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The I's that are left over.

If the word MISS is to appear then you can't really have MSIS or SMIS so treat [MISS] as a single unit in itself (no arrangements required... if you switch the S's nothing has changed). You're then left with 6 letters, 2 of which are identical. You can treat [MISS] as another individual letter...

You would only need to divide by 2! for the S's if you first multiplied by 2! to arrange them :p.

0101010001101000011000010111010000100000011100110110100001101001011101000010000001101001011100110010000001101000011010010110110001100001011100100110100101101111011101010111001100101110

Yup, that's pretty much how I solved it except I made ln(1/tanx) = -ln(tanx) (leading to the same conclusion). I had a look at the graph in graphmatica and zero would look to be the rigth answer.

Let the roots be a, b and c where a + b = 0. Using the sum of roots a + b + c = 3, ∴ c = 3 is a root. This means that (x-3) is a factor so you can divide by inspection or using polynomial division. x³ - 3x² - 4x +12 = 0 (x - 3)(x² - 4) = 0 (x - 3)(x +...

I thought that you could say Δ ABC ||| Δ DEF (AAA) i.e. triangle ABC is similar to triangle DEF because they have three angles equal. (or Δ ABC ||| Δ DEF (sides in equal ratio) ... once you've proven that that is the case)

ln[tan(π/2 -x)] = ln(cotx) as x--> π/2 , cotx ---> 0 ∴ ln[tan(π/2 -x)] ---> -∞ I think you need to use the integral properties for this one because the question hinges around the symmetry (I think) the graphs have around x=π/4. You can't really say "ln[tan(π/2...

I'd be careful with that because if you have F(x)/G(x) where F(x) --> ∞ and G(x) --> ∞ when x-->k then that doesn't necesarily mean that F(k)/G(k) = 1. e.g. as x-->0 , [1/x²]/[1/(2x²+x)] ≠ 1 You could bring something outside of the syllubus like...

If you do what antwan2bu suggested you'll get the equation of the circle x² + y² = 4 (or half of it at least). Since you have two equations containing relations of z you actually get a specific value (though this value does all on the circle). If you were just given arg...

That's what I figured. '1' by common sense.

Muons get created in the atmosphere when Pi mesons decay. The pi mesons themselves are formed by interactions involving cosmic rays in the upper atmosphere (pi mesons being a form of meson which are two quark particles). Muons are like heavy electrons.

P(x) = x⁵ - ax + 1 (I'm going to assume that a is positive) P'(x) = 5x⁴ - a, so x = ±(a/5)^1/4 when P'(x) = 0 Here I used a bit of BS intuition... it only has 2 real inflection points indicating that it has a cubic shape where P(x) --> ∞ as x-->...

My reasoning may be shaky but in the first ensemble you have 5 cellist options, 5 pianist options, 5 violist options and 5 violinist options. For the next you have 4x4x4x4 then 3x3x3x3 and so on giving: 5⁴ + 4⁴ + 3⁴ + 2⁴ + 1 = 979 (i'm not sure...

I might be wrong in this but using what they give you: ∫ ln(tanx) dx = ∫ ln(tan{π/2 -x}) dx (both integrals between 0 and π/2) ∫ ln(tanx) dx = ∫ ln(1/tanx) dx ... (since tan(π/2 - x) = 1/tanx) ∫ ln(tanx) dx = - ∫ ln(tanx) dx ... ( 'cause ln(1/tanx) =...

If Z = x + iy then Z(bar) = x - iy If Z = Z(bar) then im(Z) = y = 0, ∴ the locus of Z is y = 0

Hey, Nice work :D. I think I've got the solutions to that paper somewhere under piles and piles of other papers but post up the questions anyway and if people can't be bothered answering them I'll see if I can be bothered scanning the relevant parts of the solutions... that is, if I can even...

If you expand using the binomial theorem then: e = 1 + ⁿC₁(1/n) + ⁿC₂(1/n²) + ... = 1 + n/(1!n) + [n(n-1)]/2!n² + [n(n-1)(n-3)]/3!n³ + ... + [nⁿ .... ]/n!nⁿ as n --> ∞ If you take a...

A necklace is ussually circular so they're probably ditching the permutations which are identical via rotation. Is the answer anything like 15!/(8!7!) = 6,435 ?

Yup, that's exactly it.