CPhill

And if you want to find AC   ...

cos BDC  =  12/20  = 3/5

cos BDA  = 5/13

sin BDC = 4/5

sin BDA = 12/13

So  cos  (BDC +  BDA)  =  cosBDCcosBDA - sinBDC sinBDA =

                                            (3/5)(5/13) - (4/5)(12/13)  =

                                             15/65  -  48/65 =

                                              -33/65

So....using the  Law of Cosines

AC^2  = AD^2 + DC^2  - 2(AD * DC) cos ( BDC + BDA)

AC^2  = 5^2   +  20^2  - 2(5 * 20)(-33/65)

AC^2  =  425   + 200(33/65)

AC^2  = 6845 / 13

AC  ≈ 22.946

1 - kx  =  - 3y         sub (4, -3) into the equation

1 - k(4) = - 3(-3)     simplify

1 - 4k  =  9       subtract 1 from both sides

-4k  =  8         divide both sides by - 4

k  =  -2

10xy + 14x + 15y  = 166     simplify as 

5y ( 2x + 3)  =  166 - 14x

Since the left side is divisible by 5, then so is the right side

When  x  =  4

5y (2*4 + 3)  =  166 - 14 * 4

5y (11)  =  166 - 56

55y  = 110

y  =  2

So  x  + y  =   4 + 2  =  6

Midpoint  of  (2, 5) , (4,9)  =

[  (2 + 4) / 2, (5 + 9) / 2 ]  =   ( 3, 7)

So......  x  = 3, y  = 7

And  x +  y  =   10   =    b

Independent

P(A l B)  =  P(A)

7)The two bases of a cylinder are two parallel cross sections of a sphere. We know the radius of the sphere is 3 and the height of the cylinder is 4. Find the volume of the cylinder

I can barely make this out....but...looking at a cross-section of the cylinder ABCD, the diameter of the sphere is a "diagonal"  of the cylinder......and this is 6

The  height of the cylinder will be one leg of a right triangle and the cylinder diagonal is the hypootenuse,  so we can use the Pythagorean Theorem to find the diameter of the cylinder....so we have

_{diametercylinder} =  √ [ 6^2 - 4^2]  = √ 20  =  2√5

So...the radius,r, of the cylinder is half of this  = √5

So...the volume of the cylinder  =  pi * r^2 *height    = pi * (√5)^2 * 4  =  20 pi units^3

Here's a pic :

\(1 \le a \le 10$ and $1 \le b \le 36 \)

\(\sqrt{a + \sqrt{b}} \)

b  is  only an integer when  b  = 1, 4, 9, 16, 25 , 36

The possible pairs of (a, b)  are

(1, 9) 

(2, 4) 

(3, 36)

(4, 25)

(5, 16)

(6, 9)

(7, 4)

(8, 1)

(10, 36)

So.....nine possible ordered pairs

Here's my best attempt......

Assuming that only the ones who are strong enough to play offensive lineman only play that position...

We have 3 ways to choose these players for this position

And for the other 7, we need to choose any 3 of them to man the other positions

So..the total starting line-ups are

3 * C(7,3)  =   3 * 35   =   105

However...if  the the offensive lineman not selected for the offensive line can also play any of the other positions we have

3 ways to  choose a lineman * 9 ways to fill any of the remaining 3 positions* 8 ways to choose any remaining two positions * 7 eays to  choose the last position

C(3,1) * C(9,1) * C(8, 1) * C(7,1)  =  3 * 9 * 8 * 7  =  1512 starting lineups

No figure is shown  ...

6)The surface area of planet AoPS is 100 times that of Earth, and its volume is n times that of Earth. What is  n(Assume both planets are perfect spheres.)

Let the  surface area, S, of the Earth  be

S  = 4pi r^2     where   r is the radius of the Earth    (1)

And the surface area of AoPS  is

100S  =  4pi R^2   where R  is the radius  of AoPS   (2)

Subbing (1)  into (2)  we have

100(4pi r^2)  = 4 pi R^2

400 pi r^2  = 4 pi R^2

100r^2  = R^2

100 =  R^2 / r^2       take the square root of both sides

10  =  R / r      (1)

The volume of the Earth is  (4/3) pi r^3    

And, by (1), the radius of AoPS  is 10 times that of Earth...so...its volume is

(4/3)pi (10r)^3  =  (4/3) pi (1000)r^3 =  1000(4/3)pi r^3

So....AoPs  has  a volume of 1000 times the Earth...so....n  = 1000