(WILL MARK BRAINLIEST) 16 POINTS! GRAPHING LINEAR EQUATIONS. (picture below)
the map below shows certain locations in town. Use the map and the box of equations at the right to answer the questions below.

Step-by-step explanation:

Gas Station (2,7) -> y = x + 6 if you plug in the numbers it becomes 7 = 1 + 6

Grocery Store (6,3) -> x + 6y = 24 plug in the numbers it becomes 6 + 18 = 24

Bowling Ally (1, -3) -> 6x + 3y = -3 plug in numbers 6 -9 = -3

Movies (7, -7) -> y = -7 because the y coordinates for the movies = -7

Park(-2, -4) -> -6x + 2y = 4 plug in the numbers 12 - 8 = 4

Coffee Shop (-6, -8) -> -10x + 5y = 20 plug in the number 60 - 40 = 20

Gym (-8, -3)-> y = - x - 7 plug in the numbers -3 = 4-7

School (-6, 6) -> y = - x +4 plug in the numbers 6 = 2 + 4

6x + 3y = -3 quadrant II is (-x, y)

rearrange the equation:

3y = -6x - 3

y = -2x - 1

if x is -1,  y will be 2 -1 = 1. So another solution for this equation is (-1, 1).

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

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.