Answer:
C.35⁰
Step-by-step explanation:
C= 180⁰-145⁰= 35⁰
The question is incomplete. Here is the complete question.
Semicircles and quarter circles are types of arc lengths. Recall that an arc is simply part of a circle. we learned about the degree measure of an ac, but they also have physical lengths.
a) Determine the arc length to the nearest tenth of an inch.
b) Explain why the following proportion would solve for the length of AC below: 
c) Solve the proportion in (b) to find the length of AC to the nearest tenth of an inch.
Note: The image in the attachment shows the arc to solve this question.
Answer: a) 9.4 in
c) x = 13.6 in
Step-by-step explanation:
a)
, where:
r is the radius of the circumference
mAB is the angle of the arc
arc length = 
arc length = 
arc length = 9.4
The arc lenght for the image is 9.4 inches.
b) An <u>arc</u> <u>length</u> is a fraction of the circumference of a circle. To determine the arc length, the ratio of the length of an arc to the circumference is equal to the ratio of the measure of the arc to 360°. So, suppose the arc length is x, for the arc in (b):


c) Resolving (b):
x = 
x = 13.6
The arc length for the image is 13.6 inches.
Answer:
I didn't understand your question
Answer:
The graph in the attached figure
Step-by-step explanation:
we have

Remember that the denominator cannot be equal to zero
so
The value of x cannot be equal to x=-2
<em>Simplify the numerator</em>
----> by difference of squares
substitute

simplify

The domain is all real numbers except the value of x=-2
The y-intercept is the point (0,-6) ---> value of y when the value of x is equal to zero)
The x-intercept is the point (2,0) ---> value of x when the value of y is equal to zero)
therefore
The graph in the attached figure