In the diagram below, de is the perpendicular bisector of fg. which of the
following statements must be true?
check all that apply.
i
o
a. de is perpendicular to fg
b. fg bisects de
i
c. dhf is a right angle

The Question is Incomplete the complete question with diagram is below.

Options A, C, D and E are correct choices.

Step-by-step explanation:

We have been given that DE is the perpendicular bisector of FG. We are asked to choose the correct options about our given statement.

Let us check our given choices one by one.  

A. FH ≅ GH

We can see that at point H line DE bisects line FG, so by the definition of bisect line segment FH is congruent to line segment GH.

Therefore, option A is the correct choice.

B. DH ≅ EH

Since we have not been given that FG also bisects DE,

Therefore option B is not a correct choice.

C. Line DE is perpendicular to line segment FG.

Since we have been given that line DE is perpendicular to line segment FG,

Therefore, option C is the correct choice.

D. H is the midpoint of FG.

Since we know that bisects means divides into two equal parts,

Therefore, point H is the midpoint of FG and

option D is the correct choice.

E. ∠ DHF is a right angle.

Since perpendicular means that a line meets another line at 90 degrees, Therefore, option E is the correct choice.

F. FG bisects line DE.

Since  we have not been given that FG also bisects DE,

Therefore option F is not a correct choice.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

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.