Answer:
- The sequence of transformations that maps triangle XYZ onto triangle X"Y"Z" is <u>translation 5 units to the left, followed by translation 1 unit down, and relfection accross the x-axis</u>.
Explanation:
By inspection (watching the figure), you can tell that to transform the triangle XY onto triangle X"Y"Z", you must slide the former 5 units to the left, 1 unit down, and, finally, reflect it across the x-axys.
You can check that analitically
Departing from the triangle: XYZ
- <u>Translation 5 units to the left</u>: (x,y) → (x - 5, y)
- Vertex X: (-6,2) → (-6 - 5, 2) = (-11,2)
- Vertex Y: (-4, 7) → (-4 - 5, 7) = (-9,7)
- Vertex Z: (-2, 2) → (-2 -5, 2) = (-7, 2)
- <u>Translation 1 unit down</u>: (x,y) → (x, y-1)
- (-11,2) → (-11, 2 - 1) = (-11, 1)
- (-9,7) → (-9, 7 - 1) = (-9, 6)
- (-7, 2) → (-7, 2 - 1) = (-7, 1)
- <u>Reflextion accross the x-axis</u>: (x,y) → (x, -y)
- (-11, 1) → (-11, -1), which are the coordinates of vertex X"
- (-9, 6) → (-9, -6), which are the coordinates of vertex Y""
- (-7, 1) → (-7, -1), which are the coordinates of vertex Z"
Thus, in conclusion, it is proved that the sequence of transformations that maps triangle XYZ onto triangle X"Y"Z" is translation 5 units to the left, followed by translation 1 unit down, and relfection accross the x-axis.
Answer:
im fairly certain the answer is C
Step-by-step explanation:
you take the base problem, 9-4n and plug in a number for n each time. 9-4(1), 9-4(2), 9-4(3)…
Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula.
x=2/3, 0
Step-by-step explanation:
121630....................................
Answer:
6
Step-by-step explanation:
