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1. Points $X$ and $Y$ lie on a circle centered at $O,$ and arc $XY$ is $80^\circ.$ The circle passing through points $O,$ $X,$ and $Y$ is drawn. Find the measure of arc $XOY$ on the smaller circle.

2. Trapezoid $WXYZ$ is inscribed in a circle, with $WX \parallel YZ$. If arc $YZ$ is $30$ degrees, arc $WZ$ is $t^2 + 7t$ degrees, and arc $XY$ is $60 - 4t$ degrees, find arc $WTX$.

it says at least 2, would that change the answer?

My answer is somewhat different.  The area  GEFC = 0.75 cm²  

the answer was 2/5, but thanks anyway!

Besides those 4 small circles, it's possible to inscribe 24 more corcles.  

Using casework, you can find that there are 16 combinations that work.

f(f(9)) = 20.

Have you used the distance formula to find the distances yourself yet?

{1, 3, 3, 3, 4} | {1, 3, 3, 4, 3} | {1, 3, 4, 3, 3} | {1, 4, 3, 3, 3} | {3, 1, 3, 3, 4} | {3, 1, 3, 4, 3} | {3, 1, 4, 3, 3} | {3, 3, 1, 3, 4} | {3, 3, 1, 4, 3} | {3, 3, 3, 1, 4} | {3, 3, 3, 4, 1} | {3, 3, 4, 1, 3} | {3, 3, 4, 3, 1} | {3, 4, 1, 3, 3} | {3, 4, 3, 1, 3} | {3, 4, 3, 3, 1} | {4, 1, 3, 3, 3} | {4, 3, 1, 3, 3} | {4, 3, 3, 1, 3} | {4, 3, 3, 3, 1} (total: 20)

The set R will contains 16 rectangles and four squares, so the fraction of square is 4/20 = 1/5.

We can use a difference table on the values of the polynomials: https://brilliant.org/wiki/method-of-differences/

So the first differences are 1, 1, 1, 1, -119.  The second differences are 0, 0, 0, -120.

If we continue, we find that g(0) = -17.

If we think of the numbers as letters, there are 5! ways to arrange them but since 3 is repeated 3 times, we divide by 3! So 5!/3! = 20.

There are C(5,3) = 10 ways to choose the position of the 3s, then C(5,1) = 5 ways to choose the position of the 4, and C(5,1) = 5 ways to choose the position of the 1, so the number of sequences is \(10 \cdot 5 \cdot 5 = \boxed{250}\)

The arrangement around the table must be of the form DCDCDCDC, where D stands for dog and C stands for cat. We can place the first dog anywhere; we then have 3! = 6 ways of placing the remaining dogs and 4! = 24 ways of placing the cats. This gives a total of 3! * 4! = 144 arrangements.

no prob :)

Thank you so much!

I think that's incorrect because of the restriction about the dogs needing to sit between two cats.

2. We seat Rick first. He has 9 choices. We seat Morty next. He has 2 choices: two spots to the left of Rick or two spots to the right of Rick. Then, the remaining seven members can sit in 7! ways. This appears to give a total of 9*2*7! seatings. However, because two seatings are considered the same if one is a rotation of the other, each distinct seating is counted 9 times in this product, once for each possible location of Rick. So, we must divide by 9 to give a total of \(2\cdot 7! = 2\cdot 5040 = \boxed{10080}\) seatings.

We might have instead started by noting that it doesn't matter where Rick sits. He starts by sitting anywhere, and we count the number of ways the members sit relative to Rick. Morty still has 2 choices: two spots to the left of Rick or two to the right of Rick. The remaining seven members then can be seated in 7! ways, for a total of \(2\cdot 7! = 2\cdot 5040 = \boxed{10080}\) seatings, as before.

1. First we decide what ranks we will use. Then we will pick suits for all of the cards.

We choose the two paired ranks in \({13\choose 2}=78\) ways and the remaining rank in \({11\choose 1}=11\) ways. Then we choose the suits for these cards in \({4\choose 2}{4\choose 2}{4\choose 1}=144\) ways. This gives a total of \(78\cdot 11\cdot 144 = \boxed{123552}\) two pair hands.

There are 4!*4! = 576 ways of arranging the cats and dogs.

what was the question?