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Suppose w^3-1=(w-1)(w^2+w+1)= 0 and w \neq 1 Then w^2+w+1=0 Thus Im(w^2+w+1)&=0\\Im(w^2)+Im(w)&=0\\Im(w)&=-Im(w^2) Using the fact w^2+w+1=0 gives (1+2w+3w^2)(1+3w+2w^2)&= (2+2w+2w^2-1+w^2)(2+2w+2w^2-1+w)\\&= (-1+w^2)(-1+w)\\&=1+w^3-w^2-w-1+1\\&=3 Do something similar to the other...

Ignore what Shadowdude said as its mostly incorrect. If you did maths, it doesn't make sense to have do MATH1151 and MATH1141 at the same time if they are basically the same course. A simple check on the unsw maths website can confirm this. For anyone serious about being an actuary, you will...

Just divide each term in the numerator and denominator by the highest power of x in the denominator, cancel out any term that goes to 0 as x goes to infinity (ie. anything with powers of x in the denominator after cancellation). Anything left over will be your asymptote. If there is x left over...

Let angle (PRA)=y angle (PBA)=angle(PRA)=y (angle in same segment) let angle(OSR)=x angle(ABQ)=angle(OSR)=x (ext. angle of cyclic quad) angle(PBQ)=x+y (obvious from diagram) angle(ROS)+x+y=180 (angle sum of triangle) thus angle(PBQ)+angle(ROS)=180 and hence OBPQ cyclic

Cos you didn't bribe them afterwards.

Question 1 involves conditional probability, as it asks for the expected number of visits given that a patient has had at least one visit. So P(X=x given X>0)= P(X=x and X>1)/P(X>0) eg. P(X=1 given X>0)=P(X=1)/P(X>0)=0.2/0.5=2/5 Do the same for 2,3,4,5 and then find the expected value.

I will give tips and leave the rest to yourself 13) Let p be probability of even number. Give an expression for probability of even number appearing 5 times using binomial theorem. Do same for prob of even number appearing 4 times. Then use the fact P(even number 5 times)=2P(even number 4 times)...

Let me rephrase the question: Which university has the best toilet paper?

Refer to attached diagram. Have a go yourself and note the angle between tangent and radius is pi/2. The answer is 2pi/3 btw.

or cis(-pi/3). There should be 2 possible solutions

(i) a+b=[sum of roots of z^7-1] -1 = 0-1 =-1 (ii) Easy, do it yourself. Just expand and simplify, remembering p^7=1 and 1+p+p^2+...+p^6=0 (iii) note sum of roots = - (a+b)=1 product of roots =ab from (ii) So equation is x^2+x+ your answer from (ii) (iv) Solve quadratic in (iii) using...

yes

arg [(1+i)z] =arg [(1+i)] + arg[z] = \frac{\pi}{4} + arg[z] So you are really sketching: -\frac{\pi}{4}< arg [z] < \frac{\pi}{4} This will basically be the first quadrant rotated by \frac{\pi}{4} clockwise

you are allowed to multiply everything on either side provided both sides are non negative . Think about it intuitively, a large number multiplied by a large number will be bigger than a small number multiplied by a small number.

(a+b)=1 so (a+b)^2=(a+b)

Only do accounting if you actually plan on being an accountant and getting CA. Accounting degrees are needed just to gain entry into CA program.

Since x^2+y^2\geq 2xy holds for any x,y>=0, we can substitute x with sqrt(x), y with sqrt(y) to get: (\sqrt{x})^2+(\sqrt{y})^2\geq 2\sqrt{xy} \\ x+y\geq 2\sqrt{xy} Now I have shown you this, Q2 should be very easy to solve using the above. Have a go yourself

For both questions, I shall use the fact: x^2+y^2 \geq 2xy for x,y>0 (you should know how to prove this. Q1) Noting a+b=1 \frac{1}{a}+\frac{1}{b}\\ = \frac{a+b}{ab} \\ = \frac{(a+b)^2}{ab}\\ = \frac{a^2+b^2+2ab}{ab}\\ \geq \frac{2ab+2ab}{ab} = 4

Consider only constant terms in the quartic equation.

I will only do the ones which are incorrect or unknown. 2. The mean is very dependent on outliers and can change significantly due to one very large/small value. Thus it doesn't give a good summary of the WHOLE data. The median on the other hand is independent of outliers , and hence would...