt it wrong

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