A parallel line has the same gradient
First you will need to rearrange the equation 10x+2y=-2
Then once you do that please comment
If you don’t know how please comment
Answer:
Please check the explanation.
Step-by-step explanation:
Part a)
Given that the two parallel lines are crossed by a transversal line.
Given that
m∠2 = 2x + 54 and m∠6 = 6x - 11
Angle ∠2 and ∠6 are corresponding angles.
Corresponding angles are congruent.
Thus,
m∠2 = m∠6
2x + 54 = 6x - 11
flipe the equation
6x - 11 = 2x + 54
subtract 2x from both sides
6x - 2x - 11 = 2x - 2x + 54
4x - 11 = 54
adding 11 to both sides
4x - 11 + 11 = 54 + 11
4x = 65
dvide both sides by 4
4x/4 = 65/4
x = 16.2500 (round to 4 decimal places)
Part b)
We have already determined
x = 16.2500
Given
m∠2 = 2x + 54
substitute x = 16.2500 in the euation
= 2(16.2500) + 54
= 86.5°
As angle ∠2 and angle ∠1 lie on a straight line. Hence, the sum of their angles must be 180°.
i.e.
m∠1 + m∠2 = 180°
substituting m∠2 = 86.5° in the equation
m∠1 + 86.5° = 180°
subtracting 86.5° from both sides
m∠1 + 86.5° - 86.5° = 180° - 86.5°
m∠1 = 93.5°
Therefore, the measure of angle m∠1 is:
The sum of the measure of the exterior angles of any polygon is 360. x represents the measure of a single exterior angle.
360/56=x
The simplified fraction is 45/7
I believe it could be option 1
Answer:
$141.75
Step-by-step explanation:
189*.75
1-.25 b/c 25 percent off
