CPhill

Triangles CPD  and  APB  are similar

The scale factor  between these  is   (4/2)   = 2

So....the area of triangle  APB  = [ CPD] * 2^2  =  5 * 2^2  = 20

Next...the  height of triangle   CDP  can be found as 

5 =(1/2) 2 * h  

h =  5

So....the height of triangle APB is twice this  =  10

So.....the total height of the figure =  5 + 10  =  15

And the area of triangle  CPB  = area of triangle  CDB  - area of triangle CPD  =

(1/2) (CD) * height of figure    - [ CPD]   =  (1/2)(2)(15)  -  5   =  10

And [CPB ]  = [ DPA]

So.....the total area  of the figure =  [ CPB] + [ APB ] + [ CPB ] + [ DPA]  =

5 + 20 + 10 + 10  = 

45  =    [ABCD ] 

If I didn't miss any  !!!!

11, 12, 14, 16

21 , 23 , 25, 29

32 , 34, 38

41, 43, 47 , 49

52, 56, 58

61, 65, 67

74, 76

83, 85, 89

92, 94, 98

This is rather lengthy, Vinculum....if I keyed everything in correctly, see if this  helps

https://www.wolframalpha.com/input?i=%5B+%28+%C2%A0+-+1+-sqrt+%283%29+i+%29+%2F2%C2%A0+-+%28+sqrt+%283%29++-+1i%29++%2F2%C2%A0+%5D+%5E9

First.....we need to find the circle center

Rewrite as

x^2 - 6x   + y^2 + 8y   = 0              complete the square on  x and y

x^2 - 6x + 9  + y^2 + 8y + 16 =   9 + 16

(x - 3)^2  +  ( y + 4)^2  = 25

This is a circle with a center of ( 3, -4)     and  a radius of  5

Now.....we can use  a special formula that will tell us how  far the  center of the circle is  from our given line

We are using the equation of the line in standard  form and  filling in x  and  y with the circle's center

The squre root in the  denominator is just the coefficients of the  line (each squared)

We have that

l  3(3) + 4(-4) - 25  l                 l -32 l              32

________________   =       ________  =      __

 sqrt  [ 3^2 + 4^2 ]                 sqrt (25)            5

Subtract the radius of the circle from this

32 / 5    -  5   =

32 /5   -   25 /5   =

7/5  =  1.4   units  = the shortest distance  between the given line  and  the circle

Let  S be the number of successful shots saved  in the  first instance  and  let 3S be the total number attempted in that case

So....... in the second case we have that

S + 1

______   =     .40         solve for S

3S + 1

S + 1  =  .40 (3S + 1)

S + 1  = 1.2S + .40

1 - .40  = 1.2S - S

.6-  = .2S

S = .6/.2

S = 3

Now in the third case, let N  be the number of extra saved shots needed to reach 60%  and we  have that

S + 1 + N

___________     =  .60                   substitute for S

  3S + 1 + N

3 + 1 + N

_________   =     .60

9 + 1 + N

4 + N

_____   =   .60

10 +N

4 + N  =  .6 ( 10 + N)

4 + N  =  6 + .6N

N - .6N  = 6 - 4

.4N = 2

N =  2 / .4   =  5  = the number of additional saves necessary for a 60% save rate

\($\dfrac{5+12i}{2-8i}$\)

Multiply numerator    /  denominator    by    2 + 8i

( 5 + 12i) ( 2 + 8i)                 10 + 24i + 40i  +  96i^2             10 + 64i - 96         -86 + 64i

______________  =           ____________________  =      ___________ =   ________  =

(2 - 8i)  (2 + 8i)                       4  -   64i^2                                   68                       68

-43  + 32i

_______     

    34

y = 2x^2  + 7

This is a parabola  that turns upward and has its vertex at (0, 7)

The max of the function  occurs at x = 2  and equals   2(2)^2 + 7 =  15

So......the range is    [7 , 15 ]   on the given interval

Let's write the equation as

x^2 + 8x  +  n  =  0

We will only have no real solutions  whenever the discriminant is <  0

So

8^2  - 4(1)(n)  < 0

64 <  4n

4n >  64

n > 16

In other words, we will have real solutions for  any integer  n   from 1 - 10 inclusive

So......the probability is  0

The probability =   1  -    probability that a "3"  doesn't appear either time  =

1  - (5/6)^2   =  

1 - 25/36  =

11 / 36

7x^2 + 5x  =  -x^2  + x + h         rearrange as

8x^2  + 4x  -  h  =  0

If we have only one solution, the discriminant =  0.....so....

4^2  -  4 (-h)(8)  =  0

16 + 32h  =  0

32h =  -16

h =  -16/32  =   -1/2