will likes two brands of healthy breakfast cereal. in superfiber cereal, there are 5 grams of fiber in one cup. in fiber oats cereal, there are 4 grams of fiber in one cup. let x represent the number of cups of superfiber will ate this week and let y represent the number cups of fiber oats he ate this week. which inequality represents the situation if the cereal will ate this week contained at least 30 grams of fiber? a)5x + 4y ≥ 30b)4x + 5y ≥ 30c)4x + 5y ≤ 30

Let x  represent the number of cups of Superfiber Will ate this week and let y  represent the number cups of Fiber Oats he ate this week.

1. If in Superfiber cereal, there are 5 grams of fiber in one cup, then in x cups there are grams of fiber.

2. If in Fiber Oats cereal, there are 4 grams of fiber in one cup, then in y cups there are grams of fiber.

3. The total amount of fiber Will ate is grams.

4. This week Will ate at least 30 grams of fiber. Then

correct choice is A.

the inequality is:  a)5x + 4y ≥ 30

Step-by-step explanation:

Hi, to solve this problem we have to analyze the information given:

So, the cereal Will ate this week contained at least 30 grams of fiber. At least means more or equal ( ≤ 30 ).

If each cup of Superfiber cup has  5 grams of fiber ⇒ 5x

If each cup of Fiber Oats cup has  4 grams of fiber ⇒ 5y

Finally, the inequality is:

a)5x + 4y ≥ 30

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.