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jnney bby how u -ayscebby

When I walked onto campus early this week and found those trees cut down, it took a part of my soul away man. They've been with me through good and bad times. Now what trees shall I weep on during finals week? (I am well aware there are others, but those were irreplaceable) Only time will...

So I'm not the only one who felt sad about the eastern ave trees that got cut down :D yay! I especially miss the tree next to the bubbler in front of the chemistry building labs. eh, they'll be back.......when i've moved on from uni lyf... :cry:

I wish there were more courses like ATHK1001. Thanks for the recommendations - HPSC1000 LOOKS AMAZING!

Hey guys, it's Aysce (on my brother's account). For those of you who have done ATHK1001 (Analytical thinking), could you please offer some opinions? I'm planning to take it but I also need to do another elective that is useful to me and will also enable me to maximise my WAM. WRIT1001 is...

Hey guys :D I'm doing MX1, MX2, physics, chemistry, economics and english advanced. Despite what I say in my sig, my atar aim is 98+ (One year left to go, LETS DO THIS)

D: y = -3/4 Sub y = -3/4 into the equation of the line PF. 3x + 4(-3/4) - 3 = 0 3x - 6 = 0 x = 2 Therefore R(2,-3/4)

Got it, thanks!

a = 3/4 units F(0, 3/4) m = (3-3/4) / (-3-0) m = -3/4 Sub into the point gradient formula y - 3 = -3/4 . (x+3) 4y-12 = -3x - 9 3x + 4y - 3 = 0

Find the exact length of the tangent from (4,-5) to the circle x^2 + 4x + y^2 - 2y - 11 = 0 Thanks :)

Yeah, it didn't give me a particular form. I chose my own one, but only one specific form worked. I must've done something wrong so I'll go back and redo the working out.

Hey guys so I'm a bit confused on something. So here's two separate questions: $Solve for 0^{\circ} \leq x \leq 360^{\circ} $ \\ 4sin\theta - cos\theta =-3\\ and\\ sin\theta -cos\theta =1\\ So i've tried using the transformation method, and I've expressed the first question in the form...

Thank you very much!

Hey guys, I'm confused for both questions a) and b) Thanks in advance :)

1) dy/dx = -6/x^2 (Gradient of the tangent at any particular x-value) Finding the equation of the tangent: y-3=(-6/x^2)(x-0) Since we need to find the point of intersection, sub in y=6/x 6/x - 3 = -6/x 12/x = 3 12=3x x=4 (This is your 'a') sub into y=6/x y=6/4 = 3/2 Therefore P(3, 3/2) 2)...

Ohhh I did this method and I see where I went wrong now, thanks so much!

Hey guys, Q: Find the coordinates of the centre of the circle that passes through points (7,2), (2,3) and (-4,-1). Thank you :)

i) 104 degrees. This is because the diagonals of a rhombus bisect the angles and therefore 52 + 52 = 104. ii) 76 degrees. Opposite angles in a rhombus are equal. 180 - 104 = 76 degrees.

Here you go, if you don't understand some of my working out tell me.

y=(3x+5)^6 . (x)^\frac{1}{2} . All you do is use the product rule and when you differentiate for u' and v', you must use the product rule as well. Try it for yourself first, but if you need the working out tell me.