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hey CM its cool to have all ur questions to think about.. just wondering...cos i have assessments coming up on: locus problems of complex no. conics polynomials.. wondering if it wud be too much hassle if u can start a new thread and give questions on those topics =) if not then...

so 5/c^2, 5/b^2, 5/a^2 are the roots of the new eqn rite.. so u just let y = 5/x^2 and sub bak into the original eqn expand and simplify and u ll get the answer =)

ahhh just had dinner....which was great.. anyway another way to do q1 on page on is: P(x) = ax^4 + bx^3 + cx + d P'(x) = 4ax^3 + 3bx^2 + c P"(x) = 6x(2ax + b) :. x = -b/2a P(-b/2a) = a(-b/2a)^4 + b(-b/2a) + c(-b/2a) + d = - b^4 - 8a^2bc + 16a^3d sum of roots: -3b/2a + @...

kkz.. working on it..

how do u do the second part of the question of the first page question?

I need help wif the following conics questions: 1) The point P (a sec#, b tan#) lies one the hyperbola (x^2/a^2) - (y^2/b^2) = 1. Let d1 and d2 be the distances from P to each of the hyperbola's asymptotes. Show d1.d2 = (a^2.b^2)/(a^2 + b^2) 2) The points A (a, 0), A' (-a, 0) and P...

Thanks JOn i got another one which is hard..and i cant do as well... from the same exercise but the next 2 questions Q26. If tan X, tan Y, tan Z are the roots of the eqn x^3 - (a + 1)x^2 + (c - a)x - c = 0 show that X + Y + Z = npi + pi/4 Q27. Expand cos (2A + B) and hence prove...

4cos2@cos@ - 2cos@ + 1 = 0 4*(1/2)(cos3@ + cos@) - 2cos@ + 1 = 0 2cos3@ + 2cos@ - 2cos@ + 1 = 0 what happened in these 3 lines?

I need help...for polynomials from Fitzpatrick Ex 27 d Q 25 Show that the cubic equation 8x^3 - 6x +1 =0 can be reduced to the form cos 3theta = -1/2 by substituting x = cos theta. Deduce that: a) cos 2pi/9 + cos 4pi/9 = cos pi/9 b) sec 2pi/9 + sec 4pi/9 = 6 + sec pi/9 c)...

i dunno if your way to do it is right clerisy cos does 49C2 just mean the remaining cards..so wat about the aces.. and also at least 3 aces..so i think that means u can have 4 aces too.... but then i keep on trying to do it...but i dun get it right! -_-' Help.....

Um...well from the book the answer is.. 19/10829 so yea?? help? ><

Hey thanks for helping me out got into another question..from same book ex 28c q.6 Five cards are drawn from a pack of 52 playing cards. What is the probability of drawing at least 3 aces? how wud u approach these kinda of question?

Hey i dunno how to do this question from 3u Fitzpatrick, its a permutation question.. Q.29 In how many ways can 4 people be accommodated if there are 4 rooms available? i dun understand permutations at all :( thanks