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How do I do part ii? Thanks.

Q(IV) I don't exactly understand the solution to this question. Thanks.

A particle is projected with speed V and angle of elevation alpha from point O on the edge of a cliff of height h. When the particle hits the ground its path makes an angle arctan(2tan(alpha)) with the horizontal. Why is tan(beta)=vertical velocity/horizontal velocity? Thanks.

Thanks for the help

https://drive.google.com/open?id=1NRMNrcOf0uy0tUMH4cTgUEj_eegL2jMv Can you please provide working out. Thanks

I already calculated the coordinates of Q from tan(theta/2)tan(phi/2) = 1-e/1+e, but i don't know how to calculate the coordinates of Q from tan(theta/2)tan(phi/2) = 1+e/1-e.

https://drive.google.com/open?id=1tckKJxDql38T785s5GPrq-dPQLgbDpCy How would I do question 9b? In particular, how would I calculate secPhi?

The results you obtain in year 11 doesn't matter, but the content you learn in year 11 carries over to year 12. i.e The maths you learn in prelim is crucial in understanding the content in year 12 and the same is for the science courses etc.

Thanks mate

Why is the answer A? Thanks for your assistance.

Sketch the graph y=(x+1)^4/x^4+1. Use this graph to find the set of values of the real number k for which the equation (x+1)^4=k(x^4+1) has two real distinct roots. I sketched the graph, but I don't understand what I should do after. Thanks for your assistance.

Thanks

How would I find the asymptotes for the graph x^2-4y^2=4? Thanks.

For the graph x^3+y^3=1, why is there an asymptote at y=-x? Thanks.

Thanks for the help.

I did that previously and ended up with the mass of the sun, the mass in Kepler's law refers to the central mass. I'm just unsure of how to calculate the mass of Pluto.

Question 24. Thanks.

To find the stationary points of a curve the gradient must be equal to zero so sub 0 into y', i.e. 4(x-2)^3 =0, then sub each of the x values into the original equation to find the y coordinates.

I think it is 9P3/2!2!

y''=8x y'=4x^2+C m=tantheta m=tan45 m=1 Tangent at (-2,5) 1=4(-2)^2 +C C = -15 y'=4x^2 -15 y=4x^3/3 -15x + C 5=4(-2)^3/3 -15(-2) + C C = -43/3 Therefore y=4x^3/3 -15x -43/3