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
8.50x+100=308.25
subtract 100 from both sides
8.5x = 208.25
divide 8.5 from both sides
x= 24.5 hours worked
Answer:
the first answer or a
Step-by-step explanation:
just believe meeeee
Answer:
W = 15 ft. and L = 30 ft.
Step-by-step explanation:
Perimeter = 90 ft.
Twice as long as it is wide: L=2W
P = 2(L + W) = 2(2W + W) = 6W
90 = 6W
W = 15 ft. and L = 30 ft.
Answer: See below
Step-by-step explanation:
27. -(a-3)
28. (b-1)(b+3)
29. (c+4)(c+5)
30. d(d+5)
31. -(3/4)(2e-5)
Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.
For example: In problem 30:
(5d-20)/(d^2+d-20) * [??]/20d = 1/4
Factor:
<u>(5(d-4))</u> <u>d(d+5)</u> = 1/4
(d-4)(d+5<u>)</u> 20d
The (d-4), d+5, and d terms cancel, leaving
5/20 = 1/4