Learning Activities this lesson we shall focus on quadrilaterals that are parallelograms, their
operties, and different theorems regarding relationships about its sides and
ngles. But first, we need to remember the Quadrilateral family tree. From this we
the relationships between
the
quadrilaterals.
Quan ateral
Foralelograr
Trapezoc
roger
Rectangle
Promous
Rhombo
Bosces
Trapezo
Sobre
Figure
Figure
Description
Rectangles
-All angles are
equal
Description
Quadrilaterais
-Four-sided
polygon and
Sum of all angles
is equal to 360°
Parallelograms
pairs of
parallel sides
Trapezoids
-1 pair of parallel
sides
Rhombus
-All sides are
equal
Squares
-All sides and
oll angles are
equal
Isosceles
Trapezoids
-a pair of
legs
congruent
Trapeziums
-No pair of
parallel
sides
-Same
properties as
quadrilaterals
Rhomboids
-Some
properties OS
parallelograms
Kite
-2 pairs of
non-
opposite
congruent
sides
PIVOT​

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

konichiwa~! my name is zalgo and i am here to answer your anime-related question today. more than likely, kirishima would kill izuku, marry mino and kiss momo. all of this would be my opinion by the way. (and i am also a weeb lol)

i hope this you! : d

stay and stay proud! " - zalgo

(and do you think you can mark me as brainliest? i'd greatly appreciate it! arigato~! )

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.

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.