VCE Methods Question Thread!

[Coolgalbornin03Lo]

With differentiation of tan(0) to sec^2(0) Which = 1/cos^(0)

Does that mean if the angle were Pi on 3 a would get cos of Pi on 3 (1/2) square it and then make it the denominator of that fraction 1/1/4= 4?

[kxyaa]

Silly question but did anyone use 2B pencil in the exams in previous years but still received a good study score? Or is it alright as long as the pencil is dark enough and doesn't smudge?

[Coolgalbornin03Lo]

With differentiation of tan(0) to sec^2(0) Which = 1/cos^(0)

Does that mean if the angle were Pi on 3 a would get cos of Pi on 3 (1/2) square it and then make it the denominator of that fraction 1/1/4= 4?

Bump (if anyone sees this before the exam)

[keltingmeith]

Bump (if anyone sees this before the exam)

Yes

[hailstriker]

how did everyone go for Methods. It was kinda time consuming.

[Vatsal0305]

how did everyone go for Methods. It was kinda time consuming.

Agreed, it was rather time consuming, there wasn't much time left to do the last two 8-markers. But I think they made the last questions fairly simple and easy to understand compared to other exams since the whole exam was so time consuming. I reckon the probability part was simple, just a little shocked that there were no questions on Normal distribution.

[The Cat In The Hat]

Agreed, it was rather time consuming, there wasn't much time left to do the last two 8-markers. But I think they made the last questions fairly simple and easy to understand compared to other exams since the whole exam was so time consuming. I reckon the probability part was simple, just a little shocked that there were no questions on Normal distribution.

I was sad there was nothing on Normal distribution, I'd studied that a lot... hoping for plenty tomorrow.

Did anyone/everyone even finish the paper? I didn't.

[ABB0005]

Anyone know if derrick ha's bound reference is worth it?

[Sine]

Anyone know if derrick ha's bound reference is worth it?

I would definitely recommend it. I used those bound references for both my Methods and Specialist exams. Alongside using it as a bound reference you can definitely learn a lot of content from it and get a good understanding of the best way of setting out your answers. There are also exams inside (these exams are much harder than normal practice exams but they are really good in exposing students to all the types of little tricks that are possible to come up in exams).

[Corey King]

Hey guys,
I'm learning how to divide polynomials at the moment.
I don't understand why it is you can divide an expression by only one term in a binomial, then multiply the whole binomial by the quotient you just produced.

I feel like the explanation of the workings of this process is missing, either that or there is some assumed knowledge I am missing.

(an example of the method I do not understand)
https://gyazo.com/9ffeba1f03746e3f7dc84327f885bbc5

Any help in explaining how dividing with a binomial like this works would be much appreciated,
Corey

[a weaponized ikea chair]

The letters AABBBA are arranged in a row. How many different ways can they be rearranged?

I have no idea how to do this.

[The Cat In The Hat]

The letters AABBBA are arranged in a row. How many different ways can they be rearranged?

I have no idea how to do this.

If there's no limit (e.g. 'must have an A on one end' or something), you use the factorial (!). For this, since there are 6 letters to be arranged, it would be 6!, which is 6*5*4*3*2*1.

Someone check if I'm right, though, there may be more to this question than I'm seeing, I don't know; haven't done this for a long time.

[p0kem0n21]

The letters AABBBA are arranged in a row. How many different ways can they be rearranged?

I have no idea how to do this.

I think it is 20 ways? I get this from 6!/(3!*3!). Since there are 6 letters in general, we have 6! ways of arranging them. Then, we have to account for the repeats of A by dividing by 3! (since there are 3 repeats///3 As). We divide by 3! again due to the three repeats of B for the exact same reason as with A, and that's how we get 20.

If there's no limit (e.g. 'must have an A on one end' or something), you use the factorial (!). For this, since there are 6 letters to be arranged, it would be 6!, which is 6*5*4*3*2*1.

Someone check if I'm right, though, there may be more to this question than I'm seeing, I don't know; haven't done this for a long time.

Mmm, we have to account for the fact that the same letters appear, which would make certain orders the same. For example, if we numbered each of these letters, we could have:

A₁ A₂ A₃ B₁ B₂ B₃

We could also have:

A₂ A₃ A₁ B₁ B₂ B₃

However, these would both be A A A B B B, making them identical arrangements. We divide by the respective factorial amounts to account for these reappearing orders.

[The Cat In The Hat]

I think it is 20 ways? I get this from 6!/(3!*3!). Since there are 6 letters in general, we have 6! ways of arranging them. Then, we have to account for the repeats of A by dividing by 3! (since there are 3 repeats///3 As). We divide by 3! again due to the three repeats of B for the exact same reason as with A, and that's how we get 20.

Mmm, we have to account for the fact that the same letters appear, which would make certain orders the same. For example, if we numbered each of these letters, we could have:

A₁ A₂ A₃ B₁ B₂ B₃

We could also have:

A₂ A₃ A₁ B₁ B₂ B₃

However, these would both be A A A B B B, making them identical arrangements. We divide by the respective factorial amounts to account for these reappearing orders.

Oh, whoops, thanks for checking that... I think what you're saying is correct.

[rotaryboy]

Hi, all any help would be appreciated!
A girl travels from A to B at 5km/h and from B to A at 4km/h. The
total journey takes 30 minutes. Find the distance travelled, 2d.p.