Yes you are right in that sense. But the cost of doing those extra courses is imho outweighed by the benefits of extra education (esp in maths with the national priority scheme). What I'm trying to say is that it doesn't cost that much extra to obtain a second degree (unless its a degree which...
No need to do gen ed courses which are a waste of time
Alternate career paths available if actuary doesn't work out + Extra qualification
Can do more interesting courses in other degrees than other courses in commerce degree
Lower HECS per year (cos commerce is in the highest band. Doing...
Maths is recommended strongly, but it is not necessary for the actuarial degree. The first year maths courses are sufficient for the actuarial major. I feel that maths is probably the best match for actuarial studies in terms of job prospects as you are able to work in areas where strong...
Aside from the maths courses you do in first year actuarial, you don't do any more maths courses. Actuarial studies focuses a lot more on statistics rather than actual maths.
the maths you do in actuarial courses is not really maths in the purest sense like number theory and algebra, it is more tied to modelling, probability and statistics linked to insurance. However it is still (imho) quite demanding and a good mathematical background is necessary.
Finance, from...
that was the most woeful first half by the blues I have seen. 3 fkn soft tries in first half.
Lyon and wolfman... wtf.... first tackle and they kick for a low percentage play. Both should be DROPPED. I dunno why they kept lyon from first game.... Hopefully Morris and jennings to come in for...
$Sub into the first integral $ u = \frac{x}{\sqrt{2}}\\\\
$So $ \sqrt{2}du=dx
$then$ \displaystyle\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\, dx=\sqrt{2} \int_{-\infty}^{\infty} e^{-u^{2}}du\\\\
$Using change of variable$\displaystyle\sqrt{2}\int_{-\infty}^{\infty} e^{-u^{2}}du =...
a) kurt.physics is right but just note that f(0) > 0 so division by f(0) is allowed
b)f(x+(-x))=f(x)f(-x)
f(0)=f(x)f(-x) and noting f(0)=1 and that f(x)>0 so division by f(x) is allowed
1=f(x)f(-x)
f(-x) = 1/f(x)
c)
If statement is true for all integer n>=0 and then doing...
So the normal to ellipse (and circle) at (5,7.5) is 4x-2y=5
the normal at (5, -7.5) is 4x+2y=5
Since radius is perpendicular to tangent of circle, which is shared by ellipse, then the radius must be the normal to the ellipse, so where the 2 normals meet will be the centre.
Solving...
tangential means the circle TOUCHES the ellipse at that point and share the same tangent
the easiest way i can think of to do the question is to find the normal of the ellipse at both points. Since the circle and ellipse share the same tangents, they share the same normal. knowing that...
Mechanics is NOT THAT HARD. Its very routine based and all you pretty much have to do for the questions is to equate forces horizontally and/or vertically. If you have a decent grasp of physics and algebraic skills, it should be very easy.
back in the 80-90s the papers were more difficult so 4u was considered quite hard. The paper always had questions which separated the good and the best in maths in the state.
but now they dumbed down the exam papers so badly since 2000 (prob one of the hardest papers along with 1993), where...
^ No way is mechanics harder than harder 3u. Mechanics is really routine based and the processes are "mechanical", harder 3u is nearly impossible to be completely prepared for cos of the massive variety of topics they can test.
My ranking from easiest to hardest:
integration
complex, tie...
Discrete maths is recommended for those who wants to major in maths in the science component. Since you are planning to major in maths, it is recommended that you take this course. Also, for those doing commerce(actuarial)/science (maths or stats) check out:
School of Mathematics and...
In rational functions, to find oblique/diagonal asymptotes, you divide top and bottom by highest power in DENOMINATOR and cross out parts which goes towards zero to get diagonal asymptote.
eg f(x) = x^3/(x^2 +1)
divide top and bottom by x^2 gives f(x) = x/(1+1/x^2)
So clearly the asymptote...
For the equation x^2 + px + 1 = 0 since coefficients are real, then the roots must occur in conjugate pairs. Thus letting the roots be z1 and z2 where z2 is the conjugate of z1.
From product of roots: (z1)z2 = 1
z1z2 = |z1|^2 since z1 and z2 are conjugates (one of the rules of multiplying...