Ireally need a geniuswill give the points will be good too.

Hey there!

a) The amount of fencing needed to completely surround the football field is also known as the perimeter. In order to find the perimeter, you can multiply both the length and width by two, because in a rectangle there are four sides.

F = 2w + 2l

b) If the length of the football field is twice its width, that would mean that the length is the width multiplied by two.

l = 2w

c) If w = 12 and l = 20, we would need to use those values and plug them back into our original equation to find the perimeter, or F.

F = 2(12) + 2(20)

F = 24 + 40

F = 64 feet

d) If we have 120 feet of fencing, then that would be our F value in the original equation. And then we would also plug in 30 for our l value. Then, we would solve for the width as normal.

120 = 2w + 2(30)

120 = 2w + 60

2w = 60

w = 30 feet

Hope this helps!! :)

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.