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:
39
Step-by-step explanation:
g(x)= -2x+2 and f(x)= 3x^2+4
(g+f)(-3)
g(-3) = -2(-3) +2 = 6+2 =8
f(-3) = 3 (-3)^2 +4 = 3(9)+4 = 27+4 = 31
(g+f)(-3) = g(-3) + f(-3) = 8+31 = 39
This is Bridge Algebra 2 stuff. Well you see that 2x+y=8 and x-1=y. To slove for y you have to slove for x first to find y. So 2x+x-1=8 ( the reason I did that is because there is a y in that equation and I placed it on that equation) now lets slove this equation for x. So 3x-1=8( I add 2x with x) 3x=9 ( I made -1 a positive one and add it to 8) x=3 ( I divided 9/3 to get 3). Now that you have slove x we will use x to slove for y. So 3-1=y, 2=y. Your answer is y=2.
Answer:

Step-by-step explanation:
If
, then
. It follows that
![\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5C%5C%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7D%7Bh%7D%20%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Ccdot%20%5Bg%28x%2Bh%29%20-%20g%28x%29%5D%20%5C%5C%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%2Bh%7D%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5Cend%7Baligned%7D)
Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.
