HELP YOU ONLY NEED TO DO LIKE THREE QUESTIONS questions at bottom. Figures and cross section

Using a cereal box, find the surface area and the box's volume. The dimensions are 8 inches wide by 12 ⅛ inches tall by one ¾ inch deep. Parallel to the base, it would be a rectangle 8 inches x 1 ¾ inches.

Surface area = 2x(12 ⅛ x8) + 2 x (8 x 1 ¾) =2 x ( 12 ⅛ x 1 3/4 ) = 194 + 28 + 42 7/16 = 264 7/16 square inches.

Surface area if scaled up by a factor of 10: 10^2 = 100. 26,443 ¾ square inches.

Volume = 8 x 12 ⅛ x 1 ¾ = 169 ¾ cubic inches.

Volume scales down by 1/10
⅓^10 =3 ⅓
⅓ = 100/3 divided by 3 = 3/100

Please give measurements for Length = Width = Height =
of the cereal box. Describe the cross section of the box

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

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.

1 ?

step-by-step explanation:

u said 1+1+1+1+1+1+1+1+1+1+1+1*0 so ur adding 1 but then multiplying by 0 so its 0 but then u said 1*0+1 so now ur adding 1 so the answer is juan ?

~leah