Answer:
See attachment
Step-by-step explanation:
The given triangle has coordinates (-3,3),(1,4), and (2,2).
If we dilate with a scale factor of 2 centered at (-3, 1), then we use the rule:
![(2x,2y)\to(x'-3,y'+1)](https://tex.z-dn.net/?f=%282x%2C2y%29%5Cto%28x%27-3%2Cy%27%2B1%29)
To find the image of (-3,3), we substitute x=-3, and y=3
![(2*-3,2*3)\to(x'-3,y'+1)](https://tex.z-dn.net/?f=%282%2A-3%2C2%2A3%29%5Cto%28x%27-3%2Cy%27%2B1%29)
This implies that:
and ![y'+1=6](https://tex.z-dn.net/?f=y%27%2B1%3D6)
This implies that: ![x'=-6+3=-3, and\:y'=6-1=5](https://tex.z-dn.net/?f=x%27%3D-6%2B3%3D-3%2C%20and%5C%3Ay%27%3D6-1%3D5)
Hence the image of (-3,3) is (-3,5)
We repeat the same process to obtain:
![(1,4)\to(5,7)](https://tex.z-dn.net/?f=%281%2C4%29%5Cto%285%2C7%29)
![(2,2)\to (7,3)](https://tex.z-dn.net/?f=%282%2C2%29%5Cto%20%287%2C3%29)
We now plot the image points and join the vertices with a straight edge.
See attachment for graph
Answer:
b
Step-by-step explanation:
got it right
Answer:
The bottom table.
Step-by-step explanation:
It's the only one that shows options for heads and tails for every number.