Post by: Aqualim on December 08, 2009, 09:24:06 pm
Ok, so I've worked out that when
Plus I know that my range must be from
Post by: /0 on December 08, 2009, 09:29:05 pm
Solving
So the range is
Post by: Aqualim on December 08, 2009, 09:35:11 pm
I've linked the question below.
Post by: Aqualim on December 08, 2009, 09:51:26 pm
Then when I made
That seem right?
Meaning that would make the Range =
Post by: TrueTears on December 08, 2009, 09:55:03 pm
Are you only graphing only the positive half of the graph? Why not the negative half?I'd do what /0 did here, just work out the range of the graph and sketch it.
Solving,
. Since the coefficient of
is negative, this is an inverted parabola, so it will have a maximum at
So the range is
Post by: /0 on December 08, 2009, 09:58:34 pm
This is because it is a part of an application question which requires to know the area of a box. More than likely meaning it only includes positive numbers.
I've linked the question below.
So... is the equation
In that case, all seems right.
Still confused about what 'Area' this represents though.
Post by: Aqualim on December 08, 2009, 10:02:32 pm
This is because it is a part of an application question which requires to know the area of a box. More than likely meaning it only includes positive numbers.
I've linked the question below.
So... is the equationnot
?
In that case, all seems right.
Still confused about what 'Area' this represents though.
Equation is
and it represents the 'BARS' Area
Post by: Ilovemathsmeth on December 09, 2009, 12:24:53 pm
Post by: Aqualim on December 09, 2009, 12:35:09 pm
Exciting question, I don't remember it from Essentials or are there extra qs in the Essentials CAS book?
Essentials 1 & 2 Teacher's Edition :)
Post by: Ilovemathsmeth on December 09, 2009, 01:35:09 pm
How did you get the teacher's edition??
Post by: Aqualim on December 09, 2009, 02:01:36 pm
I see :)
How did you get the teacher's edition??
Parents friend is one of the authors of the Essentials Methods Book (or something along those lines). Just brought the CD home one day and said it might come in handy..
Post by: Ilovemathsmeth on December 09, 2009, 02:53:20 pm
Post by: Aqualim on January 05, 2010, 09:17:25 pm
In the expansion of
ok so this is as far as I got;
Since it's the second term,
Now what? Calculator couldn't even solve it :S
Yet I know purely from inserting numbers that the answer is 6
Post by: TrueTears on January 05, 2010, 09:18:11 pm
Well yeah, there is no elementary solution.
Post by: brightsky on January 05, 2010, 09:46:09 pm
Post by: TrueTears on January 05, 2010, 09:48:24 pm
Was this from essentials? I can't remember it.
Post by: brightsky on January 05, 2010, 09:52:25 pm
Post by: Aqualim on January 05, 2010, 11:26:38 pm
Post by: brightsky on January 06, 2010, 01:00:41 pm
A.
B.
C.
D.
E. Other
I've been staring at this question for the past 20 mins....:p
Post by: cipherpol on January 06, 2010, 01:09:46 pm
sub into
Post by: brightsky on January 06, 2010, 01:21:57 pm
Post by: Ilovemathsmeth on January 12, 2010, 12:11:49 am
Post by: Aqualim on January 15, 2010, 04:37:29 pm
Calculator says
Now I'm stuck..
Post by: brightsky on January 15, 2010, 04:43:46 pm
Then....
Case 1
Case 2
Post by: Aqualim on January 15, 2010, 04:47:00 pm
Post by: brightsky on January 15, 2010, 04:47:45 pm
Post by: Aqualim on January 17, 2010, 01:24:25 pm
Also how do you know what shape
As when i've drawn graphs, I've found that the square root graphs can look like linear graphs aswell.
Post by: the.watchman on January 17, 2010, 01:49:17 pm
That's what I usually do, it's more accurate than a table of values
Post by: Aqualim on January 17, 2010, 07:37:08 pm
Couldn't you sketch each function individually, then add the ordinates to get?
That's what I usually do, it's more accurate than a table of values
Which ordinates in particular though? y-intercepts, x-intercepts?
Post by: the.watchman on January 17, 2010, 08:23:08 pm
Couldn't you sketch each function individually, then add the ordinates to get
That's what I usually do, it's more accurate than a table of values
Which ordinates in particular though? y-intercepts, x-intercepts?
Look for:
When either y-value equals zero
When both y-values are the same
When one is positive and negative (cancelling out)
Then join the rest
Post by: Aqualim on January 20, 2010, 11:25:22 pm
Post by: cipherpol on January 20, 2010, 11:26:13 pm
Post by: the.watchman on January 20, 2010, 11:32:09 pm
Post by: kyzoo on January 20, 2010, 11:40:42 pm
Post by: the.watchman on January 20, 2010, 11:42:40 pm
Trust me, at the end of the year this stuff will be so ingrained into you that you won't need a reference to recall it. It's the same thing with the quotient rule, at first it's daunting, but in the end it's automatic knowledge.
+1 Very true! I'd say just sit down for 90 seconds and memorise them (look for patterns), then you'll know them for life :P
Post by: Blakhitman on January 27, 2010, 04:45:35 pm
Ok, I've attached a worked example from my textbook concerning the product rule, what I want to know is why on the third step, the
It's probably obvious, but I haven't seen it.
Post by: brightsky on January 27, 2010, 04:47:42 pm
Let
Taking out the factor
Substitute in back the values for
Post by: stonecold on January 27, 2010, 04:49:08 pm
yepp, i was gonna say that. learnt that just yesterday. it is fair confusing at first.
Post by: Blakhitman on January 27, 2010, 04:52:42 pm
And Stonecold, it's the intention that counts :p
Post by: Ilovemathsmeth on January 30, 2010, 06:34:19 pm
Post by: Aqualim on February 05, 2010, 06:23:34 pm
Plan A: a Fixed charge of $30 and 50 cents a call after 150 calls. There is no extra charge for the first 150 calls
Plan B: No fixed charge and 30 cents a call
b) the rule for the cost of plan A, of 'n' calls is;
Given that 'c' is when n is equal to or lower than 150, and that 'an+b' is when 'n' is greater than 150
Hence, sketch the graph of Plan A.
When I sketch this graph I found that the gap between the first part of the hybrid and the second, didn't match the question. It looked like something below;
Shouldn't the hybrid graph be joined, as it won't cost an extra 60 odd dollars after the 150 calls.
Post by: Stroodle on February 05, 2010, 06:35:28 pm
Post by: Aqualim on February 05, 2010, 06:43:11 pm
Post by: Stroodle on February 05, 2010, 06:55:53 pm
Edit* there's different ways to get this. I prefer to do it by solving the equations 30=150x+y and 31=152x+y
Post by: Aqualim on February 05, 2010, 10:12:38 pm
The second part of the hybrid function should be C=0.50n-45 where n is greater than 150.
Edit* there's different ways to get this. I prefer to do it by solving the equations 30=150x+y and 31=152x+y
Ok you've lost me, wouldn't you expect the +b to just be the fixed charge?
Post by: the.watchman on February 05, 2010, 10:17:34 pm
The second part of the hybrid function should be C=0.50n-45 where n is greater than 150.
Edit* there's different ways to get this. I prefer to do it by solving the equations 30=150x+y and 31=152x+y
Ok you've lost me, wouldn't you expect the +b to just be the fixed charge?
No, because it is 50 cents a call ONLY FROM n=150
Therefore, we need to find the second line with a gradient of 0.5 and which goes through (150,30)
Post by: Aqualim on February 05, 2010, 11:46:14 pm
The second part of the hybrid function should be C=0.50n-45 where n is greater than 150.
Edit* there's different ways to get this. I prefer to do it by solving the equations 30=150x+y and 31=152x+y
Ok you've lost me, wouldn't you expect the +b to just be the fixed charge?
No, because it is 50 cents a call ONLY FROM n=150
Therefore, we need to find the second line with a gradient of 0.5 and which goes through (150,30)
ahh ok, makes sense now, wouldn't have thought to go that far to work it out. Was thinking it was just a matter of substitution.
Thanks for helps guys :)
Post by: the.watchman on February 06, 2010, 09:38:36 am
What I find helps is to try to visualise the situation (eg. fixed charge THEN 50cents a call from 150 calls onwards)
Then you can see what is obviously incorrect and what answers can be automatically discarded
Post by: Christiano on February 07, 2010, 10:11:19 pm
Need help with this: A line has a gradient of 6 and pass through the points with coords (-1, 6) and (7, a). Find the value of a.
How do you go about doing this? I got like 10 more similar questions after this!
Post by: the.watchman on February 07, 2010, 10:13:21 pm
I know the feeling! :)
Post by: Christiano on February 07, 2010, 10:14:37 pm
Post by: the.watchman on February 07, 2010, 10:15:43 pm
Thanks ;D
No prob, happy to help :)
Post by: Christiano on February 07, 2010, 10:31:14 pm
Line has a gradient of -6 and passes through the points with coords (1,6) and (7,a). Find value of a
How does the answer become
EDIT: Sorry its actually gradient of -6 and passes through the cords (1,6) and (b,7). Find value of b
Post by: the.watchman on February 07, 2010, 10:35:33 pm
How on earth did the book get
Post by: superflya on February 07, 2010, 10:59:49 pm
the easy things are hard for watchman :P
Post by: the.watchman on February 08, 2010, 06:23:07 am
Post by: superflya on February 08, 2010, 05:12:31 pm
superflya, I didn't see the edit, so yes I can do this question (I think ... :P)
lol my bad.
Post by: the.watchman on February 08, 2010, 05:14:55 pm
superflya, I didn't see the edit, so yes I can do this question (I think ... :P)
lol my bad.
I'm offended... >:(
Post by: Aqualim on February 08, 2010, 10:53:01 pm
Express
and I got
Then from there I got the two asymptotes which I think are;
But they don't lie within the graph, well atleast not horizontally. (don't know how to draw a straight line vertically on the TI-Nspire calculator)
Post by: TrueTears on February 08, 2010, 10:53:45 pm
Post by: Aqualim on February 08, 2010, 11:00:21 pm
did you long divide it properly heh xD you shouldn't get another polynomial as the numerator.When I put both equations into the calculator they are exactly the same, unless i'm supposed to simplify it further? if thats what your saying, haha sorry bit slow
Post by: TrueTears on February 08, 2010, 11:02:11 pm
Post by: the.watchman on February 09, 2010, 05:50:53 am
For horizontal asymptotes, it's the top co-efficient of x over the bottom co-efficient of x (in this case,
For vertical asymptotes, let the denominator equal 0,
So to divide, pull the
There shouldn't be any x in the numerator
Moderator Action: fixed LaTeX
Post by: Aqualim on February 23, 2010, 09:27:33 pm
How would I go about showing the working out for that equation?
I already know by looking at it, that
Post by: superflya on February 23, 2010, 09:29:46 pm
Post by: m@tty on February 23, 2010, 09:33:26 pm
Post by: Aqualim on February 23, 2010, 09:44:36 pm
Post by: Aqualim on February 26, 2010, 05:22:18 pm
Show that
Would that just be as simple as;
Meaning that
Post by: Syncness on February 26, 2010, 05:35:26 pm
Well forto be defined
is not defined.
Isn't it different when they ask you to define something, and prove if it exists?
Defining is just writing it, regardless if it exists?
Proving it exists is matching the range and domains.
Post by: Aqualim on February 28, 2010, 02:21:04 pm
The graph of the cubic function cuts the x-axis at the point with coordinates
Ok, so I know that it has to be a repeated factor, since it touches one x-intercept whereas it cuts another, but how would I get
Post by: TrueTears on February 28, 2010, 02:23:24 pm
y = q(x-r_1)(x-r_2)^2
Post by: the.watchman on February 28, 2010, 02:24:16 pm
Let
Sub
So the equation is
EDIT: Beaten yet again :D
Post by: Aqualim on February 28, 2010, 02:28:40 pm
Ok:
Let
Subinto above
So the equation is
EDIT: Beaten yet again :D
hmm it really wasn't hard at all... how sad
Post by: the.watchman on February 28, 2010, 02:29:58 pm
hmm it really wasn't hard at all... how sad
There's no need to say that, now you know how to do these types of questions! :)
Post by: Aqualim on February 28, 2010, 02:31:17 pm
Very true, I suppose the letters just through me off, if they were numbers it would have clicked.hmm it really wasn't hard at all... how sad
There's no need to say that, now you know how to do these types of questions! :)
Thanks for your help guys :)
Post by: TrueTears on February 28, 2010, 02:35:14 pm
Post by: the.watchman on February 28, 2010, 02:36:04 pm
np
+1 :)
Post by: TrueTears on February 28, 2010, 02:36:43 pm
Post by: Aqualim on March 01, 2010, 04:39:03 pm
Given that A is (-8,2) and B is (-6,10);
Find the coordinates of P, where P
Post by: brightsky on March 01, 2010, 05:46:57 pm
The X-coordinates for each of them are -8 and -6 respectively.
We are looking to divide the line in 3 parts in respect to the x-axis.
If we were to draw in the lines
The equation of the line AB is easily found by:
Now to find the y-coordinate of P, just sub in the x-coordinate.
So the coordinates of P is
Hope this is right. :p
Post by: Aqualim on March 01, 2010, 06:04:21 pm
Ok so the two points A and B are in the 3rd quadrant.
The X-coordinates for each of them are -8 and -6 respectively.
We are looking to divide the line in 3 parts in respect to the x-axis.
If we were to draw in the linesand
, we would find that the distance between (-8,0) and (-6,0) is
. So if we divide that by three, we get:
So when we plot these lines on the line AB, the points would be
and
respectively, with
and
being their y coordinates. Now, AP:PB = 2:1, so clearly P has to be the point
The equation of the line AB is easily found by:
Now to find the y-coordinate of P, just sub in the x-coordinate.
So the coordinates of P is
Hope this is right. :p
Apparently the answer is
Post by: brightsky on March 01, 2010, 06:16:56 pm
The equation of the line AB is
So the y-coordinate is:
So the coordinates are
Post by: Aqualim on March 01, 2010, 06:20:24 pm
Post by: the.watchman on March 01, 2010, 06:21:18 pm
Post by: brightsky on March 01, 2010, 06:23:29 pm
ok thanks, but what does the 'AP:PB = 2:1' part mean?
Its the ratio between the lines AP and PB.
The line AB is one whole line. P is a point on that line somewhere in the middle of A and B such that the line AP is two times the length of PB.
EDIT: What the.watchman said. :p
Post by: the.watchman on March 01, 2010, 06:24:22 pm
ok thanks, but what does the 'AP:PB = 2:1' part mean?
Its the ratio between the lines AP and PB.
The line AB is one whole line. P is a point on that line somewhere in the middle of A and B such that the line AP is two times the length of PB.
Well said! :)
Post by: Aqualim on March 01, 2010, 06:27:23 pm
Post by: superflya on March 01, 2010, 06:27:31 pm
Post by: brightsky on March 01, 2010, 06:28:22 pm
Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Post by: Aqualim on March 01, 2010, 06:35:11 pm
Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
Post by: the.watchman on March 01, 2010, 06:43:33 pm
Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
Post by: Aqualim on March 01, 2010, 06:49:12 pm
Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
ahh ok so if it said something like 'AP:PB = 4:3' would that mean you divide by 7? as P is four parts away from A and 3 parts away from B?
I hope thats right :)
Post by: the.watchman on March 01, 2010, 06:49:49 pm
Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
ahh ok so if it said something like 'AP:PB = 4:3' would that mean you divide by 7? as P is four parts away from A and 3 parts away from B?
I hope thats right :)
Yep, you're getting the hang of it! :D
Post by: Aqualim on March 01, 2010, 06:53:28 pm
Post by: the.watchman on March 01, 2010, 06:57:15 pm
Excellent! :) how do you know the answer to basically ever question? lol sometimes the figures and wording throws me off completely :P
I don't, I just have a train of thought that I follow, so I don't have to show working elsewhere :)
Once you do enough questions, you should be able to see patterns in the solution methods etc.
Post by: Cthulhu on March 01, 2010, 07:02:39 pm
Post by: the.watchman on March 01, 2010, 07:03:29 pm
(He looks in the back of the book
My textbook remains at school (I'm not lying! :D)
Post by: TrueTears on March 02, 2010, 02:54:35 pm
Eh just do it algebracally, just change the ratios to fractions, just different notation, concept is the same...Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
ahh ok so if it said something like 'AP:PB = 4:3' would that mean you divide by 7? as P is four parts away from A and 3 parts away from B?
I hope thats right :)
Post by: the.watchman on March 02, 2010, 03:04:06 pm
Eh just do it algebracally, just change the ratios to fractions, just different notation, concept is the same...Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
ahh ok so if it said something like 'AP:PB = 4:3' would that mean you divide by 7? as P is four parts away from A and 3 parts away from B?
I hope thats right :)
nice idea :)
Post by: Aqualim on March 04, 2010, 10:36:26 pm
Eh just do it algebracally, just change the ratios to fractions, just different notation, concept is the same...Thanks, so how did you know that the points were in the third quandrant? wouldn't that be the 2nd quadrant (top left hand) seeing as the x point is negative whilst the y point is positive (-8,2)....
Woops, another mistake. :p Yeah it's in the second quadrant. Doesn't make a difference though. :)
Alright so dividing it by three has no correlation with the quadrant? then how did you know to divide it by three, or would you divide by three in any case? lol sorry for all these questions
You would divide the line into three parts, as from "AP:PB = 2:1", you can see that the line should be split into 2+1=3 parts. Then the point P would be two 'parts' along from A
I hope this explanation helps :)
ahh ok so if it said something like 'AP:PB = 4:3' would that mean you divide by 7? as P is four parts away from A and 3 parts away from B?
I hope thats right :)
nice idea :)
Ooo, this sounds better, anyone like to show me? sorry about this :S
Post by: TrueTears on March 04, 2010, 10:37:00 pm
a:b = a/b
Post by: Aqualim on March 04, 2010, 10:38:32 pm
Post by: TrueTears on March 04, 2010, 10:40:54 pm
AP/PB = 2/1
AP = 2PB
Thus AB = AP+2PB
Post by: the.watchman on March 04, 2010, 10:46:20 pm
AP:PB = 2:1
AP/PB = 2/1
AP = 2PB
Thus AB = AP+2PB
Yup, then follow on with the rest of the working shown earlier to find point P
Post by: Aqualim on March 05, 2010, 07:21:26 pm
Find the magnitude of
Post by: Aqualim on March 05, 2010, 08:55:08 pm
ok thanks.. This is a follow up question;
Find the magnitude ofBAC in degrees correct to one decimal place if A=(-8,2), B=(-6,10) and C=(4,11)
Anyone?
Post by: the.watchman on March 05, 2010, 09:15:00 pm
Gradient of AC:
So the angle
Post by: Aqualim on March 05, 2010, 09:55:10 pm
Do you take AB away from AC because the question is asking for the magnitude of BAC (hence B is before C)?
And also, is there a way to verify this on the cas calculator?
I've drawn both graphs on the Graphs page and have found an 'angle' option, but where on the two straight lines am I finding the angle? lol stupid question I know
Post by: the.watchman on March 05, 2010, 10:16:49 pm
Cheer mate, quick question regarding the working out process.
Do you take AB away from AC because the question is asking for the magnitude of BAC (hence B is before C)?
And also, is there a way to verify this on the cas calculator?
I've drawn both graphs on the Graphs page and have found an 'angle' option, but where on the two straight lines am I finding the angle? lol stupid question I know
No, I used the one with a greater gradient first, because the angle it makes with the positive x-direction is larger
To check it on a graphics, click any point on AB (to the right of the intersection), then the intersection point, then move the cursor around to find the angle
Post by: Aqualim on March 05, 2010, 10:20:33 pm
Cheer mate, quick question regarding the working out process.
Do you take AB away from AC because the question is asking for the magnitude of BAC (hence B is before C)?
And also, is there a way to verify this on the cas calculator?
I've drawn both graphs on the Graphs page and have found an 'angle' option, but where on the two straight lines am I finding the angle? lol stupid question I know
No, I used the one with a greater gradient first, because the angle it makes with the positive x-direction is larger
To check it on a graphics, click any point on AB (to the right of the intersection), then the intersection point, then move the cursor around to find the angle
So does that mean in general we should use the one with the greater gradient? (assuming we are finding the positive gradient)
Post by: the.watchman on March 05, 2010, 10:22:13 pm
Cheer mate, quick question regarding the working out process.
Do you take AB away from AC because the question is asking for the magnitude of BAC (hence B is before C)?
And also, is there a way to verify this on the cas calculator?
I've drawn both graphs on the Graphs page and have found an 'angle' option, but where on the two straight lines am I finding the angle? lol stupid question I know
No, I used the one with a greater gradient first, because the angle it makes with the positive x-direction is larger
To check it on a graphics, click any point on AB (to the right of the intersection), then the intersection point, then move the cursor around to find the angle
So does that mean in general we should use the one with the greater gradient? (assuming we are finding the positive gradient)
I think so, that's how I would do it
But then if it's negative gradient, then the least negative (greatest? :D) gradient first maybe?
Post by: TrueTears on March 06, 2010, 10:40:39 am
Post by: the.watchman on March 06, 2010, 10:44:49 am
or.... Sketch the graph and see! A quick sketch neva hurts
Nice! :D
Post by: Aqualim on March 14, 2010, 10:54:29 am
Using
Post by: the.watchman on March 14, 2010, 11:01:09 am
Is this question solveable without a calculator?
Usingand
, sketch on the same set of axes. Hence sketch the graph of
on the same graph
Yeah, sketch both on the same axes, find major points (eg. zeros, squares etc.) and join the dots with a smooth curve
Post by: the.watchman on March 14, 2010, 12:55:13 pm
Is this question solveable without a calculator?
Using
Yeah, sketch both on the same axes, find major points (eg. zeros, squares etc.) and join the dots with a smooth curve
On second thoughts, don't do that, just dilate
Post by: Aqualim on March 26, 2010, 10:43:21 pm
When finding equations for exponential modelling, how do you know when and when not to include the irrational variable '
In some questions I have had, you are required to have an
Post by: GerrySly on March 26, 2010, 10:53:25 pm
Post by: Aqualim on March 26, 2010, 11:02:17 pm
Example;
A highly volatile substance initially has a mass of 1200 g and its mass is reduced by 12% each second.
1) Write a formula that gives the mass of the substance (m) at time (t) seconds.
Why isn't Euler's number included in this formula, whereas the standard equation for growth or decay is
Post by: GerrySly on March 26, 2010, 11:16:22 pm
Doing a quick check with our previous points,
So it does include
Post by: Aqualim on March 26, 2010, 11:45:29 pm
But where did you pull the 1056 from?
Post by: GerrySly on March 26, 2010, 11:48:59 pm
ok because the formula in the answers sayswhich is the same as your answer.
But where did you pull the 1056 from?
"Decreasing at 12% per year"
Post by: Aqualim on March 31, 2010, 03:45:47 pm
Heres the question;
Find the values of a, b and c;
I've seen ways for solving simultaneous equations on google, but they only show worked examples for when all three variables are separated after expansion, whereas if you expand this equation you will have two variables placed together
Post by: the.watchman on March 31, 2010, 03:55:04 pm
[2] - [3]:
[5] -> [4]:
BLAH BLAH BLAH
sub both into one of the eqns to get c
Post by: Aqualim on March 31, 2010, 04:29:00 pm
Post by: the.watchman on March 31, 2010, 04:31:12 pm
Thanks for that ;) but when you took [2] from [3] what happened to the ''a's''? did you just eliminate them because we only want to find out b?
No, I divided both sides by a :P
Post by: ioaus09 on April 07, 2010, 08:42:04 pm
A) f is a one-to-one function
B) f is an onto function
C) f is a function, but is neither one-to-one nor onto
D) f is not a function
There can be more than one answer
Post by: m@tty on April 07, 2010, 10:12:57 pm
What are
Post by: brightsky on April 07, 2010, 10:15:57 pm
Post by: m@tty on April 07, 2010, 10:23:53 pm
If that
Post by: /0 on April 09, 2010, 11:37:34 pm
A function
If we assume Z is the integers and B can be anything, then I don't think any of the statements are true for all B.
See, if we have
By the way, if
Post by: Aqualim on April 14, 2010, 02:36:51 pm
Also was given an exponential modelling equation (can't post up due it being on a test) which stated there was a cooling temperature of something like 40 degrees and an initial temperature of 70 degrees, and normally the initial temperature is placed within the equation where it says;
Post by: the.watchman on April 14, 2010, 03:43:22 pm
But once you get some idea for what to do, you should notice that there is an axis of symmetry when (if |f(x)|), f(x)=0 and the 'positive' side is reflected to make up the 'negative' side
Post by: m@tty on April 14, 2010, 04:04:05 pm
Post by: Aqualim on April 14, 2010, 04:10:40 pm
Post by: the.watchman on April 14, 2010, 04:13:13 pm
Yeah, for your equation notice that every x is always positive, so all you have to do is sketch forand reflect in the y axis..
Yep, but for a more general case, stuff like
Not that that equation will pop up in methods. Ever.
ok so would it split up like so?
Yes, you can do that, but once you get more experienced, you only need the positive side, then reflect it
Post by: m@tty on April 14, 2010, 04:17:25 pm
Post by: the.watchman on April 14, 2010, 04:20:54 pm
Yeah, I was only saying for this specific case. And in your equation there would also be zeros at:P
Lol, misconfusion galore :)
And No. No. No.
Post by: TrueTears on April 14, 2010, 04:22:39 pm
Yeah, I was only saying for this specific case. And in your equation there would also be zeros at2^0 = 1
Post by: m@tty on April 14, 2010, 04:25:51 pm
Post by: Aqualim on April 15, 2010, 06:25:15 pm
It's a multiple choice question for the record
EDIT: made a mistake :)
Post by: the.watchman on April 15, 2010, 06:27:45 pm
I'd say it's both (a) and (d) because they are the same :P
For a>0, only one arm of the graph remains, and it is one-to-one
Post by: Aqualim on April 15, 2010, 06:32:09 pm
Post by: superflya on April 15, 2010, 06:34:36 pm
ok so would you just simply draw the graph in the calculator to know what the graph looks like or would you sketch it somehow?
easier to graph it on calc, but u could sketch. just visualize and ull be able to see if its one to one.
Post by: the.watchman on April 15, 2010, 06:35:13 pm
ok so would you just simply draw the graph in the calculator to know what the graph looks like or would you sketch it somehow?
You can imagine it in your head, it would look similar to ln(|x|) but steeper
You should also note that there are two arms, and each arm is one-to-one, so you need to restrict the graph to just one arm.
This is allowed by a>0
Post by: Aqualim on April 15, 2010, 06:41:52 pm
Post by: the.watchman on April 15, 2010, 06:46:23 pm
I'm assuming you are able to visualise this in your head, just by being exposed to this type of graph over a longer period of time yeah? cause I can't just visualise how a 'g of f(x)' would look off the top of my head :P
Lol, just think of it as x^4 is 'similar' to x but it increases heckloads quicker and starts off slower, right?
And the negative side is the positive side reflected :)
Post by: Aqualim on April 15, 2010, 06:49:20 pm
I'm assuming you are able to visualise this in your head, just by being exposed to this type of graph over a longer period of time yeah? cause I can't just visualise how a 'g of f(x)' would look off the top of my head :P
Lol, just think of it as x^4 is 'similar' to x but it increases heckloads quicker and starts off slower, right?
And the negative side is the positive side reflected :)
Makes sense, also just found out by plugging various powers into the graphs part of the calculator that for every odd power there is one arm and for every even power there are two arms, just as the power increases the graph is stretched more in the y-direction
Post by: Aqualim on April 15, 2010, 09:05:32 pm
show that
Post by: the.watchman on April 15, 2010, 09:11:21 pm
AND
:D
Post by: Blakhitman on April 15, 2010, 09:18:29 pm
AND
:D
:P
Post by: the.watchman on April 15, 2010, 09:19:44 pm
AND
:D
:P
Yep, oops...
Post by: Blakhitman on April 15, 2010, 09:21:49 pm
And I would have expanded the first one, because it's easier to see similarities than having to factorise the second one. Personal preference!
Post by: Aqualim on April 15, 2010, 09:34:21 pm
Write a matrix equation that would complete the transformations on (x,y) to (x',y') with the function
Post by: the.watchman on April 15, 2010, 09:37:37 pm
Then it'd be:
[x' = [0.5 0 [x + [-0.5
y'] 0 1] y] 0 ]
Because, in this scenario,
Then
Post by: Aqualim on April 15, 2010, 09:39:33 pm
Post by: Aqualim on April 17, 2010, 07:26:00 pm
The inverse of a function,
Ok so in the previous question I found that the inverse of f(x) was
Post by: brightsky on April 17, 2010, 07:34:08 pm
Post by: Aqualim on April 17, 2010, 07:46:34 pm
Post by: brightsky on April 17, 2010, 07:54:05 pm
The inverse of y = f(x) is, as you've found:
You can prove this the other way around as well:
The inverse of y = g(x) is
EDIT: Hmm..really think I'm missing something here...xD
Post by: Blakhitman on April 17, 2010, 07:56:55 pm
Post by: Aqualim on April 17, 2010, 07:57:40 pm
But f(x) is clearly not an inverse of g(x) and vice versa.
The inverse of y = f(x) is, as you've found:
You can prove this the other way around as well:
The inverse of y = g(x) is.
EDIT: Hmm..really think I'm missing something here...xD
Yeah I think it's a badly worded question, not really 100% sure what it is asking of me
Post by: Blakhitman on April 17, 2010, 08:00:44 pm
For example,
Post by: Aqualim on April 17, 2010, 08:02:24 pm
I don't think it's asking to show they're inverses of "each other". Just show that they are inverse functions.
For example,is known to be the inverse of
so simply show that through the matrix operation.
Ok how would I do that exactly? is there a formula for inverses in matrices?
Post by: Blakhitman on April 17, 2010, 08:09:16 pm
you get
Then do same for f(x) lol ;D
Post by: Aqualim on April 17, 2010, 08:57:47 pm
You posted it before :P.
you getand
then sub in (note it's the swapping x and y business) to
and you get
Then do same for f(x) lol ;D
How so?
Post by: the.watchman on April 17, 2010, 09:06:19 pm
Post by: Aqualim on April 17, 2010, 09:45:28 pm
Post by: Aqualim on April 20, 2010, 06:17:17 pm
Solve for x;
Post by: m@tty on April 20, 2010, 06:24:48 pm
Anyway,
Post by: superflya on April 20, 2010, 06:30:21 pm
Post by: Blakhitman on April 20, 2010, 06:34:13 pm
lol i think u misinterpreted what he wrote, it says literal :P or he may have modified it..
He initially typed integral! ;D
Post by: Aqualim on April 20, 2010, 06:34:49 pm
Post by: superflya on April 20, 2010, 06:36:35 pm
lol i think u misinterpreted what he wrote, it says literal :P or he may have modified it..
He initially typed integral! ;D
ahh thought so ;D
Post by: kamil9876 on April 20, 2010, 06:41:27 pm
How is that an integral?? o.0
Anyway,
What happens when
Post by: m@tty on April 20, 2010, 06:45:56 pm
What happens when?
?
I meant it as "
Post by: the.watchman on April 20, 2010, 06:50:54 pm
And yes, kamil is right, you forgot about
Post by: m@tty on April 20, 2010, 07:03:09 pm
But what? You are meant to show that
And
Yeah, I did forget about
Post by: Aqualim on April 20, 2010, 07:06:02 pm
Also quick question, what happened here? I don't see how you could cancel out (1-a) and (a-1) since they are both different?
Post by: the.watchman on April 20, 2010, 07:07:20 pm
Ahh, I must have misinterpreted his post.
But what? You are meant to show that??
And? :|
Yeah, I did forget about
I think you should show what solutions you can have for x with 'a' as these values
Is it no solutions? Or infinite solutions? :)
Also quick question, what happened here? I don't see how you could cancel out (1-a) and (a-1) since they are both different?
Post by: naved_s9994 on April 20, 2010, 07:23:31 pm
What happens when
I meant it as "". I remember seeing textbooks use the combined
, to mean the same thing, so I assumed it is mathematically correct??
Which book, page no?
Thanks!
Post by: m@tty on April 20, 2010, 07:32:29 pm
Post by: Aqualim on April 20, 2010, 08:51:27 pm
How is the answer?
Wouldn't it just be;
Post by: m@tty on April 20, 2010, 08:54:41 pm
Post by: Aqualim on April 20, 2010, 08:59:19 pm
Therefore when I'm dilating by
Post by: m@tty on April 20, 2010, 09:02:40 pm
Dilation from y-axis by factor of a:
Dilation from y-axis by factor of
Post by: the.watchman on April 20, 2010, 09:03:15 pm
so not matter what equation I'm dilating by 'a' from the y-axis, it will always end up as?
Therefore when I'm dilating by
Failproof way to do this:
When dilation by factor 'k':
- From x-axis: replace y with
- From y-axis: replace x with
So,
Post by: Aqualim on April 21, 2010, 02:57:26 pm
Another terminology question;
Domain of g(f(x))= Domain of f(x)
Would that mean Range of g(f(x)) = Range of g(x)?
Post by: the.watchman on April 21, 2010, 03:15:23 pm
So if
Then Dom gof = Dom f
And for Ran gof, let Dom g = Dom f = R+ U {0}
Then for that domain Ran g = R+ U {0}
So Ran gof = R+ U {0}
Does that make any sense?
I get the feeling it doesn't...
EDIT: lol
Post by: Aqualim on April 21, 2010, 03:29:48 pm
And for Ran gof, let Dom g = Dom f = R+ U {0}
Then for that domain Ran g = R+ U {0}
So Ran gof = R+ U {0}
Does that make any sense?
I get the feeling it doesn't...
EDIT: lol
Not really, so you have to let the domain of g equal the domain of f?
So if the domain of g was [0,10] and the domain of f was [-15, 25], you would make the domain of g equal [-15,25]? therefore the range of g(f(x)) = [-15,25]?
Post by: GerrySly on April 21, 2010, 03:39:10 pm
For example,
So therefore the domain of
(http://img200.imageshack.us/img200/6954/msp5819a8b98i403i02cb00.gif)
Graphing it you see the range is
Post by: Aqualim on April 21, 2010, 03:44:50 pm
Post by: GerrySly on April 21, 2010, 03:46:47 pm
ok so you'd have to find it by joining the two functions together, and then just sketch it. I thought there just may have been an easier way to work it out. Thank you :)
Yeah not really a shortcut unless of course it's a straight line then it's just the domain end points. Or if it's not a straight line just get the derivative find the turning point, sub in the domain end points and you'll have 3 y values and just get the highest and lowest values and you'll be sweet :)
Post by: the.watchman on April 21, 2010, 03:49:01 pm
Not really, so you have to let the domain of g equal the domain of f?
More or less, sorry about the crappy explanation... :(
Post by: m@tty on April 21, 2010, 05:16:12 pm
Post by: the.watchman on April 21, 2010, 05:25:26 pm
One 'shortcut' is to let theand then work out
on that domain, this range is the range of
.
Exactly what I meant ... but expressed WAY neater :(
Post by: Aqualim on April 22, 2010, 04:13:38 pm
Could you go any futher than this;
Post by: /0 on April 22, 2010, 04:20:52 pm
I think your second last step is as far as you can go. Even the original equation I would say looks 'simplified'. It's a matter of opinion.
(3500 posts, yay! XD)
Post by: Aqualim on April 22, 2010, 06:31:39 pm
A reflection in the y-axis is different to a reflection from the y-axis yeah?
reflection from the y-axis =
reflection in the y-axis =
If I'm correct?
Post by: Blakhitman on April 22, 2010, 06:37:08 pm
Post by: Aqualim on April 22, 2010, 06:43:36 pm
Post by: Aqualim on June 30, 2010, 03:01:06 pm
6. Find in exact form all real solutions to the equation
Let
Since
Post by: TrueTears on June 30, 2010, 03:07:19 pm
Post by: brightsky on June 30, 2010, 03:09:48 pm
Multiply a to both sides:
By the null factor law,
Hence
Hence
Post by: Blakhitman on June 30, 2010, 03:10:23 pm
Post by: Aqualim on June 30, 2010, 04:40:46 pm
When finding the rule, would I have to use the Quadratic Formula at some point?
Post by: the.watchman on June 30, 2010, 05:00:33 pm
For the functionwith the rule
find the rule for the inverse function.
When finding the rule, would I have to use the Quadratic Formula at some point?
Absolutement! Unless you want to complete the square manually :)
Post by: Aqualim on June 30, 2010, 05:03:20 pm
For the function
When finding the rule, would I have to use the Quadratic Formula at some point?
Absolutement! Unless you want to complete the square manually :)
How would you do it manually?
Post by: the.watchman on June 30, 2010, 05:04:54 pm
For the function
When finding the rule, would I have to use the Quadratic Formula at some point?
Absolutement! Unless you want to complete the square manually :)
How would you do it manually?
By completing the square...
Let
Then complete the square like normal :)
Post by: Aqualim on July 01, 2010, 01:28:45 pm
For the function
When finding the rule, would I have to use the Quadratic Formula at some point?
Absolutement! Unless you want to complete the square manually :)
How would you do it manually?
By completing the square...
Let
Then complete the square like normal :)
Ok I got two answers;
since the domain says its negative only, then I'm assuming the inverse would be quadratic with the negative in it. But after sketching on the calculator it doesn't look like its inverse?
Post by: the.watchman on July 01, 2010, 01:43:10 pm
However, if you were to restrict the domain of the original function to a specific interval, then you could pick either one or the other :)
Post by: Aqualim on July 01, 2010, 01:48:39 pm
Post by: Aqualim on July 01, 2010, 02:28:30 pm
I keep getting a different answer to the calculator..
Post by: Aqualim on July 05, 2010, 05:31:44 pm
- a translation one unit in the positive direction of the x-axis and two units in the positive direction of the y-axis
- a reflection in the y-axis
- a dilation of factor 2 from the x-axis
- a reflection in the line
a) Find an expression for the tranformed
Post by: brightsky on July 05, 2010, 06:11:19 pm
Also I need help with finding a general solution for
I keep getting a different answer to the calculator..
Looking at the sine graph, we would have solutions
Post by: Aqualim on July 05, 2010, 08:55:29 pm
A functionundergoes the following sequence of transformations:
- a translation one unit in the positive direction of the x-axis and two units in the positive direction of the y-axis
- a reflection in the y-axis
- a dilation of factor 2 from the x-axis
- a reflection in the line
a) Find an expression for the tranformedand
co-ordinates in terms of x and y.
Can anybody help me here?
Post by: Aqualim on July 06, 2010, 11:14:27 am
State the correct order the transformations necessary to obtain the graph of
Is this correct?
- Dilation by a factor of 16 from the x-axis
- Translation of 3 units in the negative direction of the x-axis
- Translation of 1 unit up along the y-axis
I have a funny feeling the middle answer is wrong.
Post by: moekamo on July 07, 2010, 12:14:37 am
so it is:
- a dilation of 2 from the x axis,
- a reflection in the y axis,
- and a translation of 1 unit in the positive y direction
Post by: Aqualim on July 07, 2010, 11:32:21 am
&
The sequence of transformations can be described through the matrix equation;
where;
Solve the matrix 'equation 1' for X. Give the answer in the matrix form
Also;
Thanks any help would be greatly appreciated :)
Post by: yejiawen on July 09, 2010, 02:02:28 pm
2x(x-6) ? :S
Post by: Stroodle on July 09, 2010, 02:17:40 pm
Post by: superflya on July 09, 2010, 02:25:02 pm
Post by: yejiawen on July 09, 2010, 03:25:11 pm
Post by: yejiawen on July 09, 2010, 04:02:03 pm
12/(x+1)^2
Thanks in advance =.='
Post by: the.watchman on July 09, 2010, 04:06:34 pm
Post by: brightsky on July 09, 2010, 04:57:11 pm
Let
The integrand becomes:
Post by: cameron_15 on July 09, 2010, 09:16:26 pm
Find the equation of the graph f(x)
At first I though y=ax^3+bx^2+etc...
Then I thought y=(x-3)^2(x-a)... or it could also be (3-x)^2(x-a)...
Now I am confused... I'm usually pretty good with all the simultaneous equation stuff.
Post by: brightsky on July 09, 2010, 09:18:13 pm
The cubic has turning point at (3,0), so:
Passes through (0,18) and (1,16) so:
Use these system of equations to solve for a, b, c, d.
EDIT: Previous post was rubbish. :p
Post by: cameron_15 on July 09, 2010, 09:30:33 pm
Use turning point form,, where
is the turning point and
are constants.
I was thinking that, how does it handle situations in which there are 2 turning points?
Post by: brightsky on July 09, 2010, 09:33:49 pm
Use turning point form,
I was thinking that, how does it handle situations in which there are 2 turning points?
Ignore that, it was rubbish. xD The form is used for stationary points of inflexion. Edited post.
Post by: cameron_15 on July 09, 2010, 09:37:43 pm
Thanks for your help!
Post by: brightsky on July 09, 2010, 09:38:45 pm
I had 3 of the equations, just not the derivative one :)
Thanks for your help!
No problem. :)
Post by: Aqualim on October 25, 2010, 05:58:36 pm
so essentially its the abs(3*(abs(x))-abs(x^3))
Post by: 7132 on October 29, 2010, 04:02:21 pm
a) Using this info find probability that a boy selected at random is greater than 170cm
b) find probability that a randomly selected boy is more than 180cm given that his height is above mean
Post by: Whatlol on October 29, 2010, 04:07:13 pm
Thats because Pr(Z < -1/2 ) = Pr(Z > 1/2)
Pr(X>180| X >175) = Pr(Z>0.5) / 0.5 = 0.3/0.5 = 3/10 x 2 = 3/5
Post by: Juddinator on October 29, 2010, 04:24:40 pm
Post by: Whatlol on October 29, 2010, 09:02:40 pm
In the 2007 Insight Exam 2, one of the answers to a question is. However I used my CAS to graph the function and found the answer to be 0.314, converting it to
. Would I still get the answer correct in a VCAA exam as it is still in exact answer form?
no you would not, your answer is not an exact answer it is an approximate answer.
Post by: m@tty on October 29, 2010, 09:36:00 pm
In the 2007 Insight Exam 2, one of the answers to a question is
Can't you get the exact answer with your calc?
Anyway, no you wouldn't get the mark, because, as Whatlol said, your fraction is just the same number as you had earlier, containing only 3 dp. Anything involving pi must have pi in the answer, unless you are specifically asked to approximate the value.
Post by: Juddinator on October 29, 2010, 09:39:27 pm
Well I tried to get an exact value on my calc but it wouldn't spit one out for me.... I have a TI CAS so if anyone knows if it's possible could you explain it? I have already tried 'converting decimal to fraction'.In the 2007 Insight Exam 2, one of the answers to a question is
Can't you get the exact answer with your calc?
Anyway, no you wouldn't get the mark, because, as Whatlol said, your fraction is just the same number as you had earlier, containing only 3 dp. Anything involving pi must have pi in the answer, unless you are specifically asked to approximate the value.
Thanks
Post by: Aqualim on October 29, 2010, 11:32:02 pm
Pr (X > 170) = Pr ( Z > -0.5 ) = 1- PR (Z>0.5) = 0.7
Thats because Pr(Z < -1/2 ) = Pr(Z > 1/2)
Pr(X>180| X >175) = Pr(Z>0.5) / 0.5 = 0.3/0.5 = 3/10 x 2 = 3/5
Shouldn't Pr(Z > 0.5) correlate to the mean, since it's the middle number? How do you know that 170 is the middle number, when it says the mean is 175? or do you just make that assumption when the question asks that?
Post by: Whatlol on October 30, 2010, 04:46:41 pm
Pr (X > 170) = Pr ( Z > -0.5 ) = 1- PR (Z>0.5) = 0.7
Thats because Pr(Z < -1/2 ) = Pr(Z > 1/2)
Pr(X>180| X >175) = Pr(Z>0.5) / 0.5 = 0.3/0.5 = 3/10 x 2 = 3/5
Shouldn't Pr(Z > 0.5) correlate to the mean, since it's the middle number? How do you know that 170 is the middle number, when it says the mean is 175? or do you just make that assumption when the question asks that?
Pr(Z>0.5) indicates 0.5 standard deviations away from the mean. Pr(Z>0) indicates you are at the mean and the probability will be 0.5
Post by: itolduso on October 30, 2010, 08:21:11 pm
Post by: Whatlol on October 30, 2010, 11:04:28 pm
if one is greater than two, is two smaller than one?
what? one is less than two lol
Post by: lachymm on October 31, 2010, 02:17:05 pm
How is
And how do you solve
Sorry about my latex skills...
Post by: 7132 on October 31, 2010, 03:16:42 pm
dilate the graph X 4 away from the x-axis?
Post by: m@tty on October 31, 2010, 03:26:45 pm
I got a couple silly questions and they are from the same exam as that probability question previously asked...
How isthe same as
its probably simple but i dont see it?
And how do you solve
Sorry about my latex skills...
Assuming
Post by: lachymm on October 31, 2010, 08:02:54 pm
Are there any other rules for manipulating Tan(x) ?
Also thanks M@tty for the help, you are shooting for a big score in methods
Post by: lachymm on October 31, 2010, 08:05:52 pm
I seem to be having trouble manipulating simple indices and stuff :(
If someone could give me a hand it would be appreciated
Post by: m@tty on October 31, 2010, 08:12:56 pm
Other rules of importance are
That's all you need to know, I think.
And I already did Methods last year.
Post by: lachymm on October 31, 2010, 08:18:22 pm
Chears for the help you going to be on here for much longer, im going through practice exams i did a few weeks ago and seeing if i dont understand some stuff, if you are just chilling i might have some questions for ya :D
Post by: m@tty on October 31, 2010, 08:19:41 pm
Post by: lachymm on October 31, 2010, 08:27:00 pm
1st one isnt really related but anyways..
As someone who has already done methods -In the exam what is the best way to use your reading time in the exams and also do you think its better to leave the multiple choice in exam 2 for last?
Post by: lachymm on October 31, 2010, 08:59:46 pm
Post by: m@tty on October 31, 2010, 09:05:15 pm
Thoroughly read the extended answer, get an idea of where they are going, what you need to do and any specific requirements of questions such as 2dp, certain forms, integers, etc. Also look out for any tricks embedded in questions, if you can get a thorough understanding of these things during reading time, I find you all but eliminate stupid errors.
I actually would discourage you from dong MC in reading time, as many are complex and you want to employ your calculator to its fullest extent, as there is no working required. Certainly, though, have a quick read of the multiple choice questions in reading time if you have a chance after scrutinising section 2.
Personally with Methods I just did MC then the short answer, but that was because I had no problem with time. If you are worried about not finishing then you should do section 2 first and then the MC. If you run out of time you can guess all of the MC, but you can make almost no progress in a minute with short answer. In addition, there are 58 marks in section two, compared to only 22 in section 1.
Post by: m@tty on October 31, 2010, 09:07:11 pm
I got a quick question, what do you guys do when you come upto a question similar to "which one of the following satisfies the functional eqution? They take me for ages and i have now begun skipping them as they waste my time. However i usually do trial and error, i get them right its just they are so time consuming...Any tips for doing them quickly?
Can you provide an example?
I'd say, and this should not be taken as truth as this was not in non-CAS, that you should just work through the alternatives, starting with the one that looks most likely to save time, assuming it's a MC.
Post by: lachymm on October 31, 2010, 09:09:35 pm
I got
Post by: lachymm on October 31, 2010, 09:10:57 pm
Post by: brightsky on October 31, 2010, 09:12:01 pm
2tan(x) = 2
tan(x) = 1
Base angle: pi/4
Some of the other angles: pi + pi/4, 2p + pi/4, 3p +pi/4,...
So the general solution is npi + pi/4 = (4npi + pi)/4 = pi(4n + 1)/4, where n is an integer.
Post by: jasoN- on October 31, 2010, 09:13:58 pm
Post by: m@tty on October 31, 2010, 09:15:09 pm
What is the general solution to?
I gotfrom the calc but it isnt the right answer
That is a correct alternate form.
That takes
So the general solution is
Post by: lachymm on October 31, 2010, 09:16:47 pm
Post by: brightsky on October 31, 2010, 09:18:45 pm
Post by: lachymm on October 31, 2010, 09:19:29 pm
Post by: lachymm on October 31, 2010, 09:23:44 pm
Or do i need to set the domains of 2 different functions then derive individually?
Post by: m@tty on October 31, 2010, 09:26:47 pm
Oh cheers, matty bet my question. Do you need to recognize thatis the same as
? geez the exam writers are shifty
So that's why it works.
Post by: brightsky on October 31, 2010, 09:28:33 pm
When i derivethe calculator gives me
what does this mean?
Or do i need to set the domains of 2 different functions then derive individually?
See signum function. 5sgn(5x + 3) = 5(5x+3)/(|5x + 3|). For methods, it's probably better to just split it into cases.
EDIT: Mogz, thanks m@tty. Again. :p
Post by: lachymm on October 31, 2010, 09:30:43 pm
I know you cant get the log of a negative number or 0. But i got the general solution of
Post by: lachymm on October 31, 2010, 09:31:51 pm
thanks brightskyWhen i derive
Or do i need to set the domains of 2 different functions then derive individually?
See signum function. 5sgn(5x + 3) = 5|x|/(x(5x + 3)). For methods, it's probably better to just split it into cases.
Post by: m@tty on October 31, 2010, 09:32:16 pm
When i derive
Or do i need to set the domains of 2 different functions then derive individually?
When you differentiate a modulus function you should split it into its components.
It makes it much easier.
The same thing is given by
The signum operation takes only the sign (+ or -) of the number, so if
Post by: brightsky on October 31, 2010, 09:33:52 pm
2x = 0, pi, 2pi, 3pi, etc.
So yes your answer is correct.
Post by: lachymm on October 31, 2010, 09:36:18 pm
sin(2x) = 0
2x = 0, pi, 2pi, 3pi, etc.
So yes your answer is correct.
but their answer is
Post by: m@tty on October 31, 2010, 09:43:16 pm
This is the first solution.
So the general solution is
Post by: lachymm on October 31, 2010, 09:47:50 pm
Post by: m@tty on October 31, 2010, 10:05:56 pm
When i derive
Or do i need to set the domains of 2 different functions then derive individually?
See signum function. 5sgn(5x + 3) = 5|x|/(x(5x + 3)). For methods, it's probably better to just split it into cases.
I doubt anyone will notice, but on the off chance that this may mislead someone
Rather, from the definition
Post by: The Detective on November 01, 2010, 11:11:44 am
Post by: lachymm on November 01, 2010, 11:49:04 am
Post by: m@tty on November 01, 2010, 12:23:57 pm
If you find yourself in need of more obscure symbols, follow the link at the bottom of that page for extensive commands.
Post by: lachymm on November 01, 2010, 05:36:01 pm
I thought you needed to anti differintiate the y function and do that
Post by: lachymm on November 01, 2010, 05:45:25 pm
Also In the formula
Post by: lachymm on November 01, 2010, 05:50:04 pm
Post by: lachymm on November 01, 2010, 05:56:36 pm
Post by: brightsky on November 01, 2010, 07:08:36 pm
How do you use the chain rule for inside functions other than the typicalkind of thing example...
i dont understand how the
and the
thingy works when inside a function
Let
Therefore
When i find there inverse ofi get
which is correct because you need to choose the negative root in the context of the question however when i went to restrict the domain i said
however the answer is
why is this?
You should be right.
also how would you go about solvingand
and
then
is? BTW that
is an upside down U that they sometimes have in probabilty..
Also In the formulaand
what does n actually represent? i have forgoten :-[
Recall that
Then substituting values we get
I don't understand what you mean by this.
Post by: nacho on November 01, 2010, 07:21:32 pm
Question:
Let M = [ -2cosx 3 ]
[ 2 cosx ]
Where x is a real number. What is the range of det( M )?
^that is a single square matrice by the way.
My workings lead me to :
cosx - root3 = 0
OR
cosx + root3 = 0
Possibly incorrect, but i'd assume that i'd find what x equals in both scenarios and the answer would be [smaller x value, larger x value]
Just do not know how to get there.
(Sorry, too lazy to learn LaTex :D)
Post by: lachymm on November 02, 2010, 04:40:37 pm
Alright some more this time from kilabah 2010..
1) How does
2) How would you go about solving
3) How would you find the inverse of
Also is there a good way to remember the binomial formula the one
I mean how do you remember the factorial part? which numbers go where on the fraction? I havent really been taught how to do the binomial formula by hand always done it on the calc.
Post by: m@tty on November 02, 2010, 05:03:38 pm
2)
Quadratic formula is the easiest, or you could complete the square.
3)
By the quadratic formula:
Post by: lachymm on November 02, 2010, 05:34:47 pm
1) It doesn't.
Thanks
2)
Quadratic formula is the easiest, or you could complete the square.
I dont understand this last step howmeans that
3)
I also dont understand how you gotfrom
By the quadratic formula:
Thanks m@tty for your time
Post by: Declanator on November 02, 2010, 07:31:05 pm
its one of the examples for the new exam one content
thanks
Solve f(x)=0 for x
Post by: fady_22 on November 02, 2010, 07:46:29 pm
and
Post by: The Detective on November 03, 2010, 09:31:56 am
Post by: lachymm on November 03, 2010, 10:04:35 am
Post by: fazida on November 03, 2010, 10:42:21 am
what does "tbh" mean i see it all the time on vcenotes
it means 'to be happy'
Post by: Whatlol on November 03, 2010, 10:51:40 am
... it means to be honestwhat does "tbh" mean i see it all the time on vcenotes
it means 'to be happy'
Post by: d-ea-6 on November 03, 2010, 10:58:21 am
... it means to be honestwhat does "tbh" mean i see it all the time on vcenotes
it means 'to be happy'
Yeah, I hope that other guy was just joking...
Post by: lovingit on November 03, 2010, 12:13:19 pm
How do you determine if it is continous or not?
Post by: m@tty on November 03, 2010, 02:13:42 pm
Quote
I dont understand this last step how
3)
I also dont understand how you got
Thanks m@tty for your time
That second one was a typo. It was meant to read:
Multiply both sides by
Post by: lachymm on November 03, 2010, 03:49:56 pm
Can someone please explain to me how you know when a connecting point in a hybrid graph is diffentiable
How do you determine if it is continous or not?
If a 2 functions are apart of a hybrid function and they both have the same y value when there domains intersect (They continue where the other left off) while having the same gradient at this point they will be continuous thus not requiring a open circle..Finally i am able to help someone else out on here, feels good
Post by: doter on November 03, 2010, 04:17:06 pm
thanks in advance!!
Post by: m@tty on November 03, 2010, 05:22:36 pm
Factor out
Post by: doter on November 03, 2010, 06:44:51 pm
understand it now :)