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never done this sort of stuff, so there might be a better way. first divide the equation* by (x + 1) = (x+1)(x^2 +x -4) + 5 so you have as the last two terms, -4(x+1) and +5 so you are left with x^2(x+1) + x(x+1) = x^3 + 2x^2 + x so you know that the coefficient for (x+1)^3 must be 1, which...

I don't get what you did there, could you please explain? and the number of pages in 800 words doesn't matter does it? btw, I put my working into the calculator wrongly the first time for that question, and we got the same answer for the other 2 questions

not sure about the first one. I would think it's 1 - (799/800)^200 - 200(799/800)^199 * (1/800) = 0.0264... second one: combinations of 2/4 boys is 4c2 = 6. Total number of combinations = 2^4 = 16, so 10/16 chance of not having 2 boys and 2 girls. Chance of all 4 families not having 2 boys, 2...

glad I could be of help :) also, I personally wouldn't consider choosing poor scaling subjects that you love, to be 'not tryng your best'. I don't think that would be a very good way of viewing things. I'm sure you'll get well over 90 no matter what subjects you choose anyway.

I didn't think you really cared if you got a top UAI or not; just if you tried your best (correct me if I'm wrong about that) . If that's the case I think you should only consider how much you enjoy the subject, not whether it scales better - even if you think that, all else considered, you...

just remember to look for difference of two squares, and if you can group eg. grouping: p^2 + pq - pr - qr = p(p + q) - r(p + q) = (p - r)(p + q) x + y + x^2 - y^2 = x + y + (x-y)(x+y) note that there is a like term of x + y = 1(x+y) + (x-y)(x+y) = (x-y+1)(x+y) and remember difference of...

x = angle of projection, t = time d = Ucosx t v = Usinx - gt, where v = -Usinx (same speed in opposite direction at end of flight_ t = 2Usinx/g d = Ucosx 2Usinx/g = U^2 sin2x/g d' = U^2/g * cos2x = 0 , at maximum distance cos2x = 0, x = 45 at max d so max d = U^2/g and if U^2 = gd d = gd/g...

what was missing?

2 particles are launched from a point. They have the same speed. P is launched at an angle of alpha, Q at an angle of Beta (measured in the same direction, and also these are in the same vertical plane - which I think means same side of the horizontal that they were projected from, perhaps also...

the question says past experience suggests 3 of 100 will be faulty. This means the equivalent of saying 1 car has 3/100 chance of being faulty, not that given 100 cars exactly 3 will be faulty.

yeah this can be done. I thought you would be able to do it quite easily, so I guess you must have interpreted the question wrongly. For hints.. remember that tan@ is the gradient of a line, and you need (I think) to use the sum to product identities and if you can't interpret the question...

a mole is the mass of 6.02*10^23 atoms/molecules/formula units so 1 molecule has a mass of 1 mol divided by 6.02*10^23

if the equation 6x^4 + 5x^3 - 24x^2 - 15x +18 = 0 were put in the form 6(x+a)(x-a)(x+c)(x+d) =6x^4 + 6(c+d)x^3 + 6(-a^2+cd)x^2 + 6(-a^2(c+d))x + -6a^2cd = 0 (not hard to expand since you can see which expanded terms will cancel out) you now get; c + d = 5/6 -a^2(c + d) = -15/6 a^2 = 3 a = +/-...

..can't you just find % of Ag in AgCl =~75%. 0.75*1.05 =~0.79. 0.79/1.58 = 0.5, so 50%? edit: nevermind, that was basically what minijumbuk did anyway

It sounds as though you don't understand what a mole is; it is a particular number of elementary units (atoms, molecules, formula units) - approx. 6.02*10^23 of them (the number of C.12 atoms in 12g of C.12) so saying 2mol H2 + 1mol O2 = 2mol H2O is equivalent to saying 2x molecules of H2 +...

1 molecule = 12 + 22 + 11 = 45 atoms, so multiply by 45 (so 1.108*10^20)

there are 12 ways of selecting two balls. there are 2 ways of not selecting red, ie 10 ways of selecting red. 2 of these ways include two red balls. so, probability that they were both red is 2/10 = 1/5 btw: the question doesn't really make sense, since you have to assume that the guy has...

^ basically this. though maths extension can be interesting sometimes too

i) by corresponding angles being equal, ii) ADE is similar to ABC (AAA). AB = AC hence AD = AE. AB = AD + DB = AC = AE + EC; so DB = EC iii) BC is common, DB = BC, angles DBC and ECB are equal, therefore congruent by SAS

you're calculating the mass of an atom; obviously it's going to be a really small number, not a really big one I think you've gotten the m and M confused - as you've written it, m is the molar mass, M is the atom's mass. You've entered them the wrong way around (ie multiplied molar mass by...