Post by: Special At Specialist on November 15, 2011, 09:49:32 pm
First question:
A committee of three people is to be selected from a group of five men and four women. In how many ways can the committee be selected if there must be at least two women on the committee?
I solved it a weird way because I don't know the "proper" way of solving questions like this:
Since there are two women out of 3 spots, that leaves only 1 spot for each of the other 7 people.
However, there are 6 ways in which two women can be chosen for those spots, so 6*7 = 42 combinations in total.
Is that correct?
Is there a better way to solve it for more complex problems?
I came up with a new way of solving it (still not sure if correct or not):
Combinations of two women are:
w1-w2, w1-w3, w1-w4, w2-w3, w2-w4, w3-w4, all with 5 spots for men each.
Then the combinations of three women are:
w1-w2-w3, w1-w2-w4, w2-w3-w4.
So the total number of combinations is 33.
Is that correct?
Post by: Mao on November 16, 2011, 11:33:29 am
Total is 34.
Another way is to use combinatorics. The total is
Post by: Special At Specialist on November 16, 2011, 06:56:55 pm
I have a new problem:
A graph of the form y = a*tan(bx) has asymptotes at x = -pi, x = pi and x = 3pi. Find the value of b.
I think it has something to do with tan(pi/2) = undefined, but I'm still not sure how to solve the problem :(
Post by: Greatness on November 16, 2011, 06:59:44 pm
So pi/b=2pi, therefore b=1/2. Lol am i missing something? because i dont know where you got that tanthingequaling undef
Post by: BlueSky_3 on November 16, 2011, 09:24:41 pm
Post by: dc302 on November 16, 2011, 11:45:08 pm
The period of the graph is 2pi and we know that period=pi/n for tan graphs.
So pi/b=2pi, therefore b=1/2. Lol am i missing something? because i dont know where you got that tanthingequaling undef
Yeah, if you wanna use the tanpi/2 = undefined method, you note that, at cos(bx) = 0, the graph is undefined, ie it has an asymptote. So when does cosbx = 0? When x = pi (as stated in the question), so cos(pi b) = 0 = cos(pi/2) so b=1/2.
Post by: Special At Specialist on November 21, 2011, 09:10:36 pm
(http://img525.imageshack.us/img525/2684/mathsproblem6.png)
Post by: b^3 on November 21, 2011, 09:22:21 pm
(x,y)->(3x-1,2y+4)
x'=3x-1
y'=2y+4
Rearrange it,
i.e.
so
Hopefully there are no mistakes.
Post by: luken93 on November 21, 2011, 09:25:14 pm
1/3(x' + 1) = x
y' = 2y + 4
1/2(y' - 4) = y
1/2(y' - 4) = 2[1/3(x' + 1)]^2 + 1
1/2(y' - 4) = 2/9(x^2 + 2x + 1) + 1
y' - 4 = 4/9(x^2 + 2x + 1) + 2
y' = 4/9x^2 + 8/9x + 4/9 + 6
=4/9x^2 + 8/9x + 58/9
hopefully haven't made any errors
Post by: Special At Specialist on November 21, 2011, 09:29:23 pm
Okay, I get everything up to here.
(x,y)->(3x-1,2y+4)
x'=3x-1
y'=2y+4
Rearrange it,
i.e.
so,
and
Hopefully there are no mistakes.
I don't get that part. How did you find the link between y and x? First you were splitting them up and then you somehow joined them together.
Post by: dc302 on November 21, 2011, 09:32:49 pm
Post by: b^3 on November 21, 2011, 09:33:24 pm
We have y=s*x2+bx+c
So we found, y in terms of y' and x in terms of x', simply we sub those into y=2*x2+1
Then what are we looking for? y', so rearrange for y'.
I hope that makes sense.
EDIT: yeh as dc302 has stated.
Post by: Special At Specialist on November 21, 2011, 09:40:38 pm
To:
i.e.
How did you get rid of the y' and the x' just like that?
Post by: b^3 on November 21, 2011, 09:46:45 pm
Okay, but what about the second last step where you went from:yeh I probably shouldn't do that. We found the image y' in terms of x, but I'm just representing it on an x-y plane, i.e. y in terms of x.
To:i.e.
How did you get rid of the y' and the x' just like that?
(I'm not explaning this well, sorry)
Post by: Special At Specialist on November 21, 2011, 09:50:09 pm
Post by: b^3 on November 21, 2011, 09:53:47 pm
I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.
EDIT: Dc or someone can probably provide a more concise wording for that last part.
Post by: Special At Specialist on November 21, 2011, 10:02:30 pm
Post by: dc302 on November 21, 2011, 10:05:40 pm
Ahh okay... so what is the difference between R --> R and R^2 --> R^2? Is there ever anything else I have to learn, like R^3 --> R^3?
R --> R means from the real line to the real line
R^2 --> R^2 means from the plane to the plane
R^3 --> R^3 means from 3-d space to 3-d space
Post by: Special At Specialist on November 21, 2011, 10:09:18 pm
Post by: dc302 on November 21, 2011, 10:10:02 pm
I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.
EDIT: Dc or someone can probably provide a more concise wording for that last part.
Yeah it's hard to explain without calling upon higher maths but think of it as a map of australia. Draw a path from melbourne to sydney.
This path is represented by a function. Now what happens if you 'stretch' your map so that it;s twice as wide? The path you have seems to have changed to a new function. This is what you are doing. You have a function on the plane, and you are basically changing the geometric construction of the plane and thus obtaining a new function.
edit:
Okay cool. Will functions always be written as R^2 --> R^2? Because I could swear I remember a function given the rule f:R-->R or some weird notation like that. And it was a 2D cartesian plane graph.
It should always be R^2 -> R^2 if it's a function from the plane to the plane.
Post by: b^3 on November 21, 2011, 10:14:04 pm
i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x2
Post by: Special At Specialist on November 21, 2011, 10:21:26 pm
I think special at spesh is thinking about the domain restrictions for functions with the R->R part.
i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x2
Yeah, that's the one :)
This path is represented by a function. Now what happens if you 'stretch' your map so that it;s twice as wide? The path you have seems to have changed to a new function. This is what you are doing. You have a function on the plane, and you are basically changing the geometric construction of the plane and thus obtaining a new function.I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.
EDIT: Dc or someone can probably provide a more concise wording for that last part.
Oh, so the function is just being moved or squished, but it never actually changes?
Post by: dc302 on November 21, 2011, 10:24:38 pm
I think special at spesh is thinking about the domain restrictions for functions with the R->R part.
i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x2
Ah yes, this may have confused him because it is only a convenience the way we represent a function on a single graph. The way it would supposed to be represented is to draw two real lines, and any point on the first real line is x, and has an image f(x) on the second real line.
When you look at a 2-d plane that's representing a function from R to R, what you're looking at is actually called the 'graph of f':
g : R --> R^2 (or graph(f)), where g is defined by g(x) = (x, f(x))
An example of an actual function from the plane to the plane is:
f(x,y) = (2x,3y)
Quote
Oh, so the function is just being moved or squished, but it never actually changes?
It's changing along with the change you make to the whole plane, so it IS still changing with respect to the original function.
Post by: Special At Specialist on November 21, 2011, 10:32:26 pm
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?
Post by: dc302 on November 21, 2011, 10:35:43 pm
I sort of see what you're saying...
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?
For a sure-fire way of doing it:
1. Find x in terms of x' and y', y in terms of x' and y' from the information provided. (remember x, y are the coords from the original plane, and x', y' are from the image of the original plane.
2. Substitute x and y (written in terms of x' and y') into the equation of the function.
3. Rearrange till you get the new function, in x' and y', in the desired form. ie y' = ax'^2 + bx' + c
4. Write down your solution but now change x' and y' back to x and y.
Post by: Special At Specialist on November 21, 2011, 10:39:12 pm
I sort of see what you're saying...
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?
For a sure-fire way of doing it:
1. Find x in terms of x' and y', y in terms of x' and y' from the information provided. (remember x, y are the coords from the original plane, and x', y' are from the image of the original plane.
2. Substitute x and y (written in terms of x' and y') into the equation of the function.
3. Rearrange till you get the new function, in x' and y', in the desired form. ie y' = ax'^2 + bx' + c
4. Write down your solution but now change x' and y' back to x and y.
Ahh now that explains everything. Thank you! I'm going to attempt a few more of those problems to see if I really get the hang of it...
Post by: Special At Specialist on November 26, 2011, 04:44:34 pm
(http://img28.imageshack.us/img28/9587/mathsproblem7.png)
It looks like such a simple algebra/logarithms question, but I don't know what to do! The best I could do was find a few solutions to c in terms of logx(y) or in terms of logx/y (y) or something along those lines. I don't know how to get rid of the x's and y's :(
Post by: brightsky on November 26, 2011, 04:57:17 pm
x^(a/b) = y
sub into second equation:
x^a = (x^(a/b)/x)^c
x^a = (x^(a/b - 1))^c
x^a = x^(c(a/b)-1))
a = c(a-b)/b
c = ab/(a-b)
Post by: Special At Specialist on November 26, 2011, 08:13:53 pm
x^a = y^b
x^(a/b) = y
sub into second equation:
x^a = (x^(a/b)/x)^c
x^a = (x^(a/b - 1))^c
x^a = x^(c(a/b)-1))
a = c(a-b)/b
c = ab/(a-b)
Very elegant solution and it didn't even require logarithms. Thank you!
Post by: Special At Specialist on April 23, 2012, 05:57:43 pm
Suppose that:
f(x) has a domain of [-10, 10]
f(x) has a range of [-20, 15]
g(x) has a domain of (-12, 8]
g(x) has a range of (-2, 11)
What is the domain and range of f(g(x))?
What is the domain and range of g(f(x))?
Post by: kamil9876 on April 23, 2012, 06:44:20 pm
for similair reasons g(f(x)) can't exist.
Post by: pi on April 23, 2012, 06:55:32 pm
f(g(x)) can't exist (since 10.1 is in the range of g but f can't 'eat' 10.1 (it is outside its domain))
for similair reasons g(f(x)) can't exist.
To format in a VCAA exam, draw a table for ease :)
| f(x) | g(x)
Ran | ... | ...
Dom | ... | ...
Then say something like "as Domf isn't a subset or equal to (use the symbol) Rang, g(f(x)) cannot exist"
:)
edit: just looked like I corrected kamil, which I wasn't trying to do! Sorry if it came out that way haha! kamil's right, it's just that assessor's like the above format :)
Post by: Special At Specialist on April 23, 2012, 10:03:46 pm
Post by: brightsky on April 25, 2012, 06:55:22 pm
x---> g-machine ---> g(x) ---> f-machine ---> f(g(x))
you grab a set of x-values, chuck them in the g-machine, and get a set of g(x) values. for f(g(x)) to exist, all g(x) values must be within the domain of f(x).
for example, if your f(x) was sqrt(x), and you input g(x) = -1 into it, you won't get any real output f(g(x)).
more formally, we say for f(g(x)) to exist, ran g must be a subset (or equal to) dom f. if f(g(x)) exists for all g(x), then it follows that it's domain is the set of all x, which is defined in this case as the domain of g.