Yes,, that is correct.

Sounds good. Why have you not done it?2. Find the equation of the tangent to the curve y = xFirst, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.^{3}- 9x^{2}+ 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?

No, it's not. Putting h= 3 into h(t) gives h(3)= 82 meters- not "meters per second", because, as you were told, h(t) is the3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8tI simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms^{2}- 80t + 250. Find the rate at which the plane is losing height at t = 3.

^{-1}. Is this question as simple as that?heightof the plane at t seconds, NOT the rate at which that height is changing.

Yes, that is correct. Why didn't you do that for the previous question?4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)tI know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t^{3}+ (1/4)t^{2}. What is the acceleration when t = 2?

^{2}+ (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms^{-2}.

Thanks in advance. Your feedback is greatly appreciated.