The new location is in Quadrant IV and The new location has a positive x-coordinate and a negative y-coordinate. Hence, option b and c are correct.
<h3>What are Coordinates?</h3>
- The horizontal and vertical addresses of any pixel or addressable point on a computer display screen are, respectively, represented by the x and y coordinates.
- The pixel (pixel 0) on the far left of the screen serves as the beginning point for the x coordinate, which is a predetermined distance along the horizontal axis of a display.
- The y coordinate is a specified range of pixels along a display's vertical axis, starting at the top pixel (pixel 0). The x and y coordinates work together to pinpoint any particular pixel location on the screen.
- As values relative to any beginning point on the screen or any portion of the screen, such as an image, x and y coordinates can also be given.
<u>Solution</u>:
If the point were to reflect across the y-axis, it would fall into the fourth quadrant.
Additionally, the coordinates will be positive for the x-axis and negative for the y-axis.
Learn more about similar numerical on coordinate:
brainly.com/question/7869125
#SPJ4
I hope you are trying to ask the question below:
<u>Question</u>
A point in Quadrant III is reflected across the y-axis. What is true about the new location? Check all that apply.
a. The new location is in Quadrant II.
b. The new location is in Quadrant IV.
c. The new location has a positive x-coordinate and a negative y-coordinate.
d. The new location has a negative x-coordinate and a positive y-coordinate.
e. The new location has negative x- and y-coordinates.
Answer:
-7
Step-by-step explanation:
You use the quadratic formula of 
to get:
a = 1
b = 7
c = -13
then you plug that into x= (-b ± √b²-4ac)/(2a) :
x= ( -7 ± √7²- 4(1)(-13) )/(2(1))
The sum of these two x values turns out to be:
As shown in the model below, there are three lines.
Answer:
90% chance that it could happen.
Answer:
The answer to your question is: (4, -1)
Step-by-step explanation:
M = midpoint (-1, 1)
A = (-6, 3)
B = (x, y)
x = 2xm - x1
x = 2(-1) + 6
x = -2 + 6
x = 4
y = 2ym - y1
y = 2(1) - 3
y = 2 - 3
y = -1
