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Pay attention to what variables are positive, real, what inequalities are initially given etc. as this information is almost certainly neccessary in deducing the inequality Recognise deducing the inequality as one or more of the following; 1) Adding, subtracting or multiplying given...

4!, 5!, 6! etc. are all multiples of 4. So each of these terms is equal to 1. There are 2017 of these terms Also; i^1 = i i^2 = -1 i^6 = -1 So the sum is 2015 + i Answer: B

Mastering probability comes down to understanding how the situations are performed. You must have a specific thought process. For example in Question 6, I would have a thought process like this (note: this is not the level of working out you need to do); a) It doesn't say that you need a...

No need to use sums and products of roots. If -4 is a root then P(-4) = 0 Put x = -4 and you will get k = -26 Additionally, -28 is the constant term. I think you mistook k as being the constant term, when it is actually (k-2), which would show that k= -26

Yes Area under the graph between 0 and 6 seconds. Be careful with the signs and direction when evaluating the answer. Answer looks horribly wrong. I believe what you are calculating is correct.

Suppose a projectile is launched with initial velocity v and angle A Initial y velocity: Vsin(A) Time taken to reach peak height (using v=u+at) : \frac{VsinA}{g} Total time of flight: \frac{2VsinA}{g} Total horizontal range (by the fact that horizontal velocity is constant)...

Suppose you have n! that you want to express as the product of smaller factorials. Let the largest of these smaller factorials be k! e.g. in the above example n = 10 and k=7 n!/k! = n(n-1)(n-2)...(n-k+1) For the example given, 10!/7! = 10 x 9 x 8 = 720 Then it is a matter of finding smaller...

Would this be correct working? Define the following conditions as follows; P: x^2 + y^2 < 1 ~P: x^2 + y^2 >= 1 Q: (x,y) is a point inside the unit circle ~Q: (x,y) is a point outside the unit circle Statement = P <-> Q = (P -> Q) and (Q -> P) = (Q or...

Contrapositive: If m or n are odd, then mn or m + n is odd Note that 'or' refers to the union of two conditions. So 'm or n are odd' could mean m is odd, n is odd or both m and n are odd.

What is a quick way to find the negation of implication and also how does this extend to the negation of an iff statement? For example, if I wanted to find the negation of How would I go about doing that, if the arrow was one-way or two-ways?

If it's parallel to the x-axis then there is no vertical component to the vector -> the coefficient of j is 0. xa + 4b = (-13x + 8)i + (20x + 60)j 20x + 60 = 0 x = -3

Will you come back to bos anytime soon?

Don't think this is true. I remember seeing a project that a multimedia student uploaded to their yt in 2018, and it very clearly showed their face and it was fine.

Gonna be devil's advocate and say that I'm leaning towards usyd...

Free periods come in clutch for when u didn't do ur hw the night before

If I drop another subject I'll be having a free period a day. I got an actual goated timetable lol.

I'm a 4u student and my fav levels of math (ranked) are 3U then 4U then 2U. But 2u is so chill it may as well be 7 free periods a fortnight lol.

Imagine grinding for the hsc and then u get a headache/stomachache in the final exam lol.

Memorise? I'll be honest, I remember basically everything on the formula sheet, but never product-to-sums. Just remember the sheet is always there for your reference

The 'degree' of a factor is called its' multiplicity A root of multiplicity one just passes through the x-intercept A root of multiplicity 2 has a tangent to the x-axis at that point, for example x=-1 and x=3 on your graph A root of multiplicity 3 has a point of inflection to the x-axis at that...