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If the largest deviation is 37.00, it should be recorded as 37 and the mean result is 40, not 37.00 and 40.00 respectively. The measurement is 40 +/- 3, the absolute error is 3 (not -3). :santa: :santa: :santa:

For your future reference Solutions for the last three Westpac Mathematics Competitions are available from http://itute.com/mathline/exams_solutions.html :santa:

f(x)=3^x - x^3 f(x) = e^(xln3) - x^3 f'(x) = (ln3)e^(xln3) - 3x^2 = -1/3 (ln3)e^(xln3) - 3x^2 + 1/3 = 0 Graph y = (ln3)e^(xln3) - 3x^2 + 1/3 Find x-intercepts 1.244 and 2.727 :santa: :santa: :santa:

Multiply top and bottom by the conjugate of the bottom. Expand and simplify using sinsq + cossq =1. Factorise and cancel common factor (2 - 2cos(pi/7)) to obtain -cos(pi/7) + isin(pi/7) :santa:

It is an arithm. series of n = 74 terms, first term (k = 4) is a = 19, common difference is d = 6. Sum of the 74 terms = (n/2)[2a + (n-1)d] = 17612 :santa:

Length = sqrt[(x2 - x1)^2 + (y2 - y1)^2] =sqrt[(2ap -2aq)^2 + (ap^2 - aq^2)^2] =sqrt[4a^2(p - q)^2 + a^2(p^2 - q^2)^2] =sqrt[4a^2(p + 1/p)^2 + a^2(p^2 - 1/p^2)^2] =a sqrt[4(p + 1/p)^2 + (p^2 - 1/p^2)^2] =a sqrt[(p + 1/p)^2 (4 + (p - 1/p)^2)] =a sqrt[(p + 1/p)^2 (p + 1/p)^2] =a(p + 1/p)^2...

x^2 < x^2 + y^2 for y =/= 0, |x|^2 < |x+iy|^2, |x| < |x+iy|, hence |re(z)| < |z|, where z = x+iy. :santa: :santa: :santa:

z = 0 does not satisfy |z-i| + |z+i| = 1. For this and similar equations, substitution of z = x + iy will lead to a wrong cartesian equation. Using geometry for the interpretation of the equation the solution is { }. :santa: :santa: :santa:

The original question did not specify that they are integers! Your assumption may not be the intention of the question. :santa:

Basically you used the same information to obtain the two equations in k and l, and they should be the same. You don't end up with two different equations. :santa:

How do you solve these equations? 3 equations for 4 unknowns or in terms of k and l? :santa:

What is |z-i| + |z+i| = 1 ? :santa:

Anyone still interested? http://www.itute.com/mathline/exams_solutions.html :santa:

Anyone still interested? http://www.itute.com/mathline/exams_solutions.html :santa:

f'(x) = cx + d implies that f(x) is quadratic. Turning pt at (2,0) gives f(x) = a(x-2)^2. Crosses the y-axis when y=4: 4 = a(0-2)^2, therefore a = 1. Hence f(x) =(x-2)^2, and f'(x) =2x-4. c = 2 and d = -4. :santa:

Correction to (4) w=4*ellipse area/(pi*L) :santa:

(1) By counting number of unit squares in the region. (2) r=sqrt(circle area/pi) (3) r=2*sqrt(sector area/pi) or r=R*sqrt(2*angle/pi), where R is radius of sector, angle of sector in radians. (4) w=ellipse area/(pi*L) (5) Please give an example to show what you meant. :santa:

Cauchy-Schwarz Inequality |a.b|<=|a||b| If a=pi+qj and b=mi+nj, then C-S I can be expressed as |pm+qn|<=rt(p^2+q^2)rt(m^2+n^2) (pm+qn)^2<=(p^2+q^2)(m^2+n^2) This can be extended to 3 or higher dimensions: (pm+qn+...)^2<=(p^2+q^2+...)(m^2+n^2+...). Proof: Since 0<=|cos$|<=1, therefore...

(3 root7)^3=3^3*(rt7)^3=27*7rt7=189rt7 (root5 + 7)^2=(rt5)^2+2*rt5*7+7^2=5+14*rt5+49=54+14*rt5 (root7 - root11)(root7 + root11)=(rt7)^2-(rt11)^2=7-11=-4 2 root3 (4 root3 + 5)=(2rt3)(4rt3)+(2rt3)*5=24+10rt3 Also some indices (4x10^3)x(2x10^5)x(7x10^-4)=4*2*7*10^(3+5-4)=56*10^4=5.6*10^5...

64(2/3)^n=2(4/3)^n 32(2/3)^n=(4/3)^n 2^5*2^n=4^n 2^(5+n)=2^(2n) 5+n=2n n=5