Maths: 2/26/2014 20:26:11 
Dis cool dude
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Can somebody good at maths with a little bit of time please reply? I need help with my homework. Questions: http://www.maths.liv.ac.uk/~mem/challenge/sch14.pdfI only need help starting the questions; I simply don't know how to go about solving them. I have solved 5, 6, 7 and have an idea of how to do 8. Thanks for any help.

Maths: 2/26/2014 22:52:47 
Wenyun
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1: He has a range of 10 degrees. You'll need to zigzag in some way to cover both. (Or maybe he should go like a sine graph.)
2: Uh, cut each group once, combine them together, unless I'm misunderstanding the initial position. (Please elaborate if the initial position isn't what I'm thinking.)
3: Cows have 1 mouth, 4 legs and 2 eyes, Chickens have 1 mouth, 2 legs and 2 eyes, Warriors have 1 mouth, 2 legs and 1 eye. Set up a system of equations. Also watch out for pairs of legs  which means you'll have to divide legs by 2 >.> (I have one pair of legs!)
4: Lookup halflives. This problem is the exact same, with girls being undecayed, and boys being decayed. (Yeah feminism (sarcasm!))
8: ABC are limited to 5 positions each; H is limited to 8; OL are limited to 6 positions combined. (When he's not chasing kids, Olaf eats with Vikings?)
Most of these are a mix between logic puzzles and math. (Yes, math. I'm on the other side of the pond.)
Edited 2/26/2014 23:19:55

Maths: 2/27/2014 00:03:03 
Dis cool dude
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Wow thanks for the fast reply!
1. I got that, I just wasn't sure how to zigzag to counter the inaccuracy. 2. Same thoughts here; surely that's too easy though. 3. Yay equations. 4. Will do. 8. I wrote that down, but I've never studied outcomes so I don't know exactly what to do. Multiply 5, 8 and 6?
Thanks again, appreciate the help.

Maths: 2/27/2014 10:04:41 
Dis cool dude
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Ok thank you.

Maths: 2/27/2014 11:48:20 
Dis cool dude
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Looking at half lives now...
One year would be one half life, right? So the formula could be:
P(n) = 1/(2^n)
However the problem I have with this is the wording of the question. It says the expected fraction... in the nth year. Could that mean just in that year? Wouldn't that be a half every time?

Maths: 2/27/2014 12:48:57 
Dis cool dude
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Done 3  it was actually very simple once I got started. I prefer raw equations to word problems :D
Still very stuck on 1 and 2.
Edited 2/27/2014 12:54:39

Maths: 2/28/2014 00:04:25 
[16] b3rz3rk3r
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For 1, I think you can see the shore as the yaxis as the cliffs. the origin as the harbor and Olafs position as (n,0) for some n.
Then you need to find a function which will lead you to the cliff independent of the deviation. My first guess is you should start sailing in a spiral, but there might be a better way.
For 2 it kinda a think outside the box thing. The first step is to take one of the pieces and separate it into 5 singles.
By for 4 think they mean of the total 100 male children they get eventually. (but the wording is a bit confusing)

Maths: 2/28/2014 01:28:25 
Memele
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nº4 > 1/2^n, it's that, don't look for more. Example: In first year there are 50 boys and 50 girls born. That means that 50 parents stop having kids. In year 2 only 50 parents have kids so 25 are boys and 25 girls. 25 more parents will stop having kids. In year 3 25 parents have kids... 12.5 each (yeah, thats imposible, but we are speakinf of expectations, not facts). As you can see is half each year... in fact the problem is rather easy when you do understand the question.
nº2 > Take 1 of the 6 initialpieces and separate it in 5 links (4 cuts = 8 coins), use these 5 links to join the other 5 pieces (5x4 = 20 coins). Total 28 coins. This is a famous riddle.
nº1 > zigzag it's ok. You go to 5º direction then change for +5º. Actually you will be moving 10º to 0º or 0º to 10º, you need to keep with this till you are close enough to realize that you are in front (no more zigzag) or a bit north/south (one more change and that's enough).
nº8 > 3 with no preferences = the others arrangements x their arregements (3 can rearrange in 3!=6 forms) so for now we nned to arrange 5 and multiply it by 6.
There are 3 "no left" so 5 positions (4 right and 1 in front). take 3/5 positions is a combination of 10, these 3 positions can be arranged in 3!=6 forms, so 60 in total. The 3 "no left" guys have 60 options available. But we have to take in account the ones that use the front one (because of Hagar) If we take 3 of 4 sits (right ones) there are 4 options, rearranging them is 4*6 = 24 options So 6024=36 options with front used.
Olan + Leif = 6 options (AB, BC, CD and the same changing positions by themselves)
Front used: 36 (no left trio) x 6 (left duo) x 4 (Hagar) x 6 (no preference trio) = 5184
Front dont used: 24 (no left trio) x 6 (left duo) x 3 (Hagar) x 6 (no preference) = 2592
Total options = 7776
I hope you can understand it... I am not very good with english and I can't write math symbols such as combinations...

Maths: 2/28/2014 12:35:39 
Dis cool dude
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Ok; only question 1 I don't understand now. I sort of understand 8 but parts of it are not clear to me.
Thanks for the help.
Edited 2/28/2014 14:01:59

Maths: 2/28/2014 14:33:01 
[16] b3rz3rk3r
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A picture with memeles idea (which works better than the spiral i suggested before) http://imgur.com/XL0C64rSo in short follow a path like the green one and you get there.
Edited 2/28/2014 14:36:10

Maths: 2/28/2014 14:48:04 
Green
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[censored]
Good point Sharpe
Edited 2/28/2014 19:58:27

Maths: 2/28/2014 19:41:34 
Dis cool dude
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Ok so finding a route that keeps me within x degrees of the proper route will eventually land me in eye sight of the harbour.
Can number 8 be reexplained?

Maths: 2/28/2014 19:55:10 
Richard Sharpe
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Shouldn't you be doing your own homework and not having random strangers do it for you? It does specifically say that "your entry must be your own work"...
Oh, and Memele's solution is off by a full order of magnitude.
Edited 2/28/2014 20:08:21

Maths: 2/28/2014 20:18:41 
x
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Back in my day we had to beat up a nerd and make him do our homework for us. You couldn't just crowdsource it. People giving away free labour on here, do you know how much damage you are doing to the nerd economy? Many scouse nerds will have to give up their lunch money, since their homework services are no longer required. Because of you.
Anyway, the answer to number 2 is that you just weld the original 6 pieces together. It doesn't mention that the original chains were endless, but it does specify that the chain of 30 links is endless. Therefore, it's fair to assume that the original 6 pieces are not looped.
6 x 4 = 24, genius, there's your working out.
Q1 doesn't even give any measurements so my answer would be "slowly" or "very slowly".
I mean, you don't know how wide the gap in the cliffs is, and you don't know that he is due east of the harbour. The harbour could be in any position, he is just "somewhere to the east of it".
If you're thinking of the coast as a yaxis, and the harbour at 0,0 because why not, he could be at literally any position that is x > 0 + "a few meters". So hopefully the ship is small and manoeuvrable, since he will need to travel west at a slow enough speed* that he can stop when he sees the cliff. Then he will need to make a 90degree turn and crawl either north or south along the coast until he sees somewhere. Now that I write it all out like that he better just stay put until the fog lifts.
*He can see for a few meters, he will need to go at a speed where he can come to a complete stop in that distance, including reacting to the sight of the cliff and executing the stop. Actually, fuck that, he'd probably get there faster by going at the speed at which he can crash into a cliff and still be seaworthy. Although he might have a sentimental attachment to the boat and not want to damage it. These complexities are why my answer would just be 'slowly'.
Edited 3/1/2014 11:11:01

Maths: 2/28/2014 21:16:07 
x
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Let's assume that he is due east of it, his compass is 5 degrees out so the longer he travels, the further away he is going to be from it. And the width of the curve he will need to move in is a few meters wide. I mean, the whole thing is absurd.
Draw an angle of 10 degrees with sides of 10cm. Let's assume he is 1km away and can see 3 meters. he will have to move in a curve within that touches the sides of your angle and he has to touch each side every 0.03cm.* And he will have to go slow anyway because he could hit a cliff at any time. This means my method would probably be quicker. With my straighttothecoast method, you could also estimate out how far you were from the harbour (if you didn't hit it first time), since you approximately know the length of the hypotenuse of the triangle, you know the angles of it are the ten degrees you approached the coast at, the right angle you intended to make with the coast, and the 80 degree angle you actually made.
This is assuming he is due east of the harbour which he obviously isn't. So this whole post is irrelevant.
Stupid question.
*How the fuck would he be able to manage this when he doesn't even have a working compass and can't see a thing, I don't know.
Edited 3/1/2014 11:10:36

Maths: 3/1/2014 01:06:39 
Daisuke Jigen
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Don't fault poor Olaf. The compass is just a tankard of sugar water, in which a cork skewered by a weakly magnetized pin bobs.

Maths: 3/3/2014 14:24:41 
x
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am i right or am i right

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