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I mean, I've seen the configurations written out (not literally seen the atom) where the outer shell had more than 8 electrons

I'm on my 4th or 5th 128 page book, and use a fair bit of scrap paper I've never even gotten to a 2nd book any other year

no idea what their reasoning is, but what I would say (is this appropriate reasoning?) is that: in a hyperbola such as x2/a2 - y2/a2 = 1, then if the distances of the intervals of a line intersecting the hyperbola and the corresponding assymptote at 2 points, is equal, then the y components of...

yeah you have 2n2 max. in a shell, with 8 max in the outer shell (I think I've seen ones with more, but that's what we're taught) the 2, 6, 10 is the max for sub shells (s, p, d) which hold an odd number of electron pairs maybe you were thinking of the 2-8-18-18-8 configuration of Xenon, or...

for b) it's y = vyt - g/2 t2 dy/dt = vy - gt x = vxt dx/dt = vx so for a constant t dy/dx = (vy - gt)/vx so for 2 different t (dy/dx)1 - (dy/dx)2 = -gt1/vx + gt2/vx in the question the initial and final angles of the 2 projectiles are the same time apart, so Ucosα(tanα - -tanα) = -g(t1 -...

the greatest coefficient will occur at the maximum value for which 8Ck>8C(k-1) so 8Ck/8C[k - 1] > 1 (9-k)/k>1 9>2k k<4.5 so the greatest value of k is 4 8!/4!4! = 70

I can't really figure out what the question is asking. I get tanα for the gradient joining the lines too. Is an "inclination to the horizontal distance from P to the foot of the wall" different from an inclination to the horizontal?

for ii) you should know your powers of 2, but to solve it without trial and error just put ln 64/ ln 2 into the calculator, which gives you 6 for i) another way to do it would be to use the fact that log5[1/x] = -log5x which gives x = 5-1/2

probs a typo and because taking the negative of both sides is the same of having the positive of both sides, and taking the negative of only one side is the same as taking the negative of only the other (multiply each side by -1)

^ you forgot the absolute values it would be: 4x-3y+2=3x+4y-7 x-7y+9=0 or 4x - 3y + 2 = -(3x + 4y - 7) 7x + y - 5 = 0

lol I just realised I didn't even solve that question fully. I only solved for a. x = cos[a] = cos[pi/60 + {0,1,2,3,4}2/5 pi] or something

yeah lol. In your solution you were up to the last step.. and then started to go backwards

Re: inequation you don't know if (2-x) is positive or negative. If it was negative you would have to change the sign around. However, (2-x)2 is always positive so this is not a problem. another way to solve this is to use the critical points method - finding where the expression changes from...

well, I've been self teaching for a while (mostly from doing questions on here, reading bits of stuff on the internet) - but not specifically for 4unit

cos5x = Re(cos[x] + isin[x])5 = c5 - 10c3s2 + 5cs4 substituting 1-c2 for s2 and expanding + simplifying gives: = 16c5 - 20c3 + 5c 32x^5 - 40x^3 + 10x - sqrt3 = 0 let x = cos[a] 2cos[5a] = sqrt3 cos[5a] = sqrt3 /2 = +-cos[pi/12 +- 2npi) a = +-pi/60 +-2/5 npi

ii) the total force *sin@ will be the acceleration (v2/rsin@) *mass N = mv2/sin2@ but it is also the vector sum of the accelerating force and the gravitational force (mg) N2 = m2v4/r2sin2 + m2g2 subbing in for N gives N2 = mv2/r *N + m2g2 solving the quadratic, and taking the positive value...

r is the radius of the hemisphere, not the radius from the centre of the circle the particle is travelling in - the radius of the circle will therefore be r sin@ (assuming you made the mistake I did) so tan@ = v2/rsin@g v2 = rg sin@tan@

well, you asked about the sine expansion, and it says your doing 3u you don't need that formula to figure out these questions, just think about it like I was saying - eg. the angle gets flipped either 180' or about the x or y axis, or they're complimentary angles, or whatever else, to figure out...

you can't expand a function that way for sine, the expansion is: sin(A+B) = sinAcosB + sinBcosA so sin(180+x) = sin180cosx + cos180sinx = -sin[x] it's the negative of sin[x] because, by adding 180 you're flipping it - so the opposite side will have the same magnitude but be pointing downwards...

the coefficients are given by nCr 3n-r so: n!/4!(n-4)! 3n-4 = n!/5!(n-5)! 3n-5 simplifying; 3/(n-4) = 1/5 n=19