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I don't blame drongoski for saying "I'll finish the rest if you hire me as your tutor. Otherwise complete the rest yourself." But if he does become like this guy ^, then yea, he'll be like a recruit Btw, it's (3t+1)^2

What's the progress on the forum posts? Is there like a % or time remaining thing?

hahahaha ^double entendre

If you need a bit more info for iv) The sum of roots is -1. 2cos2pi/7 + 2cos4pi/7 + 2cos8pi/7 = -1 Then divide by 2 on both sides to answer the question

I don't think you've seen the world enough. Noone always plays by the rules

Lol wtf? when it comes to the hsc you shouldn't care about ethics. Use w/e you have to your advantage. Though this is my opinion

Double angle formula for cos to get rid of the 1. Double angle formula for sin to use the de moivre theorem. I remember doing this in the Terry Lee book

Considering your subjects, you are right But No. It's not just memory and apply for some subjects, understanding is needed. So I have no idea why slyhunter said that

Oh..it's net force, not weight experienced

Wouldn't the force be 2645.2 + mg?

+ 1

blood is made up of mostly water. It will turn slightly acidic

Use oh exploitables straight use of the formula or... cos2x = 1-2sin^2(x) rearrange to make sin^2x the subject (1-cos10a)/2 - (1-cos6a)/2 =(cos6a -cos10a)/2 = cos(8a-2a) - cos(8a + 2a) all over 2 And you will end up with sin2sin8a

Why do people say there's not much difference in difiiculty between 2U and 3U? Comparing 2U and 3U hsc and prelim topics, you will see 3U concepts are more complex than than 2U. Hmmm come to think of it, it's probably because you guys don't have separate classes for 2U and 3U and thus unable...

You wouldn't be able to do that with hsc phyics

idk say same equipment and some other things that are left unchanged

Hmmm...getting confused by the answers made by other people Forces acting on an astronaut are the ground reaction force (more like the rocket floor reaction force) and the force on the person produced from the force on the rocket. Letting acceleration due to gravity remain constant (9.8)...

^ +1

You need to restrict the range too if you're going to show it algebraically

Uhh, I'll take your word for it. @OP for the 54b, do you have the answer to that question?