I. Cierto o Falso. Coloca en el espacio de la derecha la letra C para los enunciados ciertos y la F para los falsos. Valor 14 puntos. Un punto cada una 1. Para x+3=2 el resultado para x es 1 F

2. En el caso de expresiones racionales : el producto de es
C
3. Los productos notables son fórmulas para reducir expresiones. C
4. Decimos que una expresión racional esta simplificada si está no tiene máximo común divisor.
5. Una expresión racional puede evaluarse sólo en aquellos valores de las variables para las cuales el denominador no sea nulo.
6. Simplificar es escribir un resultado en su mínima expresión.
7. El cociente de polinomios es un polinomio.
8. Una expresión racional es una suma de términos algebraicos.
9. El numerador de un producto es el producto de numeradores.

10. El producto de

11. En todo caso x2 -2xb + b2 es (x + 2b)2
12. Para 2x+7-x=0, se tiene que x= 7
13. Un ejemplo de expresión racional es x
14. La expresión x+1=2 es un ejemplo de ecuación.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>

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.