A diagonal of a cube measures square root 294 cm. the diagonal of a face measures 14cm. what is the length of the cube to the nearest tenth.

  9.9 cm

Step-by-step explanation:

The space diagonal is the hypotenuse of the right triangle with legs that are the edge length and the face diagonal length. If x represents the edge length, then we have ...

  (√294)² = 14² + x²

  x² = 294 -196 = 98

  x = √98 ≈ 9.9

The edge length of the cube is about 9.9 cm.

Comment on the solution

There are (at least) two other ways you can work this problem.

The face diagonal length will be √2 times the edge length, so the edge is ...

  (14 cm)/√2 = 7√2 cm ≈ 9.9 cm

The space diagonal is √3 times the edge length, so the edge is ...

  (√294)/(√3) = √(294/3) = 7√2 ≈ 9.9 cm

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.

i believe b is the answer