Answer:
See Explanation
Step-by-step explanation:
Look into this site called
math-papa It is a really helpful calculator with a capability of basically anything math related (see screen shot)
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
we know that
The measurement of the external angle is the semi-difference of the arcs it comprises.
so
Step 
<u>Find the measure of the arc AJ</u>
m∠BDE=![\frac{1}{2} *[measure\ arc\ AJ-measure\ arc\ BE]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-measure%5C%20arc%5C%20BE%5D)
in this problem we have
m∠BDE=
substitute in the formula
=![\frac{1}{2} *[measure\ arc\ AJ-38\°]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-38%5C%C2%B0%5D)
=![[measure\ arc\ AJ-38\°]](https://tex.z-dn.net/?f=%5Bmeasure%5C%20arc%5C%20AJ-38%5C%C2%B0%5D)
Step 
<u>Find the measure of the arc FH</u>
m∠FGH=![\frac{1}{2} *[measure\ arc\ AJ-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-measure%5C%20arc%5C%20FH%5D)
in this problem we have
m∠FGH=
substitute in the formula
=![\frac{1}{2} *[112\°-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5B112%5C%C2%B0-measure%5C%20arc%5C%20FH%5D)
=![[112\°-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5B112%5C%C2%B0-measure%5C%20arc%5C%20FH%5D)
therefore
<u>the answer is</u>
the measure of the arc FH is 
7/10 in its simplest form is 0.7
Answer:
140°
Step-by-step explanation:
Supplementary angles sum to 180° , thus
angle = 180° - 40° = 140° ← other angle
