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Well he didn't really talk about you two. D.K

Ah, I was sitting right in front of you. You probably don't know me but I know you guys from a mutual friend that goes to your school lol.

Hoping there isn't a 5 marker on sanitising of water supply or lead/acid cell

think i sat in front of u in 4u

Don't think so. It was the same animal in Librah's pic up a few posts --^^^ Lol don't think that's a sloth

It might be easier if you note that y = 37^{42 + \ln x} y = 37^{42} * 37^{\ln x} y = 37^{42} * e^{\ln (37^{\ln x})} y = 37^{42} * e^{\ln x * ln37} y = 37^{42} * \left (e^{\ln x} \right )^{\ln37} y = 37^{42} * (x)^{\ln 37}

Was it B? $ Since equal amounts are used for both reactions, each reaction consumes $ \frac{400}{2} = 200g 200g $ of $ C_{(s)} = \frac{200}{12} = 16.67 $ mol $ $ By the mole ratio, reaction 1 produces 16.67 mol of CO_{(g)} $ $ mass of CO_{(g)} = 16.67 * (12+16) $ By the mole ratio...

Should this be 2C(s) + O2(g) --> 2CO (g)?

It helps to think about what you want. It asks for mass, and it gives you a relationship between moles and energy, that is, kJ/mol. Since they give you kJ, you can find mol by doing \frac{kJ}{\left (\frac{kJ}{mol} \right ) } = $ mol $ ,. that is, divide given energy by the energy/mol ratio...

n_{C_{5}H_{11}OH} = \frac{108}{2800} \therefore m_{C_{5}H_{11}OH} = \frac{108}{2800} * (5*12.01 + 12*1.008 + 16) = 3.40 g ( 3 sf. ) \rightarrow D

Re: HSC 2015 3U Marathon = 100 \sum_{k=2}^{100} \frac{1}{\log_{k}100!} = 100 \sum_{k=2}^{100} \frac{ \ln k }{\ln 100! } = \frac{100}{\ln 100!} \sum_{k=2}^{100} \ln k = \frac{100}{\ln 100!} ( \ln 2 + \ln 3 + \ln 4 + ... + \ln 100 ) = \frac{100}{\ln 100!} \left( \ln (2*3*...*100)...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread ^ Could have also used the idenity Tan (90+x) = tan (90-(-x))= cot (-x) = -cotx

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread If you're talking about that part it should be root13. Expanding cos(a+b) = cosAcosB - sinAsinB sinA = 2/3 and sinB = -2/(13)^(1/2) so -sinAsinB = 4/(3(13)^1/2

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $ For $ \frac{\pi}{2} < x < \pi , \pi < 2x < 2 \pi -1 < \sin 2x < 0 \sin 2x = 2 \sin x \cos x $ Note that in this domain, $ -1 < \cos x< 0 $ Since $ \sin x = \frac{3}{4}, $ then $ \cos x = \frac{ - \sqrt{4^2 - 3^2...

Re: HSC 2015 3U Marathon $ let x be the amount of red balls where x \geq 1 $ $ let y be the amount of blue balls where y \geq 1 $ $ P(both red) = 5 * P(both blue) $ \therefore \frac{x}{x+y} * \frac{x-1}{x+y-1} = 5 * \frac{y}{x+y} * \frac{y-1}{x+y-1} x(x-1) = 5y(y-1) $ P(diff...

Hahah, it had a quote on it. I thought the guy was one of the hosts because he was walking around with a sandwich

lol who was the guy in the sloth shirt?

How did you do it with stars and bars? I can't really visualise it

Re: HSC 2015 4U Marathon I saw that one and the complex number one which I liked lol. arg((i+1)(i+2)(i+3)) = arg(10i) = pi/2

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread sinx = x/200 200sinx = x Sketch y = 200sinx Sketch y = x Both on the same cartesian plane. The amount of solutions will be evident