Jordan and Taylor are collecting canned foods for the homeless. They hope to collect 10 cans per day to reach their goal. They have already collected 120 cans. If the function for this problem is ƒ(c) = 10t + 120, determine how many cans Jordan and Taylor will have collected after 8 days.

none.

step-by-step explanation:

m

1

=

5

m1=5

b

=

−

1

b=-1

find the slope and y-intercept of the second equation.

tap for more

m

2

=

0

m2=0

b

=

−

6

b=-6

compare the slopes

m

m of the two equations.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

compare the two slopes in the decimal form. if the slopes are equal, then the lines are parallel. if the slopes are not equal, then the lines are not parallel.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

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.