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/Step-by-step explanation:
Recall: SOHCAHTOA
1. Reference angle = 70°
Adjacent side = x
Hypotenuse = 6 cm
Apply CAH. Thus,
Cos 70 = adj/hyp
Cos 70 = x/6
6 × cos 70 = x
2.05 = x
x = 2.05 cm
2. Reference angle = 45°
Adjacent side = x
Hypotenuse = 1.3 m
Applying CAH, we would have the following ratio:
Cos 45 = adj/hyp
Cos 45 = x/1.3
1.3 × cos 45 = x
0.92 = x
x = 0.92 m
3. The who diagram is not shown well. Some parts are missing, however you can still solve the problem just the same way we solved problem 1 and 2.
I can’t see can you put it closer
Answer:
You find out the answer.
Step-by-step explanation:
But, the answer is one side cause three - four is one
P(x) = 2x² - 4xq(x) = x - 3
To find the answer, we plug q(x) into p(x):
p(q(x)) = 2(x - 3)² - 4(x - 3)p(q(x)) = 2(x² - 6x + 9) - 4x + 12p(q(x)) = 2x² - 12x + 18 - 4x + 12p(q(x)) = 2x² - 16x + 30
The third option is correct.
