-7 - (5.x) = 28? Is that what you were looking for?
Answer:
∠X in the pre-image will be equal to ∠L in the main image
Step-by-step explanation:
△LMN is the result of a reflection of △XYZ which means △LMN is the mirror image △XYZ
hence, the left of △XYZ will be equivalent to the right of △LMN and the right of △XYZ will be equivalent to the left of △LMN
Hence, ∠X in the pre-image will be equal to ∠L in the main image
20/t=v and 12/t=(v-4) solve both for t
t=20/v and t=12/(v-4) since t=t
20/v=12/(v-4) cross multiplying...
20v-80=12v
8v-80=0
8v=80
v=10mph
The outbound speed was 10mph
3 2/3 or three and two thirds