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Thank you so much, that helps a lot.

Thank you!

There are 2 cases: one case where the player who is fine either way is not in the two-person team, or another case where that player is.

For case 1, there are \(5\cdot4=20\) ways to select the teams, since there are 5 possibilities for a spiker and 4 for a setter.

For case 2, there are 9 ways to select the team, since the player who is fine either way can pair up with any of one of the 9 remaining people.

In total, there are \(20+9=\boxed{29}\) ways.

         478 * 9 = 4302

+      478 * 30 = 14340

-------------------------------------

         478 * 39 = 18642

I think it is like $240.00

I checked again and I think that was wrong, but I don't know the right answer 

I will do part A) as an example.

general form: f(x) = Asin(x - phi) + y

A = amplitude

phi = horizontal shift

y = vertical shift

A) amplitude = | -2 | = 2

period = 2pi/x = 2pi

THANKS!!!!!!!! You helped me! GREATLY APPRECIATED!!!!

We can complete the square.

\(x^2 + 4x + 3 = (x^2 + 4x + 4) - 1\)

\(\boxed{(x+2)^2 - 1}\)

40 deg = 2pi/9 rad

30 deg = pi/6 rad

120 deg = pi/3 rad

a) shift 2pi/9 units left, 7 units down

b) shift pi/6 units left, 8 units down

c) shift pi/3 units left, 3 units up

W = Fd = (8 * 9.8) * 6 = 470.4 N.

(my physics is still a bit rusty, but this is what i can come up with for now)

Simplify:

f(x) = x^2 + 4x + 2 = (x + 2)^2 - 2

So, the top graph is the correct graph. 

f(x) = g(x) when the graphs intersect. The intersection point is (0, 2), or x = 0

The graph is skewed to the right since most of the numbers are given at the left side (0 to 3). So, the answer is A. 

11 divided by n^3 ==> \(\frac{11}{n^3}\)

Area = (big rectangle) - 2(small rectangle) 

        = (10)(7) - 2(3)(5)

        =  70 - 30

        = 40 cm^2

One BIG shape   7 x 9 = 63  

  minus the cut out   2 x 3 = 6

63 - 6 = 57 m²

Split this figure into rectangles:
Area = 2 * 7 * 3 + 3 * 5 

         = 14 * 3 + 15

         = 42 + 15

         = 57 m²

Divide shape into 3 rectangles.

2*3*2 + 5*9 = 12+45 = 57 cm²

\(\frac{1}{2} \sqrt 5 + \frac{ (\frac{9}{2} \sqrt5)}{2}\)

\(\frac{\sqrt{5}}{2} + \frac{9\sqrt{5}}{4} = \boxed{\frac{11\sqrt{5}}{4}}\)

Volume = 6²π(26) - (2.1)²π(26)

             = 26(36π - 4.41π)

             = 26(31.59π)

             = 821.34π cm³

Divide shape into 3 rectangles.

2*3*2 + 4*7 = 12 + 28 = 40 cm²

Volume of roll = (volume of big cylinder) - (volume of little cylinder)

The volume of a cylinder with radius r and height h is \(\pi r^2 h.\)

\(V = \pi (\frac{12}{2})^2(26) - \pi (\frac{4.2}{2})^2(26) = \boxed{821.34}\) cm^3.

Suppose this is interest compounded continuously.

P = principal (initial) amount = 6970400

R = annual rate = 0.8%

n = compound (continuous)  ==> n = e = 2.718

t = time = 1 year

Pe^(rt) = 7,026,386.85 

6 teams must play 5 other teams twice     in each dividsion (2) =     6 x 5 x 2 x 2 =120

    and   6 teams each must play other division 6 teams once  =       6 x 6 x 2 = 72

      this is a total of 192 games....BUT each game involves TWO teams soooo   192/2 = 96 games total

6C2 = 15 pairings. This happens twice though so it's 15 * 2 = 30 matches in one division

Then you need to do the other division, which means you have 30 * 2 = 60 games.

Then there are 6 * 6 games between the two divisions which means your total is 60 + 6 * 6 = 60 + 36 = 96 conference games

Area of the figure = 5 x 4 + 2 x (3 + 4 + 3)

                             = 20 + 2 x 10

                             = 20 + 20

                             = 40 cm²