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oh yea..oopz...my mistake..keke forgot alreadi how to use the terms.. =( im gonna fail.. can we just say that tho..cos we know that alreadi..?

hey.. i have a feeling that our school is heaps behind than everyone else... wat topics r pplz up to in ext I and II?? for us we've done like: complex no. conics and polynomials

When do we learn this telescoping series thing? and wat topic is it under?

keke haven't done these type of max. / min. questions since yr 11 phew..i still remember some of it

let y = x ^ n y' = nx^(n - 1) = 0 if x = 0 :. turning pt at x = 0 at x = 1 y' = n ie. y'>0 :. to have minimum pt, at x = -1, y'<0 at x = -1, y' = n(-1)^(n-1) :. for y'<0, n-1= odd :. n= odd +1 which is even. if n<0 then y'>0 :. x^n has min. turning pt if n is...

ohhh i fink i get it now its outta the syllabus..from 3 u anyway cos u use affinity's method

this is from fitzp. 3u book mm..that answer says its 2n(pi) or (4/5)n +_ (pi)/5

help..i ve been asked.. whats the diff between back emf and eddy currents and are they both produced in motors?

wahhh that way is sooo much betta.. even tho its still complicated but its a good approach!! its clever how u used sum of roots to get the midpt..its like one of those tricks and short cuts...i didnt get it at first wen i learnt it @ tutoring...

when M is midpoint of PQ does that make angle PMT and QMT = 90 degrees?

the way i wud do it is like this...but that seems such an algebra bash... find midpt of PQ find eqn of OM use chord of contact to find out coordinates of T sub T into eqn to see if it fits.. this way..u have all the weird algebra or trig..so i dunno if it works cos i didn complete doing...

mmm... thats not the way i did it..but i dunno if i did correctly.. let angle PNT = @ PT/PN = tan @ @ = tan^-1 (a tan@/b) since mPN = a tan@/b :. PT/PN = a tan@/b :. (PT/PN)^2 = a^2 tan^2@/ a^2 (1 - e^2) = tan^2@/(1 - e^2) ... thats y i asked if for tan(...

q4 was just using the m1 * m2 = -1... answer : - b^2/a^2 for q5 a) was normal subbing into tangent b) i) i used the similar triangle thing so PX/XY became PC/OY where C a point where the vertical line from P cuts the X axis..so the distances became easy to calculate as its just...

for 3c instead of using PS/PS' i used PM/PM' which is the same..since PS = e.PM and it shud b easier cos its a horizontal line..

hey for q2.. does tan ( tan ^-1 ((a/b) tan @)) = (a/b) tan @ like can u just cancel out the tan..?

argh..ill do this tmw or some other time.. eyes getting tired.. oh btw for q3 how do u represent a variable X on a triangle? do u give it co ordinates? or can u prove that they r similar some other way?

the way that im approaching q2 so far: PNT is a triangle i made a line thru P that cuts X axis perpendicularly at C then i found NC^2 + PC^2 = PN^2 and similarly CT^2 + PC^2 = PT^2 and then i'm solving PT^2/PN^2 but i get this huge weird algebra.. CM am i on the rite track?

keke q1 was from cambridge.. for q2..is there a shortcut or do u have to find PT and PN and expand out?

for 1 (b) i got up to the part where u used the assumption to make LHS = sin a/ 2^k sin (a/2^k) * cos (a/2^k+1) gimme a hint of wat to do next...like wat does cos (a/2^k+1) equal?

(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2) Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4) cos(a/2)=sina/2sin(a/2) RHS = 2sin(a/2)cos(a/2)/2sin(a/2) = cos(a/2) = LHS ............(...