Which of the following represent Exponential Decay? Choose all that apply. A) y=3^-x B) y=(7/8)^x C) y=(3/2)^x D) y=1.2^x

step-by-step explanation:

d. $180

step-by-step explanation:

if the dimensions of the room is 9ft by 12 ft, we have the area of the floor as,

so the area of the floor is 108 square feet, converting that into square yards we get,

the cost of carpet is $15 per square yard, therefore the cost of carpet for 12 square yards is,

hence, the cost of carpet is $180.

assuming the dimensions of the room be,

15 ft by 12 ft

so, the area of the floor of that room is,

we can convert 180 square feet into square yards, we know that,

1 sq ft = 0.111 sq yard

so, 180 square feet is,

so the area of floor is 20 square yards.

now, the cost of carpet is 15 dollars per square yard, therefore the cost of carpet for 20 square yards is,

so, it will cost $300.

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.