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OHHHH SNAPPP But yes if it wasn't counted, I wouldn't be surprised if students (myself included) would pick 12 units and completely neglect english, which would basically defeat the point of making English compulsory.

You have 2 equations sinx = 1 \ \ x = \frac{\pi}{2} cos2x = 0 The sine equation only has one solution, the cos equation has 4 solutions between 0 and 2pi

since x-2 is a factor of P(x), then by the factor theorem, P(2) = 0, use that to solve for a.

Like it's great that you guys here can achieve these comebacks and all, but I get the feeling that this doesn't apply to everyone.

i) Construct the centre point O. Connect that point with A and C to form Triangle AOC. OA = OC = 6 (Radii of circle) Angle AOC = 30 (angle at the centre is twice the angle at the circumference) Now simply use the cos rule to find AC. ii) One way I see is using the sine rule to find the...

That's why you gotta learn latex man. \frac{1-sin(\frac{\pi}{2}-2 \theta)}{sin(2 \theta)} = \\ \\ \frac{1-cos(2 \theta)}{sin(2 \theta)} = \frac{1-(1-2sin^2 \theta)}{sin(2 \theta)} \\ \\ = \frac{2sin^2 \theta}{2 sin \theta cos \theta} = \tan \theta

Hint: after you get M you then consider xy, then also consider x/y = -1\p^2 and y/x = -p^2

Just become asian and you got this. Jokes aside, for me I basically practiced questions more often (from textbook and past papers), especially since 4U has many more different variety of questions to ponder about.

Absolutely not!

You need to use the concept a = \frac{d}{dx} (\frac{1}{2}v^2) From there you integrate with respect to x then find the constant with the initial condition then multiply both sides by 2.

you have to expand the second term first using your sum of angles and simplify before you use your usual auxiliary angle. fan96 did suggest a faster method, though that formula isn't in the syllabus anymore.

8-9 hrs on week day, 10 on weekends. I personally don't condone all-nighters but to each to their own I guess.

Can't speak for myself since I had tutoring for math back then soz. I can however speak for some of my cohort, who didn't really have tutoring for Math but yet attained a solid E4.He was pretty consistent with his math work and was nearly always up to date, our math teacher at school also gave...

You're going to be making a lot of life decisions when you're finished with school. Many people choose to go to uni, many choose to go to tafe, many choose to work, many choose a professional sporting career. Now people say that you should make a clear choice and stick to it, and that you...

n = 1 \Rightarrow \\ \\ LHS = 1*1! = 1 \ \ \ RHS = (1+1)! - 1 = 2 - 1 = 1 True for n = 1 $ Assume it's true for $ \ \ n = k \ \ \\ \\ \sum_{r=1}^{k} r*r! = (k+1)! -1 \\ \\ $Then we prove it's true for $ \ \ n = k+1 \ \ \Rightarrow \\ \\ \sum_{r=1}^{k+1} r*r! = (k+2)! -1 LHS =...

Can you please rotate your pictures next time? For the first one, \frac{1}{cos^2ax} = sec^2ax

Woah Woah, that's not even close to true.

y>0 Since y \rightarrow \infty \\ \\ \ \ $as$ \ \ x \rightarrow -2 \ \ $or$ \ \ x \rightarrow -3

The Range would be 0<y\leq \frac{1}{2} You should think about what the lowest value you could have in your denominator, in that case it would be \sqrt{4} = 2 That lowest value would give you the highest value in the function (the maximum) Now think about the largest possible value in...

Yeah dividing that amount doesn't do anything So for Q3 it goes like this, basically after every 12 months, $100 gets added into the account A_{12} = 100(1.0075)^{12} \\ \\ A_{13} = (100+A_{12})(1.0075) \\ \\ \dots \\ A_{24} = (100+A_{12})(1.0075) ^{12} = 100(1.0075)^{12} + 100(1.0075)^{24}...