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Re: Discrete Maths Sem 2 2016 Yes what your friend did is valid. Since intersection is both associative and commutative, we can do intersections in any order (like, (X cap Y) cap Z = X cap (Y cap Z) = X cap (Z cap Y) = (X cap Z) cap Y, using associativity and commutativity. I used 'cap' to...

Re: Discrete Maths Sem 2 2016 To get the final answer your friend got (which looks correct), use an absorption law at the second last line of your proof. (Unfortunately your simplification in your last line isn't valid. But if we just apply the absorption law there we'll get the answer. :))...

Re: MATH1231/1241/1251 SOS Thread To find where it meets a certain axis, we set the other two variables to 0 and solve for the last variable. E.g. to find where it hits the y-axis, we would let x = 0 and z = 0 and solve for y in the plane equation. The point where it meets the y-axis would be...

Re: MATH1231/1241/1251 SOS Thread By inspection. He was probably just looking for an easy vector in S, and the one he chose is easy to find (because we just make one of the terms 0 by making that component of the vector equal to 0 and then see what we need to do for the other entries. If we...

Re: Discrete Maths Sem 2 2016 That's correct (by definition of subset essentially) :).

Re: Discrete Maths Sem 2 2016 A is an element of B, but not a subset (assuming the b's aren't coincidentally just equal to the a's). Since A isn't a subset of B, it's not in the power set of B. But {A} is in P(B), since {A} is a subset of B, since A is an element of B.

Re: MATH1231/1241/1251 SOS Thread Why do you think this thread will get closed?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread There was a typo in the last line of leehuan's work. He meant the RHS there to have a 9 outside the bracket instead of 3 (because he squared 3(x – 3) from the previous line).

Re: MATH1231/1241/1251 SOS Thread $\noindent Yeah, it's based on the fact that $\cos\left(ix\right) = \cosh x$ and $\sin \left(ix\right) = i\sinh x $ (one way to see this is via Euler's Formula and the exponential definitions of the hyperbolic functions). (We can also show from these by...

Re: MATH1231/1241/1251 SOS Thread Often for a^2 + x^2, x = a*sinh(t) turns out to be neater than using tan. Just make sure if you're using hyperbolic ones, you're familiar with hyperbolic analogs to trig stuff, e.g. know how to integrate things like sech, simplify things like compositions of...

Yeah, so we multiply its distance from x = 2 by 3/4 to get its distance from y = -5. E.g. If |x – 2| was 8, then |y + 5| would be 6, which is (3/4)*8 = (3/4)*|x – 2|.

$\noindent The distance of the point $P\left(x,y\right)$ to the line $y=-5$ is $\left|y+5\right|$. Its distance to the line $x=2$ is $\left|x-2\right|$. So we need $\text{distance to }(y=-5) = \frac{3}{4}\text{ distance to }(x=2)\Longleftrightarrow \left|y+5\right| = \frac{3}{4}\left|x-2\right|$.$

You can use the chain rule explicitly as follows. Let y = sin(3x+1). Let u = 3x+1. So y = sin u. Now, dy/du = cos u, and du/dx = 3. So dy/dx = (dy/du)*(du/dx) = (cos u)*3 = 3*cos u = 3*cos(3x+1). With enough practice, you will probably be able to write down the answers to things like this...

Ah it's a HSC level Q. I was just about to say that if this was in HSC, they'd probably want you to use the b/(b+c) stuff. But then I remembered that conditional probability isn't taught (explicitly) in the HSC. If this 2U class was taught it, surely they'd have learnt the definition of...

Well yeah, they probably want you to say things like P(X|Y) = b/(b+c) (via their diagram). But this just uses the definition anyway and makes the proof more tedious than it needs to be).

The form of the chain rule you've quoted is for when our function is raised to some power, which isn't the case here. The form of the chain rule we'll want for this is: d/dx (f(ax+b)) = a*f'(ax+b). So here, d/dx (sin(3x+1)) = 3*cos(3x+1).

This is just showing that if P(X|Y) = P(X), then X and Y are independent. By definition, P(X|Y) = P(X,Y)/P(Y). (, meaning intersection). Since P(X|Y) = P(X) (by assumption), we have P(X,Y)/P(Y) = P(X) ==> P(X,Y) = P(X)P(Y). Also, P(Y|X) = P(Y,X)/P(X) = P(X,Y)/P(X) = P(X)P(Y)/P(X) = P(Y)...

Did your teachers teach you anything related to this when doing pracs?

$\noindent I think it's meant to be something like `$\text{measured value }\pm\text{ half the limit of reading}$'.$ $\noindent Here, the `measured value' is 13 V, and the limit of reading is 1 V, so half the limit of reading is 0.5 V. So I think it should be $13 \pm 0.5$ V (option (A)).$

Isn't Standard similar difficulty to Advanced?