Ok, so I'm not 100% sure which category these fall under, but I'm hoping I'm posting to the right one. Please correct me if I'm wrong.

The first question:

I can't draw the picture here, but there was a picture of a mystery polygon that was mostly hidden, only showing a part that looked like a quadrilateral. It gave two of the four angle measurements, said the remaining two were equal, and asked how many sides the polygon had.

Here's what I wrote in my notes about it:

The sum of the measures of a quadrilateral is 360 degrees. The shown angles = 80, leaving 280 in the quadrilateral. The remaining angles are equal, making each 140 degrees. So, if a polygon has equal angles, each measuring 140 degrees, how many sides does it have?

Now, I followed the explanation up to here (the online practice SAT explains all questions you get wrong), but after that I really didn't follow the rest. So instead of writing out the rest of what they said, I'll ask that you provide a fresh explanation here, in the hopes that I understand yours better.

....Actually, scratch that. I'll write it here because it brings up a sub-question, but please don't base your answer on the following:

"The sum of the angles of the n-gon is (n-2)x180 degrees. Therefore, 140n=180(n-2), and solving for n gives 140n=180n-360, or n=9. "

That whole entire equation just Does Not Work in my head. (;_

The second question:

I never learned the mathematical symbol of a circle with a dot in the middle in school, and I only halfway remember functions (I'm attempting to relearn them two tabs over as I take a breather to write this out), so I blatantly Do Not Understand the following question:

....*sigh*, okay, I didn't write down the question, so I will find a similar one....

This is what wikipedia has to show about this Mystery Alien Symbol Created To Confuse Wendy:

"

∘

composed with

f∘gis the function, such that (f∘g)(x) =f(g(x)).^{[8]}if f(x) := 2x, andg(x) :=x+ 3, then (f∘g)(x) = 2(x+ 3).

"

For anyone who read all that, thank you -so- much, and for anyone who can help, thank you a dozen times more!

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