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the answer to the graphs question is A i assume you know why it can't be C or D with B, the graph suggests that 1/f(x) approaches 1 as x approaches infinity. if we look at the original graph, in order for this to be true, f(x) would also need to have a horizontal asymptote at 1. however, this...

2. In C1, angle CAB = angle CPB (angles standing on the same arc are equal; arc BC) In C2, angle DAB = angle DQB (angles standing on the same arc are equal; arc BD) Since angle CAB = angle DAB, .'. angle CPB = angle DQB If we let angle CPB and angle DQB be x, In triangle PBQ, angle PBQ =...

1. a) assuming the table is round, if you group the 2 particular girls together as 1, then you'll have 7 "people" (4 boys 2 girls and the group of 2 particular girls) since it is a round table, the number of arrangements possible will be (7-1)!*2, since the girls in the group can switch...

melting point is the point when a solid turns into a liquid. the freezing point is the point when a liquid turns into a solid. the freezing point of a substance will be the same as its melting point, but the definitions of the two terms are different. in this case, if x's melting point is -119...

Re: HSC 2012 Chemistry Marathon In reaction 1, HCO3- acts as a bronsted-lowry base; it accepts a proton from the hydronium ion on the RHS to form H2CO3

Re: 2012 HSC MX2 Marathon could it also be done this way: if Q was the complex number z*conj(w), and OP was |Re(z*conj(w))|, angle OPQ would be a right angle, and therefore OQ > OP because OQ is the hypotenuse of triangle OPQ. since OQ is also the modulus of z*conj(w), and |conj(w)| =...

change the 1 on the top to tan(pi/4) and make the tan(x/2) at the bottom multiplied by tan(pi/4) then you have the expression for tan(x/2 + pi/4)

Re: 2012 HSC MX2 Marathon the complex number z and w in mod-arg form are 2cis(pi/3) and root2 cis(pi/4) respectively .'. zw = 2root2 cis(pi/3 + pi/4) =2root2cis(7pi/12) = 2root2*cos(7pi/12) + i2root2sin(7pi/12) also, zw = (1+iroot3)(1+i) = 1 - root3 + i(root3 + 1) equating real parts...

Re: 2012 HSC MX2 Marathon mine ended up being a sideways parabola, so i had one that was "concave right"

Re: 2012 HSC MX2 Marathon (i) angle NPC = angle CBP (alternate segment theorem) angle PNC = 90 (given) .'. angle NCP = 180 - 90 - angle NPC (angle sum of a triangle is 180) = 90 - angle NPC angle ACB = 90 (angle subtended at the circumference from a diameter is a right angle) since angle...

Re: 2012 HSC MX2 Marathon b) (i)gradient PQ = (p+q)/2 .'. chord PQ has equation y - ap^2 = [(p+q)/2][x-2ap] if PQ passes through (2a, 0), then x=2a y = 0 must satisfy the equation of the chord .'. 0 - ap^2 = [(p+q)/2][2a-2ap] -2ap^2 = 2ap + 2aq - 2ap^2 - 2apq divide everything through by...

LHS = (cot(t) + cosec(t))^2 = cot^2(t) + 2cot(t)cosec(t) + cosec^2(t) using the identity 1 + cot^2(t) = cosec^2(t), =cot^2(t) + 2cot(t)cosec(t) + 1 + cot^2(t) = 1 + 2cot^2(t) + 2cot(t)cosec(t) = 1 + [2cos^2(t)/sin^2(t)] + {2cos(t)/sin(t)}*{1/sin(t)} since cot(t) = cos(t)/sin(t) and cosec(t)...

asin^2(theta) + bcos^2(theta) = c a(1 - cos^2(theta)) + bcos^2(theta) = c a - acos^2(theta) + bcos^2(theta) = c [b-a]cos^2(theta) = c - a cos^2(theta) = (c-a)/(b-a) .'. cos(theta) = plus minus sqrt{(c-a)/(b-a)} do the same with with the identity sin^2(theta) + cos^2(theta) = 1 to get...

i get it now, thanks guys! :D

so does that mean vectors PQ and QP are different?

i have a couple of questions about nightweaver's answer [our school just started complex numbers], 1. what does the arrow above |QP| and |OR| mean? 2. if there any other way to do this question? i would have done it this way too but is there an alternative to this?

because the tangents are perpendicular, then the gradients of the tangents must be equal to -1 since the gradients of the tangents to the curve at P and Q are p and q respectively, therefore pq=-1

if x^2 + 2x + k = 0 has real and distinct roots with k>0, then the discriminant must be positive .'. 4- 4k > 0 4k < 4 k < 1, but k is also >0 .'. 0< k< 1 looking at the next polynomial, the discriminant of x^2 + 2kx + 1 = 0 is 4k^2 - 4 since k is between 0 and 1, when you square k, it...

the LHS isn't supposed to equal the RHS, when you sub PQ into the ellipse, you'll get: x^2 + 3*[(1/root3)*(x-2a)]^2 = 2a^2 x^2 + 3*[1/3*(x-2a)^2] = 2a^2 x^2 + (x-2a)^2 = 2a^2 x^2 + x^2 - 4ax + 4a^2 = 2a^2 2x^2 - 4ax + 2a^2 = 0 2(x^2 - 2ax + a^2) = 0 2(x-a)^2 = 0 since this eqn. has only...

Re: 2012 HSC MX1 Marathon 1. RHS = 1/cosx - sinx/cosx = [1-sinx]/cosx multiply top and bottom by 1+sinx, =[1-sin^2x]/[cosx(1+sinx)] =cos^2x/[cosx(1+sinx)] =cosx/[1+sinx] =LHS 2. LHS = cosx - cos^3x {sin(90-x) = cosx} =cosx(1-cos^2x) =cosx*sin^2x =RHS