Post by: Blakhitman on January 26, 2010, 03:54:51 pm
Quick question, say I'm differentiating:
Can I leave the answer as
Or should I expand it to get
If I leave it at the first answer, will I get full marks?
Thanks!
Post by: naved_s9994 on January 26, 2010, 05:04:29 pm
Post by: Blakhitman on January 26, 2010, 05:45:50 pm
Post by: /0 on January 26, 2010, 06:45:55 pm
Post by: Blakhitman on January 27, 2010, 01:46:03 am
Thanks.
Post by: jimmy999 on January 27, 2010, 05:41:10 pm
Post by: superflya on January 27, 2010, 05:45:22 pm
Both answers would get you the marks. However if you choose to expand it, then you might make a mistake in your expansion thus losing you a mark. If your answer ends up in factored form, leave it like that, or if it's in expanded form, leave it like that. However if the question explicitly asks you to leave it in a certain form, then you must put it in that form or you will lose a mark
+1
Edit: 500 posts :D
Post by: Blakhitman on February 06, 2010, 12:34:34 am
Show that
need help on how to start this...I've been trying and it's late and I'm going to sleep so I better find help tomorrow!!!!!! :knuppel2:
Please and thanks :) .
Post by: the.watchman on February 06, 2010, 08:57:03 am
Then again, that could just be my mistake...
Post by: Blakhitman on February 06, 2010, 09:48:42 am
Post by: the.watchman on February 06, 2010, 09:51:28 am
Because
So there is always a repeated factor of (x-3)
Post by: the.watchman on February 06, 2010, 09:55:41 am
The final term of c(x) is
Post by: Blakhitman on February 06, 2010, 10:07:50 am
Appreciate it!
Post by: the.watchman on February 06, 2010, 01:03:15 pm
Post by: Mao on February 06, 2010, 01:05:51 pm
Actually I just saw a shortcut:
The final term of c(x) is
Not necessarily. The final term can also be
Post by: the.watchman on February 06, 2010, 01:15:58 pm
Actually I just saw a shortcut:
The final term of c(x) is
Not necessarily. The final term can also be, you have to use the factor theorem to check.
Ah, I get what you mean now!
I was working after the knowledge that x-k is a factor (but i guess factor theorem is best :P)
Post by: the.watchman on February 06, 2010, 01:51:25 pm
Post by: Blakhitman on February 20, 2010, 10:55:33 pm
The cross-section of a theme park ride has a central structure as shown below.
(http://img12.imageshack.us/img12/5166/654zn.png)
The sides of the structure can be
modelled by (http://img515.imageshack.us/img515/6235/asdasdt.png)
The diameter of the top of the structure
is 20 m and the height of the structure is
40 m. The base of the structure has a
radius of 50 m.
(a) Find the value of a and b.
Post by: stonecold on February 20, 2010, 10:58:16 pm
and do simulanius solve?
Post by: Blakhitman on February 20, 2010, 10:58:43 pm
Post by: Blakhitman on February 20, 2010, 11:00:52 pm
hmmm oh well...
Thanks anyway!
EDIT: Got it...stupid me!
Post by: stonecold on February 20, 2010, 11:07:01 pm
and a= 125000/3
i screwed it yeah?
Post by: GerrySly on February 20, 2010, 11:09:31 pm
i got b=-50/3
and a= 125000/3
i screwed it yeah?
One too many zeros heh
Post by: stonecold on February 20, 2010, 11:11:33 pm
Post by: Blakhitman on February 20, 2010, 11:15:30 pm
Thanks a lot guys.
Post by: coletrain on February 21, 2010, 10:24:10 am
did you try subbing in points (50,0) and (10,40)
and do simulanius solve?
how did you get (10,40)
Post by: the.watchman on February 21, 2010, 10:28:58 am
So each side has x-coordinates -10 and 10, also, it is 40m high, so the y-coordinate of both points is 40
Post by: coletrain on February 21, 2010, 10:34:14 am
Post by: the.watchman on February 21, 2010, 12:22:09 pm
thanks
No prob, worded questions are harder to interpret :)
Post by: Blakhitman on February 26, 2010, 09:08:37 pm
Thanks!
Post by: TrueTears on February 26, 2010, 09:11:52 pm
1 - 2k+5=0
k = 3
lim_{x -> -infty} f(x) = 5
so a = 5
f(x) = 0 we need to find x.
let u = e^(x)
u^2 - 6u + 5 = 0
solve for u and then x.
Post by: TrueTears on February 26, 2010, 09:35:11 pm
Thus for every
But
But
Post by: Blakhitman on February 26, 2010, 09:49:45 pm
haha...wish I could return the favour by helping you with that...not gonna happen
Post by: TrueTears on February 26, 2010, 09:56:39 pm
Post by: kamil9876 on February 26, 2010, 10:09:20 pm
Post by: TrueTears on February 26, 2010, 11:00:24 pm
I wouldn't bother with something like that. Better of proving the basic general laws and simply then applying them to special cases like these (although Kamil doesn't like to bother with these kinds of very special cases anyway).well how would you finish off the proof, can you show me?
Post by: Blakhitman on March 01, 2010, 07:44:35 pm
The graph of y=|x| is transformed by the following sequence of transformations
- Translation of 5 units in the positive direction of the x-axis
- Translation of 6 units in the negative direction of the y-axis
- reflection in the x-axis
What's the rule of the image...
Post by: naved_s9994 on March 01, 2010, 07:51:44 pm
Modulus is slightly different to other functions for transformations.
Post by: Blakhitman on March 01, 2010, 07:55:33 pm
(a) y=6-|x-5|
(b) y=6+|x+5|
(c) y=-6+|x-5|
(d) y=6+|5-x|
(e) y=-6-|5-x|
Post by: the.watchman on March 01, 2010, 07:56:28 pm
y= - |x - 5|-6
Modulus is slightly different to other functions for transformations.
Actually, wouldn't it be
The reflection occurs after the translation...
Post by: naved_s9994 on March 01, 2010, 07:57:15 pm
Isnt it Dialation Reflection Translation..?
Post by: superflya on March 01, 2010, 07:59:10 pm
Post by: the.watchman on March 01, 2010, 07:59:57 pm
Im not sure.
Isnt it Dialation Reflection Translation..?
But they specifically GIVE the sequence, so you have to follow what the question says? (I think)
Confirmation from anyone else? :)
Post by: naved_s9994 on March 01, 2010, 08:00:36 pm
DRT or RDT, both work.
Ohh okay, wasnt sure of RDT.
I knew or DRT. Hence my answer..
Post by: naved_s9994 on March 01, 2010, 08:01:37 pm
Im not sure.
Isnt it Dialation Reflection Translation..?
But they specifically GIVE the sequence, so you have to follow what the question says? (I think)
Confirmation from anyone else? :)
I have no idea about that..
Yea, please...Could anyone confirm?
Post by: Blakhitman on March 01, 2010, 08:05:23 pm
y= - |x - 5|-6
Modulus is slightly different to other functions for transformations.
Actually, wouldn't it be?
The reflection occurs after the translation...
YES you're right...you freaking genius! :)
But how did you get to that, I don't get it, so what if the reflection occurs after the translation?
I put (e) but it was wrong.
Post by: naved_s9994 on March 01, 2010, 08:08:26 pm
Then in that case translation the reflection.
Go by whats in question, as stated by Watch.
Post by: Blakhitman on March 01, 2010, 08:20:56 pm
like I don't know how I'd end up with a positive 6 from x-axis translation if I hadn't sketched...
Post by: the.watchman on March 01, 2010, 08:23:39 pm
Oh I just tried sketching each step and yeah I get it now, but how do I do it without sketching the graph?
like I don't know how I'd end up with a positive 6 from x-axis translation if I hadn't sketched...
Whew, I'm glad I got it right :D
If you let
Then horizontal translation:
Vertical translation:
Reflection:
Hope this makes some sense!
Post by: Blakhitman on April 04, 2010, 06:02:56 pm
ok question...how do you do this?
If
Post by: /0 on April 04, 2010, 08:15:32 pm
So they are inverse functions, and are symmetrical about
Thus, if they intersect, say at
Post by: Blakhitman on April 04, 2010, 09:00:21 pm
Sorry if I sound like an idiot, and if it's really simple, I'll vow to always allow a period of 2 hours to think about the question :P
Post by: the.watchman on April 04, 2010, 09:20:45 pm
thanks heaps Watchman...just saw this...
ok question...how do you do this?
Ifhas exactly one solution, find the value of
correct to three decimal places.
Haha, more than a month later :P
I tried using /0's method, but got a bit stuck, my head's still in a daze after too much music rehearsal.
Anyone else trying this one? :)
EDIT: Maybe try solving
AND
That may work out better than what I was doing earlier, I really can't do it in this state of mind :P
Post by: Blakhitman on April 04, 2010, 09:36:27 pm
EDIT: I'm starting to think maybe my teacher forgot some info with this question.
Post by: the.watchman on April 04, 2010, 09:59:19 pm
haha I said thanks -_-...but nah it'll help for the SAC and upcoming stuff anyway.
EDIT: I'm starting to think maybe my teacher forgot some info with this question.
Don't think so, try subbing the first eqn into the second and solve for a, I think that's how to do it :)
Post by: brightsky on April 04, 2010, 10:17:01 pm
Post by: Blakhitman on April 04, 2010, 10:22:31 pm
Post by: Blakhitman on April 04, 2010, 10:26:25 pm
Post by: the.watchman on April 04, 2010, 10:26:42 pm
Hmm..three equations, three unknowns, should be able to get the answer...
Erm two, f(x)=x and f^-1(x)=x are the same
The only two you want are f(x)=x and f'(x) = [f^-1(x)]', which I have in my previous post
Can someone find the value of a then please? I still can't get it :S
I'm sorry, I would but I need to sleep :)
(Sorry to hijack, but if anyone is free, MYO chamber strings is doing a concert tomorrow at St John's Southgate at 3:00, would be great if any of you can make it, sorry again :P)
Post by: the.watchman on April 04, 2010, 10:46:37 pm
I don't think you can let y=everything, as obviously the derivative graphs DO NOT follow the relation y=x, only the originals
Anyway:
Because
Therefore subbing into (1):
Subbing this into
Not too sure about this one though
Post by: Blakhitman on April 04, 2010, 10:55:18 pm
the.watchman I think you're right though, I subbed the a values you both got into the first two expressions and sketched them, the one watchman got had a point of intersection while brightsky's didn't.
Thanks heaps!
Post by: the.watchman on April 04, 2010, 11:07:31 pm
Thanks so much guys.
the.watchman I think you're right though, I subbed the a values you both got into the first two expressions and sketched them, the one watchman got had a point of intersection while brightsky's didn't.
Thanks heaps!
No prob, that's quite a nice problem :P
Post by: brightsky on April 04, 2010, 11:10:58 pm
Post by: Blakhitman on April 04, 2010, 11:12:04 pm
But he's the best teacher...so I forgive him :D
Post by: TrueTears on April 04, 2010, 11:12:35 pm
Ifpr0 shit right herethen
So they are inverse functions, and are symmetrical about.
Thus, if they intersect, say atthe intersection must lie on the line
,
is concave up and
is concave down, so if they only intersect once then they must 'touch' and not 'cross' - this means that their derivatives must be equal at the point of intersection. (draw this to see)
Post by: the.watchman on April 05, 2010, 09:50:54 am
Ifpr0 shit right here
So they are inverse functions, and are symmetrical about
Thus, if they intersect, say at
Heck yeah, I bet 99% of MM students wouldn't have thought of that :P
Post by: Blakhitman on April 07, 2010, 11:48:49 pm
I remember reading that one of them isn't right...don't know why though ><.
Post by: TrueTears on April 07, 2010, 11:49:36 pm
So is it,both are right.or
?
I remember reading that one of them isn't right...don't know why though ><.
i fink u mean |x| =\= (rt{x})^2
Post by: Blakhitman on April 07, 2010, 11:50:16 pm
That makes sense lol.
Post by: Blakhitman on April 15, 2010, 06:34:39 pm
I got
Post by: the.watchman on April 15, 2010, 06:37:57 pm
Solve this algebraically or graphically to get
So the domain is
Post by: Blakhitman on April 15, 2010, 06:51:13 pm
Post by: Blakhitman on April 15, 2010, 07:24:29 pm
Section 2 question 4di and 4dii, would I have gotten the marks if I worked out
and same for 4dii!
I'm not equating from the start btw...
Post by: brightsky on April 15, 2010, 08:36:18 pm
http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/assessreports/mmcas2_assessrep_07.pdf
Section 2 question 4di and 4dii, would I have gotten the marks if I worked outthen work out
then say like
and same for 4dii!
I'm not equating from the start btw...
I don't see why they shouldn't give you the marks. :) Although your way is a tad bit longer :p.
Post by: Blakhitman on April 15, 2010, 09:17:15 pm
http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/assessreports/mmcas2_assessrep_07.pdf
Section 2 question 4di and 4dii, would I have gotten the marks if I worked out
and same for 4dii!
I'm not equating from the start btw...
I don't see why they shouldn't give you the marks. :) Although your way is a tad bit longer :p.
Lol but I wouldn't have thought of adding and subtracting 1 in 4di, so I wouldn't have proved it :P
Post by: the.watchman on April 15, 2010, 09:19:20 pm
http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/assessreports/mmcas2_assessrep_07.pdf
Section 2 question 4di and 4dii, would I have gotten the marks if I worked out
and same for 4dii!
I'm not equating from the start btw...
Equating from the beginning is fundamentally incorrect, because it means you have assumed it is true
This is like proving that lobsters are expensive, using the 'knowledge' that lobsters are expensive, rather than attacking the problem from information you already know ......... (END DR HE RANT :D)
Post by: Blakhitman on April 15, 2010, 09:20:36 pm
Post by: Blakhitman on April 18, 2010, 08:35:32 pm
what the equation of the image?
Post by: fady_22 on April 18, 2010, 08:38:57 pm
Post by: Blakhitman on April 18, 2010, 08:49:06 pm
Just making sure thanks a lot Fady!!!!
Who do I mean? O.o.............................................................lol
Post by: fady_22 on April 18, 2010, 08:56:21 pm
Sweet, then my friend is wrong and I'm right!
Just making sure thanks a lot Fady!!!!
Who do I mean? O.o.............................................................lol
LOL thats OK. :)
Post by: Blakhitman on June 20, 2010, 09:27:02 pm
Yeah, probably best to define that ugly thing first :-[
Post by: stonecold on June 20, 2010, 09:33:13 pm
Post by: superflya on June 20, 2010, 09:35:51 pm
Post by: stonecold on June 20, 2010, 09:36:21 pm
Post by: Blakhitman on June 20, 2010, 09:40:09 pm
Stupid N-Spire, I used to think it was amazing, gonna complain to teacher during the last part of the SAC.
Update: 5 minutes later, and it is still trying to solve it...
Yea...umm...I had to take out the lid on the back so it shuts off to get it to work again.
Post by: stonecold on June 20, 2010, 09:41:33 pm
lol I need to do this in our SAC ><.
Stupid N-Spire, I used to think it was amazing, gonna complain to teacher during the last part of the SAC.Update: 5 minutes later, and it is still trying to solve it...
Yea...umm...I had to take out the lid on the back so it shuts off to get it to work again.
LOL same. If VCAA pulls any of this shit in the exam, I might as well just pack my bags and leave now.
Should I buy a Classpad?
Post by: Blakhitman on June 20, 2010, 09:42:20 pm
Post by: brightsky on June 20, 2010, 09:45:22 pm
Post by: Blakhitman on June 20, 2010, 09:46:18 pm
Post by: TrueTears on June 20, 2010, 09:46:56 pm
Post by: the.watchman on June 20, 2010, 09:47:44 pm
Wolfram Alpha ftw! :D
Bullshit it worked on there!
Wolfram alpha is my maths homework saviour :P
Post by: stonecold on June 20, 2010, 09:48:52 pm
That is when it actually works. *sighs*
Everyone in my class has Classpads. Sometimes they are better, and other times they are worse.
i.e. They can graph relations such as circles. You can drag a function to graph it which is pretty cool.
Then it does some stupid shit like not give simplified answers sometimes, or not giving trig answers in order when you solve, which is kinda important when a question asks for the first however many positive solutions.
Why can't these tools make a calculator that just works between them.
Post by: the.watchman on June 20, 2010, 09:55:27 pm
It's simpler in my opinion
Post by: Blakhitman on June 20, 2010, 09:56:55 pm
Now if only I could find their iphone app cracked...I am not cheap btw, $3 isn't exactly "inexpensive".
Ever since the new 2.0 OS came out it's been so slow. The features it has are awesome.
That is when it actually works. *sighs*
Everyone in my class has Classpads. Sometimes they are better, and other times they are worse.
i.e. They can graph relations such as circles. You can drag a function to graph it which is pretty cool.
Then it does some stupid shit like not give simplified answers sometimes, or not giving trig answers in order when you solve, which is kinda important when a question asks for the first however many positive solutions.
Why can't these tools make a calculator that just works between them.
Thing is I'm still on 1.7(think) so it isn't because of the update.
lol these tools just want money and if there is a way that requires less time and thinking, then they will do it that way.
Post by: stonecold on June 20, 2010, 09:57:30 pm
You can also tell it to find Min, Max and inflection points too.
Post by: Blakhitman on July 15, 2010, 07:47:26 pm
therefore
A. an x-intercept
B. local minimum
C. local maximum
D. point of discontinuity
No choice for P.O.I.
Post by: Martoman on July 15, 2010, 08:42:21 pm
So i take in the iphone with wolfram. It rocks.
Post by: brightsky on July 15, 2010, 09:55:16 pm
when
when
thereforemust have:
A. an x-intercept
B. local minimum
C. local maximum
D. point of discontinuity
No choice for P.O.I.
A. It must touch the x-axis at x = 1.
Post by: Yitzi_K on July 15, 2010, 10:21:07 pm
Post by: brightsky on July 16, 2010, 12:40:12 am
Post by: Yitzi_K on July 16, 2010, 12:53:18 am
Take for example
Post by: Blakhitman on July 16, 2010, 06:52:45 am
Post by: moekamo on July 16, 2010, 07:13:45 am
Post by: Blakhitman on July 16, 2010, 05:47:37 pm
I think it'd be the missing option!