15/40 Marks 41%
a) Find, in terms of a and c, the vector MN.
Note: Ignore arrows above vectors eg write MN, not MN
ABCD is a parallelogram.
Mis the midpoint of CB.
N is the midpoint of AB.
DA = a
A
N
DC = C
+
B
b) Show that CA is parallel to MN.
1
+
a
Ем
D
С
C

pentagon

step-by-step explanation:

the coordinates   a(5, -2), b(3,-1),c(-4,-4) d(-3,8) and e(-1,4) forms the vertices of the polygon when they are connected in order.

in order to classify that which polygon is it with the given coordinates, first count the number of vertices and then count the number of sides it can make with these vertices.

since, the given coordinates have five vertices, therefore, it will make five sides. moreover, pentagon is regular in nature therefore it will be convex.

now, a polygon with five sides is known as pentagon, therefore, the given polygon is the pentagon.

answer:

a=9

step-by-step explanation: lmk if you need a 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.