Pangolin14

Case 1: 0, 0, 2, 2 and the combinations.

There are 6 ways to arrange 0,0,2,2 and 1 way for first and second, 6 for the last two, so 36 * 6 = 216.

Case 2: 0,1,1,2 and the combinations

There are 12 ways to arrange 0,1,1,2, and 1 way for the first person, 4 ways for second, 3 ways for third, and 2 ways for fourth, so 24 * 12 = 288.

Case 3: 1,1,1,1

4 * 3  * 2 *. 1 = 24

216 + 288 + 24 = 240 + 288 = 328? I'm probably wrong, this is my best attempt.

Way better answer, thanks guest

If we draw QR, as QN is a perpendicular and midpoint, QPR is isosceles with QP=QR. This means that O is the centroid of QPR, so OR = 2*OM. However, we are not sure how to find OM, but see that we can draw a perpendicular MX from M to PN, which gives us similar triangles PMX and PQN. Next, we note that as PN = 16/2 = 8, and QN = 10, PQ = 2√41. Thus, PM = √41. 

As PM/PQ = MX/QN = PX/PN, X is the midpoint of PN, and MX = 5. Thus, PX = 4. So, as MX = 5 and XR = 16 - 4 = 12, MR = 13, so OR = 13/3 * 2 = 26/3.

it says quartic, that's why I was skeptical about that solution.

Monic indicates there is only one variable used, so we can say that $p(x) = ax^4 + bx^3 + cx^2 + dx + f,$ where $a,b,c,d,$ and $f$ are the coefficients. It appears to be that when we input $x$ as $1$, we just get $a + b + c + d + f = 3,$ and with $3$, we get $81a + 27b + 9c + 3d + f = 11,$ and with $5$ we have $625a + 125b + 25c + 5d + f = 27. $

We are trying to solve for $16a - 8b + 4c - 2d + f + 7 \cdot (1296a + 216b + 36c + 6d + f).$

See if you can simplify and/or find out this equation. Another thing you might notice is that if we have p(x) = y, then $\overline{abcdf}$ is a 5 digit number in base x, which equals y in base 10!

Ah, yes. Law of cosines.

Case 1: All green

3/5 * 3/5 * 3/5 = 27 / 125

Case 2: All red

2/5 * 2/5 * 2/5 = 8 / 125

8+ 27 = 35

7/25

radius of cylinder = 2

height = 4

16 pi

6/2 = 3

3^2 = 9

add 9 to both sides:

14 + 9 = 23

9 - 9 - 8 - 1/8 = -8 - 1/8 = -65/8