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Comments:
<0> (1 - (-1)^k)(-1)^(k/2) / 2
<0> Er
<0> (1 - (-1)^k)(-1)^((k-1)/2) / 2
<1> woah, give me a second =)
<1> Kasadkad: did you just come up with that?
<0> Yeah
<1> Kasadkad: Care to explain it a bit? I don't quite get how that works
<0> Well, the (1 - (-1)^k)/2 is to get the sequence 1, 0, 1, 0, ...
<0> When k is even, (-1)^k = 1 so the whole thing goes away, when it's odd you get 2 in the denominator instead
<2> HI :)
<0> The (-1)^((k-1)/2) is so that every other odd term is -1
<2> Do you all enjoy answering math questions?
<0> Hi
<3> no, we are masochists
<2> lol
<2> Me too
<3> do you enjoy asking useless ones?
<2> yes, very much so
<3> then everybody's happy
<2> lol ok, enough sarcasm
<2> I am stuck on a homework problem. The math lab at my school is at the same level I am.
<3> oh ok, thanks for leting me know
<3> try asking the actual question, though background information is always great
<2> Sorry, small talk isn't acceptable
<2> Mental note..
<3> it's just not very helpful
<2> Linear programming
<2> Finite math...
<2> Well.. I will Just read the question..
<4> I've never met anyone who had such a strong faith that multiplication isn't distributive.
<2> The Hoover Steel Mill produces two grades of stainless steel, which is sold in 100-pound bars. The standar grade is 90% steel and 10% chromium by weight, and the premium grade is 80% steel and 20% chromium. The company has 80,000 prounds of steel and 12,000 pounds of chromium on hand. If the price per bar is $90 for the standard grade and $100 for the premium grade, how much of each grade should it produce to maximize revenue?
<5> I asked this question earlyer, but im still having trouble with it: compute the second componenet of a(b "dot product" c) for {a,b,c} E R^3 (where a, b, c are vectors), the problem im having is i thought that doing a dot product gives you a single answer and not something in seperate components.
<2> Ok... I know how to do everything, except how to write out an equation for the bars being 100 pounds. Because I come up with the "right" number, just way too many zeros..
<0> shroomloops: Right, the result of a dot product is a scalar; then you multiply that scalar by the vector a and get another scalar
<0> erm
<0> *multiply that scalar by the vector a and get another vector
<5> oh, ok yeah sorry
<5> good call
<3> shroomloops: what's hte operation between a and (b "dot product" c)
<3> is a a scalar?
<5> a is a vector as well, and its just multiplication, i was just confused
<1> Kasadkad: Ah, I see, thanks man.
<2> Is anyone working on my problem?
<5> but to check the answer would then be: a2b1c1 + a2b2c2 + a2b3c3
<5> all vectors
<5> ?
<6> what's the problem Regina
<2> The Hoover Steel Mill produces two grades of stainless steel, which is sold in 100-pound bars. The standar grade is 90% steel and 10% chromium by weight, and the premium grade is 80% steel and 20% chromium. The company has 80,000 prounds of steel and 12,000 pounds of chromium on hand. If the price per bar is $90 for the standard grade and $100 for the premium grade, how much of each grade should it produce to maximize revenue?
<1> Homework!
<2> I got everything down, but how to figure in the 100-pound bars.
<2> standar = standard*
<7> Ok another question
<7> I have the sum 1 --> n of 2^i
<7> Where i is the counter
<7> I need it to be 0 ---> n of something
<7> is that possible
<6> of course
<6> let j = i - 1 ?
<6> i think
<0> Then it goes from 0 to n-1 though
<6> yea
<7> Err I was wrong
<6> yea it does
<7> No that doesn't help
<7> I need the sum to be 0 --> n
<7> Can I change the internal parts?
<0> You could
<0> Why do you need the sum in that form?
<7> Because I need it to match one of the few idents I can use
<7> the bounds MUST be 0 --> n
<0> *shrug*
<0> So write sum(2^i, i from 0 to n) - 1
<7> :-/
<7> doesnt work
<0> Why not?
<7> Actually - the original problem is
<6> yea that might help
<7> Sum 1 --> n of 2^(i-1)
<6> just derive it:)
<7> I need 0 --> n of (something)
<6> it's the "best" way
<7> derive?
<6> deriving the formula never fails and takes less than 30 seconds
<6> yea
<7> Take derivative?
<6> come up with the formula yourself
<6> once you know how to do it, it's trivial
<7> That's what I am trying to figure out...
<6> well let's see
<0> Are you looking to use the geometric series formula?
<7> No
<8> integral of xcosx + sinx = xsinx + C
<8> right?
<7> Dont worry about that
<7> The bounds of the sum just must be 0 --> n
<0> Arcanimus: Yes
<8> ok thanks
<0> Cpudan80: So use sum(2^i, i from 0 to n)
<0> Then once you do whatever it is you're doing, subtract 1
<7> hrrmmm
<2> amy' did you try to solve my question?
<7> I dont think that works - let me check
<6> Sum 1 --> n of 2^(i-1), call that S_n, so S_n = 1 + 2 + 2^2 + ... + 2^n, so 2S_n = 2 + 2^2 + ... + 2^(n +1), so S_n - 2S_n = (1 - 2)S_n = 1 - 2^(n + 1), so S_n = (1 - 2^(n + 1))/-1 ...mmm, did I mess up? I don't think so
<6> Regina, no
<9> Regina, what was your questiion ?
<10> anyone know if there is a certain package necessary for the use of the \subtitle command in LaTeX? or a certain documentcl***?
<2> The Hoover Steel Mill produces two grades of stainless steel, which is sold in 100-pound bars. The standar grade is 90% steel and 10% chromium by weight, and the premium grade is 80% steel and 20% chromium. The company has 80,000 prounds of steel and 12,000 pounds of chromium on hand. If the price per bar is $90 for the standard grade and $100 for the premium grade, how much of each grade should it produce to maximize revenue?
<2> I can solve everything
<2> but I don't know how to take into account that each bar is 100 pounds
<9> Regina, so whats the problem ?
<9> Regina, its simple ,thei're like discreting the problem by saying you only have 80,000/100 bars
<2> hmm
<9> Regina, no sorry
<9> Regina, im wrong
<2> k
<6> spx2, what's new?
<9> Regina, if a bar is 100 pounds it can have either 80pounds steel and 20 poinds chrom or 90poinds steel and 10pounds chrom
<9> amy`, well i think i have some problems...medical problems,tommorow ill go to the doctor
<2> ok
<6> amy`, oh I'm sorry
<9> Regina, and now we have to make a function
<6> spx2, oh i'm sorry
<2> Yea
<6> second time I've done that in 3 days:<
<9> amy`, you havin problems too ?
<6> spx2, lack of sleep:<
<6> spx2, 4 exams in 4 days
<9> Regina, so we have this function wich gives us the money we get from a certain choice of bars
<9> amy`, likewise over here...but i have also other medical problems,im also doing regular physical exercise daily(8 stadium tours for endurance)
<2> Yea, but we don't plug in the money till we have our corners to the feasble region.
<6> oh wow
<7> arrgh
<11> yay for linear programming :o)
<7> I messed up
<2> :D
<7> My sum was Sum 1 --> n of 2^(i-1)
<7> Then I need
<7> Sum 0 --> n
<6> the answer to the first is (1 - 2^(n + 1))/-1, unless I messed up
<9> Regina, f(x,y) = x*90 + y*90 where x are the low quality bars(less chrom) and y are the high quality(more chrom)
<9> Regina, we also have a constraint
<9> Regina, namely 80% of x + 90% of y is only steel
<9> Regina, so we have that the steel ammount in each cannot be more than how much steel we have
<9> Regina, same thing for the chromium so 20% x + 10% y is not higher than the ammount of chrome
<2> got that :)
<9> forget the "%" because we are talking about pounds heh
<6> should be just 2^(n + 1) Cpudan80 right? looking at what I got for the first answer it's really (1 - 2^(n + 1))/-1 = 2^(n + 1) - 1 for the first one, and for your second sum, you are adding another term, namely, 2^0 = 1, so add 1 to my first answer and your second sum is 2^(n + 1)
<6> err
<6> nevermind
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