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These are questions 23 and 24 of page 114 in fitzpatrick I really dont know where to start: 23) Assume that the tides rise and fall in SHM. A ship needs 10 metres of water to pass down a channel safely. At low tide the chanel is 9m deep and at high tide 12m deep. Low tide is at 9am and high...

Ive just started yr 11 in a school thats ranked in the high 100's(didnt make the top 200 last yr though) and in the most unconvenient of times have been given an offer to go to fort street high school within this week; i am so split between wether to go or not. Fort st is about 1 hr 15 mins from...

WIth 1) it just says its in the shape of a cone (apex facing upwards), but with the way u did it it seems as though the cone was inverted, i.e standing on its apex; is that how u did it?

with 1) i understood it as inverted, did u? 2) isnt an altitude from an angle to a side? 3) im sure its the books mistake

1) A conical tank whose axis is vertical has a semi-vertical angle of 45 degree. water initially at a depth of 5 metres leaks out through a hole at the bottom at the rate of 0.2sqrth metres^3/min when the depth is h metres. find the rate at which the depth is decreasing when the depth is 4...

ok, does that mean that as long as we get the same value, the angle giving the terminal ray can be anything?

For the general solution such as that in tan (@ - pi/3) = -sqrt3 does the angle we use (giving the terminal ray) in the right hand side matter (knowing that the tan of that angle = -sqrt three) for e.g tan (@ - pi/3) = -sqrt3 tan (@ - pi/3) = tan 2pi/3 @ = npi + pi @ = pi(n + 1) OR tan...

1) Does the cambridge three unit yr 12 contain the entire extension 1 course? 2) What is the first topic that should be completed in the HSC curse? Thanks in advance

LOL, ur not following the rules, you gotta answer my Q first; ill give u a hint, if you produce the tangent at P and A to meet at a point Z, angle PZA = angle AZP.

Draw PR, PA and PM in and let centre be O since OB is perp. to PQ, PV = VQ (perpendicular from centre to chord bisects chord) now, angle AVP = angle AVQ = 90 degrees. PV = VQ and MV is common to triangles MVP and MVQ. Therefore triangle MVP is congruent to triangle MVQ (SAS) . therefore MQ =...

thanks for the help guys; would they ask for a proof from a definition in an exam?

sorry to ask for help yet again....... this stuff just doesnt click with me from the DEFINITION of nPr and nCr proove: i) n+1^P_r = nPr + r. (n^P_r-1) ii) nCr = ((n - r + 1)/r).(n^C_r-1) thanks in advance

ohh crap..... i guess i thought too much thanks for pointing that out for me :)

isnt there a way of reasoning why? like for at least 1 object you can reason that every object can be dealt with in two (it can be included or rejected) ways therefore you can deal with all the objects taking any at a time 2^n ways; but considering rejecting all objects as an unwanted...

well, i absolutely hate em, stuck on another q, In how many things can n things be shared between two people? im gettin 2^n - 1 - n but its 2^n - 2 thanks in advance

Thanks heaps guys :)

1) In how many ways can 4 people be accomodated if there are 4 rooms available answer: 256 2) In how many ways can 8 oarsmen be seated in an eight oared boat if 3 can row only on the stroke side and 3 can row only on the bow side? answer: 1152 thanks in advance

cool, the part i lost myself was not using the u substitution; ending up with a quartic equation thanks heaps! btw its pretty retarded how u can get 4 roots from one (i.e b)

LOL i already got that, this is wat ive done: ab = cd a + b + c +d = 11/6 abc + abd + acd + bcd = -11/3 abcd = 4 ab = cd = -2 ( It cant = +2 as this wont satisfy a + b + c +d = 11/6) (a + b)(c + d) = - 1/3 I cannot go further, solutions are 4/3, -3/2, 1+ sqrt3, 1- sqrt3

....... the product of the roots are equal not the sum