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yeah realised that just after i posted it. hence the deletion

no projectiles with angles of @ and 90-@ will have the same range.

do you not also have to take the 5th root of the modulus? EDIT: sorry ignore me. i didnt read the post carefully enough

i'm going to major in pure maths at sydney anyone else going to sydney? you can do maths though the adv science or just the normal Bsc course and even transfer to adv maths later is you want.

you dont need the substitution its a very basic 3u concept that cos x is the derivative of sin x. it is only a matter of applying the function of a function rule in reverse.

there is also a really cool pascals triangle like pattern that developes in three dimensional for for a tri-nomial, 4 dimensional for quad-nomial........ and n-dimensional for an n-nomial. you can then apply the formula you derived and find a number at any point in the n-dimensional pyramid...

1) assume true for n = k consider n = k + 1 we must prove: __ __ _____ ________________ z1 + z2 + ... + z(k+1) = z1 + z2 +... + z(k+1) __ __ __ ______________ (nB: z1 + z2 + ... + zk = z1 + z2 + ... + z k)...

you have to be careful though because the areas under the x axis are negative. if you just add the areas you'll get 5 instead of 3

because the integral of x is x^2/2 you can ignore the absolute value signs as the out come will be positive anyway therefore int ( 2 - abs(x) ) from -3 to 3 2 int ( 2 - x ) from 0 to 3 (even function) = 2[2x - x^2/2 ] from 0 to 3 = 2[(6 - 9/2) - (0)] = 2[3/2] = 3...

it think jago ment he checked your answer by using gaphamatica himeself

The probability that the prize will never be won limits to zero therefore the probability that it will be won eventually after its 25th time will be equal to the probability that it will not be won on the first 25 times. Your approch with binomial probability is wrong even if the question...

yes but equal angles at the circumfrenced also subtend equal chords. this is important it will come up a lot

4a) let CBA = x therefore ADC = 180 - x (cyclic quad, supp opp angles) ADC is isosolese therefore DAC = ACD = x/2 DBC = DAC = x/2 (angle subyended byh the same seg.) BCD is isosolese therefore DBC = BDC = x/2 therefore triangles BCD and ADC are congruent (AAS)...

sub in r(cos@ + isin@) and then expand and simplify you will have to use the cosine and sine addition of angle laws.

i wont actually do 2 and 3 i dont have time but for 2: trynd draw a fairy accurate diagram so that you can see wich side of the centre cirtain chords lie and which angles are acute/obtuse etc use isosoles triangles formed by two radii and a chod and angle sums of trangles and also the fact...

first one: BEA = ADC (subtented by equal chords AB and AC respectively) BEA = EAD (alternate angles on parallel lines BE and AD) therefore: ADC = EAD (both equal BEA) ADC and EAD are therefore equal and alternate angles of the lines AE and CD there fore Ae is parallel to CD

for 2 ii: by drawring the forces in for particle A we can see that: T = mlw^2 (7) where T is the tension force which is the same the whole way through the string and therefore is the same for particle B. l is the radius. from (4): T(x) = MrQ^2 from the diagram we...

for 2 i: by drawring the forces accociated with particle B we can see that: T(y) + N = Mg (1) where T(y) is the vertical component of the Tension force T. we can also see that T(y) = T cos@ (2) T(x) = T sin@ (3) and that: T(x) = MrQ^2 (4)...

at the point P: sin @ = cos^2 @ where @ is the x coordinate at P y = tan x y' = sec^2 x at point P: sec^2 @ = sin^-1 @ (cos^2 @ = sec^2 @) therefore m1 = sin^-1 @ y = cos @ y' = -sin @ therefore m2 = -sin @ m1 x m2 = sin^-1 @ x -sin @ = -1...

im pretty sure it was 4/7 or 7/4 or something