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.
9514 1404 393
Answer:
see attached
Step-by-step explanation:
I like to use a spreadsheet for repetitive calculations. The distances are computed from the distance formula:
d = √((x2 -x1)^2 +(y2 -y1)^2)
The results are shown in the second attachment. The drawing in the first attachment has the lengths rounded to the nearest tenth.
Y
=
−
2
x
+
5
y
=
-
2
x
+
5
Use the slope-intercept form to find the slope and y-intercept.
Tap for more steps...
Slope:
−
2
-
2
y-intercept:
(
0
,
5
)
(
0
,
5
)
Any line can be graphed using two points. Select two
x
x
values, and plug them into the equation to find the corresponding
y
y
values.
Tap for more steps...
x
y
0
5
5
2
0
Answer:
257
Step-by-step explanation:
5(41-(-2))+{-7(-4-2)}
5(41+2)+{-7(-6)}
5(43)+(42)
215+42
257