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$ \noindent Volume of revolution about the y-axis: $ \pi \int_{y_1}^{y_2} x^2 $ d$y \\ \\ $ So making $x$ the subject, we get $ x= e^{y} \\ \\ \therefore V=\pi \int_{1}^{3}e^{2y} $ d$y \\ \\= \left [ \frac{\pi}{2}e^{2y} \right ]_{1}^{3} \\ \\ = \frac{\pi}{2}e^2\left ( e^4-1 \right ) $ units$ ^3

\noindent $Degree of Ionisation $ = \frac{[H^+]}{[HA]}\times 100 \% \\ \\ $Where $ HA $ is the intact acid molecule $ \\ \\ \therefore 0.5 = \frac{[H^+]}{0.05} \Leftrightarrow [H^+]=0.025 \\ \\ $ pH $ = -\log [H^+]\\ = -\log 0.025 \\ = 1.6 it basically means that only 50% of all acid...

between your 6th last and 5th last step log(a+b) =/= log(a)+log(b) ie, you cant expand it like a normal algebraic term, its like how you cant expand sin(a+b) into sin(a)+sin(b) so continuing, ln(8+√48)-ln4 = ln(8+4√3)-ln4 =ln((8+4√3)/4) =ln(2+√3)

TDS is found by evaporation to dryness so the solid remaining in the basic after evaporation is the mass of the TDS, which is 44.67-43.53 = 1.14 now to find the percentage, divide this by 250 mLs and times it by 100 = 0.46% (to 2sf)

99.55 Rip english

Re: 2017 HSC Marks Thread 4u: 97 3u:99 Eng adv: 86 Chem: 94 Phys: 93

Can you guys pls shout me lunch if I get <99.95 thanks :)

interesting

85.35

https://imgur.com/a/BAU45 oops pls ignore the negative signs

Really straightforward tbh

for complex number z=a+ib, arg(z) = arctan(b/a) [-pi < arg(z) ≤ pi] and its clear that 5-5i is in the 4th quadrant, so the argument of 5-5i is arctan(-1) = -pi/4

\noindent $Modulus of $iz = |iz| = |i| \cdot |z| = r \\ ($as the modulus of i is 1 and z is r$) \\ $Argument of $iz = \arg (iz) = \arg (i)+ \arg(z) = \theta + \frac{\pi}{2} \\ $ (as the argument of i is $\frac{\pi}{2} $and z is theta$) if you havent learnt the above stuff, then let z = a+ib...

haha dont worry, im sure atleast half the people did the same thing (ie bs it to the max)

Good luck my bos fam :') We'll all hopefully smash it

99.95 mate ;)

Yes

if you want the exact value of x; take the log of both sides to get log(2^x)=log(7) now apply the rule log(a^b)=blog(a) on the LHS to get xlog(2)=log(7) hence x=log(7)/log(2) put this on the calc if you want an approximation

here's a pree cool nested root integral to try out; \noindent $ Evaluate $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x

I guess so cause that's the only induction q in that paper haha