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If you live near Epping, I'd be delighted to provide you face-to-face 1-on-1 tutoring in 4U Maths. Message me first if interested.

I "invented" this method (let's for now call it the "Drongoski Method"). A 2-hour lesson would be sufficient for an introduction to my method. So far, I've attracted no interest. I think they don't know what my approach is all about. But I believe it to be a very powerful, intuitive and very...

Ok. Posted it especially for your benefit, since you indicated multiple times you could not follow my explanation.

But is this version of any help. Do you now understand my solution?

A picture is worth a thousand words.

Hello again, clarifying my approach to integration..

Hello Hello Hello.

Wish I was half as smart as you at your age. Good on you.

Are you in Year 9 only?

\int \frac {ln ([3x^5 + 3]^{5x^4})}{x^5 + 1}dx \\ \\ = \int \frac {5x^4[ln(x^5 + 1) + ln 3]}{x^5 + 1}dx \\ \\ = \int ln(x^5 + 1)\cdot \frac {5x^4}{x^5 + 1} dx + ln 3\int \frac {5x^4}{x^5 + 1} dx \\ \\ = \int ln(x^5 + 1)dln(x^5 + 1) + ln3\int \frac {1}{x^5 + 1}d(x^5 + 1) \\ \\ = \frac...

OK. Say you take x = 5.7 say. Then this number is |5.7 + 1| = 1 + 5.7 = 6.7 away from the number -1, and |5.7 -5| = 0.7 away from the number 5; so it is = the constant 6 plus 2 x 0.7 from the 2 numbers. Remember, any x outside the closed interval [-1,5] is 6 + twice its distance from the nearer...

Say you choose any x between -1 and 5, x = 2, say. then |x+1| + |x-5| = |2+1| + |2 - 5| = 6. Choose another such number, say x = -0.7; then |x+1| + |x-5| = |-0.7 + 1| + |-0.7 -5| =0.3 + 5.7 = 6 (again). If you choose x = -3 (which is more than 0.5 to the left of -1), then |-3+1| + |-3-5| = 2 + 8...

Follow what i said. Draw a line and mark off -1 and 5. any number x between and including -1 and 5 will have |x+1| + |x-5| = 6 (a fixed sum - so we need at least 1 more for this sum) So x must be outside this interval; x now only needs to be more than 0.5 beyond the 2 numbers -1 and 5; i.e. at...

To me graphing is easier like this: Draw the number line and mark off the 2 numbers "-1" and "5". The distance between these 2 numbers = 5-(-1) = 6. Now read the inequality this way: the distance of the number x from "-1" plus the distance of x from the number "5" is greater than 7. Now x...

Yes! For someone doing MX2, your grasp of the fundamentals of trigonometry needs to be better.

It is. It is always true that: cos(\pi -\theta) = -cos \theta whatever the value of theta.

No.

In outline as LaTeXing is so tedious. \therefore b_2 = b_{1+1} = b_1(b_1 + 1) = 1(1+1) = 2\\ \\ \therefore b_2 = 2 = 1 + \sum ^1 _{r=1} b_r ^2 = 1 + 1^2 Therefore true for k = 1. Now assume true for n = k (k >= 1) \therefore b_{k+1} = 1 + \sum ^k _{r=1} b_r ^2 \\ \\ b_{k+2} = b_{k +1}(b_{k...

Australia likes to boast about world-class universities. All those university ranking bodies like the Times Higher Education World University rankings that appear annually shouldn't be taken too seriously. To give a bonus point for a Band 3 in (the so-called - what a misnomer) Advance Maths is a...

I recall having a chat with a stranger about 10 years ago who told me he was doing Electrical Engineering at USyd, but did not study Physics for his HSC. They let him do it. May well have eventually completed his B Eng(Electrical). Miracles do happen! Scott Morrison is right after all.