If the gradient of PA is equal to the gradient of PB, then P must lie on the line through A AND B. Otherwise the gradients of PA and PB will be different (you can draw up a diagram to visualise this). So in essense, the locus of P is equation of the line through A and B (as P can lie anywhere on...
Find the derivative of the parabola at each poiints and then use the point-gradient formula to find the eqn of tangents
y = 5/2 x^2
dy/dx = 5x
at x = 2, dy/dx = 10
therefore, the eqn of tangent at (2,10) is y - 10 = 10(x-2) [rearrange into the general form]
at x = -4/5, dy/dx = -4...
Re: HSC 2018 MX2 Integration Marathon
an easier way is to just use the substitution u = sqrt(x)
the integrand will be 2/sqrt(u^2+1), which is trivial to integrate (those usual log ones)
Yes, basically if you come first internally, then the assessment mark will be equal to the highest external in that subject from your school (which will be your exam mark if you get the highest external). So yes, if you come first internally, then your classmates wont affect your results at all...
Use the law of conservation of energy
PE (initial) + KE (initial) = PE (final) + KE (final)
-62.3 * 10^6 m + 1/2 mv^2 = -1.3 * 10^6 m + 0 (assuming the craft stops when it reaches point P then accelerates due to the moons gravity)
solving for v, we get 11 kms^-1
ye basically do tons of practice questions where you apply the formula, and you'll automatically remember the formula and will get the hang of applying it without referring to any formula sheet or whatever
Like the above post, understanding what's behind the formula also helps massively
Re: HSC 2018 MX2 Marathon
http://community.boredofstudies.org/14/mathematics-extension-2/356905/question-7-ekmans-compilation.html#post7213476
for the working out of integrands clever solution if anyones interested
Re: HSC 2018 MX2 Marathon
Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)