The following histogram shows the number of items sold at a grocery store at various prices: Histogram titled Items Sold with Price Range on the x axis and Number of Items Sold on the y axis. Bar 1 is 0 to 2 dollars and 50 cents and has a height of 2. Bar 2 is 2 dollars and 51 cents to 5 dollars and has a height of 0. Bar 3 is 5 dollars and 1 cent to 7 dollars and 50 cents and has a height of 4. Bar 4 is 7 dollars and 51 cents to 10 dollars and has a height of 2. Which of the following data sets is represented in the histogram? {0.75, 2.50, 2.51, 5.00, 5.01, 7.50, 7.51, 10.00} {2.50, 2.50, 7.50, 7.50, 7.50, 7.99, 9.25, 10.00} {2, 0, 4, 2} {0.50, 2.00, 5.01, 6.25, 7.45, 7.50, 7.89, 9.99}

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

no

step-by-step explanation:

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

are you missing something?

can i have brainliest? i need it : d