Select the correct answer from each drop-down menu. The length of side PQ is units. The length of side QR is about units. The perimeter of the parallelogram PQRS is about units. Reset Next

The length of side PQ is 2 units.

The length of side QR is 2.24 units.

The perimeter of the parallelogram PQRS is about 8.48 units.

Step-by-step explanation:

The given figure is as follows :

To find - The length of side PQ is ....... units.

              The length of side QR is about ......... units.

               The perimeter of the parallelogram PQRS is ........

Solution -

From the figure , we can see that

P = (-5, 3)

Q = (-3, 3)

R = (-4, 1)

S = (-6, 1)

Now,

By distance formula, we get

PQ = √(3 - 3)² + (-3 + 5)²

     = √0 + 2²

     = √4

     = 2

∴ we get

The length of side PQ is 2 units.

Now,

QR = √(1 - 3)² + (-4 + 3)²

     = √(-2)² + (-1)²

     = √4 + 1

     = √5

     = 2.236

∴ we get

The length of side QR is 2.24 units.

Now,

We know that,

Perimeter of Parallelogram = 2(a + b) where a is the base and b is the side

Here Given that,

a = 2

b = 2.24

So,

Perimeter of parallelogram PQRS = 2(2 + 2.24)

                                                         = 2(4.24)

                                                         = 8.48

∴ we get

The perimeter of the parallelogram PQRS is about 8.48 units.

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

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.