Recent content by Drongoski

1. What tutor to pick

Yes, I still take students, but only a few.
2. Maths Help

3. \frac {dy}{dx} = e^{2y}\\ \\ \therefore \frac {dy}{e^{2y}} = dx \\ \\ \therefore \int dx = \int e^{-2y} dy \\ \\ \therefore x = -\frac {1}{2}e^{-2y} + C
3. Maths Help

2. \frac {dy}{dx} = 1 - \frac {y}{3} = \frac {3 - y}{3}\\ \\ \therefore \frac {dy}{3-y} = \frac {dx}{3}\\ \\ \therefore \int \cdots dy = \int \cdots dx \\ \\ \therefore -ln |3-y| = \frac {x}{3} + C \\ \\ x = 2 \implies y = 4 \implies -ln|3 - 4| = \frac {2}{3} + C \implies C = -\frac {2}{3} \\...
4. Binomial Distribution Assignment Help

Question too bloody long!
5. How do I do this complete mindblank

You can use the "Terry Lee" method to find the square root of -60-32i. Here's how: \sqrt{-32i - 60} = 2\sqrt{-8i -15}\\ \\ \sqrt{-8i - 15} = \sqrt{1 - 8i -16} = \sqrt{1^2 + 2\times 1 \times ({-4i}) +(-4i)^2} \\ \\ \\ $Remember:$ (a + b)^2 = a^2 + 2 \times a \times b + b^2 = (a + b)^2 \\ \\...
6. confused with this q

A^4 + 4B^4 = A^4 + 4B^4 + 4A^2B^2 - 4A^2B^2 \\ \\ = (A^2 + 2B^2)^2 - (2ab)^2 = (A^2 + 2B^2 +2AB)\times(A^2 + 2B^2 - 2AB)
7. how to improve in maths

I don't think you can get an easy answer here. If you can afford it, get a good tutor to check you out or a friend who is good in maths to help you.
8. Is reverse chain rule just integral of product of function and it's derivative?

Maybe I'll show you how to do RCR later.
9. forgot how to integrate this

\int ^1 _0 \frac {\sqrt x}{1 + x} dx = \int ^1 _0 \frac {\sqrt x}{1+(\sqrt x)^2} \times 2\sqrt x d\sqrt x\\ \\ =2\int^1 _0 (1 - \frac {1}{1 + (\sqrt x)^2} )d \sqrt x \\ \\ =2 \left[\sqrt x - tan^{-1} \sqrt x \right ]^1 _0 \\ \\ = 2[(1 - \frac {\pi}{4} ) -(0 - 0)] = 2 - \frac {\pi}{2} \\ \\...

No.
11. Is reverse chain rule just integral of product of function and it's derivative?

Integrands which are derivatives(or constant multiples thereof) of composite functions are the ones usually handled by "integration by substitution". They often refer to reverse chain rule - but I've seldom seen how this reversal is achieved. Using my method, I can do the reversal of the chain...
12. is motion in a straight line

Motion, in general, does not need to be in a straight line. In your Adv & Ext 1 Maths, you are usually dealing with motion in a straight line, because this is the easiest to deal with. Motion in a straight line is often referred to as "rectilinear motion".
13. 9b, a bit confused (calculus)

\frac {dy}{dx} = -3\frac {d}{dx}(x^4-2x^2-1)^{-1} = -3\frac{d(x^4-2x^2-1)^{-1}}{d(x^4-2x^2-1)} \times \frac {d(x^4-2x^2-1)}{dx}\\ \\ = -3\times {-1}(x^4-2x^2-1)^{-2} \times (4x^3-4x) \\ \\ = \frac{3\times 4x(x-1)(x+1)}{(x^4-x^2-1)^2}\\ \\ = 0 $at x$=0, -1,1\\ \\ y(-1) = \frac {-3}{1-2-1} =...
14. Integration question

Again, if not required to use the substitution: \int ^0 _{-3} \frac {x}{\sqrt {1-x} }dx = -\int ^0 _{-3}\frac {1-x - 1}{\sqrt {1-x} }dx\\ \\ = \int ^0 _{-3} (1-x)^{0.5} d(1-x)-\int ^0 _{-3}(1-x)^{-0.5}d(1-x)\\ \\ = \left[ \frac{2}{3}(1-x)^{1.5} -2(1-x)^{0.5} \right]^0 _{-3} \\ \\ =-(\frac...
15. is this 3u

Avoiding formal substitution: \int ^1 _0 x\sqrt{1-x^2} dx = -\frac {1}{2}\int ^1 _0 (1-x^2)^{0.5} d(1-x^2) \\ \\ =-\frac{1}{2} \left[\frac {(1-x^2)^{1.5}}{1.5} \right]^1 _0 = -\frac {1}{2}\left [ (\frac{2}{3} \times(1-1)^{1.5}) -(\frac {2}{3}(1-0)^{1.5})\right ]\\ \\ = -\frac{1}{2} \times...