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a)  We can choose any 2 of the 12  = C(12,2)  = 66 ways

b) Mark is a given....choose any 1 of the remaining 11 =   11 ways

c)  Mark is  out......choose any 2 of the  other 11  =  C(11,2)   = 55 ways

The points  (-4,0) and (0,2)  are on the graph

Slope =  (2 - 0)  / ( 0  - - 4 ) =   2/4  = 1/2

Using (0,2)  we can write the equation ofthe  line as

y = (1/2)x + 2     multiply throygh by 2

2y  = x + 4

x - 2y  =   -4

Find all \(w\) so that neither \(-2(6+2w)\le -16 + 7w\) nor \(-3w\ge 18 - 5w + 3\).

4 <= 11w --> w >= 4/11 (8/22)

2w >= 21 --> w >= 21/2 (231/22)

all w with w >= 4/11 satisfy these equations

so the answer is w < 4/11 

correction: case 1: -5 < x

This is AoPS homework.  Do not post AoPS homework.

(x - 5)(x + 5)(x - 3) > 0

(x^2 - 25)(x - 3) > 0 | : (x - 3)

> case 1: if x > 3 (this case as we search the smallest integer)

x^2 - 25 > 0 | + 25

x^2 > 25 | sqrt

x > 5 and -5 > x

=> smallest integer: -4

(> case 2: if x < 3 (jst ignore; opposite of case 1)

x^2 - 25 < 0 | + 25

x^2 < 25 | sqrt

-5 < x < 5)

Thank you :_)

Here :

https://web2.0calc.com/questions/geometry_55509

Note

w^2 + l^2 + h^2 = the length we want

Sum of edges  = 

4( w + h + l)  =  44

w + h + l =  11      squaring both sides produces

(w^2 + l^2 + h^2) + (2 wh + 2 wl +  2 hl)  = 121

(w^2 + h^2 + l^2) + 2 (wh + wl + hl)   =121       (1)

Surface Area  =

2 (wh + wl + hl)  = 30               (2)

Sub (2)  into (1)

(w^2 + l^2 + h^2)  +  30  =  121

w^2 + l^2 + h^2 =  91

The horizontal distance from the center of one of the faces of the tetrahedron to the edge of the tetrahedron = (1/3)s, where s is the side length of the base of the tetrahedron

The  length of one  side  of the base of EFGH =   sqrt  [ (1/3 * s)^2  + (1/3 *s)^2 ] = 

sqrt [ (2/9) s^2 ] =   (sqrt (2) / 3)   s

So the area of the base of of EFGH = (2/9)s^2

And the height of the pyramid = (1/3) height of the  of the tetrahedron =  (1/3) h

Volume of tetrahedron  = (1/3)s^2 * h

18= (1/3) s^2 * h

s^2 * h  =  54

Volume of  pyramid  EFGH =  

(1/3) (2/9)s^2 * (1/3) h  = 

(2/81) s^2 * h  =

(2/81) (54)  =  2 (54 / 81)  =  2 (2/3) =   4/3

For the all the units digits under 9, it just won't work. Even if the first digit is divisible by 7, adding one to the units digits will only change it by 1 value, which doesn't make it divisable by seven. If you know modular arithmetic, the first number would be 0 mod 7, and the second 1 mod 7. For the value 9, it gets slightly more tricky. However, if you try a number like 759, adding 1 will give you 760, or -1 mod 7. Adding another 9, like 5599 (it there must be an even amount of 5 and 9s - can you see why?) pulled up to 5600 will give you 11. 555999 gives you 556000, or 16 (doesn't work!). The next one (55559999) is much too big. From now, we can consider numbers that end with 0. For the number 770 (769), it doesn't work. Same for for 7770, 77770, 777770, etc. 700 doesn't work, and 7000 doesn't. But 70000 works! 70000 and 69999 works! Sadly, afterwards no other number works. So your answer is one. (BTW, I'm not very sure if this is correct, because I somewhat brute forced this.)

Also, just a bonus thing I learned, the sum of the digits (if it has a nine - otherwise it doesn't work) has to be divsible by 21 (7*3) - idk who asked though

x^3 - ax^2 - b^2 x + ab^2  =

x^2 (x - a)  - b^2 ( x - a)  =

(x - a)  ( x^2 - b^2)  =

(x - a) (x + b) ( x - b)

(x- y)^2  - 3 ( y - x)  = 

(x - y)^2 + 3 ( x - y)   =

(x - y)  [  ( x + y) + 3  ]

I would like to draw a diagram for you, but I was too lazy to learn Asymptote 

But I drew a lil' diagram on my paper, and found that 3 adjecent side are the \(10\) sides. Honestly, I don't have a proof for this, but I'm sure that this means the "rectangular" prism is actually a cube. That means that the side lengths are \(\sqrt{10}\), or the remainder is \(10\sqrt{10}\). 

This person isn't very good at combinatorics, but I guess I will give this a try? 

Let's call the American skaters \(a\), the Canadian skaters \(c\), and the other skaters \(o\). 

(Quick note, here are the numbers for \(a,c,\) and \(o \).)

\(a=2\)

\(c=5\)

\(o=12\)

Than, in the top 3 skaters, we want a standing that looks something like this: \((a_1,o_6,c_2)\)

(for this problem, I'm assuming that the Americans or Canadians are different, let me know otherwise...)

We can think just think of one case, since we will be multiplying the probablities, and multiplication is commutative.

(BTW, let me know if there is AT LEAST ONE in the problem, or just one American/Canadian)

I'm just gonna take the case \((a,c,o)\). For the American, there are 2 choices: \(a_1 \) and \(a_2\). For the Canadians, likewise, there are 5 choices. And for the other 12 people, there are 12 choices. We can simply multiply these 3 numbers together and get \(2*5*12 = 120\). We can multiply this number by 6, since there are \(3! \) ways to arrange the standings, so we get \(120*6 = \boxed{720}\)

IF  2x/3    =   3     we know the value of the function is  '4'     <===== given 

    so     2x/3 = 3 

               x = 9/2 = 4.5 

for this value of 'x'   we know that  f(x) = 4     now     add '5' to the function and the value becomes  9    then divide by 11  = 9/11

   so a point on the new graph would be    (4.5, 9/11)

sin A = 1/3

sin A   =  BD  / AB

BD / AB =  1/3

BD / 24 = 1/3

BD = (1/3)(24)  = 8

sin C = 1/2

arcsin (1/2)  = 30°  = C

BD =  8

So since triangle  CBD   = 30 -60 - 90

Then BC = 2 BD  =  2  *  8   = 16

The triangle is  isosceles....let the base = (sqrt 6   - sqrt (3)

cos ( base angle)  = ( [ sqrt  6 -sqrt 3  ] / 2 )  /  2 =    [sqrt 6 - sqrt 3 ]  / 4

arccos ( [ sqrt 6 -sqrt 3 ] / 4 )  ≈ 79.7° 

Apex angle =   180 - 2(79.7)   ≈  20.6°

Angles are    20.6° , 79.7° , 79.7°

Slope =  [ b^2   - a^2  ] /  [  b - a ]  =   [ (b -a) (b + a) ] / (b -a) =  b + a  =  2

                           A

B                    32                               C

tan B   =  3/5                                tan C = 1 / 4

sin B =  3 / sqrt (34)                     sin C =  1 /sqrt (17)  =   sqrt (2) / sqrt (34)

Equating Areas

(1/2) AB  (32) [ (3)/sqrt (34) ]   = (1/2) AC (32) [sqrt (2) / sqrt (34) ]

3AB =  sqrt (2) AC

AC =  (3/sqrt 2) AB

Law of Cosines

AC^2  = AB^2 +BC^2  - 2 (AB *BC) cos B

(9/2)AB^2  = AB^2 + 32^2  - 2 (AB * 32) (3/ sqrt (34))           Let AB  = x

(7/2)x^2  =  32^2  -  (192 / sqrt 34)x

(7/2)x^2  + (192/sqrt (34)) x  -  1024 = 0

Solving this for x ≈  13.036  = AB

Area  =    (1/2) (13.036) (32) * [ 3 /sqrt (34) ] ≈  107.3

x^3 + x^2 -3x + 5 = x^3 + 2x^2

x^2 + 3x -  5   = 0

x^2 + 3x  =  5            complete the square on x

x^2 + 3x + 9/4  = 5 + 9/4

(x + 3/2)^2  = 29/4           take  both roots

x + 3/2 =  sqrt (29/4)                       x + (3/2)  = -sqrt (29/4)

x = [sqrt (29 - 3 ] / 2                        x  =   [ -sqrt (29) - 3 ] / 2

y = [-sqrt (29) - 3 ] / 2                       y =    [sqrt (29) -3 ]  / 2

(x,y)  = ( [sqrt (29 -3 ]  /2 , [-sqrt (29)-3 ] /2 ,  [-sqrt (29) -3 ] / 2  , [ sqrt (29) - 3 ] / 2 )

(x - 2)^2 + (y - 7)^2  = 25       (1)

 x + y =  13  →  y = 13  - x     (2)

Sub (2)  into (1)

(x  -2)^2  + ( 13-x -7)^2 =  25

(x-2)^2  + ( 6 -x)^2 = 25

x^2 - 4x + 4 + x^2 -12x + 36 = 25

2x^2  -16x + 40 = 25

2x^2 - 16x + 15  =   0

x = [ 16 + sqrt [ 16^2 - 4*2 * 15 ] ] /  4            and    x = 4 - sqrt (136)  /4        

x = [16 + sqrt [ 136] ] / 4 = 4 + sqrt (136)/4              y = 9 + sqrt (136)/4

y = 13 - x =   9 - sqrt (136)/4

(x , y)  =  ( 4 + sqrt (136) / 4 , 9 -sqrt (136)/4)   and  ( 4 -sqrt (136)/4 , 9 + sqrt (136) / 4)

                                     A

                             3                  5

                    B                7  D                C

Law of Sines

BC^2  = BA^2 + AC^2 - 2 (BA  * AC)  cos (BAC)

7^2 = 3^2 + 5^2  - 2 (3 * 5) cos (BAC)

[ 7^2 - 3^2 - 5^2 ] / [ -2 (3 * 5) ] =  cos (BAC)

-1/2  = cos (BAC) 

arccos (-1/2)  = BAC =  120°

cos (BAD)  =  cos (120° / 2)  = cos 60°  =   1/2