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Re: HSC 2015 3U Marathon That's true. How about this then: \frac{k+2}{k+1} = \frac{k+1}{k+1} + \frac{1}{k+1} $ Since $ k+1 \geq 2 \frac{1}{k+1} \leq \frac{1}{2} \frac{1}{k+1} + \frac{k+1}{k+1} = \frac{k+2}{k+1} \geq 1.5 \left(\frac{k+2}{k+1}\right)^{k+1} \geq 2.25

Re: HSC 2015 3U Marathon \underline{$For n=1 and n=2$} $LHS$ = 1! = 1 $RHS$ = \left(\frac{1+1}{2}\right)^{2} = 1 = LHS $Equality for n = 1$ $LHS$ = 2! = 2 $RHS$ = \left(\frac{2+1}{2}\right)^{2} = 2.25 > $LHS$ \therefore $equality for n=1 and inequality for n=2$...

Re: MX2 2015 Integration Marathon I_{n} = \int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2} - x)(\cos x+\sin x)^ndx 2I_{n} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} dx I_{n} = \frac{\pi}{4} \sqrt{2^n} \int_{0}^{\frac{\pi}{2}} (sin(x + \frac{\pi}{4})^n dx $Consider$ \int_{0}^{\frac{\pi}{2}} (\sin...

Re: MX2 2015 Integration Marathon I_{n} = \int_{0}^{\frac{\pi}{2}}x(\cos x+\sin x)^ndx I_{n} = \int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2} - x)(\cos x+\sin x)^ndx 2I_{n} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} dx I_{n} = \frac{\pi}{4} \left[x \right]_{0}^{\frac{\pi}{2}} =\frac{\pi ^2...

You can rewrite the addition/subtraction of either sine/cosine as one trigonometric function. 6 \cos 2t + 8 \sin 2t \equiv R \cos (2t - \alpha) \equiv R \cos2t \cos \alpha + R \sin 2t \sin \alpha $ Equating, $ 6 = R \cos \alpha 8 = R \sin \alpha $ Squaring and adding, $ 8^2 +...

Yes you're correct. Note that $KE$ = $hf$ - \phi $where$ \phi $ is the work function$ $let KE$ = 0, \phi = $hf$ $where h is just Planck's constant and f is the x-intercept (so 9x10^14 Hz for metal Y) nevermind already done above

Or alternatively collect the fraction together = (2x^4 - 1)^12 / x^12 and use the general term to find x^12 at the top as x^12 / x^12 gives the constant term

Meetup at parra at 3

Is this the one about adding NaOH in small amounts? I wrote that the acetic acid and sodium acetate formed a buffer solution whi ch resists changes in pH. Then I said something like adding small amounts of NaOH will cause the pH to increase slightly, but by LCP, it will use up H+ ions which will...

That sounds logical. But they gave amount of citric needed to neutralise it anyway so wouldn't this already mean it takes into account the weakness of it? C6H8O7 + 3NaOH ---> Na3C6H5O7 + 3H2O nC6H807 = 0.1*0.025 = 0.0025 nNaOH = 0.0025*3 = 0.0075 [NaOH] = 0.0075/0.0415 = 0.108mol/L I have a...

I got 0.181 (3sf.) lol Was the question something like 25mL citric at 0.100mol/L reacts with sodium hydroxide and avg sodium hydroxide was 41.5mL?

Yep

Damn I was conflicted because D had both O at the end so thought the more correct was B

What I did this and got like 9.35 x 10^-3 Lol nvm read it wrong it is C

That's what i thought but everyone is saying its D.

GG when you only memorised Leclanche dry cell and it isnt listed in the dot point

Yep lol, my bad.

Iirc, Fermi noticed that when a neutron underwent radioactive decay to form a proton and an electron, the initial momentum was not equal to the addition of momentums of proton and electron, which would violate the conservation of momentum. Thus, there must've been another particle that was...

Re: HSC 2015 3U Marathon \log _{b}a * \log _{a^2 }b = \frac{ \ln a}{ \ln b} * \frac{ \ln b}{ \ln a^2 } = \frac{ \ln a}{ \ln b} * \frac{ \ln b}{2 \ln a} = \frac{1}{2} \Rightarrow $A$

It should be valid. Make sure you integrated from 0 to x to get rid of constant