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to find point of contact solve the two equations of the parabola and the equation of the tangent simultaneously

C, because the first time it's spun doesn't matter what colour it points to, the second time it has a 1/4 prob of same colour P(E) = 1 * 1/4 = 1/4

How is this soln to 16a |z0+z1| + |z1| >, |z0| |z0+z1+z2| + |z2| >, |z0 +z1| >, |z0|-|z1| therefore |z0+z1+z2| >, |z0|-|z1|-|z2| then , |z0 +z1 +z2...Zn| + |Zn| >, |z0|-|z1|-|z2|...|Zn-1| therefore true Q15 was quite easy except induction someone post soln up for that plz

For Q 17 in three extra students go C=2(n+3) + 5 as n is the number of students already going C=2n+11 so the chose increases by 11-5 = 6 =A

cos we're secretly twin brothers.

do what you did for the trials

you really love those sigma signs don't you?

don't you sub in y=mx+b into the equation of the hyperbola? then discriminant = 0

why didn't you make on for 2u as well?

Re: 2U Maths marathon new q: prove that d/ dx(Ln(x). sqrt(cosx)) = (1-sin^2(x)/ cosx^3/2 ) - [x Ln(x) tanx sqrt(cosx)] / 2

sorry it should be < typo there, so the probability that it won't seed is 2/3, and that is 1-0.9 = 0.1 say theres n number of flowers for it to exceed 0.9 let (2/3)^n < 0.1 n Ln(2/3) < Ln(0.1) so n > Ln(0.1) / Ln(2/3)

Not possible to integrate Ln(y) at 2u level so you must use area of rectangle - area bounded by the x axis Since e^x cuts the y axis at 1, and at y = e , x = 1 Area of triangle = e x 1 = e units^2 Area of unshaded region = integrate from 0 to 1 [e^x] Area of shaded region equals...

yeap answer is 6 (2/3)^n > 0.1 logging both sides gives n >5.6 therefore n =6

it should be accepted, and your choice on whether you find the second derivative or table values should depend on the function

i've got independent chem clean version but physics written all over

will do:)

this should be near 98 without bonus points GL

shut your mouth dawg don't talk shit! ye true that though

School rank 152* looking at a solid 50+

not sure if dumb or dumb