Hej Pomocy! Mam zadania z Matematyki a nie wiem jak je zrobić. Pomożecie? Rozszerz obie liczby tak, aby można było wykonać dzielenie pisemne (dzielnik musi być liczbą naturalną): Np.: 5,40306 : 0,009 = 5403,06 : 9 (przecinek należy przesunąć o 3 miejsca) (1) 42,6 : 0,3 = 142 (2) 4,605 : 0,05 = 92.1 (3) 57,6 : 0,09 = 640 (4) 4,97 : 0,007 = 71,000 (5) 0,00204 : 0,006 = 0,34 Zadanie 1: W którym przykładzie (po rozszerzeniu liczb) dzielna jest najmniejsza? Zadanie 2: Ile razy w zapisie dzielenia pisemnego z przykładu (3) występuje cyfra 6? Ile razy w zapisie dzielenia pisemnego z przykładu (5) występuje cyfra 2? Jeśli ktoś to wykona na jutro(do 9:00) to jest wielki!

at least two

step-by-step explanation:

answer: i feel you honey.

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.