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the answer is A because citric acid is triprotic --> C3H5O(COOH)3 ---> 3H+ + C3H5O(COO)3 -3 if you were to write out neutralisation reactions for each of those 4 acids, you would see that because citric acid dissociates into 3 hydrogen ions, the mole ratio of NaOH to citric acid would be 3:1...

it's probably best to avoid dividing by an unknown when solving an equation, but if you wanted to, i suppose you could test the solution sinx=0, and if it didn't work, then you could divide the whole equation by sinx

also, if wolframalpha doesn't give you the answer you're looking for, you can also get it to differentiate your answer, and it should come up with the original function

= (k/2) [2a+dk-d] + (1/2) [2a+2dk] = [(k+1)/2] [2a+dk-d+2a+2dk] that line was incorrect, you didn't actually factorise that properly (k/2)[2a+dk-d] + (1/2)[2a+2dk] = (1/2)[k(2a+dk-d)] + (1/2)[2a+2dk] = (1/2)[k(2a+dk-d) + 2a + 2dk] = (1/2)[2ak + dk^2 - dk + 2a + 2dk] = (1/2)[2ak + dk^2 +...

8. H2SO4 is a diprotic acid, so the concentration of hydrogen ions is twice the concentration of H2SO4 .'. [H+] = 2*[H2SO4] = 2*0.0115 =0.023 mol/L then using pH = -log[H+], pH = 1.638 which is closest to C 15. with this equilibrium, we can see that there are 4 moles of gas on the LHS...

(n+1)/2 >/ root(n*1) ((n-1)+2)/2 >/ root((n-1)*2) ---> (n+1)/2 >/ root(2*(n-1)) ((n-2)+3)/2 >/ root((n-2)*3) ---> (n+1)/2 >/ root(3*(n-2)) if we keep doing this, we should eventually get n/2 inequalities similar to those above multiplying them all together: [(n+1)^(n/2)]/2^(n/2) >/...

referring to the image, perimeter is P = 2r + 2h + r*pi for some constant P .'. 2h = P - 2r - r*pi area is 2rh + (pi*r^2)/2 = r(P - 2r - r*pi) + (pi*r^2)/2 = Pr - 2r^2 - (pi*r^2)/2 to find max area, we differentiate this expression: P - 4r - r*pi .'. max area when P - 4r - r*pi = 0 P = 4r...

yeah, i felt that that would be the quickest way to answer OP's question. if there were more parts to it, like finding tangents with gradient 2, then i would've done it the same way as carrotsticks

Re: HSC 2012 Marathon :) i got -1/32 as well

if tan theta = -4/5, then you know that theta is either pi - arctan(4/5) or 2pi - arctan(4/5) since tan is negative in the second and fourth quadrants from this you can work out the points where they touch the ellipse with the parametric equations x=5cos(theta) and y=4sin(theta), then use the...

i'll probably start doing trial papers in the holidays...still assignments and other assessment tasks left to do

when you were finding the integral of cot^6(x), from the first line to the second, the negative sign should also be carried onto the integral of cot^2(x); two negatives make a positive. you should've had a bracket around -1/3*cot^2(x) - integral cot^2(x), since that whole expression is equal to...

isn't there meant to be a negative sign?

Re: HSC 2012 Marathon :) x = (u-3)^2 u-3 = rootx u = rootx + 3 dx/du = 2(u-3) dx = 2(u-3) du .'. integral of 1/(3+rootx) dx = integral (2(u-3))/(3+u-3) du = integral (2(u-3))/u du = integral (2 - 6/u) du = 2u - 6ln(u) = 2(rootx + 3) - 6ln(rootx + 3) + C

dy/dx = sec^2(x) dy/dx (when x=pi/3) = sec^2(pi/3) = 4 .'. gradient of tangent at (pi/3, root3) is 4 .'. gradient of normal at this point is -1/4 using point gradient formula, y - root3 = -1/4(x-pi/3) 4y - 4root3 = -x + pi/3 x + 4y - (4root3 + pi/3) = 0

in triangles BFC, AFE, angle AFE is common angle FAE = 90 (angles of a rectangle are right angles) similarly, angle ABC = 90 .'. angle FBC = 180 - angle ABC (adjacent supplementary angles) = 90 .'. angle FBC = angle FAE .'. triangle BFC is similar to triangle AFE (equiangular) .'. angle...

i use the whole page, but i only write notes or the theory that we get in class, and i do maths questions on scrap pieces of paper which i throw out afterwards

subbing the information into the equation you showed in b), 2 = 10(1-e^(-2*5*k)) 1/5 = 1 - e^(-10k) e^(-10k) = 4/5 -10k = ln(4/5) .'. k = -(ln(4/5))/10 when the difference in water level is 2cm, then 20-x - x = 2, since the level of the water in the compartment which initially had 20cm...

ah, alright. thanks for the clarification! :)

Hey everyone, I'm just wondering whether policies such as taxes on goods or services that are environmentally unsustainable or subsidies on more environmentally-friendly goods are microeconomic policies or macroeconomic policies. I'm thinking that they're microeconomic policies, but if so, what...