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n(moles) = m(mass)/M(molar mass) so n = 5/24.31

Example: Calculate the number of moles of 5 grams of magnesium (use the periodic table)

Re: Need some help with my decision SKDJFBSDKJFNJSSDBISFSDZHJKSDCDBKNV yeh like that. general maths will suffice, helps u with some rearranging equations and subbing in variables - chemistry calculations is about applying formulas CORRECTLY

Re: Need some help with my decision SKDJFBSDKJFNJSSDBISFSDZHJKSDCDBKNV Chemistry doesn't need maths for the calculation questions, its just numbers and calculator. The most most u'll ever do is the re-arrangement of something to find it example: n (moles) =\frac{m(mass)}{M(molar mass)} This...

For u to understand the working out of 'AnimeX', u need to notice this: \Delta Y = 2\left ( \Delta C \right ) Look at the change in consumption and income EACH YEAR

Re: HSC 2013 3U Marathon Thread ahh i see, made a mistake on my very last term (wrong power lol)

Re: HSC 2013 3U Marathon Thread I found a mistake in my working out for the RHS (-1)^n \binom{2n}{n} If i equate coefficients of x^(2n) for [3], i get \binom{2n}{2n}

y = e^{ln(3^{4x-2})} ($by applying the rule: e^{lna}=a) \\ \\ y = e^{(4x-2)ln(3)} \\\\ $\frac{dy}{dx}= \frac{d[(4x-2)ln(3)]}{dx} \cdot e^{(4x-2)ln(3)} \\ \\ \frac{dy}{dx} = 4ln(3)\cdot 3^{4x-2}\\ \\ \frac{dy}{dx} = ln(81)\cdot 9^{2x-1} ($using $(pq)^st = (p^s q^s)^t) method 2: log both sides...

"Q is South 47 East from B." So if u've drawn the vertical and horizontal (north and south) axes for B, since the bearing of "Q is South 47 East from B.", the angle beginning from the vertical (south) axis ending at the line BQ is 47 degrees, hence angle(BPQ) = 90 - 47 = 43

Either improve ur spatial ability over time OR just look at EVERY single shape (esp. triangles) one by one

Chunk it 1st chunk - the ones that are soluble with NO exceptions 2nd chunk - the ones that are soluble but WITH exceptions 3rd chunk - the ones that are insoluble WITH exceptions Note: the way i memorised it, i made each chunk have 2 rules (so total 6 rules)

Re: HSC 2013 4U Marathon \rightarrow \\ (\sqrt{1} - \sqrt{a_i})^2 \geq 0 ($where i = 1, 2, 3, \dots n) \\\\ $1+ a_i\geq 2\sqrt{a_i} \\\\ $Let i = 1, 2, 3 \dots n and multiply all of them\\\\ $(1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n \sqrt{a_1 \cdot a_2 \dots \cdot a_n} \\...

Yes because | - f | = f OR using graphing techniques (if u have time) draw y = x-1 and y = 1-x then draw, y = |x-1| and y = |1-x| u'll notice they are the same graph, so |x-1| = |1-x| the graphing technique can be done mentally

huh? so as long as i sign it, its ok? so i dont have to write 'student's signature'?

my principal signed it and a requirement is the STUDENT'S SIGNATURE

i dont have any other form of ID that satisfy UMAT requirements are YOU SURE, i can write 'student's signature'? (last time i emailed UMAT, they said i cannot write down by date of birth, someone else like my deputy principal must write it and sign....)

For the certified letter of identification, when I sign it (above my photo), should i write down 'student's signature'? or can i just leave it blankly with just my signature

y - 3 = tan(..) (x + 2) y = 1.33...x + 5.66... y = 1.3x + 5.7

Apply the one-point gradient formula or point-gradient formula (w/e its called lol) U need a POINT and the GRADIENT (-2,3) (yes! we have a point!) Now, gradient.... m.... we dont have it, so we must recall that m = tan(theta) = tan(53'8') And there u have it, a point and the gradient...

thanks!! i understand now