Recent content by rocky1989

hi there was just wondering if anyone knew how many hours of lectures and tutorials you have to do in first year commerce / engineering. thanks

i was thinking of doing something like this....but i didn't get around to it you basically make up a story about you taking a journey with some of your friends. you find your selves stuck in these these weird dark caverns and caves and stuff and then at the end of the story you end up in a...

it doesn't really matter whether you draw the graph or not because the equation y=x^-2 doesn't meet the y-axis....ever (because it is an asymptote) you have either written the question wrong or this is a horrible question

if you think about it....take any 3 of those 5 points that are all on the same line and the shape they make isn't quite a triangle, so you have to deduct the number of combinations that have three points that won't even make triangles

y=(x^2)/4 y`=x/2 sub in x=-8 for the gradient of the tangent M of Tangent=-4 then use (Tangent gradient)*(Normal Gradient)=-1 so m=1/4 then use point gradient formula y-16=0.25(x+8) x-4y+72=0

you don't need to find the value of the roots to do this question...... did you know that?

ok, if you are trying to find the value of these roots then yes you can expand (1) and then use the quadratic formula....but unless you have some real values for this polynomial then its not going to be anything impressive.

well lets say that your 4 roots are: (b) , (1/b) , (h) and (-h) then you can get the four results: (1).........(b)+(1/b)=-p (2).........1-(h^2)=q (3).........-(h^2)((b) +(1/b))=-r (4).........-(h^2)=s from these four results it is quite easy to get the 2 answers.....maybe you had some...

4unit may be fun.......but physics is phun

point taken....but would kfc really care whether the drop-outs they are employing are band 6 or band 2 in history etc... i don't think anyone would ever be discriminated based on their SC results.....so why have them?

let z=x+iy, then --> z^2=4i x^2 - y^2 + 2ixy = 4i then you resolve real and imaginary parts...so x^2 - y^2 = 0 and 2xy = 4 how convenient that these are the two equations that they gave us at the beginning....then you simply say that the roots are where the two graphs intersect.... the two...

well the porabola is always in the form x^2=4ay and for this question: 4a=-2 therefore a=-(1/2)

the porabola is concave down, so the directrix will be above the x-axis

y=-(1/2)x^2 y'=-x sub in x=4 to get the gradient of the tangent at (4,-8) therefore m=-4 then to get the eq of the tangent, use point gradient formula y+8=-4(x-4) y=-4x+8 we also know that the directrix has the equation y=-a where we can find a=-(1/2) therefore directrix: y=(1/2) now just...

do asymptotes have to be straight lines?