VCE Methods Question Thread!

[JellyBeanz]

Could someone please explain how you get the answer D? 

If the average value between the interval is zero that means there is equal area between the triangle and area under the curve.

therefore area of triangle = 25/8

There 1/2* p^2 = 25/8

Therefore p = 5/2

hence answer = D

[Mitsuka]

Which copy of a certain Methods practice exam is preferred considering they are both identical with the exception of one of them being termed CAS as well as differing in the number of questions. I'd been organizing my collection of exams for next year though was unsure which copy of the 2009 Hefferman exam 1 I should retain.

[qazser]

Which copy of a certain Methods practice exam is preferred considering they are both identical with the exception of one of them being termed CAS as well as differing in the number of questions. I'd been organizing my collection of exams for next year though was unsure which copy of the 2009 Hefferman exam 1 I should retain.

identical, yet have a different number of questions? Look at the solutions and see which one matches the solutions. Alternatively, add up the marks and see which one adds up to 40. Btw, unless you have a huge amount of time next year, you shouldn't be doing 2009 Heffernan papers. Stick to the more recent ones

[Gogo14]

Can someone explain what this question even means and how to do it?

Calculate the area of the region bounded by the x axis and the curves with equations
y=x^(1/2) and y=6-x

[MightyBeh]

Can someone explain what this question even means and how to do it?

Calculate the area of the region bounded by the x axis and the curves with equations
y=x^(1/2) and y=6-x

Can't see what  is so I'm assuming the second equation is \(y= 6-x \).

The 'area bounded' is the space between a number curves, where the curves act as the boundary lines. We evaluate this area using definite integrals; generally speaking we use \(\int_a^b \! f(x) - \! g(x)\, \mathrm{d}x\), where \(f(x)\) is the 'higher' function and \(g(x)\) is the 'lower function'. If you find it interesting at all, the area bounded by a function and the x-axis is the same, but g(x) = 0.

This problem didn't end very neatly so lmk if you have any questions still.

Edit: Might not have ended neatly because I messed it up. It's acutally the sum of 2 integrals, fixing the working now

[LPadlan]

Hello, i do not understand complementary relationships
If sin x = 0.3 cos a = 0.6 and tan @ = 0.7:
Find cos(-a)

This question is on page 514 cambridge math methods unit 1+2.
I really dont understand any of the questions and how they got the answers

[Syndicate]

Hello, i do not understand complementary relationships
If sin x = 0.3 cos a = 0.6 and tan @ = 0.7:
Find cos(-a)

This question is on page 514 cambridge math methods unit 1+2.
I really dont understand any of the questions and how they got the answers

cos(a) = cos(-a).

Cos is positive in the first and the fourth quadrant, so all the angles between (-90, 90) will yield a positive value (for cos only).

For example, cos(-60) = 1/2, whereas cos(60) also equals 1/2 (try putting it in your calculator if you want).

Using the theory above, it can be determined that cos (-a) = 0.6

[Guideme]

Just asking this question to make sure i am correct :3
A population of rabbits cann increase by 20%per month. At the start of the year, the number of rabbits in an area was 500.
Write an equation that describes this relationship between numbers and time.

Is it P=500x(1.2)^t
P=population
t=time in month.
If it is correct pls reply with correct, however if it is wrong pls provide an explanation. Thank you

[MightyBeh]

Just asking this question to make sure i am correct :3
A population of rabbits cann increase by 20%per month. At the start of the year, the number of rabbits in an area was 500.
Write an equation that describes this relationship between numbers and time.

Is it P=500x(1.2)^t
P=population
t=time in month.
If it is correct pls reply with correct, however if it is wrong pls provide an explanation. Thank you

That's correct.

[Gogo14]

So I've been using synthetic division for a long time now, but am really intrigued on why it actually works and why it doesn't work all the time. Anyone know?

[qazser]

So I've been using synthetic division for a long time now, but am really intrigued on why it actually works and why it doesn't work all the time. Anyone know?

It is essentially the same as Long Division, but without the algebraic pronumerals hence making it easier . Will post a more detailed answer later

[keltingmeith]

So I've been using synthetic division for a long time now, but am really intrigued on why it actually works and why it doesn't work all the time. Anyone know?

Hate to push in front of qazser (sorry, friend!), but Khan Academy has the solution! (also, pretttty sure synthetic divison always work? You might just be missing a step )

[Gogo14]

Hate to push in front of qazser (sorry, friend!), but Khan Academy has the solution! (also, pretttty sure synthetic divison always work? You might just be missing a step )

I don't think it works for terms in the form (ax+b).... or does it?

[Sine]

I don't think it works for terms in the form (ax+b).... or does it?

My synthetic division is quite rusty, learnt it quite early but never used it for VCE. I think you may only be able to divide by a linear term in the form ax+b where the coefficient of the x is 1. Therefore can only divide by x+b. Can't divide by quadratics, cubics etc

[keltingmeith]

I don't think it works for terms in the form (ax+b).... or does it?

Nope - BUT the term ax+b is the same as the term a(x+b/a) - so, you simply need to do the synthetic division with x+b/a, and then multiply by 1/a. Like so:

$$
\frac{f(x)}{ax+b}=\frac{1}{a}\left(\frac{f(x)}{x+b/a}\right)=\frac{1}{a}\left(q(x)+\frac{r(x)}{x+b/a}\right)=\frac{q(x)}{a}+\frac{r(x)}{ax+b}
$$

Can't divide by quadratics, cubics etc

Actually, it CAN be done, with just a few more steps!