Answer:(-4,0)
Step-by-step explanation:
Given
Point A (6,-4)
Reflection of Point A about line y=-2 is
A' (6,0)
If any Point is reflected about any line y=constant then its x coordinate remains same
Now reflection of Point A' about x=1 is
A'' (-4,0)
If any point is reflected about line x=constant then its y coordinate remains same
Angles formed when two parallel are intersected by a common transversal include, corresponding, alternate interior and exterior, same-side interior, and vertically opposite angles
The correct option is Same–Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°
The reason the selection is correct is as follows:
The given angles are;
m∠1 = (30·x - 30)°. m∠5 = (40·x + 70)°
∠1, and ∠5, are located on the same side of the transversal and are both formed on the interior side of the parallel lines
Therefore, they are same–side interior angles
Same side interior angles theorem states that same side interior angles formed between parallel lines are supplementary
Therefore, we have;
m∠1 + m∠5 = 180°
Which gives;
(30·x - 30)° + (40·x + 70)° = 180°
(70·x + 40)° = 180°
70·x = 180° - 40° = 140°
m∠1 = (30·x - 30)° = (30 × 2 - 30)° = 30°
m∠1 = 30°
m∠5 = (40·x + 70)° = (40 × 2 + 70)° = 150°
m∠5 = 150°
The correct option is therefore;
Same-Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°
Learn more about parallel lines cut by a transversal here: