PARTE 1 1. Algoritmo que sirva para llenar un formulario de contacto en donde el usuario ingresa nombre, correo y celular. Al final muestra los datos ingresados por el
usuario.
2. Desarrolla un algoritmo que te permite determinar si un número es primo.
3. Desarrolla un algoritmo que te permita encender un vehículo, considera algunas situaciones para tomar tu decisión

PARTE 2
Indicación:
Desarrolla para cada ejercicio: Psudocódigo, Diagrama de Flujo y Diagrama NS en PSeInt.

1. Algoritmo para pintar una casa.2. Una tienda hace un descuento de $10 si el total de la compra está entre $100 y $200 y hace un descuento de $20 si el total de la compra es mayor que $200. Si la compra es mejor que $100, no hay descuento. El algoritmo debe calcular el precio a pagar, basado en el valor de la compra

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

17.08

correct me if i am wrong.

where's the diagram?

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.