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Hello All... Could someone please help me with these Q's? 1. If the tangent at P(a,b) on the curve y= 6 / x cuts the y-axis at T(0,3), find the co-ordinates of P. 2. If y = 6x - 11 touches the parabola y = ax^2 + bx + 1 at the point (2,1), find the values of a and b... Thanks for...

Hello All :) Has any one studied the poem 'Before her' by Peter Skrzynecki? My sheet says it is written by him but I'm not sure... If so, could you share your ideas in relations to themes etc... Thanks in advance. :)

Thanks heaps... That's awesome, I understand it... My problem was 'flies on a bearing of theta' - I wasn't using theta as a bearing... (BTW - Your writing is very neat)

Hello All... Can someone please help me with this question? I've tried to do it several times without avail... Q. A plane leaves Adelaide and flies for 875km on a bearing of 056 degrees. It then turns and flies on a bearing of theta for 630km until it is due east of Adelaide. Evaluate...

Yeah ok. Thanks heaps... :) Might try the second method... U willing to see if I answer questions correctly?

Hehe!! :D

Cheers!! How do you become good at Latex? Is the only answer practice??!!

So Tan 2(Theta) = + and - root 3... \therefore $ over the range: $ 2\theta = \pm 60^{\circ}, \pm 120^{\circ}, \pm 240^{\circ}, \pm 300^{\circ} \therefore $ = : $ \theta = \pm 30^{\circ}, \pm 60^{\circ}, \pm 120^{\circ}, \pm 150^{\circ}

Sorry I've just changed the question... It'll make a large difference. Please forgive me.

$Solve for -180^{\circ} \leq \Theta \leq 180^{\circ} tan^2(2{\Theta}) = 3$

Yes, I follow you, Thanks. :) (My error - taking the answer from my lined paper to here!!!)

I think I'm getting the hang of it... just one more please... (After the one I believe has two solutions... ) Thanks all. :)

No worries, it's fine... All is good Drongoski:)

So for a more simple one like $Solve for 0^{\circ} \leq \Theta \leq 360^{\circ} sin({\Theta}) = - Root 3 over 2$ Would there just be two answers... 240 and 360? :)

Ok, understand you... Lucky it isn't a number like 8!! Thanks. :) +1

And \pm \frac {1}{\sqrt 2} = Sin45....

How about this one... $Solve for -180^{\circ} \leq \Theta \leq 180^{\circ} 2sin^2(3{\Theta}) = 1$ Thanks. :)

Yes, I understand what you've said. Thanks heaps... :)

Hello All... Could someone please see if I've gone through the right process for this question. Q. $Solve for 0^{\circ} \leq \Theta \leq 360^{\circ} 2cos(2{\Theta}) = 1$ So cos(2{\Theta}) = 0.5$ Then because we have 2(theta) in the middle, the range is between 0 and 180. So...

2nd Question - Same as EpikHigh's explanation. u=7x^3 -1 u'=21x^2 v= 2x^2 + 6x + 5 v'= 4x + 6 Therefore dy/dx = (2x^2 + 6x + 5).(21x^2) + (7x^3-1).(4x + 6) which is the same as 21x^2 (2x^2 + 6x + 5) + 4x+6(7x^3 -1) Then expand and simplify. Trust I've got it right. :)