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Q1 The Quotient rule is used if the denominator also contains a variable, since it's a constant here it's "ignored" and just left as 15 and 4 respectively. Q2 u is a constant here and differentiating a constant gives 0 \frac{d}{dv} (v-u) = \frac{d}{dv}v -\frac{d}{dv}u = 1-0

was there like a magic show or anything?

Gradient of tangent at t = 3 is \frac{dy}{dx} = \frac{dy}{dt}. \frac{dt}{dx} = \\ \\ 2t.\frac{1}{2} = 3 \ \ \ \ $at t = 3$ Now you can find the gradient of normal

strong thread guys Anyways here's my story. During the yr 12 school photo day, one of the teachers deemed my beard as "too long" (btw I am asian :P) so he forced me to get this 1 blade razor and some cheap shaving cream for me to shave in the bathroom. It was quite painful trying to shave my...

we can only hope things get better :P But yeah I do get how you feel, except in my situation it was just a crush and not gf sighhh.

It's the band 6 text book brah

inbox the exam to me pls

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread doesn't need part i) :P

You'll be fine, at first it may seem intimidating and you'll go through some tough moments, but if you consistently put in the effort, you'll ace it with pride :D.

Re: 4u trials CSSA CSSA math seems to be getting worse lol

I used sin^2A+cos^2A = 1

now to get LHS to RHS LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA + cos^2A}{cos^2A-sin^2A} = \frac{(sinA+cosA)^2}{(cosA+sinA)(cosA-sinA)} = \\ \\ \\ \frac{cosA+sinA}{cosA-sinA} = RHS

Re: HSC 2015 4U Marathon $Show that any root z of$ \ \ \ z^4+z+3 = 0 \ \ \ $satisfies$ \ \ \ |z| >1 \\ \\ $and that any root z of$ \ \ \ 4z^4+z+1 = 0 \Rightarrow \ \ |z| \leq 1

Re: HSC 2015 3U Marathon The property of an arithmetic series is T_n - T_{n-1} = T_{n-1} - T_{n-2} so you solve nC4 - nC3 = nC3 - nC2. The opposite equation nC2 - nC3 = nC3 - nC4 would yield the same answer as you multiplied -1 on both sides P.S: Bit curious as why you're on so early haha

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread No you have to consider the sum of all of the sides, that is what the steel frame is. So the sum of all sides is P = (3x+x)4 +4h = 16x+ 4h \\ \\ $Now we know that$ \ \ 3hx^2 = 4374 \\ \\ \\ \Rightarrow h = \frac{4374}{3x^2} \...

got mine in 7:12 z501......

I hate this so much

I can only hope that Federer keeps up this perfect form, then he would have a great chance for the 8th Wimbledon title

Re: HSC 2015 3U Marathon Since BAC = BDC it's a cyclic quadrilateral since the angles subtended from chord BD are equal DPC = APD = 48 (alternate angles of the parallel lines) APD = ACB = 48 (angles subtended from the same arc) therefore ACB is 48

Hint, make your brackets like ((1+3x) +ax^2)^n then expand the terms normally with the binomial stuff