I believe that the answers is ABCD=FGHI
Answer:
ASA
Step-by-step explanation:
We can see that the triangle has at least one similar side and one similar angle. We can also see that the triangles meet at a point, so they must have the same angle at that point where they connect. So that leads us to our answer being ASA since they have two similar angles and one similar side. This is the answer that I think is correct just by looking at the image. I haven't take this class since last year, so this is simply from memory. Correct me if I am wrong!
Answer: 5
Step-by-step explanation:
I’m pretty sure it’d be y = -x - 2
lmk if it’s wrong.
Given:
The height, in feet, of each rocket at t seconds after launch is given by the polynomial equations:
Rocket A: ![-15t^2+100t](https://tex.z-dn.net/?f=-15t%5E2%2B100t)
Rocket B: ![-14t^2+85t+3](https://tex.z-dn.net/?f=-14t%5E2%2B85t%2B3)
To find:
The equation to find the "difference" in height of Rocket A and Rocket B.
Solution:
The difference in height of Rocket A and Rocket B is:
Difference = Height of Rocket A - Height of Rocket B
![\text{Difference}=(-15t^2+100t)-(-14t^2+85t+3)](https://tex.z-dn.net/?f=%5Ctext%7BDifference%7D%3D%28-15t%5E2%2B100t%29-%28-14t%5E2%2B85t%2B3%29)
![\text{Difference}=-15t^2+100t+14t^2-85t-3](https://tex.z-dn.net/?f=%5Ctext%7BDifference%7D%3D-15t%5E2%2B100t%2B14t%5E2-85t-3)
![\text{Difference}=(-15t^2+14t^2)+(100t-85t)-3](https://tex.z-dn.net/?f=%5Ctext%7BDifference%7D%3D%28-15t%5E2%2B14t%5E2%29%2B%28100t-85t%29-3)
![\text{Difference}=-t^2+15t-3](https://tex.z-dn.net/?f=%5Ctext%7BDifference%7D%3D-t%5E2%2B15t-3)
Therefore, the difference in height of Rocket A and Rocket B is
.