Richard and gloria ran a 10-kilometer race. gloria gave richard a head start by letting him begin 2.2 kilometers in front of the start line. the distances of richard and gloria past the start line as the race progressed can be represented by linear functions and are shown in the table of values and graph below. whose time was faster, and by how much? richard’s distance past the start line time (minutes) 5 10 15 20 25 distance (kilometers) gloria’s time was 0.5 minutes faster. richard’s time was 0.5 minutes faster. gloria’s time was 2.5 minutes faster. richard’s time was 2.5 minutes faster.

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.

step-by-step explanation:

is their an equation anywhere i can take a   look at it and see if i am right.

step-by-step explanation:

assume that you have 3 different side lengths.   apply the pythagorean theorem here:   square each of the two shortest sides; call them a and b, so that their squares are a² and b².   also square the longest side, calling the result c².

now, if a² + b² = c² is true, the triangle is a right triangle; otherwise it is not a right triangle.