Francisco's credit card has an APR of 13%, calculated on the previous
monthly balance, and a minimum payment of 2%, starting the month after the
first purchase. His credit card record for the last 7 months is shown in the
table below.
End of Previous New Payment Finance Principal New
month balance charges received charges paid balance
$0.00 $5200.00 $0.00 $0.00 $0.00 $5200.00
$5200.00 $0.00 $104.00 $56 33 $47 67 $5152 33
$5152 33 $0.00 $103.05 $55.82 $4723 $5105.10
$5105.10 $0.00 $102.10 $55 31 $46 80 $5058 31
$5058 31 $0.00 $101.17 $54.80 $46 37 $5011.94
$5011.94 $0.00 $100.24 $54.30 $45.94 $4966.00
$4966.00 $0.00 $99.32 $53.80 $45.52 $4920 47
GN
OUA
How much of the $5200 charge that Francisco made during the first month
has been paid off?

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

answer:   b

step-by-step explanation:

x2 = 16

x can not be a negative number because 16 is positive

the only answer would be positive 8  

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.