Question

@e.mccormick

Answer 1

Boo.

Answer 2

1) D, yes. I like that one best too!
2) C, yep!

Answer 3

3 is right

Answer 4

Oh, sure, I scroll down to tell you and someone beat me to it! =P

Answer 5

5 is right

Answer 6

4 I don't know, I haven't taken geometry in too long

Answer 7

OK, for 4...
A(4a, 4b), B(4c, 4b) are the top because they share y values...
(0,0) and C(4d, 0) are the botom, So the ends of the midline are...
\[\left(\frac{4a+0}{2},\frac{4b+0}{2}\right), \; \left(\frac{4c+4d}{2},\frac{4b+0}{2}\right)\]
Now.... midpoint of that...

Answer 8

No idea o.o

Answer 9

\[\left(\left[\frac{4a+0}{2}+\frac{4c+4d}{2}\right]\cdot \frac{1}{2},\left[\frac{4b+0}{2}+,\frac{4b+0}{2}\right]\cdot \frac{1}{2} \right)\]

Answer 10

OK! Now to simplify... and yes, I am doing this on the fly...

Answer 11

\[\left(\left[\frac{4a+4c+4d}{2}\right]\cdot \frac{1}{2},\left[\frac{4b+4b}{2}\right]\cdot \frac{1}{2} \right)\]
\[\left(\frac{4a+4c+4d}{4},\frac{4b+4b}{4}\right)\]

Answer 12

AH HA! Now that looks like something usable.

Answer 13

Care to revise your guess?

Answer 14

YEP!

What I did is called the midpoint formula. I had to do it three times. The midpoint formula says this: To find the midpoint, take the two ends and average them. So:
\[(x_1,y_1),(x_2,y_2)\implies \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]

Answer 15

Now, if you think of the middle as half way, the over 2 part makes sense.

Answer 16

3 was right, like Peter said.

Answer 17

Yah, yah. I am with Peter on 5 too.

Answer 18

Awesome, thank you! Might have more to do in a bit if you can xD

Answer 19

I'll be here for an hour and a half

Answer 20

So all the funky points made of letters you got. Only one you had trouble with was the midpoint, or half way. It is an average problem.... and that is a pun. (Midpoint is the average of the points.)

Answer 21

I can help a bit more, but I need to head home and shower! I am still at work. Had to run some new network cable and we do not do that when people use the office.