Complete the coordinate proof of the theorem. Given: A B C D is a square. Prove: The diagonals of A B C D are perpendicular. Art: A square A B C D is graphed on a coordinate plane. The horizontal x-axis and vertical y-axis are solid. The vertex labeled as A lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as B lies on begin ordered pair a comma 0 end ordered pair. The vertices C and D are unlabeled. Diagonal A C and B D are drawn by dotted lines.

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The coordinates of square ABCD are A(0, 0), B(a, 0), C(_, a), and D(0, _).

The slope of AC___ , when simplified, is equal to
.

The slope of BD___, when simplified, is equal to −1 .

The product of the slopes is equal to
.

Therefore, AC¯¯¯¯¯ is perpendicular to BD¯¯¯¯¯.

step-by-step explanation:

so first find out how much each is worth.

quarter= 25 cents

dime= 10 cents

pennies= 1 cent

then multiply each by how much each is worth.

200                     150                   300

x                         x                     x   1

  25                       10              

                                  300

1000                     1500

+                          

4000                                      

5000                                  

then count 2 from the back and add a period.

$50.00

$15.00

$3.00

and then add them

  $50.00

+

  $15.00

+

    $3.00

$68.00

so sunghee has $68.00

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a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58

sure i'm up for it

step-by-step explanation:

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.