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for the k>2 part, you should get y1 = (k-2)a When k> 2, y1 is positive. The parabola is symmertrical, for every y value, there are 2 x values hence there will be 2 normals that intersect at that point y1 = a Find x1 Subsititute it into the normal equation Find where it intersects again by...

I find a considerable jump between 3U and 2U. Haven't finished 4U to compare it to 3U

trial and error with factors of 120 Obtain x =2 graph is symmertical about x=-2.5 Therefore the other solution is x=-5

To make simplification easier, divide both sides by n! and multiply both sides by (n-3)! Tnen you'll only have a simple quadratic

Just use the right hand rule Current flowing down. magnetic field lines are north to south

20 is a lot I havent even gone through 1 lol

just use the roots and to find the coefficients

It might help to expand it all using the binomial theorem. You should get [2 x 5C1 (x-1)^2] + [2 x 5C3(x-1)] + 2 x 5C5 Then it would simplify to 10x^2 - 8. Just becareful when you play with the minus signs

From what I remember, type your thing and then on the bottom there is a drop down box. Click the latex option and then copy paste onto your message box. Put it in the format [*tex*] blah blah blah [/*tex*] without the *s

We follow einstein physics...nothing goes faster than c. No time travel :(

No, what I said was unrelated to maths. I did apprciate what you did for me in maths. I've said thx a thousand times

A point moves anticlockwise around the circle x^2 + y^2 = 1 at a uniform speed of 2m/s. Find an expression for the rate of change of it's x coodinate in terms of x when the point is above the x-axis.

Go all the way back to the "random-1005" era. I remember you posted shit. People didn't like your shit. You kinda started it. Now go back in time by propelling yourself to a velocity approx 1.01c. I don't get it. You can go back in time by travelling at a speed over the speed of light, but...

+ 1. I've been reading your posts quite closely

Yea...you have no idea how many accounts this "genious10" guy has. He has made around 15

If only I was drongoski... I would take the 200. and still post stuff LOL

You r so sad. You're saying we need you to pass our hsc. I wonder what drongoski would say though, that's if he does lol

6/2 x (2a +(6-1)d) = -12 14/2 x (2a+(14-1)d) = 196 Simulatenously solve a and d. Then sub it in the n/2 x (2a+(n-1)d) next you go n/2 x (2a+(n-1)d) >250 Then solve for n

If that's so...then why are you still here? Oh! Is this why you're still here? Rofl what a faggot!

Lmao yea