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Fibonacci’s Rabbits – breeding like…rabbits

December14

The original problem that Fibonacci investigated (in the year 1202) was about how fast rabbits could breed in ideal circumstances.

Suppose a newly-born pair of rabbits, one male, one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits. Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month from the second month on. The puzzle that Fibonacci posed was…

How many pairs will there be in one year?
rabbits

  1. At the end of the first month, they mate, but there is still one only 1 pair.
  2. At the end of the second month the female produces a new pair, so now there are 2 pairs of rabbits in the field.
  3. At the end of the third month, the original female produces a second pair, making 3 pairs in all in the field.
  4. At the end of the fourth month, the original female has produced yet another new pair, the female born two months ago produces her first pair also, making 5 pairs.
    And so on, until you get the following result, which is the Fibonacci series:
    seriesRead the full article here
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10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020

NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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