CPhill

First one  

Let the intersection of    AD  and BC   =  E

Triangle  AEB  is similar to Triangle   DEC

Let DE  =  x     and let  AE  =  30 - x

So

AE / AB   =  DE / DC

( 30 - x) / 20   =  x / 4       cross-multiply

4 (30 - x)   =  20x     simplify

120 - 4x  =  20x       add 4x to both sides

120  =  24x       divide both sides by  24

5  =  x   =  DE

So....CE  = sqrt [ DE^2  - DC^2]  =  sqrt [ 5^2  - 4^2 ] =  sqrt [ 9]  =  3

And BE is 5 times this  =  15

So....BC  =  BE + EC  =  15 +  3   =   18

See this :

http://www.math.ucla.edu/~radko/circles/lib/data/Handout-929-991.pdf

2x^3+6x^2+6x+2  by x+1

              2x^2   +  4x    +   2

x + 1  [   2x^3  +  6x^2  +  6x   + 2  ]

              2x^3  +  2x^2

             _______________________

                           4x^2  +  6x

                           4x^2  +  4x

                          _________________

                                      2x   +  2

                                      2x   +  2

                                     ________

The answer  is  in red

The thing that might be giving you trouble is that we have to account for the missing powers  in 

x^3  -  7

To rectify this....write this as   x^3  + 0x^2 + 0x - 7

So we have

               x^2  -  2x  +  4

x + 2  [   x^3   +   0x^2  +  0x    - 7  ]

              x^3   +   2x^2

            _______________________

                       -2x^2  +  0x

                      -2x^2   -  4x

                    ____________________

                                  4x     -    7

                                  4x     +   8

                                ___________

                                           - 15

So...the result   is    x^2  - 2x  + 4  -   15 / ( x + 2)    [ Third answer ]

Excellent !!!

The last is indeed, "neither"   [ it is just a straight line ]

The third one is tricky....note that we can write

f(t)  =  1(3/4)^t

So....this is "decay "

These are easy to identify...we have the  form  on the right side of each equation

a(b)^x

If  b > 0  but < 1......we have exponential decay

If  b > 1......we have exponential growth

Can you answer the question based on this???....give it a try and I will check your answers  [ if you want me to...]

y  = log  (3x+ 4)  + 2

To find the y intercept .....let x  = 0

y  =  log (3*0 + 4) + 2

y = log (4) + 2  ≈  2.60  =  y intercept

To find the x intercept, let y  =  0

0  =  log ( 3x + 4)  +  2      subtract 2 from each side

-2  =  log (3x + 4)

Exponentially, we have that

10^(-2)  = 3x + 4       subtract  4 from both sides

10^(-2) - 4  =   3x     divide both sides by 3

[ 10^(-2) - 4 ]  / 3   =  x   ≈  -1.33  = x intercept

 b.) vertical stretch  of the parent 

b.) 4.3

y = 3^x

y = log₃ x

a.) the graphs  of  functions are symetrtic to each other over the line y= x

c.) The functions are inverses of each other

Note that triangle  BCA  is similar to triangle  BHC

Which implies that

HC / BC  =  CA / BA..... so...

h / a  =  b / ( x + y)    (1)

Now expand   ( x + h)^2 + ( y + h)^2  =  ( a + b)^2    (2)

x^2 + 2xh + h^2  + y^2 + 2yh + h^2  =  a^2 + 2ab + b^2     (3) 

And since  x^2 + h^2  =  a^2    and   y^2 + h^2  = b^2

We can subtract these equal parts from  (3)  and we are left with

2xh + 2yh  =  2ab        divide through by 2

xh + yh  =  ab      factor out h on the left

h ( x + y)  =  ab       rearrange as

h / a  =  b / ( x + y)     but, by (1)....this is true

So (2)  must be true, as well