143 / 6x+12 = 132 / 72
143 / 6x+12 = 11 / 6
crossmultiply:
66x + 132 = 858
66x = 726
x = 11
Answer:

Step-by-step explanation:
Given
Intersecting lines
Required
Find x
To do this, we make use of:
--- angle on a straight line
Collect like terms


Divide both sides by 3

Answer:
m∠BAC = 105°
m∠FAB = 75°
Step-by-step explanation:
By using the property of an exterior angle of a triangle,
Measure of an exterior angle is equal to the sum of opposite two angles of a triangle.
From the triangle given in the picture,
m∠ABC + m∠BCA + m∠CAB = 180°
(13x - 3)° = (3x + 2)° + 55°
13x - 3 = 3x + 57
13x - 3x = 57 + 3
10x = 60
x = 6
m∠FAB = (13x - 3)° = 75°
m∠ABC = (3x + 2)° = 20°
Since, ∠BAC and ∠FAB are the linear pair of angles,
m∠BAC + m∠FAB = 180°
m∠BAC + 75° = 180°
m∠BAC = 180° - 75° = 105°
The arithmetic formula is expressed as an = a1 + d *(n-1)where n is an integer. Substituting from the given a1 = -12 and a27 = 66, 66 = -12 + d *(27-1). hence , d is equal to 3. a42 thus using the formula is equal to 111. The final answer to this problem is 111.
<u>ANSWER TO PART A</u>
The given triangle has vertices 
The mapping for rotation through
counterclockwise has the mapping

Therefore



We plot all this point and connect them with straight lines.
ANSWER TO PART B
For a reflection across the y-axis we negate the x coordinates.
The mapping is

Therefore



We plot all this point and connect them with straight lines.
See graph in attachment