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You should find many of the proofs in your text books.

Don't know how to do the vector notations. So unable to post solution. Also, there's too much to type out. But vector PR = vector OR - vector OP = (2a - 3b) - (a+b) = a - 4b. You can do as suggested by TheOnePheeph or solve simultaneously any pair of related results, like (2a-3b).(a+b) = 0 and...

So do the new Maths Ext 2 textbooks cover Logic concepts like? - proposition, T and F, disjunction, conjunction, negation, exclusive disjunction, Truth Table, logical equivalence, implication, equivalence, converse, inverse, contrapositive, tautology, contradiction? If not, it does not include a...

You put up the post. I was surprised. I did not think it is in 4U. Have to check the new syllabus again.

Truth Table p q p => q Not(p => q) T T T F T F F T F T T F F F T F

Tutoring costs a lot of money. If you can cut down on a few subjects, you'll save your parents heaps. Also, if you were to do tutoring in, say, even just 3 subjects, then where are you going to find time to do your own studies. Even with tutoring, you still need time to digest what you are...

Your parents must be very rich

If you can come to Epping, I can do Ext 1 for $95/lesson (1.5 hrs). I can explain difficult concepts better than most teachers. I've been posting efficient solutions on Bored since 2009. If interested, message me on 0490 - 780 347. Dr Drongoski

But please read (well-written material, good books, old & new) widely, and, critically.

I'm sure there are many many from non-English speaking background who do extremely well in their English. Just because you speak Mandarin or Hindi, Korean, Vietnamese , Urdu or Hokkien at home doesn't mean your English is handicapped. In fact I find so many spelling, grammar and punctuation...

\int \frac{sin 2x}{4+cos^2 x} dx =\int \frac {sin 2x}{4 + \frac{1}{2} (1 + cos 2x)}dx = \int \frac {2sin 2x}{9 + cos 2x}dx \\ \\ = \int \frac {-d(9 + cos 2x)}{9 + cos 2x} = -ln|9+cos 2x| + C\\ \\ \therefore \int _0 ^{\frac {\pi}{4}} = -[ln(9 + 0) - ln(9 + 1)] = ln \frac {10}{9} Is this right??

I didn't mean to criticise your method. It is, after all, the standard method. But for simple cases, the "Terry Lee" method is really short. But I bet many will find it hard, unless they are handy with algebraic fiddling.

Well - you can compare the "normal" method with the "abnormal" method. If you can manage the shorter method, would you want to go through the rigmarole of the "normal" method, unless the question mandates this approach.

-5 + 12i = 4 + 2 \times 2 \times 3i -9 = 2^2 + 2\times 2 \times 3i + (3i)^2 = (2 + 3i)^2 \\ \\ \therefore \sqrt {-5 + 12i} = \pm(2 + 3i) Take your pick: which is shorter??

I like the method in Terry Lee's text: z^2 = 2 - \sqrt 5 i = \frac {1}{2}(4 - 2\sqrt 5 i) = \frac {1}{2}(5 - 2\sqrt 5 i -1) \\ \\ = \frac{1}{2}((\sqrt 5)^2 +2\times\sqrt 5 \times (-i) +(-i)^2)) = \frac{1}{2} (\sqrt 5 - i)^2\\ \\ \therefore z = \pm \frac {1}{\sqrt 2} (\sqrt 5 - i) @Zoee...

Aren't you lucky. You have the ideal Socratic learning situation: 1 student & 1 teacher!

Musings of an IB Maths tutor . . . . From my years of tutoring IB Maths, one thing strikes me. A large proportion of students entering Yr 11 do not have an adequate foundation to handle their Mathematics, be it Mathematics Studies, Mathematics(SL) and Mathematics(HL). Many of the maths they...

Unfortunately I don't know how to provide diagrams ( which would have helped greatly). So I need to explain as best I can. The weight 40g, can be resolved into 2 components: 40g sin(30) down slope and 40g cos(30) perp to (and against) the slope. The force P can be resolved into 2 comps: P...

Is the answer about 349.5 Newton (using g = 9.8) or 351.53 N (g = 10)??

Ask your teacher, for example, to check you out and advise you. Maybe you can handle MX2, given the right guidance. I presume you are beginning Yr 12 now - so you don't have much time to seek help/advice. People on this forum cannot advise you meaningfully because they do not know your true...