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From what I remember about MATH1081 it also has aspects of proofs in the course where we have to learn to write the negation and stuff. Essentially MATH1081 which is Discrete Maths takes slices of cake from all the following Year 12 maths units Mathematics Standard (Networks) Mathematics...

From what I know I believe you would have to prove these identities. Here we start, \sin\left(\cos^{-1}\frac{3}{11}-\sin^{-1}\frac{3}{4}\right)=\frac{19}{44}. Here \sin(\cos^{-1}\frac{3}{11})\cos(\sin^{-1}\frac{3}{4})-\cos(\cos^{-1}\frac{3}{11})\sin(\sin^{-1}\frac{3}{4}). Simplifying we will...

For the denominator 4ax^{2}+bx+a we use the quadratic formula where x=\frac{-b\pm{\sqrt{b^{2}-16a^{2}}}}{8a}. For only one asymptote in this case we would like to have \sqrt{b^{2}-16a^{2}} to not exist. Here, we know that a solution does not exist when b^{2}-16a^{2}<0 which in this case is...

Here is what we can say Suppose the roots are \alpha,\;\alpha\;,\beta\;,\gamma. Then we have 2\alpha+\beta+\gamma=p+q \alpha^{2}+2\alpha\beta+2\alpha\gamma+\beta\gamma=0 \alpha^{2}\beta+\alpha^{2}\gamma+2\alpha\beta\gamma=-\left(p-q\right) \alpha^{2}\beta\gamma=-1...

You will have a parallelogram, and here you will see that the two vectors mentioned will be at 117^{\circ} using the fact that there are "C-angles" due to the fact that co-interior lines make 180^{\circ} in a parallelogram. Afterward, use the cosine rule to find the magnitude of the vector, and...

If you dilate the whole thing by -1 then what you are really doing is flipping the graph on the x-axis which in this case is -\sqrt{x}. Theen going up two units you will then have y=2-\sqrt{x}.

Here is my take on this question. Let vector \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD} and \overrightarrow{HE}=\overrightarrow{HG}+\overrightarrow{GF}+\overrightarrow{FE} Exploit the fact that for a regular octagon suppose \overrightarrow{AB}=\vec{a} then...

I believe that in my memory this is where it comes from a\cdot{b}=|a||b|\cos{\theta}. This comes from the angle for vectors. There, what you do is replace a with b and then what you will have is b\cdot{b}=|b||b|\cos{\theta}. The interesting part is that since b lies in the same direction of b...

Not everyone here goes to UNSW and even if they go to UNSW they might not live in a college. You might have more luck asking this question at r/unsw on Reddit.

Very good stuff but you forgot which region you are looking at. Someone here might say, oh we did it with some irrational number like \sqrt{2} where \left(\sqrt{2}\right)^{2}+2=4. This is why instead of Hence, there exists no n such that n^2 + 2 is divisible by 4. It should be Hence, there...

Do you notice the fact that you used cross product in vectors. NESA thought, nah students do not have to know this, awakening the wrath of Dr Du when he used the cross product in one of his lessons to roast NESA to pieces.

LMAO.

Where is the 8 in front of the whole solution?

https://boredofstudies.org/threads/help-with-cambridge-extension-2-enrichment-question-complex-numbers.402265/ Have a look at this forum.

To start we will first divide z^{6}+z^{3}+1 with z^{3} which gives us z^{3}+\frac{1}{z^{3}}+1=2\cos{3\theta}+1 With the RHS since you got three multiples a division of z^{3} is just simply dividing z for each quadratic inside the bracket. This in turn will give us...

Notice that in \frac{4!}{2!} the division of 2! is because you have 2 A's as you assume the 2 A's are going to be the same.

Is this the first time you saw this question? If so then I understand why this question felt a bit difficult. Here let us start. Step 1: \int{\sin^{2}x\cos^{3}xdx}=\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx} Step 2...

Good attempt however this is beyond the scope of the NSW Mathematics Extension I syllabus.

It should be less than \frac{\pi}{2}+k\pi but greater than or equal to \frac{\pi}{4}+k\pi. The other solution is because when x=\frac{\pi}{2} the result is undefined or infinity so therefore, the value has to be smaller than x=\frac{\pi}{2} and the +k\pi comes from the fact the tan function...

Simple play \sin{2x}-2\sin^{2}x>0 2\sin{x}\cos{x}-2\sin^{2}x>0 2\sin{x}\left(\cos{x}-\sin{x}\right)>0 Now we use something called the auxiliary angle for \left(\cos{x}-\sin{x}\right)=R\cos\left(x+\alpha\right) Expand\;R\cos\left(x+\alpha\right)=R\cos{x}\cos{\alpha}-R\sin{x}\sin{\alpha}...