CPhill

1)

     1                     A               B

_______  =      _____  +   _____                       multiply through by  k (k + 3)

   k(k+3)               k             k +  3

1  =   A(k + 3)   + Bk        simplify

1 +  0k  =   ( A + B)k  +  3A

Equating like terms

3A  =   1                        

A = (1/3)

And

A + B  =  0

B = -1/3

So

(A, B)    =  (1/3, - 1/3)

Since the point   (0, -1) is on the graph, c   must  be  -1

Let the long diagonal  of the rhombus    =  L

By the intersecting chord theorem  we have that

(3/2)r * (1/2)r    = [ ( 1/2) L ]^2

(3/4)r^2  = (1/4)L^2       multiply both sides by 4

(3)r^2  =  L^2       take the square root

(√3) r  = L

So.....the area of the rhombus  =   (1/2) product of the diagonals   =

(1/2) (r) (√3) r  =

[√3/2] r^2

2) In terms of pi, what is the area of the circle defined by the equation 2x^2+2y^2+10x-6y-18=0

Divide every term by 2  and rearrange as

x^2 + 5x  + y^2  - 3y  =  9      complete the square on  x  and y

x^2 + 5x + 25/4  +  y^2 - 3y + 9/4   =   9 + 25/4  +  9/4        factor and simplify

(x + 5/2)^2  +  ( y - 3/2)^2   =  36/4 + 25/4 + 9/4

(x + 5/2)^2  + ( y-3/2)^2  =  70 /4  = 35 / 2   =17.5

The area  =  pi * r^2   =    pi * 17.5    =  17.5 pi

1)

Since (0,0)  is on the graph, c   = 0

So we have that

a(3)^2  + b(3)  = -3     ⇒  9a  +  3b  = -3   ⇒    -3a - b  =  1    (1)

a(1)^2  + b(1)  =  3     ⇒   a  +  b    =  3           (2)

Add(1)  and (2)  and we get that

-2a  =  4

a = -2

b = 5

100 a +  10 b +  c   =  -200 + 50  + 0   =   -150

a =  40°   = c

b =  180 - 2(40)  = 100°

e  is supplemental to  c  =  180  - 40  =  140°

So

a + b + c + e  =

40 + 100 + 40  + 140  =

320° 

2)

\( f^{-1}\left(f^{-1}(50)\times f^{-1}(10)+f^{-1}(26)\right) \)

f^-1 (50)  =  7

f^-1 (10)  = 3

f^-1  (26)   =  5

So   we are evaluating this

f^-1  (  7  x 3  +  5)  =

f^-1  ( 21  + 5)  =

f^-1 (26)  =

5

1)  

We have these equations

(-1 -h)^2  + (2 - k)^2  = r^2

(3 - h)^2  + k^2  = r^2

(9 - h)^2  + k^2  = r^2

Subtract the  second equation from  the 3rd and we have that

(9 - h)^2  - (3 - h)^2  =  0    simplify

h^2 - 18h  + 81  -  [ h^2 - 6h  + 9  ]  = 0

-12h  + 72   =0

-12h  = -72

h = 6

And  equating the first two equations

(-1 - 6)^2  + (2 - k)^2  =  (3 - 6)^2  + k^2

49  + k^2 - 4k + 4  =  9  + k^2

53  - 4k  =  9

44 - 4k  = 0

k  = 11

So

(h, k) =  (6, 11)

And  h + k  =  17

\( $f(x)=\dfrac{2}{x+1}$ \)

\( $f^{-1}\left(\frac{1}{5}\right)$ \)

Note that  we  need  to  solve this

(1/5)  =   2 / ( x + 1)          and we can write

5   =  ( x + 1) / 2             multily both sides by 2

10  = x +  1                   subtract 1 from both sides

9  =  x  =  f^-1(1/5)

So  the point (9, 1/5)  is on f(x)     and the point  (1/5 , 9)  is on f^-1 (x)  

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