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Hello! Oh no that's a silly mistake, it's meant to be the 2nd of April (i.e. this Sunday!) thanks for pointing that out, and sorry for the confusion!

Hey everyone, Justin from ACE here again. As we want to make these a mainstay in the calendar, we’re holding another round of our End of Module Wrap Up Seminars this weekend! Like always, we want these these seminars to do as much good as possible, so they are completely free and open to...

Hi everyone, Justin here from ACE. In case you missed the last post, I graduated from James Ruse in 2016 and am now the Head of Chemistry and Academic Lead at ACE. I am excited to post again to let you know that our new series of completely free online seminars which review each Module in the...

No worries 😊. For everyone else, don't forget to hop on tonight at 6:30pm and/or sign up using the registration link before the seminar ends to get access to the seminar slides for a limited time!

Unfortunately we won't be recording this seminar. While we do keep recordings of all our ACE lessons for students, when we run these free public seminars, a big "Recording" logo can create a barrier for students to speak up and get involved in the discussion and cause privacy issues.

Yes, the Mod 1 seminar will cover nuclear decay. You don’t have to register to join but it might help remind you for tomorrow. We’re only holding one session, but it’ll definitely be worth your while if you can make it, even if you can only join for a bit. However, keep your eyes peeled for...

Hi everyone, I’m Justin and I’m the Head of Chemistry and Academic Lead at ACE. I graduated from James Ruse in 2016 with an ATAR of 99.95 and 5 state ranks including 15th in Chemistry. I am excited for ACE to be offering a new series of completely free online seminars to review each module in...

How about this solution for the inequalities question? $Since $ (x-1)^2 \geq 0, $ then$ \\ (x-1)(x-1)\geq 0 \\ (x-1)(\sqrt x - 1)(\sqrt x +1) \geq 0 $ as $ x \geq 0 \\ $But, $ \sqrt x + 1 >0 $ for all $ x \geq 0 \\ $then $ (x-1)(\sqrt x-1)\geq 0 \\ $so upon expanding, $ x\sqrt x + 1 \geq x+\sqrt x

the piece in Question 3 was literally just noise. lol.

Wrong thread - but here goes you should quadratic formula to get a solution for y^2. Then discard the negative solution. Take positive/negative square root for solutions for y.

Does anyone remember the name of the piece in the melodic dictation excerpt and the subsequent excerpt? It was a really awesome piece :)