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I guess. Cracking this code would be illegal anyway, since you'd essentially be stealing the points from others (they're fairly costly too).
Any luck? Free Wii points would be useful. O.o
Here's another video I made later on the cosine rule;
http://www.youtube.com/watch?v=m7_WaHlLIr4
View at your own leisure!
k i have another one only this time its a rectangular solid.
The given is that l=8, W=4 and H= 2. I need to find the diagonal... which in this case is d. and the answer is written as:
d= √ then the answer.
In case my solution wasn't clear, I created a video;
http://www.youtube.com/watch?v=tSuD5a6rf38 (the example used is the problem you gave me)
If you need a bit of a re-cap on Pythagoras' Theorem, you might want to see my other video;
http://www.youtube.com/watch?v=vS_fbzvPtmU
I will also be creating sine, cosine and tangent videos, where I'll show in detail how to find angles and sides in right-angled triangles using sine, cosine and tangent. I'll also move on to 3-D trigonometry if you're interested, which is more of a GCSE topic.
EDIT: I have just created another video on the sine rule. You can see it here;
http://www.youtube.com/watch?v=DWKpQ4l94v8
It contains two examples of problems that I made up. I will include harder examples later, and more videos using the cosine and tangent rules, along with the inverse trig functions.
PS: Sorry for the terrible handwriting, I had to use a mouse.
Sometimes questions like that can be hard to imagine, so imagine in our rectangular solid we want to find the diagonal on the bottom of the solid. To do that, we need the length and the width (8 and 4) - by Pythagoras' Theorem, a² + b² = c²;
8² + 4² = c²
80 = c²
Square rooting both sides;
√80 = c
At this point, don't calculate that square root - you'll need the most precise possible value in order to complete the problem. Now that we have the diagonal that runs along the bottom, we can use that to find the diagonal that connects the two opposite corners in the rectangular solid that you are referring to. Here's why they gave us the height - 2. We use Pythagoras' Theorem again;
a² + b² = c²
2² + √80² = d²
4 + 80 = 84 (because the square root of something squared means they'll cancel out !)
d = √84
So the solution to the problem is that the diagonal is √84. If this is hard to picture in your head (which it is usually for most people without paper) then I can provide a video if need be.
x = √(8²+10²) = √164 = 12.8 (3 s.f.)
As for the answers, use the tangent rule twice;
Angle Y: arctan(8/10) = 38.66° (probably 39° by your choices)
Angle Z: arctan(10/8) = 51.34° (also could have simply used the 180° rule of triangles)
If you didn't understand what I did above: remember SOHCAHTOA? Sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent? Well, these can also be applied to angles too; you have to use the inverse trig functions (arcsin, arccos, arctan - sometimes denoted sin[sup]-1[/sup], cos[sup]-1[/sup], tan[sup]-1[/sup], or asin, acos, atan - in order to find the angles.
For Angle Y: 8 is opposite the angle Y, so it is my OPPOSITE side. Adjacent to that is 10, and it couldn't possibly be the hypotenuse because we know that it would be opposite the right-angle in this case. So if tan = opposite/adjacent, then tan(Y) = 8/10... we have to take the inverse tangent of both sides.
If you still need further help on this topic, I will make a video on my YouTube channel.
Hi guys,
A Wacom Tablet might be a good idea... but quite expensive! In order to save money, my plan is to use a webcam (or some sort of video recording device) to film a sheet of paper that I'll be writing on in a standard felt-tip pen, whilst verbally giving explanations. People that do this are Allen Lin, Edugratis.com, and PatrickJMT (youtube.com/patrickjmt), who recently was the subject of an article. Tutoring is on the rise, and what makes it really easy to do is that it is relatively low-risk! The only costs would be transport and advertising costs, which aren't too expensive anyway (advertising a small business via poster usually only costs around 80p a week). And Easter is coming up, which is always the most popular time for tutoring.
I wonder if I could use Microsoft Mike.
@JaneFairfax; Yes, the videos will all be done free of charge with no intention of profit.
@Bobbym; Thanks, but I thought they were terrible... this is just something to "test out" before I get a microphone and working recorder. I already use the P2PMaths name in my tutoring business, but I'm just looking for some basic ways to promote and expand this business for now.
He gave us the answer... but I'm still not convinced that it's unsolvable.
Don't know if this is really allowed since this is a form of advertising, but I've started a YouTube Channel devoted to tutoring mathematics. Unfortunately, I do not have a microphone or a working video recorder, so at present I have to use MS Paint and CamStudio. As is evident, the first video was not very good, so I will be looking for ways to improve it.
The channel is located here: http://www.youtube.com/user/P2PMaths
Only doing M1 module at the moment, but I'll move on to the rest of the Mechanics modules, then the core/pure core, etc... I'll do D1 soon since there aren't many D1 videos on YouTube.
I believe anything that I think has enough evidence to be believed in, hence my love for physics - but by that statement, you could say that I should believe in theories of a 'God'; but the presence of a God would cause many questions to surface... and one of those questions would be questioning the purpose of life. But do things need a purpose? And, if one were to know one's purpose of living, would one's lifestyle change as a result?
I approve of anything that gives someone hope, but I disapprove of anyone's statement that states that their belief is wholly true. That's why it's a belief.
6) A project is to be completed in 13 days. The activities involved in the project and their durations in days are given in the list
A - 3
B - 8
C - 7
D - 5
E - 8
F - 4
G - 5
H - 4
I - 4
J - 4
To determine how many workers are required;
(a) Apply the first-fit algorithm
(b) Apply the first-fit decreasing algorithm
(c) Is it possible to obtain a better solution than either (a) or (b)?
7) A small ferry that sails between two of the islands in the Hebrides has three lanes 20 metres long on its car deck. The vehicles waiting to be loaded are;
Petrol Tanker - 13m
Small Truck - 6m
Car - 4m
Small Van - 3m
Coach - 12m
Lorry - 11m
Truck - 7m
Car - 3m
(a) Can you use the first-fit decreasing algorithm to load all the vehicles on to the ferry?
(b) Can all te vehicles be taken in one trip?
8) A project consists of eight activities whose durations in hours are as follows;
A - 2
B - 4
C - 3
D - 1
E - 5
F - 4
G - 2
H - 3
Use full-bin combinations to determine the minimum number of workers needed to finish the project in 12 hours.
8) A certain kind of pipe is sold in 10m lengths. For a particular job the following lengths are required;
2m, 8m, 4m, 5m, 2m, 5m, 4m
By looking for full-bin combinations, or otherwise, find the number of 10m lengths required for the job.
9) Joan decided that she wanted to record a number of programmes on the video recorder. The lengths of the programmes were;
45 min, 1h, 35min, 15min, 40min, 30min, 50min, 55min and 25min.
Help Joan decide how many 2h tapes she requires, using;
(a) The first-fit algorithm
(b) The first-fit decreasing algorithm
(c) Full-bin combinations
10) 120, 78, 100, 90, 60, 38, 80, 26, 150
(a) The list of numbers above is to be sorted into descending order. Perform a quick-sort to obtain the required list. Give the state of the list after each rearrangement and indicate clearly the pivot elements used.
(b) (i) Use the first-fit decreasing algorithm to fit the data into bins of size 200.
(ii) Explain how you decided into which bin to place the number 78.
...and twenty-thousand years later, when code-breakers are anxiously awaiting the final permutation of Bob Farey's code, it is cracked. The message reads:
rofl i juts wast3d ur ti3m lols!!!!!!!!11
Handful of easy questions designed for students wishing to catch up on the D1 module, or simply for those interested who never studied Decision Mathematics. Questions will all be relatively easy.
1) The marks obtained in an examination by five students were;
Anne (68), Barry (42), Clare (70), David (45), Eileen (50) and Greg (55).
Use the bubble-sort algorithm to sort these marks
(a) in ascending order
(b) in descending order
2) Use the quick-sort algorithm to sort the list
6, 8, 4, 5, 10, 2, 9
in ascending order.
3) The members of a club have surnames
Monro, Jones, Malik, Wilson and Shah
Use the bubble-sort algorithm to sort these into alphabetical order.
4) Use the bubble-sort algorithm to sort these into alphabetical order;
4, -3, -5, 0, 2, 5, -1
5) The times recorded by seven competitors for a given distance were;
9.8, 9.2, 9.6, 9.7, 9.1, 8.9, 9.0
(a) Use the bubble-sort algorithm to sort these times into order.
(b) Use the quick-sort algorithm to sort the times into ascending order.
(c) Which of the methods is more efficient in this case?
Hi;
That too is interesting. Does a Cambridge graduate rate higher than a Tutorial Fellow in Mathematics at Wadham College and author of a couple of books? Possibly.
Hodge's classic work could be an exaggeration of the story.
If I recall correctly his information was also based on Simon Singh - but the best source would be someone who was actually alive at the time.
@Bob Farey - is this your code?
fhmxvwzwqcdicnxiyzwtyelazwczsyewdevyzmxhicuhvmhssl
qvvhwewjzzchvcsjcylenucxijsvlezicjzeyykplhvztjhzcz
uetvwzoecjhooujbxpysczgwtwfkwgsvfwvikxhxcecvlfxwne
iweujtjwywjtrczewcmrdidejeikourfzwjyiwfcocypwjishs
ecfehsdziyescqfwthwclwpgszvxwxjgvwzmzefykwgizhbijc
zeeiwsqlvyviwjxoxelejvhzjkwcvryehyewzjmewwhsajzjuz
ggmepidujviamzeavwwyzcwdyywwcfmufyzshmwwwvgipzwhbm
hnzcsclaledziwlehwswzzsizzjjdlveujvwszfsgjiaeimhvx
ejibyvtedycwfhwpcuiwczfhhfvavjdkzxsvfexsvjmwjwdumq
kzzhpzwiwxlfcvxyyczwhixwjwpzzrhfjwvwqsxwhiwzjhszee
So we know that you use a 26-letter wheel with letters arranged in a random (by a human or a computer - what method was used to pick the numbers?) order, and that the order in which the letters are read could be significant. Solvable by a computer? Eventually, yes. Solvable by a human? Most probably not.
Interesting.
You can't really prove anything.
Except if it's a mathematical proof.
Well, my only source was from a Cambridge graduate (working in the codebreaking faculty). I've not done much of my own research.
I downloaded the Jan 2010 D1 paper (which was oddly leaked less than 24 hours from the exam), and a good portion of the question paper is just checking to see if you understand the basic concepts. That is what exam papers generally do, but this is blatantly obvious, particularly in question 2;
2. Prims algorithm finds a minimum spanning tree for a connected graph.
Explain the terms
(a) connected graph,
(b) tree,
(c) spanning tree.
Hi Devante;
I have never actually seen one. What was it like to use it?
Initially difficult, but I was shown how to use it. Incredibly slow encrypting a message and it felt very 'bulky' - but it was definitely a worthwhile experience. If I remember correctly, the arrangement is not the usual "QWERTY" setup, since the German language has different frequency letters; arrangement is "QWERTZUIO" (1st row), "ASDFGHJK" (2nd row), and "PYXCVBNML" (3rd row). However, though the enigma machine was difficult to crack, and initially very successful for the Germans -- the British did eventually crack the code. They always ended their coded messages with "Heil Hitler", and, not only that, but they became rather careless with their alteration of the machine's three wheels; when you encrypt a code, you spin the wheels several times such that the order is completely different. Instead, the Germans decided to turn only the first wheel by 1 unit, meaning a pattern emerged in the enigma machine's code. They weren't particularly careful with the code configuration either; they published the new configurations monthly, which were accessible to the public -- which the British later discovered. It was the carelessness of the Germans that led to their codes being decrypted so quickly.
Good fun.
GameFAQs.
That is all.
Hmm... I think I remember talking to a Cambridge under-grad about an uncrackable cipher, and replied that it was indeed possible to do so. Can't remember how, though.
Speaking of enigma machines - I feel honoured to have used one. It's amazing what luxuries history holds.
650 = 8[sup]3[/sup] + 2[sup]7[/sup] + 2[sup]3[/sup] + 4[sup]½[/sup]