CPhill

4.  Here's the first.....the asymptotes  are also shown

https://www.desmos.com/calculator/krknmtrhap

6.       

                 x - 8

        __________               factor the denominator

        x^2 + 5x  - 6

                x  - 8

            __________

            (x + 6) ( x - 1)

The two  values that  make the denominator  = 0  are    x  = -6   and  x  = 1

So.....these are the points of discontinuity  (last answer )

OK

We have that

y  = k wx  / z      and we know that  y  = 320   when  w  = 6, x = 20   and z  = 3

So   we need to  find  the proportionality constant, k......so we have.....

320  = k (6) (20)  / 3

320  = k *120 / 3

320  = k * 40       divide both sides by 40

8 = k

So......the correct answer is

y  = 8 wx /  z

Here's the second one

xy  + 16  = 0 

https://www.desmos.com/calculator/r8slfagryd

Remember, Manuel.....

Perimeter  =  2W + 2L        and Area   = LW

So  

40  =  2W + 2L         divide through by 2

20  = W + L     subtract W  from both sides

20 - W  = L         (1)  

And

96  = LW       sub in (1)   for    L.....and we have

96  = (20 - W) * W       simplify

96 = 20W  - W^2       add W^2 to both sides, subtract 20W from both sides

W^2 - 20W  +  96  = 0        factor.....-8 and  - 12  are what we need

(W  - 8) ( W - 12)   = 0

Set each factor to 0  and solve for W

W  = 8    or   W  = 12

So, Manuel....we can let  W  = 8....so

 L = 20 - 8  = 12

So....Width  = 8   and Length   = 12

Check  - perimeter =  2W  + 2L  =   2(8)  + 2(12)   =  16 + 24  = 40

area  = 8 * !2   =  96     

Hope that helped......!!!!

B + D  =  180

So

116  + x  = 180     subtract 116 from both sides

x  = 84

So...A  =   2(84) - 40  = 128°

7) If AB = 3, AC = 8 and BS = 21, find the area of the circle.

AS * AB  =  AP * AC

(AB + BS)  * AB  =  (AC + CP)  * AC

24 *  3  =  (8 + CP) * 8

72  =  64 + 8*CP

8  = 8*CP

PC  = 1

AP  = PC + AC

AP = 1 + 8   = 9

SP  = √  ( AS^2  - AP^2)   =  √ ( 24^2  - 9^2 ) =  √495

Draw SC

So.....  SC   =  √ ( SP^2  + PC^2 )  =√ (495 + 1)  =   √  496

And  SPC is right....so  SC  is a diameter.of the circle.....so the radius  is  √496/2

So.....the area of the circle  is

pi *  (√496 / 2)^2   =     496 * pi  / 4    units^2 = 124 pi units^2 ≈  389.56 units^2

5)Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?

This is known as the secant-tangent theorem.....specifically

PY^2  = AP * PB

9^2  =  15  * PB

81  =  15  * PB

PB  = 81/15  =   9/ 5

So.....

PB + AB  = PA

9/5  +  AB  = 15

AB  =  15  - 9/5

AB  =   [ 75 - 9 ] / 5     =    66/5

Thanks, Kennedy....I do what I can to make this stuff a little more understandable  !!!

Lastly  :

B + D   = 180

(2x + 3)  +  ( 4x + 3)  = 180

6x + 6  = 180

6x  = 174

x  =   29

So.... C  =  2*29 + 1    =   59°

Third one :

Here....I'm assuming that  YX  is tanglent to the circle

This is known as the  tangent-secant theorem....it says that

YX^2 = ZY  * VY    ....so......

YX^2   =  (28 ) * 9

YX^2  =  252        take the positive root

YX  = √ 252   =  √ [ 36 * 7  ]  =  6√7   =  15.87 m

Second one :

Itersecting chord theorem :  When two chords meet in a circle.....the products of their respective individual segments are equal

This implies that

DP *  BP  = AP * PC

4 * BP  =   3.5 * 6   simplify

4 * BP  = 21   divide both sides by  4

BP =  21 / 4  =   5.25