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i get (9πa2)/4 not sure if im reading the question right though :/

oh good point :S would have made things interesting

agreed, first one doesnt need induction. second one let z1.z2......zk = x + iy and zk+1= a +bi to solve

using question 5 from page2.jpg will allow you to complete question 5 from page 1 and question 6 from page 2 so let I = Integral{a->0} f(x) dx let u = a - x -du = dx so I = Integral{0->a} f(a - u) (-du) = Integral{a->0} f(a- u) du now the variable does not matter so replacing u...

red pen for axis, pencil for everything else

for product of roots (a-b)2(a+b)2= 25(a-b)2 =25[(a+b)2-4ab] = 25[25-4.7] = -75

using the circumference-centre angle property of a circle it can be shown the circle is x2 + (y-2)2 = 8 (in the diagram W = π/4)

Question 2 z = (3 + iq)2/(3-iq)(3+iq) = (9 + 6qi - q2)/( 9 + q2) z = (9-q2)/(9+q2) + 6qi/(9+q2) let z = x + iy x = (9-q2)/(9+q2) y = 6qi/(9+q2) the question says it should be a circle so square both sides x2 = (9-q2)2/(9+q2)2 x2 = ((9+q2)2 - 36q2)/(9+q2)2 x2 = 1 -...

i think people are getting confused between the number of 'unique' arrangements that questions involving words usually look for (ie. most past hsc questions) and probability ie number of combinations over total arrangements

he realised he was wrong and changed it............do u actually have the solution or just the answer?

you missed one :) ABAANNS i would suggest from krabby's working that the answer is 4/7 10x2 placements of A (for mirroring) 3! ways of arranging A 4! ways of arranging the other letters 2x10x3!x4! = 2880 total combinations = 7! = 5040 P(E) = 2880/5040 = 4/7

it should be x= -50tcosα + 110 as it is traveling in the opposite direction to the first particle

multiply by Cisθ where θ is the angle in radians hence rotation by Π/2 anti clockwise is the multiplication by i as Cis(Π/2) = i edit: Trev was too quick :P

how can you times both sides by two different things?

like i said there are other ways of doing it a more verbose way of saying it would be let α be the intercept of the two curves thus α is a solution to mx^3 - 3x^2 + 1 = 0 but at α the gradient of each line will be equal thus dy/dx of each curve is equal at x =...

Tcos@ = mg Tsin@ = mrw^2 but from the basic diagram r = l sin@ therefore Tcos@ = mg T = mlw^2 divide the 2 cos@ = g/(lw^2) lcos@ = g/w^2 but again from the basic diagram h = l cos@ thus h = g/w^2 //(A) now remember w = 2fπ where f is the frequency thus for 75...

i get (7g)/(100pi) where g is gravity

since its a tangent we know that at the point of intersection that the x,y values are equal thus mx = 3 - 1/x^2 mx^3 - 3x^2 + 1 = 0 (A) and the gradients are equal, thus the solution of (A) will be shared with its derivitive hence 3mx^2 - 6x = 0 3x(mx-2) = 0 hence mx - 2 = 0 is...

1st in state with 36? thats unexpectedly low. i got 35/40 which was still a HD (only just i would suspect)

haha, I'm doing this exercise now, pretty tedious I must say.