The graph shows the solution to which system of inequalities? a) y > -3x + 3 and y > 2x b) y ≤ -3x + 3 and y ≤ 2x c) y ≤ -3x + 3 and y ≥ 2x d) y ≥ -3x + 3 and y ≥ 2x

The graph shows the solution to the system inequalities

y ≥ -3x + 3 and y ≥ 2x ⇒ answer D

Step-by-step explanation:

* Lets study the graph to solve the question

- There are two solid lines, that means the sign of the inequalities

  are ≤ or ≥

- One of the two line passes through the origin and the other

  intersect the y-axis

- The shaded region is over the two lines, then the signs of the

  inequalities are ≥ (greater than or equal)

* Lets make the equations of the two lines

∵ The form of the equation is y = mx + c, where m is the slope of the

   line and c is the y-intercept

- The rule of the slope of the line wich passes through points (x1 , y1)

  and (x2 , y2) is m = (y2 - y1)/(x2 - x1)

- Y-intercept means the line intersect the y-axis at point (0 , c)

* Now lets find from the graph two points on each line to make the

 equations of the lines

# The line which passes through the origin

∵ Points (0 , 0) and (5 , 10) are on the line

- Let (0 , 0) is (x1 , y1) and (5 , 10) is (x2 , y2)

∴ m = (10 - 0)/(5 - 0) = 10/5 = 2

∴ The equation of the line is y = 2x + c

∵ The line passes through the origin

∴ c = 0

∴ The equation of the line is y = 2x

- The shaded is over the line

∴ The inequality is y ≥ 2x

# The line which intersect the y-axis at point (0 , 3)

∵ Points (0 , 3) and (1 , 0) are on the line

- Let (0 , 3) is (x1 , y1) and (1 , 0) is (x2 , y2)

∴ m = (0 - 3)/(1 - 0) = -3/1 = -3

∴ The equation of the line is y = -3x + c

∵ The line passes through the point (0 , 3)

∴ c = 3

∴ The equation of the line is y = -3x + 3

- The shaded is over the line

∴ The inequality is y ≥ -3x + 3

∴ The graph shows the solution to the system inequalities :

   y ≥ -3x + 3 and y ≥ 2x

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the correct answers i think are b. and d.

give me brainliest if it is correct!