> Write an equation in slope-intercept form for a line that passes
through (-3,-4) and (-2, -8).

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

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.