1. angles 7 and 3, angles 2 and 6
2. angle 1
3. they’re alternate exterior angles
4. they’re consecutive interior angles
5. angle 2
6. angle 4
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.
Option A
Must click thanks and mark brainliest
Sorry if i am wrong
Alright, buckle in because this will be a bit of a bore to listen to.
First, find the areas of all surfaces that you have the length and width of, which i assume you already have.
Next, use the Pythagorean theorem to find the missing measure of that triangle to find the measure of it and the back. make sure to find ALL of the sides, including the ones on the backside of the figure. Too lazy to do it myself, but at least you know how i'd do it.
