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Worse. Now we have English.

Re: Business Studies Section III: Business Report I was chilling through this, thought question 28 was going to be just as easy. :o

That's such a bad idea. You gave away an easy 10 marks. Always tackle the easiest marks first. I guessed a few short answers. Went horribly on last question.

Re: Business Studies Section IV: Extended Response This. fml.

Psht, I'm not going to waste time writing them down :p I'll remember them though ;) We can take the paper out, can we not?

The answer was right. Remember, no value for y exists when x is pi/2. Tan(90) is still undefined, so tan(tan^-1(pi/2)) is undefined. The graph would look like:

Can you type the first one up on this? Online LaTeX Equation Editor Second: 2sin(^2)x + cos x -2 = 0 2(1 - cos(^2)x) + cosx - 2 = 0 2 - 2cos(^2)x + cosx - 2 = 0 2 cos(^2)x - cosx = 0 cosx(2cosx - 1) = 0 cosx = 0 or cosx = 1/2 Solve for x

More information needed? :confused: h = -t sqrt(t) + c Don't you need to be supplied the initial height if this is to work?

3-2[cos(^2)x]-2=0 1 - 2[cos(^2)x] = 0 2[cos(^2)x] = 1 cos(^2)x = 1/2 cosx = +- sqr.root (1/2) Solve from there.

You wrote the question wrong. Proof: Let b = 3a (i.e. 2nd root 3 times the first). Your equation: 2(x^2) + pq + q a + 3a = 0 (no x^1) 4a = 0 Therefore a=0, 3a = 0, two roots are 0, not true. Invalid question =p ---- edit: p & q are constants or not? If not, then p(q + 1) = 0 p = 0 or q = -1...

x+y = 25 y = 25-x f(x) = xy = x(25-x) = 25x - x^2 f'(x) = 25 - 2x when f'(x) = 0, x=12.5 f''(12.5) = -2, therefore max. y = 25-12.5 = 12.5 hence x = 12.5, y=12.5 i. a + b + c = 4 ii. 1/a + 1/b + 1/c = (ab + ac + bc) / abc = 0 Question:

Have you checked the resources page?

Are you sure the answer is 77 not 17?

Well the mean would just be the sum of the series divided by the number of terms. So for an AP, if we let a=first time, l=last term: Sn = 1/2n(a + l) Sn/n = 1/2(a+l) = 1/2(5+51) = 28 (assuming by 5&51, you mean 5 is first term, 51 is last term) Same logic applies to a GP. Sn = a(r^n -...

y%20=%20ln(sinx^x)%20=%20x%20ln(sinx) And then it's normal chain rule. f'(x) = ln(sinx)(1) + cotx(x) = ln(sinx) + xcotx

It's an X2 question. Don't worry about it.

@untouchablecuz: dont have pen+paper, doing phys review. tried doing your q, and failed first time on ms word. takes too long, ill try later with paper. answer me this though. i get it down to y-y1=m(x-x1), when x=0, y=0, hence y1 = mx1. Am I meant to solve for m? --- ^^^ I'm not that great...

You missed a minus in your answer. Been a while since I did probability. I'm assuming by different she can't pick the same square twice? i. 32P3/64P3 = 5/42 ii. 2 x 5/42 = 5/21 iii. 1- 5/21 = 16/21 Let me know if I'm wrong. --- Using the substitution u = 2x + 1 or otherwise, integrate...

e^2x+3e^x-10=0 Let u = e^x u^2 + 3u - 10 = 0 (u-2)(u+5) = 0 u = 2 or -5 e^x = 2 or e^x = -5 x = ln(2) or x = ln(-5) [not solution] Therefore x = ln2 Differentiate y = cos(ln(x) + 1)

In my trials I got 65/84, so I guess that's where I'm at. I heard as a rough guideline 60 is a top band? Obviously changes year to year.