wolfson

Well in the Uk, we can do "work experience" thats to help build experience in a subject, i.e. we can visit ICI and do the techs job, for a period usually 1month, and if you get on well, the school/college will allow you to have say 2-days a week at the specific area.

As always anytime

As I said and explained the value of the e- is not insignificant. Show me how it would significantly alter my calculation(s). How could you learn chemistry without realizing the significance of oxidation states; Fe is not the same as Fe(III). Did i say it was NO, i said the student(s) should and do learn the significant value, of oxidation and reduction. Despite the plausible existence of teachers who are ignorant enough to underemphasize this point, this subtle detail should be pointed out Well I teach first and second year students (part of a course), and they do well, and I can be sure i am not ignorant, some teachers though are slightly under achieving. And I'm sure you would agree by now that pointing out this aspect of the problem was not in error, certainly not stupid. I already said that i didnt mean to sound angry, and if you think i didnt know the term "oxidation" you are seriously mistaken.

Im not saying they should not learn about OIL/RIG, but i am just saying that in the above calculations using the value of e- would be insignificant, i am not saying that they should not learn the concepts of OIL/RIG, most students (A-G grade) will know about OIL/RIG and the formations, but i am saying that nearly all students will not use the e- as part of there calculations.

I didnt mean to sound angry, all i am meaning is that a mass of an electron is 9.1093897e-28g, so unless i was aiming to have a value to 31d.p., but if I or anyone used a 32d.p. value then they would be marked down, as this makes no alteration to the mass 151.909g do you understand where i am coming from?

And yes, it is General Chemistry even high school chemistry where professors will use subtle problems as this to mark off the A students from the B. 1. I doubt this person is at A-Level grade, 2. It would confuse them, and three the AQA and EDE syllabus mentions nothing to that specification, that is the UK A-Level guides. and i was refering to your comment being stupid not you.

Or i could teach you trig.

You have not done trig so just imagine it as if you have a degree on the horizontal you NEED to use cosine.

Thats just stupid. At his/her level of chemistry you do not do anything of the sort.

You need to use |cos| to calculate the force, it is used as you have a angle, and the only way we can calculate 400N on a 30 degress elavation is by using cosine, (you may also need to use |sine| in other parts of physics), have you come accross trigonometry?, same concept here. So by using cos (seen an your calculator as cos), we can use Equation 1. Work = F * Cos theta EQUATION 1 thus We can find out the work done (SI unit joules) do you understand?

Work = F * Cos degrees Thus: Work = (400 N) * (30m) * (Cos 30) = 10392.30 J = 10.39 KJ

I know its a nightmare, we all make silly mistakes though.

Oh sorry didnt realise the answer was posted sorry

probability you have the illness: 0.05 prob that it comes up accurate: 0.98 prob that it comes up not accurate: 0.02 therefore, prob it says you have the illness:- 0.05 x 0.98 = 0.049 probability you don't have the illness and its says of have the illness = 0.95 x 0.02 = 0.019 therefore, probability it comes up positive= 0.019+0.049 =0.068 and the probability that you actually have the illness, given that it comes up positive:-0.049/0.068 = 0.72 get a 2nd opinion? urm...I would.

If we label the columns and rows from 0 to 7, then we have queen at position (i,j), where j = (3 + 2i) mod 8, for 0 _< < 4 _ j = (2i - 2) mod 8, for 4 _< i Hope that helps chess.zip

Good explanation bloodhound a browny point

My favourite solution (there are quite a few) has 180degrees rotational symmetry

Kinetic energy = 1/2 mv^2 m= mass v= velocity Units of Joule PE = mgh m= mass g= gravity (9.81ms^-2) h = height (metres) Unity joule Average power = workdone / time (secs) Work done = force (9.81) * distance moved. If a force F acting on a particle has accelerated the particle to a velocity v , the instantaneous power expended is: power = F * v There's a few off the top of my head.

um just a few What level of chem you at Gampin?

Have you tried "work experience" based applications.

If all sixty-four squares of the chess board are numbered consecutively from one to sixty-four, all 92 distinct solutions can be represented by 92 integer sequences, where each queen placement represents a term. See: http://en.wikipedia.org/wiki/8_queens_problem 6, 9, 21, 26, 40, 43, 55, 60. Since each queen must belong to one and only one row and column (and diagonal), and there are the same number of queens as there are rows and columns, the sum of any such sequence must be fixed. With the integer sequence representation above, the sum must be 8 + sum(1..7) + 8 * sum(1..7) = 260, the magic constant of Eight queens puzzle, for all 92 distinct solutions Thus: # Return a list of solutions to the n-queens problem on an # n-by-width board. A solved board is expressed as a list of # column positions for queens, indexed by row. # Rows and columns are indexed from zero. def n_queens(n, width): if n == 0: return [[]] # one solution, the empty list else: return add_queen(n-1, width, n_queens(n-1, width)) # Try all ways of adding a queen to a column of row new_row, returning # a list of solutions. previous_solutions must be a list of new_row-queens # solutions. def add_queen(new_row, width, previous_solutions): solutions = [] for sol in previous_solutions: # Try to place a queen on each column on row new_row. for new_col in range(width): # print 'trying', new_col, 'on row', new_row if safe_queen(new_row, new_col, sol): # No interference, so add this solution to the list. solutions.append(sol + [new_col]) return solutions # Is it safe to add a queen to sol at (new_row, new_col)? Return # true if so. sol must be a solution to the new_row-queens problem. def safe_queen(new_row, new_col, sol): # Check against each piece on each of the new_row existing rows. for row in range(new_row): if (sol[row] == new_col or # same column clash sol[row] + row == new_col + new_row or # diagonal clash sol[row] - row == new_col - new_row): # other diagonal return 0 return 1 for sol in n_queens(8, 8): print sol (APP (MATHS) 2000-2004 ©)

Thats they way

http://www.innermind.co.uk/hypnosis.html Try the above.

RunningManiac the information you needed has been put up on the site: http://masterdragon1.proboards19.com/index.cgi Look under past papers