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C30N9

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  1. ?

    Also how is doubling time calculated in general.

    1. Show previous comments  7 more
    2. Memfis

      Memfis

      let me try...

      ---

      1/x < 2

      we can see that 1/x = 2 when x = 1/2, and that increasing x will decrease 1/x, so the answer is

      x > 1/2 I guess

      or maybe you can just multiply both parts by x and then divide by 2, but dunno if you're allowed to do that??? maybe it will get tricky since x might be either positive or negative, dunno

      ---

      classic method here, multiply numerator and denominator by something so you can use the formula (x-y)*(x+y)=x^2 - y^2

      lim x→∞ (x − √ (x^2 + x − 1)) = lim x→∞ ((x^2 - x^2 - x + 1) / (x + √ (x^2 + x - 1)) = lim x→∞ ((1 - x) / (x + √ (x^2 + x − 1))) = lim x→∞ ((-1 + 1/x) / (1 + √ (1 + 1/x - 1/x^2))) = -1 / (1 + 1) = -1/2.

      ---

      x^3 - x - 1 = 0

      no idea, either I'm missing something trivial or you just have to take a lucky guess?

    3. C30N9

      C30N9

      Memfis said:

      let me try...

      ---

      1/x < 2

      we can see that 1/x = 2 when x = 1/2, and that increasing x will decrease 1/x, so the answer is

      x > 1/2 I guess

      or maybe you can just multiply both parts by x and then divide by 2, but dunno if you're allowed to do that??? maybe it will get tricky since x might be either positive or negative, dunno

      ---


      I actually solved this before (x < 0 or x > 1/2), but I wanted to double check.



      x^3 - x - 1 = 0

      no idea, either I'm missing something trivial or you just have to take a lucky guess?


      There is no "regular" way to solve this, instead there is the cubic general formula.

    4. Memfis

      Memfis

      Oh ok, in Russia they usually don't teach the cubic formula. Also lol at my fail at the first problem, let's say that was just me being too sleepy. (dunno how people can sleep in stuffy and noisy trains...)

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