Answer: There is an asymptote at x = 4 and 2
and hole at x=2
Explanation:
To find vertical asymptote simply equate denominator to zero and find the value of x which is the vertical asymptote.
and hole is the point where numerator and denominator gives equal value
So, to find hole find fractions of numerator and denominator separately the common factor is the hole.
Here after factorisation the given function
Here, (x-2) is the common factor in numerator and denominator is the hole and value of x at denominator after equating to zero is x=2 ,4
10+2h because 10 dollars is coming whether you like it or not then you have to pay 2 dollars per each hour you use it....i would just carry the kid.
Answer:
The measure of angle LMW is 
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the measure of arc MW
we know that
The inscribed angle measures half that of the arc comprising
so
![m\angle MLK=\frac{1}{2}[arc\ MW+arc\ WK]](https://tex.z-dn.net/?f=m%5Cangle%20MLK%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20MW%2Barc%5C%20WK%5D)
substitute the given values
![65\°=\frac{1}{2}[arc\ MW+68\°]](https://tex.z-dn.net/?f=65%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20MW%2B68%5C%C2%B0%5D)
![130\°=[arc\ MW+68\°]](https://tex.z-dn.net/?f=130%5C%C2%B0%3D%5Barc%5C%20MW%2B68%5C%C2%B0%5D)
step 2
Find the measure of arc LK
we know that
-----> by complete circle
substitute the given values
step 3
Find the measure of angle LMW
we know that
The inscribed angle measures half that of the arc comprising
so
![m\angle LMW=\frac{1}{2}[arc\ LK+arc\ WK]](https://tex.z-dn.net/?f=m%5Cangle%20LMW%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20LK%2Barc%5C%20WK%5D)
substitute the given values
![m\angle LMW=\frac{1}{2}[66\°+68\°]=67\°](https://tex.z-dn.net/?f=m%5Cangle%20LMW%3D%5Cfrac%7B1%7D%7B2%7D%5B66%5C%C2%B0%2B68%5C%C2%B0%5D%3D67%5C%C2%B0)
