Specialist's Methods Thread

[Special At Specialist]

Hello my name is Special At Specialist and this is going to be my mathematical methods question thread. Any help at all with these questions would be highly appreciated. Thanks!

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?

[Mao]

You missed one. w1-w3-w4.

Total is 34.

Another way is to use combinatorics. The total is

[Special At Specialist]

Thanks, Mao!

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

[Greatness]

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

[BlueSky_3]

Nah you're right Swarley, I think he means that in general tan(x) has asymptotes at multiples of x=pi/2 which is where the graph is undefined like what special guy said. But I think he probably forgot that the tan graph is dilated by a factor of 1/b from y-axis.

[dc302]

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.

[Special At Specialist]

Please help me with this transformations problem:

[b^3]

(x,y)->(3x-1,2y+4)
x'=3x-1

y'=2y+4



Rearrange it,

i.e.
so , and

Hopefully there are no mistakes.

[luken93]

x' = 3x - 1
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

[Special At Specialist]

(x,y)->(3x-1,2y+4)
x'=3x-1

y'=2y+4

Okay, I get everything up to here.

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.

[dc302]

He just used the original equaton, y = 2x^2 + 1

[b^3]

Basically we want the equation of the image, i.e. we want y' in terms of x'.
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.

[Special At Specialist]

Okay, but what about the second last step where you went from:

To:

i.e.

How did you get rid of the y' and the x' just like that?

[b^3]

Okay, but what about the second last step where you went from:

To:

i.e.

How did you get rid of the y' and the x' just like that?

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.
(I'm not explaning this well, sorry)

[Special At Specialist]

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?