<h2>its D because u can </h2><h2> and add / mutiply *</h2><h2> </h2>
The function will simply get reflected about the y-axis.
Let's approach this through what we know. Since we know that the x values are mirrored, we know that the points in Quadrant I and IV will be reflected over to the negative side, Quadrants II and III, because they simply change in signs.
However, we also know that the function y-values do not change. This is because whatever the x values are don't change the range and y-values of an even function.
To be more specific, if we have an even function, we are most likely dealing with quadratics or variants/transformations of the quadratic function.
If we were to have 2, and -2, and we wanted to plug them into the equation:

, the signs do not change the y-values of the function.
Hence, we know that it ONLY gets reflected across the y-axis.
The third tree diagram is correct.
The probabilities for the first draw are 3/7 for green and 4/7 for red.
If you draw a green pear first, then your probabilities for the second pear would be 2/6 for green and 4/6 for red (since it is done without replacement, the number of pears to choose from changes).
If you draw a red pear first, then your probabilities for the second pear would be 3/6 for green and 3/6 for red.
Answer:
what
Step-by-step explanation:
sure?
That’s 12 since we calculate the sides