Harper is working two summer jobs, making $19 per hour lifeguarding and making $8 per hour walking dogs. In a given week, she can work at most 7 total hours and must earn a minimum of $80. If x represents the number of hours lifeguarding and y represents the number of hours walking dogs, write and solve a system of inequalities graphically and determine one possible solution.

1 1/4 hours

step-by-step explanation:

2/4 + 2/4 + 1/4 = 5/4 or 1 1/4

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.

step-by-step explanation:

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00 and therefore the awnser is c.