Answer:
domain: {x ≠ 0}
Step-by-step explanation:
Answer:
The answer is 4
Step-by-step explanation:
just plug the numbers where the letters are
1+1+2= 3
I believe it’s 6.4?? Maybe?
Simple....
you have:
1.)
![-9r^{2} -16](https://tex.z-dn.net/?f=%20-9r%5E%7B2%7D%20-16)
2.)
![-45b^{2} -125](https://tex.z-dn.net/?f=%20-45b%5E%7B2%7D%20-125)
3.)
![-x^{2}+7x-18](https://tex.z-dn.net/?f=%20-x%5E%7B2%7D%2B7x-18%20)
Factoring them out completely--->>>
1.)
![-9r^{2}-16](https://tex.z-dn.net/?f=-9r%5E%7B2%7D-16%20)
Factor out a -1
![-( 9r^{2} +16)](https://tex.z-dn.net/?f=-%28%209r%5E%7B2%7D%20%2B16%29)
2.)
![-45b^{2} -125](https://tex.z-dn.net/?f=%20-45b%5E%7B2%7D%20-125)
Factoring out -5
![-5( 9b^{2} +25)](https://tex.z-dn.net/?f=-5%28%209b%5E%7B2%7D%20%2B25%29)
3.)
![- x^{2} +7x-18](https://tex.z-dn.net/?f=-%20x%5E%7B2%7D%20%2B7x-18)
Factoring out a -1
![-( x^{2} -7x+18)](https://tex.z-dn.net/?f=-%28%20x%5E%7B2%7D%20-7x%2B18%29)
Thus, your answer.
a.) The orientation of ABCD is clockwise, as is the orientation of A'B'C'D'. This means the transformation involves a even number of reflections (may be 0). The orientation of AB is North, and the orientation of A'B' is West, so a rotation of 90° CCW (or equivalent) is involved. We can find the point of intersection of the perpendicular bisectors of AA' and BB' (at (-1, -1)) to determine a suitable center of rotation.
ABCD can be transformed to A'B'C'D' by ...
- rotation 90° CCW about the point (-1, -1)
b.) Rotation by 90° can also be accomplished by reflection across a diagonal line. Since we want the orientation to remain unchanged, we need another reflection to put the figure into its final position. A suitable alternate sequence for mapping ABCD to A'B'C'D' is ...
- reflection across the line y=x
- reflection across the line x=-1