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From experience, it is a very powerful method because all you gotta do is to cover one column except the direction with your finger. Works similarly to the Heaviside cover-up method for partial fractions.

Okay, in that case, we will start. \left(\sqrt{x}-\sqrt{y}\right)^{2}\geq{0} and then the next step \frac{x+y}{2}\geq{\sqrt{xy}}. This is done by expanding and bringing the square root term on the other side. Well, here we can let x=\frac{a+b}{2}, y=\frac{c+d}{2}. After subbing the values of x...

Okay, @Run hard@thehsc can you show us how you answered this question? We might be of help.

Instead to find a vector that is perpendicular to two different vectors make one component in the vector zero eg let x_{1} or y_{1} or z_{1} be zero and then you have to find the dot product between your vector and the two given vectors.[/TEX][/TEX][/TEX]

I have made a small mistake in part c from my previous post but instead part c can be done very easily because of one reason. Since a+b+c=dthen we can exploit the fact that a^{2}bc+b^{2}ac+c^{2}ab=abc\left(a+b+c\right). Well that is simply abcd. Once that is out of the way we can now combine...

Okay @grainybred what are you planning to do for your future study, don't tell me you're doing Acturial studies. This is going to be a bad decision because already going into uni you need 4U to keep afloat. I am saying this to confirm if Extension II is worthwhile to keep given your performance.

For part b I have an antidote. Do you notice how part b started with "Hence ... " So using the same idea as part a you will see that \left(ab-bc\right)^{2}\geq{0} \left(bc-ac\right)^{2}\geq{0} \left(ac-ab\right)^{2}\geq{0} Expanding they become a^{2}b^{2}-2ab^{2}c+b^{2}c^{2}\geq{0}...

Who in their right mind would do Terry Lee questions if they are just doing Mathematics Advanced. That is overkill. I believe Cambridge is sufficient enough for this course and just spam AceHSC, THSC and others. If you are a very accomplished student, ignore MIF unless you need to route back and...

I think we can bring this topic back up at the end of 2023.

No, you can do all this without a calculator. Quadratic equation, moving fractions.

Okay, here is my approach, since we do not know the length of the diagonals we simply will start off with the fact that the lengths of the diagonals will be x, x-2 centimetres. Remember a rhombus will give us 4 right-angled triangles. Thus we now find the half of x, x-2 respectively which is...

Thanks, now I see why I cannot solve it at all. All because of a typo. Very unfortunate.

Wait wrong forum.

I think your friend has been playing too much gambling. What makes him think that the 2022 maths advanced exam will be easier? Instead of reckoning that the 2022 one will be slightly easier, he should work as hard as possible on mathematics (not so much that he suffers from burnout). You know so...

Some people love doing guesswork huh. Very well, The technique is to find the value where the absolute value is 0 and then find what happens before it and then find the value where the absolute value is 0 on the other side. Afterwards if there two absolute value functions find what happens in...

This is for anything between -1 and 2 less than -1 is -3 and greater than 2 is simply 3. Justifies my working.

Probably @=)(= had some difficulty with the signs.

This should give you c.

I have a different way of tackling this question \cos{\theta}=\frac{\overrightarrow{QP}\cdot{\overrightarrow{QR}}}{|\overrightarrow{QP}||\overrightarrow{QR}|} Note that \overrightarrow{QP}=\begin{pmatrix}3\\2\\-1\end{pmatrix}, and \overrightarrow{QR}=\begin{pmatrix}1\\3\\-2\end{pmatrix} So...

Welp that was helpful @Lith_30, nicely done.