You want to make 3 pairs of earrings that are in the shape of isosceles triangles. each earring will have a base of 2 cm. the wire to make the earrings is 50 cm long. if you want to maximize the use of the wire, which of the following would represent the length of each of the other two sides?

3.1 cm

Step-by-step explanation:

We are given that

Total pairs of earrings=3

Total number of isosceles triangles=

Length of base=2 cm

Length of wire used=50 cm

We have to find the length of each of the other two sides of isosceles triangle.

Let x be the length of each of two equal sides of isosceles triangle.

Perimeter of one  isosceles triangle=Sum of length of three sides =

Perimeter of 6 isosceles triangle=

Perimeter of 6 isosceles triangle Length of wire used

Substitute the values

Hence, the length of each of the other two sides of isosceles triangle=3.1 cm

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.

after going and makeing graphs i would go with a,

there is nothing that is matches the graph. if there was i would show you. but therefore my answer is a.

hope i

-lavira: )