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Triangles   ABC and   EDC  are similar

So

AC  = 9/17   of   AE

CE =  8/17  of   AE

Therefore     

CD =  (8/17)  (20) =   160/17

Simplify as

x^2  +  30x  +  _____

Take 1/2 of 30   =15

Square  it  = 225  is the constant we  need

(x +15)^2  = x^2 +  30x  + 225

23.52  ( 1 + 1.1)  =   $49.39   =   price

And

49.39  ( 1 +  .1475)  ≈  $56.68  =  what he paid

-x^2  + 6x + 20  =  b + x^2 - 4x  + 8      rearrange as

2x^2  -  10x   +  (b - 12 )  = 0

It will have no real roots  when  the discriminant   is <  0

So

10^2   -  4(2) (b - 12)  < 0

100  - 8b  + 96  <   0

196  <  8b

8b > 196

b >   196/ 8

b >  49 /  2

It will have no real roots  when    b >  49  / 2

Just a counting prob

Number of committees   =  

C ( 7,2)  * C(7 ,3) * C ( 9, 4)  * C(7 , 5)    = 

1,944,810       (that's quite a few  !!!!  )

If  x  is   "very negative," the second trem will go to 0  and the range max  will be  almost 3

If x is "very postive,"    the  function  will approach  -inf

So the range is    (-inf, 3)

Write the second equation   as   y  =  17 - 7x

Sub this  into the first equation 

9x  - 4 ( 17 - 7x)  =  30

9x - 68 + 28x  = 30

37x  =  30 + 68

37 x =  98

x =  98  / 37

And y  =   17 - 7 (98/37)  =  -57 / 37

(x , y)  =  ( 98 / 37  ,   -57 / 37  )

(-4,0)  is a root  ....so.....

If we  shifted  this   parabola to  the right by 4 units  the vertex is (0,0) and  we  would get rid of the  bx  and c   terms   

And now..... ( 1 + 4, -7)  becomes (5 , -7)  

So

y=  ax^2

-7 =  a (5)^2

-7 = 25a

a = -7  / 25

 y= -2x^2+8x-15+3x^2-6x+22     rearrange  as

y = x^2 + 2x + 7

The x coordinate of  the  vertex     =   -2 / [ 2 * (1) ]    =  -1

The y coordinate is      (-1)^2  + 2(-1) + 7 =   6

xy = 20    ⇒   y = 20/x

xz = 35   ⇒    z = 35 /x  

yz = 14

So

yz =   14

(20/x) (35/x)  = 14

700  =  14 x^2

700  / 14   = x^2

x^2   = 50

x=   5sqrt (2)

y = 20  / [ 5sqrt 2]  =   4 / sqrt 2

z=  35 / [ 5sqrt 2]  =  7 /sqrt 2

So

xyz =   5sqrt 2 *   4 /sqrt 2  *   7/ sqrt 2  =     140 / sqrt 2  =  70sqrt 2

P(Jack) =  4/ 52 =  1/13

P(Heart)  = 13/52 =  4/13

P  ( heart and then jack with replacement)  = 1/13 * 4 /13  =    4 /  169

(x + 7) (x - 11)  = 0

x^2  - 4x  - 77   =   0

5x^2   -11x   +  4  = 0

Note that      1/a + 1/b =     (a + b)  / (ab)

The sum of the roots  =  11/5  = a + b     

The product of the rooots = ab =  4/5  

So

1/a + 1/b =  11/5  / 4/5   =  11  / 4

                       B

                              4

              7                     M

                       3                     4

A                                                      C

Triangle   ABM   would have sides of  AB = 7    AM = 3   and   BM = 4

But, by the triangle inequality, this is  impossible......so what you have described is impossible

There is only one integer satisfying k

The slope between  (1,7)  and (13,16)  = 9/12 = 3/4

The  equation of the line containing this segment is

y = (3/4) (x - 1)  + 7  ⇒    3x - 4y  + 25  =  0

The distance  shortest distance  between  this line and ( 6 , k)  is  given by

l   3(6)  - 4k  +  25 l  /  sqrt ( 3^2 + 4^2)

l 43 - 4k l  /  5  or    l 4k   - 43 l   / 5

Note  that  the distance must be poisitive

So 

 4k  - 43  >  0

4k >  43

k > 43 / 4

So....since k must be an integer. >  43/4...then....

Ceiling (43/4) =   11   =  k

Surface area =    2 ( ab + bc + ac)  =  56  (1)

Length of the surface edges =  4 ( a + b + c)  =  56

So     a + b + c  =  14

Square both sides of this

a^2 + 2 ab + 2 ac + b^2 + 2 bc + c^2  = 196       rearrange as

a^2 + b^2 +c^2  + 2 (ab + bc + ac) =  196    (2)

Sub (1)  into (2)

a^2 + b^2  + c^2   +  56  =  196

a^2 + b^2 + c^2  =  140

Length   of the dagonal  =   sqrt [ a^2 + b^2 + c^2 ] =   sqrt (140) =  2sqrt (35)

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