One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel
<h3><u>Solution:</u></h3>
Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.
We have to prove that the lines are parallel.
If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.
Now, the 1st angle will be 1/6 of right angle is given as:

And now, 15 degrees is 11 times smaller than the other
Then other angle = 11 times of 15 degrees

Now, sum of angles = 15 + 165 = 180 degrees.
As we expected their sum is 180 degrees. So the lines are parallel.
Hence, the given lines are parallel
Answer:
If A and B are two rational numbers such that A<B
Then X = A + (B-A)/π would be an example of an irrational number
such that A < X < B
Hope this helps
Please give me brainliest
Line a intersects both lines c and d.
Line a is a transversal to lines c and d.
Angles 10 and 14 are corresponding angles.
Since corresponding angles are congruent, lines a and c are parallel.
Line b has nothing to do with this question.
Answer is choice A.
H=-16t² +v(0)t+h(0)
v(0)=192
h=-16t² +192t+h(0)
h(0) should be 0, because the mortar sits on the ground.
h= - 16t²+192t
This function will have maximum because it has minus before x², and parabola is looking down.
h=-(16t²-192t)=-(16t² -2*4t*24+24²)+24²
h=-(4t-24)²+24²
h=-(4t-24)²+576
vertex (24 feet, 576 feet)
24 feet horizontally from the mortar and 576 feet up