Astore owner was buying uniforms for his employees. if each of his stores needed 79 uniforms and he had 64 stores which expression shows about how many uniforms he’d need? a.80x70 b. 70x70 c. 80x60 d.70x60

C)  The total number of  about uniforms needed is  60 x 80 uniforms.

Step-by-step explanation:

The Number of total stores the owner has = 64 stores

The number of employees uniform needed in 1 store  = 79 uniforms

Now, 64 is Rounded off to about 60.

and 79  is Rounded off to about 80.

⇒ The total number of uniforms in 60 stores = 60 x ( uniform in 1 store)

= 60 x  (80 uniforms)  

or, the expression of total uniforms needed = 60 x  (80 uniforms)  

Hence, the total number of  about uniforms needed is 60 x 80 uniforms.

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.

b

step-by-step explanation: