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Thank you so much for your reply.

Thank you so much for your help, Blamocur, The Highlander, and Dr. Peterson. So, if you bought 100 tickets, for example, the answer for a) would be 100/200 = 1/2, and the answer for b) would be (100/200) X (99/199) = 99/398 Correct???

I have a problem with the following question: A small lottery consists of 200 tickets. There are two prizes. The first prize ticket is drawn and then, without replacement, the second prize ticket is drawn. If you buy two tickets, what are your chances of winning a) first prize b) both prizes...

I have finally worked out the second part of the qustion as follows: [(z +1)/(z -1)]5 = 1 = cis [0 + n2π] where n = 0, 1, 2, 3.... So, (z + 1)/(z -1) = cis [(1/5)*(0 + n2π)] = cis (n2π/5) So, z + 1 = (z - 1)*cis (n2π/5) So, z - z*cis (n2π/5) = - 1 - cis (n2π/5) So, z [1 - cis(n2π/5)] = - 1 -...

I have made some progress as follows, however, what I have ended up with is slightly different to icot(nπ/5) - the only difference is that I've got " - " sign in front of icot(nπ/5) - even though I have checked my working below many times and there seems to be no error : [(z +1)/(z -1)]5 = 1 =...

Do you mean that "n" can take 0? Is that the reason why you think that there are 5 roots, not 4? But the qustion that I have here clearly says that "n" takes only 1, 2, 3, 4. (and not 0) I do not think that "n" can take 0 because, as you can see from my incomplete working above, z= [cis (n2π/5)...

With regard to the second part of the question, the following is all that I can do: [(z +1)/(z -1)]5 = 1 = cis [0 + n2π] where n = 0, 1, 2, 3.... So, (z + 1)/(z -1) = cis [(1/5)*(0 + n2π)] = cis (n2π/5) So, z + 1 = (z - 1)*cis (n2π/5) So, z - z*cis (n2π/5) = - 1 - cis (n2π/5) So, z [1 -...

Thank you very much for your help, Dr. Peterson. With your help, I was able to finally work it out as follows: 2cos[(x-y)/2]cis[(x+y)/2] = 2cos[(x-y)/2]*[cos[(x+y)/2] + isin[(x+y)/2]] = 2cos[(x+y)/2]*cos[(x-y)/2] + i2sin[(x+y)/2]]*cos[(x-y)/2] = cos[(x+y)/2 + (x-y)/2] + cos[(x+y)/2 - (x-y)/2]...

I have expanded the right-hand side of the equation like this, but it seems that I am not getting anywhere... 2cos[(x-y)/2]cis[(x+y)/2] = 2cos[(x-y)/2]*[cos[(x+y)/2] + isin[(x+y)/2]] = 2cos[(x+y)/2]*cos[(x-y)/2] + 2isin[(x+y)/2]]*cos[(x-y)/2] = 2cos(x/2 + y/2)*cos(x/2 - y/2) + 2isin(x/2 +...

I have ERRATA for the book with this question in it, which does not say that it is an error. So it is not a typo.

That is only the first part of the following question: "Prove that cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2]. Hence show that [(z+1)/(z-1)]5 = 1 has solutions of the form z = i cot [n(pi)/5] for n = 1, 2, 3, and 4. Honestly, this is an extremely hard qustion for me and I cannot even...

This is a very odd unrealistic question as you say. Yes, I agree with you. I wonder why anyone would use this kind of backward method using a stick to measure the height of the hill when they are acually using laser to accurately measure the distance from the stick to the top of the hill? It...

The name of this textbook that you should not use! The name of the textbook is: New Zealand Mathematics second edition 10 Published by Haese & Harris Publications Ltd (Auckland New Zealand) This is a second edition published long time ago back in 2002, so the current one might be better with...

Thank you for your help. Your method should be perfectly appropriate for this question because it says "How high, correct to the nearest metre, is their ESTIMATE of the height of the hill?" So we are not expected to give a precise answer but it has to be an estimate. Having said that, our...

Thank you for your deep insight. Oh, I did not notice that the answer would be, as you say, 720 meters, if the distance of 1500 m referred to the horizontal distance from the eye to the hill. However, it is so clear from the question that this distance is not the horizontal one because it...

Thank you, but not entirely satisfied... Thank you for letting me know that your answer is also the same as mine. But I must let you know that this is a question designed for Year 10 students in NEW ZEALAND! It is not for US or UK or even Australia where they learn things faster and do more...

I have a problem with the following question: Two surveyors estimate the height of a nearby hill. One stands 5 m away from the other on horizontal ground holding a 3 m stick vertically. The other surveryor finds a "line of sight" to the top of the hill, and observes this line passes the...

Thank you. Thank you for your advice. Finally I worked it out but I have ended up with more than 2 pages of long workings.

Still hard and I am stuck again. Please open the attachment to see my working. I am stuck again and I do not think that I am getting any closer to the RHS.

Resubmission of the question. What I put did not make sense because inside each bracket on LHS is not multiplied by 2 but what I wanted mean is like (cos (...))^2. This equation is very difficult to read if you try to express it with ^ , brackets and slash, and that is the reason why I asked...