15(t+2)+9t=6 Please help!
2 answers:
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Answer:
The correct option is O B'
Step-by-step explanation:
We have a quadrilateral with vertices A, B, C and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units away from the line, C is 7 units away from the line and D is 9 units away from the line.
Now we perform the reflection and we obtain a new quadrilateral A'B'C'D'.
In order to apply the reflection to the original quadrilateral ABCD, we perform the reflection to all of its points, particularly to its vertices.
Wherever we have a point X and a line of reflection L and we perform the reflection, the new point X' will keep its original distance from the line of reflection (this is an important concept in order to understand the exercise).
I will attach a drawing with an example.
Finally, we only have to look at the vertices and its original distances to answer the question.
The vertice that is closest to the line of reflection is B (the distance is 4 units). We answer O B'
