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Re: HSC 2016 3U Marathon Are those meant to be binomial coefficients?

Re: First Year Uni Calculus Marathon And another related Q. $\noindent Does there exist a continuous function $f \colon \mathbb{R}\to \mathbb{R}$ with $\lim_{x \to a}f^\prime \left(x\right)$ existing (for some real number $a$) but not equal to $f^\prime (a)$? (seanieg89 has just shown above...

Re: HSC 2016 3U Marathon You should probably say that opposite sides of a parallelogram/rectangle are equal.

Re: HSC 2016 3U Marathon It probably is OK, but the book's way is probably better because it lists the vertices in matching order (like matching with the corresponding vertices in the other congruent triangle). Always writing these things in matching order will also help you keep track of...

Re: First Year Uni Calculus Marathon $\noindent Let $f\colon\mathbb{R}\to \mathbb{R}$ be a function that is continuous on an open interval $I$ containing $a\in \mathbb{R}$ and is differentiable on $I$, except possible at $a$. Suppose $\lim_{x\to a}f^{\prime}\left(x\right)$ exists. Does...

Sources online seem to confirm your idea of how to find %(w/v), so you should probably just find the mass of chloride ion and use the formula you gave. To do it the answers' method, they should have found the mass of chloride ion in 250 mL (i.e. 0.25 L) by doing the calculation as: (0.0248...

Are the answers from a source you expect to be reliable?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread I think it should be what you got (4*sqrt(91) cm). I.e. 2*sqrt(20^2 – 6^2), which follows from Pythagoras's Theorem and the fact that diagonals of a rhombus bisect each other at right angles.

Let's say f is twice continuously differentiable and let y = g(x) := 1/f(x). Then assuming f(a) =/= 0, we have g'(a) = -f'(a)/(f(a)^2). So sign of g'(a) = – (sign of f'(a)). So anywhere where f would be increasing (f'(x) positive in a neighbourhood of a), g would be decreasing. So increasing...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent a) Note that $\angle AGD = \angle HGB$ (vertically opposite angles). Also, $\angle DAG = \angle BHG = 90^\circ$ (angles of a rectangle). Finally, $AD = HB$ (corresponding sides of congruent rectangles are equal)...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Here's one way to do it. Join $AC$ and $BD$ and let their point of intersection be $X$. It follows from part (b) that $AX = BX$ (equal sides in isosceles triangle $ABX$). Let $\angle XAB = x = \angle XBA$. Let...

Re: MX2 2016 Integration Marathon Are there any useful sufficient conditions that'd make the arc length of the limit curve equal the limit of the arc lengths?

Re: MX2 2016 Integration Marathon No need to even take limits, because the integrand is a continuous function on the domain of integration (if we want we can define appropriate left- or right-hand values that make the function continuous there; there isn't any problem because the denominator...

Re: MX2 2016 Integration Marathon $\noindent Also, for $a\in \mathbb{R}$, find $\int _0 ^\frac{\pi}{2} \frac{1}{1+\tan ^a x}\text{ d}x$ (determining for which real $a$ we have convergence).$ (Essentially same method as previous Q., except also need to determine for which a the integral...

Re: MX2 2016 Integration Marathon NEW QUESTION $\noindent Find $\int _0^1 x^2 \sqrt{1 -x^2}\text{ d}x$.$

It wasn't on this thread I think, some other thread (from a week or a few days ago or something).

Maybe eyeseeyou meant to say first-year Physics, rather than Maths? Because I remember he made a previous comment saying this before (about uni Physics).

Re: MX2 2016 Integration Marathon This is done using LaTeX. You can find a Short Guide to LaTeX here, which outlines how to use it on this site: http://community.boredofstudies.org/14/mathematics-extension-2/234259/short-guide-latex.html .

For the first, let's consider the sequence of "Fibonacci last digits". This is obtained by just taking the last two numbers in the sequence, adding them, and only recording the last digit. So it is: 1, 1, 2, 3, 5, 8, 3 (only record last digit of 13), 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1...

For the second one, we can break it up into cases. The numbers we want must have at least three digits, and can have at most six digits (as 300 000 has six digits). Three digits Only possibility is clearly 456. So 1 possibility. Four digits The numbers must be of the form: • 456 _ ; 10...