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Use that to our advantage we can say that we can go with first term as 4 and then second term as 8 and etc. If you have mastery of 4 times tables you will notice that the 25th term is 100 but for this question it is instead 99 so therefore this will be the 26th term because we are using 4n-1 to...

It has to be UWS. Right?

Here are some hints a which is the start is 3 and the common difference is 4.

Wait @Hivaclibtibcharkwa which uni course is this and which uni is this from?

The hint is already given. Well, 16\cos^{5}\theta=\cos{5\theta}+5\cos{3\theta}+10\cos{\theta} Then part iv is just simplified into \left(\cos{\theta}\right)\left(16\cos^{4}{\theta}-1\right)=0 from 16\cos^{5}\theta-\cos{\theta} using part ii. Thus, \cos{\theta}=0...

Well, time to go back to drawing up an actual square if that is what's happening.

Resisted motion and banked tracks, pendulum Qs, projectile motion (2020-present), pushing the mass of ...kg which is measuring forces. Check the old 3U syllabus as well. Most of them moved in with Mechanics.

Do you have your working out? Once you have that then we talk.

ci) First you are told that you have two right angles at A and also note that AB=AC=\sqrt{3}OA. How is that useful because now you can write 1 for OA and \sqrt{3} for AB and AC which will come in handy. There use Pythag to find the length of z_{2}, z_{3} which will come in handy. There we will...

WITh the diagram instead of 12 it should be square root of 12.

Follow the instruction from the diagram and then use the fact that the formula of the circle is \left(x-a\right)^{2}+\left(y-b\right)^{2}=r^{2} as the equation of the intersection.

*Before starting this part of the question break it down into bite-sized pieces.* Part ii in summary is Sphere S_{2} has equation \left(x-2\right)^{2}+\left(y-2\right)^{2}+\left(z-5\right)^{2}=1 Sphere S_{1} has equation \left(x-2\right)^{2}+\left(y-2\right)^{2}+\left(z-2\right)^{2}=12 Both...

Long division, hmmm I think perhaps we can do something else about this, instead, you should have \left(z^{2}-6z+13\right)\left(...\right) using sum and product of roots. Step 2 look at what we have here \mathbf{z^{4}}-4z^{3}-3z^{2}+50z-\mathbf{52}. In this case, your head should immediately go...

You mean x-10 because there are two margins and each margin is 5cm. Double that and you will have 10cm.

Q6iii Yesterday in my sleep I had a very good idea to finish the question. On the LHS use the fact that \cos\left(A-B\right)+\cos\left(A+B\right)=2\cos{A}\cos{B} and then slowly close up the present as you repeat that process three times.

Check the gradient of f(x) that is the first step. Step 2 find minimum and maximum turning points. Essentials

@ExtremelyBoredUser Given your post that is why I named these files like this.

Q1) Show that \sin\left(f\left(x\right)\right)+\cos\left(f\left(x\right)\right)=\sqrt{2}\sin\left(f\left(x\right)+\frac{\pi}{4}\right) Q for reference If you had learnt your Algebra from Year 7 there is an intrinsic technique here. Remember x by itself is simply 1 times x...

I reckon that is what should be done because they used the identity from part i.