Search results

___________________________________________________ Quote freaking_out is this transformation a famous one or somethin'??if not how do we meant to come up with that? ___________________________________________________ Don't worry. In the HSC there'll be a part i) show that 1 + x + x^2 +...

Explain why the polynomial b_0+b_1x+...+b_nx^n has at least one real root if, b_0/1+b_1/2+...+b_n/(n+1)=0.

__________________________________________________ freaking_out: u've been a bit "off the ball" recently.... __________________________________________________ Yeah, I know. Permutation/combination problem, then this, oldman might be showing his age. btw, archman, well done. What made...

wogboy is absolutely right. The upper limit should be int(k/2)=the highest integer <=k/2. I have edited the problem accordingly.

Show that the coeff. of x^k in the expression, (1+x+x^2+x^3)^n is SUM(j:0--->int(k/2))[nCj.nC(k-2j)] where int(k/2) is the highest integer <= to k/2.

___________________________________________________ quote : freakin_out correct me if i'm wrong, but shouldn't the way u slot the vowels be: 8C4/2!, since u have to keep the vowels in alphabetical order? ___________________________________________________ oops! forgot the alphabetical...

Funny (or disconcerting) to see that some perm/comb ext1 trial questions are harder than some ext2 problems. I remember a similar CSSA trial (1997?) on the word EXTENSION, no two vowels together. Again I'll be stepping on the backs of die_imortales and malachite's initial analyses to solve...

Student's attitude toward examiners should be : Examiners are fair, bonza and dinkum! Working from that attitude, students should be fair, bonza and dinkum back. ND : just do it! sub y=pi-x I (0-->pi/2) e^(-sinx)dx = -I (pi--->pi/2) e^(-siny)dy =I (pi/2-->pi) e^(-sinx)dx Now that...

Parallel to Affinity's acceleration triad below: 3.) a = d^2x/dt^2 = v*(dv/dx) = (1/2)*[d(v^2)/dx] is the angular acceleration triad, d^2@/dt^2=w*(dw/d@)=(1/2)*[d(w^2)/d@ where w (omega) = d@/dt.

underthesun: you got the distance and speed right, but the time might be worth 1/2 pt. The time of 3:08 is the most rapid decrease in distance, the time 3:24 gives the most rapid increase in distance. Otherwise, well done. A slightly simpler approach : D=sqrt(25-24cos@) where @ is the...

Not after you get your FIAA.:)

Try the following. Let @= angle between minute and hour hands, measured positive clockwise (of course!) Then D=sqrt(25-24cos@) and proceed to differentiate wrt time. There are valuable lessons in this question for students who seriously try.

A clock's minute and hour hands are lengths 4 and 3 respectively. At the moment when the distance between the two tips is increasing most rapidly: i) what is the distance? ii) what is the speed? iii) what is the time after 3 o'clock does this first happen?

_________________________________________________ Quote Fosweb: OLDMAN: thanks. I did say above that i knew this was wrong (with 3 or 2 or 1 exactly pairs) for this reason, i just couldnt work out how to get rid of the other pairs that could occur also. However: it this still the easiest...

Consider "formula" for " exactly one same letter pairing": (1) 4C1 7!/2!2!2! This is obviously wrong as it already equals 2520=8!/2!2!2!2! ie. total number of ways. There is massive overcounting in (1). In fact the corresponding formula for "exactly" two pairings (2) 4C2 6!/2!2! is...

Check Affinity's answer using "4-unit" methodology. The key to problems of this type :mutual exclusivity, and geometric series. Consider all possible outcomes. Di wins 1st toss.............................1/36.............Dot losses Neither 7 nor 12 1st...

A test on recursive probability tree. Di and Dot bet on the total roll of two standard dice. Di bets that a 12 will be rolled first. Dot bets that two consecutive 7's will be rolled first. They keep rolling until one wins. What is the probability that Di will win? Sorry about the silly...

_________________________________________________ Quote So, let Bob throw a coin twice, and penny once. There is a 1/2 chance of bob getting more heads than penny? I still don't get it.. ____________________________________________________ Underthesun: Required probability is...

Your teacher is right. There is a distinction between 1 kg mass and 1 kg wt. The later is a force equivalent to 1*9.8 =9.8 Newtons. Thus if a force of 1 kg wt is applied to a mass of 1 kg, an acceleration of 9.8 m/s^2 results.

Method 3 P(Bob wins w/o revisiting start node)= 1/4+1/2*1/4=3/8 P(Penny wins w/o revisiting start node)= 1/4 P(Bob wins)=(3/8)/(3/8+1/4) P(Penny wins=(1/4)/(3/8+1/4)