The rule for the reflection is explained below. The correct option will be x-axis (x, y) = (x, -y).
In real life, we think of a reflection as a mirror image, like when we look at own reflection in the mirror.
This idea of reflection correlating with a mirror image is similar in math.
<h3>Reflection Over the X Axis </h3>
A reflection of a point, a line, or a figure in the X axis involved reflecting the image over the x axis to create a mirror image.
In this case, the x axis would be called the axis of reflection.
The rule for reflecting over the X axis is to negate the value of the y-coordinate of each point, but leave the x-value the same.
When point P with coordinates (5,4) is reflecting across the X axis and mapped onto point P’, the coordinates of P’ are (5,-4).
Notice that the x-coordinate for both points did not change, but the value of the y-coordinate changed from 4 to -4.
<h3>Reflection Over the Y Axis </h3>
A reflection of a point, a line, or a figure in the Y axis involved reflecting the image over the Y axis to create a mirror image.
In this case, the Y axis would be called the axis of reflection.
The rule for reflecting over the Y axis is to negate the value of the x-coordinate of each point, but leave the y-value the same.
when point P with coordinates (5,4) is reflecting across the Y axis and mapped onto point P’, the coordinates of P’ are (-5,4).
Notice that the y-coordinate for both points did not change, but the value of the x-coordinate changed from 5 to -5.
Hence the correct option is r x-axis(x, y) = (x, -y).
For more details on reflection follow the link:
brainly.com/question/4998744
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
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