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When I was in my Primary and Secondary school, classical (Euclidean) Geometry was an important part of our Maths curriculum. If you are good in this elementary geometry, it will help your other maths greatly. Proofs in school geometry are your first exposure to Deductive Reasoning - i.e. how to...

y = 8x + 5 So, take any point on the line like (1, 13) or (0, 5) Gradient m = 8 means x : y = 1 : 8 so a direction vector is \binom 1 8 So a vector equation of y = 8x + 5 is: \binom x y = \binom 1 {13} + \lambda \binom 1 8 Correction: I thought it was for a vector equation. For my...

Given line equation does not look correct. Maybe intended to be: 2i + 4j+ l(-2i + 7j) ???

\frac{d ln(ln x)}{dx} = \frac{d ln(ln x)}{d lnx}\times \frac{dln x}{dx} = \frac{1}{ln x} \times \frac{1}{x} = \frac {1}{xln x}

Your question too open.

Suppose to the contrary: rational number + irrational number = rational number . . . . (*) i.e. R1 + IR = R2 i.e. m/n + IR = r/s where m,n,r,s are integers .: IR = r/s - m/n = (nr - ms)/ns = integer/integer (a rational number) .: an irrational number is a rational number This is a...

I didn't realise until a few years ago that a Sign Diagram can be a very efficient way to solve such inequalities. It is not normally taught in the HSC.

i see your problem. I'm unaware of present day challenges. The world has changed dramatically since my time as a kid.

Just read widely and read a wide variety of books.

Neither unless e = 0. x^5 and x^3 terms, are odd; sum of odd = odd. e = even. Sum of odd and even terms is neither even nor odd. f(x) = 3x^5 - 11x^3 + 7. \\ \\ f(-x) = -3x^5 + 11x^3 + 7 \neq f(x) $ and $ \neq -f(x) If e = 0: f(x) = 3x^5 - 11x^3 \implies f(-x) = -3x^5 + 11x^3 = -(3x^5...

See how far you can go without tutoring. If you can do very well without, then maybe you don't need tutoring. You can then send me the money you have now saved.

Why would they set such a question if they do not teach you the cross product?

Just relax. You can't learn a whole lot of stuff in one day. If you come to me, you won't need an Entrance Exam.

Not really trial-&-error; there is a logical approach to this.

I did a similar type of equality a few years ago: |x+1| + |x-2| = 3 If you look at region 2: every point x where -1 <= x <= 2 satisfies this equality BUT NOT the points OUTSIDE this region.

I have a simpler geometric way to do this that I investigated 2 to 3 years ago. Let me work this one out.

Correction