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So basically for cylindrical shells you are making cylinders perpendicular to your non rotational axis. Thus parallel to your rotation axis. The key here is to note which axis it's perpendicular to because essentially thats how many shells you are making. technically speaking, you are trying...

Who me? Why did i work at a biscuit factory and then change to a maths textbook printing place?

Perfect, Thanks dude

First you draw the picture Then prove the two triangles are similar and what you come to realise is the smaller triangle is 1/3 of the bigger triangle. Thus the length between the man and the lamp is x and the length of the shadow must be x/2. Ultimately that means if the man is traveling...

Title says it all - now do them! Even for your trials and stuff

apparently it is, thats why i deleted my post.

Sorry i couldn't up load the data file for you but if you know how to do this maybe you can show me via an example

[1] 65 15 111 62 35 30 14 11 11 17 12 40 23 16 39 11 16 10 [19] 6 68 49 65 70 11 84 25 124 52 86 26 155 10 2 32 95 1 [37] 10 1 12 16 48 67 70 6 200 6 26 26 38 20 19 20 72 5 [55] 42 9 26 8 116 1 5 4 13 24 44 9 69 64...

Does anyone know how to bin a set of data? for example how would you bin 40 different data sets in a vector into groups of 4?

Does anyone know how to bin a set of data? for example how would you bin 40 different data sets in a vector into groups of 4?

Dear entity beyond the radiating frames. Are you looking for a tutor that can get you 100uai? Or get you top ranks in MATHS? Or get you to understand MATHS like never before? Before you read on i want to advise you to read the whole paragraph instead of leaving after the first or second...

There are 13 letters with 3 Es 3 Ts 3 Ns So there are 13!/(3!3!3!) arrangements. If all two Es are together then then we shall treat them as one letter so then there are 12!/(2!3!3!) arrangments. Hence probability.

Ah, sorry i didn't post it - cause its quite tedious. But i dont get why you need the estimate the second bit and not the first part with the >=1

can someone explain to me how HSC 1997 Q6a iii) is done and the need of the condition it gives you.

Imo, it depends how good you are at maths and how much time you will spend on it. But if you are willing to do well in the HSC and thus in your school subjects then i suggest 4unit maths as its imo the best scaled subject. (some people argue extension languages scale better.. true it does...

Ahaha you made that up in your head LHS>b a>b say LHS=5 b=3 a=6 5>3 true! 6>3 true! now you want to replace b with a so that LHS>a lets see 5>6 hmmmmm in what world does that work? Okay you can argue that a could be 5 and lhs could be 6 but theres no point cause if one example dosnt...

Ah thanks i got my numbers stuffed up The third is done by 2/5+(2/5)^2.(3/5)+(2/5)^3.(3/5)^2 Because if he is to win in less than 4 turns he has to win the first three wins and also B has to lose in the first three turns so add up the probability he wins on the first turn, second turn and...

In a game, two players take turns at drawing, and immediately replacing, a marble from a bag containing two green and three red marbles. The game is won by player A drawing a green marble, or player B drawing a red marble. Player A draws first. Find the probability that: i. A wins on her first...

ABS value makes the included term positive and x^2+4 cannot be negative hence it doesnt need ABS value. x^2-1 needs it because when x=0 the term equates to -1 which needs ABS value to change it to positive 1. However if you give |x^2-1| you can say ABS value for -1<x<1 but this is unneccessary.

if there is give me a clear example and explain