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first find dy/dx of y=logx dy/dx = 1/x i'm assuming you mean at the point where x=e^2, so sub this into dy/dx to find the gradient of the tangent to y=logx where x=e^2 .'. gradient of tangent = 1/e^2 now when x=e^2, y = log(e^2) = 2loge = 2 so the point on the curve where x=e^2 is (e^2, 2) now...

it is meant to be (x^2/a^2) + (1)^2 = 4y/a this is from the line before where he had (p+q)^2 + (p-q)^2 = 4y/a p+q = x/a, so (p+q)^2 = x^2/a^2 and from the condition, p-q = 1, so (p-q)^2 = 1^2 substituting those back in, you get the equation (x^2/a^2) + (1)^2 = 4y/a

Q2: acceleration = d(1/2*v^2)/dx = -2 integrating with respect to x, 1/2 * v^2 = -2x + C when t=0, x=8 and v=4 .'. 1/2 * (4^2) = -2*8 + C 16/2 = -16 + C 8 = -16 + C C = 24 .'. 1/2 * v^2 = -2x + 24 v^2 = -4x + 48 v = +- sqrt(-4x + 48) = +- 2sqrt(12-x)

99.30 ...a bit higher than my aim, lol

the solution in the conical flask isn't more basic; there's just more moles of base in there, and hence you'll need more acid to neutralise all the base

97; mid 20s

the nCk should not be there; what that's there for normally is to account for the varying ways in which you can have 2 days raining and 5 days not raining out pf 7 days the (2/3)^2*(1/3)^5 by itself (without the 7C2) shows the probability of having 2 days raining and 5 days not raining for only...

y = 3sec(theta) - 4 y+4 = 3sec(theta) sec^2(theta) = (y+4)^2/9 ----- 1 x=1+2tan(theta) tan(theta) = (x-1)/2 tan^2(theta) = (x-1)^2/4 ----- 2 1 - 2: sec^2(theta) - tan^2(theta) = (y+4)^2/9 - (x-1)^2/4 1 = (y+4)^2/9 - (x-1)^2/4

the first one the answer is D because from each year, you can see that consumption and savings have both increased by 100 each, so the MPC and the MPS are both 0.5 i.e. for every extra dollar earned, 50% of it is used for consumption and 50% is saved. since the multiplier = 1/MPS, .'. multiplier...

are you sure it's 2n^2? if it was twice the square of a number, or twice a squared number, i would agree, but twice a number squared sounds more like(2n)^2 to me :/

what raw mark would you need to just scrape a band 5?

tanx/tan2x = (sinx/cosx)/(sin2x/cos2x) = (sinxcos2x)/(sin2xcosx) = (sinx(1-2sin^2(x))/(2sinxcos^2(x)) = (1-2sin^2(x))/(2cos^2(x)) = (1-2sin^2(x))/(2(1-sin^2(x)) = (2(1-sin^2(x))-1)/(2(1-sin^2(x)) = 1 - 1/(2(1-sin^2(x)) = 1 - 1/2cos^2(x) = 1 - 1/2(sec^2(x)) integrating that just gives x...

oh, ok, thanks guys! and to bored_of_HSC, i was looking at a question that asked about economic effects of an increase in the participation rate, and the thought sort of came to me...

i think that's hard-core unemployment...i'm referring to people who just choose not to actively seek work, and thus aren't counted in unemployment figures

Do the hidden unemployed receive unemployment benefits? They aren't actually counted as unemployed, but does that mean they don't receive these benefits?

Re: MX2 Integration Marathon let u=arcsin(x) I = integral(u^2) dx = integral(u^2) du * dx/du u=arcsin(x) x=sinu dx/du = cosu .'. I = integral (u^2 * cosu) du integrating by parts, m=u^2 m'=2u n'=cosu n=sinu I = u^2*sinu - integral(2u*sinu) du = u^2*sinu - 2*integral(u*sinu)...

i did it differently to the answers, but got the same values for part i, if the computer chooses both your numbers, then the number of combinations where this can happen are 2C2 * 38C11, since 2 of the 13 numbers chosen must be the numbers you chose, and then the computer chooses the other 11...

p is the sum of the roots 2 at a time; for product of roots, you're finding the value of q

i think nightweaver's one is right

i don't think the answer is purely 4/7, because the question says if at least one gold coin is selected. this doesn't mean that the gold coin was selected first; it could have been selected second, and all we know is that at least one was chosen, so i reckon barbenator's answer is correct